ePhysics 11

Pearson Heinemann An imprint of Pearson Education Australia A division of Pearson Australia Group Pty Ltd 20 Thackray Road, Port Melbourne, Victoria 3207 PO Box 460, Port Melbourne, Victoria 3207 www.pearsoned.com.au/schools Offices in Sydney, Melbourne, Brisbane, Perth, Adelaide, and associated companies throughout the world. Copyright © Carmel Fry, Keith Burrows, Rob Chapman, Doug Bail, Geoff Miller 2008 First published 2008 by Pearson Education Australia 2011 2010 2009 2008 10 9 8 7 6 5 4 3 2 1 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that that educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL) under the Act. For details of the CAL licence for educational institutions contact Copyright Agency Limited (www.copyright.com.au). Reproduction and communication for other purposes Except as permitted under the Act (for example any fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All enquiries should be made to the publisher at the address above. This book is not to be treated as a blackline master; that is, any photocopying beyond fair dealing requires prior written permission. Publisher: Malcolm Parsons Editor: Marta Veroni Text Designer: Rebecca Harrison Copyright & Pictures Editor: Caitlin O’Brian, Katherine Wynne Project Editor: Jane Sunderland Production Controller: Aisling Coughlan Illustrator/s: Guy Holt, Margaret Hastie, Brent Hagan, Chris Hurley, Pat Kermode, Cynthia Nge Printed in China National Library of Australia Cataloguing-in-Publication entry Heinemann physics 11 / Carmel Fry ... [et al.]. 3rd ed. 9781740819367 (pbk.) Includes index. For secondary school age. Physics--Textbooks. Other Authors/Contributors: Fry, Carmel. Dewey Number: 530 Pearson Australia Group Pty Ltd ABN 40 004 245 943 Acknowledgements We would like to thank the following for their contributions to our text book. The following abbreviations are used in this list: t = top, b = bottom, c = centre, l = left, r = right. Airbus Industries: p. 474. Alamy Limited: pp. 53, 90,135, 536. Anglo-Australian Observatory: pp. 338b, 340l; David Malin Images: pp. 338t, 346. ANSTO: pp. 454, 458b. Arthur Wigley / Royal Melbourne Hospital: p. 546t Australian Associated Press Pty Ltd: pp. 73, 144, 181, 189, 194, 201b, 242, 355t. Australian Science Media Centre / Daniel Mendelbaum: p. 158. Carmel Fry: pp. 542b, 545. Coo-ee Picture Library: pp. 267l, 499, 507. Corbis Australia Pty Ltd: pp. 24t, 56, 117, 143l, 198c, 281, 336b, 336t, 365l, 385, 432, 440c, 445r.

CSIRO: pp. 492, 505. David Malin Images / Akira Fujii: p. 394; Royal Observatory Edinburgh: p. 418. Doug Bail: p. 512. EFDA-JET: p. 466. European Space Agency: pp. 393, 416; John Bahcall: p. 429. European Space Observatory: pp. 377r, 424(b); Lars Lindberg Christensen: p. 424(a). Getty Images Australia Pty Ltd: pp. 1, 4, 109, 110, 111b, 113, 127tl, 152, 165, 178, 179t, 199, 211, 215, 317, 435, 444, 542t, 357b. Imaginova Corporation / Starrynight.com: pp. 337r, 347. iStockphoto: pp. 70, 77, 80, 93, 95, 111t, 131, 132l, 143br, 143tr, 156, 161, 162, 179b, 191c, 191t, 198b, 198t, 201t, 207, 210bl, 210br, 210cl, 210tl, 225, 237, 291l, 299, 343, 401, 404l. Keith Burrows: pp. 57, 59, 69, 98, 99, 255, 384, 397, 409l. Malcolm Cross: pp. 291r, 303, 375. Mark Fergus: pp. 61, 258b, 258t, 292, 312. Meade Instruments: p. 373. NASA: pp. 151b, 286, 334, 337c, 337l, 369, 376, 379, 382l, 387, 420, 424(d), 476; Astronomy Picture of the Day/ Robert Gendler: p. 383; Joe Gurman, Simon Plunkett, Steele Hill and Stein Vidar Haugan: p. 407; Hubble Site / Space Telescope Science Institute: p. 419; European Space Agency / J. Hester and A. Loll (Arizona State University): p. 421; Wikisky: p. 424(c); WMAP Science Team: p. 433; European Space Agency, M. Robberto (Space Telescope Science Institute/ ESA) and the Hubble Space Telescope Orion Treasury Project Team: p. 390. Nationwide News Pty Ltd: p. 132r. Official 2008 Melbourne Marathon Course Map: p. 112. PASCO Scientific: p. 119r. Pearson Asset Library: p. 245l. Pearson Education Australia / Dale Mann: p. 47; Advanced Science Medical Physics: pp. 531, 533. Photodisc: pp. 143cr, 229b, 335, 399, 404r, 409r, 428, 504. Photolibrary Pty Ltd: pp. 26, 36, 183, 245tr, 249, 254, 366c, 404t, 405, 406, 439, 447, 511, 530b, 534; Digital Vision: p. 465; Gianni Tortoli: p. 370; Mary Evans Picture Library: p. 402; Oxford Scientific: pp. 319, 472; Photo Researchers: pp. 23, 24b, 146, 245br, 324, 403, 431, 547; Science Photo Library: pp. 2, 6, 8, 13, 31, 37, 41, 52, 101, 118, 119l, 142, 154, 206, 226r, 234, 262, 272, 280, 287, 306, 357t, 361, 362, 365r, 366t, 377l, 381, 382r, 386, 398, 410c, 423, 425, 440b, 440t, 445l, 528, 532, 535, 544, 553, 554, 555. Quasar Publishing: p. 355b. RMIT: p. 475; Craig Mills: p. 259. Shutterstock: pp. cover, 18, 127bl, 143c, 151r, 164, 173, 188, 191b, 192, 223, 224, 226l, 229t, 244, 506, 515, 520, 523. Sport: The Library: p. 126. State Library of South Australia/Mountford-Sheard Collection: p. 340r. Track & Field News: p. 127r. University of Michigan: p. 410t. Virtual Hospital: pp. 546b, 549. Wikipedia / Tao’olunga: p. 350. Yerkes Observatory: p. 366b. Every effort has been made to trace and acknowledge copyright. However, should any infringement have occurred, the publishers tender their apologies and invite copyright owners to contact them. Disclaimer/s The selection of Internet addresses (URLs) provided for this book were valid at the time of publication and chosen as being appropriate for use as a secondary education research tool. However, due to the dynamic nature of the Internet, some addresses may have changed, may have ceased to exist since publication, or may inadvertently link to sites with content that could be considered offensive or inappropriate. While the authors and publisher regret any inconvenience this may cause readers, no responsibility for any such changes or unforeseeable errors can be accepted by either the authors or the publisher.

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Introduction vi Unit 1 Area of study 1 Nuclear physics and radioactivity 1 Chapter 1 Nuclear physics and radioactivity 2 1.1 Atoms, isotopes and radioisotopes 1.2 Radioactivity and how it is detected 1.3 Properties of alpha, beta and gamma radiation 1.4 Half-life and activity of radioisotopes 1.5 Radiation dose and its effect on humans Chapter review Area of study review Nuclear physics and radioactivity

3 8 15 20 26 32 34

Area of study 2 Electricity 36 Chapter 2 Concepts in electricity 37 2.1 Electric charge 2.2 Electrical forces and fields 2.3 Electric current, EMF and electrical potential 2.4 Resistance, ohmic and non-ohmic conductors 2.5 Electrical energy and power Chapter review

Chapter 3 Electric circuits

38 45 51 59 67 75

77

3.1 Simple electric circuits 78 3.2 Circuit elements in parallel 84 3.3 Cells, batteries and other sources of EMF 89 3.4 Household electricity 97 Chapter review 102 Area of study review Electricity 104

Unit 2 Area of study 1 Motion Chapter 4 Aspects of motion

109 110

4.1 Describing motion in a straight line 4.2 Graphing motion: position, velocity and acceleration 4.3 Equations of motion 4.4 Vertical motion under gravity Chapter review

111 122 130 135 140

Chapter 5 Newton’s laws

142

5.1 Force as a vector 5.2 Newton’s first law of motion 5.3 Newton’s second law of motion 5.4 Newton’s third law of motion Chapter review

143 150 156 164 175

Chapter 6 Momentum, energy, work and power 178 6.1 The relationship between momentum and force 6.2 Conservation of momentum 6.3 Work 6.4 Mechanical energy 6.5 Energy transformation and power Chapter review Area of study review Motion

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179 187 191 198 209 217 219

Chapter 11 Astrophysics 390

Area of study 2 Wave-like properties of light Chapter 7 The nature of waves 7.1 Introducing waves 7.2 Representing wave features 7.3 Waves and wave interactions Chapter review

223

11.1 The stars—how far, how bright? 391 11.2 Our favourite star 401 11.3 We know the stars by their light 409 11.4 Whole new worlds 423 11.5 The expanding universe 431 Chapter review 437

224

Chapter 12 Energy from the nucleus 439

225 232 240 247

Chapter 8 Models for light 249 8.1 Modelling simple light properties250 8.2 Refraction of light 258 8.3 Critical angle, TIR and EMR 8.4 Dispersion and polarisation of light waves 280 Chapter review 285

Chapter 9 Mirrors, lenses and optical systems 286 9.1 Geometrical optics and plane mirrors 287 9.2 Applications of curved mirrors: concave mirrors 291 9.3 Convex mirrors 299 9.4 Refraction and lenses 306 9.5 Concave lenses 312 9.6 Optical systems 317 Chapter review 327 Area of study review Wave-like properties of light 329

Units 1 & 2 Area of study 3 Detailed studies Chapter 10 Astronomy The story continues ... 10.1 Motion in the heavens 10.2 The Sun, the Moon and the planets 10.3 Understanding our world 10.4 The telescope: from Galileo to Hubble 10.5 New ways of seeing Chapter review

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333 334 335 337 347 357 369 379 388

12.1 Splitting the atom— nuclear fission 12.2 Aspects of fission 12.3 Nuclear fission reactors 12.4 Nuclear fusion Chapter review

440 447 453 463 468

Chapter 13 Investigations: flight 470 13.1 13.2 13.3 13.4

The four forces of flight Modelling forces in flight Investigating flight Investigation starting points

471 481 486 489

Chapter 14 Investigations: sustainable energy sources492 14.1 14.2 14.3 14.4

Energy transformations Renewable or sustainable—the key to our future Investigating alternative energy sources Starting points

493 498 499 503

Chapter 15 Medical physics515 15.1 Ultrasound and how it is made 15.2 Ultrasound interactions 15.3 Scanning techniques 15.4 Diagnostic X-rays 15.5 Radiotherapy, radioisotopes in medicine and PET Chapter review

Appendix A Appendix B Appendix C Solutions Glossary Index

516 523 528 537 550 558

560 563 565 576 599 608

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The complete package for Units 1 and 2 VCE Physics Heinemann Physics 11 3rd edition is the most up to date and complete package for VCE Physics. The 3rd edition has been fully revised and upgraded to match the content and focus of the new 2009 VCE Physics Study Design. Successful features of the 2nd edition have been retained while significant improvements and innovations will make the books even easier and more stimulating to use. Heinemann Physics 11 3rd edition covers Units 1 and 2 and Heinemann Physics 12 3rd edition covers Units 3 and 4. Heinemann Physics 11 3rd edition textbook Includes the ePhysics 11 student CD

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Heinemann ePhysics 11 includes: UÊ >ÊV«iÌiÊV«ÞÊPhysics 11 third edition UÊ interactive tutorials UÊ iÃÃ>ÀÞ UÊ ICT toolkit

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Heinemann Physics 11 3rd edition Teacher’s Resource and Assessment Disk The Teacher’s Resource and Assessment Disk contains a wealth of support material and makes effective implementation of the Study Design easy. Included: • Detailed answers and worked solutions to all questions in the textbook • Extensive range of short and long practical activities all with teacher notes and suggested outcomes and answers • Sample assessment tasks with marking guidelines • Complete electronic copies of the textbook and ePhysics 11 student CD

Heinemann Physics 11 3rd edition Companion Website www.pearsoned.com.au/physics The Companion Website includes further support for teachers including weblinks

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The Heinemann Physics series is now in its third edition. The first edition was published in 1996 and since then the original author team has remained together and has continually strived to build on and improve the series. Over that time, the authors have not only remained highly involved in the teaching of Physics but have also contributed to Physics and Physics education as members of professional organisations, supported curriculum development and have regularly presented professional development to their colleagues. The third editions of Heinemann Physics 11 (Units 1 and 2) and Heinemann Physics 12 (Units 3 and 4) represent the authors’ ongoing commitment to Physics teachers and students. The series has been fully revised and upgraded to match the content and focus of the new 2009 VCE Physics Study Design. Successful features of the second edition have been retained, while significant improvements and innovations have been added. These include: • New full colour design • All Detailed Studies in the textbook • Exact match to the structure and sequence of the study design • Chapters divided into student-friendly sections • Clear explanations and development of concepts consistent with the intent and scope of the Study Design • Extension and enrichment material clearly designated • Numerous well-graded end-of-section questions and chapter reviews • Exam-style questions that are exam style! • Extensive glossary • ePhysics interactive CD with each text

Heinemann Physics 11 3rd edition The authors have written a text that will support students’ learning in Physics while making the subject interesting, enjoyable and meaningful. The book uses clear and concise language throughout. All concepts have been fully explored first in general and then illustrated in context. Illustrative material is fresh, varied and appealing to a wide range of students. Each chapter has been divided into a number of self-contained sections. At the end of each section is a set of homework-style questions that are designed to reinforce the main points. More demanding questions are included at the end of the chapter. At the end of each Area of Study is a set of exam-style questions. These can be used for revision. The large number of questions is designed to assess students’ understanding of basic concepts as well as giving them practice at problem solving. Answers are supplied at the end of the text and extended answers and fully worked solutions are available on the Teacher Resource and Assessment Disk.

vi

Within each section, the concept development and worked examples occupy the main column. The minor column has been set aside for some of the numerous photographs and diagrams, as well as small snippets of ‘Physics File’ information. The longer pieces of high interest and context material are contained in the page-width ‘Physics in Action’ sections. Both Physics in Action and Physics File sections are clearly distinguishable from remaining material, yet are well integrated into the general flow of information in the book. These features enhance students’ understanding of concepts and context. The authors have written the text to follow the sequence, structure and scope of the Study Design. Material outside the scope of the Study Design is clearly marked. This includes entire sections and a sub-sections.

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This material has been included for a number of reasons, including as important background to core concepts, important physics in its own right and as extension material for more able students. Teachers should consider whether they wish to incorporate this material into their work program.

Optional sub-section.

The third edition includes all Detailed Studies in the textbook. Chapters 10–15 are the Detailed Studies. Students will undertake one Detailed Study in each Unit. The Detailed Study chosen for Unit 1 must be different from the Detailed Study chosen for Unit 2. The textbook includes an interactive CD, ePhysics 11, which will enhance and extend the content of the texts, and includes: • Fully interactive tutorials that allow students to explore important concepts which may be too difficult, dangerous or expensive to do first-hand in the class room

ephysics 11 3rd ed

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vii

Doug Bail

about the authors

Is an experienced physics educator and writer with a particular interest in the development and integration of new technologies into science teaching. He has previously been a Head of Science and senior physics teacher, and maintains a passion for making physics relevant, stimulating and accessible to all students. Doug now runs his own company developing and distributing products for Physics education.

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He led the development of the new practical activities that form part of the teacher support material. These activities were extensively trialled throughout Australia and include a range of activities from teacher demonstrations to discovery-based investigations, suiting a range of learning styles and needs. This includes many short activities for when time is limited!

Keith Burrows Has been teaching senior physics in Victorian schools for many years. He is a member of the Australian Institute of Physics Victorian Education Committee and was actively involved with the VCAA in the design of the new course. Keith was a VCAA representative involved in the introduction of the new VCE course to physics teachers in Victoria and running the workshop sessions for teachers. He is particularly keen to portray ‘The Big Picture’ of physics to students. Keith would like to acknowledge Maurizio Toscano of the Melbourne University Astrophysics Group who has provided invaluable help and advice in the preparation of the Astronomy and Astrophysics detailed studies.

Rob Chapman Has taught physics for many years from HSC onwards. Rob has been enthusiastic in exploring the possibilities presented by changing technologies over the years. He has been Science Coordinator at St Columba’s College in Essendon, where he was instrumental in introducing the use of datalogging technology to junior science and senior physics classes. Rob is currently teaching Senior Physics at PEGS (Penleigh and Essendon Grammar School). He has

written a wide variety of curriculum support materials, including physics units for the CSFII. Rob has also produced physics trial examination papers and is the author of the acclaimed Physics 12—A student guide.

Carmel Fry Has 19 years’ involvement in development of text, CD and on-line curriculum materials for VCE Physics and Science.  She is Head of Science at Ivanhoe Girls’ Grammar School, where she continues her interest in providing high-quality curriculum resources and learning experiences for students. Carmel is the author of numerous texts, multimedia resources and teacher-resource materials developed for senior physics. These materials are currently in use in many parts of Australia and overseas. She led the development of the Interactive Tutorials. Carmel is particularly passionate about providing physics curriculum materials that involve a variety of approaches to learning, and that support independent learning through stimulating and appealing contexts and activities. Carmel would like to acknowledge the on-going support of her husband and children over her many years of publishing.

Review panel The publisher and authors would like to acknowledge and thank the expert review panel consisting of experienced VCE teachers and educators: Luke Bohni, Mike Davies, Barry Homewood, Chris Hourigan, John Joosten,  Terry Trevena, Steve Treadwell, Lyndon Webb and Chris Ward.

Acknowledgments The publisher would like to acknowledge and thank the author team for their ongoing commitment and passion for this project. It is a huge and complex task and the demands including short timelines are great. Carmel, Keith, Rob and Doug it has been a pleasure and privilege to work with you.

Unit

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area of On completion of this able to study, you should be vant explain and model rele ribe the physics ideas to desc clear sources and uses of nu ivity and reactions and radioact things, the their effects on living dustry. environment and in in

chapter 1

d n a s c i s y y h p r t a i e v l i c t Nu radioac

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any people think that they never come into contact with radioactive materials or the radiation that such materials produce. They are wrong to think this way. Human beings have always been exposed to radiation from a variety of natural sources. The ground that we walk on is radioactive. Every time we inhale, we take minute quantities of radioactive radon into our lungs. Even the food we eat and the water we drink contain trace amounts of radioactive isotopes. It is now accepted that exposure to higher than normal levels of high-energy radiation leads to the development of cancerous tumours and leukaemia. However, radiation and radioactive elements can also be used in a variety of applications that are of real benefit to people in industry and in medicine; for example: • Radioactive substances are used in the diagnosis and treatment of cancer. The photograph shows an image taken with a gamma ray camera. Technetium-99m (a radioactive isotope) was injected into the bloodstream of a patient. This allows the blood-flow patterns within the brain to be studied. • Smoke detectors usually contain a small sample of the radioactive element americium-241. • Geologists and archaeologists determine the age of rocks, artefacts and fossils by analysing the radioactive elements in them. • In industry, the thickness of manufactured sheet metal is accurately measured and controlled using radiation. In this chapter, we will examine radioactivity and discuss the associated dangers and benefits of its many applications. An understanding of this topic will help you to develop an informed opinion on this important issue.

by the end of this chapter you will have covered material from the study of nuclear physics and radioactivity including: • the origin, nature and properties of α, β and γ radiation • the detection of α, β and γ radiation • stable, unstable, natural and artificial isotopes • production of artificial radioisotopes • the half-life of a radioactive isotope • radiation doses from internal and external sources • effects of α, β and γ radiation on humans and other organisms • nuclear transformations and decay series.

1.1 Atoms, isotopes an

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Atoms

Physics file

In order to understand radioactivity, it is necessary to be familiar with the structure of the atom. The central part of an atom, the nucleus, consists of particles known as protons and neutrons. Collectively, these particles are called nucleons, and are almost identical in mass and size. However, they have very different electrical properties. Protons have a positive charge, whereas neutrons are electrically neutral and have no charge. The nucleus contains nearly all of the mass of the atom, but accounts for less than a million millionth (10−12) of its volume. Most of an atom is empty space that is only occupied by negatively charged particles called electrons. These are much smaller and lighter than protons or neutrons and they orbit the nucleus of the atom at high speed. The simplest atom is hydrogen. It consists of just a single proton with a single electron orbiting at a distance of about 5 × 10−11 m. Compare this with a uranium-238 atom. Its nucleus contains 92 protons and 146 neutrons. Its 92 electrons orbit the nucleus. Uranium-238 is the heaviest atom found in the Earth’s crust.

Two important terms that are used to describe the nucleus of an atom are its: • ATOMIC NUMB…R (Z)—the number of protons in the nucleus of an atom. • MASS NUMB…R (A)—the total number of protons and neutrons in the nucleus. A particular atom can be identified by using the following format: mass number atomic number

A Z

X

To gain an idea of the emptiness of atoms and matter, consider this example. If the nucleus of an atom was the size of a pea and this was placed in the centre of the MCG, the electrons would orbit in a three-dimensional space that would extend into the grandstands.

nucleus consisting of neutrons ( ) and protons ( ) cloud of electrons

Figure 1.1 The nucleus of an atom occupies about 10−12 of the volume of the atom, yet it contains more than 99% of its mass. Atoms are mostly empty space!

(a)

element symbol

The atomic number defines the element. Atoms with the same number of protons will all belong to the same element. For example, if an atom has six protons in its nucleus (i.e. Z = 6) then it is the element carbon. Any atom containing six protons is the element carbon, regardless of the number of neutrons. In an electrically neutral atom, the number of electrons is equal to the number of protons. Any neutral atom of uranium (Z = 92) has 92 protons and 92 electrons. The complete list of elements is shown in the periodic table in Figure 1.6.

Isotopes All atoms of a particular element will have the same number of protons, but may have a different number of neutrons. For example, lithium exists naturally in two different forms. One type of lithium atom has three protons and three neutrons. The other type has three protons and four neutrons. These different forms of lithium are isotopes of lithium. Isotopes are chemically identical to each other. They react and bond with other atoms in precisely the same way. The number of neutrons in the nucleus does not influence the way in which an atom interacts with other atoms. The

(b)

Figure 1.2 (a) Hydrogen is the simplest atom. It consists of just one proton and one electron. (b) Uranium-238 is the heaviest naturally occurring atom. Its nucleus contains 238 nucleons—92 protons and 146 neutrons.

Chapter 1 Nuclear physics and radioactivity

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difference between isotopes lies in their physical properties. More neutrons in the nucleus will mean that these atoms have a higher density.

(a)

ISOTOP…S are atoms that have the same number of protons but different numbers of neutrons. Isotopes have the same chemical properties but different physical properties. When referring to a particular nucleus, we talk about a nuclide. In this case, we ignore the presence of the electrons. For example, the nuclide lithium-6 has three protons and three neutrons. Stable isotopes can be found for most of the elements and, in all, there are about 270 stable isotopes in nature. Tin (Z = 50) has ten stable isotopes, while aluminium (Z = 13) has just one.

(b)

Radioisotopes Figure 1.3 Isotopes of lithium. (a) The nucleus of a lithium-6 atom contains three protons and three neutrons. (b) The nucleus of a lithium-7 atom contains three protons and four neutrons.

Figure 1.4 This symbol is used to label and identify a radioactive source.

Most of the atoms that make up the world around us are stable. Their nuclei have not altered in the billions of years since they were formed and, on their own, they will not change in the years to come. However, there are also naturally occurring isotopes that are unstable. An unstable nucleus may spontaneously lose energy by emitting a particle and so change into a different element or isotope. Unstable atoms are radioactive and an individual radioactive isotope is known as a radioisotope. By way of illustration, carbon has two stable isotopes, carbon-12 and carbon-13, and one isotope in nature that is not stable. This is carbon-14. The nucleus of a radioactive carbon-14 atom may spontaneously decay, emitting highenergy particles that can be dangerous. If you look at the periodic table in Figure 1.6, you will see that every isotope of every element with atomic mass greater than that of bismuth (Z = 83) is radioactive. Most of the elements found on Earth have naturally occurring radioisotopes; there are about 200 of these in all. As well as these, about 2000 radioisotopes have been manufactured. During the 20th century, an enormous number of radioisotopes were produced in a process known as artificial transmutation.

Artificial transmutation: how radioisotopes are manufactured

Figure 1.5 Artificial radioisotopes for medical and industrial uses are manufactured in the core of the Lucas Heights reactor in Sydney. This is Australia’s only nuclear reactor facility and has been operating since 1958. The original reactor was replaced by the OPAL (Open Pool Australian Light-water) reactor in 2007.

4

Nuclear physics and radioactivity

Natural radioisotopes were used in the early days of research into radiation. Today, most of the radioisotopes that are used in industrial and medical applications are synthesised by artificial transmutation. There are now more than 2000 such artificial radioisotopes. In the periodic table, every element with an atomic number greater than 92 (i.e. past uranium) is radioactive and is produced in this way. One of the ways that artificial radioisotopes are manufactured is by neutron absorption. (In Australia, this is done at the Lucas Heights reactor near Sydney.) In this method, a sample of a stable isotope is placed inside a nuclear reactor and bombarded with neutrons. When one of the bombarding, or irradiating, neutrons collides with a nucleus of the stable isotope, the neutron is absorbed into the nucleus. This creates an unstable isotope of the sample element.

Group 1 2

3

4

5

6

7

8

9

10

11

12

Group 13 14

15

16

17

1

2 3 4 5 6 7

2

H

Period 1 3

Li

6.94 11

He

1.01

4

18

5

Be

B

6

C

7

N

8

O

9

F

4.00 10

Ne

9.01 12

10.81 13

12.01 14

14.01 15

16.00 16

19.00 17

20.18 18

22.99 19

24.31 20

28.09 32

30.97 33

32.06 34

35.45 35

39.95 36

39.10 37

40.08 38

44.96 39

47.90 40

72.59 50

74.92 51

78.96 52

79.91 53

83.80 54

85.47 55

87.62 56

88.91 57

91.22 72

Na Mg

Al

Si

P

S

Cl

Ar

24

25

26

27

28

29

30

26.98 31

50.94 41

52.00 42

54.94 43

55.85 44

58.93 45

58.71 46

63.54 47

65.37 48

69.72 49

92.91 73

95.94 74

(99) 75

101.07 102.91 76 77

106.4 78

107.87 112.40 114.82 118.69 121.75 127.60 126.90 131.30 79 80 81 82 83 84 85 86

132.91 137.34 138.91 178.49 180.95 183.85 87 88 89 104 105 106

186.2 107

190.2 108

192.2 109

195.09 196.97 200.59 204.37 207.19 208.98 110 111 112 113 114 115

(210) 116

(223)

(264)

(277)

(268)

(271)

(272)

(277)

(289)

60

61

62

63

64

65

K

Rb Cs Fr

Ca Sr

Ba

21

Sc Y

(227)

23

V

Cr Mn Fe Co Ni Cu Zn Ga Ge As

Zr Nb Mo Tc

La Hf

Ra Ac

(226)

22

Ti

Ta

W

Ru Rh Pd Ag Cd

Re Os

Ir

In

Pt Au Hg Tl

Sn

Pb

Sb Bi

Se

Te

(262)

(263)

Lanthanides 58

Ce

Every isotope of these elements is radioactive

59

(289)

66

67

68

I

Kr

Xe

Po At Rn

Rf Db Sg Bh Hs Mt Ds Rg Uub Uut Uuq Uup Uuh

(261)

Br

69

(210) 117

(222) 118

Uuo (293)

70

71

Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

140.12 140.91 144.24

(145)

150.35 151.96 157.25 158.92 162.50 164.93 167.26 168.93 173.04 174.97

Actinides 90

91

232.04

(231)

Th Pa

92

U

238.03

93

94

95

96

97

(237)

(242)

(243)

(247)

(247)

Np Pu Am Cm Bk

98

Cf

(249)

99

100

101

102

103

(254)

(253)

(256)

(254)

(257)

Es Fm Md No Lr

Figure 1.6 The periodic table of elements.

Physics file

cobalt-59: stable

cobalt-60: radioactive

Figure 1.7 The artificial radioisotope cobalt-60 is used extensively in the treatment of cancer. It is produced by bombarding a sample of cobalt-59 with neutrons.

This is how the radioisotope cobalt-60 (widely used for cancer treatment) is manufactured. A sample of the naturally occurring and stable isotope cobalt-59 is irradiated with neutrons. Some of the cobalt-59 nuclei absorb neutrons and this results in a quantity of cobalt-60 being produced: 1 n + 59 Co → 60 Co. This nuclear transformation is illustrated in Figure 1.7. 0 27 27

The heaviest stable isotope in the universe is 209 Bi. Every isotope of every 83 element with more than 83 protons, i.e. beyond bismuth in the periodic table, is radioactive. For example, every isotope of uranium (Z = 92) is radioactive. Technetium (Z = 43) and promethium (Z = 61) are the only elements with an atomic number below bismuth (Z = 83) that do not have any stable isotopes. Uranium is the heaviest element that occurs naturally on Earth. All the elements with atomic numbers greater than 92 have been artificially manufactured.


5

Worked example 1.1A Use the periodic table in Figure 1.6 to determine: a the symbol for element 9542X b the number of protons, nucleons and neutrons in this isotope.

Solution a From the periodic table, the element with an atomic number of 42 is Mo, molybdenum. b The lower number is the atomic number, so this isotope has 42 protons. The upper number is the mass number. This indicates the number of particles in the nucleus, i.e. the number of nucleons, so this atom has 95 nucleons. The number of neutrons can be found by subtracting 42 from the mass number. This isotope has 53 neutrons.

Physics in action

Quarks and other subatomic particles! Our understanding of the atom has changed greatly in the past 100 years. It was once thought that atoms were like miniature billiard balls: solid and indivisible. The word ‘atom’ comes from the Greek ‘atomos’ meaning indivisible. That idea was changed forever when the first subatomic particles—the electron, the proton and then the neutron—were discovered in the period from 1897 to 1932. Since World War II, further research has uncovered about 300 other subatomic particles! Examples of these include pimesons, mu-mesons, kaons, tau leptons and neutrinos. For many years, physicists found it difficult to make sense of this array of subatomic particles. It was known that one family of particles called the leptons had six members: electron, electron-neutrino, muon, muon-neutrino, tau and tau-neutrino. Then in 1964 Murray Gell-Mann put forward a simple theory. He suggested that most subatomic particles were themselves composed of a number of more fundamental particles called quarks. Currently, it is accepted that there are six different quarks, each with different properties (and strange names!): up, down, charmed, strange, top and bottom. The latest quark to be identified was the top quark, whose existence was confirmed in 1995. The proton consists of two up quarks and one down quark, while neutrons consist of one up quark and two down quarks. Subatomic particles that consist of quarks are known as hadrons. Leptons are indivisible point particles; they are not composed of quarks. A significant amount of effort and money has been directed to testing Gel-Mann’s theory—both theoretically and experimentally. This has involved the construction of larger and larger particle accelerators such as Fermilab in Chicago and CERN in Geneva. Australia built its own particle accelerator—a synchrotron—next to Monash University. This began operating in 2007.

6


Figure

1.8 energy p This particle ac ce h of light in ysics. It accelera lerator is at the CERN Eu te under 2 ro 0 secon s protons from ds! rest to 9 pean centre for high9.99995 % of the speed While the current theory suggests that quarks and leptons are the ultimate fundamental particles that cannot be further divided, the nature of scientific theories and models is such that they can change as new experimental data are obtained. Are quarks and leptons made of smaller particles again? Time will tell!

1.1 summary Atoms, isotopes and radioisotopes • The nucleus of an atom consists of positively charged protons and neutral neutrons. Collectively, protons and neutrons are known as nucleons. Negatively charged electrons orbit the nucleus. • The atomic number, Z, is the number of protons in the nucleus. The mass number, A, is the number of nucleons in the nucleus, i.e. the combined number of protons and neutrons. • Isotopes of an element have the same number of protons but a different number of neutrons. Isotopes

of an element are chemically identical to each other, but have different physical properties. • An unstable isotope—a radioisotope—may spon taneously decay by emitting a particle from the nucleus. • Artificial radioisotopes are manufactured in a process called artificial transmutation. This commonly takes place as a result of neutron bombardment in the core of a nuclear reactor.

1.1 questions Atoms, isotopes and radioisotopes 1 How many protons, neutrons and nucleons are in the following nuclides? a 45 Ca 20 197 b 79Au c 235 U 92 230 d 90Th 2 How many protons and neutrons are in these atoms? Use the periodic table to answer this question. a Cobalt-60 b Plutonium-239 c Carbon-14 3 What is the difference between a stable isotope and a radioisotope? Give three examples of stable isotopes. 4 Can a natural isotope be radioactive? If so, give an example of such an isotope. 5 Which of these atoms are definitely radioactive? 24 Mg, 59 Co, 195 Pt, 210 Po, 238 U 12 27 78 84 92 Explain how you made your choice. 6 a A proton has a radius of 1.07 × 10−15 m and a mass of 1.67 × 10−27 kg. Using the fact that the volume of mass a sphere is V = 43 πr3 and density = : volume i calculate the volume of a proton ii calculate the density of a proton. b If we assume that the density of an atomic nucleus is equal to that of a proton, determine the mass of 1 cm3 of nuclear material.

c How many 1 tonne cars would it take to balance 1 cm3 of nuclear material? d What does this tell you about the density of normal matter compared to the density of atomic nuclei? 7 The nucleus of a gold atom has a radius of 6.2 × 10−15 m while the atom itself has a radius of 1.3 × 10−10 m. Using the volume formula from the previous question, determine the value of the fraction:

volume of nucleus volume of atom

8 As part of a science project, a student wanted to make a scale model of a gold atom using a marble of radius 1.0 cm as the nucleus. Calculate the radius of the sphere to be occupied by the electrons in this model. Use the information in question 7 to assist your calculations. 9 Krypton-84 is stable but krypton-89 is radioactive. a Discuss any differences in how these atoms would interact chemically with other atoms. b Describe the difference in the composition of these two atoms. 10 A particular artificial radioisotope is manufactured by bombarding the stable isotope 27Al with neutrons. The radioisotope is produced when each atom of 27 Al absorbs one neutron into the nucleus. Identify the radioisotope that is produced as a result of this process.


7

1.2

cted

is dete t i w o h d n a y Radioactivit

Through the Middle Ages, alchemists had tried without success to change lead into gold. They thought that it would be possible to devise a chemical process that would change one element into another. We now know that it is extremely difficult to change one element into another and that chemistry is not the way to do it. About 100 years ago, Ernest Rutherford and Paul Villard discovered that there were three different types of emission from radioactive substances. They named these alpha, beta and gamma radiation. Further experiments showed that the alpha and beta emissions were actually particles expelled from the nucleus. Gamma radiation was found to be high-energy electromagnetic radiation, also emanating from the nucleus. When these radioactive decays occur, the original atom spontaneously changes into an atom of a different element. Nature was already doing what the alchemists had so fruitlessly tried to do!

Alpha decay 42a Figure 1.9 Marie Curie performed pioneering work on radioactive materials. In fact, Marie Curie coined the term ‘radioactivity’ and is the only scientist to have been awarded two Nobel prizes. She received one for chemistry and one for physics.

When a heavy nucleus undergoes radioactive decay, it may eject an alpha particle. An alpha particle is a positively charged chunk of matter. It consists of two protons and two neutrons that have been ejected from the nucleus of a radioactive atom. An alpha particle is identical to a helium nucleus and can also be written as 42He2+, α2+, 42 α or simply α. alpha particle

uranium-238: unstable

thorium-234

Figure 1.11 When the nucleus of uranium-238 decays, it will spontaneously eject a high speed alpha particle that consists of two protons and two neutrons. The remaining nucleus is thorium-234. Kinetic energy, carried by the thorium-234 and alpha particles, is released as a result of this decay.

Figure 1.10 Ernest Rutherford was born in New Zealand and is considered to be one of the greatest experimental physicists that ever lived. His discoveries form the foundation of nuclear physics.

8


Uranium-238 is radioactive and may decay by emitting an alpha particle from its nucleus. This can be represented in a nuclear equation in which the changes occurring in the nuclei can be seen. Electrons are not considered in these equations—only nucleons. The equation for the alpha decay of uranium-238 is: 238 U → 234 Th + 42α + energy 92 90 or α 234 238 U→ Th 92 90 In the decay process, the parent nucleus 238 U has spontaneously emitted 92 an alpha particle (α) and has changed into a completely different element, 234 Th. Thorium-234 is called the daughter nucleus. The energy released is 90 mostly kinetic energy carried by the fast-moving alpha particle. When an atom changes into a different element, it is said to have undergone a nuclear transmutation. In nuclear transmutations, electric charge is conserved—seen as a conservation of atomic number. In the above example 92 = 90 + 2. The number of nucleons is also conserved: 238 = 234 + 4.

!

In any nuclear reaction, including radioactive decay, atomic and mass numbers are conserved. Energy is released during these decays.

Beta decay –10β Beta particles are electrons, but they are electrons that have originated from the nucleus of a radioactive atom, not from the electron cloud. A beta particle can be written as −10e, β, β− or −10β. The atomic number of −1 indicates that it has a single negative charge, and the mass number of zero indicates that its mass is far less than that of a proton or a neutron. Beta decay occurs in nuclei in which there is an imbalance of neutrons to protons. Typically, if a light nucleus has too many neutrons to be stable, a neutron will spontaneously change into a proton, and an electron and an – uncharged massless particle called an antineutrino ν are ejected to restore the nucleus to a more stable state. Consider the isotopes of carbon: 126C, 136C and 146C. Carbon-12 and carbon‑13 are both stable, but carbon-14 is unstable. It has more neutrons and so undergoes a beta decay to become stable. In this process one of the neutrons changes into a proton. As a result, the proton number increases to seven, and so the product is not carbon. Nitrogen-14 is formed and energy is released. antineutrino N

carbon–14: unstable

nitrogen–14: stable

beta particle –10B

Figure 1.12 The nucleus of carbon-14 is unstable. In order to achieve stability, one neutron transforms into a proton, and an electron and antineutrino are emitted in the process. The emitted electron is a beta particle, and it travels at nearly the speed of light.

The nuclear equation for this decay is: – 14 C → 147N + −10β + ν + energy 6 The transformation taking place inside the nucleus is: – 1 n → 11p + −10e + ν 0 Once again, notice that in all these equations the atomic and mass numbers are conserved. (The antineutrino has no charge and has so little mass that both its atomic and mass numbers are zero.)

Physics file A different form of beta decay occurs in atoms that have too many protons. An example of this is the radioactive decay of unstable nitrogen-12. There are seven protons and five neutrons in the nucleus, and a proton may spontaneously change into a neutron and emit a neutrino and a positively charged beta particle. This is known as a β+ (beta-positive) decay and the positively charged beta particle is called a positron. The equation for this decay is: 12 7

N→

C + +10e + ν + energy

12 6

Positrons, +10β, have the same properties as electrons, but their electrical charge is positive rather than negative. Positrons are an example of antimatter.

Gamma decay γ Generally, after a radioisotope has emitted an alpha or beta particle, the daughter nucleus holds an excess of energy. The protons and neutrons in the daughter nucleus then rearrange slightly and off-load this excess energy by releasing gamma radiation (high-frequency electromagnetic radiation). Gamma rays—like all light—have no mass and are uncharged and so their symbol is 00γ. Being a form of light, gamma rays travel at the speed of light. A common example of a gamma ray emitter is iodine-131. Iodine‑131 decays by beta and gamma emission to form xenon-131 as shown in Figure 1.13.


9

Physics file Gamma decay alone occurs when a nucleus is left in an energised or excited state following an alpha or beta decay. This excited state is known as a metastable state and it usually only lasts for a short time. An example of this is the radioactive decay of iodine‑131, usually a two-stage process. First, a beta particle is emitted and the excited nuclide xenon-131m is formed. Then, the nucleus undergoes a second decay by emitting a gamma ray: I→

131 5 3

131m 54

Xe + −10e

Xe → 13514Xe + γ The ‘m’ denotes an unstable or metastable state. Cobalt-60 and technetium-99 also exist in metastable states. 131m 54

beta particle –10B

iodine-131: unstable

xenon-131

gamma ray 00G

Figure 1.13 In the beta decay of iodine-131, a high-energy gamma ray photon is also emitted. This high-energy electromagnetic radiation has no electric charge—just energy. The beta particle and xenon nucleus also carry energy.

The equation for this decay is: I → 131 Xe + −10e + γ 54

131 53

or

b, g

I → 131 Xe 54 Since gamma rays carry no charge and have almost no mass, they have no effect when balancing the atomic or mass numbers in a nuclear equation. The chart in Figure 1.14 identifies the 272 stable nuclides, as well as some radionuclides and decay modes. 131 53

Worked example 1.2A Strontium-90 decays by radioactive emission to form yttrium-90. The equation is: 90Sr → 90Y + X 38 39 Determine the atomic and mass numbers for X and identify the type of radiation that is emitted during this decay.

Solution By balancing the equation, it is found that X has a mass number of zero and an atomic number of −1. X is an electron and so this must be beta decay. The full equation is 90Sr → 90Y + 0e. 38 39 −1

Worked example 1.2B Iodine-131, a radioisotope that is used in the treatment of thyroid cancer, is produced in a two-stage process. First, tellurium-130 (130 Te) is bombarded with neutrons inside the core 52 of a nuclear reactor. This results in the formation of the very unstable tellurium-131 and a gamma ray. a Write down the balanced nuclear equation for this process. b Tellurium-131 decays by beta emission to produce a daughter nuclide and an antineutrino. Identify the daughter nuclide.

Solution a 13052 Te + 10n → 13152 Te + γ b Both the atomic and mass number of the antineutrino are zero. The beta particle has a

10


mass number of zero and an atomic number of –1. 131 Te → 131 X + 0β + – ν 52 53 −1 Balancing the nuclear equation gives the unknown element an atomic number of 53 and a mass number of 131. The periodic table reveals the daughter nuclide to be iodine‑131.

140 Bismuth, Z = 83 130

120

110

100

Number of neutrons (N)

90

80

‘Line of stability’ 70

60

50

40

30

20

10

0

0

10

20

30

A

A

AAA

AAAA AAAAAAA

N=Z

Promethium, Z = 61

Technetium, Z = 43

Key

A

40

A

AAAA AAAA AAAAAA AAAAAA

AAAAAA AAAAAA

AAAAAA AAAAAA AAAAAAA

AAAAAA

AAAAA

AAAAA

AAAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAAAA

AAAAAA

AAAAAA

AAAAAA

A AAAAAA

AAAAAA

AAAAAAA AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA A AAAA A A A AAA AA AAA A AAA A AAA AAA AAAA AAAA AAAA A AAAA AAAAA AAAAA AAA AA AA AAAAAAAAA AAAAAAAAAA

A A A A A AAAA AAA AAA AAA AAA AAA AAA A A A A A AAAA AAA AAA AA AA AA A A A A AA AA A A

50

60

stable nuclide

B emitter B emitter emitter

70

80

90

100

Atomic number (Z)

Figure 1.14 From this table of stable isotopes and radioisotopes, it is evident that for larger nuclei there is a distinct imbalance between the number of protons and neutrons. The ‘line of stability’ of the stable nuclides can be seen as a line that curves away from the N = Z line. Notice that every element, up to and including bismuth, has stable isotopes, except for technetium and promethium. Also notice that every isotope of every element beyond bismuth is radioactive.


11

Why radioactive nuclei are unstable

Physics file

Inside the nucleus there are two completely different forces acting. The first is an electric force of repulsion between the protons. On its own, this would blow the nucleus apart, so clearly a second force must act to bind the nucleus together. This is the nuclear force, a strong force of attraction between nucleons, which acts only over a very short range. In a stable nucleus, there is a delicate balance between the repulsive electric force and the attractive nuclear force. For example, bismuth-209, the heaviest stable isotope, has 83 protons and 126 neutrons, and the forces between the nucleons balance to make the nucleus stable. Compare this with bismuth-211. It has two extra neutrons and this upsets the balance between forces. The nucleus of 211Bi is unstable and it ejects an alpha particle in an attempt to attain nuclear stability. Figure 1.14 shows all the stable nuclei with their proton and neutron numbers. It is evident that there is a ‘line of stability’ along which the nuclei tend to cluster. Nuclei away from this line are radioactive. For small nuclei with atomic numbers up to about 20, the ratio of neutrons to protons is close to one. However, as the nuclei become bigger, so too does the ratio of neutrons to protons. Zirconium (Z = 40) has a neutron to proton ratio of about 1.25, while for mercury (Z = 80) the ratio is close to 1.66. This indicates that for higher numbers of protons, nuclei must have even more neutrons to remain stable. These neutrons dilute the repelling forces that act between the extra protons. Elements with more protons than bismuth (Z = 83) simply have too much repulsive charge and additional neutrons are unable to stabilise their nuclei. All of these atoms are radioactive.

Neutrinos are particles with the lowest mass in nature, and they permeate the universe. Neutrinos have no charge and their mass has only recently been discovered to be about one-billionth that of a proton, i.e. about 10−36 kg. While you have been reading these sentences, billions of neutrinos have passed right through your body, and continued on to pass right through the Earth! Fortunately neutrinos interact with matter very rarely and so are completely harmless. It has been estimated that if neutrinos passed through a piece of lead 8 light-years thick, they would still have only a 50% chance of being absorbed!

Interactive tutorial 1 Atomic stability

How radiation is detected

thin mica window

positively charged electrode + –

argon gas

negatively charged aluminium tube

Figure 1.15 A radioactive emission that enters the tube in a Geiger counter will ionise the argon gas and cause a pulse of electrons to flow between the electrodes. This pulse registers as a count on a meter.

12


Our bodies cannot detect alpha, beta or gamma radiation. Therefore a number of devices have been developed to detect and measure radiation. A common detector is the Geiger counter. These are used: • by geologists searching for radioactive minerals such as uranium • to monitor radiation levels in mines • to measure the level of radiation after a nuclear accident such as the accident at Chernobyl • to check the safety of nuclear reactors • to monitor radiation levels in hospitals and factories. A Geiger counter consists of a Geiger–Muller tube filled with argon gas as shown in Figure 1.15. A voltage of about 400 V is maintained between the positively charged central electrode and the negatively charged aluminium tube. When radiation enters the tube through the thin mica window, the argon gas becomes ionised and releases electrons. These electrons are attracted towards the central electrode and ionise more argon atoms along the way. For an instant, the gas between the electrodes becomes ionised enough to conduct a pulse of current between the electrodes. This pulse is registered as a count. The counter is often connected to a small loudspeaker so that the count is heard as a ‘click’. People who work where there is a risk of continuing exposure to lowlevel radiation usually pin a small radiation-monitoring device to their clothing.

This could be either a film badge or a TLD (thermoluminescent dosimeter). These devices are used by personnel in nuclear power plants, hospitals, airports, dental laboratories and uranium mines to check their daily exposure to radiation. When astronauts go on space missions, they wear monitoring badges to check their exposure to damaging cosmic rays. Film badges contain photographic film in a lightproof holder. The holder contains several filters of varying thickness and materials covering a piece of film. After being worn for a few weeks, the film is developed. Analysis of the film enables the type and amount of radiation to which the person has been exposed to be determined. Thermoluminescent dosimeters are more commonly used than film badges. TLDs contain a disk of lithium fluoride encased in plastic. Lithium fluoride can detect beta and gamma radiation as well as X-rays and neutrons. Thermoluminescent dosimeters are a cheap and reliable method for measuring radiation doses.

Figure 1.16 Film badges are used by doctors, radiologists, dentists and technicians who work with radiation, to monitor their exposure levels.

PRACTICAL ACTIVITY 1 Detecting radiation with a Geiger– Muller tube

Physics in action

How technetium is produced Technetium-99m is the most widely used radioisotope in nuclear medicine. It is used for diagnosing and treating cancer. However, this radioisotope decays relatively quickly and so usually needs to be produced close to where it is to be used. Technetium-99m is produced in small nuclear generators that are located in hospitals around the country. In this process, the radioisotope molybdenum-99, obtained from Lucas Heights, is used as the parent nuclide. Molybdenum‑99 decays by beta emission to form a relatively stable (or metastable) isotope of technetium, technetium-99m, as shown below: 99 42

Mo →

Tc +

99m 43

0 −1

β + –ν

Technetium-99m is flushed from the generator using a saline solution. The radioisotope is then diluted and attached to an appropriate chemical compound before being administered to the patient as a tracer. Technetium-99m is purely a gamma emitter. This makes it very useful as a diagnostic tool for locating and treating cancer. Its decay equation is: Tc →

99m 43

99 43

Tc + γ

Figure

1.17 radioiso Technetium g en to and gam pes. The genera erators are use d tor has ma radia a thick le in hospitals th tion. at re ad shield that abs quire orbs the beta


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1.2 summary Radioactivity and how it is detected • Radioactive isotopes may decay, emitting alpha, beta and gamma radiation from their nuclei. • An alpha particle, α, consists of two protons and two neutrons. It is identical to a helium nucleus and can be written as 42α, α2+ or 42He. • A beta particle, β, is an electron, −10e, that has been emitted from the nucleus of a radioactive atom as a result of a neutron transmuting into a proton. • A gamma ray, γ, is high-energy electromagnetic

radiation that is emitted from the nuclei of radioactive atoms. Gamma rays usually accompany an alpha or beta emission. • In any nuclear reaction, both atomic and mass numbers are conserved. • Radiation can be detected using a device such as a Geiger counter. People can monitor their exposure to radiation with film badges and thermoluminescent dosimeters.

1.2 questions Radioactivity and how it is detected 1 From which part of a radioisotope, the nucleus or the electron cloud, are the following particles emitted? a alpha particles b beta particles c gamma rays 2 Discuss the physical differences between α, β and γ rays. 3 Identify each of these particles. a −10A b 11B c 42C d 10D 4 Determine the atomic and mass numbers for the unknown elements, X, in these decay equations, then use the periodic table to identify the elements. a, g a 218 Po → X + α + γ b 235 U→X 84 92 b, g c 228 Ra → X + β + γ d 198 Au → X 88 79 5 Determine the mode of radioactive decay for each of the following transmutations. x, g a 218 Rn → 214 Po + X + γ b 234 Pa → 234 U 86 84 91 92 x, g 214 239 235 c 214 Pb → Bi + X + γ d Pu U → 82 83 94 92 60m 60 e 27 Co → 27Co + X 6 When the stable isotope boron-10 is bombarded with neutrons, it transmutes by neutron capture into a different element X and emits alpha particles. The equation for this reaction is: 10 5

B + 10n → X + 42He.

Identify the final element formed. 7 Identify the unknown particles in these nuclear transmutations. a 147N + α → 178O + X b 27 Al + X → 27 Mg + 11H 13 12 14 14 1 23 26 c 7N + X → 6C + 1p d 11Na + X → 12Mg + 11H 8 Carbon-14 decays by beta emission to form nitrogen-14. The equation for this is 146C → 147 N + −10e + –ν. It can be seen that the carbon nucleus initially has six protons and eight neutrons.

14


a List the particles that comprise the decay side of this equation. b Analyse the particles and determine which particle from the parent nucleus has decayed. c Write an equation that describes the nature of this decay. d Energy is released during this decay. In what form does this energy exist? 9 Use the chart in Figure 1.14 to answer these questions. a List all the stable nuclides of calcium, Z = 20. b How many stable nuclides does niobium, Z = 41, have? c 48 K has a large imbalance of neutrons over protons 19 and so is radioactive. Find potassium-48 on the chart and determine whether it is an alpha or beta emitter. d Write the decay equation for potassium-48 and determine whether the daughter nucleus is itself stable or radioactive. e Calculate the ratio of neutrons to protons for each of potassium-48 and its daughter nucleus. f 217 Fr is a radioisotope. Is it an alpha or beta 87 emitter? g Determine the decay processes that francium‑217 undergoes before it becomes a stable nuclide; identify this nuclide. 10 Gold has only one naturally occurring isotope, 197Au. If a piece of gold foil is irradiated with neutrons, neutron capture will occur and a radioactive isotope of gold will be produced. This radioisotope is a beta emitter. Write an equation that describes the: a neutron absorption process b decay process.

1.3

Proper ties of alp and gamma ra ha, beta diation

Alpha particles, beta particles and gamma rays all originate from the same place—the nucleus of a radioisotope. Each type of radiation has enough energy to dislodge electrons from the atoms and molecules that they smash into. This property is what makes radiation dangerous, but it also enables it to be detected. The properties of alpha, beta and gamma radiation are distinctly different from each other. During early investigations of radioactivity, the emissions from a sample of radium were directed through a magnetic field. As shown in Figure 1.18, the emissions followed three distinct paths, suggesting that there were three different forms of radiation being emitted.

magnet

α

N

γ

β

Figure 1.18 When radiation from radium passes through a magnetic field, the radiation splits up and takes three different paths. One path is undeflected. The other two paths deviate in opposite directions and to different extents. This suggests that there are three different forms of radiation being emitted from radium.

Alpha particles Alpha particles, α, consist of two protons and two neutrons. Because an alpha particle contains four nucleons, it is relatively heavy and slow moving. It is emitted from the nucleus at speeds of up to 20 000 km s−1 (2.0 × 107 m s−1), just less than 10% of the speed of light. Alpha particles have a double positive charge. This, combined with their relatively slow speed, makes them very easy to stop. They only travel a few centimetres in air before losing their energy, and will be completely absorbed by thin card. They have a poor penetrating ability. (a)

S

Ra

PRACTICAL ACTIVITY 2 A model of alpha scattering

~0.1c

4 2A

(b)

~0.9c

0 –1 B

(c)

c 0 0G

ray α

Figure 1.19 The relative speeds of alpha, beta and gamma radiation. (a) Alpha particles are the slowest of the radioactive emissions. Typically they are emitted from the nucleus at up to 10% of the speed of light. (b) Beta particles are emitted from the nucleus at speeds up to 90% of the speed of light. (c) Gamma radiation, being high-energy light, travels at the speed of light (3.0 × 108 m s−1).

β γ

aluminium

Beta particles Beta particles, β, are fast-moving electrons, created when a neutron decays into three parts—a proton, an electron (the beta particle) and an antineutrino. Beta particles are much lighter than alpha particles, and so they leave the nucleus with far higher speeds—up to 90% of the speed of light.

lead

Figure 1.20 Gamma rays can pass through human tissue and sheets of aluminium quite readily. A 5 cm thick sheet of lead is needed to stop 97% of the gamma rays in a beam. By comparison, alpha particles are not capable of penetrating through a sheet of paper or beyond the skin of a person.


15

PRACTICAL ACTIVITY 3 The diffusion cloud chamber Physics file Some types of radiation such as radio waves are harmless. Other types, however, are dangerous to humans. Known as ionising radiation, these interact with atoms, having enough energy to remove outer-shell electrons and create ions. Alpha particles, beta particles and gamma rays are all ionising. So too is electromagnetic radiation with a frequency above 2 × 1016 Hz. Thus, X-rays and ultraviolet-B radiation are ionising. When ionising radiation interacts with human tissue, it is the ions that it produces that are harmful and which lead to the development of cancerous tumours. Lower energy electromagnetic radiation such as radio waves, microwaves, infrared, visible light and ultraviolet-A are non-ionising. We are exposed to significant amounts of such radiation each day with no serious consequences. Non-ionising radiation does not have enough energy to change the chemistry of the atoms and molecules that make up our body cells.

Physics file X-rays and gamma rays are ionising radiations. They are both high-energy forms of electromagnetic radiation (released as high-energy photons), but gamma rays have higher energies. This means that gamma rays are more highly penetrating than X-rays. The defining distinction between X-rays and gamma rays is the method of production. X-rays are created from electron transitions within the electron cloud, whereas gamma rays are emitted from the nuclei of radioactive atoms. Gamma rays and X-rays have similar properties, but X-rays are not the result of radioactive decay.

Beta particles are more penetrating than alpha particles, being faster and with a smaller charge. They will travel a few metres through air but, typically, a sheet of aluminium about 1 mm thick will stop them.

Gamma rays Gamma rays, γ, being electromagnetic radiation with a very high frequency, have no rest mass and travel at the speed of light—3.0 × 108 m s−1 or 300 000 km s−1. They have no electric charge. Their high energy and uncharged nature make them a very penetrating form of radiation. Gamma rays can travel an almost unlimited distance through air and even a few centimetres of lead or a metre of concrete would not completely absorb a beam of gamma rays.

The ionising abilities of alpha, beta and gamma radiation When an alpha particle travels through air, its slow speed and double positive charge cause it to interact with just about every atom that it encounters. The alpha particle dislodges electrons from many thousands of these atoms, turning them into ions. Each interaction slows it down a little, and eventually it will be able to pick up some loose electrons to become a helium atom. This takes place within a centimetre or two in air. As a consequence, the air becomes quite ionised, and the alpha particles are said to have a high ionising ability. Since the alpha particles don’t get very far in the air, they have a poor penetrating ability. Beta particles have a negative charge and are repelled by the electron clouds of the atoms they interact with. This means that when a beta particle travels through matter, it experiences a large number of glancing collisions and loses less energy per collision than an alpha particle. As a result, beta particles do not ionise as readily and will be more penetrating. Gamma rays have no charge and move at the speed of light, and so are the most highly penetrating form of radiation. Gamma rays interact with matter infrequently, when they collide directly with a nucleus or electron. The low density of an atom makes this a relatively unlikely occurrence. Gamma rays pass through matter very easily—they have a very poor ionising ability but a high penetrating ability.

The energy of α, β and γ radiation The energy of moving objects such as cars and tennis balls is measured in joules. However, alpha, beta and gamma radiations have such small amounts of energy that the joule is inappropriate. The energy of radioactive emissions is usually expressed in electronvolts (eV). An electronvolt is the energy that an electron would gain if it were accelerated by a voltage of 1 volt.

One …L…CTRONVOLT is an extremely small quantity of energy equal to 1.6 × 10−19 J, i.e. 1 eV = 1.6 × 10−19 J.

16


Alpha and beta particles are ejected from unstable nuclei with a wide range of energies. Alpha particles typically have energies of 5–10 million electronvolts (5–10 MeV). This corresponds to speeds of about 16 000 km s−1, about 5–10% of the speed of light. Beta particles are usually ejected with energies up to a few million electronvolts. For example, sodium-24 emits beta particles with a maximum energy of 1.4 MeV. This is equivalent to 2.24 × 10−13 J. These particles are travelling at speeds quite close to the speed of light. Gamma rays normally have less than a million electronvolts of energy. They may even have energy as low as 100 000 electronvolts. For example, the gamma rays emitted by the radioactive isotope gold-198 have a maximum energy of 412 000 eV (412 keV) or 6.6 × 10−14 J. Increasing the energy of a gamma ray does not increase its speed; it increases the frequency of the radiation.

Interactive tutorial 1 Atomic Stability

Table 1.1 The properties of alpha, beta and gamma radiations Property Mass Charge

α particle

b particle

g ray

heavy

light

none

+2

−1

none

Typical energy

~5 MeV

~1 MeV

~0.1 MeV

Range in air

a few cm

1 or 2 m

many metres

~10−2 mm

a few mm

high

high

reasonable

poor

Penetration in matter Ionising ability

Worked example 1.3A Uranium-238 emits alpha particles with a maximum energy of 4.2 MeV. a Explain why a sample of this radioisotope encased in plastic is quite safe to handle yet, if inhaled as dust, would be considered very dangerous. b Calculate the energy of one of these alpha particles in joules.

Solution a The alpha particles have a poor penetrating ability and so would be unable to pass through the plastic casing. However, if the radioactive uranium was on a dust particle and was inhaled, the alpha-emitting nuclei would be in direct contact with lung tissue and the alpha particles would damage this tissue. b 4.2 MeV = 4.2 × 106 eV = 4.2 × 106 × 1.6 × 10−19 J = 6.7 × 10−13 J

Physics file The energy released during any nuclear reaction (including radioactive decay) is many times greater than that released in a typical chemical reaction. For example the chemical reaction of a sodium ion capturing an electron releases about 1 eV of energy. Na+ + e− → Na + 1 eV Nuclear reactions such as alpha, beta and gamma decays typically release energies of the order of megaelectronvolts, MeV, i.e. nuclear reactions release about a million times more energy than chemical reactions.


17

Physics in action

Smoke detectors Each year, dozens of people in Australia die as a result of domestic fires. Evidence has shown that the installation of a smoke detector can reduce the risk of dying in a house fire by about 60%. For this reason, new houses are required to contain at least one smoke detector. Domestic smoke detectors contain a small radioactive source. The radioisotope most commonly used is americium-241, an artificial isotope which is produced in the core of a nuclear reactor. Americium‑241 emits alpha particles and low-energy gamma rays. The penetrating ability of the alpha particles is so poor that they are stopped by the case of the detector. Some gamma rays will escape into the room, but they have such low energy (~60 keV) that exposure to them is insignificant when compared with the level of background radiation. As well as this, the detectors are usually located in the ceiling, some distance from people, and this distance further reduces the intensity of the radiation. A smoke detector contains a pair of oppositely charged low-voltage metal electrodes. When the alpha particles pass between these electrodes, they ionise the air molecules that are present. These ions are then attracted to the electrodes. However, when smoke (or steam) is present, the ions attach themselves to the smoke particles. The flow of charges to the electrodes reduces greatly because these charged smoke (or steam) particles are much bigger and so much less mobile than the ionised air molecules. It is this reduction in the flow of charges reaching the electrodes that triggers the alarm.

Figure

1.21 radioacti Smoke detectors ve co the chan material. When u ntain a small qu a ces of be s ing killed ed correctly, the ntity of y greatly or injured reduce in a hous e fire.

Physics in action

Monitoring the thickness of sheet metal Beta particles can be used to monitor the thickness of rolled sheets of metal and plastic during manufacture. A beta particle source is placed under the newly rolled sheet and a detector is placed on the other side. If the sheet is being made too thick, fewer beta particles will penetrate and the detector count will fall. This information is instantaneously fed back to the rollers and the pressure is increased until the correct reading is achieved, and hence the right thickness is attained. Would alpha particles or gamma rays be appropriate for this task? Alpha particles have a very poor penetrating ability, so none of them would pass through the metal. Gamma rays usually have a high penetrating ability and so a thin metal sheet would not stop them. Workers would also need to be shielded from gamma radiation. You can see that the penetrating properties of beta rays make them ideal for this job. The thickness of photographic film and coatings on metal surfaces are also monitored in this way.

18


control box

rollers Geiger counter

β source

Figure 1.22 The thickness of a sheet of metal is monitored using a strontium-90 isotope. A beam of beta particles is directed into the metal and those penetrating the metal sheet are counted by a detector on the other side. This count gives an indication of the thickness of the metal sheet. The thicker the sheet, the lower the count in any given time period.

1.3 summary Properties of alpha, beta and gamma radiation • Alpha particles, α, are ejected with a speed of about 5–10% of the speed of light. Alpha particles have a double positive electrical charge and are relatively heavy. They are a highly ionising form of radiation, but their penetrating ability is poor. • Beta particles have a single negative electrical charge and are much lighter than alpha particles. They are a moderately ionising and penetrating form of radiation.

• Gamma rays are high-energy electromagnetic radiation and so have no electrical charge. They have a high penetrating ability, but a weak ionising ability. • The energy of alpha, beta and gamma radiation is usually measured in electronvolts (eV). • 1 eV = 1.6 × 10−19 J

1.3 questions Properties of alpha, beta and gamma radiation 1 As part of an experiment, a scientist fires a beam of alpha, beta and gamma radiation at a brick. If the three radiation types are of equal energy, arrange them in order of: a increasing penetrating ability b increasing ionising ability. 2 Which one of the following correctly explains how penetrating ability relates to the ionising ability of a radioactive emission? A Emissions with more ionising ability have greater penetrating ability. B Emissions with less ionising ability have more penetrating ability. C There is no relationship between the ionising ability and penetrating ability of a radioactive emission. 3 An external source of radiation is used to treat a brain tumour. Which type of radioactive emission is best suited for this treatment? 4 A radiographer inserts a radioactive wire into a breast cancer with the intention of destroying the cancerous cells in close proximity to the wire. Should this wire be an alpha, beta or gamma emitter? Explain your reasoning. 5 Cancer patients being treated with an external source of radiation have to wear lead aprons to protect their other tissue from exposure. Which forms of radiation is the lead apron shielding them from?

6 Calculate the energy in joules of: a an alpha particle with 8.8 MeV of energy b a beta particle with 0.42 MeV of energy c a gamma ray with 500 keV of energy. 7 Alpha particles travelling through air ionise about 100 000 atoms each centimetre. Each time they ionise an atom, the alpha particles lose about 34 eV of energy. a How much energy will alpha particles lose as they pass through 1 cm of air? b Calculate the approximate distance that an alpha particle with 5.6 MeV will travel in air before it loses all of its energy. 8 Which one of the following has the greatest penetrating ability? A An alpha particle with 5.3 MeV of energy B A beta particle with 1.2 MeV of energy C A gamma ray with 700 keV of energy D A gamma ray with 0.81 MeV of energy 9 Which radiation identified in question 8 will be the most damaging to human tissue should irradiation occur? 10 A radioactive sample is emitting alpha, beta and gamma radiation into the air. A Geiger counter held about 20 centimetres from the sample would be most likely to detect: A alpha, beta and gamma radiation. B gamma radiation only. C alpha radiation only. D beta and gamma radiation only.


19

topes o s i o i d a r f o y t ife and activi

1.4 Half-l (a) Po

Po 30 minutes later

(b) Ra

Ra 10 years later

Figure 1.23 (a) The emissions from polonium-218 only last for a relatively short time. Its activity decreases very rapidly. (b) The emissions from a sample of radium-226 remain steady for a very long time. Its activity does not change significantly.

PRACTICAL ACTIVITY 4 An analogue experiment of radioactive decay

Different radioisotopes will emit radiation and decay at very different rates. For example, a Geiger counter held close to a small sample of polonium-218 will initially detect a significant amount of radiation, but the activity will not last for very long. After half an hour or so, there will hardly be any radiation detected at all. Compare this with a similar sample of radium-226. A Geiger counter directed at the radium will show a sustained but low count rate—much lower than that of the polonium-218 sample. Furthermore, the activity will remain relatively steady for a very long time. In fact, no change in the count rate would be noticed for decades! To explain this, you need to know that radionuclides are unstable but to different degrees. Consider again the sample of polonium-218. If the sample initially contains 100 million undecayed polonium-218 nuclei, as shown in Figure 1.24, after 3 minutes about half of these will have decayed, leaving just 50 million polonium-218 nuclei. A further 3 minutes later, half of these remaining polonium-218 nuclei will decay, leaving approximately 25 million of the original radioactive nuclei, and so on. Key:

1 million 218Po nuclei

Initially: 100 million 218 Po nuclei

After 3 minutes: After 6 minutes: ~ 50 million ~ 25 million 218 218 Po nuclei Po nuclei

Figure 1.24 During one half-life, the number of nuclei of the radioisotope sample decreases by half (i.e. by 50%). After two half-lives, only one-quarter (25%) of the original radioisotope nuclei will remain.

N = N0( 12) n

100 Percentage remaining

The time that it takes for half of the nuclei of a radioisotope to decay is known as the half-life of that radioisotope. The half-life of polonium-218 is 3 minutes.

where n = no. of half-lives N0 = original amount N = final amount

The decay rate of a radioisotope is measured in terms of its half-life (t1/2). The HALF-LIF… of a radioisotope is the time that it takes for half of the nuclei of the sample radioisotope to decay spontaneously.

50

25

12.5 0

1 2 Number of half-lives

3

Figure 1.25 The amount of the original isotope halves as each half-life passes. This is an exponential relationship and the mathematical relationship that describes it is shown.

20


As time passes, a smaller and smaller proportion of the original radio isotope remains in the sample. The graph in Figure 1.25 shows this. It is important to appreciate that although the behaviour of a large sample of nuclei can be predicted, it is impossible to predict when any one particular nucleus will decay. The decay of the individual nuclei in a sample is random. It is rather like throwing dice. If 60 dice are thrown, on average, 10 will roll up ‘6’. You just don’t know which ones! Furthermore, the half-life of a radioisotope is constant and is largely unaffected by any external conditions such as temperature, magnetic field or the chemical environment. It is related only to the instability of the nucleus of the radioisotope.

Look at Figure 1.23 once again. It is evident that radium-226 has a very long half-life when compared with polonium-218. In fact, the half-life of radium-226 is about 1600 years. Clearly, a sample of radium-226 will emit particles and decay for centuries. The half-lives of some common radioisotopes are shown in Table 1.2. This table also illustrates that the halflife of a radioisotope is a factor in its application. For example, most medical applications using a radioisotope as a tracer require a short half-life. This is so that radioactivity does not remain in the body any longer than necessary. On the other hand, the radioisotope used in a smoke detector is chosen because of its long half-life. The detector can continue to function for a very long time, as long as the battery is replaced each year.

Table 1.2 Some common radioisotopes and their half-lives Emission

Half-life

Polonium-214

α

0.00016 seconds

Strontium-90

β

28.8 years

Cancer therapy

Radium-226

α

1630 years

Once used in luminous paints

Carbon-14

β

5730 years

Carbon dating of fossils

Uranium-235

α

700 000 years

Nuclear fuel, rock dating

Uranium-238

α

4.5 billion years


Thorium-232

α

14 billion years

Fossil dating, nuclear fuel

Technetium-99m

γ

6 hours

Medical tracer

Sodium-24

γ

15 hours

Medical tracer

Isotope

Application

Natural Nothing at this time.

Artificial

Iodine-131

γ

8 days

Medical tracer

Phosphorus-32

β

14.3 days

Medical tracer

Cobalt-60

γ

5.3 years

Radiation therapy

Americium-241

α

460 years

Smoke detectors

Plutonium-239

α

24 000 years

Physics file Try this activity to improve your understanding of half-life and the rate of decay of a radioactive sample. Get 50 dice and toss them all at once. Each die represents an atom of a radioisotope. Let those that come up ‘evens’ (i.e. 2, 4 or 6) be the atoms that have decayed, and remove these from the sample. Because, on average, half of the sample will be removed after each roll, the half-life for this experiment will be ‘one roll’. Now, toss the remaining dice and once more remove the ‘decayed atoms’. Continue this process for say six or seven rolls until almost all of the ‘atoms’ have decayed, and plot a graph of the sample size against the number of rolls—the decay curve of your radioisotope. Now repeat the experiment using the same dice, but this time only those that roll up a ‘6’ have decayed. What is the half-life now? You will also notice that it is impossible to predict when a particular die will ‘decay’—just as it is impossible to predict when a particular atom in a real sample will decay.

Interactive tutorial 2 Radioactive decay and half-life


Activity A Geiger counter records the number of radioactive decays occurring in a sample each second. This is the activity of the sample.

ACTIVITY is measured in becquerels, Bq. 1 Bq = 1 disintegration per second Over time, the activity of any sample of a radioisotope will decrease. This is because more and more of the radioactive nuclei have decayed and will no longer emit radiation. So, over one half-life, the activity of any sample will be reduced by half. If the sample of polonium-218, discussed previously, has an initial activity of 2000 Bq, then after one half-life (i.e. 3 minutes) its activity will be 1000 Bq. After 6 minutes, the activity of the sample will have reduced to 500 Bq and so on.


21

Short-lived radioisotopes have an initially high activity. Their nuclei decay at a fast rate and so the sample lasts only for a short time. Highactivity samples are extremely dangerous and must be handled with great caution.

Decay series Generally, when a radionuclide decays, its daughter nucleus is not completely stable, and is itself radioactive. This daughter will then decay to a grand-daughter nucleus, which may also be radioactive, and so on. Eventually a stable isotope is reached and the sequence ends. This is known as a decay series. The Earth is 4.5 billion years old (4.5 gigayears)—enough to have only four naturally occurring decay series remain active. These are: • the uranium series in which uranium-238 eventually becomes lead-206 • the actinium series in which uranium-235 eventually becomes lead-207 • the thorium series in which thorium-232 eventually becomes lead-208 • the neptunium series in which neptunium-237 eventually becomes bismuth-209. (Since neptunium-237 has a relatively short half-life, it is no longer present in the crust of the Earth, but the rest of its decay series is still continuing.) Geologists analyse the proportions of the radioactive elements in a sample of rock to gain a reasonable estimate of the rock’s age. This technique is known as rock dating. 238 A 4.5 r 109 years

236

Th Pa U

234

B 24 days

232

Th

230

A 2.5 r 105 years

A 8 r 104 years

228 Mass number (A )

U B 6.7 h

226

Ra

224

A 1.6 r 103 years Rn

222

A 3.8 days

220 218 216 214

A 3 min Po B 19 min Pb Bi Po A 160 Ms

212 B 27 min

Pb Bi Po B 20 years B 2.6 r 106 years 208 210 206

Pb A 138 days 82

84

86

88

90

92

Atomic number ( Z )

Figure 1.26 The uranium decay series. The half-life and emissions are indicated on each of the decays as radioactive uranium-238 is transformed into stable lead-206.

22


Worked example 1.4A A sample of the radioisotope thorium-234 contains 8.0 × 1012 nuclei. The half-life of 234Th is 24 days. How many thorium-234 atoms will remain in the sample after: a 24 days? b 48 days? c 96 days?

Solution a Initially, there were 8.0 × 1012 thorium-234 nuclei. 24 days is one half-life, so half of these will decay leaving 4.0 × 1012 thorium-234 nuclei. b 48 days is two half-lives. This means that there will be: 1 1 × × 8.0 × 1012 = 2.0 × 1012 thorium-234 nuclei 2 2 c 96 days corresponds to four half-lives. In this time the number of atoms of the original radioisotope will have halved four times. This means that: 1 1 1 1 1 × × × = 2 2 2 2 16 or one-sixteenth of the original 234Th nuclei remain; i.e. 5.0 × 1011 nuclei.

Worked example 1.4B In 2 hours, the activity of a sample of a radioactive element falls from 240 Bq to 30 Bq. What is the half-life of this element?

Solution During each half-life, the activity of the radioisotope will fall by half. The activity of this element has decreased from 240 → 120 → 60 → 30 counts per second, so it has decayed through three half-lives in this 2 hour (120 minute) period. Thus the half-life must be 120/3 = 40 minutes.

Physics in action

Radiocarbon dating Carbon dating is a technique used by archaeologists to determine the age of fossils and ancient objects that were made from plant matter. In this method, the proportion of two isotopes of carbon—carbon-12 and carbon-14—in the specimen are measured and compared. Carbon-12 is a stable isotope whereas carbon-14 is  radioactive. Carbon-14 only exists in trace amounts in nature. In fact, carbon-12 atoms are about 1 000 000 000 000 (1012) times more prevalent than carbon-14 atoms. Carbon-14 has a half-life of 5730 years and decays by beta emission to nitrogen-14. Its decay equation is:

C → 147N + −10β

14 6

Both carbon-12 and carbon-14 can combine with other atoms in the environment, for example with oxygen to form carbon dioxide. While plants and animals are alive, they take in carbon-based molecules and so all living things will contain the same percentage of carbon-14. In the environment, the production of carbon-14 is matched by its decay and so the proportion of carbon-14 atoms to carbon-12 remains constant.

Figure 1

.27 C show that arbon-dating tech niq th around th e Shroud of Turin w ues were used to e 14th cen as most p robably m tury. ade


23

After a living thing has died, the amount of carbon-14 will decrease as these atoms decay to form nitrogen-14, and are not replaced. The number of atoms of carbon-12 does not change as this is a stable atom. So, over time, the proportion of carbon-14 to carbon-12 atoms falls. By comparing the proportion of carbon-14 to carbon-12 in a dead sample with that found in living things, and knowing the half-life of carbon-14 (5730 years), the approximate age of the specimen can be determined. Consider this example. The count rate from a 1 g sample of carbon that has been extracted from an ancient wooden spear is 10 Bq. A 1 g sample of carbon from a living piece of wood gives a count rate of 40 Bq. We then assume that this was also the initial count rate of the spear. For its count rate to have reduced from 40 to 10 Bq, the spear must be (40 → 20 → 10) two half-lives of carbon-14 old, i.e. about 11 500 years old. In 1988, scientists used carbon-dating techniques to show that the Shroud of Turin was probably a medieval forgery. Carbon-dating tests on samples of the cloth the size of a stamp established that there was a high probability that it was made between 1260 and 1390 ad, not around the time of Christ. Radiocarbon dating is an important aid to anthropologists who are interested in finding out about the migration patterns of early peoples—including the Australian Aborigines. This technique is very powerful since it can be applied to the remains of ancient campfires. It is accurate and reliable

Figure

1.28 found in This baby mamm no oth fossil w dating ha rthern Russia in 2007. Ca as s shown rbo th extinct 1 1 000 ye at mammoths be n ars ago. came

for samples up to about 60 000 years old. Carbon dating cannot be used to date dinosaur bones as they are more than 60 million years old, but it can be used to determine the age of more recently extinct mammoth fossils.

Physics in action

How old is the Earth? Carbon dating is useful when examining samples that were once alive—such as wood or bones. However, this technique cannot be used to date the age of specimens that were never alive, such as rocks. There are a large number of dating procedures that are now used for this purpose. The oldest dating technique analyses remnants of uranium and lead that are found in the rock that is being examined. Uranium has two naturally occurring isotopes: uranium-235 and uranium-238. As was discussed earlier in Decay Series (p. 22), uranium-235 decays through a number of steps and finishes up as lead-207. Uranium-238 undergoes a different series of decays to finally become lead-206. Scientists can compare the proportions of each isotope present using a mass spectrometer and, knowing the half-lives involved, determine the age of the rock. If, for example, a rock sample was quite young (i.e. it had crystallised relatively recently), it would contain higher levels of uranium and lower levels of lead because there has not been time for many uranium atoms to complete the decay process. The oldest rocks that have been found on Earth have been dated at almost 4 billion years. Most rocks are much younger than this as a result of remelting and reforming over the ages. When rocks brought back from the Moon were analysed, they were found to be 4.2 billion years old. Furthermore, when different meteorites were analysed, they were all found to be exactly the same age of 4.56 billion years. These observations can be explained by assuming that the meteorites are parts of asteroids that have drifted into Earth’s orbit. The current

24


Figure 1.

29 All meteorit es have been fo the same age— und to be exac 4.56 billion ye tly ars. This result scientists to fix has enabled the age of the Ea about 4.6 billion rth and the so lar system at years.

theory suggests that the solar system was formed all at once and that the age of the asteroids gives a reliable estimate of the age of the solar system. In other words, the age of the Earth and the rest of the solar system is about 4.6 billion years. In all, there are about forty different dating techniques and they have been found to give very consistent and reliable results.

1.4 summary Half-life and activity of radioisotopes • The rate of decay of a radioisotope is measured by its half-life. The half-life, t1/2 , of a radioisotope is the time that it takes for half of the nuclei in a sample of the radioisotope to decay. • The activity of a sample indicates the number of radioactive decays that are occurring in the sample

each second. Activity is measured in becquerels (Bq) where 1 Bq = 1 disintegration per second. • The activity of any radioactive sample will decrease with time. Over a half-life, the activity of a sample will halve.

1.4 questions Half-life and activity of radioisotopes

2 A radioactive element has a half-life of 15 minutes. If you start with a 20 g sample of this element, how much of the original radioisotope will remain after: a 15 minutes? b 30 minutes? c 45 minutes? d 1.5 hours? 3 A Geiger counter measures the radioactive disintegra tions from a sample of a certain radioisotope. The count rate recorded is shown below. Count rate (Bq)

400

280

200

140

100

70

Time (minutes)

0

10

20

30

40

50

a Plot a graph of count rate against time. b Use your graph to estimate the activity of the sample after 15 minutes. c What is the half-life of this element? Use both your graph and the table to determine your answer. d Determine the activity of the sample after 60 minutes have elapsed. 4 The activity of a radioisotope changes from 6000 Bq to 375 Bq over a period of 1 h. What is the half-life of this element? 5 Gold-198 is a radioisotope with a half-life of 2.7 days. Consider one particular nucleus in a small sample of this substance. After 2.7 days this nucleus has not decayed. What is the probability that it will decay in the next 2.7 day period? 6 A hospital in Alice Springs needs 12 mg of the radio isotope technetium-99m, but the specimen must be ordered from a hospital in Sydney. If the half-life of 99mTc is 6 hours and the delivery time between hospitals is 24 hours, how much must be produced in Sydney to satisfy the Alice Springs order?

7 Radioactive materials are considered to be relatively safe when their activity has fallen to below 0.1% of their initial value. a How many half-lives does this take? b Plutonium-239 is a by-product of nuclear reactors. It has a half-life of about 24 000 years. For what period of time does a quantity of 239Pu have to be stored until it is considered safe to handle? 8 Uranium-235 has a half-life of 700 000 years, while the half-life of uranium-238 is many times longer at 4.5 × 109 years. a If you had 1 kg of each of these radioisotopes, which one would have the greater activity? b The uranium that is mined in Australia and other parts of the world is 99.3% 238U and only 0.7% 235U. Explain why 235U currently exists in trace amounts only. 9 A Geiger counter measures the radioactive disintegra tions from a sample of a certain radioisotope. The graph of the count rate is shown below. 1000 800

Activity (Bq)

1 A radioactive isotope has a half-life of 1 hour. If a sample initially contains 100 mg of this isotope, which one of the following correctly gives the amount of the radioisotope remaining after 2 hours have elapsed? A none B 50 mg C 25 mg D 100 mg

600 400 200 0

0

5

10

15 Time (min)

20

25

30

a Determine the half-life of the isotope. b What would be the activity of the isotope after 40 min? 10 A geologist analyses a sample of uranium ore that has been mined at Roxby Downs in South Australia. You may refer to Figure 1.26, the decay series graph for uranium, when answering this question. a Explain why the sample would be expected to contain significant traces of lead. b Explain why the geologist would be unlikely to find any 214Po in the sample.


25

its d n a e s o d n o i Radiat 1.5 effect on humans Physics file The wicks or mantles used in old-style camping lamps are slightly radioactive. They contain a radioisotope of thorium, an alpha-particle emitter. They have not been banned from sale because they contain only small amounts of the radioisotope and could be used safely by taking simple precautions such as washing hands and avoiding inhalation or ingestion. A scientist from the Australian National University has called for the banning of these mantles on the grounds that they tend to crumble and turn to dust as they age. If this dust were to be inhaled, alpha particles could settle in someone’s lung tissue, possibly causing cancers to form. Several years ago, a schoolboy in the United States used thousands of lamp mantles to construct a crude nuclear reactor. It raised the background level of radiation in his street by a factor of 9000!

Ionising radiation The term radiation is widely used and widely misunderstood. There are many different forms of radiation and the degree of danger that they present depends on their ability to interact with atoms. Some radiation has enough energy to interact with atoms, removing their outer-shell electrons and creating ions. For this reason, these radiations are known as ionising radiations. As was discussed in Section 1.3, alpha particles, beta particles and gamma rays are all ionising. The electromagnetic spectrum consists of a variety of electromagnetic radiations: radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays. Electromagnetic radiation with a frequency above 2 × 1016 Hz is ionising. Thus gamma rays, X-rays and ultraviolet B and C radiations are ionising and you would be well advised to avoid exposure to them. When they interact with the tissue in an organism, they create ions which can lead to the development of cancerous tumours. Non-ionising radiation includes radio waves, microwaves, visible light and UV-A radiation. We are exposed to significant amounts of such radiation each day without serious consequences. The level of exposure to radiation from the environment is called the background level.

Table 1.3 Summary of the different ionising and non-ionising types of radiation

Figure 1.30 Radiologists administer very precise doses of ionising radiation that are designed to destroy cancer cells. This treatment is successful because rapidly dividing cancer cells are more susceptible than normal body cells to damage from ionising radiation.

Ionising radiation (high energy)

alpha particles, beta particles, gamma rays, X-rays, UV-B and UV-C radiation

Non-ionising radiation (low energy)

radio waves, microwaves, visible light, infrared, UV-A radiation

The background level of ionising radiation to which we are continually exposed is not a significant health problem. However, exposure to aboveaverage levels of ionising radiation is dangerous. It may lead to long-term problems such as cancer and genetic deformities in future generations. Extremely high levels of exposure can cause death, and in extreme cases this can happen within just a few hours. It is important that people who work with radiation in fields such as medicine, mining, nuclear power plants and industry are able to monitor closely the amount of radiation to which they are exposed. Furthermore, radiologists, who administer courses of radiation treatment to cancer patients, also need to be able to measure the amount of radiation that they are applying.

Measuring radiation exposure 1 Absorbed dose When a person is exposed to high-energy radiation, the energy of the radiation acts to break apart molecules and ionise atoms in the person’s body cells. The severity of this exposure depends on the amount of radiation

26


energy that has been absorbed by the individual’s body. This quantity is known as the absorbed dose. The absorbed dose is the radiation energy that has been absorbed per kilogram of the target material.

13% cosmic rays

energy absorbed by tissue mass of tissue ABSORB…D DOS… is measured in joules/kilogram (J kg−1) or grays (Gy), i.e. 1 Gy = 1 J kg−1.

20% ground and building

ABSORB…D DOS… =

17% food and drink

To illustrate this, if a 25 kg child absorbed 150 J of radiation energy, then the absorbed dose would be 6 Gy. This is a massive dose and would be enough to kill the child within a few weeks. However, an adult, being much larger, would be less severely affected by this radiation. If a 75 kg adult absorbed 150 J of radiation energy, the absorbed dose would be just 2 Gy. This dose would give the adult a severe case of radiation sickness but would probably not be fatal. You can think of dose in the same way as one administers medicine. The small mass of a child means that taking just half a tablet might be equivalent to an adult taking two tablets.

37% natural radioactivity in air 10.4% medical

1.6% coal burning

2 Dose equivalent Different forms of radiation have different abilities to ionise, and so cause different amounts of damage as they pass through human tissue. Alpha particles are the most ionising form of radiation. Their low speed, high charge and large mass mean that they interact with and ionise virtually every atom that lies in their path. This means that an absorbed dose of alpha radiation is much more damaging to human tissue than an equal absorbed dose of beta or gamma radiation. In fact, it is about 20 times more damaging. This weighting of the biological impact of the radiation is called the quality factor. A list of quality factors is shown in Table 1.4. By way of contrast with alpha particles, gamma rays and X-rays, having no charge and moving at the speed of light, fly straight past most atoms and interact only occasionally as they pass through a substance. This is reflected in their low quality factor. A measure of radiation dose that takes into account the absorbed dose and the type of radiation will give a more accurate picture of the actual effect of the radiation on a person. This is the dose equivalent. Dose equivalent is measured in sieverts (Sv), although millisieverts (mSv) and microsieverts (mSv) are more commonly used.

0.5% nuclear weapons fallout

0.5% air travel, etc.

0.001% nuclear power

Figure 1.31 Humans are exposed to radiation from many different sources. Almost 90% of our annual exposure is from the surrounding environment.

Table 1.4 Quality factors

DOS… …QUIVAL…NT = absorbed dose × quality factor Dose equivalent is measured in sieverts (Sv).

Radiation

For example, an absorbed dose of just 0.05 Gy of alpha radiation is biologically equally as damaging as an absorbed dose of 1.0 Gy of beta radiation. While the energy carried by the alpha particles is lower than that of beta particles, each alpha particle does far more damage. In each case, the dose equivalent is 1 Sv, and 1 Sv of any radiation causes the same amount of damage.

Quality factor

Alpha particles

20

Neutrons* (>10 keV)

10

Beta particles

1

Gamma rays

1

X rays

1

* Radiation from neutrons is only found around nuclear reactors and neutron bomb explosions.


27

Physics file The level of background radiation varies around the world as Table 1.5 shows. Locations of greater latitude and greater altitude receive a larger dose of cosmic rays. Aberdeen has a high reading because it is built on large deposits of granite that release radon, a radioactive gas. The soil in Chennai is slightly radioactive and is responsible for the higher than average doses received there.

Table 1.5 Background radiation levels around the world Location

Annual background radiation dose (mSv)

Australia (average level)

2000

New York, USA

1000

Paris, France

1200

Aberdeen, Scotland

5000

Chennai, India

8000

It is important to appreciate that 1 Sv is a massive dose of radiation and, while not being fatal, would certainly lead to a severe case of radiation illness. In Australia the average annual background radiation dose is about 2.0 mSv, or 2000 mSv. A microsievert is a millionth of a sievert. Use Table 1.6 to estimate your annual dose.

Table 1.6 Annual radiation doses in Australia Radiation source Cosmic radiation Rocks, air and water Radioactive foods and drinks Manufactured radiation

Medical exposures

Physics file Cosmic rays—atomic particles and gamma rays that continually bombard the Earth—emanate from the Sun and from deep space. We are, to a large extent, protected from these by the shielding effect of the Earth’s atmosphere and magnetosphere. Most people receive a dose of around 300 mSv each year due to cosmic radiation. However, when we travel at high altitudes, the atmosphere’s shielding effect is diminished. Passengers taking a return flight to Perth from Melbourne would be exposed to a dose of about 30 mSv.

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Average Local variations annual dose (mSv) 300

1350

350

Plus 200 mSv for each round-the-world flight. Plus 20 mSv for each 10° of latitude. Plus 150 mSv if you live 1000 m above sea level. Plus 1350 mSv if you live underground. Plus 1350 mSv if your house is made of granite. Minus 140 mSv if you live in a weatherboard house. Plus 1000 mSv if you have eaten food affected by the Chernobyl fallout.

60

Plus 60 mSv if you live near a coal-burning power station. Plus 30 mSv from nuclear testing in the Pacific. Plus 20 mSv if you watch 20 hours of TV on a CRT television set each week.

–

Plus 30 mSv for a chest X-ray. Plus 300 mSv for a pelvic X-ray. Plus 1000 mSv if you have had a ‘barium milkshake’ ulcer examination. Plus 40 000 000 mSv for a course of radiotherapy using cobalt-60.

Worked example 1.5A A 10 g cancer tumour absorbs 0.0020 J of energy from an applied radiation source. a What is the absorbed dose for this tumour? b Calculate the dose equivalent if the source is an alpha emitter. c Calculate the dose equivalent if the source is a gamma emitter d Which radiation source is more damaging to the cells in the tumour?

Solution energy absorbed a Absorbed dose = =

0.020 J = 0.20 Gy. mass of tissue 0.010 kg b Dose equivalent = absorbed dose × quality factor = 0.20 × 20 = 4.0 Sv for the alpha emitter. c Dose equivalent = absorbed dose × quality factor = 0.20 × 1 = 0.20 Sv for the gamma emitter. d The alpha particle source is more damaging. It causes more ionisation in the cells and so has a higher dose equivalent.

3 …ffective dose

Table 1.7 The ICRP weighting values, W

The different organs of the body have different sensitivities to radiation doses. For example, if a person’s lung was exposed to a dose of 10 mSv, it would be more than twice as likely that cancers would develop than if the same 10 mSv dose was delivered to the liver. The weightings assigned by the International Commission of Radiological Protection (ICRP) to the various organs in shown in Table 1.7. Effective dose is used to compare the risk of a non-uniform exposure to ionising radiation with the risks caused by a uniform exposure of the whole body. It is found by calculating a weighted average of the dose equivalents to different body parts, with the weighting factors, W, designed to reflect the different radiosensitivities of the tissues.

Body part

…FF…CTIV… DOS… = S(dose equivalent × W) Effective dose is measured in sieverts (Sv).

Weighting, W

Ovaries/testes

0.20

Bone marrow

0.12

Colon

0.12

Lung

0.12

Stomach

0.12

Bladder

0.05

Breast

0.05

Liver

0.05

Oesophagus

0.05

Thyroid

0.05

Rest of body

0.07

Total

1.00

Worked example 1.5B During therapy for cancer, a patient’s lungs receive 2500 mSv and her thyroid gland receives 1000 mSv. Use Table 1.7 to calculate the effective dose of radiation to which this woman has been exposed.

Solution Effective dose = S(dose equivalent × W) = (2500 × 0.12) + (1000 × 0.05) = 350 mSv This means that the cancer risk as a result of her whole body receiving a uniform dose of 350 mSv is the same as when her lungs receive 2500 mSv and her thyroid receives 1000 mSv (and her other organs receive no exposure).

The effects of radiation If at all possible, exposure to ionising radiation should be avoided. When alpha, beta or gamma radiation passes through a body cell, it may turn one of the molecules in the cell into an ion pair; for example, if the radiation ionises a water molecule, then a hydrogen ion and a hydroxide ion will be formed. These ions are highly reactive and can attack the DNA that forms the chromosomes in the nucleus of the cell. This can cause the cell to either die or divide and reproduce at an abnormally rapid rate. When the latter occurs, a cancerous tumour may form. The effects of a dose of ionising radiation can be divided into two groups: the somatic (short-term) effects and the long-term genetic effects.

H

G ray

O H

H O

H

Figure 1.32 Ionising radiation has enough energy to break the bonds within a water molecule and create a pair of ions.

Somatic effects Somatic effects arise when ordinary body cells are damaged, and depend on the size of the dose. Very high doses lead to almost immediate symptoms, lower doses could lead to symptoms developing years later.

Genetic effects When cells in the reproductive organs (ovaries or testes) are damaged, the body suffers genetic effects. Cells in the reproductive organs develop


29

Table 1.8 The somatic effects of radiation doses Whole body dose (Sv) Symptom

Physics file Our bodies can tolerate low levels of exposure to ionising radiation, but higher doses can affect the body in many different ways. The International Commission for Radiological Protection recommends that the whole body dose from artificial sources for members of the general public not exceed 1000 mSv per year. This dose is in addition to the annual dose of about 2000 mSv that we all receive from natural sources.

<1

Non-fatal Only minor symptoms such as nausea White blood cell level drops

2

Death unlikely Radiation sickness, i.e. nausea, vomiting and diarrhoea Skin rashes Hair loss Bone marrow damage

4

50% likelihood of death within 2 months Severe radiation sickness High probability of leukaemia and tumours

8

Almost certain death within 1 or 2 weeks Acute radiation sickness—convulsions, lethargy

into ova and sperm, so if the DNA in the chromosomes of these cells is damaged, this genetic change could be passed on to a developing embryo. These changed or defective cells are known as mutations. There are many different ways in which genetic defects can show up in future generations, including poor limb development, harelips and other birth abnormalities. They may surface in the next generation or lie dormant for several generations. In other words, if you suffer damage to your reproductive cells, your children may be quite normal but your grandchildren may be genetically weakened. For these reasons, when a patient is undergoing radiotherapy, it is most important that their reproductive organs are well shielded from the radiation. These organs are among the most radiosensitive organs (i.e. easily damaged by radiation) in the body. A developing foetus is also very sensitive to radiation and so pregnant women should avoid having X-rays. For this reason foetal images are now gathered using ultrasound techniques.

Physics in action

Detecting cancer with radioactive tracers Cancers that form on the skin, testes or in the breasts can often be detected by a simple external examination. However, in order to diagnose the presence of cancerous growths at specific sites inside the body, a variety of radioisotopes tagged to particular drugs are used. The radioisotope is known as a radioactive tracer. These drugs, radiopharmaceuticals, can be administered by swallowing (ingestion), inhalation or injection. The radioisotope used in the radiopharmaceutical depends on the site of the suspected tumour. The body naturally distributes different elements to different organs. For example, iodine is sent to the thyroid gland by the liver. So if a radiopharmaceutical containing radioactive iodine is taken, most of this iodine will end up in the thyroid. When the tracer has reached the target organ, a radiation scan is taken with a gamma ray camera. An unusual pattern

30


on the scan indicates a possible cancerous tumour. The radioisotopes used for this type of diagnosis need to be gamma ray emitters so that the radiation has enough penetrating ability to pass out of the body to reach the detector—the gamma ray camera. The isotope should have a relatively short half-life so that the patient is not subjected to any unnecessary long-term exposure to radiation. The most commonly used radioactive tracer is technetium99m. It is produced on site at hospitals with small nuclear generators. Technetium-99m is a gamma emitter with a halflife of 6 hours and is used to monitor the state of many organs in the body. Radioactive tracers are also used to monitor other bodily functions. Some examples are shown in Table 1.9.

Table 1.9 Some radioactive tracers and their target organs Radioactive tracer

Function monitored

Iodine-123

Function of thyroid gland

Xenon-133

Function of lungs

Phosphorus-32

Blood flow through body

Iron-59

Level of iron uptake by spleen

Technetium-99m

Blood flow in brain, lungs and heart Function of liver Metabolism of bones Figure 1

.33 A perform a gamma ray came ra is being bo us with the ra ne scan. This patie nt has bee ed to dioisotop e techneti n injected is a γ emit um-9 ter detects th with a half-life of 6 9m. This isotope e emitted  h. The cam ga image tha e t can be s mma rays and pro ra duces an een on th e compute r screen.

1.5 summary Radiation dose and its effect on humans • Exposure to background radiation is a natural and un avoidable part of our existence. Unnecessary exposure to high-energy (or ionising) radiation can be dangerous and should be avoided. • Absorbed dose is a measure of the radiation energy that our bodies absorb per kilogram of irradiated tissue. Absorbed dose is measured in grays (Gy); 1 Gy = 1 J kg−1. • The quality factor of radiation is a weighting that indicates its damaging effect on body tissue. Alpha particles have a quality factor of 20, while beta and gamma radiation typically have a quality factor of 1. • Dose equivalent gives a measure of the degree of biological damage that a dose of radiation causes. Dose equivalent = absorbed dose × quality factor. The

• • •

•

units for dose equivalent are sieverts (Sv). A typical background dose in Australia is about 2 mSv per year. Effective dose takes into account the radiosensitivity of the organ that has been exposed to ionising radiation. Effective dose = S(dose equivalent × W) and is meas ured in sieverts (Sv). When ionising radiation passes through human tissue, it may ionise atoms and molecules in the body cells, which can lead to the development of cancerous cells. Exposure to ionising radiation can lead to both somatic and genetic effects. Depending on the radiation dose, somatic effects can vary from feelings of nausea to severe illness and even death. If a person’s reproductive cells are damaged by radiation, genetic abnormalities may arise in future generations.

1.5 questions Radiation dose and its effect on humans 1 a Which of following types of radiation is electro magnetic in nature? (One or more answers.) A Radio waves B Visible light C Ultraviolet radiation D Beta particle E Gamma ray

b Which of the following types of radiation is ionising? (One or more answers.) A Radio waves B Visible light C Alpha particles D X-rays E Beta particles


31

2 Use Table 1.4 to answer this question. Calculate the dose equivalent from a radiation source if the absorbed dose is 0.50 mGy and the radiation is: a alpha radiation b beta radiation c gamma radiation.

c The record for time spent in space is held by cosmonaut Sergei Krikalev, who was on the Mir and the International Space Stations for a total of 803 days (and 9 hours and 39 minutes). How much radiation (in mSv) was the cosmonaut exposed to in this time?

3 An 80 kg tourist absorbs a gamma radiation dose of 200 mGy during a return flight to London. a Calculate the dose equivalent that has been received. b Determine the amount of radiation energy that has been absorbed.

6 Discuss some strategies that you could employ to minimise your exposure to ionising radiation.

4 a Which one of the following is the most damaging radiation dose? A 200 mGy of gamma radiation B 20 mGy of alpha radiation C 50 mGy of beta radiation b Which one of these is the most damaging radiation dose? A 200 mSv of gamma radiation B 20 mSv of alpha radiation C 50 mSv of beta radiation 5 When in space, astronauts usually receive a radiation dose of about 1000 mSv per day. The maximum allow able annual dose for people working with radiation is 50 mSv. a The normal annual background dose per year on Earth is 2 mSv. How many days does it take for astronauts to exceed this dose? b How long would astronauts have to be in space before they exceeded the maximum annual dose for radiation workers?

7 In the immediate aftermath of the Chernobyl accident, 29 people died and more than 200 received hospital treatment for acute radiation sickness. Discuss the radiation dose to which these people must have been exposed. 8 a To treat cancer of the uterus, a radioactive source is implanted directly into the affected region. If the uterus receives a dose of 0.40 Gy per hour from the source, how many hours should it be left there to deliver a dose of 36 Gy? b Explain why caesium-137, a beta emitter with a half-life of 33 years, is well suited for this task. 9 Which one of the following is most appropriate for use as a radioactive tracer to detect the presence of a brain tumour? A Radon-222: α emitter, half-life = 3.8 days B Sulfur-35: β emitter, half-life = 97 days C Cobalt-60: γ emitter, half-life = 5.3 years D Technetium-99m: γ emitter, half-life = 6 hours 10 Calculate the effective radiation dose for a woman whose organs received the following exposures during a course of radiotherapy. Her ovaries and bladder each received a dose of 35 mSv and her colon received a dose of 50 mSv.

chapter review 1 Determine the number of protons, neutrons and nucleons in these isotopes. a 35 Cl b 226 Ra 17 88 2 Consider this list of different types of radiation: alpha particles, X-rays, infrared radiation, beta particles, microwaves, gamma rays. Which of these: a is a form of electromagnetic radiation? b has a positive electrical charge? c consists of four nucleons? d is a fast-moving electron?

32


e is able to ionise matter? f has the greatest penetrating ability? 3 Find the value of x and y in each of these radioactive decay equations. a 208 TI → yxPb + β b 180 Hg → yxPt + α 81 80 4 Identify the emitted particle in each of these radioactive decays. a 40 Ca → 40 Sc + X b 150 Yb → 146 Er + X 20 21 70 68 140 140 c 60Nd → 59Pr + X

5 A radioactive isotope has a half-life of 8 h. Sketch the decay curve of a 60 g sample of this radioisotope over a 2 day period. 6 A scientist has a 120 g sample of the radioisotope polonium-218. The first three steps in the decay series of polonium-218 are an alpha emission followed by two beta particle emissions. These decays have half-lives of 3, 27 and 19 minutes respectively. Use the periodic table when answering this question. a How much polonium-218 remains after 15 minutes? b Write the equation for the alpha particle decay. c List the isotopes formed as a result of these three decays. d Which of these isotopes is predominant in the sample after 15 minutes? Explain your reasoning. 7 An archaeologist analyses an ancient bone and finds that it contains 20 g of carbon. The carbon from the bone was examined with a Geiger counter and gave a count rate of 80 disintegrations per minute. Carbon from the bones of recently deceased animals has a count rate of 16 disintegrations per gram per minute. a Write the decay equation for carbon-14, a beta particle emitter. b How many disintegrations would be detected each minute from a 1 g sample of the carbon from the ancient bone? c If the half-life of carbon-14 is 5730 years, what is the approximate age of the bone? 8 The decay curve for a sample of the radioisotope technetium‑99m is shown below. It emits gamma rays with 140 keV of energy and has an initial activity of 4.0 × 106 Bq. 4x

Activity (MBq)

1

x

x x x

0

10 A woman exposed to a large whole-body radiation dose was later found to suffer from anaemia (low red blood cell count). Is this a genetic or a somatic effect? 11 In the actinium decay series, 235 U decays to produce eventually 92 207 stable 82Pb. How many alpha decays and beta decays are there in this decay series? 12 Protactinium-234 is a radioactive element with a half-life of 70 s. If a sample of this radioisotope contains 6.0 × 1010 nuclei, how many nuclei of this element will remain after: a 70 s? b 140 s? c 210 s? d 7 minutes? 13 A laboratory produces 60 g of a radioisotope that has a half-life of 1 h. How much of the radioisotope will remain after 2 h? A none B 60 g C 30 g D 15 g 14 When a sample of beryllium-9 is irradiated with protons, an alpha particle is released and a stable isotope is formed. Determine the identity of this isotope. 15 A Geiger counter was used to compare the activity of two samples of the same radioisotope. Sample A had double the activity of sample B. a How do the half-lives of these two samples compare? b After two half-lives have passed, how will the activity of sample A compare with that of sample B? 16 The decay process of an atom of the radioisotope nitrogen-12 is: 12 N → 126C + β+ + ν 7 a Discuss the changes that have taken place in the nucleus of this atom. b Energy is produced as a result of this decay. What form does this energy take and which particles are carrying it?

3

2

c Which of these radiations would be stopped if the sample was stored in a steel box? d Energy is produced as a result of this decay. What form does this energy take and which particles are carrying it?

12 Time (h)

24

a Calculate the energy of the gamma rays in joules. b Use the graph to determine the half-life of technetium‑99m. c What is the activity of this sample after one half-life? d If the sample is produced in a hospital at 4 pm, what will its activity be when it is used at 10 am the next day? 9 The radioisotope sodium-24 decays by emitting a beta particle and a gamma ray. a Write the decay equation for sodium-24 and identify the nuclide that is produced. b Discuss and compare the penetrating abilities of beta and gamma radiation in air.

17 A small nuclear power station produces about 1000 MW of power. Two months after the reactor had been shut down, it was still generating about 5 MW of power due to the energy released from radioactive decay. Assuming that each decay releases 1 MeV of energy, calculate the activity of the reactor in becquerels. 18 A worker in an X-ray clinic takes an average of 10 X-ray photo graphs each day and receives an annual radiation dose equivalent of 7500 mSv. a Estimate the dose that the worker receives from each X-ray photograph. b How does this dose compare with the normal background radiation dose? 19 Patient A receives a radiation dose of 5000 mSv to the stomach and 4000 mSv to the colon. Patient B receives a uniform whole body radiation dose of 1000 mSv. Who is at greater risk of developing cancer from these radiation doses—patient A or patient B? Explain.


33

area of study review Nuclear physics and radioactivity 1 Explain what beta particles are and where they come from. 2 A small sample of radioactive material is located in a lead container. The radioactive material emits radiation into a region of uniform magnetic field. The radiation is deflected as shown in the diagram. Identify the radiation (together with a brief description) associated with each of the paths i, ii, and iii. ii

The following information applies to questions 10–13. A Geiger counter measures the radioactive disintegrations from a sample of a certain radioisotope. The count rate recorded is shown in the following table. Activity (Bq)

800

560

400

280

200

140

Time (min)

0

5.0

10

15

20

25

10 Plot a graph of activity versus time.

i

11 Use your graph to estimate the activity of the sample after 13 minutes. 12 What is the half-life of this element?

iii

13 Determine the activity of the sample after 30 minutes have elapsed. 3 Which one of the following best describes the part of an atom from which beta particles originate? A the electron cloud B a decayed neutron C a decayed proton D none of the above 4 From which part of a radioisotope—the nucleus or the electron cloud—are the following particles emitted? a alpha particles b beta particles c gamma rays 5 A certain radioisotope K-40 with a half-life of 1.3 × 109 years decays to a stable isotope Ar-40. Copy and complete the following table. Time (× 109 years) 0

No. of K nuclei

No. of Ar nuclei

1000

0

Ratio K:Ar

1.3 2.6 3.9 The following information applies to questions 6 and 7. Gold-197 is stable but gold-198 is radioactive.

14 If a particular atom in the sample has not decayed during the first half-life, which one of the following statements best describes its fate? A It will definitely decay during the second half-life. B It has a 50% chance of decaying during the second half-life. C There is no way of determining the probability that it will decay. D If it does not decay during the first half-life, it will not decay at all. 15 Explain why gamma rays have very low ionising ability and therefore high penetrating ability. 16 Explain why alpha particles have very high ionising ability and poor penetrating ability. 17 A small nuclear power station in North Dakota produced about 2.0 GW of power. Some time after had it been decommissioned, the reactor was still generating about 8.0 MW of power due to the energy released from radioactive decay. Assuming that each decay releases 500 keV of energy, calculate the activity of the reactor in becquerels. The following information applies to questions 18–21.

6 Discuss any differences in the chemical properties of these atoms.

Tritium (hydrogen-3) is radioactive and its decay equation is shown below. 3 H → X + –10Y 1

7 Describe the difference in the composition of these two atoms.

18 How many protons and neutrons are in each tritium nucleus?

8 A 73Li nucleus is bombarded with a high-speed proton resulting in the production of two identical particles. Write the nuclear equation that describes this reaction.

19 Which element is the daughter nuclide X?

9 Calculate the energy of these particles in MeV: a an alpha particle with energy 1.4 × 10–12 J b a beta particle with energy 6.7 × 10–14 J c a gamma ray with energy 8.0 × 10–14 J

34


20 Which of the following best describes the nature of Y in the decay equation? A It is a positron. B It is an electron. C It is a proton. D It is a neutron. 21 One of the nucleons in tritium has spontaneously transformed during this decay. Which one and what has it transformed into?

The following information applies to questions 22–24. Gold-185 is an artificial radioisotope of gold. It is an alpha emitter. 22 Write a decay equation for gold-185. Use a periodic table to help you. 23 Describe the nucleons that are in each gold-185 nucleus. 24 If you held a speck of pure gold-185 in your hand, would you suffer from the radiation exposure? Discuss. 25 Radiotherapy treatment of brain tumours involves irradiating the target area with radiation from an external source. Why is cobalt60—a gamma emitter with a half-life of 5.3 years—generally used as the radiation source for this treatment? 26 An airline pilot of mass 90 kg absorbs a gamma radiation dose of 300 mGy during a return flight to New York. Calculate the dose equivalent that has been received. The following information applies to questions 27–29. The graph below shows the data obtained in an experiment to determine the half-life of sodium-26.

Activity (Bq)

4000 3000 2000 1000 0

50 100 150 200 250 300 Time (s)

27 Use the graph to work out the half-life of sodium-26. 28 If the initial sample contained 150 g of sodium-26, how much of this radioisotope will remain after 5 minutes? 29 Sodium-26 is a beta emitter. Write the nuclear equation for its decay. The following information applies to questions 30 and 31. A man received a uniform full-body radiation dose equivalent of 1200 mSv. 30 What would the somatic effects of this dose be? 31 Which organs cells would be at most risk of developing cancer during this exposure? 32 The reason that an alpha particle has a higher quality factor than an X-ray is: A alpha particles travel faster than X-rays. B alpha particles have less ionising power than X-rays. C alpha particles have more ionising power than X-rays. D alpha particles can penetrate flesh further than X-rays.

33 During a course of radiotherapy, a man’s bone marrow was exposed to a radiation dose of 6000 mSv and his testes to a dose of 4000 mSv. What is the effective dose for this man? The following information applies to questions 34 and 35. Consider the following nuclear equation. It describes an interaction between a beryllium-7 nucleus and an electron. Be + –10e → 73Li + g

7 4

34 Explain why it is not correct to write the electron in the equation as a beta particle. 35 A nuclear physicist was bombarding a sample of beryllium-7 with a beam of electrons in an effort to smash the electrons into the nuclei. Why would it be difficult for a collision between the electrons and the nuclei to occur? The following information applies to questions 36 and 37. A nuclear scientist has prepared equal quantities of two radioisotopes of bismuth, 211Bi and 215Bi. These isotopes have half-lives of 2 minutes and 8 minutes respectively. Assume when answering these questions that each sample has the same number of atoms. 36 Which one of the following statements best describes the activities of these samples? A The samples start with an equal activity, then bismuth-211 has the greater activity. B Bismuth-211 initially has four times the activity of bismuth215. C Bismuth-215 initially has four times the activity of bismuth211. D Bismuth-211 initially has twice the activity of bismuth-215. 37 How will the activity of these samples compare after 8 minutes? 38 In a major incident in a nuclear reactor, a 75 kg employee received a full-body absorbed radiation dose of 5.0 Gy. The radiation was gamma rays. a Calculate the amount of energy that was absorbed during this exposure. b Calculate the dose equivalent for this person. c Describe some of the somatic effects that this person would experience. The following information applies to questions 39 and 40. A worker in an X-ray clinic takes an average of ten X-ray photographs each day (she works 250 days a year) and receives an annual radiation dose equivalent of 0.03 Sv. 39 Estimate the dose that the worker receives from each X-ray photograph. 40 How does this worker’s annual dose compare with the normal background radiation dose?

Area of study review

35

Unit

1

area o f stud y2

y t i c i r t Elec

Unit

outcome

area of study, On completion of this investigate you should be able to rcuit model and apply a basic DC ci ated devices, to simple battery-oper C) electrical car and household (A e the safe systems, and describ ectricity by and effective use of el mmunity. individuals and the co

1

Conc

epts

in ele

I

n 1752 Benjamin Franklin flew a kite in a thunderstorm and showed that a key attached to the string became electrified. Don’t try this yourself; the following year the German physicist Georg Richmann was killed while doing a similar experiment! Interestingly, the connection between lightning and electricity, something most of us now take for granted, was not discovered until the mid 18th century when Benjamin Franklin did his famous experiment. Franklin had made his fortune in publishing and inventions. His curiosity was aroused by a demonstration of electrostatic effects which he happened upon in 1746. Fascinated, he obtained the necessary equipment and went to work. Five years later he had developed a theory of electricity and published a book on the subject. It was he who introduced the idea of conservation of charge—the notion that charge could not be created or destroyed, only transferred from one object to another. Having found that lightning was a huge electric spark, and also that a conductor with a sharp point lost its charge more rapidly than one without, he put the two ideas together and produced his most famous invention—the lightning rod. He argued: Would not these pointed rods probably draw the electrical fire silently out of a cloud before it came nigh enough to strike, and thereby secure us from that most sudden and terrible mischief? In 1752 the first ‘Franklin Rod’ was installed on Mr West’s shop in Philadelphia. Some time later, it was hit by a direct stroke of lightning but suffered no damage, thus proving its effectiveness.

by the end of this chapter

you will have covered material from the study of electricity including: • the concepts of electric charge and electric forces • the concepts of current, EMF and electric potential • resistance in ohmic and non-ohmic conductors • electric energy and power.

ctrici ty

2.1 Electric charge electron

neutron proton nucleus

When a plastic pen rubbed with a dry cloth is brought near some small pieces of paper, the paper may ‘dance’. Some of the bits of paper may even jump onto the pen and then jump off again, seemingly at random. The pen has gained what we call an electrostatic charge, which creates an electric field around it. This field will cause the paper to experience a force. Why? As you will know, present theory suggests that all material matter in the universe is constructed from about 100 different types of atoms. Further, all atoms are made up of just three fundamental particles: the proton, the electron and the neutron. You will have investigated the properties of these particles in Chapter 1 ‘Nuclear Physics and Radioactivity’.

Atomic charge Figure 2.1 The way we draw an atom is not meant

to be a representation of what it would ‘really look like’. For a start, if it was drawn to scale both the nucleus and electrons would be so small they would be invisible! Always remember, this is a model, not reality.

Physics file The word ‘atom’ is derived from the Greek word ‘atomos’ meaning indivisible. However, during the past century, physicists have not only shown that there is an internal structure to the atom, but that there is even an internal structure to the protons and neutrons. As well as their mass, the charge (or lack of it) carried by these atomic particles is an intrinsic feature of each of them. It is impossible to remove the charge from, say, an electron. One cannot have an electron without negative charge or a proton without positive charge! It is not possible to somehow ‘charge’ a neutron. When an object becomes charged it is because there has been a transfer of charged subatomic particles, normally electrons, either to or from the object.

38

Electricity

Inside the atom, the heavier protons and neutrons reside together in an extremely dense, positive region called the nucleus. The nucleus contains almost all the mass of the atom, but occupies only a tiny fraction of its volume. Orbiting at relatively large distances are the tiny negative electrons. These particles are very light indeed. Typically, the electrons only contribute about 1/4000 of the mass of the atom. However, their orbits define the size of the atom, and, importantly, they balance the positive charge of the nucleus. Ever since the time of Benjamin Franklin’s experiments with electricity it has been realised that electric charge appears to be indestructible. Franklin was the first to suggest that all matter is made up of equal amounts of positive and negative charge, normally in balance. Electrification, he suggested, is the transfer of some of this charge from one object to the other, resulting in an imbalance between the charges. To demonstrate this he had two people stand on insulated stools. One used a cloth to rub a glass rod held by the other. Afterwards, when they each took the charge from their cloth and glass respectively and brought their fingers close, a spark jumped between them and they both lost their charge. This type of experiment led to the idea of conservation of charge. That is, charge cannot be created or destroyed, only transferred from one object to another. If, for example, a glass rod is rubbed by a cloth and the rod acquires a positive charge, then the cloth will have acquired an equal amount of negative charge. Overall the charge is still zero. The principle of conservation of charge is now regarded, like conservation of energy, as one of the central principles of modern physics.

Electric charge is conserved. This means that it cannot be created or destroyed, only transferred from one object to another.

Electrostatic charge In all chemical reactions it is the outer electrons orbiting the atom that are either swapped or shared between atoms. Remember that the orbiting negative electrons are held to the nucleus by the attraction of the positive protons. The atoms of different elements ‘hang on’ to their electrons to varying degrees and these differences are responsible for the huge variety of chemical reactions that occur around us.

When two different materials are rubbed together, this tendency for electrons to move between atoms normally results in one of the materials gaining electrons at the expense of the other. The one that gains electrons will thus attain an overall negative charge and the other, now with fewer electrons than protons, will become positively charged.

An excess of electrons causes an object to be negatively charged, and a deficit in electrons will mean the object is positively charged. Perspex and polythene are two common materials that are easily charged by rubbing with cloth. In the process of rubbing, polythene tends to gain electrons and so becomes negatively charged. However, perspex tends to lose electrons and become positively charged. If a charged polythene strip is brought near another charged polythene strip, the two strips will repel. However, when a charged polythene strip is brought near a charged perspex strip, the two strips attract. In both cases, the effect is greater the closer the strips are to each other.

– + + +–+ +– + – – – +– +–+ –– + +– – – – + +– + – –+ ++– –+ + + –

– +–+ +– + – +– +–++––– +– – + – – + + – + –+– + – –+ + +– +– +

+ + + – + + + – + –– +– – – – + +– + +– – + + + + – – –+ + ++– –+ + – +

– – – +–+ +– ++– – +– +–+ –– +– – +– – – + + – + –+ – –– + + – – ++––++ –

–

Figure 2.2 If as a result of being rubbed together some electrons are transferred from one object to another, the first will become positively charged and the second negatively.

Like charges repel and unlike charges attract. The closer the charges are to each other, the stronger the force. The Van de Graaff generator is often used as a source of electrostatic charge in the laboratory. In effect, it deposits the charge produced by the contact between a plastic roller and a rubber belt onto a metal dome. While the belt is running, the concentration of charges on the dome becomes greater and greater. However, anyone who has watched a Van de Graaff generator in action knows that the charge does not keep building up for ever. Eventually the concentration of charge becomes so great that charges start to jump off the dome—either as a bright spark across to an earthed object or as tiny crackling sparks into the air.

A unit for charge

Figure 2.3 Like charges repel. Unlike charges

attract.

PRACTICAL ACTIVITY 5 Electrostatics with a Van de Graaff generator

In order to measure the actual amount of charge on a charged object, a ‘natural’ unit would be the charge on one electron or proton. This fundamental charge is often referred to as the elementary charge and is given the symbol e. The proton therefore has a charge of +e and the electron −e. Despite many experiments designed to look for smaller charges, no charge smaller than e has ever been found in nature. All larger charges are understood to be whole number multiples of e.

metal dome metal roller collector brush (connected to dome)

insulating column rubber belt (carries charge to dome)

Figure 2.4 The Van de Graaff generator. In the base the rubber belt passes around a plastic

roller. The close contact between rubber and plastic results in electrons being transferred to the rubber. The electrons are replaced by others which flow onto the roller from a metal foil or comb that is electrically connected to ‘earth’. The electrons on the belt are carried to the top where they are picked up by another foil or comb and allowed to flow onto the dome. There, because they repel each other, they spread out over the conductive dome. A very large concentration of charge can build up.

plastic roller charging brush electric motor

Chapter 2 Concepts in electricity

39

Physics file The nuclear particles, protons and neutrons, are thought to be made up of fundamental particles called quarks. The theoretical charges on quarks are 1 2 positive and negative 3 e and 3 e. While experimental evidence for these particles is strong, theory predicts that they can exist only in combinations which produce particles with exactly 1e and not by themselves. So we are unlikely to ever see particles with charges that are not whole multiples of e.

Physics file The number of elementary charges in a coulomb (6.242 × 1018) is a result of the original definition of electric current. As we shall see, charge and current are closely related, but in the early days of electrical experimentation the unit for current (the ampere) was defined before the coulomb. When dealing with electrostatics, the coulomb is a huge unit of charge. For this reason smaller units are often used: 1 mC = 10−3 C 1 µC = 10−6 C 1 nC = 10−9 C 1 pC = 10−12 C

Physics file Chemistry students will know that, as the atomic mass of aluminium is 27, there will be 6 × 1023 (Avogadro’s number) atoms in 27 g of the metal. As there are 13 protons and electrons in each atom, the total number in the 700 g dome can be found from a little simple arithmetic.

Before going further, it is interesting to reflect on the nature of the proton and the electron for a moment. They have significantly different mass and size. The proton can be imagined to have a radius of around 10−15 m while the electron is so small it is meaningless to give it a size. The mass of the proton is almost 2000 times that of the electron. Incredibly though, while being opposite in sign, the magnitudes of their charges are absolutely identical! Atoms containing equal numbers of both are always exactly neutral.

The …L…M…NTARY CHARG…, e, is the magnitude of the charge on a proton or electron. It is the smallest charge found in nature. The elementary charge is clearly a very small unit of charge. Even the small charge rubbed on the pen for the ‘dancing paper’ experiment would involve many billions of electrons being either lost or gained. For practical purposes a much larger unit of charge is used. The SI unit is the coulomb (symbol C). It is equivalent to 6.242 × 1018 elementary charges. The reciprocal of this number is therefore the charge, in coulomb, on a proton or electron.

The elementary charge, e, the charge on a proton, is equal to 1.602 × 10−19 C. The charge on an electron is −e.

Worked example 2.1A a It has been stated that the charge on a rubbed pen would involve many billions of electrons. What is the charge in coulomb carried by 10 billion electrons? charge on a school Van de Graaff generator might be around −3.0 µC (1 µC = 1 microcoulomb = 10−6 C). How many extra electrons are on the dome?

b The

Solution a In coulomb, the charge on 10 billion electrons is 10 × 109 × 1.602 × 10−19 = 1.6 × 10−9 C. This can be referred to as 0.0016 µC or as 1.6 nC (nanocoulombs). b The number of electrons in a charge is the magnitude of the charge divided by the charge on one electron, i.e. ne = q/e. The negative 3.0 µC charge on the Van de Graaff dome consists of ne = q/e

−6 = 3 × 10 = 1.9 × 1013 electrons (19 000 billion electrons) 1.6 × 10−19 Any normal electrostatic charge involves huge numbers of electrons!

It is interesting to compare the number of extra charges on a Van de Graaff dome with the number of charges in the metal itself. In Worked example 2.1A it was found that a charged Van de Graaff dome might have an excess of nearly 20 million million electrons (2 × 1013). This is a huge number of electrons! But, assuming the aluminium dome has a mass of about 700 g, there would be a total of about 2 × 1026 electrons and the same number of protons in the aluminium atoms of the dome. So the number of extra electrons on a fully charged dome is actually an extremely small fraction (10−13) of the total number of electrons in the metal of the dome!

40

Electricity

You might like to confirm that this would be the equivalent of adding just a litre or so of water to Port Phillip Bay.

Electrostatic induction Recalling the dancing pieces of paper, how then did the charged plastic pen cause them to dance? We now know that the pen was negatively charged. It had created around it an electric field in which negative charges will move away from the pen and positive ones towards it. Any electrons in the paper that are free to move will therefore move to the opposite edge of the paper, leaving the edge closer to the pen with an excess of protons and thus a positive charge. While the paper is still neutral overall, this positive charge is closer to the pen than is the negative charge on the other edge and will therefore be attracted more strongly than the negative charge is repelled. If the pieces of paper are small enough, and the charge on the pen is great enough, the paper will be lifted from the table and may even jump onto the pen. This process is called electrostatic induction. The charges in the paper are ‘induced’ to move by the presence of the charged object, thus creating ‘induced charges’ of opposite sign on opposite sides of the paper. Electrostatic induction will occur regardless of the sign of the charge on the pen. If the pen were to be made positive, electrons in the paper would move towards it, causing the closer side to become negative and the further side to become positive. The ‘lightning rod’, invented by Benjamin Franklin, is a good example of an application of the principle of electrostatic induction. A tall pointed metal rod on the highest part of a building is well connected to the ground by a heavy wire. When a charged thundercloud moves overhead, charges of the opposite sign will be induced in the ground below, particularly in taller objects. The charge concentration on the end of the lightning rod can become so intense that the air molecules nearby become ionised and form a conducting path towards the cloud. This normally will have the effect of discharging the cloud sufficiently so that a lightning strike will not occur. If a strike does occur it will be conducted to ground through the lightning rod rather than the building, thus protecting the building and its occupants.

–– – – – –– – – – –– –– –– – – – ––– – ++ – – – –– – – ++ –– + + + – + ++ – ––– – ++ +++ + + –– + + ++ + + – – – – – –– ––– – – – –– –

Figure 2.5 A negatively charged pen induces a positive charge on the nearer side of the paper and a negative charge on the opposite side. Because the positive side is closer, the paper is attracted.

Physics file The effect of a lightning rod can easily be demonstrated by placing a nail (vertically) on top of a Van de Graaff generator. Because of the intense electric field at the point of the nail, the charge dissipates into the air and the spark from the dome will not be nearly as impressive as that normally obtained.

Conductors and insulators Any attempt to produce an electrostatic charge by rubbing a metal rod instead of a plastic or glass rod is normally unsuccessful. Charge transferred to the metal rod will flow away through the rod and your hand. (If the metal is mounted on plastic a charge can be produced.) Unlike plastic and glass, metals are conductors: they allow the movement of charge through their structure. The structure of metals is such that the outermost electrons of the atoms are free to move around in the fixed crystal lattice made up of the atoms. Any excess of electrons in one place will soon be dispersed as the electrons flow away from each other. Materials such as plastic and glass do not allow the flow of electrons. They are called insulators.

Figure 2.6 This type of lightning rod did not become popular!


41

Figure 2.7 The distinction between insulators and conductors was discovered by Stephen Gray in 1729. He showed that the human body is a conductor of electric effects. He suspended a boy from silk cords and brought a charged glass rod near his legs. The electric effect was transmitted through the boy’s body to his hands and face—as shown by the dancing paper. While the boy acted as a conductor, the silk acted as an insulator, preventing the charge from escaping to ground.

Table 2.1 Some common conductors and insulators Conductors

Insulators

Good All metals, especially silver, gold, copper and aluminium Any ionic solution

Plastics Polystyrene Dry air Glass Porcelain Cloth (dry)

Moderate Water Earth Semiconductors, e.g. silicon, germanium Skin

Wood Paper Damp air Ice, snow

There is not always a clear distinction between insulators and conductors. Wood, for example, will conduct electrostatic effects reasonably well, but certainly cannot be used as a conductor for household appliances! Wood can be classed as a poor conductor or a poor insulator depending on the situation. Another important group of materials is the semiconductors. Most notably these include silicon and germanium, the basis of the modern electronics industry. Pure semiconductors are not nearly as conductive as metals, but can be modified by ‘doping’ them with small amounts of certain elements so that they will conduct quite well. They are the materials from which transistors and integrated circuits are made. Returning once more to the dancing paper experiment, you will probably now realise why the paper might jump off the pen after a little while. Because paper is not a very good insulator, it will slowly allow charge to move through it. As it picks up some of the charge from the pen it gradually becomes charged with the same charge as the pen. Once the charge builds up sufficiently, the paper will be repelled by the pen and fly off.

Physics in action

Robert Millikan and the elementary charge By the late 19th century it was well established that electrons and protons had an identical, but opposite, charge. The big question was how this charge related to the coulomb, which was defined in terms of large-scale electrical phenomena such as magnetism and electrolysis. An experiment was needed which linked the tiny forces on atomic charges to the more macroscopic electrical measurements. Robert Millikan, at the University of Chicago, took up the challenge in 1907 and published his results 2 years later. His experiment was remarkably accurate and is now famous, both for the importance of the results and for the sheer elegance of its design. In essence, Millikan measured the electrostatic force on tiny drops of oil sprayed from an atomiser. The drops acquired an electrostatic charge simply as a result of their motion through the air. His apparatus was basically two horizontal plates held 1.6 cm apart which could be given opposite charges by an 8000 V battery. The oil drops were allowed to

42

Electricity

fall through the air in the space between these plates (they took more than 20 seconds to fall 1 cm). The rate at which such drops fall depends on the balance between the upward air-resistance force and the downward weight force. As the air resistance depends on the size, Millikan was able to calculate the weight of the drops from measurements of their speed. (The same theory applies to the rate at which a balloon falls through still air.) When the plates were charged by the battery, the speed changed as a result of the added electric force on the drops. Some drops fell faster, others almost stopped or even rose. As the speed at which the drops fell was directly related to the total force on them, he was able to calculate the strength of the electric force. From this force he could calculate the electric charge on the drop. He found that this charge only came in whole number multiples of a certain smallest amount. This charge he assumed to be the charge on a single electron. He was able to

show that one elementary charge was equal to 1.64 × 10−19 C, which, considering the difficulties of the experiment, was very close to the currently accepted value of 1.60219 × 10−19 C. In 1923 Millikan was awarded the Nobel Prize for his work on the elementary charge, along with his later experimental work on Einstein’s interpretation of the photoelectric effect.

oil spray

air oil

microscope

switch high voltage

oil drop

metal plates

Figure 2.8 Millikan’s apparatus consisted of an atomiser which sprayed a fine mist of oil into the chamber. Some of the drops fell into the space between plates that could be charged via the switch from the battery.

Physics in action

Lightning Lightning is undoubtedly one of nature’s greatest spectacles. No wonder it was for so long thought of as the voice of the gods. When Benjamin Franklin showed that it was basically the same sort of electrical phenomenon as could be achieved by rubbing a glass rod with wool, he didn’t so much demystify it as move the mystery into another realm. How indeed can such enormous voltages be created in a cloud, something normally associated with the moisture that makes electrostatic experiments hard to perform!

T = –65°C

12 km T = –15°C

5 km

T = 10°C

2 km

Figure 2.9 A thundercloud can be several kilometres wide and well over 10 km high. Strong updrafts drive the electrical processes that lead to the separation of charge. The strong negative charge will induce positive charges on tall objects on the ground. This may lead to a discharge, which can form a conductive path for lightning.

A thundercloud normally has three charged regions. In the lower centre there is a strong negatively charged region, often less than a kilometre in thickness, but possibly several kilometres in width. The top of the cloud is mostly positively charged. There is normally also a smaller positively charged region at the bottom owing to positive charges attracted up from the ground. There will be strong electric fields between these regions of opposite charge. If they become sufficiently strong, electrons can be stripped from the air molecules (they become ionised). Because of the electric field, the free electrons and ions will gain kinetic energy and collide with more molecules, thus precipitating an ‘avalanche of charges’. This is the lightning flash seen either within the cloud or between the Earth and the cloud. Most flashes are within the cloud; only a relatively small number actually strike the ground. A typical lightning bolt to the ground bridges a potential difference of hundreds of millions of volts and transfers 10 or more coulombs of negative charge to the ground in a brief current pulse of up to 10 000 A. A moderate thundercloud with a few flashes per minute generates several hundred megawatts of electrical power, the equivalent of a small power station. The exact mechanism responsible for the charge buildup in a cloud is still not entirely clear, but it is thought that charge is transferred in collisions between the tiny ice crystals that form as a result of the cooling of the upwardflowing moist air and the larger, falling, hailstones. As a result of small temperature differences between the crystals and hailstones the crystals become positive and the hailstones negative. The crystals carry their positive charge to the top of the cloud while the negative charge accumulates in the lower region. As a result, an electric field (a downward force on positive charge) builds up within the cloud.


43

This electric field, however, enhances the process greatly because it then induces a charge on the hailstones, positive at the bottom, negative at the top. This results in a greater transfer of positive charge to the crystals as they collide with the lower, positive part of the hailstone. Hence there is a greater build-up of charge in the cloud and a still stronger field. The whole process is a self-reinforcing cycle—the stronger the field, the more effective the charging process. Eventually the positive and negative concentrations of charge become so great that huge voltages are formed and sparks fly!

(a)

(b)

+

hailstone

– ice crystals

–

+

+ + ++ + + + ++ + + + E – –– –– + +

+ ++

– – – – – – –– –– –– – –– – –

Figure 2.10 (a) When falling hailstones hit ice crystals of a different temperature, charge is transferred. (b) As the field in the cloud builds up, electrostatic induction further enhances the process.

2.1 summary Electric charge • Matter is made up of vast numbers of positive and negative charges (protons and electrons respectively). Normally there is an equal number of each. • Like charges repel and unlike charges attract. • Charge cannot be created or destroyed, but it can be transferred from one object to another. • An electrostatic charge involves an imbalance of positive and negative charges.

• The charges on a proton and electron are equal in magnitude but opposite in sign. The magnitude of this charge is referred to as one elementary charge. • One coulomb of charge is equal to 6.242 × 1018 elementary charges, or one elementary charge is equal to 1.602 × 10−19 C. • If a charged object is placed near a conductor, an opposite charge will be induced on the side of the conductor nearer the charge and a like charge on the side further away from the charge.

2.1 questions Electric charge 1 Why was Benjamin Franklin’s kite-flying experiment so dangerous? 2 What are some of the ways in which our lives would be different if we did not have electricity? 3 List some of the electric motors in your household. How many do you think would be found in the average household?

6 Plastic strip A, when rubbed, is found to attract strip B. Strip C is found to repel strip B. What will happen when strip A and strip C are brought close together? 7 Some early theoreticians thought that electricity was a sort of fluid that could be transferred from one material to another. How could this model account for the fact that there were two types of charge?

4 Both simple electrostatic and magnetic experiments show forces that act through space between two different types of object. What similarities and differences can you think of in the results obtained from simple electrostatic experiments and those performed with magnets?

8 Why does a Van de Graaff generator have a smooth round aluminium dome on the top rather than, say, an aluminium cube that would be easier to make?

5 Why do you think that early experimenters such as Franklin concluded that there were only two types of electric charge and not more?

10 We could use the terms ‘red’ and ‘black’ to describe the two different types of charge. What advantages do the terms ‘positive’ and ‘negative’ have over the use of colours as labels for charge?

44

Electricity

9 Why is it not safe to stand under an isolated tree in a thunderstorm? What should you do if caught out in a thunderstorm?

2.2 Electrical force sa

nd fields

Coulomb’s law Electricity is clearly one of nature’s fundamental forces. It was Charles Coulomb, in 1785, who first published the quantitative details of the force that acts between two electric charges (see Physics in action page 48). The force between any combination of electrical charges can be understood in terms of the force between the simplest possible arrangement of charges: two so-called ‘point charges’ separated by a certain distance. The expression ‘point charges’ simply means that the two charges are regarded as being very much smaller than the distance between them. Remember that between like charges there will be repulsion and between unlike charges attraction. Coulomb found that the force, whether repulsive or attractive, between two charges q1 and q2 a distance r apart was proportional to the product of the two charges, and inversely proportional to the square of the distance between them. This can be expressed by the simple equation below (where k is the proportionality constant).

COULOMB’S LAW for the force between two charges q1 and q2 at a distance of r: kq q F = r12 2 It is not surprising that the force between two charges depends on the product of the two charges. Imagine that we found a force of 10 N between two particular charges A and B. If charge A was then doubled, for example by adding another identical charge, we would be surprised if the force between A and B did not increase to 20 N. If charge B was then also doubled, would we not expect the force between A and B now to increase to 40 N? While this argument is quite feasible, it certainly does not prove that the force is proportional to the charge. The story of physics is full of discoveries that were not as we might have expected! However, many experiments have confirmed Coulomb’s law to a high degree of accuracy. The force being inversely proportional to the square of the distance means that, for example, if the distance between A and B is doubled, the force will decrease to one-quarter of the previous value. There are a number of important inverse square laws in physics. The reason for this is suggested in the Physics file on the next page. The constant k has a value (in SI units) close to 9.0 × 109 N m2 C−2 in air or a vacuum. This means that the force between two charges of 1 C each, placed 1 m apart, would be almost 1010 N—equivalent to the weight of about ten large battleships! This suggests that a 1 C charge is a huge amount of charge. Imagine, for example, the repulsive force between two halves of any 1 C charge on an ordinary sized object. It would blow itself to pieces with enormous energy! In practice, the amount of charge that can be placed on ordinary objects is a tiny fraction of a coulomb. Even a highly charged Van de Graaff dome will have only a few microcoulombs (1 µC = 10−6 C) of excess charge.

Physics file While an understanding of the basic nature of electrical forces and fields may not be a specific requirement of the curriculum, it is not really possible to gain a satisfactory understanding of the basic electrical concepts such as potential (voltage), current and resistance without it.

q1

F

q2

++ ++ + +++ + ++ ++

++ ++ + +++ + ++ ++

F

r

q1

q2

++ ++ + +++ + ++ ++

–– –––– –– –– –– – – –

F

F

Figure 2.11 The nature of the force, F, between two point charges q1 and q2 a distance r apart was discovered by Charles Coulomb in 1785.


45

Physics file Both Coulomb’s law for the force between electric charges and Newton’s universal law of gravitation, for the gravitational force between any two masses, are examples of ‘inverse square’ laws. Another is the law for the intensity of light around a ‘point source’ of light. In all these cases something, whether it is physical (light), or simply an ‘influence’ (a force) can be imagined to be spreading out evenly from a point. The intensity of this ‘something’ at a certain distance will therefore be inversely proportional to the area of the sphere over which it is spread. As the area of this sphere increases with the square of the distance (A = 4πr 2), the intensity will therefore decrease with the square of the distance. This, of course, does not constitute a ‘proof’ that these laws should be inverse square. It simply suggests that it is reasonable that they would be.

Another way to get a feel for the magnitude of electrical forces is to realise that all matter is held together by the electrical forces between atoms. For example, Mount Everest is supported by the electrostatic repulsion between the atoms underneath it. The strength of the hardest steel is due to the electrical forces between its atoms. In comparison to the Earth’s gravitational forces on atoms, the electrical forces between them are totally overwhelming—by a factor of about a billion billion! In fact, only in the last stages of collapse of a giant dying star can the gravitational forces overwhelm the electrical forces between atoms and cause them to collapse into the super-dense state of matter that exists in what is called a neutron star.

Worked example 2.2A Two Van de Graaff machines are placed 50 cm apart and switched on. If they both attain a charge of 3 µC of the same sign, what will be the force between them? (Ignore the size of the machines for the moment.) How would this force change if: a one of the machines sparks and loses half its charge? b the machines are moved to a distance of 1 m apart? c the charges were of opposite sign instead of the same?

Solution

kq1q2 where r 2 k = 9 × 109 N m2 C−2, q1 = q2 = 3 × 10−6 C and r = 0.5 m. The force between them is given by F =

9 × 109 × (3 × 10−6)2 = 0.3 N, not a large force, but possibly noticeable. 0.52 As both machines have charge of the same sign the force will be a repulsive one. a If one machine loses half its charge we do not need to repeat the calculation, we know that the force will halve to 0.15 N. b If the distance in an inverse square law doubles, the force will reduce to one-quarter. Given the original charge of 3 µC, the force will decrease to 0.3/4 = 0.08 N. c If the charges were opposite, the force would be attractive rather than repulsive. (It is worth noting that if two Van de Graaff machines with the same sign were actually placed this distance apart, the force would probably be less than that calculated because the charges would repel and move to the opposite sides of the domes. On the other hand, if the charges were opposite, they would attract and move to the closer sides of the dome, thus increasing the force.)

Thus the force F = Figure 2.12 The inverse square law. Three metres from a point source of light, the light will be spread over an area nine times as large as that at 1 m. The light will therefore appear only one-ninth as bright.

Electric fields It is often very difficult to use Coulomb’s law directly to calculate the force on a charged object because, for example, the force may originate from many charges spread around on a conductor. However, in many cases it is possible to measure or calculate the electric field. The electric field, like the gravitational field, is the amount and direction of the force on one unit— one unit of (positive) charge in the case of the electric field, one unit of mass in the case of the gravitational field. Just as the gravitational field (g) is the force on 1 kg, the electric field (E) is the force on 1 coulomb (C). If we find that there is a gravitational force of 19.6 N on a 2 kg mass, we know the gravitational field is 9.8 N kg−1. If we were to find a force of 10 N on a 2 C charge, we would know that the electric field is 5 N C−1.

46

Electricity

The electric field … in any region of space is defined as the electric force per unit charge: F … = q An electric field has both direction and strength, and so is a vector quantity. The electric field is often shown by lines that represent the direction of the field. The closeness of the lines can normally be taken to represent the relative strength of the field. The shape of the field around some charged objects is shown in Figure 2.13. The shape of the field around a small charge is radial, pointing outward in the case of a positive charge and inward in the case of a negative charge.

The force on a charge q in an electric field … is given by F = q….

Physics file We have used simple numbers to make the point that the electric field is the force per coulomb. In practice we would never have a net charge of 1 C in one place—the coulomb repulsive force would be so great it would blow itself to smithereens in a huge explosion! The electric field is the ratio of force to charge. The field between a pair of parallel plates such as those illustrated in Figure 2.13 and connected to a simple 12 V battery might be, say, 500 N C−1. Any ordinary sized object, say a small scrap of paper, that we might put in such an arrangement would only have a static charge of less than about 0.1 µC, and so the force on it would be:

A single positive charge

This may seem insignificant, but it could well be about the same weight as the paper—so this simple electric field is balancing the pull of gravity of the entire Earth on our small piece of paper. As we shall see later, the electric force on an electron in such a field is a million million times as strong as the gravitational force on it!

+

Two positive charges

+

=q = 0.1 × 10−6 × 500 = 5 × 10−5 N

+

A positive and a negative charge

+

–

Parallel oppositely charged plates

+ + + + + + + + + + + + +

– – – – – – – – – – – – –

Figure 2.13 In these photographs, grass seeds suspended in oil make the electric field around charged objects visible. Electric fields can be represented by lines showing the direction of the field. The closeness of the lines suggests the strength of the field.


47

As shown in Figure 2.13, the electric field lines between two charged parallel plates are mostly parallel and uniform. This means the strength of the field has the same value everywhere. We will see later that there is a very simple way to calculate the strength of the field in such a region. In fact, a pair of parallel plates is often used as a convenient way of obtaining a uniform electric field of known strength. Millikan used such plates in his experiment described on page 42.

Worked example 2.2B Robert Millikan measured the charge on an electron by finding the electric force on tiny oil drops in a known electric field. If the force on an oil drop was due to the charge of one single extra electron on the drop, and found to be 8.0 × 10−14 N upwards, what was the strength and direction of the electric field he was using?

Solution One elementary charge, the charge on an electron, is equal to 1.6 × 10−19 C. The strength of the field is therefore given by: … = F q 8.0 × 10−14 = 1.6 × 10−19 = 5.0 × 105 N C−1 The direction of the field is downwards. Remember that the direction of the field is the direction of the force on a positive charge. A negative charge experiences a force in the opposite direction to the field. It is interesting to note that Millikan achieved this field by connecting a battery providing 8000 V across two parallel plates 1.6 cm apart.

Physics in action

Charles Coulomb and the force between two electric charges Charles Coulomb (1736–1806) was an engineer who investigated variations in the Earth’s magnetic field. In order to do this he invented a ‘torsion balance’, a device that could measure very small forces. He adapted his device to the problem of determining the force between two charges (Figure 2.14). The principle of a torsion balance is simple. A carefully balanced horizontal rod is suspended by a thin fibre. First the force required to twist the rod through a given angle is found. Then two charges are placed, one on a small fixed sphere as shown and the other on a similar sphere on one end of the rod (which is an insulator). By finding the amount of twist caused by the electric force between the charges, the magnitude of the force can be calculated. By varying the amount of charge on the spheres as well as the distance between them, he was able to show that the force between them was proportional to the amount of charge on each, and decreased with the square of the distance between them.

48

Electricity

Figure 2.14 A sketch of Coulomb’s original torsion balance. The torsion force in the fibre is measured by the degree of rotation of the knob at the top needed to compensate for the electrical force between the fixed and movable spheres.

Physics in action

Photocopiers In 1938, American inventor Chester Carlson used an electrostatic effect to transfer an image from one piece of paper to another. He called it ‘xerography’ from the Greek words for ‘dry’ and ‘writing’. But it wasn’t until 1959 that the Xerox Corporation produced the first truly successful office copier, which became the basis of the current multibillion dollar industry. When light is shone on some materials that have been given an electrostatic charge, they become conductive (‘photoconductivity’) and lose the charge. If the light is in the form of the image of some writing, an ‘electrostatic image’ can be formed. If fine carbon particles with the opposite charge are then sprinkled on the image they will adhere to the places that are charged. If they can then be ‘fixed’ in place, we have a ‘photocopy’ of the original. A modern photocopier is a very complex piece of machinery, but the basis of its operation is still this same photoelectric effect that Carlson used. At its centre is an aluminium drum that has two very thin (about 0.05 mm) coatings. The first is a layer of photoconductive material such as cadmium sulfide. This is coated with an even thinner transparent insulating film.

As the drum rotates, the various parts undergo different processes. First, it is given an electrostatic charge from a ‘corona wire’. This is a fine wire charged to about 5000 V. Because it is very thin, the electric field near it is intense and it ionises the air, creating ions (charged atoms) and free electrons. (This is the same effect that Benjamin Franklin used for his lightning rods.) If the wire is negative, the electrons will charge the surface of the drum and the underside of the thin insulating layer will become positively charged by induction as charges move through the photoconductive layer. Next, the image of the work to be copied is projected onto the drum. Where it is bright, it increases the conductivity of the photoconductive layer and allows the charge to escape. The dark parts remain charged. Next the image is ‘developed’ by allowing positively charged toner particles to be attracted to the still negatively charged dark parts of the image on the drum. In order to transfer the toner powder, and hence the image, to the paper, the paper is given a strong negative charge from another corona wire and brought into contact with the drum. The paper is then taken through two hot rollers where adhesive in the toner melts and ‘fixes’ the image on the paper. At the same time the drum is cleaned and made ready for a fresh charge from the first corona wire.

book

lens strong light

– 5000 V charging corona

light from book

discharging light

–––––– ++ + + ++ +

– + + +

dark area

insulating layer

–

cleaning blade

+

–

– + –– ++

light area –

+ + +

++

+

toner

+ + ++

.. .

.

+

photoconductive layer

....

cleaning roller

blade

+ –– + – +

drum hot rollers

– – –

............. + ++

paper – 5000 V transfer corona

Figure 2.15 There are several types of photocopy machine, but the basic principles are illustrated here. The drum is given an electrostatic charge before being exposed to light from the image. The light enables the charge to dissipate. Black toner powder is attracted to the still charged dark parts of the image. The toner is then transferred to the paper and fixed by hot rollers.


49

2.2 summary Electrical forces and fields • Coulomb’s law for the force between two charges q1 and q2 at a distance of r is: kq1q2 r 2 • The constant, k, in Coulomb’s law has a value of 9 × 109 N m2 C−2, indicating that an electrostatic charge of 1 C would be enormous. F=

• The electric field E in any region of space is the electric force per unit of charge in that space: E = F/q. Conversely, the force on a charge q in an electric field E is given by F = qE.

2.2 questions Electrical forces and fields 1 A charge of +q is placed a distance r from another charge also of +q. A repulsive force of magnitude F is found to exist between them. Describe the changes, if any, that will occur in the force when: a one of the charges is doubled to +2q b both charges are doubled to +2q c one of the charges is changed to −q 1 d the distance between the charges is changed to 2 r.

the domes of the machines and then proceed to use Coulomb’s law to calculate the force they expect to find between them. a What force do they expect to find between the machines? b Assuming that the charge on each machine is 5 µC, why do they find that the measured force is less than they expect?

2 What force would exist between them if we could place two 1 C charges 100 m apart?

5 Some small charged spheres are to be placed in an electric field which points downwards and has a strength of 5000 N C−1. a What force would be experienced by charges of +2 µC and −5 µC? b A sphere with an unknown charge is found to experience an upwards force of 1 × 10−3 N in this field. What was the charge on the sphere?

3 How practical would it be to set up the situation described in the previous question? 4 Danielle and Daniel set up two Van de Graaff machines exactly 80 cm apart (centre to centre) on frictionless trolleys that allow them to measure the force between them. They read that the manufacturer states that it is possible to obtain a charge of 5 µC on

50

Electricity

Electric curre 2.3 and electrica nt, EMF l potential When a battery is connected to a conductor (such as a torch bulb) one end of the conductor becomes positively charged and the other end becomes negatively charged. This sets up an electric field along the length of the conductor. As a result, the mobile charges (electrons in a wire for example) will move along the conductor. We call this movement of charge along a conductor an electric current. In this section we first discuss the measurement and nature of an electric current, and then look at the way in which the battery supplies the energy to drive this current.

0.5 C of charge I t=1s

q 0.5 C I = – = = 0.5 A t 1s

Electric current Just as a current in a river involves the flow of water, electric current is the flow of electric charge. Any moving charge constitutes an electric current. Whether it consists of electrons moving through the atomic structure of a metal, or protons from the Sun flying through space, moving charges make up a current. The magnitude of the current is defined simply as the rate of transfer of charge. We can think of it as the amount of charge that flows past any point in a conductor in 1 second. A current of 1 ampere flows when 1 coulomb of charge flows past a point in 1 second. So 1 ampere = 1 coulomb per second. Electric current is given the symbol I.

Electric current is the rate of transfer of charge: q I= t where q is the charge transferred t is the time taken If the charge is measured in coulombs and the time in seconds, the current is measured in amperes. Conversely, the charge, in coulombs, carried by a current of I amperes in t seconds is given by q = It.

Figure 2.16 If a charge of 0.5 C passes a point in a conductor in 1 s, a current of 0.5 A is flowing.

Physics file The ampere is named in honour of André Marie Ampère (1775–1836). You may know that a magnetic force exists between two wires carrying an electric current. It was Ampère who first worked out the mathematics of this force. The ampere is actually defined as the current which, when flowing in two long straight parallel wires 1 metre apart, produces a force of exactly 2 × 10−7 N between them. This force was typical of that produced by the current that could be obtained from the batteries of that time. The coulomb was later defined as the amount of charge carried by a current of 1 ampere in 1 second.

1 ampere (A) = 1 coulomb per second (C s−1), so 1 coulomb (C) = 1 ampere second (A s).

Worked example 2.3A Determine the charge that has flowed through a torch battery producing a current of 300 mA if it has been left on for 20 minutes.

Solution I = q/t, so q = It where I = 300 × 10−3 = 0.300 A and t = 20 × 60 = 1200 s. Thus q = 0.300 × 1200 = 360 C Figure 2.17 It is not practicable to measure the force between two very long wires 1 m apart, but, based on the definition, the force between two current-carrying coils can be calculated and used to set up a primary standard of current upon which other instruments can be calibrated. In a standard current balance the magnetic force is balanced against a known weight.


51

The direction of current

Table 2.2 Typical values for electric current Situation

Current

Lightning

10 000 A

Starter motor in car

200 A

Fan heater

10 A

Toaster

3A

Light bulb

400 mA

Pocket calculator

5 mA

Nerve fibres in body

1 mA

(a)

There is sometimes confusion about the direction of an electric current. In the case of a river, the direction of the current is clear; it is the direction of flow of the water. Unlike water, however, electric charge can be either positive or negative. So what is the direction of an electric current? The direction of an electric current is the direction of transfer of positive charge. However, positive charge can be transferred to the right, let’s say, either by moving positive charge to the right, or by moving negative charge to the left.

I t=0

A current starts to flow between two neutral objects.

t later

I

– – – – –– – – –q

+ + + + + + + + +q

where q = It

A short time later the object to which the current is flowing will become positive and the other negative. (b)

I +

+ +

+

I –

– –

–

I + –

+

–

Figure 2.19 (a) If two objects are electrically connected and one is becoming more positive while the other becomes more negative, then we say there is a positive current flowing towards the one which is becoming more positive. (b) This might result either from positive charges moving in the direction of the current, from negative charges moving in the opposite direction, or both at once.

52

Figure 2.18 In an electrical circuit, an ammeter is used to measure the current. The ammeter is connected in series with the device; that is, in such a way that the current flows through it as well as the device. Typically, a meter will require the user to select a current range. If the current is too large it could damage the meter.

Electricity

To get a feel for the meaning of this, consider the situation shown in Figure 2.19. Both objects are neutral to start with. Remember that a neutral object comprises huge, but equal, numbers of positive and negative charges. Imagine that a current then starts to flow from left to right as shown. Because current is the transfer of positive charge, we expect the object on the right to become more positive and that on the left more negative. Now it is important to realise that this flow of positive charge to the right can be accomplished either by moving positive charge to the right, or negative charge to the left. In a metal wire, for example, a positive current to the right is carried by electrons moving to the left. On the other hand, in a fluorescent tube a positive current to the right is carried both by positive ions moving to the right and by electrons moving to the left. Because the electrons in a wire actually move in the opposite direction to the current, the terms electron current and conventional current are sometimes used to distinguish between them. It is important to remember, however, that current is the rate of transfer of positive charge.

(b)

– –

– – –

+

–

–

I I lil

A

ilil

–

+

–

–

+

–

PRACTICAL ACTIVITY 6 Connecting circuits

+

... +

–

i lililil ilil

–––––––– – ––

(a)

– –

Figure 2.20 (a) Current flows around a circuit from the positive terminal of the battery to the

negative. In the connecting wires the current is carried by electrons travelling in the opposite direction. In the battery itself, and in the salt solution, the current will involve the movement both of positive ions in the direction of the current and negative ions moving in the opposite direction. (b) The beam of electrons travelling down a cathode ray tube in a television set produces a positive current in the opposite direction.

EMF and electric potential In order to drive a current around an electric circuit the charges must be given energy. A battery or generator is the usual source of this energy. Another increasingly common source of electrical energy is the photovoltaic cell, or solar cell. These devices are all referred to as sources of EMF. The letters stand for ElectroMotive Force, but that is an inaccurate term because the EMF involves the energy given to the charges rather than the force on them. In order to understand the meaning of an EMF it is helpful to consider another source of EMF, the Van de Graaff machine (see Figure 2.4). The source of energy in this case is very obvious. The motor is pushing the charges on the rubber belt up against the electrostatic repulsion of the charges already on the dome. The more charge already on the dome, the greater the force, and hence the greater the work that has to be done to bring more charge to the dome. In fact, you may hear the motor slow down as the concentration of charge on the dome builds up. In many ways the EMF can be visualised as a ‘concentration of charge’. The more charges put on the dome, the more concentrated they become and the greater the force of repulsion between them. The work done pushing the charges together (by the motor in this case) is stored as electrical potential energy. Just as the compressed spring in a jack-in-the-box contains potential energy, so do all the ‘concentrated’ charges. And just as the spring energy can be recovered when it is allowed to expand, so the electrical energy can be recovered when the charges are allowed to fly apart again. When a spark flies from the Van de Graaff generator we see the result. The potential energy is rapidly converted into kinetic energy, and as the charges collide with the air molecules it is turned into heat, light and sound energy. EMF is defined as the amount of work done for each unit of charge in this process of charge concentration. Because it is, therefore, actually the ‘electric potential energy per unit charge’, this quantity is most often

Physics file Whenever the term ‘current’ is used in this book, the direction is assumed to be that of the transfer of positive charge as defined by the equation I = q/t. This is sometimes called conventional current. The term electron current, which you may also come across, is equivalent to defining current by the equation I = −q/t and will not be used in this book as it may lead to confusion. Simply remember that positive charge transfer occurs either as a result of positive charges moving in the direction of the current, by negative charges moving in the opposite direction (electrons in metals), or by both at once.

Figure 2.21 A Van de Graaff machine does work pushing the charges on the belt up against the repulsive force from those already on the dome. If a person touches a Van de Graaff machine while standing on an insulated chair, the charge will spread from the dome and over them. The charged hairs repel each other.


53

Physics file As EMF is defined as the work done in ‘concentrating’ charges, it is related to the area under the Coulomb force– distance graph for many pairs of charges. In practice this would amount to adding the area under the graphs for a huge number of charge pairs. With the aid of some calculus, however, it can be shown that the potential is actually the sum of all the kq/r terms, where r is the separation between the charges. So the closer the charges, and the greater the number of them concentrated in an area, the greater the potential. This is the sense in which we can think of voltage as representing ‘charge concentration’.

abbreviated simply to electric potential or just potential. The EMF is then the electric potential given to charges by the device. A battery uses chemical potential energy to give the charges in the circuit this potential energy. A generator uses the kinetic energy of rotation, and solar cells use the energy in sunlight. The SI unit for potential is joule per coulomb. One joule per coulomb is given the name volt, in honour of Alessandro Volta, the inventor of the first chemical battery. It is easy to see why the terms ‘potential’ and ‘voltage’ are often used interchangeably. A normal dry cell has an EMF of 1.5 V. This means that every coulomb of charge on the positive terminal has 1.5 J of potential energy more than those at the negative terminal. The symbol used for EMF is usually E (a script E). So, for a dry cell, E = 1.5 joules per coulomb, that is, 1.5 V.

A source of EMF gives charges electrical potential energy. The EMF, E, is the energy per unit of charge. 1 volt = 1 joule per coulomb (1 V = 1 J C−1).

Worked example 2.3B Physics file The EMF of a charged Van de Graaff generator, or any source of high voltage, can be estimated from the length of the spark it will produce. Between smooth spheres, in dry air, a spark will jump about 1 cm for every 25 000 V. Between pointed conductors it will jump 1 cm for every 10 000 V. Under good conditions a Van de Graaff generator might produce a spark of up to 15 cm, which corresponds to nearly 400 000 V.

The alternator of a car being driven at night with the headlights on, is producing a 50 A current at an EMF of 12 V. a How many coulombs of charge flow from the alternator each second? b How many joules of energy does each coulomb of charge obtain? c How many joules of energy does the alternator produce each second? d Where does this energy go?

Solution a The 50 A current means that 50 C of charge flow each second (q = It). b The 12 V EMF means that each 1 C of charge is given 12 J of energy. c Each second, 50 C of charge each with 12 J of energy flow from the alternator, so the energy produced is 50 × 12 = 600 J.

d This energy will go to the headlights, the ignition system and any other electrical devices in operation. Some may also be used to recharge the battery.

plastic or cardboard outer covering positive end

negative end zinc can (anode) asphalt electrolyte seal carbon paste of MnO2 rod (cathode) and NH4Cl

Figure 2.22 A modern dry cell is a complex mixture of chemicals designed to drive electrons from the top (positive) terminal to the bottom (negative) one.

54

Electricity

Electric circuits and potential difference Any electric circuit consists of at least one source of EMF, conductors that carry current (hopefully with very little loss of energy) and the various circuit elements. The circuit elements are the working parts of the circuit: light bulbs, motors, loudspeakers, heating elements and so on. In these, the electric potential energy is converted into heat, light, motion or whatever other form of energy is required. A torch consists of a battery, a switch and a bulb all connected by conductors to form an electric circuit. The battery may consist of one or more cells. The current from the battery goes from the positive terminal to the bulb and then via the switch back to the base of the battery, which acts as the negative terminal. When a battery is connected in a circuit, it produces a potential differ ence; that is, a difference in the potential energy of the charges in the conductors connected to its terminals. The potential difference produced by the battery can be imagined as a difference in ‘charge concentration’ on the

(a)

(b) switch contact ON

positive terminal

negative terminal

Figure 2.23 (a) A battery and light bulb connected by conductors in a torch constitute an electric circuit. (b) This circuit can be represented by a simple circuit diagram.

conductors connected to either side of the battery. While the switch is off, all the conductors connected to the positive terminal, including the bulb and one side of the switch, will have a uniform positive charge concentration. All those connected to the negative terminal will have a uniform negative concentration. This initial distribution of charge, which ensures that there is no electric field in the conductors, will take place in the first fraction of a microsecond after the battery is put in place. Remember that there is no current flowing in the conductors at this stage. All the conductors on one side of the switch will have the same positive potential and all those on the other side the same negative potential. Any difference in charge concentration along a conductor will create an electric field. So when the switch is turned on an electric field is created around the circuit, which will cause charges to start moving. Where the conductor is wide, the charges flow easily and there will not be much change of charge concentration. Where the conductor is narrow, in the light bulb filament for example, a high charge concentration difference will occur, resulting in a stronger field. This stronger field means two things. First, it pushes the charges harder so that the current through the filament is the same as that in the rest of the circuit—if it isn’t, the charges build up, the concentration difference increases, the field becomes stronger, and the charges move faster (so the current increases). Second, the stronger field means that the charges lose more potential energy. That is, they will gain more kinetic energy, which they transfer to the filament as light and heat through constant collisions with the atoms. The EMF is the potential energy given to each unit of charge by the battery. The potential difference across a circuit element is the potential energy lost by each unit of charge in that element. The potential difference, sometimes called potential drop or just p.d., across a circuit element is written ∆V, the ∆ representing the fact that there is a change of potential. In practice, the ∆ is often omitted as there is rarely a need to refer to anything other than a change of potential. As ∆V is the energy lost by one unit of charge, a charge of q coulomb will lose q∆V joules of potential energy as it goes through a potential difference of ∆V.

A charge q moving through a potential difference ∆V will lose potential energy given by ∆U = q∆V, or simply ∆U = qV.

Physics file Strictly speaking, the ‘dry cells’ we buy should not be called batteries. The term ‘battery’ really refers to a group of cells connected together, as in the torch shown. When a battery is connected in this way (in series) the total EMF is equal to the sum of the EMF values of the individual cells.

(a) +12

+12

+12

+12 open switch 0

+12 No E field

0 +12

+

–

0V

12 volts

(b) +8+7+6 +5 +4

+9

+3

closed switch +2

Higher E field +10

Lower E field

+11

+1 +12

+

–

0V

12 volts

Figure 2.24 The electric field and the potential around a circuit are related. (a) In an open circuit there is no potential drop around the circuit and no electric field (except at the switch). (b) In the closed circuit charge flows as the result of an electric field. As it flows it loses potential. Where the conductor is narrow the field is ‘squeezed’ and becomes stronger, thus the charges are pushed harder. The potential also drops more quickly through the narrow region as more work is done and potential energy lost.


55

Worked example 2.3C The potential difference across a torch bulb is found to be 2.7 V. The current flowing through it is 0.2 A. a How much charge flows through the torch in 1 minute? b How much energy is lost by this charge?

Solution a q = It = 0.2 A × 60 s = 12 C b Each coulomb lost 2.7 J of energy. ∆U = qV = 12 × 2.7 = 32.4 J

A useful analogy (a)

(b)

I

Figure 2.25 (a) It can be useful to compare charge going around a circuit with water in a dam. The current in the conductors in the torch is like the water in the dam or river, while the bulb is like the spillway of the dam at which the potential energy of the water becomes kinetic energy (the noise and turbulence). (b) Just as the water will lose its potential energy to kinetic energy, so the electrons in a wire will lose potential energy to kinetic energy as they go through the thin wire of the filament. This kinetic energy appears as heat and light (red colour).

56

Electricity

It may be helpful to compare an electric circuit to water flowing from a high dam or lake over a spillway or waterfall to a river below. The water in the dam has potential energy relative to the water in the river below. The Sun is the source of the potential energy. It caused the water to evaporate, the water eventually raining on the hills. The energy it gave to every kilogram of water can be seen as being rather like the ‘EMF’ of the system—but this time it is gravitational potential energy per unit of mass, instead of electrical potential energy per unit of charge. As the dam fills, the water flows until it is all at the same level; that is, it equalises the potential energy. This is rather like what happens when batteries are put in a torch, with the switch off. Charges will flow around the conductors until all points connected to the positive terminal have the same (positive) potential energy and all those connected to the negative terminal have the same (negative) potential energy. The switch is rather like the dam wall: on one side the potential (or water level) is higher than on the other side, in the case of the switch on our torch, by 3 V. What happens when the switch is closed? This is somewhat like opening the sluice gate that lets water over the dam spillway, as in Figure 2.25. As the water flows down the spillway its potential energy is transformed into kinetic energy. The bulb in the electrical circuit is the equivalent of the spillway. Because the filament is very thin compared to the other connecting wires, the charges have more trouble moving through it and give up most of their kinetic energy to the atoms they collide with; this energy then appearing as heat and light. The current in the other conductors is rather like the water in the dam or river. Just as water will slowly move through the dam itself as a result of the barely perceptible height difference created by the loss of water over the spillway, so charge will slowly move around the conductors as a result of the very small potential difference created along the conductors by the movement of charge through the filament. However, the energy lost in these conductors is very small compared to that lost in the filament. We can see and hear the potential energy being turned into kinetic energy as the water falls. The charges in the circuit also gain kinetic energy as they ‘fall’ through the 3 V potential difference that the battery will maintain across the bulb. This kinetic energy is transferred to the atoms in the filament as the electrons collide with them. A typical incandescent light bulb is designed so that the heat energy produced raises the temperature of the filament to about 2200°C. At this temperature only about 5% of the energy is actually radiated as visible light; the rest is lost as infrared radiation and heat conducted away through the filament supports and connections.

The water that has come over the spillway will now slowly flow along a broad river to the sea. The charges, having done their work in the bulb, will make a leisurely return to the battery through the (relatively) broad conductors in the torch. On reaching the sea, some of the water will gain enough energy from the Sun to rise in the atmosphere and become a cloud. On reaching the battery, some of the charges will eventually be given potential energy, by the chemical reactions, and find themselves on the positive terminal again. It is important to remember that there are many differences between electric currents and flowing water! Any analogy is only useful to a limited extent. For a start there is no such thing as ‘negative water’ while there is both positive and negative charge. As well, while the water stored in the dam represents a large amount of gravitational potential energy, the battery does not store up electrical potential energy. It uses its stored chemical energy to produce an electric field that transfers energy to the charges already in the conductors. The analogy is useful, however, in that the ‘electric potential’ of charges can be compared to the ‘height’ of the water. Just as the actual kinetic energy released over the spillway of the dam depends on both the height and the amount of water, so the energy produced in the filament of the bulb, or any other device, depends on both the potential difference and the amount of charge that flows. This is a very important concept in our progress towards a good understanding of electrical ideas.

Physics in action

What is an electric current? It is interesting to consider the nature of current flowing in the different states of matter. For any current to flow, mobile charge carriers are required. In the solid state, metals are the best conductors owing to the presence of many ‘free’ electrons. Metals, being ductile, can be drawn into wires, and so metal wires are the most familiar means for transferring electricity. Liquids in which there are free ions will also conduct an electric current. Although water molecules themselves are neutral, there is always a very small number of charged ions present (both positive and negative) and these ions allow even pure water to conduct electricity, although poorly. The addition of impurities, such as dissolved salts, raises the number of charged ions considerably and increases the conductivity of water. The electrolyte in a car battery (a solution of sulfuric acid) is an example of a good liquid conductor. In any liquid, the current is made up both of positive ions flowing in the direction of the current and negative ions flowing in the opposite direction. Gases too can carry electric current as long as enough of the gas atoms become ionised. For example, lightning will occur when the electric field within a charged cloud is strong enough to strip electrons from the gas molecules in the air. Fluorescent lamps incorporate mercury vapour, which ionises relatively easily. A plasma is a gas heated to the point at which it ionises and becomes conductive. Again, conduction in gases involves both positive and negative charge carriers moving in opposite directions.

Figu

re 2.26 A photograp n electric current in ho a top of a Va f the corona from ir! This is a a pin plac n de Graaff ed on the of the ion machine. ise As be seen. Th d air molecules, ma well as the glow ny e earthed d bright spark (right) small sparks can ome was occurred brought n when the ear.


57

2.3 summary Electric current, EMF and electrical potential • A battery establishes an electric field in a conductor connected to its terminals. This electric field results in the movement of charge, i.e. an electric current. • Electric current is the time rate of transfer of charge: I = q/t (1 A = 1 C s−1) • The direction of an electric current is the direction of the transfer of positive charge. This can occur as a result of positive charge movement in that direction, negative charge movement in the opposite direction, or both.

• In metals the charge carriers are electrons moving in the direction opposite to the conventional current. • A source of EMF gives charges electric potential energy. The EMF, E, is the electric potential energy per unit of charge. 1 volt = 1 joule per coulomb. • A potential drop, V, is the loss of potential energy (joules) per unit charge (coulomb) as charge flows through a circuit element. The potential energy lost is given by ∆U = qV.

2.3 questions Electric current, EMF and electrical potential 1 What current flows in a light bulb through which a charge of 30 C flows in: a 10 seconds? b 1 minute? c 1 hour? 2 A car headlight may draw a current of 5 A. How much charge will have flowed through it in: a 1 second? b 1 minute? c 1 hour? 3 a In a solution of salt water a total positive charge of +15 C was seen to move past a point to the right in 5 s, and in the same time a total negative charge of −30 C was seen to move to the left. What was the current through the solution during this time? b Some time later it was found that in 5 seconds a total of +5 C had moved to the right while −15 C had moved to the right as well. What was the current this time? 4 Using the values given in Table 2.2, find the amount of charge that would flow through a: a pocket calculator in 10 min b car starter motor in 5 s c light bulb in 1 h. 5 Do the values for the charge that you obtained in Question 4 indicate the amount of energy required to operate the devices for those times? Explain. 6 The dome on a fully charged Van de Graaff machine may carry something of the order of 50 million million extra electrons. When running well it may take about 3 s to charge up. If we assume no loss of charge in this time, what is the current flowing up the belt to the dome? 7 When water runs through a hose at the rate of 0.5 litre per second, it can be calculated that 1.7 × 1026 electrons pass any point in the hose each second.

58

Electricity

a What electric current (in amps) does this represent? b Is this the actual electric current in the hose? Explain. 8 Although you will not normally get a shock if you put your hands on the terminals of a car battery, you will if you touch the spark plugs while the engine is running. Why is this? 9 a What is the EMF of a battery that gives a charge of 10 C: i 40 J of energy in 1 second? ii 40 J of energy in 10 seconds? iii 20 J of energy in 10 seconds? b What current flowed in each case? 10 The negative terminal of a 12 V car battery is connected to the car frame, which can be regarded as ‘ground’, at a potential of 0 V. What is the potential of the other terminal? A 0 V B −12 V C +6 V D +12 V 11 A charge of 5 C flows from a battery through an electric water heater and delivers 100 J of heat to the water. What was the potential of the battery? 12 How much energy will each coulomb of charge flowing from a 9 V transistor radio battery possess? 13 How much charge must have flowed through a 12 V car battery if 2 kJ of energy was delivered to the starter motor? 14 In comparing the electrical energy obtained from a battery to the energy of water stored in a hydroelectric system dam in the mountains, to what could the EMF of the battery be likened?

Resistance, o hmic and 2.4 non-ohmi c conductors When a potential difference is applied across an electrical device in a circuit a current will flow. Generally speaking, the greater the potential applied, the greater the current that will flow. The actual relationship between the current and the potential difference applied is the subject of this section. In order to measure the current through and the potential difference across a circuit element we must introduce two invaluable tools for any exploration of electric circuits: the voltmeter and the ammeter.

Ammeters and voltmeters PRACTICAL ACTIVITY 7 Using electrical meters

Figure 2.27 Voltmeters and ammeters come in a wide variety of shapes and forms. Some are digital and some are analogue. Multimeters combine several functions in one meter.

As its name implies, a voltmeter is used to measure the potential difference across any circuit element or source of EMF. A voltmeter could be compared to a pressure gauge used to measure tyre pressure. Just as the pressure gauge measures the difference between the tyre pressure and atmospheric pressure, the voltmeter measures the difference in electrical potential between two points in a circuit. This is why the voltmeter connections are always placed across the circuit element. The ammeter could be compared to the paddle-wheel water meter you might see on an irrigation channel. It is used to measure the current flowing in a circuit and is therefore placed so that the current flows through the ammeter as well as the circuit element. Never place an ammeter across a circuit element (as you would a voltmeter) because it would effectively be a short circuit that could damage the meter, the circuit or both.

Current–voltage graphs Whether the device is a simple light bulb or a complex electronic component, a knowledge of the relationship between the current and the voltage, the so-called I–V characteristic, is necessary in order to predict the behaviour of the device or the power it will consume. This is often given in the form of a graph. The voltage is plotted on the horizontal axis because it is usually

Physics file Ideally a voltmeter should draw no significant current from the circuit it is being used to measure. In practice, just as a pressure gauge may take a little air out of the tyre and reduce the pressure, a voltmeter will draw some current and change the circuit a little. It is important to ensure that any current drawn by the voltmeter is much less than that flowing in the circuit. A water meter should not significantly slow down the rate of flow of the water that it is measuring. Likewise, an ammeter should offer no significant resistance to the current flowing through it and so it is important to ensure that the resistance of the ammeter is much less than that of the other components in the circuit.


59

+

the ‘independent variable’, the quantity we set by using a certain battery or power supply. Some examples of I–V graphs for some common electrical devices are shown in Figure 2.29. We can divide these devices into two groups: those that have a straight I–V graph and those that do not. The resistor is in the first group, the light bulb and the diode are in the second. Those with a straight I–V characteristic are called ohmic conductors and those which don’t are (rather logically) called non-ohmic conductors.

–

i lili lil il il

A

ili l

light bulb

lil

(a)

ammeter lil

V

diode

light bulb

4

4

4

ilil

i lililil ilil

resistor

3

I (A)

2

(b) light bulb resistor or load

3

3

I (A)

voltmeter

I (A)

2

2

1

1

1

2

4

V (V)

6

8

2

4

6

V (V)

8

10

12

0.5

1.0

V (V)

1.5

ohmic resistor diode

Figure 2.28 (a) A simple circuit of one battery and one circuit element, together with an ammeter and voltmeter to measure the current and potential difference. (b) Four examples of symbols for possible circuit elements: a light bulb, a resistor or other circuit element, an alternative symbol for a resistor used to specify an ohmic resistor in particular, and a diode.

Figure 2.29 Examples of the relationship between current and applied potential difference for three common electrical devices.

Resistance Georg Ohm (1789–1854) found that if the temperature of a metal wire was kept constant, the current flowing through it was directly proportional to the potential placed across it: I ∝ V. This is known as Ohm’s law.

OHM’S LAW states that I ∝ V. PRACTICAL ACTIVITY 8 Resistance and temperature

Physics file It is interesting to consider the reason for the increase in resistance of the light bulb. At 1 V the filament is barely glowing dull red, if at all. At 12 V it is shining very brightly and at a temperature of around 2200°C. At this temperature the atoms of the filament are vibrating much more rapidly. It is not surprising then that the electrons have a more difficult time getting through. In effect, their average drift speed (the speed due to the current as distinct from their thermal speed) is slowed. As a result, more voltage is needed to achieve the same current, i.e. the resistance is greater.

60

Electricity

Rather than writing this as I = kV (where k is the slope of the I–V graph) this relationship is normally written the other way around as V = IR, where R is called the resistance (it is the inverse of the gradient). So Ohm basically said that the resistance of a metal wire (at a certain temperature) is constant. Even if the resistance of a conductor is not constant (the graph is not straight) it is still defined as the ratio of the potential difference across a conductor to the current flowing in it at that potential.

R…SISTANC… is the ratio of potential difference to current: V R = or V = IR I

The expression V = IR is sometimes referred to as Ohm’s law, but that is only correct if R is constant. This expression is basically the definition of resistance. Ohm’s law effectively says that R is constant for some types of conductors—and they are called ohmic conductors. We can see from Figure 2.29 that while the resistance of the resistor is constant, that of the light bulb increases with increasing potential, whereas the resistance of the diode decreases with increasing potential. The resistor obeys Ohm’s law, but the others do not.

Ohmic conductors Many conductors do obey Ohm’s law quite closely and so their I–V characteristic is completely specified by a single number, the resistance R. The unit for resistance is volts per ampere and is given the name ohm (symbol Ω, omega). It helps to think of the resistance as the number of volts needed to make a current of 1 ampere flow through the conductor. But remember that the resistance is a ratio—the actual resistor may go up in smoke if 1 A actually flowed through it! Ohmic conductors are often simply referred to as ‘resistors’: 1 ohm = 1 volt per ampere (1 Ω = 1 V A−1)

Worked example 2.4A A resistor of 5 Ω is supplied with a potential that can vary from 1 V to 100 V. a What will be the range of currents that will flow in it? b How much energy will be dissipated in the resistor each second?

Solution a 5 V is required to make 1 A flow in this resistor. Therefore, at 1 V the current will be 15 A or

0.2 A. More formally: 1 At 1 V, I = V/R = 5 = 0.2 A 100 At 100 V, I = 5 = 20 A (or simply 100 times the previous answer). b At 1 V, 0.2 C flow through the resistor each second. The energy is given by: ∆U = qV = 0.2 × 1 = 0.2 J At 100 V, ∆U = 20 × 100 = 2000 J. Notice the very large increase in energy as the voltage is increased. As the voltage increased by 100 times, the energy released each second increased by 10 000 times, because both voltage and current increased by 100 times.

(a)

(b)

Figure 2.30 (a) An assortment of ohmic resistors, including a variable one. (b) Some non-ohmic devices including diodes, a lightdependent resistor, thermistors, light-emitting diodes, a neon indicator lamp and a torch bulb.


61

Non-ohmic conductors

PRACTICAL ACTIVITY 9 Ohmic and non-ohmic conductors

A light bulb is a common example of a non-ohmic conductor. Typically, a car headlamp bulb may draw about 1 A at 1 V, but as the voltage increases, the current will not increase in proportion, as you can see in Figure 2.29. At 12 V the current might be 4 A; so while the resistance at 1 V is 1 Ω, at 12 V the resistance has increased to 3 Ω. While it may sometimes be useful to know the resistance of the bulb at its operating voltage of 12 V, it cannot be used to calculate the current flowing at other voltages. The bulb does not obey Ohm’s law. To quote the resistance of the diode in Figure 2.29 would be almost meaningless: it decreases very rapidly once the voltage reaches about 0.5 V. The important thing to know about the diode is that once the voltage exceeds a certain level the current increases, apparently without limit. In practice there will be a limit to the current because the power dissipated will reach a level at which the diode will become too hot and burn out. Other non-ohmic conductors include devices whose resistance changes with light or temperature. These are particularly useful as detectors in sensors that need to respond to changes in light levels or temperature.

Worked example 2.4B The graph represents the I–V characteristic of a 240 V, 60 W light bulb. What is the resistance at: a 24 V? b 120 V? c 240 V?

200

I (mA) 100

Solution 100

200

V (V)

Resistance is given by R = V/I at any point on the graph. Note that the current is given in mA (100 mA = 0.1 A). a At 24 V R = 24/0.10 = 240 Ω b At 120 V R = 120/0.20 = 600 Ω c At 240 V R = 240/0.25 = 960 Ω Resistance increases as the filament becomes hotter. Notice that we cannot use the inverse slope of the graph; resistance is simply the ratio V/I at a particular voltage.

Resistance and resistivity What are the factors that determine the resistance of a conductor? Given that the resistance of a piece of metal wire is a measure of the ability of the wire to somehow impede the flow of electrons along its length, it is reasonable to expect that: 1 If the wire is made longer there will be a greater resistance as there is more to impede the flow of the electrons.

62

Electricity

2 If the wire is made thicker there will be less resistance as there is more pathway for the charge flow and so less impedance. 3 If a different metal is used it will most likely have a different structure and so a different impedance to the current. This would suggest a simple relationship: rL R= A where L is the length of the wire A is the cross-sectional area ρ (rho) is the proportionality constant The value of ρ will depend on the particular material used. When experiments are done it is found that this relationship does indeed hold, at least as long as the temperature is held constant. The proportionality constant, ρ, is called the resistivity, and it is seen as a measure of the inherent resistance of the material without regard to its shape or size. The value of ρ varies over a huge range, about 23 orders of magnitude from around 10−8 Ω m for good conductors such as copper to more than 1015 Ω m for good insulators. Table 2.3 gives some typical values. Silver is the best conductor, followed closely by copper and then gold.

Physics file The value of ρ normally increases with the temperature of the material. For example, at 300°C the resistivity of copper is about twice its value at room temperature. This is because at higher temperatures the atoms of the metal are vibrating more vigorously and will therefore impede the progress of the electrons to a greater extent.

Table 2.3 Resistivity of some common materials at 20°C Material

Resistivity, ρ (Ω m)

Silver

1.6 × 10−8

Copper

1.7 × 10−8

Gold

2.4 × 10−8

Aluminium

2.8 × 10−8

Worked example 2.4C

Tungsten

5.5 × 10−8

Normal household wiring uses 1.8 mm diameter copper wire. What is the resistance of a 10 m long piece of this copper wire? What voltage drop will there be along it if a current of 10 A is flowing through it?

Tungsten (2000°C)

70 × 10−8

Nichrome

100 × 10−8

Doped silicon*

1 × 10−3

Solution

Pure silicon

3 × 103

Pure water

5 × 103

Soils and rock

103 → 107

Wood

108 → 1011

Glass

1010 → 1014

Fused quartz

1016

The resistivity of copper is found from Table 2.3: ρ = 1.7 × 10−8 Ω m. The cross-sectional area of the wire is given by: A = πr2 = 3.14 × (0.9 × 10−3)2 = 2.5 × 10−6 m2 The resistance is therefore found from: rL R = A 1.7 × 10−8 × 10 = = 0.068 Ω 2.5 × 10−6 If the wire is carrying a 10 A current there will be a voltage drop along the length of the wire of V = IR = 10 × 0.068 = 0.68 V—not a serious problem in a 240 V system.

* Doped silicon is used for transistors and integrated circuits.

Physics in action

What really travels along a wire? It is worth reflecting for a moment on the nature of an electric current in a metal. If it were actually possible to see all the atoms and electrons in a metal wire we would see a constant blur of activity. The atoms would be vibrating madly and the free electrons would be rushing around at random with enormous speeds—rather like a game of air hockey gone wild! Now imagine that a current suddenly starts to flow in this wire. What difference would it make? The answer is none

at all! At least not to any perceptible extent. However, if we could watch a single electron for a few seconds we would notice that, on top of its wild random dance, it had moved a few millimetres in one direction. If you think you might have noticed that, consider that in those few seconds of its random dance that same electron would have collided with millions of billions of atoms and covered thousands of kilometres!


63

+

B`

B E

+ + + + ++ + +++ + + + + + +

+

–

E–

–

– – – –– – – –– – – ––––

–

Figure 2.32 A potential difference applied to the ends of the metal conductor creates an electric field along the length of the conductor. A

Figure 2.31 This diagram represents the path of an individual electron

in a wire as it might appear without (AB) and with (AB′) an electric field (…). The motion due to the field, however, has been greatly exaggerated. Drawn to scale it would be impossible to notice any difference. If the electrons travel so slowly around the wires, why does the light come on almost instantaneously? In fact, electricity travels along a wire at close to the speed of light. Less than a millisecond after the switch is closed at the power station in the Latrobe Valley, the voltage appears on the power line in Melbourne! What travels down the power line is not really charge, but a wave of electric field. This wave of electric field pushes the charges a little closer (or a little further apart). It is this change in the concentration of the charges that gives rise to the potential, the voltage, that travels down the wire so quickly. A simple analogy can be drawn with a garden hose. Provided it is full of water to begin with, water will come out the other end virtually as soon as the tap is turned on. The actual water that goes into the hose when the tap is turned on may take quite a few seconds to emerge from the other end. Again, what travels so quickly down the hose is a pressure pulse, not the actual water. The water then flows because the pressure at the tap end of the hose is higher than that at the other end. Likewise, if the potential at one end of a wire is greater than that at the other, a steady current will flow along the wire.

Resistance and the electric field in a wire We can gain some insight into the nature of resistance from our simple model of the mechanism of charge movement in wires. For a current to flow in a wire, an electric field must be established in order to produce a force on the electrons in the wire. This electric field is set up by the source of EMF, which produces a different ‘charge concentration’ (potential) at one end of the wire relative to that at the other end. This is shown diagrammatically in Figure 2.32. The concentration of charge along the length will gradually change from being strongly positive at one end to strongly negative at the other end. It is this changing concentration that establishes the uniform electric field. At any point an electron ‘sees’ more positive charge to the left than to the right and will therefore experience a force to the left. (Remember that an electron experiences a force in the direction opposite to the field.) If in fact the field was stronger at one place than another, electrons would move more quickly there and this would weaken the field until it became uniform.

64

Electricity

It is this electric field that ‘drives’ the current. It creates a steady force on the free electrons that results in their slow movement along the conductor. In an average conductor the electrons might take around a minute to move a millimetre. Don’t forget, however, that there are huge numbers of electrons moving and so the number passing any point in 1 second will also be huge. In a current of 1 A there will be 6.2 × 1018 electrons flowing past any point each second. The reason the electrons move so slowly is simple: they keep colliding with the atoms in the wire and giving up their energy (which is why the temperature of the wire rises). The stronger the field, the faster the electrons move, however, and therefore the larger the current. Not surprisingly, it turns out that the speed is directly proportional to the strength of the field and that the field is directly proportional to the potential difference applied. Thus the current is proportional to the potential difference, as Georg Ohm discovered. Why is the electric field proportional to the potential difference between the ends of the wire? If we imagine a small positive test charge, +q, moving along the wire in a steady field , it will lose potential energy given by ∆U = qE∆x (work = force × distance) where ∆x is the distance it has travelled. If it travels the whole length, l, the energy lost will be ∆U = qEl. Note that the kinetic energy gained by the electron is given up almost immediately to the atoms the electron collides with. Now the energy transferred per unit of charge is just what is meant by the potential difference, V. That is, V = ∆U/q. From the previous expression we see therefore that V = El. This is the electrical equivalent of the expression ‘work = force × distance’. We simply need to remember that V represents the ‘work per unit charge’ and is the ‘force per unit charge’. In this context remember that the work done is equal to the change of potential energy. Rearranging, we can write E = V/l. The electric field can thus be expressed as the ‘potential gradient’ along a wire. Indeed it is usually given units of volts per metre. It is now easy to see why the simple relationship between resistance, length and cross-sectional area of a conductor holds. The longer the wire (with a constant potential difference), the weaker the electric field and hence the lower the current that will flow. The field created in the wire is dependent on the potential difference and the length, not on its width, and so if one piece of wire has twice the crosssectional area of another it will carry twice the current. The resistance will therefore decrease with increasing crosssectional area.

2.4 summary Resistance, ohmic and non-ohmic conductors • The current–potential difference (I–V) relationship for an electrical device describes the electrical behaviour of the device. • A device with a linear I–V graph is known as an ohmic conductor. It obeys Ohm’s law: the current is directly proportional to the voltage. • While the current in a non-ohmic conductor will be dependent on the voltage, it is not directly proportional to it. • The resistance of a conductor, whether ohmic or not, is defined as the ratio of the potential difference to the current (1 Ω = 1 V A−1). Thus V = IR. The resistance of an ohmic conductor is constant.

• A current flows in a metal because of an electric field established by a potential difference along it. The amount of current depends on the field, the crosssectional area and the length as well as the nature and temperature of the material. • The resistance of a conductor is given by rL A

R=

where ρ is the resistivity, a characteristic of the material. The resistivity normally depends on the temperature.

2.4 questions Resistance, ohmic and non-ohmic conductors lil

M1

lil

lil

ilil

M2

lil

M3

i l il il il

i l il

ilil

i l il il il

i l il

i l il il il

i l il

ilil

i l il il il

i l il

ilil

M4

1 Andy wishes to measure the I–V characteristic of a light bulb. He has set up a circuit as shown. In which of the positions, shown M1–M4, can he place: a a voltmeter? b an ammeter? 2 A student obtains a graph of the current–voltage characteristics of a piece of resistance wire. 10

Current (A)

3 A student finds that the current through a resistor is 3.5 A while a voltage of 2.5 V is applied to it. a What is the resistance? b The voltage is then doubled and the current is found to increase to 7.0 A. Is the resistor ohmic or not? 4 Rose and Rachel are trying to find the resistance of an electrical device. They find that at 5 V it draws a current of 200 mA and at 10 V it draws a current of 500 mA. Rose says that the resistance is 25 Ω, but Rachel maintains that it is 20 Ω. Who is right and why? 5 Nick has an ohmic resistor to which he has applied 5 V. He measures the current at 45 mA. He then increases the voltage to 8 V. What current will he find now?

8 6 4 2 0

c What assumptions did you make in answering part b? d What is the resistance of this wire at 5 V, and at 20 V?

0

5

10 15 Voltage (V)

20

25

a What current flows in this wire at a voltage of 7.5 V? b What voltage would be required to make a current of 20 A flow in this wire?

6 Lisa finds that when she increases the voltage across an ohmic resistor from 6 V to 10 V the current increases by 2 A. a What is the resistance of this resistor? b What current does it draw at 10 V? 7 The resistance of a certain piece of wire is found to be 0.8 Ω. What would be the resistance of: a a piece of the same wire twice as long? b a piece of wire of twice the diameter?


65

8 A strange electrical device has the I–V characteristic shown.

9 Two students have measured the I–V characteristics of two electrical resistors and have found them to be straight lines with different slopes. Elsa says that the one with the steeper slope has a greater resistance, but Cathryn says the one with the lower slope has the greater resistance. Who is right and why?

1.5

1.0 I (A) 0.5

0

10 V (V)

20

a Is it an ohmic or non-ohmic device? Explain. b What current is drawn when a voltage of 10 V is applied to it? c What voltage would be required to double the current drawn at 10 V?

66

Electricity

d What is the resistance of the device at: i 10 V? ii 20 V?

10 In Worked example 2.4C the resistance of a piece of copper wire 1.8 mm in diameter and 10 m long was found to be 0.068 Ω. a What would be the resistance of a piece of aluminium wire of the same dimensions? (The resistivity of copper is 1.7 × 10−8 Ω m, aluminium is 2.8 × 10−8 Ω m.) b What voltage drop would occur along the aluminium wire if 10 A were flowing in it?

2.5 Electrical ener gy

and power

Anyone who has been anywhere near a lightning strike knows the enormous electrical energy that nature can unleash in a fraction of a second. You are quite likely using a tamed version of that energy by which to read this book. One only needs to look around at all the devices that rely on it, in one form or another, to be reminded that electrical power is central to our modern way of life.

Electrical energy Electrical potential energy is produced whenever charges are pushed close together. This energy can be transmitted long distances from power stations or simply produced on demand from the chemical energy stored in batteries. The EMF, or voltage, of a power source is a measure of the number of joules of energy stored for each coulomb of charge. As the charges move through the circuit they lose the energy given to them by the source. The potential energy lost by a charge q moving through a potential difference V is given by ∆U = qV. This energy is released in the circuit as heat or some other useful form of energy E and so: E = qV As the current is the rate at which charge is moving, the total charge q can be expressed as q = It. The total energy produced is therefore given by:

Electrical energy (joules) = potential drop (volts) × current (amps) × time (seconds) … = VIt

Worked example 2.5A How much energy is used in 1 h by a 240 V heater drawing 5 A?

Solution Each coulomb of charge gives 240 J of heat energy, and in 1 h the number of coulombs used is given by q = 5 A × 3600 s. The total energy used is thus … = VIt = 240 × 5 × 3600 = 4.3 × 106 J = 4.3 MJ.

Electric power Power is the rate of energy use: P = E/t. (Remember that the SI unit for power is the watt, where 1 watt = 1 joule per second.) Dividing the previous expression by t gives: E/t = VIt/t or P = VI This is an important expression. A simple example might help you see what it really means—as distinct from simply knowing the formula! Think of the energy delivered to a 12 V headlight bulb drawing 5 A from a car battery. Each second, 5 C of charge pass through the lamp filament. Each of these coulombs carries 12 J of energy from the battery. This means that every second 12 × 5 = 60 J of energy is delivered to the lamp. The power is


67

the energy delivered per second and so this lamp is operating at a power of 60 J per second, that is, 60 watts. This can be summarised: Total energy Energy provided Number of supplied to bulb = by each unit of charges supplied × each second charge each second For the car bulb: 60 J s–1 = 12 J C–1 × 5 C s–1, which is the same as 60 W = 12 V × 5 A.

Power (watts) = voltage (volts) × current (amps) P = VI This relationship applies in all electrical situations. Fundamentally it is another expression of the principle of conservation of energy: the energy we can obtain from an electric current (each second) is equal to that put into it by the source of the voltage.

Worked example 2.5B Two different torch bulbs are rated as 2.8 V, 0.27 A, and 4.2 V, 0.18 A. a Which will be the brightest? b Could they be interchanged? c What are the resistances of the two bulbs at their operating voltages?

Solution a The brightness is indicated by the power used—although only about 5% becomes light energy. P = VI and so their powers are 2.8 × 0.27 = 0.76 W (or 760 mW) and 4.2 × 0.18 = 0.76 W. The bulbs will be the same brightness. b Although the power of the bulbs is similar, using them in the wrong torch will either result in the bulb burning out or running dim. The bulbs are designed to work at a certain voltage. If a greater voltage is used, too much current will flow in the bulb, which will result in it burning out. c The resistance is given by R = V/I = 2.8/0.27 = 10.4 Ω for the first bulb and R = 4.2/0.18 = 23 Ω for the second bulb. Clearly, if the first bulb was subject to 4.2 V much more than the 0.27 A would flow—and more power than it could handle would be produced. The higher resistance of the second bulb would mean that insufficient current would flow at 2.8 V.

Another unit for electrical energy Physics file It is easy to confuse the energy and power units for electricity because of the presence of the ‘kW’ in both. Remember: Units 1 1 Units 1

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of energy MJ = 103 kJ = 106 J kW h = 103 W h = 3.6 MJ of power MW = 103 kW = 106 W

Electricity

The total amount of energy used by an appliance depends on the time for which it is switched on. The total energy is given by the product of the power and the time: E = Pt (1 joule = 1 watt × 1 second). When discussing domestic appliances, time is more likely to be measured in hours than seconds. Just as one ‘watt second’ is one joule, a ‘watt hour’ is also a unit of energy and will be equal to 3600 joules (as there are 3600 seconds in an hour). Similarly one ‘kilowatt hour’ will be 3 600 000 joules (3.6 MJ). The watt hour (W h) and kilowatt hour (kW h) are the amounts of energy used in one hour by a device using a power of 1 watt or 1 kilowatt respectively.

Worked example 2.5C How much energy does a 100 W light bulb use in half an hour?

Solution Here P = 100 W and t = 0.5 h. So … = 100 W × 0.5 h = 50 W h or 0.05 kW h This could also be given as 100 W × 1800 s = 180 000 J = 180 kJ. Electricity supply companies install a meter in every home which measures the power consumed in kW h. We are charged around 15 cents for each kW h used on the normal tariff. In homes where off-peak electricity is used, for example for a storage hot water heater that is only on in the early hours of the morning, off-peak electricity is charged at a lower rate. A typical power bill is shown in Figure 2.33.

Interactive tutorial 3 Kilowatt hours

Figure 2.33 Although we often refer to our ‘power bill’, it is really an ‘energy bill’. Note that in this case, while a total of 299 kW h was used from the grid, 317 kW h (that is, 410 – 93) was generated by photovoltaic (PV) panels on the roof. The power company will only credit up to the amount used, and so the credit is for only 299 kW h. The amount paid for the solar electricity is a little more than that charged for grid electricity, but service charges are added. In some countries, all solar electricity generated is paid for at a rate of up to four times that charged for grid electricity.


69

Worked example 2.5D At a rate of 15 cents per kW h, how much will it cost each week to run a 200 W television set for 4 hours per day? Compare this to the cost of running an electronic clock rated at 5 W.

Solution The energy required for the television set each day is … = Pt = 200 × 4 = 800 W h = 0.8 kW h. In one week the total will be seven times this, or 5.6 kW h. That will cost 84 cents. The clock will use 5 × 24 × 7 = 840 W h, or 0.84 kW h in 1 week. This will cost 13 cents.

Electric power production and transmission Figure 2.34 A typical old-style household electricity energy meter. The dial on the right reads the number of unit kW h while each one to the left reads one power of ten higher. When reading a meter, be careful to note that each alternate dial turns in the opposite direction and to read the number which the pointer has most recently passed. The small dial reads tenths of a kilowatt hour and the large horizontal disk, which drives the meter, spins at a rate which you will find stated on the meter (400 revs per kW h in this case). These meters are gradually being replaced by new ‘smart meters’ that, among other things, can be read remotely.

When electric power is generated on a large scale it is almost always AC, or alternating current power. What does this mean? A battery, a solar cell and a Van de Graaff generator all produce what is known as a DC EMF. The letters stand for direct current but could just as easily mean direct voltage in many situations. In fact, although it is something of a contradiction in terms, the expression ‘DC voltage’ is commonly used. A DC source of EMF always pushes charges in one direction. The top (red or +) terminal of a dry cell is always positive, as the chemicals inside push electrons from the top terminal to the bottom. In any device connected to the dry cell, current will flow from the positive terminal of the cell to the negative terminal. As the name suggests, an alternating current continually changes direction. The AC mains voltage we use in our homes reverses direction 50 times every second. The active terminal (often coloured red or brown) might have a positive potential at one moment, but it will have a negative potential 0.01 s later. A detailed study of alternating current remains for Year 12, but for most purposes we can assume that the mains 240 V AC power will have the same effect as a 240 V DC source.

Physics file In fact, the potential of the active terminal of an Australian AC supply varies between +340 V and −340 V during one cycle. The neutral terminal remains at zero potential. The 240 V quoted as ‘mains voltage’ is actually an average potential. It is the DC potential that would be required to provide the same amount of power.

Direct current

Alternating current

E e

E e

e

e

e

e

Figure 2.36 In a wire carrying a DC current, the electric field (…) always points in the same direction and the electrons are slowly drifting in the opposite direction. In an AC current the field is alternating in direction 50 times per second and the electrons hardly move at all—they just vibrate back and forth.

+340 240 V DC

V (V)

0V

24 AC

–340

Figure 2.35 Voltage of Australian AC supply.

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Electricity

t

Most of the power used in cities is generated by large power stations a long way from where it is eventually used. The power is generated at about 20 kV (20 000 V) but then ‘transformed’ to a much higher voltage, typically 500 kV, for transmission to the city. There it is transformed down to 22 kV for distribution. Transformers on the poles along our streets reduce it further to 240 V for domestic use. We won’t be studying the internal operation of transformers this year, but the law of conservation of energy tells us that all the power that goes into them must come out. In fact, a little will be converted into heat, but in a good transformer more than 99% continues on as electrical energy but at a different voltage.

The power relation P = VI tells us that if a transformer changes the voltage but does not alter the power, the current must also change. If the voltage was doubled, for example, the current would halve. This is exactly why the voltage is changed. At 500 000 V the current needed to transmit power is clearly far less than the current needed at lower voltages. As the size of the cables needed to transmit power depends on the current, considerable cost savings are achieved by using high voltages. For a transformer: Pin = Pout so VinIin = VoutIout or Iout/Iin = Vin/Vout In other words, if the voltage is increased by a certain ratio, the current decreases by the same ratio.

Worked example 2.5E If the power from a generator operating at 20 kV and producing a current of 10 000 A is transformed up to 500 kV, what current will flow at the higher voltage? How much power is being produced? (Assume the transformer is 100% efficient.)

Solution The output voltage of 500 kV is 25 times higher than the input voltage of 20 kV. This means that the output current will be 25 times lower than the input current of 10 000 A. The output (high voltage) current will therefore be 10 000/25 = 400 A. The power can be calculated at either voltage: P = VI = 20 kV × 10 000 A = 200 MW (megawatts) or 500 kV × 400 A = 200 MW

  Table 2.4 Typical values for daily domestic electrical energy consumption Appliance

Typical power (W)

Average use (h day–1)

Energy use (average W h day–1)

Min.

Max.

Min.

Max.

15

100

1.0

3.0

15

300

Refrigerator

100

260

6.0

12.0

600

3120

Electric stove

800

3000

0.25

1.00

200

3000

Microwave oven

650

1200

0.2

0.3

120

360

Toaster

600

600

0.03

0.1

18

60

15

100

0.3

1.0

5

100

500

1000

0.1

0.4

50

400

Min.

Max.

Kitchen Lights

Laundry Lights Iron

500

900

0.2

0.3

100

300

1800

2400

0.2

0.5

360

1200

15

75

0.07

0.07

1

5

Lights

15

200

1.0

4.0

15

800

Television

25

200

0.5

5.0

13

1000

Washing machine Dryer Sewing machine Living

100

100

0.5

5.0

50

500

Stereo

60

200

0.5

5.0

30

1000

Radio

10

40

0.3

3.0

3

120

Vacuum cleaner

100

1000

0.1

0.3

10

300

Computer

100

300

1.0

5.0

100

1500

Video recorder


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Appliance

Typical power (W)

Average use (h day–1)

Energy use (average W h day–1)

Min.

Max.

Min.

Max.

11

100

0.5

2.0

6

200

11

100

0.2

1.0

2

100

11

100

0.2

2.0

2

200

200

800

0.2

0.2

20

160

1750

2500

4.0

6.0

7000

15 000

Fan heaters

2000

7000

6.0

12.0

12 000

84 000

Strip heaters

500

1500

0.5

1.0

250

1500

1000

3000

8.0

14.0

8000

42 000

Min.

Max.

Bedrooms Lights Bathroom Lights Garage/external Lights Power tools Hot water Electric storage Electric heating*

Oil-filled heaters *During winter months

Physics in action

Electric power and greenhouse gases In 1960 the average Victorian used about 8 kW h of electrical energy each day (including industry and commerce). Today, that figure is about 29 kW h. Every kilowatt-hour of electric energy we use results in the release of about 1.4 kg of carbon dioxide into the atmosphere. This is because 85% of our electricity is generated by burning brown coal, one of the Biogas 0.2% Hydroelectricity 0.2% Biomass 2%

Solar energy 0.1% Wind energy 0.1%

Oil 32% Brown coal 49%

Figure 2.37 By far the largest source of primary energy in Victoria,

brown coal is almost entirely used to generate electricity. Some gas is also used for electricity, but as you can see, the amount of renewable energy used is very small.

Electricity

Transport 1% Water etc. 1% Mining 2%

Agriculture 1% Manufacturing 38%

Electricity generation & distribution 16%

Natural gas 17%

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worst fossil fuels for CO2 emission—but a fuel of which Victoria has plentiful supplies. As electricity generation accounts for a little over half of our primary energy use, the problem of reducing greenhouse emissions from electricity production is a very significant one.

Commerce 20%

Residential 21%

Figure 2.38 The largest user of electricity in Victoria is the manufacturing industry, with commerce and residential using about 20% each. As you can see, a significant proportion of the electric power generated is actually used, or lost in transmission, by the electricity industry itself.

The next generation ‘advanced pressurised fluid bed’ power stations may be able to reduce the CO2 emissions from 1.4 kg per kW h down to about 0.8 kg per kW h, but the only hope of a sizeable reduction in emissions from coal-burning power stations is to capture and bury the CO2 (geosequestration) before it is released into the atmosphere. However, neither of these advances is likely before the 2020s. Likewise, even if nuclear energy was approved, it would be well over a decade before it could begin generating power. Serious attention to greenhouse gas emission requires action on a much faster timescale if we are to avoid dangerous climate change. Fortunately, there are ways in which electricity can be generated with little CO2 produced. Wind and solar energy are widely used in Europe, with Denmark, for example, generating about 20% of its electricity from wind farms. Solar electricity from photovoltaic (PV) panels is becoming more cost-effective and also has the advantage that it supplies energy when the peak demand from air-conditioners is greatest. It is often said that the problem of supplying ‘base-load’ electricity means that these renewable energy sources will remain marginal. However, when different systems (including hydro, geothermal and tidal) are linked over large distances by high voltage DC transmission lines, a very large proportion of our electrical energy could be supplied by sustainable means.

Figure 2.39 On e proposal for pr oducing solar en glass-roofed gr ergy is the ‘Pow eenhouse capt er Tower ’. A vast ures the Sun’s en air produced ru , ergy in dark-col shes up the 1 km oured rocks. Th tall chimney in generate electri e hot the centre, driv city. Because th ing e rock stores he still be produced at during the da turbines which at night. y, electricity ca n


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2.5 summary Electrical energy and power • The potential energy stored when charge is concentrated can be converted into other forms of energy in nature or in man-made devices. • The rate at which this energy is released (the power) is given by the product of the potential (voltage) and the current, P = VI. In SI units, 1 watt is equal to 1 volt × 1 ampere (1 W = 1 A V).

• The direction of an EMF may be either constant or alternating. A constant EMF gives rise to a direct current (DC) and an alternating EMF gives rise to alternating current (AC). • Power is transmitted to cities at high voltages in order to keep the current needed relatively low.

• Total energy produced is given by E = Pt (1 J = 1 W s). The kilowatt hour (kW h) is an alternative energy unit and is equal to 3.6 MJ. Matter is made up of huge numbers of positive and negative charges.

Some charges are free to move in conductors but not in insulators.

Charge cannot be created or destroyed. An electrostatic charge is the result of transfer of charge from one object to another.

Like charges repel, unlike charges attract.

Electrostatic effects are transmitted by conductors but not by insulators.

The coulomb force between charges is given by kq1q2 F= . r2

Systems of charges will create an electric field: F E= . q

A source of EMF will create an electric field in a conductor which will result q in a current, I = . t

The current depends on the resistance of the particular conductor. In ohmic conductors the resistance is constant, V = IR.

A source of EMF gives charges electric potential energy (U) by ‘concentrating’ ∆U them: EMF = . q

As charges move through a circuit they lose their potential energy. Energy released E = qV.

The power produced as a current I flows in an electrical device with a potential V across it is given by P = VI.

74

Electricity

2.5 questions Electrical energy and power 1 A 4.5 V battery is used to power the motor of a toy car. a How much energy (in joules) is given to each coulomb of charge that flows through the battery? b How much electrical potential energy is released in the motor by each coulomb of charge that flows through it? c Will all of the electrical potential energy released in the motor appear as kinetic energy of the car? Explain. 2 How much energy does one electron receive when it travels through a 1.5 V cell? 3 What is the power used by a: a 3 V torch bulb drawing 0.2 A? b car starter motor which takes 200 A from a 12 V battery? c mains-powered (240 V) toaster rated at 3 A? 4 How much current is used by a: a 60 W, 240 V light globe? b 1200 W mains-powered heater? c 90 W car windscreen wiper motor?

5 What is the voltage of a: a 100 W spotlight which draws 4 A? b 200 mW radio operating with a current of 23 mA? c 7500 W (10 HP) industrial motor using 18 A? 6 A large power station generator is rated as 500 MW with a 24 kV output. What current would it be generating? 7 How much energy is used by a 5 W digital clock in 1 week? Answer in kW h as well as in joules. If electric energy costs 15 cents per kW h, how much will it cost to run the clock for a year? 8 A step-down transformer is used to run a 12 V model railway from the 240 V mains. If the model engine operates at a power of 18 W, and the transformer can be assumed to be 100% efficient, what is: a the current used by the engine? b the input current to the transformer from the mains? 9 Briefly describe the difference between AC and DC electric power. Give examples of some sources of each. 10 Why is electric power transmitted to cities at very high voltages?

chapter review 1 Matter is said to be made up of huge, but equal, numbers of negative and positive particles. What are these particles called, where are they normally located, and how can matter become charged? 2 Before the nozzle of the fuel pump is inserted into the fuel tank of an aircraft, an ‘earth wire’ is always connected to the metal frame of the plane. Why do you think this is done? 3 i

A

B

A

B

iii

A

B

iv

A

B

i Two metal blocks A and B are in contact and near a positively charged object C. ii The blocks are then separated while C remains present. iii C is then removed. iv Finally the blocks are brought together again.

Describe the charge present on the two blocks (A and B) at each stage (i–iv) of this process.

C

4 Would it be possible for lines of electric field, such as those shown in Figure 2.13, to cross each other?

C

5 Do you think a world in which like charges attracted and unlike charges repelled would be possible?

ii

6 What would be the force between two 100 µC charges placed 1 m apart? 7 If the two 100 µC charges in the last question were then moved together until they were 10 cm apart, how would the force change?


75

8 A charge of 5 µC is placed near a system of charges and experiences a force of 2 N. What is the strength of the electric field at that point? 9 It was stated that a lightning bolt can transfer 10 C of charge to the ground with a current of up to 10 000 A. On the basis of these values, how long does the lightning bolt last? In fact, it is said that a lightning bolt may last up to half a second. Can you find an explanation for this apparent discrepancy?

a What is the resistance of the extension cord? b How much voltage drop will occur in the cord while the heater is running? c How much power will be dissipated in the extension cord? 17 The graph shows the I–V characteristic of a device known as a thermistor. 1.0

10 Wood is sometimes regarded as a conductor and yet in other circumstances is referred to as an insulator. Why is this? Give some examples to illustrate your answer.

0.8 0.6

11 How much electrical energy can be obtained from each coulomb of charge taken from a 240 V mains power point? 12 A torch like that in Figure 2.23 has two fresh 1.5 V cells in place but is switched off. What is the potential difference between: a the bulb and the spring at the bottom? b the bulb and the positive terminal of the battery? c the bulb and the metal strip that makes contact with the bulb when the torch is turned on?

I (A)

0.4 0.2

10

20

V (V)

13 When the torch shown in Figure 2.23 is turned on, in what part of the circuit would you find the: a most abrupt change of potential? b strongest electric field?

a Is this an ohmic device? Explain your answer. b What is the resistance of the thermistor at voltages of 8 and 16 volts? c What power would be dissipated in the thermistor at these voltages?

14 A 3 V torch with a 0.3 A bulb is switched on for 1 minute. a How much charge has travelled through the filament in this time? b How much energy has been used? c Where has this energy come from?

18 The power from a 10 000 V (10 kV), 200 A generator is transformed up to 400 kV by a loss-free transformer for transmission. a What is the power being produced by the generator? b What current will flow in the power line?

15 The resistance of a certain heating element when tested with a multimeter is found to be 40 Ω. It is then used as a heater supplied with 240 V and the current is found to be 5 A. a What is the resistance of the element when used as a heater? b Why is this resistance greater than the 40 Ω obtained with the meter? c What is the power of this heating element?

19 An underground aluminium power cable specified for a house is to be 20 m long with a diameter of 3 mm, giving a cross-sectional area of 7.1 mm2. The electrician finds that a new alloy is available which has half the resistivity of aluminium and so decides to use a 1.5 mm diameter cable of the new alloy. (Resistivity of aluminium = 2.8 × 10−8 Ω m.) a What would be the resistance of the cable specified originally? b Will the new cable have the same resistance as the one specified? Explain.

16 An electrical extension cord used to run a 2400 W mainspowered heater is found to have the I–V characteristics shown in this graph.

10 Current (A)

8 6 4 2 0

76

Electricity

0.2

0.4 0.6 Voltage (V)

0.8

20 It is found that on an average day a refrigerator motor, rated at 200 W, comes on for 15 minutes every hour. How much energy does it use in a year? Give your answer in kilowatt hours as well as in joules.

Elect ric ci

rcuits

I

f you have ever looked inside a radio or other electronic device you will realise that electric circuits can be quite complex. However, the basic principles that apply to these circuits are the same as those that apply to a simple torch, or for that matter to an electric train using megawatts of power. An understanding of these basic principles will prepare us for whatever electrical or electronic device might come our way. Perhaps more importantly, once we have developed an insight into the nature of electrical circuits and devices, we will have gained a feeling for the essential nature of a technology without which most of our modern lifestyle would be impossible. Whether we ever again tackle a circuit with a soldering iron, or even look inside an electrical device, a good understanding of electricity and its uses can help us to gain an awareness of the underlying nature of our technological world. Many questions can be asked about the wise use of much modern technology. For example, the use and generation of electricity involve fundamental questions for both developed and developing societies. There is a lot of talk of ‘alternative’ forms of energy, but much of it is ill informed. This discussion needs to be taken seriously by a well-informed public. An understanding of the basics of electrical technology is a very good place to start for anyone who wishes to help find sensible answers to these questions.

by the end of this chapter you will have covered material from the study of electricity including: • series and parallel circuits • characteristics of cells, batteries and power supplies • household electricity supply, circuits and appliances • electric shocks and safety • electricity in cars and in isolated situations.

3.1 Physics file Superconductors are materials that have zero resistance at very low temperatures. Mercury and lead become superconducting at temperatures of 4.2 K and 7.2 K respectively. Curiously, most metals that are good conductors at normal temperatures, gold and copper for example, do not become superconducting at all. In 1986 a new class of so called ‘warm superconductors’ was discovered. It was found that various obscure compounds of metal oxides with rare earths became superconducting around the relatively high (but still rather cold!) temperature of liquid nitrogen (77 K). The significance of this was that liquid nitrogen is about as cheap as milk and so it made practical applications look much more feasible. So far, the applications are mainly in the production of very strong magnets. This is because once an electric current is set up in a superconductor it simply keeps going as there is no resistance to absorb its energy. As you will probably know, a strong electric current creates a strong magnetic field.

its u c r i c c i r t c e l Simple e Any electric circuit consists of one or more sources of EMF connected by good conductors to various combinations of circuit elements. By ‘good conductors’ we mean that any potential drop (p.d.) in them will be very small compared with the potential drops across the circuit elements. Only in a superconductor is there no p.d. at all. For example, the p.d. along an extension cord used to run a 240 V heater might amount to 2 or 3 V. In order to draw clear diagrams of electrical circuits, a range of more or less standard symbols and conventions is used. Some of these are given in Figures 3.1 and 3.2. Device

wires crossed not joined

Symbol

or

wires joined, junction of conductor

Device

Symbol

cell (DC supply)

battery of cells (DC supply)

resistor or other load

AC supply

fixed resistor

ammeter

A

filament lamp

voltmeter

V

diode

fuse

Physics file The symbols for the circuit devices shown in Figure 3.1 are commonly used, but are not the only ones you will see representing these devices. There is no ‘official’ set of symbols, although widespread usage of symbols such as these tends to mean that they become the unofficial standard. It is important, however, to keep a flexible approach and interpret symbols in the context of the particular circuit.

earth or ground

switch

Figure 3.1 Some commonly used electrical devices and their symbols.

Two circuit rules The principle of conservation of charge and the principle of conservation of energy can be used to establish two very important rules that apply to all electric circuits.

In any electric circuit the sum of all currents flowing into any point is equal to the sum of the currents flowing out of it. For example, if at a junction of three wires there is 2 A flowing in on one wire and 3 A flowing in on another, then there must be a current of 5 A flowing out on the third. As current is simply flowing charge, this ‘rule’ is basically a consequence of the principle of conservation of charge.

78

Electricity

2A

(a)

5A bulb

switch (b)

element

3A

switch

mains lead

element

AC

Figure 3.3 The current flowing into any junction must be equal to the current flowing out of it. Each little symbol could be imagined as the movement of 1 C of positive charge in 1 s.

5 volts + 12 V

switch (c)

starter motor

battery motor

switch

Figure 3.2 Some electrical devices and their circuit diagrams. (a) A torch. (b) A single bar radiator. (c) A car starter motor. 2 volts

Sometimes this rule is abbreviated simply to ‘the sum of all currents at a point is zero’. Remember that currents flowing into the point will be positive and those flowing away from the point negative. The second rule is an expression of the principle of conservation of energy.

3 volts

Figure 3.4 The little creatures represent coulombs of charge loaded with various amounts of energy from the battery, which they give up as they travel around the circuit. They will use up a little in the conductors, but that would be very small compared with the amount given to the light bulbs.

The total potential drop around a closed circuit must be equal to the total EMF in the circuit. In the torch, for example, if the battery supplies an EMF of 3.0 V and we measure a 2.8 V p.d. across the bulb, there must be a 0.2 V drop somewhere else in the circuit—possibly across the switch contacts if they are a little dirty. As the EMF is the energy given to each unit of charge and the total potential drop is the energy obtained from each unit of charge, they must be equal. These two rules are known as Kirchhoff’s laws after Gustav Kirchhoff who first stated them in 1845. In terms of our water cycle analogy (Section 2.3) the first rule is equivalent to saying that the total water flowing into a river junction is the same as that flowing out from it. The second is equivalent to saying that the height gained by water evaporated by the Sun is equal to the distance it falls as it returns to the sea. While the first might be a little questionable for water, as a result of some evaporation or seepage


79

into the ground, the second obviously has to be exactly true. As electric charge does not ‘evaporate’ or seep into the insulation, both laws are very accurate for electric circuits. These two simple rules, or ‘laws’, are the basis for the analysis of any electrical circuit. In the rest of this chapter we will use them to look at some practical circuits.

(a)

(b)

Two ways of connecting circuits

Figure 3.5 Two circuit elements (a) in series and (b) in parallel.

PRACTICAL ACTIVITY 10 Series circuits

Figure 3.6 Mains-powered Christmas tree lights are usually wired in series so that lower voltage globes can be used. If 20 globes are used in series the voltage across each is just 12 V. The problem is that, if one globe burns out, all the bulbs go out, although some bulbs are made to short circuit when they burn out so that the current continues to flow. EMF

V1

V2

Figure 3.7 In a series circuit the sum of the two potential drops V1 and V2 will be equal to the total supplied potential difference.

80

Electricity

No matter how complex a circuit, it can always be broken up into sections in which circuit elements are combined either in series or in parallel; that is, one after another, or beside each other. Charge flowing in a series circuit flows through one element and then through the other. Charge flowing through a parallel circuit flows through one element or the other. Christmas tree lights are often wired in series. Normal house lights and power points are wired in parallel. We will consider series circuits in this section and parallel circuits in the next.

Series circuits Consider the case of a battery in a circuit with a torch bulb and a resistor in series, as in Figure 3.7. The current flowing in this circuit will pass through both the bulb and the resistor. The current that flows through the bulb will also flow through the resistor. Notice that, as Kirchhoff’s first law tells us, no current is ‘lost’ as it goes around the circuit. What is ‘lost’ is the potential energy carried by the charges. This energy is converted to heat, light and so on in the various devices in the circuit. This will result in a potential drop across each device and, as Kirchhoff’s second law tells us, the sum of the potential drops will be equal to the total EMF supplied from the battery. So what characterises a series circuit is that the same current flows through each device, and the total of the voltage drops across the devices adds up to the total EMF in the circuit. Note that when adding the EMFs it is important to take into account the direction. If one cell is reversed, for example, that EMF is counted as negative. It is the I–V characteristics (see Section 2.4) that define the electrical behaviour of a circuit. For an ohmic resistor the I–V characteristic can be given simply by its resistance. For any other circuit device, such as the bulb in this circuit, the I–V graph is usually needed. If the characteristic is known then it is possible to determine the current and power at any particular voltage. The question, then, is if the I–V characteristics of the two individual devices are known, can the I–V characteristic of the series combination be found? When circuit elements are placed in series, because the charge has to be ‘pushed’ through one element and then the next, common sense would lead us to expect that a greater total voltage would be needed to achieve a particular current flow. This is indeed the basis for finding the combined characteristic. As the current in both must be the same, we find the voltages required to cause that current to flow in each of them, and then add them to find the total required voltage. If the two individual I–V characteristics are drawn on a graph (see Figure 3.8) this can be done, in effect, by adding the two graphs ‘sideways’. That is, choose several current values and for each find the two voltages and add them to obtain the I–V characteristic for the series combination. This is best explained by an example.

Worked example 3.1A 5

The resistor and the light bulb whose I–V characteristics are given in the graphs in Figure 3.8 are to be combined in series. a What is the resistance of the resistor? b Draw the I–V characteristics of the series combination. c The light bulb is designed to operate at a current of 4 A. It is necessary to run the light bulb from a power supply that can only produce voltages between 15 V and 25 V. Suggest a way in which this could be done using the resistor in conjunction with the bulb.

Solution a The resistance is found from V = IR, that is:

I (A)

t igh

3

lb

bu

l

2

0

2

or

ist

s re

1

4

6 8 V (V)

10

Figure 3.8 The I–V characteristics of a fixed resistor and a light bulb.

R = V I

= 10 5 = 2 Ω b In order to find the characteristics of the series combination the graphs are added sideways; that is, the voltages are added at constant currents. For example, to obtain a current of 1.0 A in the circuit requires 2.0 V across the resistor, but 0.5 V across the bulb; a total of 2.5 V is required. A table of values obtained this way follows. Current (A)

Vresistor (V)

Vlight bulb (V)

Vtot (V)

0

0

0

0

1.0

2.0

0.5

2.5

2.0

4.0

2.0

6.0

3.0

6.0

4.0

10.0

4.0

8.0

9.5

17.5

These values are graphed in Figure 3.9. c The bulb requires 9.5 V to operate at 4 A. In order to operate it from a power supply greater than that, a resistor will be required in series to lower the potential across the bulb to 9.5 V. As can be seen from the graph of the series combination, to obtain a current of 4 A through both, the total voltage needs to be set at 17.5 V. At this point there is a p.d. of 8 V across the resistor and 9.5 V across the bulb.

5 4 I (A)

4

3 2

t igh

lb

bu

es

seri

r

to

is es

l

r

1 0

2

4

6

8 10 12 14 16 18 V (V)

Figure 3.9 The resistor and light bulb graphs have been added at constant current, that is, sideways. The voltage of the series graph at any current is the sum of the two individual voltages.

V1

V2

R1

R2

I

V I Re

Resistors in series When ohmic resistors are placed in series with each other the situation is somewhat simpler. Because the I–V characteristic of a resistor is a straight line, they will add together to give a combined characteristic with another straight line. The use of Kirchhoff’s laws enables us to find the effective resistance of a combination of resistors in series. The effective resistance is the value of the single resistor that would have the same I–V characteristic as the two in series. If two resistors are placed in series, we expect the effective resistance to be higher than that of either resistor separately. (Remember that the resistance can be thought of as the number of volts required to achieve a current of 1 A.) Figure 3.10 shows two resistors of values R1 and R2 in series and another resistor of value Re , the value of the single resistor that could replace the

Figure 3.10 Re is the value of the single resistor that could replace the series combination of R1 and R2.

Physics file While the term ‘potential drop’ is more strictly correct, the simpler term ‘voltage’ is very commonly used. In either case, remember that the term always refers to the difference in potential across a circuit element or source of EMF.


81

series combination of the first two. As the two resistors are in series, the sum of the voltages across them must add to the total voltage, that is the voltage across Re. (V1 is the voltage across R1 and V2 that across R2.) The voltage across Re is V, the sum of V1 and V2. That is: V = V1 + V2 For two resistors in series the current, I, through each is the same. As the voltage across a resistor is always given by V = IR we can write: IRe = IR1 + IR2 Dividing both sides by I we have: Re = R1 + R2 Not surprisingly, the effective resistance is simply the sum of the two resistances. If there are more than two resistors in series it is easy to show that the effective resistance is simply the sum of all of them: Re = R1 + R2 + R3 + …

Worked example 3.1B Two pieces of nichrome wire (as used in heater elements) have resistances of 10 Ω and 20 Ω. a What current would flow through them, and what power will be produced in them, if they are separately connected to a 12 V battery? b If they are connected in series what is their total resistance? c When placed in series across the 12 V battery, what current will flow through them and what power will be produced?

Solution a The current will be given by I = V/R, so for the two wires separately the currents will be PRACTICAL ACTIVITY 11 Resistance in a combination circuit

12/10 = 1.2 A and 12/20 = 0.6 A. The power is found from P = VI and so will be 12 × 1.2 = 14.4 W and 12 × 0.6 = 7.2 W, a total of 21.6 W. b When connected in series the total resistance will be 10 + 20 = 30 Ω. c The current that flows from the 12 V battery will be V R = 12 30 = 0.4 A The total power will be 12 × 0.4 = 4.8 W, considerably less than the previous situation.

82

Electricity

I =

3.1 summary Simple electric circuits • An electric circuit normally consists of a source of EMF, conductors to carry current with little loss of energy, and circuit elements that turn the potential energy of the charges into some form of useful energy. • Circuit diagrams are drawn using a set of conventions (see Figure 3.1). • Kirchhoff’s two rules enable us to analyse any circuit: 1 In any electrical circuit the sum of all currents flowing into any point is equal to the sum of the currents flowing out of it. 2 The sum of all EMF values around a circuit is equal to the sum of all potential drops around the circuit.

• All circuits are combinations of circuit elements in series (one after another) or in parallel (beside each other). • In a simple series circuit the same current goes through each element in turn. The total voltage drop is the sum of the individual drops. • The combined I–V characteristics of a series combin ation of circuit elements can be found by adding the individual characteristics ‘sideways’, i.e. adding the voltages at constant currents. • For ohmic resistors in series the effective resistance is given by Re = R1 + R2 + …

3.1 questions Simple electric circuits 1 State the two circuit rules (Kirchhoff’s laws) that apply to all circuits. What are the two basic principles that they are based upon?

and through the other 0.5 A. When they are combined in series, what current will flow through them and what is their effective resistance?

2 An autoelectrician finds that while the EMF of the car battery is 12 V there is only 10 V across one of the taillight bulbs in a car. Can you suggest a reason for this?

8 A 400 Ω resistor and a 100 Ω resistor are placed in series across a battery with an EMF of 5 V. a How much current will flow from the battery? b What will be the voltage across each resistor?

3 Emily has found a point in her car where four wires are attached together. She finds currents of +2.5 A and +1.0 A in two of the wires, and −4.2 A in a third (+ means current into the point and – means current out of the point). What is the current in the fourth wire?

9 Two ohmic resistors, whose I–V graphs are shown, are to be combined in series.

5 Bill has bought two 12 V headlamps for his truck but finds that it has a 24 V battery. He decides that the simplest way to overcome the problem is to wire them in series. a Will the headlamps work correctly? b Do you see any problem with his scheme? 6 Two equal resistors are placed in series and found to have a combined resistance of 34 Ω. What is the resistance of each one? 7 A 10 V power supply is used across two separate resistors. The current through one is found to be 0.4 A

I (A)

4 Two torch bulbs are placed in series with each other and a 4.5 V battery. The current through one bulb is found to be 0.25 A and the voltage across it is 2.1 V. a What is the current through the other bulb? b What is the voltage across the other bulb?

R2

6 4

R1

2

0

5

10 V (V)

15

20

a Draw a sketch graph of the I–V characteristic of the two resistors combined in series. b What is the effective resistance of the combination? 10 It is found that there is 60 V across one resistor of a pair in series and 20 V across the other. If the smaller resistance of the two is 5 Ω, what is the value of the other resistor?


83

3.2 EMF I1

Itot

I2

Figure 3.11 The total current flowing through a parallel circuit is the sum of the two currents in the individual elements: Itot = I1 + I2.

el l l a r a p n i s t n Circuit eleme When two circuit elements (resistors, light bulbs etc.) are put in parallel, as in Figure 3.11, the charges moving around the circuit will go through one or the other. This is in contrast to the series circuit where the charges go through one and then the other. In other words, the current divides at some point. The total current remains the same, but some of it flows through one element and the rest through the other. Because the ends of the two devices are connected together, the voltage across both of them is the same. What characterises a parallel circuit is that the same voltage is across each circuit element and the total of the currents through them adds up to give the total current flowing in the circuit. So putting two elements in parallel increases the paths available to the current and will therefore increase the total current in the circuit. This is in contrast to a series circuit where adding more elements in series decreases the current that can flow.

The I–V graph of parallel circuit elements PRACTICAL ACTIVITY 12 Parallel circuits

When circuit elements are placed in parallel, the charge can flow through either conductor, so common sense would suggest that a greater total current will flow at any particular voltage. This is indeed the basis for finding the combined characteristic: as the voltage across each conductor is the same, we find the current flowing in each and add them to find the total current flowing. This time, the I–V characteristic for the combination is obtained by adding the individual characteristics ‘upwards’. That is, for several different voltages add the two currents together to obtain the total current that will flow in the circuit and plot that point on the graph to obtain the I–V characteristic for the parallel combination. Here is an example.

Worked example 3.2A The resistor and the light bulb whose I–V characteristics were given in Figure 3.8 (and repeated in Figure 3.12) are to be combined in parallel. a Draw the I–V characteristics of the two elements in parallel. b The light bulb is designed to operate at a current of 4 A. If the two are placed in parallel across a power supply, what voltage setting will be required and what total current will flow from the power supply?

9 8 7

pa ral

5

I (A)

4

lb

t bu

ligh

3

r

to

is es

2

r

1 0

Solutions a To find the parallel characteristic requires that the graphs are added upwards; that is,

lel

6

2

4

6

es

seri

8 10 12 14 16 18 V (V)

Figure 3.12 The resistor and light bulb graphs

have been added at constant voltage. The current of the parallel graph at any voltage is the sum of the two individual currents.

the currents at a particular voltage are added. At 2.0 V, for example, the total current is given by 1.0 + 2.0 = 3.0 A. The other values are given below: Voltage (V)

Iresistor (A)

Ilight bulb (A)

0

0

0

2.0

1.0

2.0

3.0

4.0

2.0

3.0

5.0

6.0

3.0

3.5

6.5

8.0

4.0

3.8

7.8

10.0

5.0

4.0

9.0

The combined characteristics are shown in Figure 3.12.

84

Electricity

Itot (A)

0

b The bulb requires 10 V to operate at 4 A. In a parallel circuit the same voltage is applied across both elements; therefore, the voltage should be set at 10 V. As can be seen from the graph (or the table) this will mean that there is also 5 A flowing in the resistor, giving a total current drawn from the power supply of 9 A.

Resistors in parallel

I1

When ohmic resistors are placed in parallel with each other we can expect a greater flow of current in the circuit at any particular voltage. This is because adding each resistor adds a new path for the current without affecting the other paths. So adding more resistors (in parallel) actually decreases the total effective resistance in the circuit. The effective resistance of a combination of two parallel resistors will therefore be less than that of either of them alone. The combined characteristic will again be a straight line. As was the case with the series combination of resistors, the effective resistance of two resistors in parallel can be found by using Kirchhoff’s laws and a little algebra. Remember that the effective resistance is the value of the single resistor that would have the same characteristics as the parallel combination of the two. If the two resistors are placed in parallel it is the sum of the two currents that must add to get the total current: I = I1 + I2 Again, by re-expressing this relationship in terms of the voltages and resistances we can find what we are looking for. This time, replace I by V/R, so I = I1 + I2 becomes: V V V = + Re R1 R2 Dividing both sides by V we have: I I I = + R e R1 R2 This time it is the reciprocals of the resistances that are to be added. Again, this is what might be expected. Re will always be somewhat smaller than either of the other two. If, for example, both resistances are the same, say I 1 2 R, the above expression yields = or Re = 2 R. As might be expected, Re R putting two similar resistances in parallel halves the total resistance in the circuit—as it has doubled the available paths for the current. If there are more than two resistors in parallel the effective resistance is found by adding all the reciprocals: I I I I = + + … R e R1 R2 R3

Worked example 3.2B As in Worked example 3.1B, two pieces of nichrome wire (as used in heater elements) are found to have resistances of 10 Ω and 20 Ω. a If they are connected in parallel what is their effective resistance? b What total current will flow through them and what power will be produced if the combination is placed across a 12 V battery?

R1

I I2

R1 V I Re

Figure 3.13 Re is the value of the single resistor that could replace the parallel combination of R1 and R2.

Physics file The reciprocal of the resistance is sometimes called the ‘conductance’ for the simple reason that a large value would mean a good conductor. It is given the symbol G, i.e. G = 1/R. The unit is then Ω−1, sometimes called a ‘mho’ (ohm backwards). The expression for two resistors in parallel then becomes G = G1 + G2. It is not surprising that the total conductance of two conductors in parallel is simply the sum of the individual conductances. (Combining two conductances in series yields an effective conductance given by 1/G = 1/G1 + 1/G2.) In most circuits we deal with, the resistances have values in the hundreds or thousands of ohms and so it is normal to use resistance rather than the corresponding very small values of conductance. On the other hand, electric power engineers sometimes deal with conductance because their cables need to have very low resistances.


85

Solution a The effective resistance is found from

I

I

I

= + Re R1 R2

1 1 + 10 20 3 = 20 20 Thus Re = = 6.7 Ω 3 b The total current is given by V I = R 12 = 6.7 = 1.8 A The power is therefore P = VI = 12 × 1.8 = 21.6 W (You will find that this answer was also obtained in part a of Worked example 3.1B. Can you see why?)

=

Power in ohmic resistors Quite often we need to calculate the power dissipated in an ohmic resistor. In some cases the resistor may be a heating element specifically designed to produce heat. Because the temperature may affect the resistance, it is important to measure the resistance at the appropriate temperature. In other situations the power (and therefore heat) may be a nuisance, in the confined space of a compact electronic device for example. The power lines that carry electricity from the power stations to the city are ohmic resistors (with very low values of resistance). The power lost in them represents not only wasted generating capacity but extra fuel used and extra greenhouse gases produced. We have already seen (in Chapter 2) that the power dissipated in any electrical device is given by P = VI. This is true whatever the characteristics of the device. The power may appear as any form of energy: kinetic energy from a motor, sound energy from a loudspeaker or heat from a simple resistance wire. In the case of an ohmic resistor the power will produce heat. Because of the simple relationship between current, voltage and resistance, it is possible to find two simple relationships between the power produced and these quantities. By substituting either V = IR or I = V/R into the power equation P = VI, we can find the following expressions.

The power produced in a resistor is given by: 2 P = VI or P = I 2R or P = V R

86

Electricity

Worked example 3.2C In Worked examples 3.1B and 3.2B the two wires with resistance 10 Ω and 20 Ω were placed in series and in parallel across a 12 V battery. Use these two new expressions for power to find the power produced in each wire when they are: a in series b in parallel.

Solution a When the wires were in series a current of 12/30 = 0.4 A flowed. The power in each can be found directly from the expression P = I 2R. For the first wire P = 0.42 × 10 = 1.6 W, and for the second P = 0.42 × 20 = 3.2 W. The total power is thus 4.8 W as found in Worked example 3.1B, part c. b When in parallel the expression P = V 2/R can be used to find the power without the need to find the current first. For the first wire P = 122/10 = 14.4 W, and for the second P = 122/20 = 7.2 W, as was found in Worked example 3.2B, part a. Of course we can obtain all these answers without the use of these new expressions. They are simply useful short-cuts in certain situations. Physics in action

High power—low power Simple heaters of various sorts often have a ‘three heat’ switch. An electric blanket will usually have ‘low’, ‘medium’ and ‘high’ settings, for example. Rather than making three different heating elements the manufacturer can use two elements in different series and parallel combinations to obtain the three heat settings. If the two elements are placed in series the total resistance is relatively high and therefore

V2 ). For the medium R setting one of the elements will be used by itself. The high setting is then achieved by placing both elements in parallel. It is a simple matter to work out the relative power being used for the three settings. If it is assumed that the resistance of both elements is the same (R) and does not change appreciably with temperature, the effective resistance in the three cases will be given by: Low heat (two elements in series): Re = R + R = 2R Medium heat (one element only): Re = R 1 High heat (both in parallel): Re = R 2 As the power is inversely proportional to the resistance 2 V (P =  ), if we call the high setting 100%, then the others will R be 50% and 25%.

element 1 resistance 480 Ω A 240 V N

F

OF

element 2 resistance 480 Ω

the power will be a minimum (as P =

To obtain the various settings the following circuits are used: OFF Not connected to power Resistors connected in series, R = 960 Ω, I = 0.25 A, P = 60 W Only one resistive element is connected, R = 480 Ω, I = 0.5 A, P = 120 W Resistors connected in parallel, R = 240 Ω, I = 1.0 A, P = 240 W

Figure 3.14 An example of the use of series and parallel combinations of resistors to achieve three heat settings for an electric blanket.

3.2 summary Circuit elements in parallel • In a simple parallel circuit the same voltage drop occurs across both elements. The total current that flows is the sum of the two individual currents. • The combined I–V characteristics of a parallel combination of circuit elements can be found by adding the individual characteristics ‘upwards’; that is, adding the currents at constant voltages.

• For ohmic resistors in parallel, the effective resistance is given by: I I I = + +… R e R1 R2 • The power produced in an ohmic resistance can be V 2 expressed as P = VI or P = I 2R or P = . R


87

3.2 questions Circuit elements in parallel 1 Two torch bulbs are placed in parallel with each other across a 3.0 V battery. The current through the battery is 0.55 A. The current through one of the bulbs is 0.25 A. a What is the current through the other bulb? b What is the voltage across the other bulb? 2 What is the effective resistance of two 10 Ω resistors placed in: a series? b parallel? 3 Two ohmic resistors, whose I–V graphs are shown, are to be combined in parallel.

R2

I (A)

6 4

R1

2

0

5

10 V (V)

15

a What is the effective resistance of the combination? b What is the voltage across the pair of resistors? c How much current will flow in each resistor? 7 How much power will be dissipated in each of the two resistors in the previous question? 8 Three resistors of 900 Ω, 1.5 kΩ and 2.0 kΩ are to be used in a circuit. What is their effective resistance if they are all placed: a in series? b in parallel? 9 Two resistors and three ammeters are arranged as shown in the circuit diagram. The battery is 10 V. a What are the readings on each of the three ammeters? b How much power will be produced in each of the resistors? c What is the total effective resistance in the circuit? d What power would be produced if a single resistor of the value determined in part c replaced the two resistors?

20

a Draw a sketch graph of the I–V characteristic of the two resistors combined in parallel. b What is the effective resistance of the combination?

10 V A1 2Ω

4 Two equal resistors are placed in parallel and found to have a combined resistance of 34 Ω. What is the resistance of each one? 5 A 10 V power supply is used across two separate resistors. The current through one is found to be 0.4 A and through the other 0.5 A. When they are combined in parallel, what current will flow through them and what is their effective resistance? 6 A current of 3 A is found to be flowing through two resistors of 20 Ω and 10 Ω in parallel as shown. 20 7

3A 10 7

88

Electricity

5Ω

A2

A3

10 While overseas Ann has bought a hair dryer labelled 220 V, 800 W, and she is going to use it on the Australian 240 V mains. She tells her flatmate Betty that as the voltage difference is less that 10% it will only produce about 10% more heat and that should not matter. Betty, however, claims that it is more likely to produce about 20% more heat, which may cause it to burn out. Who is correct and why?

3.3 Cells, batteries and other sources

of EMF

The source of the energy that the charges on the dome of a Van de Graaff machine obtain is obvious. The motor that drives the belt is literally doing work on the charges by pushing them up closer to other like charges. Although it is a source of EMF, in practice the power it can deliver is limited by the extremely small current it can maintain. Like the Van de Graaff generator, any source of EMF uses some form of energy to create electrical potential energy.

Chemical cells

Physics file

In a battery, or cell as we should say, the source of EMF is the chemical energy stored in the materials used. As we saw at the beginning of the last chapter, atoms have varying abilities to attract electrons—chemists call it electronegativity. A cell basically consists of two materials with different electronegativities, between which there is what we shall call a ‘go-between’ material. Such a cell is represented in Figure 3.15. high electronegativity

low electronegativity A

–

+

B

C

The potential that a good Van de Graaff generator can develop is limited mainly by the size of the dome. The smaller the dome, the more intense the electric field in the air around it and the greater the possibility of ionisation that allows the charges to escape. For the same reason, any sharp points or projections on the dome will reduce the potential. A good school Van de Graaff generator can reach 400 000 volts!

–

+ – +

I

I ions migrating e–

Figure 3.15 A cell consists of two materials that attract electrons to different degrees, along with a ‘go-between’ that acts as a conductor. In this diagram material A attracts electrons most (high electronegativity) and material C attracts electrons least. Material B acts as the go-between.

The chemistry of electrical cells can be very complex, but for our purpose it is sufficient to realise that electrons will flow from the material of lower electronegativity (for example zinc) to the one of higher electronegativity (for example copper) through the external circuit connected to the terminals. In the diagram material A has the highest tendency to attract electrons and material C the lowest. Material B acts as the ‘go-between’. It is effectively a conducting material that allows the other two materials to ‘trade’ electrons by undergoing a chemical reaction with each of them which either replaces the electrons lost (material C) or takes up the electrons gained (material A). In the course of this process the positive and negative ions in material B migrate towards A and C respectively. As a result of these reactions, material B is eventually used up and the product of the reactions replaces it. At this stage the cell stops working and we say it has gone ‘flat’.

Electric circuits

89

The properties of the materials chosen for A, B and C are very important. Materials A and C must have as different electronegativities as possible but also undergo suitable reactions with material B. Material B must allow the migration of the ions (charged atoms) formed in the reaction and so is normally a liquid or a moist paste. In a dry cell, C is the outer zinc casing, and B is a paste of ammonium chloride and other special substances (see Figure 2.22). Material A is not actually a metal but manganese dioxide powder, which is mixed with the ammonium chloride paste. The carbon rod in the centre of the cell is there to collect the current. In a charged car battery, A is lead dioxide coated on a lead plate, B is a solution of sulfuric acid and C is a lead plate. As current from the car battery is used, lead from the lead plate is converted into lead sulfate, which remains as a coating on the plate. The lead dioxide is also converted into lead sulfate, which remains on the other plate. Fortunately this process can be reversed by forcing an electric current through the battery in the opposite direction, and so the battery can be recharged. This is one of the functions of the car’s alternator. There are many other types of cells in use. Some are single use and some (so-called ‘Ni-Cads’ for example) can be recharged. There is now considerable incentive for manufacturers to develop smaller, more efficient rechargeable batteries for use in electric cars and portable electronic devices.

Batteries in series and parallel Cells can be combined in series, parallel or both in order to provide a battery with the required characteristics. The EMF of a set of cells joined in series, ‘head-to-tail’, will be equal to the sum of the individual EMF values. In this case each cell takes the potential of the previous cell and adds to it its own potential. One could imagine each cell taking the charges from the previous cell and pushing them a little closer—’concentrating’ them as we said in the last chapter—in particular, giving them more and more potential energy. Cells combined in parallel must have the same EMF and be joined + to + and − to −. In this case, charge moving around the circuit will go through either one or the other of the cells but not both. As a result, the EMF will not be increased, but each battery will only need to provide half of the total current and so they will last twice as long.

Figure 3.16 Two car batteries are connected in parallel when jumper leads are used to start a car with a flat battery. It is very important to ensure that the batteries are in parallel (+ to + and − to −) and not series in this situation!

90

Electricity

Notice that if, in an attempt to combine the cells in parallel, they were actually joined head-to-tail, instead of head–head and tail–tail, it would create a closed circuit with two cells in series and no significant resistance. This is what is referred to as a short circuit. The batteries would run flat very quickly and could even become hot enough to cause a fire. When jumper leads are used to connect two car batteries in parallel, it is particularly important to ensure that they are not inadvertently connected in series. If they were, an extremely large current could flow and burn the leads or even cause a battery to explode.

PRACTICAL ACTIVITY 13 Internal resistance of a dry cell

Internal resistance of cells

Worked example 3.3A

V Ri

I

IR i single cell battery

load resistor

Figure 3.17 A battery is represented in a circuit diagram as an ‘ideal’ source of EMF in series with a resistor (called the internal resistance Ri). The terminal voltage is given by V = E − IRi, where I is the current drawn by the rest of the circuit. If no current is being drawn V will be equal to E.

Imax

fre

sh

I (A)

Any chemical cell will have a certain amount of internal resistance due to the fact that the current must flow through the various chemicals and the electrodes. Although the chemical reactions always give the charges a certain amount of energy (the EMF of the cell), some of this energy will be lost as the charges move through the cell. As the cell gets ‘flat’ this loss of energy shows up as an increase in internal resistance. If the battery is fresh and being used in a situation in which the current drawn does not exceed its design limits, the internal resistance can be assumed to be negligible relative to the other resistances in the circuit. Generally, to make a battery with a very low internal resistance, the battery has to be large in order to provide plenty of ‘reaction surface’ for many chemical reactions to take place at the same time. A 12 V car battery, for example, might have an internal resistance of about 0.01 Ω. On the other hand, the small 9 V batteries often used in electronic devices have an internal resistance of about 10 Ω. When a current is flowing through the battery there will be a voltage drop across the internal resistance. This results in the actual voltage at the terminals of the battery being somewhat less than the EMF. A battery can be represented as in Figure 3.17; that is, as a source of EMF in series with a resistor. The EMF can be measured by a voltmeter, which draws no significant current, as then there will be no voltage drop across the internal resistance Ri. However, when the battery is in use and a current is flowing, the actual voltage at the terminals will be lower than the EMF by the drop occurring across Ri. This drop will depend on both the current drawn and the value of Ri. The terminal voltage will therefore be given by the expression V = E − IRi. As the battery becomes flat the value of Ri rises and so the terminal voltage decreases. Another advantage of using cells in parallel to make up a battery will now become clear. Because of the lower current in each cell, less energy is wasted in warming the battery (remember that P = I2R) and the voltage at the terminals is closer to the EMF of the cells. Almost any source of EMF will involve some internal resistance. In designing a generator, for example, there is always a compromise between making the wire in the coils thin enough to keep the overall size manageable and the added internal resistance that this will result in.

Imax

used

ba

tte

batt

ry

ery

V (V)

Figure 3.18 As the current drawn (I) from a battery increases, the voltage drop across the internal resistance increases, leading to a lower terminal voltage, V. As the battery goes flat the internal resistance increases, leading to a lower available current, Imax.

Michael and Mary-Ann have made small model electric motors. Mary-Ann is using two D cells in series to drive her motor. Michael, not to be outdone, decides to use a 9 V battery from a transistor radio, thinking that it should get better results than the 3 V battery


91

PRACTICAL ACTIVITY 14 Electrical power

Mary-Ann is using. Unfortunately Michael finds that his motor hardly turns with the 9 V battery and yet when he uses Mary-Ann’s battery it goes well. He tests the two batteries with a voltmeter and finds that they do indeed read 3 V and 9 V as expected. Mary-Ann then suggests reading the voltage while the motor is operating and they find that the voltage of her battery reads 2.5 V while Michael’s reads only 2.0 V. Both motors were found to be drawing 0.5 A while this reading was taken. a Why did Michael’s battery not drive his motor properly despite the higher voltage? b What was the value of the internal resistance of each battery? c When he placed the 9 V battery back into the radio it worked perfectly, the battery showing a voltage of 8.0 V. How much current was the radio using?

Solution a Michael’s battery had a relatively high internal resistance and the motor was drawing too much current. As a result, the voltage drop across the internal resistance was very large. b A current of 0.5 A caused a voltage drop of 0.5 V in Mary-Ann’s battery (3.0 V − 2.5 V) and 7 V (9 V − 2.0 V) in Michael’s battery. The internal resistance is given by Ri = ∆V/I = 0.5/0.5 = 1 Ω (Mary-Ann’s) or Ri = 7.0/0.5 = 14 Ω (Michael’s). c When used in the radio the voltage drop across the 14 Ω resistance was 1.0 V. Thus the current drawn was DV I = R 1.0 = 14 = 0.071 A = 71 mA

Other sources of EMF By far the most common source of EMF is a generator or alternator (they are much the same for our present purposes). Very large generators at power stations can produce hundreds of megawatts of power at about 20 000 V. The alternator in a modern car is capable of producing about 500 W of power at 12 V. They both convert kinetic energy into electrical energy. We shall look at the means by which they convert this energy into electrical energy in Heinemann Physics 12.

Figure 3.19 An electric ray can deliver a brief 600 W electric pulse to stun or kill its prey. It is also a very effective defence against predators!

92

Electricity

There are a number of other interesting sources of EMF. Perhaps the most curious are animals such as the electric ray (Torpedo nobilinia) and the South American electric eel (Electrophorus). They can stun or even kill their prey with pulses of EMF reaching several hundred volts producing a current of up to 1 A. Electric eels use large numbers of cells called electroplaques, each producing 0.15 V at 1 mA. They have about 4 million of these cells arranged as 1000 parallel sets of about 4000 in series along the length of their bodies, giving a total of 1 A at 600 V. A thermocouple simply consists of two different metal wires joined at both ends. Iron and constantan (a copper–nickel alloy) are commonly used. When two different metals are in contact, electrons will have a tendency to move from one to the other. Normally this effect cannot be used as a source of EMF because any circuit will require two such joins and the two electromotive forces generated will oppose each other. However, this effect is temperature dependent and so if the two joins are held at different temperatures, one EMF will be greater than the other. This will result in a current flowing around the circuit. Indeed, this effect is often used to measure temperature, particularly where the temperature is too high for normal thermometers. By arranging many such joins in a series, a useable EMF can be produced. Such a device is called a thermopile and is used in situations where a reliable source of electric power is required and where a source of heat is available. Spacecraft that travel a long way from the Sun use the heat generated by a strong radioactive source to drive a thermopile to power their electrical systems. Gas-powered thermopiles were once used widely to power remote weather stations and the like, but solar cells have almost completely taken over that role.

Solar cells Solar cells, or more correctly photovoltaic cells, have been used widely to produce electricity from sunlight in areas where it is impractical or too expensive to use mains power. Producing electrical energy from solar cells on the roofs of our houses would be an ideal solution to many of the world’s energy problems. (a)

(b) metal current collector fingers n-type region

diffused layer

~ 0.2 Mm ~ 300 Mm

p-type region silicon base material metal base

Figure 3.20 (a) An array of silicon photovoltaic cells. (b) Most of the cell is p-type silicon, but the very thin upper layer has n-type doping material diffused into it. Metal fingers on the top and a metal base collect the current.


93

Electrically, a photovoltaic cell is somewhat like a semiconductor diode. Both comprise two layers of silicon that is ‘doped’ with very small amounts of elements that either tend to add extra ‘free’ electrons to the structure or to leave a gap, thus creating ‘holes’ that can move as electrons move into them (rather like the bubbles of air from a fish tank aerator are moving ‘holes’ in the water). The layer with the extra electrons is called an n-type semiconductor and that with the holes a p-type semiconductor. At the junction of the two layers, electrons from the n-type tend to fall into the holes in the p-type. But this means that electrons have been lost from the n-type layer and gained by the p-type layer, making the p-type layer in the region a little negative relative to the n-type layer. When a photon of light falls on the cell it may knock an electron out of place thus forming a new free electron and hole pair. If this occurs near the junction, because of their relative charges, the electron will tend to go towards the now positive n-type layer and the hole towards the more negative p-type layer. This charge movement results in the p-type region gaining an overall positive potential, while the n-type region gains a negative potential. If an external resistance (a so called ‘load resistance’) is now connected, current will flow through it from the p-type region to the n-type. As long as the light keeps creating the electron–hole pairs the process continues. The overall effect is that the charges are given potential energy by the Sun and the cell acts as a source of EMF. The characteristics of a typical cell are shown in Figure 3.21. The voltage and current obtained depend on the load being used. The power obtained is the product of the voltage and current. If the cell is being used as a source of energy it is important to try to operate the cell at the point where this product is a maximum. If the load resistance is too low, a high current will be obtained but with little voltage. On the other hand, if the load resistance is too high the full voltage will be obtained but with little current. Inherently, solar cells produce low-voltage DC power, but many applications require 240 V AC power. Fortunately, very efficient electronic ‘invertors’ have been developed in the last few years, which are able to take the low-voltage DC and convert it into ‘mains’ power. They are called invertors because electronic devices often require low-voltage DC and need to transform the 240 V AC down and convert it to DC with a combination of a transformer and so called ‘rectifier’. Invertors perform an operation that is the ‘inverse’ of this process. 1.6

Output current (A)

1.4

Figure 3.21 The characteristics of a typical 8 cm diameter solar cell. The three points shown (Pm) represent the operating conditions for maximum power in three different levels of sunlight.

94

Electricity

1.2 1.0

bright summer Sun winter Sun Pm cloudy

0.8 0.6 0.4 0.2 0

0.1

0.2 0.3 0.4 Output voltage (V)

0.5

0.6

0.8

Physics in action (extension)

Solar electricity: the big picture The Sun’s energy falls on the Earth’s surface at the rate of about 1 kW on each square metre. This means that an average roof of about 200 m2 receives 1000 kW h of energy in 5 hours of sunlight. Given that an average household might use around 20 kW h of electrical energy in a day, you can see the possibilities! Commercially available solar cells can convert about 15–25% of the energy falling on them into electricity and so a whole roof of solar cells could feasibly produce well over 100 kW h of energy per day.

The bad news The obvious problem for solar energy is that it is not available at night and it is hard to store electrical energy. Cost is another problem. Most cells are made from single crystals of very high-purity silicon. The cost to produce these cells is around $5 per watt. To collect 20 kW h of energy over an average period of 4 hours of full sun would require solar panels capable of producing 5 kW. That would cost about $25 000. Added to that, there is the cost of storing the energy and converting it into a useful voltage. Although the cost is high, for people in areas well away from mains power it is feasible. Another question is the energy cost. In order to produce the very pure silicon crystals very high temperatures are needed and so considerable energy is used in the manufacture. At present it is estimated that it takes about 2–4 years for the cells to ‘pay back’ this energy. However, given that the cells should last at least 25 years, this cost seems reasonable. Further developments promise to reduce this period to well under 2 years.

The good news There is good news, however. Scientists are developing ‘thin film PV’. For example, at the University of New South Wales, a very thin layer of silicon is deposited on glass—so called CSG (crystalline silicon on glass) technology. This reduces

the energy (and dollar) cost of production considerably as it avoids the need for high temperatures to make large pure crystals. The layer is very thin and can be vacuum deposited. Although these cells are less efficient, a little below 10% at present, the lower production costs should reduce the cost of solar electricity to less than $2.50 per watt. This cost is still more than coal-produced electricity, but if the carbon cost of coal energy is taken into account it is a very attractive proposition—and eventually the cost should fall further. While the normal household demand for power is greater at night (except in hot weather), the total demand from industry and commerce is greater during the day. So rather than trying to store the electricity produced from solar cells, the more feasible alternative for a household is to sell electricity back to the grid during the day and buy it back at night. As well as reducing the total demand from coal-fired generators, this has the great advantage of evening out the demand on those ‘base load’ generators. It is estimated that each household producing 10 kW h of electricity from the Sun each day would save about 5 tonnes of carbon dioxide from entering the atmosphere each year. When enough solar power is generated, excess power produced during the day could be stored for use at night—for example by pumping water back up into hydro power station dams. This is already done as a way of storing the excess energy generated by coal-fired stations when demand is low. Over the last decade it has become clear that if we are to avoid dangerous climate change we need a huge reduction in the amount of greenhouse gases released into the atmosphere. When combined with other forms of renewable energy such as wind and geothermal power, solar electricity could be the key to a safer future. If our transport needs could be met by a much improved electric train and tram system, as well as electric-powered cars, we may yet see the end of the ‘fossil fuel era’.

ricity is cheap

solar elect to Figure 3.22 Once roof could be used

ehold enough, each hous y. Most of it would be sold da e th g rin du it collect bought back panies and then m co r we po e th to dy installing ea alr e ar s usehold at night. Many ho such schemes.


95

3.3 summary Cells, batteries and other sources of EMF • A source of EMF uses some other form of energy to produce electrical potential energy. Batteries use chemical energy, generators and Van de Graaff machines use mechanical energy, solar cells use the energy in sunlight, and thermopiles convert heat directly into electricity. • If several sources of EMF are combined in series, the total EMF will be given by the sum of the individual sources.

• Only sources of EMF that provide the same voltage should be combined in parallel. Cells in such a battery only need to provide half the current they would otherwise. • The actual voltage on the terminals of a cell is often less than the ideal EMF because the cell will have some internal resistance. The terminal voltage is given by V = E − IR i .

3.3 questions Cells, batteries and other sources of EMF 1 Why could we not use zinc for both terminals of a dry cell? 2 In order to start a car with a flat battery sometimes another car battery is connected to the flat battery. How must this connection be made and what could happen if it was done the wrong way around? 3 What is meant by a ‘short circuit’? What are the consequences of connecting batteries in such a way that they short circuit? 4 If a set of four 1.5 V dry cells was placed in series with a small 9 V battery what would be the total EMF of the combination? 5 Some torches use a battery of eight D cells (1.5 V each) arranged as two parallel sets of four cells in series. a What is the EMF of the battery? b What is the advantage of this arrangement over one with four cells in series?

more power from it by lowering the load resistance. Unfortunately she finds that this only lowers the power available. Can you explain why she obtains this result? 11 Another student using the solar cell of Figure 3.21 in winter sunlight to operate a small electric motor, finds that he is drawing a current of 0.8 A. a What power is the motor using? b He then uses a different motor (same winter sunlight) and finds that it is using 1.2 A, but only producing the same amount of power as the first one. What is the voltage across the motor? c Are either of these motors using the power available from the cell in the most efficient way? Explain. 12 A student tests a dry cell under different loads. He obtains the points shown on the graph. 300

7 When being used to start a car, a 12 V battery with an internal resistance of 0.02 Ω is supplying a current of 100 A. What will be the actual voltage across the terminals of the battery? 8 A 1.5 V torch battery is found to provide only 1.3 V when connected to a bulb that is drawing 0.2 A. What is the internal resistance of the cell? 9 What is the maximum possible current that a battery of EMF 6.0 V and internal resistance 0.5 Ω could provide, even if short circuited? 10 While using a solar cell with characteristics such as that shown in Figure 3.21, a student attempts to get

96

Electricity

I (mA)

6 Can you explain why the headlights of a car dim when the starter motor is in operation?

200 100

0.5

1.0 V (V)

1.5

a What is the EMF of this cell? b What is the internal resistance of the cell? c How much power is available from this cell at 1.5 V, 1.0 V and 0.5 V? d How much power is being wasted in the cell at each of these voltages?

3.4 Household elec tr

icity

A modern house without electricity is virtually unthinkable. The electrical wiring is always installed as the house is built, and electrical appliances are often built in as well. Building codes specify the type of wiring, the number of fuses and much more to ensure our safety. However, while electricity is a very safe form of energy when used wisely, it does have the potential to kill. As we saw in the last chapter, electricity is supplied to our homes as an alternating voltage (AC) that varies between +340 and −340 V. The figure of 240 V is the voltage of a DC supply that would provide the same average power to a resistor such as a light globe or heater element. There are many things that a 240 V DC supply could not do, however. Transformers, for example, will not work on DC, so any device that relies on one could not be used on DC. Many electric motors are designed to be used on AC and would not work at all on a DC supply; in fact they would burn out! Any circuit containing coils, such as those in transformers and motors, will behave very differently on AC and DC voltages.

House wiring There are normally two cables carrying the electricity from the power lines in the street to our houses. One of these is the so-called ‘active’ wire, the other is the ‘neutral’. The potential of the active varies between +340 V and −340 V at a frequency of 50 times per second (50 Hz). The neutral will be very close to zero potential. It is electrically connected to the ground at every house and at every street transformer. At the meter box the active wire goes to the meter and the main switch and then to the circuit breakers (or fuses). The neutral is connected to a brass strip, called the neutral bar, which you can sometimes see in the meter box. The cables running through the walls carry an active wire, a neutral wire and an earth wire. The active wire comes from the fuse, while the neutral and earth wires come from the neutral bar. Each fuse will supply a group of lights or power points (but not both). At the power point the active wire first goes to the switch and then to the upper left terminal. This is so that when the switch is off there is no active

Physics file The reason for the difference in behaviour of circuits containing coils on AC and DC is to do with the magnetic effects of electric currents. A currentcarrying wire wound around an iron rod will cause it to become a magnet. Faraday discovered that the converse is true: a moving magnet, or a changing magnetic field, near a conductor will induce a voltage (and hence current) in it. In a transformer, the changing field from the AC in the ‘primary’ coil induces changing voltages in the ‘secondary’ coil, the voltage depending on the ratio of the number of turns in each coil. The changing fields also induce a ‘back EMF’ in the primary coil which opposes the mains voltage. This reduces the current flowing in the primary coil. If DC was used this effect would not be present and too much current would flow—burning out the coil. A similar effect is true of AC motors.

meter box active fuses 240 V AC mains supply neutral

main switch

meter

neutral bar power point

copper stake

red

switch N

A green/yellow earth

black

E

Figure 3.23 The mains power from the street poles comes into the house via the meter box and then is distributed to power points and lights through cables carrying three conductors.

Electric circuits

97

Figure 3.24 This three-pin plug shows the correct colour code used: brown for active, blue for neutral and green/yellow for earth.

element switch

active neutral earth

Figure 3.25 The element of this toaster is connected between the active and neutral conductors. For safety the switch breaks both active and neutral conductors. The metal case is permanently connected to the earth wire.

neut activ ral e earth

Figure 3.26 A very dangerous situation! The toaster has been wired with the switch in the neutral instead of the active. It will work perfectly, but the element is still live when the switch is off. The person will receive a very bad shock as he is contacting the active with one hand and the earthed case with the other.

98

Electricity

voltage in the terminal sockets. The neutral and earth are connected to the upper right and lower terminals respectively. The plug on an appliance cord must be wired so that the correct conductors are connected to their appropriate terminal sockets. Figure 3.24 shows the colour code used to identify the three conductors. Note that the fixed wiring in the house uses different colours: red (active), black (neutral) and green/yellow for earth. The wiring of a typical appliance, a toaster, is shown in Figure 3.25. The element is connected between the active and neutral conductors. The earth wire is connected to the metal case. Ideally the switch should break both active and neutral conductors as near as possible to where they enter the appliance. Often, however, the switch will only break the active wire, or worse, only the neutral wire. If the switch only breaks the neutral conductor it means that the element of the appliance remains ‘live’ even if it is not switched on. This is one reason why an appliance should always be completely unplugged before any work (including cleaning) is done on it. Even if the appliance was designed to have the switch in the active, the cable could have been wired incorrectly by a careless or ill-informed home handyperson and the active and neutral conductors interchanged.

Electrical safety A number of basic safety features are built into our mains electric supply system. 1. The fuse or circuit breaker In the event of a short circuit, for example as the result of a broken active conductor coming into contact with a neutral or earth, a large current will flow through all conductors, switches and plugs. This can easily cause them to become hot and even burn. To prevent this the fuse is designed to be the thinnest piece of conductor in the system. If the current becomes too large the fuse will melt and disconnect the rest of the circuit from the active. Circuit breakers automatically switch off an excessive current by detecting its magnetic field. It is important to realise that these devices will not save a person from electrocution if they touch a live wire. Household fuses will blow only when more than 8 A (light circuits) or 16 A (power-point circuits) flows. However, a current of even 0.1 A flowing through a person can be fatal. 2. Switches Power-point switches and those on appliances should always cut the active conductor, but as we have seen this is not always the case. Although a switch in the neutral will turn the element off, it will leave the element connected to the active wire. This can be dangerous if the element is touched by someone believing the appliance to be safe because it is off. For this reason, never touch any inner part of an electrical appliance unless it is unplugged completely. Don’t try to remove burnt toast until the toaster plug is out of the power point! 3. The earth wire The separate earth wire, which parallels the neutral wire, is purely to provide protection against a fault occurring in the appliance. If, for example, the active wire breaks and contacts the case, the whole case would become active—with dire consequences for anyone who touched it. Provided the earth wire is connected to the case, however, this condition will cause a short circuit that will blow the fuse and disconnect the circuit.

As well as these basic safety features which are an inherent part of the household wiring system, there are a number of others that have been introduced over the last decades. 4. Double insulation Many appliances are now designed so that there are two effective barriers between the active wire and a person using the appliance. If, as well as the active wire being insulated and protected inside the appliance, the case is made of plastic, there is very little chance of the user touching an active part. This is actually safer than using an earthed case as it is still possible for the case to become live if damage to the cable also breaks the earth wire and it then comes into contact with the active wire. Double-insulated appliances have no earth wire and are characterised by their two-pin plugs and special symbol. 5. Earth leakage system (also known by the less enlightening term residual current device or RCD) In a properly operating system the current in the active and the neutral conductors should be exactly equal (but in opposite directions). An RCD is designed to detect any current lost from the active–neutral circuit. The most likely reason for such a loss is that it is going to earth through a fault or a person! The RCD uses the magnetic effects of an electric current to detect any difference between the active and neutral currents. In general, the magnetic effects of two equal currents flowing in opposite directions will cancel each other. If, however, the currents are not equal a magnetic effect will occur. The RCD uses this effect to switch off the supply within about 20 milliseconds. Undoubtedly, the installation of these devices in households, schools and factories has saved many lives. In modern houses, regulations require that they be used.

Figure 3.27 The double square symbol indicates that this appliance has double insulation and does not use an earth connection.

Electric shock Because our bodies are controlled by electrical impulses along the nerves, any current from an external source that flows in the body may interfere with our vital functions. In particular, any current flowing from one arm to the other may cause the chest muscles to contract and breathing to stop. Current through the heart region can cause ventricular fibrillation, which means that the muscles become uncoordinated and the heart function stops. Depending on the actual path of the current through the body, even a brief current greater than 80 mA may cause fibrillation, which is the main cause of electrical fatalities. Despite all the safety features of modern electrical systems and appliances, each year approximately 50 Australians are killed in electrical accidents. About half of these are the result of industrial accidents and the others are domestic or commercial. Many of these accidents could have been prevented either by the use of residual current devices or by other simple precautions such as keeping equipment well maintained. The unavoidable fact is that the amount of current that will flow through the body if it is in good contact with a 240 V source is well above the level that will cause death. That some people do survive an electric shock is because it was very brief, the contact was not good or the current flowed through ‘non-essential’ parts of the body. The amount of current that will flow depends on the total resistance between the active wire and the neutral or earth. A person who touches a live wire while standing on carpet


99

Table 3.1 The likely effect of a halfsecond electric shock. The actual current that flows will depend on the voltage and skin resistance Current (mA)

Effect on the body

1

Able to be felt

3

Easily felt

10

Painful

20

Muscles paralysed—cannot let go

50

Severe shock

90

Breathing upset

150

Breathing very difficult

200

Death likely

500

Serious burning, breathing stops, death inevitable

Table 3.2 The likely effect on the human body of a 50 mA shock for various times Time of 50 mA current Likely effect Less than 0.2 s

Noticeable but usually not dangerous

0.2–4 s

Significant shock, possibly dangerous

Over 4 s

Severe shock, possible death

has considerably more resistance than someone standing barefoot on wet concrete or grass. Hot sweaty hands or skin will conduct better than cold dry skin. One square centimetre of skin will have a resistance that can vary from about 100 kΩ if it is dry, to less than 1 kΩ if it is wet. As a rough guide, 1.5 kΩ can be taken as the resistance from one hand to the other of a normal perspiring worker. At 240 V this means that a current of 240 V/1.5 kΩ = 160 mA will flow across the chest and heart region of the body. As you can see from Table 3.1 this would have very serious consequences. A multimeter with a resistance scale can be used to get an idea of the resistance of skin and the body. Indeed, the so called lie detector is often no more than a resistance meter connected to terminals held by the subject. The theory is that if the person is lying he or she will sweat more and conduct a (harmless) current better. The time for which the current flows is crucial. The shock from a Van de Graaff machine charged to 100 000 V is harmless because it only lasts for about a microsecond. On the other hand the shock from a voltage source of only 100 V can be fatal if the contact is good and it lasts for a few seconds. Table 3.2 shows the effect of a 50 mA current for various times. Another potentially fatal result of the interference with nerve function of an electric current is that involuntary contraction of muscles may make it impossible to let go of the object causing the shock. This can occur at currents as small as 10 mA, which means that situations that otherwise might be harmless may become very dangerous. If there is any suspicion at all about an electrical appliance do not touch it with an open hand; use the back of the hand so that any contraction will pull the hand away. Another wise precaution is to keep one hand well away from any possible earth, so as to avoid providing a good path to earth through your arms and chest. Normally shoes will provide some resistance and so a shock from one hand to earth through the feet may not be quite as dangerous. It would be particularly dangerous, for example, to hold the earthed case of a toaster with one hand while trying to extract burnt toast with a metal knife held in the other hand! Always unplug the toaster before doing anything like that. Never try to repair any active part of an electrical appliance or install household wiring. Not only is it illegal, it could be potentially fatal. Many deaths from accidental electrocution have occurred as the result of faulty wiring. Any suspect electrical device should be unplugged and taken to a qualified electrician.

Worked example 3.4A It has been stated that the resistance of 1 cm2 of skin could vary from 1 kΩ to 100 kΩ. Assuming contact over this area, what voltages would be needed to produce a current of 50 mA?

Solution The voltage required is found from V = IR. With the low resistance contact (1000 Ω) the voltage needed is 0.05 × 1000 = 50 V. The high resistance value is 100 times as great, so 100 times the voltage would be needed, i.e. 5 kV.

100

Electricity

Physics in action

Preventing electric shocks With wise use, electrical equipment can be perfectly safe. The reasons for the following precautions will be obvious when one understands the nature of electricity and the body’s response to it. • Never use electrical appliances when barefoot, particularly when outdoors. • Be extremely careful with any electrical appliance anywhere near water. Never use a hair dryer while wet or near a bath. • If there is any reason to suspect an appliance, touch it only with the back of the hand and keep the other hand well away from it. • At the first sign of any shock, no matter how small, have the appliance checked by a qualified electrician. • Never tamper with electrical equipment. Keep it in good order and have any damaged cords or elements repaired by a qualified person.

If the worst happens In the event that you find a person who has suffered an electric shock: • Look for the reason for the shock and pull out any plugs or turn off any switches. • If that can’t be done try to push the person away from the source by using an insulating object such as a wooden pole or thick wad of dry clothing. • Do not touch the person until you are sure that the source of voltage has been removed. Check with the back of one hand before making good contact. • If power lines have fallen, keep well away from them. If the lines have fallen on a car, do not touch the car and tell the occupants to stay inside it until you can be sure it is not live. The occupants will be safe as long as they stay right inside. They will be at a high potential, but there will be no potential difference across them unless they touch something at earth potential. • Check to see if the victim is conscious by talking loudly or gentle shaking. If so, reassure them and treat any

k that the air ma position. Chec Figure 3.28 The co the head back. mouth and tilting

ar by opening the passages are cle

burns with cold water. If the person is not conscious, place them in the coma position (see Figure 3.28) and check for breathing and pulse. • If either is missing, send for urgent medical aid and begin resuscitation procedures. Ideally everyone should learn basic first-aid techniques. St John Ambulance and the Red Cross run regular courses.

3.4 summary Household electricity • Mains electricity is an alternating (AC) voltage varying between +340 V and −340 V. A 240 V DC supply would produce the same power. • The two cables coming into our houses from the street are the ‘active’ (±340 V) and the neutral (0 V) cables. • Each power-point circuit and light circuit is supplied by an active, a neutral and an earth. The neutral and the earth are connected together at the meter box.

• The element of an electrical appliance is connected between the active and neutral. The case is connected to the earth for safety. • Severe electric shock will be experienced by a person if a current of more than about 50 mA flows through them for more than a second or so. The current will depend on the voltage and their contact resistance.


101

3.4 questions Household electricity 1 Australian houses are supplied with 240 V AC. The 240 V is the: A the maximum value of the alternating voltage. B the average value of the alternating voltage. C the value of a DC voltage that would supply the same power. D none of these. 2 Why is it that there are only two cables coming into the house from the street and yet power points always have three connections? 3 The function of a fuse is to burn out, and thus turn off the current, if the circuit is overloaded. Why is it always placed in the active wire at the meter box rather than the neutral one, given that this function could be fulfilled if it was in either? 4 What is the function of the ‘earth stake’ that will normally be found near a meter box? 5 A toaster cable with conductors coloured red, black and green is to be joined to another cable with brown, blue and green/yellow conductors. Peter has joined the red and blue, black and brown, and green and

green/yellow. Will the toaster work normally when it is plugged in and turned on? Why is the way he has connected the cables dangerous? 6 An appliance was mistakenly wired between the active and earth instead of between the active and neutral. Explain why that is a very dangerous thing to do, even though the appliance will appear to work normally. 7 What is the main advantage of a ‘double-insulated’ electrical appliance over a normal earthed one? 8 Why is the shock received when a finger touches a live wire likely to be less severe than the shock received by a person who touches a live wire with a pair of uninsulated pliers? 9 How much current would flow through a person with dry hands and a total contact resistance of 100 kΩ when they touch a 240 V live wire? 10 Why is it normally more dangerous to use an electric device outdoors? What precautions are particularly necessary outdoors?

chapter review 1 Many of the electrical devices in a car appear to have only one electric cable attached to them. Why is this when we know that all electric devices require a closed circuit to operate?

The following information applies to questions 5–8. The I–V characteristics of a diode and a resistor to be used in a simple circuit with a variable voltage power supply are shown in the graphs.

2 A set of 20 Christmas tree lights which are wired in series is to be operated from the 240 V mains. What voltage rating will each bulb need?

4 Two identical heater elements are connected in series in a 240 V fan heater. In this configuration they produce a total power of 480 W. a What current is flowing through each element? b What is the resistance of each element? c What power will be produced if the elements are connected in parallel instead of series?

102

Electricity

resistor I (mA)

3 Celeste finds that at a junction of three wires, two of the wires have currents of +3 A and −6 A (+ means towards the junction). The third has an 8 A fuse in it. Is the fuse likely to burn out?

800 600 400 200

0

1

2

V (V)

3

4

switch

800 diode

M1

M2

I (mA)

600 400 200

0

1

V (V)

2

5 What is the resistance of the resistor? 6 A voltage of 1.5 V is applied across the diode. a What current will flow through it? b What is its effective resistance at this voltage? c Why is it meaningful to speak of the effective resistance of the diode at 1.5 V and yet not at 2 V or at 3 V? 7 The diode and resistor are then placed in parallel and a variable voltage applied to them. a If a voltage of 1.0 V is applied to the combination, what will be the total current flowing? b If the voltage is increased to 2.0 V will the current double? Explain. 8 The diode and resistor are then placed in series and a variable voltage applied to them. a If a voltage of 4.0 V is applied to the combination, what current will flow through them both? b If the voltage is increased to 6.0 V what current will flow? c What current will flow if a voltage of 10 V is applied? 9 Jo has two resistors labelled 27 kΩ and 36 kΩ. What will the effective resistance of the two be if she combines them: a in parallel? b in series?

a Explain why he gets this result. b He then removes the voltmeter (M1) from the circuit and replaces it with a direct connection. What is likely to happen when he presses the switch this time? c How should he connect the two meters correctly so that they each perform their appropriate function? The following information applies to questions 13–16. A student sets up the circuit shown. A1 and A2 are ammeters and V1 and V2 are voltmeters. X, Y and Z are various circuit elements. The reading on A1 is initially found to be 4.5 A and that on V1 is 6.0 V. A2 gives a reading of 1.3 A while V2 reads 4.9 V. (Assume ‘ideal’ meters.) A1 V1

X

Y

A2

V2 Z

13 What is the current through: a X? b Y? c Z?

10 What is the conductance of a 10 Ω resistor? If it is placed in parallel with another identical resistor, what will be the combined conductance?

14 What is the potential difference across element: a X? b Y? c Z?

11 A particular 9 V battery has an internal resistance of 2 Ω. What is the maximum possible short-term current that could be obtained from it?

15 Now element Y is taken out of the circuit and not replaced by any other conductor. Meter V1 is still found to read 6.0 V. What are the readings on the other three meters?

12 Josh wires up a circuit as shown in the diagram. He places a voltmeter in position M1 and an ammeter at position M2. When he presses the switch he finds that the voltmeter reads 1.5 V but the ammeter reads 0 and the lamp doesn’t light.

16 Element Y is replaced and this time element X is taken out and not replaced (V2 is left in place). What will the four meters read this time?


103

17 What is the effective resistance of the two combinations of 10 Ω resistors shown in the diagram? (a)

10 7

10 7

10 7

10 7

(b)

19 A farmer is designing a power line to carry current from a 240 V diesel generator to a farm house. He calculates that the maximum current required will be 40 A and that the minimum satisfactory voltage at the house is 230 V. What is the maximum value for the total resistance of the power line in order to achieve this?

10 7 10 7

a What is the current in the 10 Ω resistor? b What is the resistance of the second resistor?

10 7 10 7

18 Two resistors are connected in parallel across a battery. It is found that there is a total current through the battery of 9 A. One of the resistors is 10 Ω and the voltage across the other is 40 V.

20 The farmer in the previous question was able to buy a power line cable which had a total resistance of 0.20 Ω. When under the full load of 40 A, with the generator producing 240 V, what was: a the power lost as heat in the power line? b the voltage at the house?

area of study review Electricity 1 When a voltmeter is used to measure the potential difference across a circuit element it should be placed: A in series with the circuit element. B in parallel with the circuit element. C either in series or in parallel with the circuit element. D neither in series nor in parallel with the circuit element. 2 When an ammeter is used to measure the current through a circuit element it should be placed: A in series with the circuit element. B in parallel with the circuit element. C either in series or in parallel with the circuit element. D neither in series nor in parallel with the circuit element. 3 The following figure describes a simple electric circuit in which a length of resistance wire is connected to a battery (e = 1.60 × 10−19 C). R = 50 Ω

+

= 100 V

a How many electrons pass through the wire every second? b How much electrical energy does each electron lose as it moves through the wire? c What happens to the electrical energy of the electrons as they move through the wire?

104

Electricity

d The EMF of the battery is quoted as being 100 V. What does this mean? e How much power is being dissipated in the resistance wire? f What is the total energy being supplied to all the electrons passing through the wire each second? g Determine the power output of the battery. h Discuss the significance of your answers to parts e and g. 4 In discussing what is meant by the EMF of a battery, Alf claims that a battery only has an EMF if it is connected in a circuit, as otherwise the charges in it are not moving and therefore have no energy. Bert, on the other hand, claims that the EMF is greater when the battery is not connected as the charges are not losing energy as they move through it. They are both wrong. Can you explain why? 5 Assume that when the dome of a Van de Graaff generator is fully charged it is at a potential of +400 000 V and that the current from the belt onto the dome is a continuous 2 µA. a Explain why the potential of the dome remains at 400 000 V even though the belt continues to supply charge at the rate of 2 µA. b How many electrons is the belt moving each second? Are these electrons being carried onto or off the dome? c At what power must the motor be working to keep supplying the current to the fully charged dome? d If the dome is at a potential of 400 kV, what is the potential energy (in J) of each elementary charge on it?

6 The diagram shows the current–voltage graph for a section of platinum wire. A potential difference of 9.0 V is established across the section of wire.

10 Three resistors—R1 = 100 Ω, R2 = 200 Ω and R3 = 600 Ω—in a circuit are connected in parallel with a battery of E = 120 V and zero internal resistance.

R1

50 I (mA)

R2 R3

0

I2

I

I3

2.5

V (V)

a Determine the resistance of the section of wire. b How much energy does an electron lose in travelling through the wire? c Calculate the power dissipated in the wire. d How many electrons pass through the wire in 10 s? 7 When two circuit elements are placed in series: A the power produced in both must be the same. B the current in both must be the same. C the voltage across both must be the same. D the resistance of the combination will double. 8 Consider the circuit shown in which three resistors—R1 = 80 Ω, R2 = 10 Ω and R3 = 8.0 Ω—are connected in series with a battery of EMF E = 100 V and internal resistance Ri = 2.0 Ω. R1

I1

R2

R3

Ri = 2 Ω = 100 V

a What is the total or effective resistance, RT, in the circuit (including the internal resistance of the battery)? b Determine the current, I, in the circuit. c Calculate the potential difference across: i R1 ii R2 iii R3 d What is the terminal voltage, VT, of the battery? e Calculate the total power produced by the battery. 9 When two circuit elements are placed in parallel: A the power produced in both must be the same. B the current in both must be the same. C the voltage across both must be the same. D the resistance of the combination will halve.

I

E = 120 V

I

a Calculate the total resistance in the circuit. b Determine the total current in the circuit. c Calculate the branch currents: I1, I2, and I3. d Determine the power output of the battery. e Calculate the total power consumed by all the resistors in the circuit. 11 A student needs to construct a circuit in which there is a voltage drop of 5.0 V across a resistance combination, and a total current of 2.0 mA flowing through the combination. She has a 4.0 kΩ resistor which she wants to use and proposes to add another resistor in parallel with it. a What value should she use for the second resistor? b Determine the effective resistance of the combination. 12 Belinda needs to operate a portable tape recorder requiring 12 V at 100 mA. She has some 1.5 V dry cells which each have an internal resistance of 0.3 Ω, and a small 9 V battery with an internal resistance of 5 Ω. Belinda decides to connect two of the dry cells in series with the 9 V battery to provide an EMF of 12 V for her tape recorder. Assume the batteries are all fresh. a She measures the voltage of her new battery with a voltmeter to check that it will be satisfactory. What does she find? b She then connects the battery to the tape recorder and again measures the voltage at the terminals and finds that while it operates the recorder satisfactorily it is less than she had hoped. What voltage does she find? c What problems do you see with Belinda’s arrangement? 13 a In some ways the energy released in a light bulb in an electric circuit can be compared to the energy released as water falls down a waterfall. What power would be released by water falling 0.6 m down a garden waterfall at the rate of 20 litres (20 kg) per minute? b What current would need to flow through a 3 V torch bulb to produce the same amount of power as the waterfall in part a?


105

The following information applies to questions 14–17. A heater element is found to have the I–V characteristic shown in the graph.

4

2

20 A special type of circuit device X has the volt–amp characteristics depicted in the following graph.

1 0

100

V (V)

2.5

200

2.0

14 What current flows if a p.d. of 100 V is applied to the element? 15 What is the resistance of the element at 50 V, 100 V, 150 V and 200 V?

1.5

I (mA)

I (A)

3

a What current will flow in the globe when 2 V is applied to it? Would you expect twice the current to flow if the voltage is increased to 4 V? b What is the resistance of the globe at 1 V and at 3 V? c Explain why this graph tells us that the globe is not an ohmic device. d How much power is the globe using when it is operating at 3 V?

1.0 0.5

16 What power will be produced at 150 V? 17 Why do you think that the resistance at 200 V would be greater than at 100 V? 18 The circuit shown contains a battery with E = 400 V and internal resistance Ri = 2.0 Ω. The switch S is closed initially. R1 = 100 Ω

R2 = 80 Ω

R3 = 18 Ω

V2

V3

V1

0.0

20

40

60 80 V (V)

90

100

The circuit device X is connected into the circuit as shown in the next diagram, in which the reading on the milliammeter is 2.0 mA.

I

X

R 50 k7

V1

V2

V4 A Ri = 2.0 Ω

S mA

a Calculate the total resistance (including Ri) in the circuit. b Determine the reading on the ammeter. c Determine the reading on the voltmeters V1, V2, V3, and V4. d Determine the power produced by the battery. e Calculate the power consumed by the three resistors in the circuit. f What is the terminal voltage of the battery? g What is the terminal voltage of the battery if S is open? 19 The I–V characteristics of a small flashlight globe are shown in the graph.

I (A)

0.3 0.2 0.1 0

106

Electricity

1

V (V)

2

3

250 V

a How would you describe the device X and what is its purpose? b Describe the resistance of such a device in the voltage range 60–100 V. c Determine the readings on voltmeters V1 and V2. d How much power is being consumed by the circuit device X? e How much power is being consumed by R? f What is the power output of the 250 V battery? The following information applies to questions 21 and 22. A student performs an experiment in which an electric motor is used to lift a 200 g weight through 2 m, thus increasing its potential energy by 4 J. From measurements of the rate at which the weight is lifted the efficiency of the motor is to be determined. Two different voltages were used and the current was measured. 21 In the first experiment at 6 V, a current of 0.25 A was measured and the weight took 5 s to rise the 2 m. What was the efficiency of the motor?

22 In the next experiment the voltage was increased to 8 V. The current was found to be 0.30 A and the efficiency worked out to be 60%. How long did the motor take to lift the weight the 2 m this time?

a What voltage is required to operate the light bulb correctly? b How many volts will there be across R when the circuit is operating correctly? c What is the required value of R?

The following information applies to questions 23–25. The 240 V supply cables to a certain house have a total resistance of 0.5 Ω. The maximum likely power use at any one time is estimated to be 10 kW, while the normal minimum load is estimated at 120 W.

The following information applies to questions 31–36. Three identical light-emitting diodes (LEDs) L1, L2, and L3, are connected into the circuit as shown in the diagram. The battery has an EMF of 3.0 V and zero internal resistance. All LEDs are operating normally. The brightness of an LED increases when the current through it increases. The voltage–current characteristics for the LEDs are shown in the following graph. Assume that all connecting wires have negligible resistance.

23 What is the maximum current likely to be used by the household? 24 What will the voltage drop along the supply cables be under minimum load?

E = 3.0 V

25 What will the voltage at the house switchboard be under maximum load? The following information applies to questions 26–28. Bill and Mary are discussing the lighting for their living room. At present they have four 60 W, 240 V light bulbs in parallel. Bill suggests that it might be cheaper to replace these with four bulbs wired in series.

V L1

L2

L3

26 If this was to be done, what would be the voltage and power rating of each of the new bulbs they would need in order to produce the same amount of power?

28 Bill says that they would save on electricity bills because the current is going through all four bulbs and therefore being used more effectively. Mary says this is not right and that the power bill would be exactly the same. Who is correct and why? 29 What is the effective resistance of the four resistors shown in this diagram? 37

47

57

67

30 A light bulb rated as 6 W and 0.5 A is to be operated from a 20 V supply using a dropping resistor, R. 20 V

120 100 80 60 40 20 0

I (mA)

27 What would be the total current flowing in the circuit and how would this compare to the total current flowing when the original parallel bulbs were used?

S

0.5

1 V (V)

1.5

2

31 When switch S is closed, the brightness of L1 will: A be greater than previously. B be less than previously but greater than zero. C remain the same. D be equal to zero. 32 When switch S is closed, the brightness of L3 will: A be greater than previously. B be less than previously but greater than zero. C remain the same. D be equal to zero. 33 When switch S is closed, the reading on voltmeter V will: A be greater than previously. B be less than previously but greater than zero. C remain the same. D be equal to zero.

R

6W 0.5 A


107

34 When switch S is closed, the power output of the battery will: A be greater than previously. B be less than previously but greater than zero. C remain the same. D be equal to zero.

37 Which of the following statements is true? A I1 = IA B I1 > IA C I1R1 = IA(R1 + R2) D I1R1 = IAR2

35 What is the effective resistance of an LED at a voltage of: a 1.0 V? b 1.5 V?

38 If another resistor R3 was connected between X and Y, what would happen to the meter readings? A The reading on V would increase; the reading on A would decrease. B The reading on V would decrease; the reading on A would increase. C They would both decrease. D They would both increase.

36 Determine the current in the circuit when: a the switch is open b the switch is closed. The following information applies to questions 37–39. In the circuit shown, the current through R1 = I1 and the current through ammeter A = IA. The voltmeter V has extremely high resistance and the ammeter A has negligible resistance. V R1 X

R2 Y A

108

Electricity

39 If another resistor R3 was connected between X and Y, what would happen to the power output of the battery? A It would increase. B It would decrease. C It would remain the same. 40 A car battery has an EMF of 12.0 V and an internal resistance of 0.050 Ω. The resistance of the leads between the battery and the starter motor is 0.009 Ω. When the starter motor is running it draws a current of 80 A. a Determine the terminal voltage of the battery. b Calculate the voltage across the terminals of the starter motor.

Unit

2

area o f stud y1

Unit

Motion

outcome

area On completion of this able of study, you should be e and to investigate, analys motion mathematically model in terms of particles and bodies n and of Aristotelian, Galilea Newtonian theories.

chapter 4

Asp

n o i t o m f o s t ec

P

ilots of fighter planes sometimes have to bail out of their aircraft at high altitude. Should this happen today, they would usually reach the ground safely. This has not always been the case, however. About 40 years ago, the United States Air Force conducted a series of experiments to investigate the design of parachutes that would return high-altitude pilots safely to the ground. Joe Kittinger, a US Air Force Captain, was part of this experiment. His contribution involved jumping out of a balloon from a height of 31 km! In August 1960, after 18 months of preparations, Joe Kittinger took off in a helium balloon from New Mexico. He ascended for one and a half hours, eventually reaching a height of 31 km. At this altitude, there was no air to breathe, the temperature was 35° below zero, the sky above was pitch black and he was virtually in space. He then waited at this altitude for the instruction to jump. It came after 12 minutes. When he stepped out of the balloon’s gondola, it was like falling through space. There was no sound or sensation of wind or air resistance because the atmosphere was so thin. As he rolled over to look upwards, he was amazed at how quickly he was accelerating away from the balloon. He free-fell for about four and a half minutes, reaching a maximum speed of 1150 km h−1. In doing so, he became the first person to break the sound barrier without an aeroplane! Soon after this, he started to notice the effects of the atmosphere which began to slow him down. He then opened the main parachute and reached the ground 13 minutes and 45 seconds after stepping out of the gondola. A number of space entrepreneurs and extreme-sport groups are currently planning to break Kittinger’s record. They hope to bail out from the nose cone of a purpose-built rocket at the apex of its flight and freefall to Earth. The first of these ‘space dives’ was planned for 2009 from a height of 37 km. The motion of Joe Kittinger as he fell—his speed, acceleration due to gravity, and the effects of air resistance—are some of the ideas that will be covered in this and subsequent chapters.

by the end of this chapter you will have covered material from the study of movement including: • a graphical description of motion • instantaneous and average velocities • motion with constant acceleration described using graphs and equations of motion • vertical motion under gravity.

4.1 Describing mo tio

n in a straigh

t line

Motion, from the simple to the complex, is a fundamental part of everyday life. The motion of a gymnast performing a routine and that of a mosquito trying to avoid your desperate attempts to swat it would be considered complex forms of motion. Far simpler examples are a tram travelling in a straight line along a road, and a swimmer doing a length of a pool. In this chapter, the simplest form of motion—straight line motion—will be analysed. In this section, terms that are useful in describing the motion of an object—position, distance, displacement, speed, velocity and acceleration— will be discussed.

Physics file The centre of mass of Formula 1 cars is very close to the ground. This makes them very stable and means that they can turn corners at high speeds; speeds at which normal cars would roll over.

Centre of mass When analysing motion, things are often more complicated than they first seem. For example, as a freestyle swimmer travels at a constant speed of 2 m s−1, the trunk of his body will move forwards with this speed. The motion of his arms is much more complex: at times they move forwards faster than 2 m s−1 and at other times they are actually moving backwards through the water. It is beyond the scope of this course to analyse such a complex motion, but we can simplify this by treating the swimmer as a simple object located at a single point—his centre of mass. The centre of mass is the balance point of an object. For a person, the centre of mass is located near the waist. The centres of mass of some everyday objects are shown in Figure 4.1. (a)

(b)

(c)

(d)

Figure 4.2 Formula 1 racing cars have a low centre of mas s.

PRACTICAL ACTIVITY 15 Locating the centre of mass Physics file

Table 4.1 World record distances 2008 Activity Men’s pole vault

Figure 4.1 The centre of mass of each object is indicated by a cross. When analysing their

motion, the total mass of each object can be considered to be located at these points.

Position and distance travelled Consider a swimmer, Sophie, doing laps in a 50 m pool. To simplify this situation, we will treat Sophie as a simple point object. The pool can be treated as a one-dimensional number line with the starting blocks chosen to be the origin. The right of the starting blocks is taken to be positive. The position of Sophie is her location with respect to the origin. For example, her position as she is warming up behind the starting block in Figure 4.4a is −10 m. The negative sign indicates the direction from the origin, i.e. to the left. At the starting block Sophie’s position is 0 m, then after half a length she is +25 m or 25 m to the right of the origin.

Record (m) 6.14

Women’s javelin

72.3

Paper plane flight (indoors)

58.8

Golf drive

419

Ski jump

239

Bungee jump

1012

Paper clip chain

1628

Figure 4.3 Austra lia’s Kym Howe wo n the Commonwealt h pole vault gold medal at Melbour ne in 2006 with a leap of 4.62 metres.

Chapter 4 Aspects of motion

111

(a)

PRACTICAL ACTIVITY 16 The ticker timer

PRACTICAL ACTIVITY 17 Introduction to the air track

-10

0

10

20

30

40

50

60 m position

0

10

20

30

40

50

60 m position

0

10

20

30

40

50

60 m position

(b)

-10 (c)

-10

Figure 4.4 In this situation, the position of the swimmer is given with reference to the starting block. (a) While warming up, Sophie is at −10 m. (b) When she is on the starting block, her position is zero. (c) After swimming for a short time, she is at a position of +25 m.

Distance travelled Distance travelled is a measure of the actual distance covered during the motion. For example, if Sophie completes three lengths of the pool, the distance travelled during her swim will be 50 + 50 + 50 = 150 m.

DISTANC… TRAV…LL…D, d, is how far a body travels during motion. Distance travelled is measured in metres (m). The distance travelled does not distinguish between motion in a positive or negative direction. For example, if Sophie completes one length of the pool travelling from the starting block, i.e. in a positive direction, the distance travelled will be 50 m. If she swam one length from the far end back to the start, the distance travelled will also be 50 m.

Displacement Displacement is a term related to position and distance travelled, but it has a different meaning. Displacement, x, is defined as the change in position of an object. Displacement takes into account only where the motion starts and finishes; whether the motion was directly between these points or took a complex route has no effect on its value. The sign of the displacement indicates the direction in which the position has changed.

Figure 4.5 In completing the 2008 Melbourne

marathon, the athletes started just west of the MCG and ran down Beach Rd to Sandringham. They turned around and returned along Beach Rd to the MCG. Their distance travelled was more than 42 km, but their displacement was just a few hundred metres east.

112

Motion

DISPLAC…M…NT is defined as the change in position of a body in a given direction. Displacement x = final position − initial position Displacement is measured in metres (m). Consider the example of Sophie completing one length of the pool. During her swim, the distance travelled is 50 m, and the displacement is: x = final position − initial position = 50 − 0 = 50 m, i.e. 50 m in a positive direction

If Sophie swims two lengths, she will have travelled a distance of 100 m, i.e. 50 m out and 50 m back. However, her displacement during this swim will be: x = final position − initial position =0−0 =0 Even though she has swum 100 m, her displacement is zero because the initial and final positions are the same. Displacement only considers the starting and finishing positions of the motion; it does not indicate anything about the route taken by the person or object in getting from the initial to the final position.

Scalars and vectors Physical quantities requiring a number only to fully describe them are known as scalars. Distance is a scalar quantity. Other scalar quantities include mass, time, speed and refractive index. Some physical quantities require a number (magnitude) and a direction to fully describe them. These are called vectors. Displacement is a vector quantity. Other vector quantities are velocity, acceleration and force. Vectors are represented in bold italic type; for example, x, v, a. Scalars and vectors are discussed in detail in Appendix A.

Speed and velocity For thousands of years, humans have tried to travel at greater speeds. This desire has contributed to the development of all sorts of competitive activities, as well as to major advances in engineering and design. The records for some of these pursuits are given in Table 4.2. Speed and velocity are both quantities that give an indication of how fast an object moves or, more precisely, of how quickly the position of an object is changing. Both terms are in common use and are often assumed to have the same meaning. In physics, however, these terms are defined differently. • Speed is defined in terms of the distance travelled and so, like distance, speed is a scalar. Thus, a direction is not required when describing the speed of an object. • Velocity is defined in terms of displacement and so is a vector quantity. The SI unit for speed and velocity is metres per second (m s−1); kilometres per hour (km h−1) is also commonly used.

Table 4.2 Some world speed records (2008) Speed activity

Record speed (m s–1) (km h–1)

Luge

39

140

Train

160

575

Tennis serve

68.4

246

Waterskiing

63.9

230

Cricket delivery

44.7

161

Horse racing

19

70

Instantaneous speed and velocity Instantaneous speed and instantaneous velocity give a measure of how fast something is moving at a particular moment or instant in time. If the speedometer on a car shows 60 km h−1, it is indicating the instantaneous speed of the car. If another car is detected on a police radar gun and registers 120 km h−1, it indicates that this car’s instantaneous speed is above the speed limit.

Figure 4.6 In 2003, Patrick Johnson became the first Australian to break the 10-second barrier for the 100-m sprint.


113

Displacement

C B

A

Time

PRACTICAL ACTIVITY 18 The kinematics of a student

Figure 4.7 The instantaneous velocity at point A is the gradient of the tangent at that point. The average velocity between points B and C is the gradient of the chord between these points on the graph.

Average speed and velocity Average speed and average velocity both give an indication of how fast an object is moving over a time interval. For example, the average speed of a car that takes 1 hour to travel 30 km from Dandenong to St Kilda is 30 km h−1. However, this does not mean that the car travelled the whole distance at this speed. In fact, it is more likely that the car was moving at 60 km h−1 for a significant amount of time, but some time was also spent not moving at all.

distance travelled d = time taken Dt Speed is measured in metres per second (m s−1). AV…RAG… SP……D vav =

displacement x = time taken Dt Velocity is measured in metres per second (m s−1) and requires a direction. AV…RAG… V…LOCITY vav =

A direction (such as north, south, up, down, left, right, positive, negative) must be given when describing a velocity. The direction will always be the same as that of the displacement.

Worked example 4.1A S

0

N

start

20

finish

100 m 0 45

start

Consider Jana, an athlete performing a training routine by running back and forth along a straight stretch of running track. She jogs 100 m north in a time of 20 s, then turns and walks 50 m south in a further 25 s before stopping. a Calculate Jana’s average speed as she is jogging. b What is her average velocity as she is jogging? c What is the average speed for this 150 m exercise? d Determine the average velocity for this activity. e What is the magnitude of Jana’s average velocity in km h−1?

finish

Solution a Her average speed when jogging is: 50 m

114

Motion

vav =

distance travelled d 100 m = = = 5.0 m s−1 time taken Dt 20 s

b Her average velocity when jogging is: displacement x 100 m north = = = 5.0 m s−1 north time taken Dt 20 s Note that speed has been treated as a scalar and velocity as a vector. c Jana has covered a distance of 150 m in 45 s. Her average speed is: distance travelled 150 m vav = = = 3.3 m s−1 time taken 45 s d She has finished 50 m to the north of where she started, i.e. her displacement is 50 m north. Her average velocity is: x 50 m north vav = = = 1.1 m s−1 north Dt 45 s Jana could have ended up at the same place in the same time by travelling with this average velocity. e Her average velocity is: 11 × 3600 1.1 m s−1 north = = 4.0 km h−1 north 1000 The magnitude is 4.0 km h−1.

vav =

Acceleration If you have been on a train as it has pulled out of the station, you will have experienced an acceleration. Also, if you have been in a jumbo jet as it has taken off along a runway, you will have experienced a much greater acceleration. Acceleration is a measure of how quickly velocity changes. Consider the following velocity information for a car that starts from rest at an intersection as shown in Figure 4.8.

t=0s

1s

2s

3s

Figure 4.8 The velocity of the car increases by 10 km h−1 each second, and so its acceleration is

said to be +10 kilometres per hour per second.

Each second, the velocity of the car increases by 10 km h−1. In other words, its velocity changes by +10 km h−1 per second. This is stated as an acceleration of +10 kilometres per hour per second or +10 km h−1 s−1. More commonly in physics, velocity information is given in metres per second. The athlete in Figure 4.9 takes 3 s to come to a stop at the end of a race. v = 6 m s–1

4 m s–1

2 m s–1

0 m s–1

Physics file When converting a speed from one unit to another, it is important to think about the speeds to ensure that your answers make sense. 100 km h−1 is a speed that you should be familiar with as it is the speed limit for most freeways and country roads. Cars that maintain this speed would travel 100 km in 1 hour. Since there are 1000 m in 1 km and 60 × 60 = 3600 s in 1 hour, this is the same as travelling 100 000 m in 3600 s. 100 km h−1 = = = =

100 × 1000 m h−1 100 000 m h−1 100 000 ÷ 3600 m s−1 27.8 m s−1

So km h−1 can be converted to m s−1 by multiplying by 1000/3600 (i.e. ÷ 3.6). A champion Olympic sprinter can run at an average speed of close to 10 m s−1, i.e. each second the athlete will travel approximately 10 metres. At this rate, in 1 hour the athlete would travel 10 × 3600 = 36 000 m, i.e. 36 km. 10 m s−1 = 10 × 3600 m h−1 = 36 000 m h−1 = 36 000 ÷ 1000 km h−1 = 36 km h−1 So m s−1 can be converted to km h−1 by multiplying by 3600/1000 (i.e. × 3.6). v 3.6 km h–1

t=0s

1s

2s

m s–1

3s

Figure 4.9 The velocity of the athlete changes by −2 m s−1 each second. The acceleration is

r 3.6

−2 m s−2.


115

Each second the velocity of the athlete changes by −2 m s−1, and so the acceleration is −2 metres per second per second. This is usually expressed as −2 metres per second squared or −2 m s−2. Acceleration is defined as the rate of change of velocity. Acceleration is a vector quantity whose direction is that of the velocity change. A negative acceleration can mean that the object is slowing down in the direction of travel as is the case with the athlete above. What would happen to the athlete in the next few seconds if the trend continued? The athlete’s velocity would be –2 m s−1, –4 m s−1 and so on. This too is a negative acceleration, which can also mean speeding up in the opposite direction.

AV…RAG… ACC…L…RATION is the rate of change of velocity: change in velocity Dv v - u aav = = = time taken Dt Dt where v is the final velocity (m s−1) u is the initial velocity (m s−1) Δt is the time interval (s) Acceleration is measured in metres per second squared (m s−2).

Worked example 4.1B A cheetah running at 20 m s−1 slows down as it approaches a stream. Within 3.0 s, its speed has reduced to 2 m s−1. Calculate the average acceleration of the cheetah.

Solution The average acceleration of the cheetah is: Dv v − u aav = = Dt Dt 18 2 − 20 = =− = − 6.0 m s−2 3.0 3.0 That is, each second, the cheetah is slowing down by 6.0 m s−1.

Finding velocity changes When finding the change in any physical quantity, the initial value is taken away from the final value. Thus, a change in velocity is the final velocity minus the initial velocity: ∆ = − In algebra, a subtraction is equivalent to the addition of a negative term, e.g. x − y = x + (−y). The same rationale can be used when subtracting vectors. Vector subtraction is performed by adding the opposite of the subtracted vector: ∆ = − = + (− ) The negative of a vector simply points in the opposite direction, i.e. if u is 5 m s−1 north, then −u is 5 m s−1 south.

Worked example 4.1C A golf ball is dropped onto a concrete floor and strikes the floor at 5.0 m s−1. It then rebounds at 5.0 m s−1. a What is the change in speed for the ball? b Calculate the change in velocity for the ball.

116

Motion

Solution a Both the initial and final speed of the ball are 5.0 m s−1, so the change in speed for the ball is: ∆v = v − u = 5.0 − 5.0 = 0 As speed is a scalar quantity, the direction of motion of the ball is not a consideration. b To determine the change in velocity of the ball: ∆v = v − u = 5.0 m s−1 up − 5.0 m s−1 down Let up be the positive direction, so: ∆v = +5.0 − (−5.0) = +5.0 + 5.0 = 10 m s−1 up As can also be seen in the diagram, a vector subtraction gives the change in velocity of the ball, in this case, 10 m s−1 up. Velocity is a vector quantity and the change in direction of the ball is responsible for its velocity change.

u = 5 m s–1

v = 5 m s–1

Dv = v – u =5 –5 =5 +5 = 10 m s–1 = 10 m s–1 up

Physics in action

Breaking the speed limit! Over the past 100 years, advances in engineering and technology have led to the development of faster and faster machines. Today cars, planes and trains can move people at speeds that were thought to be unattainable and life-threatening a century ago. The 1-mile land speed record is 1220 km h−1 (339 m s−1). This was set in 1997 in Nevada by Andy Green driving his jet-powered Thrust SSC. The fastest combat jet is the Sr-71 Blackbird. It reached a speed of 3900 km h−1 in 1976, which is more than three times the speed of sound. The fastest speed recorded by a train is 575 km h−1 (160 m s−1) by the French TGV Atlantique in 2007, although it does not reach this speed during normal operations. The world record speed for racing dragsters is almost as fast as this, although dragsters only race 400 m. A piston engine (as opposed to rocket-powered!) dragster can cover the 400 m in 4.4 s and reach a maximum speed of 475 km h−1. It can achieve a peak acceleration of 56 m s−2 during its trip and a parachute has to be used to slow it down. In the 1950s, the United States Air Force used a rocket sled to determine the effect of extremely large accelerations on humans. It consisted of an 800 m Figure 4.10 In 2007, M long railway track and a sled with nine rocket motors. arkus Stoe record for ckl of Austr mountain One volunteer—Lieutenant Colonel John Stapp—was ia set a ne biking. He racing dow w reached a n a ski slop strapped into the sled and accelerated to speeds of over speed of 21 speed e in C h ile. 0 km h –1 1000 km h−1 in a very short time. Then water scoops were used to stop the sled in just 0.35 s. This equates to a deceleration of 810 m s−2.


117

Table 4.3 World record times and speeds for men and women in 2008 Event

Distance (m)

Time (h:min:s)

Average speed

Event

Distance (m)

Time (h:min:s)

100

0:9.69

10.3 m s

200

0:19.30

400

Average speed

100

0:10.49

9.5 m s−1

10.4 m s−1

200

0:21.34

9.4 m s−1

0:43.18

9.1 m s−1

400

0:47.60

8.4 m s−1

800

1:41.11

7.9 m s−1

800

1:53.28

7.1 m s−1

1500

3:26.00

7.2 m s−1

1500

3:50.46

6.5 m s−1

Marathon (42.2 km)

2:04:26

5.6 m s−1

Marathon

2:15:25

5.2 m s−1

50 freestyle

0:23.97

2.1 m s−1

50 freestyle

0:21.28

2.3 m s−1

1500 freestyle 15:42.54

1.6 m s−1

1500 freestyle 14:34:56

1.7 m s−1

Women

Men Running

Swimming Cycling

56.4 km

−1

Running

Swimming

1:00.00

Downhill skiing

56.4 km h

−1

251 km h * −1

Cycling

46.1 km

1:00.00

46.1 km h−1

Downhill skiing

243 km h−1*

*Instantaneous speed.

Figure 4.11 These photos show the face of Lieutenant Colonel John Stapp while he was travelling in a rocket-powered sled. As the sled blasted off, it achieved an acceleration of 120 m s−2. The effect of this is evident on his face.

Physics in action

Measuring speed in the laboratory A variety of methods can be used to determine the speed of an object in a motion experiment. Common techniques include ticker timers, ultrasound transducers, photogates and multiflash photography. A ticker timer has a hammer that vibrates with a frequency of 50 Hz and produces a series of dots on a piece of ticker tape that is being dragged along by a moving body. Since the hammer strikes the paper at regular intervals, the distance between the dots gives an indication of the speed of the body. Where the dots are widely spaced, the body is moving faster than when the dots are close together. Precise values of speed can be determined by measuring the

118

Motion

A B

C

1 cm 2 cm

D 3 cm

E 3 cm

F 3 cm

Figure 4.12 Ticker tape was commonly used to analyse the motion of objects. If the frequency of the timer is known and the distance between the dots has been measured, the average speed of the object can be determined. distances between the dots. Consider the section of tape shown in Figure 4.12. The tape had been attached to a student to measure walking speed.

Figure 4.13 A multiflas h photogra

ph of this golf swing all motion of the club an ows the d ball to be analysed in detail. Three images the ball in flight can be of seen. Given that the fla sh frequency is 120 Hz and the scale of the ph , otograph is 1:50, you should be able to show that the initial speed of the golf ball is appro ximately 100 m s−1.

The average speed of the student is calculated by measuring the distance travelled and taking account of the time elapsed. Since the hammer strikes the tape 50 times per second, each interval between the dots represents 1/50 s (i.e. 0.02 s). Thus the average speed between A and F, a distance consisting of five intervals, is: 12 cm distance travelled vav = = = 120 cm s−1 or 1.2 m s−1 5 × 0.02 s time taken The instantaneous speed gives a measure of the speed at one particular time. This can be estimated with reasonable accuracy by calculating the average speed for the interval one dot either side of the point being analysed. For example, the instantaneous speed at point B can be estimated by calculating the average speed between points A and C: distance travelled 3 cm = time taken 2 × 0.02 s = 75 cm s−1 or 0.75 m s−1

vinst(B) ≈ vav(A to C) =

Multiflash photography is a useful method for analysing more complex motion. A photograph is taken by a camera with the shutter open and a strobe light that flashes at a known frequency. This is analysed in a similar manner to ticker tape. If the frequency of the flash is known, the time

sensors can

sonic motion Figure 4.14 Ultra ovement of an

the m be used to analyse s are y sound impulse nc ue eq -fr gh the object. Hi m fro sensor, reflected lay emitted from the sp di A . or ns d by the se object and receive be en th n ca t jec e ob of the motion of th ter screen. pu m co a on en se

between each flash (i.e. the period of the flash, T) is easily found using . For example, a flash with a frequency of 20 Hz has a period of 0.05 s. By measuring the appropriate distance on the photograph, average speed can be calculated and instantaneous speed estimated. A photogate consists of a light source and sensor that triggers an electronic timing device when the light beam is broken. Photogates are designed to measure time to millisecond accuracy, and so give very accurate speed data. Some are calibrated to give a direct reading of speed. Others will simply give a measure of the time interval between two light beams being broken. The average speed of a falling mass that passes between two photogates can be calculated by considering the distance between the photogates and the time that the mass took to pass between them. An ultrasonic motion sensor gives a direct and instantaneous measure of the speed of a body. These devices emit a series of high-frequency sound pulses that are reflected from the moving object, giving an indication of its position. The data are then processed to give a measure of the speed. Ultrasonic sensors allow complex motions such as a sprinter starting a race, or a ball bouncing several times, to be analysed in great detail.


119

4.1 summary Describing motion in a straight line • The average speed of a body, vav , is defined as the rate of change of distance and is a scalar quantity:

• The average acceleration of a body, aav , is defined as the rate of change of velocity. Acceleration is a vector:

distance travelled d = time taken Dt • The average velocity of a body, vav , is a vector and is the rate of change of displacement: displacement = = av Dt time taken • The SI unit for both speed and velocity is metres per second (m s−1). • Instantaneous velocity is the velocity at a particular instant in time. vav =

D Dt Position defines the location of an object with respect to a defined origin. Distance travelled, d, tells how far an object has actually travelled. Distance travelled is a scalar. Displacement, , is a vector and is defined as the change in position of an object in a given direction. Displacement = final position − initial position. Vector quantities require a magnitude and a direction, whereas scalar quantities can be fully described by a magnitude only. aav =

• • •

•

4.1 questions Describing motion in a straight line 1 A somewhat confused ant is moving back and forth along a metre ruler. A 0

10

20

30

B

C

40

50

D 60

70

top floor

E 80

90

100 cm

Determine both the displacement and distance travelled by the ant as it moves from: a A to B b C to B c C to D d C to E and then to D

50 m

2 During a training ride, a cyclist rides 50 km north then 30 km south. a What is the distance travelled by the cyclist during the ride? b What is the displacement of the cyclist for this ride? 3 A lift in a city building carries a passenger from the ground floor down to the basement, then up to the top floor. a Determine the displacement as the passenger travels from the ground floor to the basement. b What is the displacement of the lift as it travels from the basement to the top floor? c What is the distance travelled by the lift during this trip? d What is the displacement of the lift during this trip?

120

Motion

ground floor 10 m basement

4 Which of these physical quantities are vectors: mass, displacement, density, distance, temperature? 5 If 1 is 20 m south and 2 is 10 m north, which of the vectors A–E represents: a 1 + 2? b 2 + 1? c 3 2? d − 1?

A

B 20 m

C 20 m

D

10 m

E

30 m N

10 m

6 Liam, aged 7, buried some ‘treasure’ in his backyard and wrote down these clues to help find it: start at the clothes line, walk 10 steps south, then four steps east, 15 steps north, five steps west, and five steps south. a What distance (in steps) is travelled when tracing the ‘treasure’? b Where is the ‘treasure’ buried? c What is your displacement (in steps) after you have followed the instructions? 7 Estimate the speed: a at which you walk b of a snail crawling c of an elite 100 m sprinter d of a ten-pin bowling ball. 8 Toni rides her bicycle to school and travels the 2.5 km distance in a time of 10 min. a Calculate her average speed in kilometres per hour (km h−1). b Calculate her average speed in metres per second (m s−1). c Is Toni’s average speed a realistic representation of her actual speed? Explain. 9 A sports car, accelerating from rest, was timed over 400 m and was found to reach a speed of 120 km h−1 in 18.0 s. a What was the average speed of the car in m s−1? b Calculate the average acceleration of the car in km h−1 s−1.

c What was its average acceleration in m s−2? d If the driver of the car had a reaction time of 0.60 s, how far would the car travel while the driver was reacting to apply the brakes at this speed of 120 km h−1? 10 A squash ball travelling east at 25 m s−1 strikes the front wall of the court and rebounds at 15 m s−1 west. The contact time between the wall and the ball is 0.050 s. Use vector diagrams, where appropriate, to calculate: a the change in speed of the ball b the change in velocity of the ball c the magnitude of the average acceleration of the ball during its contact with the wall. 11 A bus travelling north along a straight road at 60 km h−1 slows down uniformly and takes 5.0 s to stop. a Calculate the magnitude of its acceleration in km h−1 s−1. b Calculate its acceleration in m s−2. 12 During a world record 1500 m freestyle swim, Grant Hackett completed 30 lengths of a 50 m pool in a time of 14 min 38 s. a What was his distance travelled during this race? b What was his average speed (in m s−1)? c What was his displacement during the race? d What was his average velocity during his record-breaking swim?


121

ition, s o p : n o i t o m Graphing 4.2 velocity and acceleration Position 0

10

20

30

40

50 m

Figure 4.15 This swimmer will travel to the 50 m mark, then return to the 25 m mark. Her position is shown in Table 4.4.

At times, even the motion of an object travelling in a straight line can be complicated. The object may travel forwards or backwards, speed up or slow down, or even stop. Where the motion remains in one dimension, the information can be presented in graphical form. The main advantage of a graph compared with a table is that it allows the nature of the motion to be seen clearly. Information that is contained in a table is not as readily accessible nor as easy to interpret as information presented graphically.

Graphing position A position–time graph indicates the position of an object at any time for motion that occurs over an extended time interval. However, the graph can also provide additional information. Consider once again Sophie swimming laps of a 50 m pool. Her position– time data are shown in Table 4.4. The starting point is treated as the origin for this motion.

50

Position (m)

45 40 35 30 25 20 15 10

Table 4.4 Positions and times of a swimmer completing 1.5 lengths of a pool Time (s)

5 0

5 10 15 20 25 30 35 40 45 50 55 60 Time (s)

Figure 4.16 This graph represents the motion of a swimmer travelling 50 m along a pool, then resting and swimming back towards the starting position. The swimmer finishes halfway along the pool.

Position (m)

gradient velocity

$x

$t Time (s)

Figure 4.17 From the units of the rise and run, it can

be seen that the units for the gradient are m s−1, confirming that this is a measure of velocity.

122

Motion

0

5

10

15

20

25

30

35

40

45

50

55

60

Position (m) 0

10

20

30

40

50

50

50

45

40

35

30

25

An analysis of the table reveals several features of the swim. For the first 25 s, Sophie swims at a constant rate. Every 5 s she travels 10 m in a positive direction, i.e. her velocity is +2 m s−1. Then from 25 s to 35 s, her position does not change; she seems to be resting, i.e. is stationary, for this 10 s interval. Finally, from 35 s to 60 s she swims back towards the starting point, i.e. in a negative direction. On this return lap, she maintains a more leisurely rate of 5 metres every 5 seconds, i.e. her velocity is −1 m s−1. However, Sophie does not complete this lap but ends 25 m from the start. These data can be shown conveniently on a position–time graph. The displacement of the swimmer can be determined by comparing the initial and final positions. Her displacement between 20 s and 60 s is, for example: x = final position − initial position = 25 − 40 = −15 m By further examining the graph in Figure 4.16, it can be seen that during the first 25 s, the swimmer has a displacement of +50 m. Thus her average velocity is +2 m s−1, i.e. 2 m s−1 to the right. This value can also be obtained by finding the gradient of this section of the graph.

V…LOCITY is given by the gradient of a position–time graph. A positive velocity indicates that the object is moving in a positive direction, and a negative velocity indicates motion in a negative direction. If the position–time graph is curved, the velocity will be the gradient of the tangent to the line at the point of interest. This will be an instantaneous velocity. Dimensional analysis can be used to confirm that the gradient of a position–time graph is a measure of velocity: rise D gradient = = run Dt The units of this gradient are metres per second (m s−1), i.e. gradient is a measure of velocity.

A car driven by a learner driver travels along a straight driveway and is initially heading north. The position of the car is shown in the graph. a Describe the motion of the car in terms of its position. b What is the displacement of the car during the first 10 s of its motion? c What distance has the car travelled during the first 10 s? d Calculate the average velocity of the car during the first 4 s. e Calculate the average velocity of the car between t = 6 s and t = 20 s. f Calculate the average velocity of the car during its 20 s trip. g Calculate the average speed of the car during its 20 s trip. h Calculate the magnitude of the instantaneous velocity of the car at t = 18 s.

Solution a The car initially travels 10 m north in 4 s. It then stops for 2 s. From t = 6 s to t = 20 s, the

Position (m)

Worked example 4.2A 10 5 0 –5

–10

2 4 6 8 10 12 14 16 18 20 Time (s)

PRACTICAL ACTIVITY 19 Analysing motion with a motion sensor

car travels towards the south, i.e. it reverses. It passes through its starting point after 14 s, and finally stops 2 m south of this point after 20 s. b The displacement of the car is given by its change in position. From the graph, we can see that the car started from zero, and after 10 s its position is 5 m, so its displacement is +5 m or 5 m north. c The distance travelled is an indication of the ground covered by the car. During the first 10 s the car travels 10 m north, then 5 m south. The distance travelled is 15 m. d The average velocity is given by the gradient during the first 4 s: rise 10 gradient = = = +2.5 m s−1 or 2.5 m s−1 north run 4 e Again, the average velocity is given by the gradient: rise 15 gradient = = − = −1.1 m s−1 or 1.1 m s−1 south run 14 f The average velocity for the 20 s can be found by calculating the gradient of the line from the start to the end of the motion: x −5 vav = = = −0.25 m s−1 or 0.25 m s−1 south t 20 g The car travels a distance of 10 m + 10 m + 5 m = 25 m in 20 s. distance 25 Its average speed vav = = = 1.25 m s−1 ≈ 1.3 m s−1 time 20 h The graph is curved at this time, so to find the instantaneous velocity it is necessary to draw a tangent to the line and calculate the gradient of the tangent: rise −5 gradient = = = = −0.56 run 9

i.e. vinst = 0.56 m s−1 south

The magnitude of the instantaneous velocity is 0.56 m s−1.

Graphing velocity A graph of velocity against time shows how the velocity of an object changes with time. This type of graph is useful for analysing the motion of an object moving in a complex manner, for example a ball bouncing up and down. A velocity–time graph can also be used to obtain additional information about the object. Consider the example of a small girl, Eleanor, running back and forth along an aisle in a supermarket. A study of the velocity–time graph in Figure 4.18 reveals that Eleanor is moving with a positive velocity, i.e. in a positive direction, for the first 6 s. Between the 6 s mark and the 7 s mark,


123

+

Velocity (m s–1)

3 2 1 0 –1

Time (s) 1

2

3

4

5

6

7

8

9

–2

Figure 4.18 This graph shows the straight line motion of a girl running back and forth along a supermarket aisle.

10

Velocity (m s–1)

_

she is stationary, then she runs in the reverse direction, i.e. has negative velocity, for the final 3 s. The graph in Figure 4.18 shows Eleanor’s velocity at each instant in time. She moves in a positive direction with a constant speed of 3 m s−1 for the first 4 s. From 4 s to 6 s, she continues moving in a positive direction but slows down, until 6 s after the start she comes to a stop. Then during the final 3 s, when the line is below the time axis, her velocity is negative; she is now moving in a negative direction. A velocity–time graph can also be used to find the displacement of the body under consideration. In the first 6 s of Eleanor’s motion she moves with a constant velocity of +3 m s−1 for 4 s, then slows from 3 m s−1 to zero in the next 2 s. Her displacement during this time can be determined from the v–t graph: = , so Dt x = v × ∆t = height × base = area under v–t graph.

3 2 1 0 –1

–3 gradient = _ = –1.5 m s–2 2 = acceleration

Area = +12 m = displacement

Time (s) 1

2

3

4

5 6 1 _ Area = x 2 x 3 2

7

8

9

10

11

∴ x = +3 m

Figure 4.19 The displacement of the girl is given by the area under the graph. During the first 6 s, her displacement is +15 m.

From Figure 4.19, the area under the graph for the first 4 s gives the displacement of the girl during this time, i.e. +12 m. The displacement from 4 s to 6 s is represented by the area of the shaded triangle and is equal to +3 m. Thus the total displacement during the first 6 s is +12 m + 3 m = +15 m.

DISPLAC…M…NT is given by the area under a velocity–time graph (or the area between the line and the time axis). It is important to note that an area below the time axis indicates a negative displacement, i.e. motion in a negative direction. The acceleration of an object can also be found from a velocity–time graph. Consider the motion of the girl in the 2 s interval between 4 s and 6 s. She is moving in a positive direction but slowing down from 3 m s–1 to rest. Her acceleration is: Dv (v − u) (0 − 3) a= = = = −1.5 m s−2 Dt Dt 2 Since acceleration is the velocity change divided by time taken, it is also given by the gradient of the v–t graph. As can be seen from Figure 4.19 once again, the gradient of the line between 4 s and 6 s is −1.5 m s−2.

124

Motion

AV…RAG… ACC…L…RATION is the gradient of a velocity–time graph of the object over the time interval. If the acceleration is changing, the velocity–time graph will be curved, and so the gradient of the tangent will give an instantaneous acceleration.

Worked example 4.2B The motion of a radio-controlled car travelling in a straight line across a driveway is represented by the graph below.

8

Velocity (m s–1)

6

gradient = –2 m s–2

4 2 0

–2

Area = +16 m 1

2

Area = –4 m 3

4

gradient = –2 m s–2

5

6

7

8

9

Time (s)

Area = –12 m

–4

Use this graph to help you to: a describe the motion of the car in terms of its velocity b calculate the displacement of the car during the first 4 s c calculate the average velocity of the car during the first 4 s d determine the displacement for the 9 s shown e find the acceleration during the first 4 s f find the acceleration from 4 s to 6 s.

Solution a The car is initially moving in a positive direction at 8 m s−1. It slows down and comes to

Physics file To determine the area under a graph, there are a number of techniques available. These can involve some degree of estimation. 1. If the graph area is a combination of simple shapes such as rectangles and triangles, use your mathematical skills to combine these areas, remembering that areas under the horizontal axis are negative. 2. A useful technique of finding the area under a complex or curved graph is that of ‘counting squares’. To determine the area under a graph by counting squares: • calculate the area of one grid square • use a pencil to check off the number of complete squares under the graph • if the graph is curved or contains part squares, estimate the combined total of these incomplete squares • add these two amounts to determine the total number of squares • multiply this value by the area of each square to determine the area under the graph. For example, in the graph in Worked example 4.2B, the area of each grid square is 2 × 1 = 2 m. Up to 4 s, in the shaded triangular area, the complete and part squares combine to make 8 squares. The total displacement during this time is 8 squares × 2 m = +16 m.

a stop after 4 s, then reverses and travels in a negative direction. From 4 s to 6 s the car gains speed in the negative direction, then maintains a constant velocity of −4 m s−1 for the final 3 s. b The displacement is given by the area under the graph; in this case the triangular area as shown. The car’s displacement during the first 4 s is +16 m. 16 Dx c Average velocity vav = t = + 4.0 = +4.0 m s–1 In this situation, since the car had a constant acceleration for the first 4 s, it would also be appropriate to simply find the average of the initial velocity u and final velocity v using: (u + v) (8 + 0) vav = = = +4 m s–1 2 2 d The displacement for the complete motion is given by the total area under the graph: +16 − 4 − 12 = 0, i.e. the car finishes where it started. e The acceleration is given by the gradient of the line. For the first 4 s, this is −2 m s−2. This indicates that the car is slowing down by 2 m s−1 each second while travelling in a positive direction.


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f The gradient of the line from 4 s to 6 s is also −2 m s−2. This now indicates that the car

Physics file

is speeding up by 2 m s−1 each second while travelling in a negative direction.

The area under a velocity–time graph is a measure of displacement. When the units on the axes are multiplied when finding the area, a displacement unit results. From Figure 4.20a: area units = m s−1 × s = m i.e. a displacement The gradient of a velocity–time graph is the acceleration of the object. When finding the gradient, the units are divided. From Figure 4.20b: gradient units = m s−1/s = m s−2, i.e. an acceleration. (a)

v (m s –1)

area = displacement

Graphing acceleration An acceleration–time graph simply indicates the acceleration of the object as a function of time. The area under an acceleration–time graph is found by multiplying an acceleration and a time value: Area = a × ∆t = ∆ The area will give the change in velocity (∆v) of the object. In order to establish the actual velocity of the object, its initial velocity must be known. Consider the toy car from Worked example 4.2B once again. The change in velocity during the first 6 s can be determined from the acceleration–time graph. As shown in Figure 4.21, the velocity changes by −12 m s−1. This can be confirmed by looking at the velocity–time graph in Worked example 4.2B. It shows that the car slows down from +8 m s−1 to −4 m s−1, a change of −12 m s−1, during this time.

Acceleration (m s–2)

t (s) gradient = acceleration

v (m s –1)

(b)

rise

2 1 0 –1 –2

1

2 3 4 5 Area = –12 m s–1 = $v

6

7

8

9 Time (s)

run t (s)

Figure 4.20 (a) The units on the axes of a v–t graph confirm that the area under the graph represents a displacement. (b) The gradient of the line is the acceleration.

Figure 4.21 The acceleration–time graph for the toy car travelling across the driveway. It was drawn by taking account of the gradient values of the velocity–time graph. The change in the car’s velocity is given by the area under the graph.

Physics in action

Timing and false starts in athletics Until 1964, all timing of events at the Olympic Games was recorded by handheld stopwatches. The reaction times of the judges meant an uncertainty of 0.2 s for any measurement. An electronic quartz timing system introduced in 1964 improved accuracy to 0.01 s, but in close finishes the judges still had to wait for a photograph of the finish before they could announce the placings. Currently the timing system used is a vertical linescanning video system (VLSV). Introduced in 1991, this is a completely automatic electronic timing system. The starting pistol triggers a computer to begin timing. At the finish line, a high-speed video camera records the image of each athlete and indicates the time at which the chest of each one crosses the line. This system enables the times of all the athletes in the race to be precisely measured to one-thousandth of a second. Another feature of this system is that it indicates when a runner ‘breaks’ at the start of the race.

126

Motion

Figu

re 4.22 At th stopwatches e 1960 Rome Olympic G am to measure th e times of sw es, the judges used hand held immers and athletes.

Each starting block is connected by electronic cable to the timing computer and a pressure sensor indicates if a runner has left the blocks early. Since 2002, to ensure that a runner has not anticipated the pistol, a reaction time of 0.10 s is incorporated into the system. This means that a runner can still commit a false start even if their start was after the pistol. A start that is less than 0.10 s after the pistol has fired is registered as being false.

MEN 100 m Quarter-Final

Heats 2

Attempts 2

Lane 1 0.158 s

Lane 2 0.131 s

Lane 3 0.151 s Lane 4 0.052 s Fault Lane 5 0.086 s Fault Lane 6 0.144 s

in lse started the USA fa thletics f o d n o m A Jon Drum the World Figure 4.23 nt quarter-final at ed. fi li a u q ri is p s d the 100 m 3 and was r 20 minutes! 0 0 2 in s k fo ship champion by lying on the trac d te s te ro  p He

Figure 4.24 A press ure pad in each starti ng block registers the starting time of each athlete. The cable leading from each sta rting block connects to a computer which ins tantly indicates the false start. The loudspeakers en sure that all the ath letes hear the starting pistol simult aneously.

–0.4 –0.3 –0.2 –0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Figure 4.25 These charts show the pressure readings from the starting blocks for the eight sprinters in the heat. The sprinters in lanes 4 and 5 were both disqualified for false starts, although the sprinter in lane 4, Jon Drummond, did not break the 100 ms limit. The most controversial false start of recent times occurred at the World Athletics championships in 2003. It was a quarter-final heat. American Jon Drummond was in lane 4 and Asafa Powell of Jamaica was in lane 5. Australia’s Patrick Johnson was in lane 6. There had already been a false start in this heat and, since 2002, the rule for false starts in athletics events has been that after one false start, the next athletes to false start are disqualified. The athletes went under starter’s orders a second time and again it was a false start. The officials examined the computer read-out from the pressure pads on the blocks and determined that both Drummond and Powell were to be disqualified. Asafa Powell immediately left the track. Jon Drummond protested his innocence and proceeded to lie and sit down on the track for the next 20 minutes. He was widely criticised for his actions, but an analysis of the pressure pad readings reveal that he may have been a little unlucky.


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4.2 summary Graphing motion: position, velocity and acceleration • A position–time graph can be used to determine the location of a body directly. Additional information can also be derived from the graph. • Displacement is given by the change in position of a body. • The velocity of a body is given by the gradient of the position–time graph. • If the position–time graph is curved, the gradient of the tangent at a point gives the instantaneous velocity.

• Velocity–time and acceleration–time graphs can also be analysed to derive further information relating to the motion of a body. • The gradient of a velocity–time graph is the acceleration of the object. • The area under a velocity–time graph is the displacement of the object. • The area under an acceleration–time graph is the change in velocity of the object.

4.2 questions Graphing motion: position, velocity and acceleration 1 The graph shows the position of a dancer moving across a stage. B

Position (m)

A

C

D

12 8

4 When did the car return to its starting point?

4 0

5

10 15 20 25 Time (s)

a What was the starting position of the dancer? b In which of the sections (A–D) is the dancer at rest? c In which of the sections is the dancer moving in a positive direction? d In which of the sections is the dancer moving with a negative velocity? e Calculate the average speed of the dancer during the first 20 s. he following information relates to questions 2–6. The T graph represents the straight line motion of a radiocontrolled toy car.

5 What was the velocity of the car: a during the first 2 s? b after 3 s? c from 4 s to 8 s? d at 8 s? e from 8 s to 9 s? 6 During its 10 s motion, what was the car’s: a distance travelled? b displacement? 7 This position–time graph is for a cyclist travelling along a straight road.

500 450 400

8

350

6

300

Position (m)

Position (m)

4 2 0

–2

1

–4

2

3

4

5

6

7

8

9

10

Motion

250 200 150 100

Time (s)

2 Describe the motion of the car in terms of its position.

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3 What was the position of the car after: a 2 s? b 4 s? c 6 s? d 10 s?

50 0

10

20

30

Time (s)

40

50

v

A

t

v

B

C

v

t

D

v

t

E

v

t

t

he following information relates to questions 9–12. T The graph shows the motion of a dog running along a footpath. In this problem, north is considered to be positive. Velocity (m s–1)

3

A

B

C

D

E

F

2 1 0

1

2

3

4

5

6

7

8

9

10

–1 –2

Time (s)

9 Describe the motion of the dog in terms of its velocity during these sections of the graph. a A b B c C d D e E f F 10 Calculate the displacement of the dog after: a 2 s b 7 s c 10 s

12 Plot a position–time graph of the dog’s motion. 13 The straight line motion of a high speed intercity train is shown in the graph. 60

Velocity (m s–1)

8 Which of the velocity–time graphs A–E best repre sents the motion of: a a car coming to a stop at a traffic light? b a swimmer moving with constant speed? c a cyclist accelerating from rest with constant acceleration? d a car accelerating from rest and changing through its gears?

11 What is the average velocity of the dog between: a t = 2 s and t = 4 s? b t = 2 s and t = 7 s?

50 40 30 20 10 0

10 20 30 40 50 60 70 80 90 100 110 120 Time (s)

a How long does it take the train to reach its cruising speed? b What is the acceleration of the train 10 s after starting? c What is the acceleration of the train 40 s after starting? d What is the displacement (in km) of the train after 120 s? 14 The velocity–time graphs for a bus and a bicycle travelling along the same straight stretch of road are shown below. The bus is initially at rest and starts moving as the bicycle passes it. Velocity (m s–1)

a Describe the motion of the cyclist in terms of speed and distance. b What was the velocity of the cyclist during the first 30 s? c What was the cyclist’s velocity during the final 10 s? d Calculate the cyclist’s instantaneous velocity at 35 s. e What was the average velocity of the cyclist between 30 s and 40 s?

bus

12 8

bicycle

4 0

2

4

6

8

10 12 14 16 18 20 22 24 Time (s)

a Calculate the initial acceleration of the bus. b When does the bus first start gaining ground on the bicycle? c At what time does the bus overtake the bicycle? d How far has the bicycle travelled before the bus catches it? e What is the average velocity of the bus during the first 8 s? 15 a Draw an acceleration–time graph for the bus discussed in question 14. b Use your acceleration–time graph to determine the change in velocity of the bus over the first 8 s.


129

4.3 Equations of motion Equations for uniform acceleration A graph is an excellent way of representing motion because it provides a great deal of information that is easy to interpret. However, a graph is time-consuming to draw and, at times, values have to be estimated rather than precisely calculated. The previous section used the graph of a motion to determine the different quantities needed to describe the motion of a body. In this section, we will examine a more powerful and precise method of solving problems involving constant or uniform acceleration. This method involves the use of a series of equations that can be derived from the basic definitions developed earlier. Consider a body moving in a straight line with an initial velocity u and a uniform acceleration a for a time interval t. After time t, the body is travelling with a final velocity v. Its acceleration will be given by: D ( − ) = = Dt Dt This can be rearranged as: v = u + at . . . . . . (i) The average velocity of this object is: av

=

displacement = t time taken

When acceleration is uniform, average velocity vav can also be found as the average of the initial and final velocities, i.e. (u + v) vav = 2 So: This gives:

x (u + v) = t 2

(u + v) t . . . . . . (ii) 2 A graph describing this particular motion is shown in Figure 4.26. The displacement x of the body is given by the area under the velocity–time graph. The area under the velocity–time graph in Figure 4.26 is given by the combined area of the rectangle and the triangle:

v

Velocity

v–u Area = _1 (v – u)t 2

As

u

then u

Area = ut t 0

Time

t

Figure 4.26 The area of the shaded rectangle and triangle represents the displacement x of the body for each time interval.

130

Motion

x=

1

Area = x = ut + 2 × t × ∆v Dv a= t ∆ = at, and this can be substituted for ∆v: 1

x = ut + 2 × t × at 1 x = ut + 2 at2

. . . . . . (iii)

Now making u the subject of equation (i) gives: u = v − at. You might like to derive another equation yourself by substituting this into equation (ii). You should get: 1 x = vt - 2 at2 . . . . . . (iv) Rewriting equation (i) with t as the subject gives: (v − u) t= a

Now if this is substituted into equation (ii): (u + v) (u + v) (v - u) v2 - u2 x= t= × = 2a 2 2 a Finally, transposing this gives: v2 = u2 + 2ax . . . . . . (v) Equations (i)–(v) are commonly used to solve problems in which acceleration is constant. They are summarised below. v = u + at (u + v) x= t 2 1 x = ut + 2 at2 1

x = vt − 2 at2 v2 = u2 + 2ax These equations can also be used with the scalar quantities speed and distance. When solving problems using these equations, it is important that you think about the problem and try to visualise what is happening. The following steps are advisable. Step 1 Draw a simple diagram of the situation. Step 2 Neatly write down the information that has been given in the question, using positive and negative values to indicate directions. Convert all units to SI form. Step 3 Select the equation that matches your data. Step 4 Use the appropriate number of significant figures in your answer. Step 5 Include units with the answer and specify a direction if the quantity is a vector.

Worked example 4.3A A snowboarder in a race is travelling at 10 m s−1 as he crosses the finishing line. He then decelerates uniformly until coming to rest over a distance of 20 m. a What is his acceleration as he pulls up? b How long does he take to come to rest? c Calculate the average speed of the snowboarder as he pulls up.

Solution a Draw a simple diagram of the situation (see graph). When the snowboarder stops, his velocity is zero. u = 10 m s−1, v = 0, x = 20 m, a = ? v2 = u2 + 2ax 0 = 102 + 2 × a × 20 a = −2.5 m s−2 b u = 10 m s−1, v = 0, a = −2.5 m s−2, x = 20 m, t = ? v = u + at 0 = 10 − 2.5 × t t = 4.0 s distance 20 c vav = time = 4.0 = 5.0 m s−1 As the snowboarder’s acceleration is uniform, this could also have been determined using: (u + v) (20 + 0) vav = = = 5.0 m s−1 2 2

Velocity (m s–1) 10 average 5 speed 4

Time (s)


131

Physics in action

How police measure the speeds of cars Road accidents account for the deaths of about 1600 people in Australia each year. Many times this number are seriously injured. Numerous steps have been taken to reduce the number of road fatalities. Some of these include random breath and drug testing, speed cameras, mandatory wearing of bicycle helmets and the zero alcohol level for probationary drivers. One of the main causes of road trauma is speeding. In their efforts to combat speeding motorists, police employ a variety of speed-measuring devices.

vehicle has a lower frequency. This change in frequency or ‘Doppler shift’ is processed by the unit and gives an instant measurement of the speed of the target vehicle. Camera radar units are capable of targeting a single vehicle up to 1.2 km away. In traffic, the units can distinguish between individual cars and take two photographs per second. The photographs and infringement notices are then mailed to the offending motorists.

Laser speed guns These devices are used by police to obtain an instant measure of the speed of an approaching or receding vehicle. The unit is usually handheld and is aimed directly at a vehicle using a target sight. It emits a pulse of infrared radiation of frequency 331 THz (3.31 × 1014 Hz). As with camera radar units, speed is determined by the Doppler shift produced by the target vehicle. The infrared pulse is very narrow and directional—just one-sixth of 1° wide. This allows vehicles to be targeted with great precision. Handheld units can be used at distances up to 800 m. If the speed registers as over the limit, the police are then likely to apprehend the offending driver.

Speed camera radar Camera radar units are usually placed in parked, unmarked police cars. These units take flash photographs of speeding vehicles, and also emit a radar signal of frequency 24.15 GHz (2.415 × 1010 Hz). The radar antenna has a parabolic reflector that enables the unit to produce a directional radar beam 5° wide, thus allowing individual vehicles to be targeted. The radar signal allows speeds to be determined by the Doppler principle, whereby the reflected radar signal from an approaching vehicle has a higher frequency than the original signal. Similarly, the reflected signal from a receding (a)

(b)

field of vision of the camera radar range

(c)

Figure

4.27 (a unit emits ) A speed camera ra a target car radar beam 5° wid dar unit. (b) The ew an the film is d gives a measure hich reflects from a of de plate and veloped, the time o its speed. (c) When the speed f day, the of the offe n clearly ind nding veh umber icated. icle are

132

Motion

Fixed speed cameras Victoria’s first fixed speed cameras were located on the Monash Freeway and in the Domain and Burnley tunnels. These cameras obtain their readings by using a system of three strips with piezoelectric sensors in them across the road. The strips respond to the pressure as the car drives over them and create an electrical pulse that is detected. By knowing the distance between the strips and measuring the time that the car takes to travel across them, the speed of the car can be determined. When a speeding car is detected, a digital photograph is taken and a fine is issued.

Point-to-point cameras The first point-to-point cameras in Victoria were installed on the Hume Freeway between Craigieburn and Broadford. These towns are about 50 km apart and there are several camera sites along the freeway between them. These do not measure the instantaneous speed of the car like the other cameras do. They record the time at which a vehicle passes the camera at, say, Craigieburn and then compare the time at which it passes the next camera about 10 km away. This allows the average speed of the vehicle to be determined. If this is higher than the speed limit, a penalty is issued. It is likely that point-to-point cameras will be placed on many more roads in the future.

Set distance

1st sensor

2nd sensor

3rd sensor

Figure 4.28 Fixed speed cameras record the speed of a car twice by measuring the time the car takes to travel over a series of three sensor strips embedded in the roadway.

4.3 summary Equations of motion • Equations of motion can be used to analyse problems involving constant acceleration. These equations are:

( + ) 2 v = u + at u+v x= t 2

av

=

(

)

1

x = ut + 2 at2 1

x = vt − 2 at2

where is displacement (m) u is the initial velocity (m s–1) v is the final velocity (m s–1) a is the acceleration (m s–2) and t is time (s).

v2 = u2 + 2ax


133

4.3 questions Equations of motion 1 A Prius hybrid car starts from rest and accelerates uniformly for 8.0 s. It reaches a final speed of 16 m s−1. a What is the acceleration of the Prius? b What is the average velocity of the Prius? c Calculate the distance travelled by the Prius. 2 A new model Subaru can start from rest and travel 400 m in 16 s. a What is its average acceleration during this time? b Calculate the final speed of the car. c What is its final speed in km h−1? 3 During its launch phase, a space-rocket accelerates uniformly from rest to 160 m s−1 in 4.0 s, then travels with a constant speed of 160 m s−1 for the next 5.0 s. a Calculate the initial acceleration of the rocket. b How far does the rocket travel in this 9 s period? c What is the final speed of the rocket in km h−1? d What is the average speed of the rocket during the first 4.0 s? e What is the average speed of the rocket during the 9.0 s motion? 4 A diver plunges headfirst into a diving pool while travelling at 28.2 m s−1. Upon entering the water, the diver stops within a distance of 4.00 m. Consider the diver to be a single point located at her centre of mass and assume her acceleration through the water to be uniform. a Calculate the average acceleration of the diver as she travels through the water. b How long does the diver take to come to a stop? c What is the speed of the diver after she has dived through 2.00 m of water? 5 When does a car have the greatest ability to accelerate and gain speed: when it is moving slowly or when it is travelling fast? Explain. 6 A stone is dropped vertically into a lake. Which one of the following statements best describes the motion of the stone at the instant it enters the water? A Its velocity and acceleration are both downwards. B It has an upwards velocity and a downwards acceleration.

134

Motion

C Its velocity and acceleration are both upwards. D It has a downwards velocity and an upwards acceleration. 7 While overtaking another cyclist, Cadel increases his speed uniformly from 4.2 m s−1 to 6.3 m s−1 over a time interval of 5.3 s. a Calculate Cadel’s acceleration during this time. b How far does he travel while overtaking? c What is Cadel’s average speed during this time? 8 A car is travelling along a straight road at 75 km h−1. In an attempt to avoid an accident, the motorist has to brake to a sudden stop. a What is the car’s initial speed in m s−1? b If the reaction time of the motorist is 0.25 s, what distance does the car travel while the driver is reacting to apply the brakes? c Once the brakes are applied, the car has an acceleration of −6.0 m s−2. How far does the car travel while pulling up? d What total distance does the car travel from the time the driver first notices the danger to when the car comes to a stop? 9 A billiard ball rolls from rest down a smooth ramp that is 8.0 m long. The acceleration of the ball is constant at 2.0 m s−2. a What is the speed of the ball when it is halfway down the ramp? b What is the final speed of the ball? c How long does the ball take to roll the first 4.0 m? d How long does the ball take to travel the final 4.0 m? 10 A cyclist, Anna, is travelling at a constant speed of 12 m s−1 when she passes a stationary bus. The bus starts moving just as Anna passes, and it accelerates uniformly at 1.5 m s−2. a When does the bus reach the same speed as Anna? b How long does the bus take to catch Anna? c What distance has Anna travelled before the bus catches up?

4.4 Vertical motion u

nder gravity

Theories of motion: Aristotle and Galileo Aristotle was a Greek philosopher who lived in the 4th century bc. He was such an influential individual that his ideas on motion were generally accepted for nearly 2000 years. Aristotle did not do experiments as we know them today, but simply thought about different bodies in order to arrive at a plausible explanation for their motion. He had spent a lot of time classifying animals, and so adopted a similar approach in his study of motion. His theory gave inanimate objects, such as rocks and rain, similar characteristics to living things. Aristotle organised objects into four terrestrial groups or elements: earth, water, air and fire. He said that any object was a mixture of these elements in a certain proportion. According to Aristotle, a body would move because of a tendency that could come from inside or outside the body. An internal tendency would cause ‘natural’ motion and result in a body returning to its proper place. For example, if a rock, which is an earth substance, is held in the air and released, its natural tendency would be to return to Earth. This explains why it falls down. Similarly, fire was thought to head upwards in an attempt to return to its proper place in the universe. An external push that acts when something is thrown or hit was the cause of ‘violent’ motion in the Aristotelian model. An external push acted to take a body away from its proper place. For example, when an apple is thrown into the air, a violent motion carries the apple away from the Earth, but then the natural tendency of the apple takes over and it returns to its home. Aristotle’s theory worked quite well and could be used to explain many observed types of motion. However, there were also many examples that it could not successfully explain, such as why some solids floated instead of sinking. Aristotle explained the behaviour of a falling body by saying that its speed depended on how much earth element it contained. This suggested that a 2 kg cat would fall twice as fast and in half the time as a 1 kg cat dropped from the same height. Many centuries later, Galileo Galilei noticed that at the start of a hailstorm, small hailstones arrived at the same time as large hailstones. This caused Galileo to doubt Aristotle’s theory and so he set about finding an explanation for the motion of freely falling bodies. (a)

(b) fire

air

water earth

Figure 4.29 (a) The Aristotelian terrestrial world consisted of earth, water, air and fire. According to this model, any type of matter has an inherent and natural tendency to return to its own state. (b) An artist’s representation of Aristotle’s view of the universe.


135

(a)

(b)

Physics file In 1971, David Scott went to great lengths to show that Galileo was correct. As an astronaut on the Apollo 15 Moon mission, he took a hammer and a feather on the voyage. He then stepped onto the lunar surface, held the feather and hammer at the same height and dropped them together. As Galileo would have predicted 400 years earlier, in the absence of any air resistance, the two objects fell side by side as they accelerated towards the ground.

Aristotelian view

Galilean view

Figure 4.30 (a) Up until the 17th century, it was commonly thought that a heavy object would fall faster than a light object. (b) After research by Galileo Galilei it was shown that if air resistance can be ignored, all bodies fall with an equal acceleration.

A famous story in science is that of Galileo dropping different weights from the Leaning Tower of Pisa. This story may or may not be true, but Galileo did perform a very detailed analysis of falling bodies. Galileo used inclined planes because freely falling bodies moved too fast to analyse. He completed extensive and thorough experiments that showed conclusively that Aristotle was incorrect. By using a waterclock to time balls as they rolled down different inclines, he was able to show that the balls were accelerating and that the distance they travelled was proportional to the square of the time, i.e. d ∝ t2. Galileo found that this also held true when he inclined the plane at larger and larger angles, allowing him to conclude that freely falling bodies actually fall with a uniform acceleration.

Analysing vertical motion Even today, many people think that heavy objects fall faster than light ones. The cause of confusion is usually related to the effects of air resistance. Some falling objects are greatly affected by air resistance, for example a feather and a balloon. This is why these objects do not speed up as they fall. However, if air resistance can be ignored, all free-falling bodies near the Earth’s surface will move with an equal downwards acceleration. In other words, the mass of the object does not matter. This is clearly shown in the multiflash photograph in Figure 4.31 where a baseball of mass 0.23 kg can be seen to fall at the same rate as a shotput of mass 5.4 kg. Given that the flash rate is 15 Hz and the markings are 10 cm apart, you should be able to calculate the acceleration of these objects and obtain a value close to 9.8 m s−2. This value of 9.8 m s−2 is the acceleration of bodies falling due to gravity and is commonly represented as g. Figure 4.31 This multiflash photograph is taken with a frequency of 15 Hz and the scale markings are 10 cm apart. The photograph shows the relative motions of a baseball and a shotput in free-fall. Even though the shotput is over 20 times heavier than the baseball, both objects fall with an acceleration of 9.8 m s−2 down. The increasing speed of both objects is also evident.

136

Motion

At the Earth’s surface, the acceleration due to gravity is g = 9.8 m s−2 down. Free fall simply implies that the motion of the body is affected only by gravity, i.e. there is no air resistance and there are no rockets firing. It is also important to note that the acceleration of a freely falling body is always 9.8 m s−2 down, and does not depend on whether the body is falling up or down. For example, a coin that is dropped from rest will be moving at 9.8 m s−1 after 1 s, 19.6 m s−1 after 2 s, and so on. Each second, its speed increases by 9.8 m s−1.

v=0

9.8 m s–1

t = 0: at rest

9.8 m s–1

1s

19.6 m s–1

v=0

2s

1s

19.6 m s–1 2s

t=0

Figure 4.32 These coins are both moving with an acceleration of 9.8 m s−2 down. (a) The speed of a coin falling vertically increases by 9.8 m s−1 each second, i.e. it has an acceleration of 9.8 m s−2 down. (b) The speed of a coin thrown upwards decreases by 9.8 m s−1 each second. It too has an acceleration of 9.8 m s−2 down.

However, if the coin was launched straight up at 19.6 m s−1, then after 1 s its speed would be 9.8 m s−1, and after 2 s it would be stationary (Figure 4.32) . In other words, each second it would slow down by 9.8 m s−1. Since the acceleration of a freely falling body is constant, it is appropriate to use the equations for uniform acceleration. It is often necessary to specify up or down as the positive or negative direction when doing these problems (see Physics file).

Worked example 4.4A A construction worker accidentally knocks a brick from a building so that it falls vertically a distance of 50 m to the ground. Using g = 9.8 m s−2, calculate: a the time the brick takes to fall the first 25 m b the time the brick takes to reach the ground c the speed of the brick as it hits the ground.

Solution Down will be treated as the positive direction for this problem since this is the direction of the displacement. a u = 0, x = 25 m, a = 9.8 m s−2, t = ? x = ut + 12 at2 25 = 0 + 12 × 9.8 × t2 t2 = 5.1 t = 2.3 s b u = 0, a = 9.8 m s−2, x = 50 m, t = ? x = ut + 12 at2 50 = 0 + 12 × 9.8 × t2 t2 = 10.2 t = 3.2 s Notice that the brick takes less time, only 0.9 s, to travel the final 25 m. This is because it is accelerating. c u = 0, a = 9.8 m s−2, x = 50 m, t = 3.2 s, v = ? v = u + at v = 0 + 9.8 × 3.2 = 31 m s−1

Physics file It is important that simple diagrams be used with vertical motion problems and that you clearly show whether up or down is the positive direction. You need to decide this. You could simply use the mathematical convention of up being positive, which would make the acceleration of the object in free-fall negative 9.8 m s–2. Alternatively, you may wish to use the direction of the initial displacement as positive. For example, where an object is dropped, this would result in down being the positive direction, which would make the object’s acceleration positive 9.8 m s–2.

PRACTICAL ACTIVITY 20 A reaction timer


137

Worked example 4.4B

The acceleration of a falling object near the Earth’s surface is approximately 9.8 m s−2. This value, denoted , can be used when describing large accelerations. For example, an acceleration of 19.6 m s−2 is 2 . An astronaut will experience an acceleration of about 4 (39.2 m s−2) at take-off. The forces involved give a crushing sensation as if the astronaut had four identical astronauts lying on top of him or her! Space missions are designed so that the acceleration does not exceed 6 . Sustained accelerations greater than this can lead to the astronauts losing consciousness. Have a look at Figure 4.11 to see the effect of a 12 acceleration!

+

_

Physics file The acceleration due to gravity on Earth varies according to location. The strength of gravity is different on different bodies in the solar system depending on their mass and size.

On winning a tennis match the victorious player, Michael, smashed the ball vertically into the air at 30 m s−1. In this example, air resistance can be ignored and the acceleration due to gravity will be taken as 10 m s−2. a Determine the maximum height reached by the ball. b Calculate the time that the ball takes to return to its starting position. c Calculate the velocity of the ball 5.0 s after being hit by Michael. d Determine the acceleration of the ball at its maximum height. e Draw an acceleration–time graph of the ball’s motion. f Draw a velocity–time graph of the ball’s motion.

Solution In this problem, up will be taken as positive since it is the direction of the initial displacement. a At the maximum height, the velocity of the ball is momentarily zero. u = 30 m s−1, v = 0, a = −10 m s−2, x = ? v2 = u2 + 2ax 0 = (30)2 + 2(−10)x ∴ x = +45 m, i.e. the ball reaches a height of 45 m. b To work out the time for which the ball is in the air, it is often necessary to first calculate the time that it takes to reach its maximum height. u = 30 m s−1, v = 0, a = −10 m s−2, x = 45 m, t = ? v = u + at 0 = 30 + (−10 × t) ∴ t = 3.0 s The ball takes 3.0 s to reach its maximum height. It will therefore take 3.0 s to fall from this height back to its starting point and so the whole trip will last for 6.0 s. c u = 30 m s−1, a = −10 m s−2, t = 5.0 s, v = ? v = u + at v = 30 +(−10 × 5.0) = −20 m s−1 The ball is travelling downwards at 20 m s−1. d The ball moves with an acceleration of 10 m s−2 down throughout its entire flight. Thus at its highest point, where its velocity is zero, its acceleration is still 10 m s−2 down. e Since the acceleration of the ball is a constant −10 m s−2, its acceleration– 10 time graph will be as shown. Acceleration (m s–2)

Physics file

Table 4.5 The acceleration due to gravity at various locations around the solar system Location

Acceleration due to gravity (m s−2)

5 0

–5

f The

velocity–time graph shows an initial velocity of 30 m s−1 reducing to zero after 3 s, then speeding up to –30 m s−1 after 6 s. The gradient of this graph is constant and is equal to −10, i.e. the acceleration of the ball.

9.832

Equator

9.780

Moon

1.6

Mars

3.6

–20

24.6

–30

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Motion

Velocity (m s–1)

South Pole

0.67

3

4

5

6 Time (s)

1

2

3

4

5

6 Time (s)

30

9.800

Pluto

2

–10

Melbourne

Jupiter

1

20 10 0 –15

4.4 summary Vertical motion under gravity • Ideas about motion have changed through the ages. Aristotle’s model of the universe organised objects into combinations of four elements: earth, water, air and fire. • According to Aristotle, an object would have a natural tendency to return to its place in the universe according to which element it was composed of. He also stated that heavy objects would fall to the ground faster than light objects. • Galileo conducted experiments and determined that falling objects move with constant acceleration. He also showed that the acceleration of a falling object did not depend on its mass.

• If air resistance can be ignored, all bodies falling freely near the Earth will move with the same constant acceleration. • The acceleration due to gravity is represented by g and is equal to 9.8 m s−2 in the direction of the centre of the Earth. • The equations for uniform acceleration can be used to solve vertical motion problems. It is often necessary to specify a positive and negative direction.

4.4 questions Vertical motion under gravity For these questions, ignore the effects of air resistance and assume that the acceleration due to gravity is 9.8 m s−2 down. 1 Vickie holds a paper clip in one hand and a brick in the other. She raises both objects so that they are 2 m above the ground and drops them at the same time. a What would Aristotle have predicted about the subsequent motion of the objects? b What would Galileo have predicted about the subsequent motion of the objects? c What actually happens as the objects are released? 2 Phung is swimming in his pool. He dives to the bottom of the pool and exhales the air from his lungs, noticing that the bubbles of air rise to the surface of the water. How would Aristotle have explained the motion of the bubbles? 3 An ostrich inadvertently lays an egg while standing up and the egg falls vertically towards the ground. Which one of the following statements is correct? As the egg falls: A its acceleration increases. B its acceleration is constant. C its velocity is constant. D its acceleration decreases. 4 Chris is an Olympic trampolinist and is practising some routines. Which one or more of the following statements correctly describes Chris’ motion when

he is at highest point of the bounce? Assume that his motion is vertical. A He has zero velocity. B His acceleration is zero. C His acceleration is upwards. D His acceleration is downwards. 5 A builder working at the MCG knocks a large bolt off a scaffold. The bolt falls 50 m vertically towards the ground. a Without using a calculator, determine the speed of the bolt after: i 1.0 s ii 2.0 s iii 3.0 s b Calculate the speed of the bolt after it has fallen: i 10 m ii 20 m iii 30 m c What is the bolt’s average speed during a fall of 30 m? 6 A golf ball is thrown vertically into the air and returns to the thrower’s hand a short time later. Assume that up is the positive direction. Ignoring air resistance, sketch the following graphs for the ball’s motion. a Distance–time b Displacement–time c Speed–time


139

d Velocity–time e Acceleration–time 7 A book is knocked off a bench and falls vertically to the floor. If the book takes 0.40 s to fall to the floor, calculate: a its speed as it lands b the height from which it fell c the distance it falls during the first 0.20 s d the distance it falls during the final 0.20 s. 8 While celebrating her 18th birthday, Bindi pops the cork off a bottle of champagne. The cork travels vertically into the air. Being a keen physics student, Bindi notices that the cork takes 4.0 s to return to its starting position. a How long does the cork take to reach its maximum height? b What was the maximum height reached by the cork? c How fast was the cork travelling initially?

d What was the speed of the cork as it returned to its starting point? e Describe the acceleration of the cork at each of these times after its launch: i 1.0 s ii 2.0 s iii 3.0 s

9 At the start of a football match, the umpire bounces the ball so that it travels vertically and reaches a height of 15.0 m. a How long does the ball take to reach this maximum height? b One of the ruckmen is able to leap and reach to a height of 4.0 m with his hand. How long after the bounce should this ruckman endeavour to make contact with the ball? 10 A hot-air balloon is 80 m above the ground and travelling vertically downwards at 8.0 m s−1 when one of the passengers accidentally drops a coin over the side. How long after the coin reaches the ground does the balloon touch down?

chapter review

The following information relates to questions 1–3. During a game of mini-golf, a girl putts a ball so that it hits an obstacle and travels straight up into the air, reaching its highest point after 1.5 s. 1 Which one of the following statements best describes the acceleration of the ball while it is in the air? A The acceleration of the ball decreases as it travels upwards, becoming zero as it reaches its highest point. B The acceleration is constant as the ball travels upwards, then reverses direction as the ball falls down again. C The acceleration of the ball is greatest when the ball is at the highest point. D The acceleration of the ball is constant throughout its motion. 2 What was the initial velocity of the ball as it launched into the air? 3 Calculate the maximum height reached by the ball.

140

Motion

4 Theories, such as those put forward by Aristotle and Galileo, are not usually replaced unless the theory no longer works or a better theory is proposed. Discuss some of the problems with Aristotle’s theories that led to them being replaced by new theories proposed by Galileo and, later, by Isaac Newton. The following information relates to questions 5–8. The graph shows the position of a motorbike along a straight stretch of road as a function of time. The motorcyclist starts 200 m north of an intersection.

Position (m)

For the following questions, the acceleration due to gravity is 9.8 m s−2 down and air resistance is considered to be negligible.

500 400 300 200 100 0 –100 –200

10

20

30

40

50

60 Time (s)

5 During what time interval is this motorcyclist: a travelling in a northerly direction? b travelling in a southerly direction? c stationary?

7 Calculate the instantaneous velocity of the motorcyclist at each of the following times. a 15 s b 35 s 8 For the 60 s motion, calculate the: a magnitude of the average velocity of the motorcyclist b average speed of the motorcyclist. The following information relates to questions 9 and 10. A skier is travelling along a horizontal ski run at a speed of 10 m s−1. After falling over, the skier takes 10 m to come to rest. 9 Which one of the following best describes the average acceleration of the skier? A −1 m s−2 B −10 m s−2 C −5 m s−2 D zero 10 Calculate the time it takes the skier to come to a stop. The following information relates to questions 11 and 12. An athlete in training for a marathon runs 15 km north along a straight road before realising that she has dropped her drink bottle. She turns around and runs back 5 km to find her bottle, then resumes running in the original direction. After running for 2.0 h, the athlete reaches 20 km from her starting position and stops.

14 Calculate the average speed of the cart during: a section A b section B c its total journey. 15 a What was the instantaneous speed of the cart when dot X was made? b Calculate the magnitude of the acceleration of the cart during section A. The following information relates to questions 16 and 17. Two physics students conduct the following experiment from a very high bridge. Thao drops a 1.5 kg shot-put from a vertical height of 60.0 m while at exactly the same time Benjamin throws a 100 g mass with an initial downwards velocity of 10.0 m s−1 from a point 10.0 m above Thao. 16 Calculate the time that: a the shot-put takes to reach the ground b the 100 g mass takes to reach the ground. 17 At what time will the 100 g mass overtake the shot-put? The graph relates to questions 18–20. The velocity–time graph is for an Olympic road cyclist as he travels north along a straight section of track. Velocity (m s–1)

6 When does the motorcyclist pass back through the intersection?

11 Calculate the average speed of the athlete in km h−1. 12 Calculate her average velocity in: a km h−1 b m s−1.

13 A jet-ski starts from rest and accelerates uniformly. If it travels 2.0 m in its first second of motion, calculate: a its acceleration b its speed at the end of the first second c the distance the jet-ski travels in its second second of motion. The following information relates to questions 14 and 15. A student performing an experiment with a dynamics cart obtains the ticker tape data as shown below. The ticker timer has a frequency of 50 Hz. A

2 cm

1.8

B 1.6

x

1.4

14 12 10 8 6 4 2 0

1 2 3 4 5 6 7 8 9 10 11 12 Time (s)

18 What is the average velocity of the cyclist during this 11 s interval? 19 Which one or more of the following statements correctly describes the motion of the cyclist? A He is always travelling north. B He travels south during the final 2 s. C He is stationary after 8 s. D He returns to the starting point after 11 s. 20 Calculate the acceleration of the cyclist at each of the following times. a 1 s b 5 s c 10 s

1.2 1.0 1.0 1.0 1.0


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chapter 5

s w a l s ’ n o t w Ne

A

lthough he did not know it at the time, Isaac Newton’s work in the 17th century signalled an end to the transition in the way the world was conceived and understood. The transition was begun by Copernicus and Galileo, but Newton was able to use mathematics to develop laws and theories that could account for the motion of the heavens. These showed the universe to be a mechanism that could readily be understood, one that was regulated by simple natural laws. The universe taught by Aristotle, and accepted up until the time of Newton, was one in which objects were classified into categories and their motion depended the category to which they belonged. Isaac Newton was born in rural England in 1642, the year Galileo died. He so impressed his mentors that he was made Professor of Mathematics at Cambridge University at the age of 26. His interests spanned light and optics, mathematics (he invented the calculus), astronomy and the study of mechanics. His greatest achievement was the formulation of the law of universal gravitation. This, along with a complete explanation of the laws that govern motion, is laid out in his book Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy), which was published in 1687. The Principia is one of the most influential publications in natural science. Newton’s framework for understanding the universe remained intact right up to the advent of Einstein’s relativity more than 200 years later. Newton died in 1727 a famous man. He remained self-critical and shy. That he could ‘see so far’ was only because he was ‘able to stand on the shoulders of giants’. In saying this, he was referring to the work of Galileo, Copernicus and many others who had paved the way for his discoveries.

by the end of this chapter you will have covered material from the study of movement including: • vector techniques in two dimensions • forces in two dimensions • Newton’s laws of motion • problems in mechanics including weight and friction.

5.1 Force as a vect or The previous chapter developed the concepts and ideas needed to describe the motion of a moving body. This branch of mechanics is called kinematics. In this chapter, rather than simply describe the motion, we will consider the forces that cause the motion to occur. Treating motion in this way falls within the branch of mechanics called dynamics. In simple terms, a force can be thought of as a push or a pull, but forces exist in a wide variety of situations in our daily lives and are fundamental to the nature of matter and the structure of the universe. Consider each of the photographs in Figure 5.1 and identify each force—push or pull—that is acting. (a)

(b)

Figure 5.1 (a) At the moment of impact by a tennis racquet, a tennis ball is distorted to a significant extent. (b) The rock climber is relying on the frictional force between his hands and feet and the rock-face. (c) A continual force causes the clay to deform into the required shape. (d) The gravitational force between the Earth and the Moon is responsible for two high tides each day. (e) The globe is suspended in mid-air because of the magnetic forces of repulsion and attraction.

In each of the situations depicted in Figure 5.1, forces are acting. Some are applied directly to an object and some act on a body without touching it. Forces that act directly on a body are called contact forces, because the body will only experience the force while contact is maintained. Forces that act on a body at a distance are non-contact forces. Contact forces are the easiest to understand and include the simple pushes and pulls that are experienced daily in people’s lives. Examples of these include the forces between colliding billiard balls, the force that you exert on a light switch to turn it on, and the forces that act between you and your chair as you sit reading this book. Friction and drag forces are other contact forces that you should be familiar with. Non-contact forces occur when the object causing the push or pull is physically separated from the object that experiences the force. These forces are said to ‘act at a distance’. Gravitation, magnetic and electric forces are examples of non-contact forces. The action of a force is usually recognised through its effect on an object or body. A force may do one or more of a number of things to the object. It may change its shape, change its speed or change only the direction of its motion. The tennis racquet in Figure 5.1a has applied a force to the tennis ball, and, as a consequence, the speed of the ball changes along with its direction. The ball also changes shape while the force acts!

(c)

(d)

(e)


143

The amount of force acting can be measured using the SI unit called the newton, which is given the symbol N. The unit, which will be defined later in the chapter, honours Sir Isaac Newton (1642–1727), who is still considered to be one of the most significant physicists to have lived. A force of one newton, 1 N, is approximately the force you have to exert when holding a 100 g mass against the downward pull of gravity. In everyday life this is about the same as holding a small apple. Table 5.1 provides a comparison of the magnitude of some forces.

Table 5.1 A comparison of the magnitude of various forces Force

Magnitude (N)

Force on the electron in a hydrogen atom

10−7

Holding a small apple against gravity

1

Opening a door

10

Pedalling a bicycle

300

Thrust of a Boeing 747 at take-off

106

Gravitational force between the Earth and the Sun

1022

Force: a vector quantity

Figure 5.2 The netball will only go through the hoop if a force of the right magnitude and direction is applied. Force is a vector; it can only be completely specified if both the direction and magnitude are given.

W

E

95 N

Figure 5.3 When drawing force diagrams, it is important that the force is shown to be acting at the correct location. In this example, the force on the ball acts at the point of contact between the ball and the foot.

144

Motion

In Chapter 4, quantities associated with motion were classified as being either vectors or scalars. Scalar quantities such as time and mass do not have a direction. Only their size or ‘magnitude’ should be given. Quantities that require a direction as well as a magnitude are called vectors. Force is a vector quantity because the direction in which a force acts is always significant. In this text, vectors are set in bold italics.

FORC… is measured in newtons (N) and is a vector quantity. It requires a magnitude and a direction to describe it fully. If a question only requires the magnitude of a vector, the direction can be ignored. In this text, italics will be used to show this. In a diagram, a force is usually shown as an arrow whose length represents the magnitude of the force and whose direction is indicated by the arrow. Consider the case of a soccer player who kicks the ball horizontally with a force of 95 N towards the east. The horizontal forces acting on the ball can be illustrated by a vector diagram as shown in Figure 5.3. If there are two or more forces acting on the same object, these forces can be shown on the same diagram. If one force is larger, it should be represented by a longer vector. If, for example, the soccer ball just discussed was sitting in thick mud so that a frictional force of 20 N towards the west was acting as it was kicked, this could be represented as shown in Figure 5.4. The subsequent motion of the soccer ball will be different in the two situations described above. When there is a large frictional force acting on the ball, its speed will be significantly reduced. The muddy ground will act to make the ball travel more slowly as it leaves the boot. To analyse the horizontal motion of the ball, it is necessary to add all the horizontal forces

that are acting on it at this instant. The ball is simply treated as a point mass located at its centre of mass. If more than one force acts on a body at the same time, the body behaves as if only one force—the vector sum of all the forces—is acting. The vector sum of the forces is called the resultant or net force, ΣF (shown as a doubleheaded arrow).

The N…T FORC… acting on a body experiencing a number of forces acting simultaneously is given by the vector sum of all the individual forces acting: ΣF = F1 + F2 + ... + Fn

95 N 20 N

Because force is a vector quantity, the addition of a number of forces must be undertaken with the directions of the individual forces in mind. Vector addition is shown in Figure 5.5. 95 N

+

20 N

3F= 75 N

Physics file

20 N

added, the resultant or net force is 75 N towards the east. The ball will move as though this resultant force is the only force acting on it.

If the forces that are acting are perpendicular (or any other angle) to each other, the resultant force must still be found by performing a vector addition. Consider the example of a shopping trolley that is being simultaneously pushed from behind by one person and pushed from the side by another. This situation is illustrated in Figure 5.6. To find the magnitude of the resultant force, Pythagoras’s theorem must be used: ΣF = √802 + 602 = √10 000 = 100 N Person 2 60 N

Person 1

Figure 5.4 The two forces being considered are acting at different locations and have different strengths. The larger force is shown as a longer vector.

95 N

=

Figure 5.5 When the forces (95 N acting towards the east and 20 N acting to the west) are

(a)

E

W

Remember, when adding vectors, the tail of the second vector is placed at the head of the first. The resultant vector is from the tail of the first vector to the head of the second. A full explanation of one-dimensional and twodimensional vector addition is included in Appendix A.

North

80 N

View from above (b)

80 N 80 N

+

60 N

=

3F = 100 N

60 N

Figure 5.6 (a) Two perpendicular forces are acting on the trolley. (b) The vector addition of these two forces gives the resultant force (ΣF) that is acting on the trolley to be 100 N at 127°T. The trolley is treated as a point mass located at its centre of mass.


145

To find the direction of the resultant force, trigonometry must be used: 60 tanθ = = 0.75 80 θ = 37° This is a direction of 37° south of east, which is equivalent to a bearing of 127°T. Hence, the net force acting on the trolley is 100 N at a bearing of 127°T.

F

Worked example 5.1A

Figure 5.7 The golf ball moves in the direction of the applied force and is in the direction of the line joining the centre of the club-head with the centre of the ball. The force will be very large.

A motorboat is being driven west along the Yarra River. The engine is providing a driving force of 560 N towards the west. A frictional force of 180 N from the water and a drag force of 60 N from the air are acting towards the east as the boat travels along. a Draw a force diagram showing the horizontal forces of this situation. b Determine the resultant force acting on the motorboat.

Solution a

Physics file

560 N

For the vector addition, treat the boat as a point mass located at its centre of mass. 560 N

3F =

=

180 N

560 N 60 N

180 N

+

+

The resultant or net force acting on the motorboat is ΣF = 320 N due west.

Worked example 5.1B While playing at the beach, Sally and Ken kick a stationary beachball simultaneously with forces of 100 N south and 150 N west respectively. The ball moves as if it were only subjected to the net force. In what direction will it travel, and what is the magnitude of the net force on the ball?

Solution

E

220°T or S40°W

F

100 N

Motion

+

150 N

=

ΣF 100 N

S

Figure 5.8 The direction 220ºT lies 220º clockwise from north. This direction can also be written as S40ºW meaning 40º west of south.

60 N

3 F = 320 N

The net force is found by treating the beachball as a point mass and is given by: ΣF = FSally + FKen

N

W

b

E

W

180 N

There are two methods for describing the direction of a vector in a twodimensional plane. In each case, the direction has to be referenced to a known direction. A ‘full circle bearing’ describes north as ‘zero degrees true’—written as 0°T. In this convention, all directions are given as a clockwise angle from north. 90°T is 90° clockwise from north, i.e. due east. A force acting in a direction 220°T is acting in a direction of 220° clockwise from north. This is the method most commonly used in industry. An alternative method is to provide a quadrant bearing, where all angles are between 0° and 90° and so lie within one quadrant. The particular quadrant is identified using two cardinal directions, the first being either north or south. In this method, 220°T becomes S40°W, literally ‘40° west of south’.

146

60 N

150 N

ΣF = √1002 + 1502 = 180 N 150 tan θ = = 1.5 100 θ = 56° This is a quadrant bearing of 56° west of south, which is equivalent to a true bearing of 236°T. Hence, the net force acting on the beachball is 180 N in the direction 236°T.

Vector components It is often helpful to divide a force acting in a two-dimensional plane into two vectors. These two vectors are called the components of the force. This can be done because the force can be considered to act in each of the two directions at once. Consider, for example, the pulling force of 45 N acting on the cart shown in Figure 5.9. This pulling force is acting through the rope and is known as tension or a tensile force. The force is acting at an angle of 20° to the horizontal, so it has some effect in the horizontal direction and some effect in the vertical direction. The amounts of force acting in each direction are the components of the force. It is usual to construct a right-angled triangle around the force vector. The force vector is the hypotenuse of the triangle, and the adjacent and opposite sides become the components of the force. The horizontal and vertical components of the pulling force can then be determined using trigonometry. It is important to remember that there is only one pulling force acting on the cart, but this force can be treated as two component forces. So, the cart will move as though a horizontal force of 42 N pulling the cart along and a vertical pulling force of 15 N upwards were acting on it simultaneously. When the components are added together, the original 45 N force is the resultant force. Is this the most effective way of using a 45 N force to move the cart forwards? No, it would be slightly more effective if the 45 N force was acting in the horizontal direction. This would make the cart travel faster, but it may be impractical or inconvenient to apply the force in this way.

F = 45 N 20°

Figure 5.9 The pulling force acting on the cart has a component in the horizontal direction and a component in the vertical direction.

45 N

Fv = 45 sin 20° = 15 N

20° Fh = 45 cos 20° = 42 N

Figure 5.10 The magnitudes of the vector components Fh and Fv can be calculated using trigonometry.

Worked example 5.1C A stationary hockey ball is struck with a force of 100 N in the direction N30°W. What are the northerly and westerly components of this force?

Solution

westerly component of the force = 50 N N 30n

W

force from the hockey stick = 100 N

northerly component of the force = 87 N 30n

E

S


147

FW = 100sin30° = 50 N FN = 100cos30° = 87 N The ball moves as though forces of 50 N west and 87 N north were acting on it simultaneously.

Q

Worked example 5.1D F

Fv = 600 N down

Q Fh = 400 N

When walking, a person’s foot pushes backwards and downwards at the same time. While playing basketball, Kate’s foot pushes back along the court with a force of 400 N, and down with a force of 600 N. What is the actual force applied by Kate’s foot?

Solution 400 N horizontally and 600 N vertically downwards are the components of the force supplied by Kate’s foot. Therefore, the force she supplies will be F = Fhorizontal + Fvertical and a vector diagram is needed. Using Pythagoras’ theorem: F = √Fh2 + Fv2 = √4002 + 6002 = √520 000 = 721 N 600 θ = tan−1 = tan−11.5 = 56° 400 So Kate supplies a force of 721 N backwards at 56° down from the horizontal.

5.1 summary Force as a vector • A force is a push or a pull. Some forces act on contact while others can act at a distance. • Force is a vector quantity whose SI unit is the newton (N). • A vector can be represented by a directed line segment whose length represents the magnitude of the vector and whose arrowhead gives the direction of the vector. • The net force acting on a body that experiences a number of forces acting simultaneously is given by the vector sum of all the individual forces acting: ΣF = F1 + F2 + … + Fn

148

Motion

• A vector addition may be calculated using a sketch vector diagram that can be solved using trigonometry. • A force F acting at an angle θ to a given direction will have components F cos θ parallel to the reference direction, and F sin θ perpendicular to that reference direction.

5.1 questions Force as a vector 1 a Which one or more of the following quantities are vectors? A mass B velocity C temperature D force

7 Use trigonometry if necessary to add the following forces: a 3 N east and 4 N west b 60 N east and 80 N south

b Calculate the resultant force in each of the following vector additions: i 200 N up and 50 N down ii 65 N west and 25 N east iii 10 N north and 10 N south iv 10 N north and 10 N west

8 A small car is pulled by two people using ropes. Each person supplies a force of 400 N at an angle of 40° to the direction in which the car travels. What is the total force applied to the car? 400 N 40° 40°

2 If the force you have to exert when holding a small apple is about 1 N and holding a kilogram of sugar is 10 N, estimate the force required for: a using a stapler b kicking a beachball c lifting your school bag d doing a chin-up exercise.

400 N

9 Resolve the following forces into their perpendicular components around the north–south line. In part d, use the horizontal and vertical directions. a 100 N south 60° east b 60 N north c 300 N 160°T d 3.0 × 105 N 30° upward from the horizontal

3 Estimate the maximum force that you can exert when pulling horizontally on an anchored rope. What would be the approximate force that could be exerted by a ten-person tug-of-war team? 4 Which one or more of the following directions are identical? A 40°T and S40°E B 140°T and S40°E C 200°T and S20°W D 280°T and N80°W

10 What are the horizontal and vertical components of a 300 N force that is applied along a rope at 60° to the horizontal used to drag a Christmas tree across the backyard?

5 Convert the following into full circle bearings (i.e. °T). a N60°E b N40°W c S60°W d SE e NNE 6 Use the vectors below to determine the forces rep resented in the following situations. Scale: 1 cm represents 20 N a

b

c F

F

F


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5.2 (a)

Ideally you would expect the ball to reach this height.

(b)

Ball ideally would roll further and reach this position.

(c)

The ball ideally will keep moving with a constant velocity.

Figure 5.11 Galileo used a thought experiment to derive his law of inertia. This became Newton’s first law and stated that the natural state of bodies was to maintain their original motion. This contradicted Aristotle’s idea that the natural state of bodies was at rest.

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Motion

n

motio f o w a l t s r fi Newton’s Aristotle and Galileo

The first attempt to explain why bodies move as they do was made more than 2000 years ago by the Greek philosopher Aristotle. As discussed in Chapter 4, Aristotle and his followers felt that there was a natural state for matter and that all matter would always tend towards its natural place where it would be at rest. Aristotle’s thesis was based on the everyday observation that a moving body will always slow down and come to rest unless a force is continually applied. Try giving this book a (gentle) push along a table top and see what happens. Aristotle’s ideas were an attempt to explain the motion of a body as it was seen, but they do not help to explain why a body moves as it does. It was not until the early 17th century that Galileo Galilei was able to explain things more fully. Galileo performed experiments that led him to conclude that the natural state of a moving body is not at rest. Significantly, Galileo introduced the idea that friction was a force that, like other forces, could be added to other forces. A generation later, Newton developed Galileo’s ideas further to produce what we now call the first law of motion. To understand Newton’s first law, follow the logic of this thought experiment, similar to one used by Galileo. Consider a steel ball and a smooth length of track. In Figure 5.11a, the ball is held at one end of an elevated track; the other end of the track is also elevated. When the ball is released, it will roll downhill, then along the horizontal section and then up the elevated section. In reality, it will not quite reach the height it started at, due to friction. Now, imagine there were no friction. The ball, in this ideal case, would slow down as it rolled uphill and finally come to rest when it reached the height at which it started. Now consider what would happen if the angle of the elevated section was made smaller as in Figure 5.11b. The ball would now roll further along the track before coming to rest because the track is not as steep. If we could again imagine zero friction, the ball would again slow down as it rolls uphill and reach the same height before stopping, but travelling further in the process. What would happen if we made the angle of this elevated section progressively smaller? Logically, we would expect that, if we had enough track, the ball would travel even further before stopping. Now consider Figure 5.11c. Here the end of the track is not elevated at all. As the ball rolls along, there is nothing to slow it down because it is not going uphill. If we can ignore friction, what will the ball do as it rolls along the horizontal track? Galileo reasoned that it would not speed up, nor would it slow down. Ideally, the ball should keep travelling horizontally with constant speed and never reach its starting height. According to Galileo, the natural state of a body was to keep doing what it was doing. This tendency of objects to maintain their original motion is known as inertia. As the ball travels along the horizontal track, there is no driving force, nor is there any retarding force acting. The net force on the ball is zero and so it keeps moving with a constant velocity. This is the breakthrough in

understanding that Newton was able to make. Any body will continue with constant velocity if zero net force (ΣF = 0) acts upon it.

N…WTON’S FIRST LAW OF MOTION states that a body will either remain at rest or continue with constant speed in a straight line (i.e. constant velocity) unless it is acted on by an unbalanced force. A good example of inertia and Newton’s first law is illustrated by the air-track. With the air turned on, give a glider a gentle push along the track. It will travel along the track with a constant velocity as described by Newton’s first law. There are no driving or retarding forces acting on the glider, so it simply maintains its original motion. Aristotle’s laws would not be able to explain the motion of the glider.

Figure 5.12 An air-track glider moves with a constant velocity because there is zero net force acting on it. This is an example of inertia.

The motion of a spacecraft cruising in deep space is another good example of a body moving with constant velocity as required by Newton’s first law. As there is no gravitational force, and no air in space to retard its motion, the spacecraft will continue with constant speed in a straight line. The absence of air explains why there is no need to make a space probe aerodynamic in shape.

Physics file At the time of the Roman Empire some 2000 years ago, it cost as much money to have a bag of wheat moved 100 km across land as it did to transport it across the whole Mediterranean Sea. One of the reasons for this stemmed from the enormous friction that acted between the wheel and the axle in the cart of the day. Some animal fats were used as crude lubricants, but the effect was minimal. It has only been during the last century that engineering has provided a mechanical solution. Today’s wheels are connected to the axle by a wheel bearing. An outer ring is attached to the wheel, and an inner ring is attached to the axle. Separating the rings are a number of small ball bearings, which are able to roll freely between the rings. In this way, the area of contact and the friction between the wheel and axle is reduced dramatically.

Figure

5.13 B friction an all bearings reduc e de very effic nable wheels to w iently. ork

Physics file Several decades before Newton, Galileo Galilei had concluded that objects tended to maintain their state of motion. He called this tendency inertia, so this conclusion is also known as Galileo’s law of inertia. Inertia is not a force; it simply describes the property of bodies to continue their motion.

Figure 5.14 Two Voyager spacecraft were launched from Cape Canaveral in 1977 with the mission to investigate the outer planets of the solar system at close hand. Both craft completed the mission successfully, passing Saturn in 1981, Uranus in 1986 and Neptune in 1989. Voyager 1 and 2 have now left the solar system and since they have effectively zero net force acting on them, they continue to travel away from the Earth with a constant velocity.


151

Σ F= 0 velocity doesn’t change

FT

Fg

Figure 5.15 The forces acting on this book are balanced—i.e. ΣF = 0—so the velocity of the book will not change; the book will continue to stay at rest. That is, the book’s velocity will be constant at 0 m s−1.

Physics file Friction had a lot to do with Australia’s Stephen Bradbury winning a surprise gold medal at the 2002 Winter Olympics. His opponents did not experience enough friction as they skated around the home turn. Their inertia, in the absence of any horizontal forces, sent them crashing into the side wall. Stephen stayed upright, his skates cutting into the ice and producing enough friction to allow him to turn the corner and win the gold medal.

Forces in equilibrium Newton’s first law states that a body will travel with a constant velocity (or remain at rest) when the vector sum of all the forces acting on it is zero, i.e. when the net force is zero. When the net force is zero, the forces are said to be in equilibrium or balance. • If a body is at rest and zero net force acts on it, it will remain at rest. This applies to any stationary object such as a parked car or a book resting on a desk, as shown in Figure 5.15. In these cases, the velocity is zero and it is constant. The forces that are acting are balanced and so the net force is zero. • If a body is moving with a constant velocity and zero net force acts on it, it will continue to move with the same constant velocity. An example of this is a spacecraft with its engines off travelling at a great distance from the Earth. If gravitation is ignored, there is nothing to slow the craft down or to speed it up, and so it will continue with a constant velocity. The net force acting on the spacecraft is zero, so it will move with a constant velocity. Similarly, if a car is travelling along a road with a constant velocity, the vector sum of the forces acting on the car must be zero. The driving forces must balance the retarding forces, i.e. ΣF = 0. (a)

(b) no forces acting <3F = 0 velocity doesn’t change

forces are balanced <3F = 0 velocity doesn’t change

Fr Ff

Fd Fg

Figure 5.17 (a) There are no forces acting on the spacecraft. The net force is zero and so it continues to move with constant velocity. (b) The forces acting on the car are in equilibrium, so the net force is zero. The car continues to move with constant velocity.

Worked example 5.2A During a car accident, a passenger travelling without a fastened seatbelt may fly through the windscreen and land on the road. Explain, using Newton’s first law of motion, how this will occur. e dbury won th

ephen Bra arned Figure 5.16 St opponents le

his gold medal as inertia and friction (or out ab d an -h st fir of friction!). rather a lack

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Motion

Solution During the accident, the car is brought to rest suddenly. Any occupants of the car will continue to travel with the original speed of the car until a force acts to slow them down. If the seatbelt were fastened, this would provide the necessary force to slow the passenger within the car. In the absence of an opposing force, the passenger continues to move— often crashing through the windscreen. (The injuries received as a consequence of not wearing a seatbelt are usually far more serious than those received if the person were fixed in the car during an accident. This is why laws require seatbelts to be worn.)

lift F1

F2 drag

applied force

upwards force from ground on roller

Fthrust

weight, Fg friction

weight of roller

Fdrag

Fg F2

applied push Flift

Fg

Fg

F1

Fthrust

Fg + F1 + F2 = 0

upward force from ground

friction weight of roller

Fthrust + Fg + Fdrag + Flift = 0

Figure 5.18 In these examples, the forces acting on the objects are in equilibrium—the net force is zero. The body will either remain at rest, like the picture hanging on the wall, or continue with constant velocity, like the aircraft. A more complex situation involves the groundsman pushing the heavy roller with constant velocity. The horizontal component of the force he applies along the handle exactly balances the frictional force that opposes the motion of the roller in the horizontal direction. The vertical component of the applied force acts downwards, and adds to the weight of the roller, but these two downward forces are balanced by an upward force provided by the ground.

Worked example 5.2B A cyclist keeps her bicycle travelling with a constant velocity of 8.0 m s−1 east on a horizontal surface by continuing to pedal. A force (due to friction and air resistance) of 60 N acts against the motion. What force must be supplied by the rear wheel of the bicycle? v = 8 m s–1

Fdrag = 60 N

Fapplied = 60 N ΣF = Fapplied + Fdrag = 0

PRACTICAL ACTIVITY 21 Force and equilibrium

Solution If the cyclist is to continue at a constant 8.0 m s−1 east, then the forces that act on the bicycle must be in equilibrium, i.e. ΣF = 0. This means that the forces due to air resistance and friction are exactly balanced by the pedalling force. A force of 60 N east must be produced at the rear wheel. (The cyclist will actually have to produce more than 60 N as the gearing of the bike is designed to increase speed, not reduce the force that has to be applied.)


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Physics in action

Galileo Galilei—revolutionary Galileo Galilei was born into an academic family in Pisa, Italy, in 1564. Galileo made significant contributions to physics, mathematics and the scientific method through intellectual rigour and the quality of his experimental design. But more than this, Galileo helped to change the way in which the universe was understood. Galileo’s most significant contributions were in astronomy. Through his development of the refracting telescope he discovered sunspots, lunar mountains and valleys, the four largest moons of Jupiter (now called the Galilean moons) and the phases of Venus. In mechanics, he demonstrated that projectiles moved with a parabolic path and that different masses fall at the same rate (the law of falling bodies). These developments were most important because they changed the framework within which mechanics was understood. This framework had been in place since Aristotle had constructed it in the 4th century bc. By the 16th century, the work of the Greek philosophers had become entrenched, and it was widely supported in the universities. It was also supported at a political level. In Italy at that time, government was controlled by the Catholic Church. Today one would think that Galileo would have been praised by his peers for making such progress, but so ingrained and supported was the Aristotelian view that Galileo actually lost his job as a professor of mathematics in Pisa in 1592. Galileo was not without supporters, though, and he was able to move from Pisa to Padua where he continued teaching mathematics. At Padua, Galileo began to use measurements from carefully constructed experiments to strengthen his ideas. He entered into vigorous debate in which his ideas (founded as they were on observation) were pitted against the philosophy of the past and the politics of the day. The most divisive debate involved the motion of the planets. The ancient Greek view, formalised by Ptolemy in the 2nd century ad, was that the Earth was at the centre of the solar system and that all the planets, the Moon and the Sun were in orbit around it. This view was taught by the Church and was also supported by common sense. As such, it was accepted as the establishment view. In 1630 Galileo published a book in which he debated the Ptolemaic view and the new Sun-

Fi

gure 5.19 Ga red hair. Ga lileo Galilei (1564–1 642) was a lileo made sig sh of the force s that act o nificant contribution ort, active man with s to our un n moving b Newton wa ders odie sq This portra uick to acknowledge s. In his book Princip tanding it was draw his debt to ia, n 8 years b G efore Galile alileo’s genius. o’s trial. centred model proposed by Copernicus. On the basis of his own observations, Galileo supported the Copernican view of the universe. However, despite the book having been passed by the censors of the day, Galileo was summoned to Rome to face the Inquisition for heresy. The finding went against Galileo, and all copies of his book had to be burned and he was sentenced to permanent house arrest for the term of his life. Galileo died in 1642 in a village near Florence. He had become an influential thinker across Europe and the scientific revolution he had helped start accelerated in the freer Protestant countries in northern Europe. For its part, the Catholic Church under Pope John Paul II began an investigation in 1979 into Galileo’s trial, and in 1992 a papal commission reversed the Church’s condemnation of him.

5.2 summary Newton’s first law of motion • Aristotle theorised that the natural state of matter was to be at rest in its natural place. • Galileo performed experiments and from these developed the idea of inertia. • Newton developed Galileo’s ideas further and devised the first law of motion, stated as ‘A body

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Motion

will either remain at rest or continue with constant velocity unless it is acted on by a non-zero net force (or an unbalanced force).’ • Where the net force on a body is zero, i.e. ΣF = 0, the forces are said to be balanced and are in equilibrium.

5.2 questions Newton’s first law of motion 1 In just a few sentences, distinguish between the understandings held by Aristotle and Newton about the natural state of matter. Describe an experiment that might help support each of these views. 2 A billiard ball is rolling freely across a smooth horizontal surface. Ignore drag and frictional forces when answering these questions. a Which of the following force diagrams shows the horizontal forces acting on the ball according to the theories of Aristotle? b Which of the following force diagrams is correct for the ball according to the theories of Newton? c Which force diagram correctly describes this situation? A

B F

F

C

D F

F

3 If a person is standing up in a moving bus that stops suddenly, the person will tend to fall forwards. Has a force acted to push the person forwards? Use Newton’s first law of motion to explain what is happening. 4 What horizontal force has to be applied to a wheelie bin if it is to be wheeled to the street on a horizontal path against a frictional force of 20 N at a constant 1.5 m s−1? 5 When flying at constant speed at a constant altitude, a light aircraft has a weight of 50 kN down, and the thrust produced by the engines is 12 kN to the east. What is the lift force required by the wings of the plane, and what drag is acting?

6 A young boy is using a horizontal rope to pull his go‑kart at a constant velocity. A frictional force of 25 N also acts on the go-kart. a What force must the boy apply to the rope? b The boy’s father then attaches a longer rope to the kart because the short rope is uncomfortable to use. The rope now makes an angle of 30° to the horizontal. What is the horizontal component of the force that the boy needs to apply in order to move the kart with constant velocity? c What is the tension force acting along the rope that must be supplied by the boy? 7 Use Newton’s first law of motion to help explain the reasons for wearing a seatbelt in a car or aircraft. 8 Consider the following situations, and name the force that causes each object not to move in a straight line. a The Earth moves in a circle around the Sun with constant speed. b An electron orbits the nucleus with constant speed. c A cyclist turns a corner at constant speed. d An athlete swings a hammer in a circle with constant speed. 9 A magician performs a trick in which a cloth is pulled quickly from under a glass filled with water without the glass falling over or the water spilling out. a Explain the physics principles underlying this trick. b Does using a full glass make the trick easier or more difficult? Explain. 10 Which of these objects would find it most difficult to come to a stop: a cyclist travelling at 50 km h−1, a car travelling at 50 km h−1 or a fully laden semitrailer travelling at 50 km h−1? Explain.


155

tion

of mo w a l d n o c e s 5.3 Newton’s

Figure 5.20 This sprinter is about to leave the starting blocks. The starting blocks stop his foot from slipping backwards, increasing the size of the forward force acting on him and increasing his forward acceleration.

Physics file From Newton’s second law, it can be seen that the unit of a newton (N) is equivalent to the product of the mass unit (kg) and the acceleration unit (m s−2). In other words: N = kg m s−2. When writing the value of a force, either unit is correct; but newton is the SI unit and is obviously more convenient to use!

Newton’s first law of motion states that when all the forces on a body are balanced, the body can only remain at rest or continue with constant velocity. Newton’s second law of motion deals with situations in which a body is acted on by a non-zero net force, i.e. ΣF ≠ 0; in other words, when the forces are unbalanced. When there is a non-zero net force acting on a body, the body will accelerate in the direction of the net force. Newton explained that the rate of this acceleration will depend on both the size of the net force and the mass of the body. Experiments show that the acceleration produced is directly proportional to the size of the net force acting: a ∝ ΣF Experiments also show that the acceleration produced by a given net force depends on the mass of the body. We know that a greater mass has a greater inertia, so it will be more difficult to accelerate. Not surprisingly, experiments reveal that the acceleration produced by a particular force is inversely proportional to the mass of the body: 1 a∝ m If the two relationships are combined, we get: 1 a ∝ ΣF × m or SF a∝ m The relationship can be converted into an equality by including a constant of proportionality, so: SF a=k m By definition, 1 newton is the force needed to accelerate a mass of 1 kg at 1 m s−2. In the SI system of units, this makes the constant k equal to 1. The relationship is therefore simplified to ΣF = ma, a mathematical statement of Newton’s second law of motion.

N…WTON’S S…COND LAW OF MOTION states that the acceleration of a body, a, is directly proportional to the net force acting on it, ΣF, and inversely proportional to its mass, m: ΣF = ma The SI unit, the newton (N), will be required for force when the mass of the accelerating body is given in kilograms (kg) and its acceleration is provided in metres per second squared (m s−2). ΣF = ma is a vector equation in which the direction of the acceleration is in the same direction as the net force. If only one force acts, the acceleration will be in the direction of that force.

Worked example 5.3A Determine the size of the force required to accelerate an 80 kg athlete from rest to 12 m s−1 in a westerly direction in 5.0 s.

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Motion

Solution First, determine the acceleration of the athlete: v = u + at v-u a = t (12 - 0) = 5.0 a = 2.4 m s−2 west The net force can now be found using Newton’s second law: ΣF = ma = 80 × 2.4 = 190 N west If more than one force acts on a body, the acceleration will be in the direction of the net force, i.e. the vector sum of all of the forces.

PRACTICAL ACTIVITY 22 Newton’s second law

Worked example 5.3B A swimmer whose mass is 75 kg applies a force of 350 N as he begins a race. The water opposes his efforts to accelerate with a drag force of 200 N. What is his initial acceleration?

Solution The net force on the swimmer in the horizontal direction will be: ΣF = Fapplied + Fdrag A vector addition gives ΣF = 150 N forwards. So, SF 150 a = = m 75 = 2.0 m s−2 in the direction of the applied force

Fapplied + Fdrag

350 N 200 N ΣF

150 N

(It is worth noting that the drag applied by the water will increase with the swimmer’s speed.)

Worked example 5.3C A 150 g hockey ball is simultaneously struck by two hockey sticks. If the sticks supply a force of 15 N north and 20 N east respectively, determine the acceleration of the ball, and the direction in which it will travel.

Solution Remember to work in kilograms. Calculate the net force acting on the ball by performing a vector addition: ΣF = F1 + F2 ΣF = √F12 + F22 = √152 + 202 = √225 + 400 = √625 = 25 N SF 25 a = = m 0.15 = 170 m s−2 Looking at the vector diagram showing the addition of the forces, we can see that θ will be given by: F 20 tan θ = 2 = = 1.33 F1 15 so θ = tan−1 1.33 = 53° The ball will travel in the direction N53°E or (053ºT).

F1 = 15 N north

F2 = 20 N east F2

N

W

E

S

F1 ΣF θ


157

Physics file Mass is usually considered to be an unchanging property of an object. This is true in Newtonian mechanics where the speed with which an object is considered to travel matches everyday experience. However, at very high speeds, Newton’s laws of motion do not apply, and the theory of relativity must be used. In 1905, Albert Einstein showed that a body with a rest mass m0 (i.e. mass when stationary) will experience an increase in mass as it gets faster. This increase is usually undetectable except when the object nears the speed of light. At these very high speeds, the mass will become greater and greater, tending to infinity as the speed approaches the speed of light.

Figure 5.21 Scien tists working at th e Australian synchr otron in Clayton ha ve to take account of these relativistic effects on the ele ctrons they acce lerate to extreme speeds . Table 5.2 gives the mass of a 1 kg block if it were to travel at speeds of 0.1c (10% of the speed of light or 3 × 107 m s−1), 0.8c and 0.99c.

Table 5.2 The mass of a 1 kg block at different speeds Speed

1.0050 kg (i.e. 5 g increase)

0.8c

1.6667 kg (667 g increase) 7.1 kg (over 700% increase)

Relativistic mass increase provides the reason why no object can travel at the speed of light. To do so would require an infinite quantity of energy, since the mass of the body would itself be infinite. Only objects with no rest mass (such as light ‘particles’) can travel at the speed of light.

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Mass of a body To this point, the idea of the mass of an object has been taken for granted. However, the concept of a body’s mass is rather subtle and, importantly in physics, the mass of a body is a fundamentally different quantity from its weight—even though people (even physics teachers) tend to use these expressions interchangeably in everyday life. In earlier science courses, mass may have been defined as ‘the amount of matter in an object’. To understand what mass really is, this description says very little. The international standard for the kilogram is not very helpful either. Since the time of the French Revolution (late 1700s), the kilogram has been defined in terms of an amount of a standard material. At first, 1 litre of water at 4°C was used to define the kilogram. More recently an international mass standard has been introduced. This is a 1 kg cylinder of platinum–iridium alloy that is kept in Paris. Copies are made from the standard and sent around the world. Newton’s second law can help to provide a better understanding of mass through the effect of a force on a massive body. Think about a mass resting on a frictionless surface. If a force is applied to the mass in the SF horizontal direction, an acceleration is produced that is given by a = . m The greater the mass, the smaller will be the acceleration. If the mass is reduced, the acceleration will increase. Here the mass can be seen as the property of the body resisting the force. Mass is the closest quantity in physics to the concept of inertia. If the above experiment is repeated at another location, the same net force acting on the body will give the same acceleration regardless of where the experiment is performed. This is because—on Earth, on the Moon, in space—the mass of the body remains the same. Mass is a property of the body, and is not affected by its environment. In fact, for any situation at this level in physics, the mass of a body will be a constant value. As discussed in the adjacent Physics file, Albert Einstein was able to show in his theory of relativity that the mass of an object does change as its speed changes.

The (inertial) MASS of a body is its ability to resist acceleration when the body is acted on by a net force. Mass is a constant property of the body.

Mass

0.1c 0.99c

Mass and weight

Motion

on Earth a = 1 m s–2

in deep space F=1N

1 kg F=1N

1 kg

a = 1 m s–2

Figure 5.22 Regardless of the external conditions, the inertial qualities of a mass remain the same. A net force of 1 N will always produce an acceleration of 1 m s−2 for a 1 kg mass. In this way, mass can be understood as the resistance to a force. The greater the mass of the body, the smaller the acceleration caused by the force.

Weight of a body In the late 1500s, Galileo was able to show that all objects dropped near the surface of the Earth accelerate at the same rate, g, towards the centre of the Earth. The force that produces this acceleration is the force of gravity. In physics, the force on a body due to gravity is called the weight of a body, Fg or W. Consider a television of mass 50 kg and a banana of mass 0.10 kg that are dropped together from several metres above the surface of the Earth.

a = 9.8 m s–2

PRACTICAL ACTIVITY 23 Newton’s second law II

a = 9.8 m s–2

Figure 5.23 If air resistance is ignored, the TV and banana will fall side by side with an acceleration of 9.8 m s−2 towards the Earth.

The TV and the banana will fall with a uniform acceleration that is equal to 9.8 m s−2. The acceleration of a freely falling object (i.e. one where the only force that is acting is gravity) does not depend on the mass of the object. If we took the TV and the banana to the Moon and dropped them, they would fall with an acceleration of just 1.6 m s−2. Gravity is weaker on the Moon because it is much less massive than Earth.

a = 1.6 m s–2

a = 1.6 m s–2

Figure 5.24 The TV and banana will fall side by side with a uniform acceleration of 1.6 m s−2

towards the Moon.

The acceleration of a freely falling object due to gravity is known as g.


159

We will now use Newton’s second law to analyse the motion of the TV as it falls to the Earth. The only force acting on the TV is the force of gravity or weight, Fg. Hence: ΣF = Fg ma = Fg The acceleration of the TV is 9.8 m s−2 or g, so Fg = mg.

a = 9.8 m s–2

Fg

Figure 5.25 The TV is in free-fall. The only force acting on it is gravity, Fg, and it accelerates at 9.8 m s−2 towards the ground.

The W…IGHT of a body, W or Fg, is defined as the force of attraction on a body due to gravity: W = Fg = mg where m is the mass of the body (kg) g is the acceleration due to gravity (m s−2) g is also known as the GRAVITATIONAL FI…LD STR…NGTH. The unit of the gravitational field strength is newton/kg or N kg−1. The acceleration of a mass due to gravity is numerically identical to the gravitational field strength, g. These two quantities have different names and different units but are numerically equal. It can be shown that 1 m s−2 is equal to 1 N kg−1. As a consequence of this, the weight of a body will change as it is placed in different gravitational fields. On the Earth a 50 kg TV will have a weight of 50 × 9.8 = 490 N downwards. On the Moon, the gravitational field strength is lower at 1.6 N kg−1, and so the TV will be easier to lift since its weight is now only 50 × 1.6 = 80 N. In deep space, far from any stars or planets, where g = 0, the TV would be truly weightless, although its mass would still be 50 kg.

Why do heavy and light objects fall with equal acceleration? While Galileo was able to show that heavy and light objects fell at the same rate, he was not able to explain why. Newton, however, after stating his laws of motion, was able to show why this happens. We can use Newton’s second law to analyse the motion of the TV and the banana as they fall towards Earth.

50 kg 0.10 kg

Fg = 490 N

Fg = 490 N

Figure 5.26 The force dragging the TV to the ground is much larger than the force that is acting on the banana. However, the mass of the TV is much greater than the mass of the banana. The acceleration that results in both cases is identical: 9.8 m s−2 down.

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Motion

The force of gravity acting on the 50 kg TV is: W = Fg = mg = 50 × 9.8 = 490 N down This is the only force acting on the TV so the net force, ΣF, is also 490 N down. The acceleration of the TV can be calculated: SF 490 a= = = 9.8 m s−2 down m 50 The force of gravity acting on the 0.10 kg banana is: W = Fg = mg = 0.10 × 9.8 = 0.98 N down This is the only force acting on the banana so the net force, ΣF, is also 0.98 N down. The acceleration of the banana can be calculated: SF 0.98 a= = = 9.8 m s−2 down m 0.10 The TV has a large force dragging it towards the ground, but this large force is acting on a large mass. The banana has a small force acting on it, but this small force is moving a small mass. The acceleration produced in each case is exactly the same: 9.8 m s−2 down.

Worked example 5.3D A 1.5 kg trolley cart is connected by a cord to a 2.5 kg mass as shown. The cord is placed over a pulley and allowed to fall under the influence of gravity. a Assuming that the cart can move over the table unhindered by friction, determine the acceleration of the cart. b If a frictional force of 8.5 N acts against the cart, what is the magnitude of the acceleration now?

Figure 5.27 The weight of this boulder is the force it experiences due to gravity given by Fg = mg. This is approximately 2.5 × 105 N directed to the centre of the Earth. The mass of the boulder is approximately 25 000 kg. If the boulder were taken to outer space where the gravitational field strength was zero, the boulder would still have the same mass but no weight.

Solution a The cart and mass experience a net force equal to the weight of the falling mass. So ΣF = Fg = mg = 2.5 × 9.8 = 24.5 N down. This force has to accelerate not only the cart but the falling mass, and so the total mass to be accelerated is 1.5 + 2.5 = 4.0 kg. 1.5 kg

2.5 kg

SF m 24.5 = = 6.1 m s−2 to the right 4.0 b In analysing the forces that now act on the cart, the net force is: ΣF = 24.5 − 8.5 = 16 N to the right, SF 16 and a = = = 4.0 m s−2 to the right. m 4.0 a =

24.5 N 8.5 N


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Physics in action

Terminal velocity Galileo was able to show more than 400 years ago that the mass of a body does not affect the rate at which it falls towards the ground. However, our common experience is that not all objects behave in this way. A light object, such as a feather or a balloon, does not accelerate at 9.8 m s−2 as it falls. It drifts slowly to the ground, far slower than other dropped objects. Parachutists and skydivers also eventually fall with a constant speed. Why is this so? Skydivers, base-jumpers and air-surfers are able to use the force of air resistance to their advantage. As a basejumper first steps off, the forces acting on him are drag (air resistance), a, and gravity, g. Since his speed is low, the drag force is small (Figure 5.28a). There is a large net force downwards, so he experiences a large downwards acceleration of just less than 9.8 m s−2, causing him to speed up. This causes the drag force to increase because he is colliding harder with the air molecules. In fact, the drag force increases in proportion to the square of the speed: ∝ v2. This results in a smaller net force downwards a (Figure 5.28b). His downwards acceleration is therefore reduced. It is important to remember that he is still speeding up, but at a reduced rate. As his speed continues to increase, so too does the magnitude of the drag force. Eventually, the drag force becomes as large as the weight force (Figure 5.28c). When this happens, the net force is zero and the base-jumper will fall with a constant velocity. Since the velocity is now

(a)

(c)

Fa ΣF

Fa v

v

Fg

Fg (b)

(d) Fa

v

Fa

ΣF

ΣF = 0 v

Fg

ΣF = 0

Fg

Figure 5.28 As the base-jumper falls, the force of gravity does not change, but the drag force increases as he travels faster. Eventually these two forces will be in equilibrium and he will fall with a constant or terminal velocity.

constant, the drag force will also remain constant and the motion of the jumper will not change (Figure 5.28d). This velocity is commonly known as the terminal velocity. For skydivers, the terminal velocity is usually around 200 km h−1. Opening the parachute greatly increases the air resistance force that is acting, resulting in a lower terminal velocity. This is typically around 70 km h−1.

Figure 5.29 Skydivers can change their speed by changi ng their body profile. If they spreadeagle they will fall slo assume a tuck position the wer. This enables them to me y will fall faster and if they et and form spectacular pat terns as they fall.

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Motion

5.3 summary Newton’s second law of motion • Newton’s second law of motion states that the acceleration a body experiences is directly pro portional to the net force acting on it, and inversely proportional to its mass: ΣF = ma where m is measured in kilograms (kg), a is measured in metres per second squared (m s−2), and ΣF in newtons (N).

• The mass of a body can be considered to be its ability to resist a force. Mass is a constant property of the body and is not affected by its environment or location. • The weight of a body W or Fg is defined as the force of attraction on a body due to gravity. This will be given by W = Fg = mg where m is the mass of the body and g is the strength of the gravitational field.

5.3 questions Newton’s second law of motion Use g = 9.8 m s−2 when answering these questions. 1 State whether the forces are balanced (B) or unbalanced (U) for each of these situations. a A netball falling towards the ground b A stationary bus c A tram travelling at a constant speed of 50 km h–1 d A cyclist slowing down at a traffic light 2 During a tennis serve, a ball of mass 0.060 kg is accelerated to 30 m s–1 from rest in just 7.0 ms. a Calculate the average acceleration of the ball. b What is the average net force acting on the ball during the serve? 3 Use Newton’s laws to explain why a 1.0 kg shot-put can be thrown further than a 1.5 kg shotput. 4 In a game of soccer, the ball is simultaneously kicked by two players who impart horizontal forces of 100 N east and 125 N south on the ball. Determine: a the net force acting on the ball b the direction in which the ball will travel c the acceleration of the ball if its mass is 750 g. 5 When travelling at 100 km h−1 along a horizontal road, a car has to overcome a drag force due to air resistance of 800 N. If the car has a mass of 900 kg, determine the average force that the motor needs to apply if it is to accelerate at 2.0 m s−2. 6 Mary is paddling a canoe. The paddles are providing a constant driving force of 45 N south and the frictional forces total 25 N north. The mass of the canoe is 15 kg and Mary has a mass of 50 kg. a What is Mary’s mass? b Calculate Mary’s weight.

c Find the net horizontal force acting on the canoe. d Calculate the magnitude of the acceleration of the canoe. 7 On the surface of the Earth, a geological hammer has a mass of 1.5 kg. Determine its mass and weight on Mars where g = 3.6 m s−2. 8 What is the average force required of the brakes on a 1200 kg car in order for it to come to rest from 60 km h−1 in a distance of 150 m? 9 Consider a 70 kg parachutist leaping from an aircraft and taking the time to reach terminal velocity before activating the parachute. Draw a sketch graph of the net force against time for the parachutist in the period from the start of the jump until terminal velocity has been reached. Explain your reasoning. 10 A 0.50 kg metal block is attached by a piece of string to a dynamics cart as shown below. The block is allowed to fall from rest, dragging the cart along. The mass of the cart is 2.5 kg. 2.5 kg

0.50 kg

a If friction is ignored, what is the acceleration of the block as it falls? b How fast will the block be travelling after 0.50 s? c If a frictional force of 4.3 N acts on the cart, what is its acceleration?


163

on

5.4 Newton’s third law of moti

Newton’s first two laws of motion describe the motion of a body resulting from the forces that act on that body. The third law is easily stated, and seems to be widely known by students, but is often misunderstood and misused! It is a very important law in physics, and assists our understanding of the origin and nature of forces. Newton realised that all forces exist in pairs and that each force in the pair acts on a different object. Consider the example of a hammer gently tapping a nail. Both the hammer and nail experience forces during this. The nail experiences a small downwards force as the hammer hits it and this moves it a small distance into the wood. The hammer experiences a small upwards force as it hits the nail causing the hammer to stop. These forces are known as an action–reaction pair and are shown in Figure 5.31(a). (a)

Figure 5.30 A hammer hitting a nail is a good

example of an action–reaction pair and Newton’s third law.

(b)

force that nail exerts on hammer

force that hammer exerts on nail

force that nail exerts on hammer

force that hammer exerts on nail

Figure 5.31 (a) As the hammer gently taps the nail, both the hammer and nail experience small forces. (b) When the hammer smashes into the nail, both the hammer and nail experience large forces.

Now consider what will happen if the hammer is smashed into the nail. The nail now experiences a large downwards force as the hammer smashes into it and this moves it a larger distance into the wood. The hammer itself experiences a large upwards force as it hits the nail, again causing the hammer to stop. You should notice that the forces acting on the hammer and nail are both larger, as shown in Figure 5.31b. This is what Newton realised. If the hammer exerted a downwards force of 25 N on the nail, the nail would exert an upwards force of 25 N on the hammer.

N…WTON’S THIRD LAW OF MOTION states that for every action force (object A on B), there is an equal and opposite reaction force (object B on A): F(A on B) = −F(B on A) When body A exerts a force F on body B, body B will exert an equal and opposite force –F on body A. It is important to recognise that the action force and the reaction force in Newton’s third law act on different objects and so should never be added together. Figure 5.32 shows some examples of action–reaction forces. It is also important to understand that even though action–reaction forces are always equal in size, the effect of these forces may be very different. A good example of this is the collision between the bus and the car shown in 5.32c. Because of the car’s small mass, the force acting on the car will

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Motion

(a)

(b)

force that foot exerts on ball

force that floor exerts on girl

force that girl exerts on floor

force that ball exerts on foot

(c)

(d) gravitational force that Earth exerts on brick (Fg)

force that bus exerts on car gravitational force that brick exerts on Earth force that car exerts on bus

Figure 5.32 Some action–reaction force pairs. Notice that these can be contact or non-contact forces. (a) The girl exerts a downwards force on the floor and the floor exerts an equal upwards force on the girl. (b) The foot exerts a forwards force on the ball and the ball exerts an equal size backwards force on the foot. (c) The bus exerts a backwards force on the car and the car exerts an equal size forwards force on the bus. (d) The Earth exerts a downwards gravitational force on the brick and the brick exerts an equal size upwards force on the Earth.

cause the car to undergo a large acceleration backwards. The occupants may be seriously injured as a result of this. The force acting on the bus is equal in size, but is acting on a much larger mass. The bus will have a relatively small acceleration as result and the occupants will not be as seriously affected.

Figure 5.33 The soccer ball and the player’s head exert equal forces on each other during this collision, but only the player will experience pain!

Worked example 5.4A In each of the following diagrams, one of the forces is given. (a)

i ii iii

(b)

(c)

Describe each given force using the phrase ‘force that exerts on ’. Describe the reaction pair to the given force using the phrase ‘force that exerts on ’. Draw each reaction force on the diagram, carefully showing its size and location.


165

Solution a i force that bat exerts on ball ii force that ball exerts on bat iii see diagram at right

b

i ii iii

c iii ii iii

force that ball exerts on floor force that floor exerts on ball see diagram at right

gravitational force that Earth exerts on astronaut gravitational force that astronaut exerts on Earth see diagram at right

Motion explained

F (reaction)

Fforwards

θ Fup F (action)

θ

θ

F (reaction)

Figure 5.34 Walking relies on an action and reaction force pair in which the foot will push down and backwards with an action force. In response, the ground will push upwards and forwards. The forward component of the reaction force is actually friction. This is responsible for the body moving forward as a whole, while the back foot remains at rest.

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Motion

Newton’s third law also explains how we are able to move around. In fact, the third law is needed to explain all locomotion. Consider walking. Your leg pushes backwards with each step. This is an action force acting on the ground. As shown in Figure 5.34, a component of the force acts downwards and another component pushes backwards horizontally along the surface of the Earth. The force is transmitted because there is friction between your shoe and the Earth’s surface. In response, the ground then pushes forwards on your leg. This forwards component of the reaction force enables you to move forwards. In other words, it is the ground pushing forwards on your leg that moves you forwards. It is important to remember that in Newton’s second law, ∑F = ma, the net force ∑F is the sum of the forces acting on the body. This does not include forces that are exerted by the body on other objects. When you push back on the ground, this force is acting on the ground and may affect the ground’s motion. If the ground is firm, this effect is usually not noticed, but if you run along a sandy beach, the sand is clearly pushed back by your feet. The act of walking relies on there being some friction between your shoe and the ground. Without it, there is no grip and it is impossible to supply the action force to the ground. Consequently, the ground cannot supply the reaction force needed to enable forward motion. Walking on smooth ice is a good example of this, and so mountaineers will use crampons (basically a rack of nails) attached to the soles of their boots in order to gain purchase in icy conditions. The situation outlined above is fundamental to all motion.

Table 5.3 All motion can be explained in terms of action and reaction force pairs Motion

Action force

Reaction force

Swimming

Hand pushes back on water

Water pushes forwards on hand

Jumping

Legs push down on Earth

Earth pushes up on legs

Bicycle/car

Tyre pushes back on ground

Ground pushes forwards on tyre

Jet aircraft and rockets

Hot gas is forced backwards out of engine

Gases push craft forwards

Skydiving

Force of gravitation on the skydiver from Earth

Force of gravitation on Earth from skydiver

Worked example 5.4B A front-wheel drive car of mass 1.2 tonne accelerates from traffic lights at 2.5 m s–2. a Discuss and identify the horizontal force that enables the car to accelerate forwards. b Draw this force on the diagram at right. Label it force A. c Carefully draw the reaction pair to this force as described in Newton’s third law. Label it force B. d Discuss and identify the reaction force that you have drawn.

1200 kg

Solution a This is the forwards force that the road exerts on the front tyre and could be called a frictional force.

b force A

c force B

d This is the backwards force that the front tyre exerts on the road. If the road surface was ice, both of the forces in (b) and (c) would be very small and the car would not be able to drive forwards.

The normal force When an object, say a rubbish bin, is allowed to fall under the influence of gravity, it is easy to see the effect of the force of gravity. As shown in Figure 5.35a, the only force acting is the weight, so the net force is the weight, and the bin therefore accelerates at g. When the bin is at rest on a table, the force of gravity (Fg = W = mg) is still acting. Since the bin is at rest, there must be another force (equal in magnitude and opposite in direction) acting to balance the weight. This upwards force is provided by the table. Because of the weight of the bin, the table is deformed a little, and being elastic, it will push upwards. The elastic force it provides is perpendicular to its surface and is called a normal reaction force FN or N, (often abbreviated to the normal force).

The NORMAL FORC… is a reaction force supplied by a surface at 90° to its plane.


167

This means that there are two forces that act on the bin which will completely balance each other so that the net force is zero. In Figure 5.35b, the bin is empty so its weight is small and the normal force is also small. However, when the bin fills up, its weight increases and so too does the normal force. Figure 5.35 (a) When the bin is in mid-air, there is an unbalanced force acting on it so it accelerates towards the ground. (b) When the empty bin is resting on the table, there is a small upwards force from the table acting on it. The bin remains at rest, so the forces are balanced. (c) The forces acting on the full bin are also balanced. The weight of the bin is greater, so the normal force exerted by the table is larger than for the empty bin.

(a)

(b)

(c)

Fg

FN

FN

Fg

Fg

The normal force and the weight force in these examples are equal and opposite. However, this does not mean that they are an action–reaction pair as described by Newton’s third law! The weight force and the normal force act on the same body (the bin) so they cannot be an action–reaction force pair. Remember that in Newton’s third law, one force acts on one object and the other force acts on the other object. Let us identify the reaction pair to each of the forces shown in Figure 5.36. (a)

Figure 5.36 (a) The reaction force pair to the weight of the bin is the gravitational force of attraction that the bin exerts on the Earth. (b) The reaction force pair to the normal force on the bin is the force that the bin exerts on the table. The force pairs are equal and opposite but they do not cancel out because they are not acting on the same object.

ΣF = Fg + FN = 0 FN

Fg

(b) Fg =

FG =

force Earth exerts on bin

force bin exerts on Earth

FN =

force table exerts on bin

force bin exerts FT = on table

In Figure 5.36a the action force shown is the force of gravity Fg on the bin. This is the attractive force that the Earth exerts on the bin. The reaction force, therefore, is the attractive force that the bin exerts on the Earth. This is shown as FG. In Figure 5.36b, the action force shown is the normal force on the bin, FN. This is the force that the table exerts on the bin. So the reaction force is the force that the bin exerts on the table. This is shown as FT.

Worked example 5.4C An 8.0 kg computer rests on a table. a Identify the forces that act on the computer. b Determine the magnitude and direction of the force that the computer exerts on the table. c If a 3.0 kg monitor is placed on the computer box, determine the new normal force acting on the computer.

Solution a The weight of the computer will be Fg = mg = 8.0 × 9.8 = 78 N down, so that if the net force on the computer is zero, the normal force supplied by the table must be 78 N upwards: ΣF = Fg + FN = 0.

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Motion

b

Since the force that the table exerts on the computer has been found to be 78 N up, then, according to Newton’s third law, the force that the computer exerts on the table must be 78 N down. c If a 3.0 kg monitor is placed on top of the computer, the table must supply a further 3.0 × 9.8 = 29 N, bringing the total normal force to 107 N. (The computer will also have to provide a normal force of 29 N upwards to balance the weight of the monitor.)

The inclined plane The Guinness Book of Records identifies the steepest road in the world as being at an angle of 20° to the horizontal. It is located in Dunedin, New Zealand. Living on such a road requires the residents to ensure that the handbrake in their car is always in good repair! To determine the force required by the handbrake of a car parked on this steep road, the physics of forces acting on a body on an inclined plane must be used. Start by thinking of a body at rest on a horizontal surface. Two forces act on the body: the weight of the body Fg and the normal force FN supplied by the surface. The weight force always acts downwards and is given by Fg = mg. The normal force is supplied by the surface and will vary depending on the situation, but it will always act upwards and perpendicular to the surface. This means that the net force on the body will be the sum of these two forces, and in this case it has to be zero since the body does not move. If the surface is tilted so that it makes an angle to the horizontal, the weight force remains the same: Fg = mg. However, the normal force continues to act at right angles to the surface and will change in magnitude, getting smaller as the angle increases. If there is no friction between the body and the surface, the two forces will not balance and a non-zero net force will be directed down the incline as shown in Figure 5.37.

PRACTICAL ACTIVITY 24 Acceleration down an incline

FN

FN Fg

FN

body remains at rest Fg

You may have observed that the mass of an object can be used in two different contexts. First, mass is a measure of the ability of an object to resist being accelerated by a force. This mass can be determined from Newton’s second law and is known as inertial mass. Second, mass can give an indication of the degree to which an object experiences a gravitational force when in the presence of a gravitational field. This mass is known as gravitational mass. Some very accurate experiments have shown that the inertial and gravitational masses are equal, although there is no theoretical reason why this should be the case.

ΣF = Fg + FN

ΣF = Fg + FN =0 FN

Physics file

Fg θ

ΣF θ

Fg

Figure 5.37 (a) Where the surface is perpendicular to the weight force, the normal force will act directly upwards and cancel the weight force. (b) On an inclined plane, FN is at an angle to Fg and as long as no friction acts, there will be a net force down the incline. The body will accelerate.

From the vector diagram of the forces: ΣF = Fg + FN = Fg sin θ = mg sin θ From Newton’s second law, the net force is: ΣF = ma so, ma = mg sin θ or a = g sin θ This means that the acceleration down an incline is a function of the angle of the incline alone, and not the mass of the body. Ignoring any friction, a car rolling down the steep street in Dunedin will accelerate at a = g sin θ = 9.8 × sin 20° = 3.4 m s−2—quite a rate!


169

Worked example 5.4D

Physics file In February 2003, a train driver pulled into Broadmeadows station and went for a toilet break. Unfortunately, he forgot to put on the handbrake. When he returned, the train was rolling away from the platform, heading for Jacana. It is a slight downhill incline from Broadmeadows to Jacana and the train simply rolled off down the hill, accelerating gradually. After Glenroy, the incline of the track is greater and the train’s acceleration increased. It is estimated that it reached speeds of around 100 km h–1 at times. Fortunately, no-one was injured as it hurtled through level crossings and stations on its way into the city. At Spencer Street, it crashed into a stationary V/Line train at around 60 km h−1. The express trip from Broadmeadows to Spencer Street took about 16 minutes. Broadmeadows Jacana Glenroy Oak Park Pascoe Vale Glenbervie Strathmore Moonee Ponds

Essendon Ascot Vale

3F = Fg + FN

FN

FN 30n

Fg

30n

Fg

Solution ΣF = Fg + FN From the vector diagram: ΣF = Fg sin θ So, ma = mg sin θ a = g sin 30° = 9.8 × sin 30° = 4.9 m s−2 down the road If friction exists between a body on an inclined plane and the surface, its direction will be along the incline but against the motion. If a frictional force is great enough to balance the sum of the normal force and the weight of the body, the net force is zero and the body will either travel with a constant velocity or remain stationary. Worked example 5.4E illustrates this.

Worked example 5.4E

Newmarket Kensington North Melbourne Spencer Street Station

0

Driver error allows a 5 tonne truck to roll down a steep road inclined at 30° to the horizontal. As it is a high-technology vehicle, there is negligible friction between the wheels of the truck and the wheel bearings. Find the acceleration of the truck if the acceleration due to gravity is taken as 9.8 m s−2.

5 km

Figure 5.38 An empty three-carriage train took 16 minutes to roll downhill from Broadmeadows to Spencer Street (now Southern Cross) station. The track was like a long inclined plane and the train accelerated along it after the driver forgot to put the handbrake on!

Kristie is a 60 kg skier. At the start of a ski slope which is at 35° to the horizontal, she crouches into a tuck. The surface is very icy, so there is no friction between her skis and the ice. Ignore air resistance when answering these questions. a If Kristie starts from rest, what is her speed after travelling a distance of 80 m on the ice? b The snow conditions change at the end of the ice patch so that Kristie continues down the slope with a constant velocity. What is the force due to friction that must be acting between Kristie’s skis and the snow?

Solution a The net force on Kristie will be ΣF = Fg + FN. This is a vector addition. From the vector diagram: ΣF = Fg sin θ So, a = g sin 35° = 9.8 sin 35° = 5.6 m s−2

3F FN Fg 35n

ice patch Fg

80 m 35n

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Motion

FN

If this acceleration continues over 80 m, Kristie’s final speed would be: v2 = u2 + 2ax so v2 = 0 + 2 × 5.6 × 80 = 900 v = 30 m s−1 (110 km h−1) b Kristie is travelling with a constant velocity, so ΣF = 0, i.e. the force of friction Ff would balance the component of the weight force parallel to the incline. So Ff = Fg sin 35 = 340 N up the incline.

Tension Another force that is experienced in everyday life is the tension force that is found in stretched ropes, wires, cables and rubber bands. If you stretch a rubber band, it will exert a restoring force on both your hands. This force is known as a tensile force and is present in any material that has been stretched. Consider the situation in which a person hangs from a cable that is tied to a beam as shown in Figure 5.39a. We will assume that the mass of the cable is negligible. (a)

(b) tension FT

tension FT

FT = 490 N

3F = 0

M = 50 g

Fg = 490 N

Figure 5.39 (a) The stretched cable exerts an upwards force on the person and an equal size downwards force on the beam. (b) The forces acting on the person are balanced, so the tension force is equal in magnitude to the weight force (490 N).

At the top end of the cable, the tension force FT pulls down on the beam. At the other end, the cable exerts an upwards force FT on the person, holding them in mid-air. In other words, the same size tension force acts at both ends of the cable, but in opposite directions. If a second identical person also hung from the cable, the tension acting at both ends of the cable would double and the cable would become tauter (and more likely to snap!) If the mass of the person in Figure 5.39a is known, the size of the tension can be determined. If the mass of the person is 50 kg, then the forces acting on them are, as shown in Figure 5.39b, an upwards tension force and the downwards pull of gravity of 490 N. If the person is at rest, the forces acting are balanced and so the upwards tensile force acting from the cable must also be 490 N. Calculations involving tension are illustrated in the following example. A naughty monkey of mass 15 kg has escaped backstage in a circus. Nearby is a rope threaded through an ideal (frictionless and massless) pulley. Attached to one end of the rope is a 10 kg bag of sand. The monkey climbs a ladder and jumps onto the free end of the rope.


171

The system of the rope and the monkey is now subjected to a net force of: ΣF = 15 × 9.8 − 10 × 9.8 = 49 N down on the side of the monkey. As a consequence, both objects and the rope will accelerate at: SF 49 a= = = 2.0 m s−2 m 25 (a)

(b)

(c) FT = 117 N

FT = 118 N

15 kg a

147 N

ΣF = 15 × 2.0 = 30 N

ΣF = 10 × 2.0 = 20 N

10 kg 98 N

Fg =147 N

Fg = 98 N

ΣF = 49 N

Figure 5.40 The monkey has a greater weight than the sandbag, and so the rope will accelerate in the direction of the monkey. The tension in the rope is found by considering the forces that act on each weight.

While all of this is occurring, the rope is under tension. To find the amount, we look at the forces acting on one of the masses (Figure 5.40c). Take the monkey: the net force on the monkey will be ΣF = Fg + FT. So, FT = ΣF − Fg = 15 × 2.0 down − 15 × 9.8 down = 30 N down + 147 N up = 117 N up To check, find the tension acting on the sandbag. Again, ΣF = Fg + FT So, FT = ΣF − Fg = 10 × 2.0 up − 10 × 9.8 down = 20 N up + 98 N up = 118 N up Note that the small difference between the two results is due to rounding error. The tension is equal on both sides of the pulley, regardless of how the masses are arranged (provided the pulley is frictionless). Intuitively, you might have thought that the tension would have been (15 + 10) × 9.8 = 245 N since the two weights are pulling in opposite directions. This is not the case because the system is allowed to accelerate, causing a reduction in tension.

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Motion

Physics in action

Frictional forces Friction is a force that opposes movement. Suppose you want to push your textbook along the table. This simple experiment can reveal a significant amount of information about the nature of friction. As you start to push the book, you find that, at first, the book does not move. You then increase the force that you apply, and suddenly, at a certain critical value, the book starts to move relatively freely. The maximum frictional force resists the onset of sliding. This force is called the static friction force, . Once the book has begun to slide, a much lower force s than s is needed to keep the book moving. This force is called the kinetic friction force, and is represented Figure 5.41 by k. Th track, so th is magnetic levitation This phenomenon can be understood when it is e frictional train in Chin forc a magnetic fo realised that even the smoothest surfaces are quite rce to a cru es are negligible. The rides 1 cm above the ising speed train is pro jagged at the microscopic level. When the book is pelled by a of about 43 0 km h −1. resting on the table, the jagged points of its bottom surface have settled into the valleys of the surface of the table, and this helps to resist attempts to try to slide the problem. Consider the moving parts within the engine of a book. Once the book is moving, the surfaces do not have any car. Friction can rob an engine of its fuel economy and cause time to settle into each other, and so less force is required it to wear out. Special oils and lubricants are introduced in keep it moving. order to prevent moving metal surfaces from touching. If the Another fact that helps to explain friction arises from moving surfaces actually moved over each other they would the forces of attraction between the atoms and molecules of quickly wear, producing metal filings that could damage the the two different surfaces that are in contact. These produce engine. Instead, both metal surfaces are separated by a thin weak bonding between the particles within each material; layer of oil. The oils are chosen on the basis of their viscosity before one surface can move across the other, these bonds (thickness). For example, low viscosity oils can be used in must be broken. This extra effort adds to the static friction the engine while heavier oils are needed in the gear box and force. Once there is relative motion between the surfaces, the differential of the car where greater forces are applied to the bonds cannot re-form. moving parts. In everyday life, there are situations in which friction is At other times, we want friction to act. When driving to desirable (e.g. walking) and others in which it is a definite the snowfields, if there is any ice on the road, drivers are Applied force required to fit chains to their cars. When driving over a patch maximum of ice, the chain will break through the ice, and the car is static friction again able to grip the road. Similarly, friction is definitely Fs required within the car’s brakes when the driver wants to slow down. In fact modern brake-pads are specially designed kinetic friction to maximise the friction between the pads and the brake Fk drum or disk. When a car is braking in a controlled fashion, the brakepads grip a disk that is attached to the wheel of the car. The retarding force, applied through friction, slows the disk and Time hence the car will come to rest. If the brakes are applied too strongly they may grab the disk, locking up the wheels in the process. The car then slides over the road, with two undesirable consequences. First, the car usually takes about motion Fk Fs 20% longer to come to rest. This is because the car is relying on the kinetic frictional force between the tyres and the road to stop. As was seen when pushing the book over the desk, this force is less than the static friction force. The other consequence is that the car has lost its grip with the road, and so the driver can no longer steer the car. Most cars now overcoming friction constant velocity employ anti-lock braking systems (ABS) to overcome the possibility of skidding. This is achieved by using feedback Figure 5.42 To get things moving, the static friction between an object systems that automatically reduce the pressure applied by the and the surface must be overcome. This requires a larger force than brake-pads regardless of the pressure applied by the driver to that needed to maintain constant velocity. the brake pedal.

PHYSICS 11

PHYSICS 11


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5.4 summary Newton’s third law of motion • Newton’s third law of motion states that for every action force, there is an equal and opposite reaction force: F(A on B) = −F(B on A) • Action and reaction forces act on different objects and so should never be added together. • Whenever a force acts against a fixed surface, the surface provides a normal force, FN or N, at right angles to the surface. The size of the normal force depends on the orientation of the surface to the contact force.

• All locomotion is made possible through the existence of action and reaction force pairs. • On smooth inclined planes at an angle of θ to the horizontal, objects will move with an acceleration of a = g sin θ. • Materials that have been stretched, such as ropes, cables and rubber bands, exert tensile forces on the objects to which they are attached. These forces are equal and opposite.

5.4 questions Newton’s third law of motion 1 Determine the action and reaction forces involved when: a a tennis ball is hit with a racquet b a pine cone falls from the top of a tree towards the ground c a pine cone lands on the ground d the Earth orbits the Sun. 2 A 70 kg fisherman is quietly fishing in a 40 kg dinghy at rest on a still lake when, suddenly, he is attacked by a swarm of wasps. To escape, he leaps into the water and exerts a horizontal force of 140 N on the boat. a What force does the boat exert on the fisherman? b With what acceleration will the boat move initially? c If the force on the fisherman lasted for 0.50 s, determine the initial speed attained by both the man and the boat. 3 A 100 kg astronaut (including the space suit) becomes untethered during a space walk and drifts to a distance of 10 m from the mother ship. To get back to the ship, he throws his 2.5 kg tool kit away with an acceleration of 8.0 m s−2 that acts over 0.50 s. a How does throwing the tool kit away help the astronaut in this situation? b How large is the force that acts on the tool kit and the astronaut? c With what speed will the astronaut drift to the mother ship? d How long will it take for the astronaut to reach the ship?

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Motion

4 A 2.0 kg bowl strikes the stationary ‘jack’, which has a mass of 1.0 kg, during a game of bowls. It is a head-on collision, and the acceleration of the jack is found to be 25 m s−2 north. What is the acceleration of the bowl? 5 a A ball rolls down an incline as shown in diagram (a). Which one of the following best describes the speed and acceleration of the ball?

(a)

(b)

A The speed and acceleration both increase. B The speed increases and the acceleration is constant. C The speed is constant and the acceleration is zero. D The speed and acceleration are both constant. b A ball rolls down the slope shown in diagram (b). Which one of the following best describes its speed and acceleration? A Its speed and acceleration both increase. B Its speed and acceleration both decrease. C Its speed increases and its acceleration decreases. D Its speed decreases and its acceleration increases.

6 During the Winter Olympics, a 65 kg competitor in the women’s luge has to accelerate down a course that is inclined at 50° to the horizontal. a Name the forces acting on the competitor. b Ignoring friction (because it’s an icy slope), deter mine the magnitude and direction of the forces that act. c Determine the magnitude of the net force on the competitor. d What acceleration will the competitor experience? 7 A cyclist is coasting down a hill that is inclined at 15° to the horizontal. The mass of the cyclist and her bike is 110 kg, and for the purposes of the problem, no air resistance or other forces are acting. After accelerating to the speed limit, she applies the brakes a little. What braking force is needed for her to be able to travel with a constant velocity down the hill? 8 Discuss and compare the size of the normal force that acts on the skater as he travels down the halfpipe from A to B to C as shown.

x

x x

9 Two students, James and Tania, are discussing the forces acting on a lunchbox that is sitting on the laboratory bench. James states that a weight force and a normal force are acting on the lunchbox and that since these forces are equal in magnitude but opposite in direction, they comprise a Newton’s third law action–reaction pair. Tania disagrees saying that these forces are not an action–reaction pair. Who is correct and why? 10 A rope is allowed to move freely over a ‘frictionless’ pulley backstage of a theatre. A 30 kg sandbag, which is at rest on the ground, is attached at one end. A 50 kg work-experience student, standing on a ladder, grabs onto the other end of the rope to lower himself. a What is the net external force on the rope? b With what acceleration will the system move? c What is the tension in the rope?

A

B

C

chapter review 1 A boxer receives a punch to the head during a training session. His opponent is wearing boxing gloves. Which of the following is correct? A The force that the glove exerts to the head is greater than the force that the head exerts on the glove. B The force that the glove exerts to the head is less than the force that the head exerts on the glove. C The force that the glove exerts to the head is always equal to the force that the head exerts on the glove. D None of the above is correct. 2 When pushing a shopping trolley along a horizontal path, James has to continue to provide a force of 30 N just to maintain his speed. If the trolley (and shopping) has a mass of 35 kg, what is the total horizontal force that he will have to provide to accelerate the cart at 0.50 m s−2?

3 A force of 25 N is applied to a 15 kg ten-pin bowling ball for 4.0 s. If the ball was initially at rest, what is its final speed? 4 Identify the action and reaction pairs for three situations that involve Newton’s third law. 5 Jane has a mass of 55 kg. She steps into a lift which goes up to the second floor. The lift accelerates upward at 2.0 m s−2 for 2.5 s, then travels with constant speed. a What is the maximum speed that the lift attains as it travels between floors? b What is Jane’s weight: i when the lift is stationary? ii when the lift is accelerating upwards? 6 a What is the mass of an 85 kg astronaut on the surface of Earth where g = 9.8 m s−2? b What is the mass of an 85 kg astronaut on the surface of the Moon where g = 1.6 m s−2?


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c What is the weight of an 85 kg astronaut on the surface of Mars where g = 3.6 m s−2? 7 The series of photographs shows a stack of smooth blocks in a tall pile. One of the blocks in the pile is struck by a hammer and the blocks above it fall onto the block below, and the pile remains standing. Explain this in terms of Newton’s laws of motion.

8 A force of 120 N is used to push a 20 kg shopping trolley along the line of its handle—at 20° down from the horizontal. This is enough to cause the trolley to travel with constant velocity to the north along a horizontal path. a Determine the horizontal and vertical components of the force applied to the trolley. b What is the value of the frictional force acting against the trolley? c How large is the normal force that is supplied by the ground on which the trolley is pushed? d Why is it often easier to pull rather than push a trolley? 9 A 100 g glider is at rest on a horizontal air track, and a force is applied to it as shown in the following graph. What will be its speed at the end of the time interval? Applied force (N) 0.5

0.2

0

1 t (s)

2

10 The following diagrams show force vectors on a puck travelling at constant speed across an air table in a games arcade. The puck experiences no friction as it moves across its cushion of air. Which diagram A–D correctly shows the forces that act on the puck? A

FN

B

direction of travel

FN

direction of travel

F Fg

air

C

FN

air direction of travel

Ff

D

Fg direction of travel

F air

Fg

air

Fg

11 Two shopping trolleys with masses 30 kg and 50 kg stand together. A force of 120 N is applied to the 30 kg trolley.

120 N

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Motion

30 kg

50 kg

a With what acceleration will the trolleys move? b Calculate the size of the contact force that the 30 kg trolley exerts on the 50 kg trolley. 12 A young girl of mass 40 kg leaps horizontally off her stationary 10 kg skateboard. Assuming that no frictional forces are involved, determine the following ratios: horizontal force on the girl horizontal force on the skateboard acceleration of the girl b acceleration of the skateboard final velocity of the girl c final velocity of the skateboard a

13 A rope has a breaking tension of 100 N. How can a full bucket of mass 12 kg be lowered using the rope, without the rope breaking? 14 The force diagram below shows the forces acting on a full water tank.

FN

Fg

a Are these forces an action–reaction pair as described by Newton’s third law? b Justify your answer to part a.

15 A car begins to roll down a steep road that has a grade of 1 in 5 (i.e. a 1 m drop for every 5 m in length). If friction is ignored, determine the speed of the car in km h−1 after it has travelled a distance of 100 m if it begins its journey at rest. 16 A small boy’s racing set includes an inclined track along which a 1 car accelerates at 2 g (i.e. 4.9 m s−2). At what angle is the track to the horizontal? 17 When skiing down an incline, Eddie found that there was a frictional force of 250 N acting up the incline of the mountainside due to slushy snow. The slope was at 45° to the horizontal, so if Eddie had a mass of 70 kg, what was his acceleration? 18 On a sketch, draw vectors to indicate the forces that act on a tennis ball: a at the instant it is struck b an instant after it has been struck. 19 Two masses, 5.0 kg and 10.0 kg, are suspended from the ends of a rope that passes over a frictionless pulley. The masses are released and allowed to accelerate under the influence of gravity. What is the acceleration of the system, and what is the tension in the rope? 20 Kevin and Therese are discussing a couple of physics problems over dinner. a First, they discuss a collision between a marble and a billiard ball. Kevin argues that since the billiard ball is much heavier than the marble, it will exert a larger force on the marble than the marble exerts on it. Therese thinks that the marble and billiard ball will exert equal forces on each other as they collide. Who is correct? Explain. b Then they discuss a basketball as it bounces on a concrete floor. Kevin claims that the ball must exert a smaller force on the floor than the floor exerts on it, otherwise the ball would not rebound. Therese thinks that the ball and the floor will exert forces on each other that are equal in magnitude. Who is correct this time? Explain.


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chapter 6

k r o w , y g r e n e , m u r t e n e w o m p o d M an

I

s skydiving on your list of things to do in your future? Basejumping? Mountain boarding? Are you a person who would love to experience the exhilaration of taking that leap out of a plane, or do you question why someone would choose to jump out of a perfectly safe aeroplane in mid-flight? When an aeroplane is climbing to the required height for the thrill-seekers, we say that the aeroplane’s engines are doing work against gravity. When the parachutist takes the jump, we say that gravity is doing work on the parachutist. The search for the ultimate extreme-sport thrill is often about ‘taking on gravity’. Whether it is snowboarding, base-jumping, or bungee jumping, the participant is experimenting with the conversion of gravitational potential energy into kinetic energy. Throughout this chapter you will be able to see the common thread of energy conversion that is present in so many of our activities. In our everyday lives we try to harness and transform various forms of energy as efficiently and cleverly as possible. We burn up our own personal energy stores as we climb up the steps to the classroom. We make sure the tennis ball hits the ‘sweet spot’ on our racquet when we hit it back to our opponent. In more thrillseeking adventures we also make the most of our understanding of energy transformations. Converting lots of gravitational potential energy into kinetic energy can be an extreme experience and great fun, as shown in the photograph. Don’t be deceived though, the laws of physics cannot be switched off!

by the end of this chapter you will have covered material from the study of movement including: • momentum and impulse • work done as a change in energy • Hooke’s law • kinetic, gravitational and elastic potential energy • energy transfers • power.

The relations 6.1 momentum hip between and force Consider a collision between two footballers on the football field. From Newton’s second law, each force can be expressed as: Σ net = m and using the relationship for acceleration: − m( − ) = Σ = Dt Dt where a is the acceleration during the collision (m s−2) ∆t is the time of contact (s) u is the velocity of either one of the footballers before the collision (m s−1) v is the velocity of the footballer after the collision (m s−1). Rearranging: Σ ∆t = m( − ) or: Σ ∆t = m∆ This relationship introduces two important ideas. • The product of net force and the contact time is referred to as impulse, I. The idea of impulse is commonly applied to objects during collisions when the time of contact is small. This concept will be further explained later. • The product of the mass of an object and its velocity is referred to as momentum: = −1 where is momentum (kg m s ) m is the mass of the object (kg) is the velocity of the object (m s−1).

Figure 6.1 When two footballers collide, they exert an equal and opposite force on each other. The effect this force will have on the velocity of each footballer can be investigated using the concept of momentum.

Momentum Momentum can be thought of as the tendency of an object to keep moving with the same speed in the same direction. It is a property of any moving object. As it is the product of a scalar quantity (mass) and a vector quantity (velocity), momentum is a vector quantity. The direction of the momentum of an object is the same as the direction of the velocity of that object. The unit for momentum is kg m s−1, which is readily determined from the product of the units for mass and velocity. Momentum often indicates the difficulty a moving object has in stopping. A fast-moving car has more momentum than a slower car of the same mass; equally so, an elephant will have more momentum than a person travelling at the same speed (just as a greater force is needed to cause the same acceleration). The more momentum an object gains as its velocity increases, the more it has to lose to stop and the greater the effect it will have if involved in a collision. A football player is more likely to be knocked over if tackled by a heavy follower than a light rover, since the product p = mv will be larger for the heavy follower. Although he used different language, Newton understood this idea, and his second law of motion can be stated in terms of momentum.

Figure 6.2 The enormous mass of a large ship endows it with very large momentum despite its relatively slow speed. After turning off its engines the ship can continue against the resistance of the water for more than 4 km if no other braking is applied.

Chapter 6 Momentum, energy, work and power

179

That is:

Dp Dt where Σ F is the average net force applied to the object during the collision, in newtons (N) ∆p is the change in momentum during contact for a time ∆t. ΣF =

The change in momentum of a body is proportional to the net force applied to it: ∆p ∝ ΣF An unbalanced net force is required to change the momentum of an object, to increase it, decrease it or change its direction. This force might result from a collision or an interaction with another object. The change in momentum (∆p) of the object will be given by:

Change in momentum = final momentum − initial momentum ∆p = pf − pi

Worked example 6.1A A footballer trying to take a mark collided with a goal post and came to rest. The footballer has a mass of 80 kg and was travelling at 8.2 m s−1 at the time of the collision. a What was the change in momentum during the collision for the footballer? b Estimate the average force the footballer experienced in this collision.

Solution a Prior to the collision the footballer’s momentum was given by: p = mv = 80 × 8.2 = 656 kg m s−1 towards the pole After the collision the momentum was zero since the footballer stopped moving. So: ∆p = 0 − 656 = −656 kg m s−1 towards the pole or: ∆p = 6.6 × 102 kg m s−1 away from the pole b The negative value for the change in momentum indicates that the direction of the momentum, and hence the force applied to the footballer, is opposite to the direction in which the footballer was travelling. The time that the footballer took to stop has not been given but a reasonable estimate of the force can be made by estimating the stopping time. Keeping to magnitudes of 10 for easy working, it would be reasonable to assume that the stopping time in this sort of collision would be less than 1 s (100) and greater than 0.01 s (10−2). Something in the order of 0.1–0.5 s (10−1) would make sense on the basis of observations of similar situations. Then using: Dp F= Dt 656 F = -1 = 6560 ≈ 7 × 103 N away from the pole, i.e. a retarding force. 10

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Motion

Impulse Think about what it feels like to fall onto a concrete floor. Even from a small height it can hurt. A fall from the same height onto a tumbling mat is barely felt. Your speed is the same, your mass hasn’t changed and gravity is still providing the same acceleration. So what is different about the fall onto the mat that reduces the force you experience? Remember that, according to Newton’s second law of motion, the velocity of an object only changes when a net force is applied to that object. A larger net force will be more effective in creating a change in the velocity of the object. The faster the change occurs (i.e. a smaller time interval ∆t), the greater the net force that is needed to produce that change. Landing on a concrete floor changes the velocity very quickly as you are brought to an abrupt stop. When landing on a tumble mat the change occurs over a much greater time. The force needed to produce the change is smaller. Another illustration of this could be a tennis player striking a ball with a racquet. At the instant the ball comes in contact with the racquet the applied force will be small. As the strings distort and the ball compresses, the force will increase until the ball has been stopped. The force will then decrease as the ball accelerates away from the racquet. A graph of force against time will look like that in Figure 6.3. The impulse affecting the ball at any time will be the product of applied force and time, i.e. I = Fav∆t. The total impulse during the time the ball is in contact will be I = Fav × t, where Fav is the average force applied during the collision and t is the total time the ball is in contact with the racquet. This is equivalent to the total area under the force–time graph. The total impulse for any collision can be found in this way.

(a)

The IMPULS… affecting an object during a collision is the product of the net average applied force and the time of contact and is equivalent to the area under a force–time graph.

The relationship between impulse and momentum

Impulse = Fav = area under graph

(b)

F (N)

The concept of impulse is appropriate when dealing with forces during any collision since it links force and contact time, for example a person hitting the ground, as described above, or a ball being hit by a bat or racquet. If applied to situations where contact is over an extended time, the average net force involved is used since the forces are generally changing (as the ball deforms for example). The average net applied force can be found directly from the formula for impulse. The instantaneous applied force at any particular time during the collision must be determined from a graph of the force against time.

0

From the derivation earlier in the chapter, the impulse is also equal to the change of momentum for an object. Previously we had: mD F= Dt

0.03 0.06 0.09 0.12 0.15 t (s)

Figure 6.3 (a) When a tennis player hits a ball, an unbalanced force is applied to the ball, producing a change in its momentum; hence an impulse is applied to the ball. The magnitude of the force will change over time. (b) The impulse can be found from the area under the force–time graph since area = x-axis × y-axis = Fav × ∆t = impulse.


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Multiplying by the time interval ∆t: Fav∆t = m∆ or: I = Fav∆t = ∆p The units for impulse (N s) and the units for momentum (kg m s−1) have each been introduced separately. Since we have just seen that impulse is equal to the change in momentum: 1 N s = 1 kg m s−1

Worked example 6.1B A tennis racquet applies a force to a tennis ball for a period of 0.15 s, bringing the ball (momentarily) to a halt. The tennis ball has a mass of 58 g and was originally travelling towards the racquet at 55 m s−1. a Find the change in momentum as the ball is momentarily brought to a halt by the racquet. b Find the magnitude of the impulse during this part of the collision. c Find the average force applied during the time it takes to stop the ball.

Solution a Initial momentum: pi = mu = 0.058 × 55 = 3.19 kg m s−1

Final momentum: pf = mv = 0.058 × 0 = 0 kg m s−1 Change in momentum: ∆p = 0 − 3.19 = −3.19 kg m s−1 in the direction of travel, i.e. ~3.2 kg m s−1 in the opposite direction. b Impulse = change in momentum: I = 3.19 ≈ 3.2 N s c Using I = Fav∆t I then Fav = Dt 3.19 so Fav = = 21.27 N ≈ 21 N in the opposite direction to the ball’s travel. 0.15

Physics in action

Forces during collisions (a)

(b)

Figure 6.4 A simple example of the effect on the applied force of the stopping time during a collision can be achieved with nothing more complicated than a hammer, a can of fruit and your finger. (a) A rigid object, such as the hammer, will stop quickly. The applied force will be large. (b) A can will experience the same change in momentum, but, having a simple crumple zone, will stop more slowly, thus reducing the applied force to a tolerable amount.

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Motion

Try this simple experiment. Grab a can of fruit or similar relatively soft non-corrugated steel can. Place your finger flat on a bench top and, carefully avoiding the can’s seam, bring the side of the can crashing down on your finger. (We take no responsibility for you using the wrong part of the can!) Were you actually game to try it? If you did, how much did it hurt? Not nearly as much as you expected, right? Why? Bringing a rigid hammer down on your finger in similar circumstances would have caused considerable damage to the finger. Yet the can crumpled in around your finger and, even though it had a similar mass to the hammer and travelled at a reasonable velocity, it caused no damage to the finger and little pain. This observation can be explained with the concept of impulse. By assuming a mass of 500 g for both hammer and can and an impact speed of 20 m s−1, the magnitude of the change in momentum, and hence impulse, can be estimated.

I = ∆p = m∆v = 0.5 × 20 = 10 N s

The hammer, being rigid, will quickly come to a stop. This time can be estimated at about 0.1 s, so: I 10 F= = = 100 N Dt 0.1 —a considerable force that could be expected to do some damage to the finger! The can is able to compress and so the stopping time will be somewhat longer, say around 0.5 s. The average force will be: I 10 F= = = 20 N Dt 0.5 Increasing the stopping time by five times has reduced the average force applied to the finger to one-fifth that applied by a rigid object to something which, while perhaps not totally pain-free, is quite tolerable and will do no real damage. The applied force is inversely proportional to the stopping time. Increase the stopping time and the applied force is decreased. Try it on a friend and see if you can prove this bit of physics to them! This simple idea is the basis upon which the absorbency systems of sports shoes, crash helmets, airbags and crumple zones of cars, and other safety devices are designed.

Walking, running and sports shoe design

Table 6.1 Relative size of forces associated with some common movements in sport Movement

Footwear

Ratio of normal to weight force

Standing still

Barefoot or shoes

1.0

Walking

Barefoot

1.6

Jogging

Barefoot

2.9

Jogging

Running shoes

2.2

Sprinting

Barefoot

3.8

Fast bowling

Cricket spikes

4.1

Long jump take-off

Athletic spikes

7.8

They offer good cushioning but are more responsive, allowing faster take-offs than grass.

Vehicle safety design Designing a successful car is a complex task. A vehicle must be reliable, economical, powerful, visually appealing, secure and safe. Public perception of the relative importance of these issues varies. Magazines and newspapers concentrate on appearance, price and performance. The introduction of airbag technology into most cars has altered the focus towards safety. Vehicle safety is primarily about crash avoidance. Research shows potential accidents are avoided 99% of the time. The success of accident avoidance is primarily attributable to accident avoidance systems such as antilock brakes. When a collision does happen, passive safety features come into operation, for example the air bag. Understanding the theory behind accidents involves an understanding primarily of impulse and force.

As athletes walk or run, they experience action–reaction forces due to gravity, the surface of the track and the air around them. These forces have been investigated in some detail in the previous chapter. The force the ground exerts on a runner creates a change in momentum as the runner’s feet strike the ground. This force can be quite large and cause considerable damage to the runner’s ankles, shins, knees and hips as it is transmitted up the bones of the leg. Jogging in bare feet can increase the forces experienced to nearly three times that applied when simply standing still. Table 6.1 lists the relative size of some forces associated with common movements in sport. An understanding of the forces generated and the elastic properties of materials are used by designers in the development of athletics tracks, playing surfaces and sports shoes. Elastic materials can reduce the forces developed between foot and track by increasing the stopping time. Based on an understanding of impulse, sports shoes are designed with soles that include gels, air and cushioning grids, which extend the stopping time and thus reduce the force applied to the runner’s body. Sophisticated modern sports shoes, properly fitted to suit the wearer, have substantially reduced the size and effect of forces on the runner, with consequential benefits for the runner’s knees and hips. The running surface can also be designed to minimise the forces and produce fewer injuries. Cushioned Figure 6.5 The surfaces reduce the impact considerably. Grass is considerab forces developed be tween track le damage actually quite effective but can sometimes be too to a an increasing the stoppin runner’s body unles d foot can do spongy. The extra time spent rebounding from the g time thro s they are track shoe redu ugh s in surface slows the runner down. The response must stopping ti corporate sophistica cushioning of the foo ced by me and thu ted design t. Modern be quick if good running times are to be achieved. s decrease p the forces rinciples to increase As a result artificial surfaces such as polyurethane, generated. ‘AstroTurf’ and ‘Rebound Ace’ have become popular.


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The air bag Seat belts save lives. They also cause injuries. Strangely, the number of people surviving an accident but with serious injury has increased since the introduction of safety belts. Previously many of today’s survivors would have died instantly in the accident. A further safety device is required to minimise these injuries. The air bag in a car is designed to inflate within a few milliseconds of a collision to reduce secondary injuries during the collision. It is designed to inflate only when the vehicle experiences an 18–20 km h−1 or greater impact with a solid object. The required deceleration must be high or accidental nudges with another car would cause the air bag to inflate. The car’s computer control makes a decision in a few milliseconds to detonate the gas cylinders that inflate the air bag. The propellant detonates and inflates the air bag while the driver collapses towards the dashboard. As the body lunges forwards into the air bag, the bag deflates, allowing the body to slow in a longer time as it moves towards the dashboard. Injury is thus minimised. Calculating exactly when the air bag should inflate, and for how long, is a difficult task. Many cars have been crash tested and the results painstakingly analysed. High-speed film demonstrates precisely why the air bag is so effective. During a collision the arms, legs and head of the occupants are restrained only by the joints and muscles. Enormous forces are involved because of the large deceleration. The shoulders and hips can, in most cases, sustain the large forces for the short duration. However, the neck is the weak link. Victims of road accidents regularly receive neck and spinal injuries. An air bag reduces the enormous forces the neck must

withstand by extending the duration of the collision, a direct application of the concept of impulse. The extent of injuries during a collision is not only dependent on the size of the force but also the duration and deflection resulting from the applied force. An increase in localised pressure will result in a greater compression or deflection of the skull. The air bag reduces the localised pressure by increasing the contact surface area and decreasing the force. The effect can be seen by the relationship: F P= A where P is the pressure (N m−2) F is the force (N) A is the contact area (m2). An air bag has a contact surface area of about 0.2 m2 compared with 0.05 m2 for a seat belt. This reduces injuries caused by seat belts, such as bruising and broken ribs and collar bones, since it increases the stopping time. It also supports the head and chest, preventing high neck loads caused by the seat belt restraining the upper torso. Most importantly, it prevents the high forces caused by contact of the head with the steering wheel. The air bag ensures that the main thrust of the expansion is directed outwards instead of towards the driver. The deflation rate, governed by the size of the holes in the rear of the air bag, provides the optimum deceleration of the head for a large range of impact speeds. The air bag is not the answer to all safety concerns associated with a collision, but it is one of many safety features that form a chain of defence in a collision.

(a)

Figure 6.6 (a) The air bag is one of a number of passive safety features incorporated into the design of modern cars. It extends the stopping time, significantly reducing the forces on the head and neck during a collision. It also distributes the force required to decelerate the mass of the driver or passenger over a larger area than a seat belt. (b) The deflation rate of the bag is governed by the size of holes in the rear of the air bag, and is designed to provide the optimum deceleration of the head for a large range of impacts.

Applied force (N)

(b)

no air bag with air bag

0

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20

Motion

40

60 80 Time (ms)

100

120

140

6.1 summary The relationship between momentum and force • The momentum of a moving object is the product of its mass and its velocity: p = mv where p is in kg m s−1 m is in kg v is in m s−1 Momentum is a vector quantity. • The change in momentum (∆p) = final momentum (pf) − initial momentum (pi). • Impulse is the product of the net force during a collision and the time interval ∆t during which the

force acts: I = Fav∆t. It can also be found from the area under a force–time graph and is measured in newton seconds (N s). Impulse is equal to the change in momentum, ∆p, caused by the action of the net applied force: Fav∆t = ∆p • Extending the time over which a collision occurs will 1 decrease the average net force applied since Fav ∝ . Dt This is the principle behind many safety designs.

6.1 questions The relationship between momentum and force Use g = 9.8 m s−2 where required. 1 What is the momentum, in kg m s , of a 20 kg cart travelling at: a 5.0 m s−1? b 5.0 cm s−1? c 5.0 km h−1?

the ball is 65 g and it is actually in contact with the racquet for 0.032 s: a what momentum does the ball have on leaving the racquet? b what is the average force applied by the racquet on the ball?

−1

3 Which object has the greater momentum—a medicine ball of mass 4.5 kg travelling at 3.5 m s−1 or one of mass 2.5 kg travelling at 6.8 m s−1? 4 Calculate the momentum of an object: a of mass 4.5 kg and velocity 9.1 m s−1 b of mass 250 g and velocity 3.5 km h−1 c that has fallen freely from rest for 15 s and has a mass of 3.4 kg d that experiences a net force of magnitude 45 N, if the net force is applied for 3.5 s. 5 A tennis ball may leave the racquet of a top player with a speed of 61 m s−1 when served. If the mass of

6 A 200 g cricket ball (at rest) is struck by a cricket bat. The ball and bat are in contact for 0.05 s, during which time the ball is accelerated to a speed of 45 m s−1. a What is the magnitude of the impulse the ball experiences? b What is the net average force acting on the ball during the contact time? c What is the net average force acting on the bat during the contact time? 7 The following graph shows the net vertical force generated as an athlete’s foot strikes an asphalt running track.

Force (N)

2 The velocity of an object of mass 8.0 kg increases from an initial 3.0 m s−1 to 8.0 m s−1 when a force acts on it for 5.0 s. a What is the initial momentum? b What is the momentum after the action of the force? c How much momentum is the object gaining each second when the force is acting? d What impulse does the object experience? e What is the magnitude of the force?

1400 1200 1000 800 600 400 200 0 10 20 30 40 50 60 70 Time (ms)

a Estimate the maximum force acting on the athlete’s foot during the contact time.


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b Estimate the total impulse during the contact time. 8 A 25 g arrow buries its head 2 cm into a target on striking it. The arrow was travelling at 50 m s−1 just before impact. a What change in momentum does the arrow experience as it comes to rest? b What is the impulse experienced by the arrow? c What is the average force that acts on the arrow during the period of deceleration after it hits the target?

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Motion

9 Crash helmets are designed to reduce the force of impact on the head during a collision. a Explain how their design reduces the net force on the head. b Would a rigid ‘shell’ be as successful? Explain. 10 Describe, with the aid of diagrams, a simple collision involving one moving object and one fixed in position. Estimate, by making reasonable estimates of the magnitudes of the mass and velocity of the moving object, the net force acting on the objects during the collision.

6.2 Conser vation o fm

omentum

The most significant feature of momentum is that it is conserved. This means that the total momentum in any complete system will be constant. For this reason momentum is very useful in investigating the forces experienced by two colliding objects—as long as they are unaffected by outside forces. The law of conservation of momentum, as it is known, is derived from Newton’s third law. From Newton’s third law, the force applied by each object in a collision will be of the same magnitude but opposite in direction: F1 = −F2 From Newton’s second law, ΣF = ma, so the forces could be expressed as: m1a1 = −m2a2 v - u and using a = we get: Dt v - u1 v - u2 m1 1 = −m2 2 Dt Dt where a1 and a2 are the respective accelerations of the two objects during the collision in m s−2 ∆t is the time of contact (s) u1 and u2 are the velocities of the objects prior to collision (m s−1) v1 and v2 are the velocities after collision (m s−1) Since the time that each object is in contact with the other will be the same, ∆t will cancel out: m1(v1 − u1) = −m2(v2 − u2) or: m1u1 + m2u2 = m1v1 + m2v2 In other words, the total momentum before colliding is the same as the total momentum after the collision.

TH… LAW OF CONS…RVATION OF MOM…NTUM states that, in any collision or interaction between two or more objects in an isolated system, the total momentum of the system will remain constant; that is, the total initial momentum will equal the total final momentum: Σpi = Σpf or m1u1 + m2u2 = m1v1 + m2v2 It is most important to realise that momentum is only conserved in an isolated system; that is, a system in which no external forces affect the objects involved. The only forces involved are the action–reaction forces on the objects in the collision. Consider two skaters coming together on a nearfrictionless ice rink. In this near-ideal situation it is realistic to apply the law of conservation of momentum. The only significant horizontal forces between the two skaters are those of the action–reaction pair as the two skaters collide. If the skaters were to skate through a puddle of water as they come together then friction would become noticeable. This force is an external force since it is not acting between the two skaters. The interaction between

Physics file The principle of conservation of momentum was responsible for the interpretation of investigations that led to the discovery of the neutron. Neutral in charge, the neutron could not be investigated through the interactions of charged particles that had led to the discovery of the proton and electron. In 1932 Chadwick found that in collisions between alpha particles and the element beryllium, the principle of conservation of momentum only held true if it could be assumed that there was an additional particle within the atom, which had close to the same mass as a proton but no electric charge. Subsequent investigations confirmed his experiments and led to the naming of this particle as the neutron.

Physics file If you release an inflated rubber balloon with its neck open, it will fly off around the room. In the diagram below, the momentum of the air to the left is moving the balloon to the right. Momentum is conserved. This is the principle upon which rockets and jet engines are based. Both rockets and jet engines employ a high-velocity stream of hot gases that are vented after the combustion of a fuel–air mixture. The hot exhaust gases have a very large momentum as a result of the high velocities involved, and can accelerate rockets and jets to high velocities as they acquire an equal momentum in the opposite direction. Rockets destined for space carry their own oxygen supply, while jet engines use the surrounding air supply.

air


balloon

187

6.0 m s–1

6.0 m s–1

Figure 6.7 When two skaters collide on a near-frictionless skating rink they exert equal and opposite forces on each other. The total momentum of the two skaters before the collision will equal the total momentum of the two skaters after the collision. Because no other large horizontal forces are involved other than those in the collision, this can be considered as an isolated system.

PRACTICAL ACTIVITY 25 Conservation of momentum in explosions

a skater and the puddle would constitute a separate isolated system (as would that between puddle and ice, ice and ground etc.). Momentum would still be conserved within this other separate system. It is virtually impossible to find a perfectly isolated system here on Earth because of the presence of gravitational, frictional and air-resistance forces. Only where any external forces are insignificant in comparison to the collision forces is it reasonable to apply the law of conservation of momentum. Also important is that in any collision involving the ground, Earth itself must be part of the system. Theoretically, any calculation based on conservation of momentum should include the Earth as one of the objects, momentum only being conserved when all the objects in the system are considered. In practice, the very large mass of the Earth in relation to the other objects involved means that there is a negligible change in the Earth’s velocity and it can be ignored in most calculations. Of course, a collision with a fast-moving asteroid would be another matter!

Worked example 6.2A Skater 1 in Figure 6.7, with mass 80 kg, was skating in a straight line with a velocity of 6.0 m s−1 while the skater 2, of mass 70 kg, was skating in the opposite direction, also with a speed of 6.0 m s−1. a The two skaters collide and skater 1 comes to rest. Assuming that friction can be ignored, what will happen to the skater 2 after the collision? b What would happen if the two skaters had hung on to each other and stayed together after the collision?

Solution a From conservation of momentum: Σpi = Σpf or m1u1 + m2u2 = m1v1 + m2v2 and m1 = 80 kg, m2 = 70 kg As both velocity and momentum are vector quantities, a positive direction should be established and taken into consideration. Adopting the direction of motion of skater 1 as the positive direction: u1 = 6.0 m s−1, u2 = −6.0 m s−1, v1 = 0 m s−1, v2 = ? Substituting into equation: 80 × 6.0 + 70 × (−6.0) = 80 × 0 + 70 × v2 and v2 = +0.86 m s−1. The 70 kg skater bounces back in the opposite direction with a speed of 0.86 m s−1. b Treating the two skaters as one mass after the collision: 80 × 6.0 + 70 × (−6.0) = (80 + 70) × v2 and now v2 = +0.4 m s−1. A different outcome after the collision results in a different velocity for each skater. There is no unique answer when applying the idea of conservation of momentum. The final velocity of any object depends on what happens to all the objects involved in the collision.

Figure 6.8 Newton’s cradle, or Newton’s balls to some, is an instructive ‘executive toy’ based on the principle of conservation of momentum extended over a number of objects.

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Motion

The law of conservation of momentum can be extended to any number of colliding objects. The total initial momentum is found by calculating the vector sum of the initial momentum of every object involved. The total final momentum will then also be the vector sum of each separate momentum involved. Separation into two or more parts after the ‘collision’ (interaction is a better word since it does not have to be destructive), for example the firing of a bullet, can also be dealt with in the same manner.

Physics in action

Collisions and pedestrians A car is designed to keep its occupants safe. Unfortunately, however, very little can be done to protect a pedestrian from the onslaught of a 1400 kg car travelling at 60 km h−1. Bull bars in residential areas are currently under review because of the enormous damage they inflict on a pedestrian. The effect on the pedestrian will depend on the person’s height and mass, the height of the front of the oncoming vehicle, and the speed, mass and shape of the vehicle. Consider the following possibilities: • a pedestrian being struck by a truck moving at 30 km h−1 • a pedestrian being struck by a car moving at 30 km h−1 • a pedestrian being struck by a cyclist moving at 30 km h−1. The injuries to the pedestrian are due largely to the change in the pedestrian’s momentum. Being hit by the cyclist will obviously result in the least injury to the pedestrian because of the lower momentum of the cycle and its rider. The mass of the other vehicles is such that they have a far larger momentum to impart to the pedestrian. Consider the following situation. A 1400 kg car is travelling at 60 km h−1 when it strikes a stationary 70 kg pedestrian. The pedestrian lands on the bonnet of the car and travels with the car until it finally comes to a halt. Assuming that frictional forces are minimal: total momentum before the collision = total momentum after the collision Before the collision: =m car 1000 = 1400 × (60 × ) 3600 4 −1 = 2.3 × 10 kg m s = 0 (pedestrian is stationary) pedestrian After the collision: so and

total

= 2.3 × 104 kg m s−1 = (mcar + mpedestrian)

Leg and hip injuries are general in this form of pedestrian– vehicle collision. Head injuries result from the pedestrian colliding with the road. Very little of a car’s design will alter the severity of head injuries received. Vehicles can be designed with a low, energy-absorbing bumper bar to reduce knee and hip damage. If the pedestrian’s knee strikes the bumper bar, knee damage is very likely. Knees do not heal as well as broken legs. A lower, energy-absorbing bar is, for this reason, preferable. (b) When a vehicle is moving very fast at the point of impact, the pedestrian’s inertia acts against rapid acceleration. The pedestrian does not initially move forward with the same velocity as the vehicle. If the pedestrian remains in approximately the same place, he or she will go either over or under the car. The pedestrian will be either run over or run under (i.e. the car goes under the pedestrian). Being run over usually results in serious injury or fatality. Massive head injuries occur as the pedestrian’s head strikes the ground. The relative height of the vehicle’s bumper bar and the height of the pedestrian determines whether they will be run over or under. Most bumper bars are below adult waist level. Small children, however, have much more chance of being run over as the height of the bumper bar is relatively much higher. If the pedestrian is run under, ‘passive’ safety features of modern car design come into play. Removal of protruding hood ornaments is essential since they can easily penetrate the body of a person and cause enormous injuries. The bonnet of a car acts as a good impact absorber, particularly in comparison with the hard surface of the road. Bull bars, however, can block the path of the pedestrian, making it more likely that they be run over. Further, they have little impactabsorbing ability. It is, therefore, logical to ban bull bars in residential areas.

1470 = 2.3 × 104 kg m s−1 total

≈ 16 m s−1 or 57 km h−1

This means that the pedestrian accelerates from rest to a speed of 57 km h−1 in the short duration of the collision. A similar collision between the pedestrian and a cyclist travelling at 30 km h−1 would result in a final speed of 5.2 m s−1 (19 km h−1). The speed of the car changes very little. The speed of the cyclist is almost halved. Antilock brakes, excellent road handling and reduced speed limits in some areas reduce the likelihood of a vehicle striking a pedestrian. Unfortunately accidents can still happen. There are essentially two possibilities that can occur when a pedestrian is struck by a car. (a) The pedestrian bounces off the front of the car and is projected through the air. This type of motion tends Figure 6.9 This to happen when the vehicle is travelling relatively slowly. designed to Nissan car has a pop -up re The pedestrian is rapidly accelerated forwards to near by making sult in less damage to bonnet that has bee s n p a pede a ce the velocity of the vehicle. Injuries occur to the pedestrian between th e bonnet a strian in a collision, nd the eng when the car strikes and again when they land on the ground. in

e.


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6.2 summary Conservation of momentum • The law of conservation of momentum states that in any collision or interaction between two or more objects in an isolated system the total momentum of the system will remain constant. The total initial momentum will equal the total final momentum: Σpi = Σpf

• Conservation of momentum can be extended to any number of colliding objects within an isolated system.

6.2 questions Conservation of momentum 1 A white billiard ball of mass 100 g travelling at 2.0 m s−1 across a low-friction billiard table has a head-on collision with a black ball of the same mass initially at rest. The white ball stops while the black ball moves off. What is the velocity of the black ball? 2 A girl with mass 50 kg running at 5 m s−1 jumps onto a 4 kg skateboard travelling in the same direction at 1.0 m s−1. What is their new common velocity? 3 A man of mass 70 kg steps forward out of a boat and onto the nearby river bank with a velocity, when he leaves the boat, of 2.5 m s−1 relative to the ground. The boat has a mass of 400 kg and was initially at rest. With what velocity relative to the ground does the boat begin to move? 4 A railway car of mass 2 tonnes moving along a horizontal track at 2 m s−1 runs into a stationary train and is coupled to it. After the collision the train and car move off at a slow 0.3 m s−1. What is the mass of the train alone? 5 A trolley of mass 4.0 kg and moving at 4.5 m s−1 collides with, and sticks to, a stationary trolley of mass 2.0 kg. Their combined speed in m s−1 after the collision is: A 2.0 B 3.0 C 4.5 D 9.0 6 Superman stops a truck simply by blocking it with his outreached arm.

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Motion

a Is this consistent with the law of conservation of momentum? Explain. b Using reasonable estimates for the initial speed and mass of the truck and Superman demonstrate what will happen. Use appropriate physics concepts. 7 A car of mass 1100 kg has a head-on collision with a large four-wheel drive vehicle of mass 2200 kg, immediately after which both vehicles are stationary. The four-wheel drive vehicle was travelling at 50 km h−1 prior to the collision in an area where the speed limit was 70 km h−1. Was the car breaking the speed limit? 8 A 100 g apple is balanced on the head of young master Tell. William, the boy’s father, fires an arrow with a mass of 80 g at the apple. It reaches the apple with a velocity of 35 m s−1. The arrow passes right through the apple and goes on with a velocity of 25 m s−1. With what speed will the apple fly off the boy’s head? (Assume there is no friction between apple and head.) 9 A space shuttle of mass 10 000 kg, initially at rest, burns 5.0 kg of fuel and oxygen in its rockets to produce exhaust gases ejected at a velocity of 6000 m s−1. Calculate the velocity that this exchange will give to the space shuttle. 10 A small research rocket of mass 250 kg is launched vertically as part of a weather study. It sends out 50 kg of burnt fuel and exhaust gases with a velocity of 180 m s−1 in a 2 s initial acceleration period. a What is the velocity of the rocket after this initial acceleration? b What upward force does this apply to the rocket? c What is the net upward acceleration acting on the rocket? (Use g = 10 m s−2 if required.)

6.3 Work Aristotle, Galileo and Newton each made significant contributions to our developing understanding of the relationships between the forces that are applied to objects and their resultant behaviour. Aristotle and those who followed him were often locked into philosophical views involving, for example, the natural resting places of objects. Unlike his predecessors, Galileo, over 400 years ago, based his proposed theories largely on the observations that he made. Although he could not be completely free from the influences of his era, observational scientific experimentation had been born through him. The implications of this change in approach were immeasurable. In this chapter our study of the interactions between forces and objects focuses on the resultant displacement that objects experience, rather than the resulting velocities and accelerations discussed earlier in our study of Newton’s laws. This leads to the examination of the concepts of work and energy. The notion of energy was not developed until a relatively short time ago, and was only fully understood in the early 1800s. Today, the concept has become one of the most fundamental in science. We will see that in physics an object is said to have energy if it can cause particular changes to occur. Energy is a conserved quantity and is useful not only in the study of motion, but in all areas of the physical sciences. Before discussing energy, it is necessary to first examine the concept of work. In common usage the term ‘work’ has a variety of meanings. Most convey the idea of something being done. At the end of a long, tiring day we might say that we have done a lot of work. This could also be said because the person feels that their reserves of energy have been used up. Imagine lifting a heavy book up onto a high shelf. The heavier the book, the more force must be applied to overcome its weight. The higher the shelf, the greater the displacement over which the force must be applied. A very heavy book lifted to a high shelf will require a considerably greater effort than moving a few pieces of paper from floor to table. Thus there are two features that constitute the amount of work done: the amount of effort required and the displacement involved. In physics work is done on an object by the action of a force or forces. The object is often referred to as the load. Many interactions are complex and there is often more than one force present. As work can only be done in the presence of a force, it is imperative that any time the work done in a particular situation is being discussed, the relevant force, forces or net force should be clearly stated. For clarity, the item upon which the work is done, the load, should also be specified. Clearly specified examples of work are: • the work done by gravity on a diver as she falls • the work done by arm muscles on a schoolbag lifted to your shoulder • the work done by the heart muscle on a volume of blood during a contraction • the work done by the net force acting on a cyclist climbing a hill. Always being clear about the particular forces and objects examined will prevent considerable confusion in this area of study. For work to be done on a body, the energy of the body must change. Thus the work done is measured in joules, which is also the unit of energy.

Figure 6.10 In each situation involving work, a load can clearly be specified.


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Physics file The symbol W is a little over-used in this area of physics. In the area of motion and mechanics it can stand for work or the abbreviation of watt. Be careful to read the context when you come across the symbol!

The different forms of energy are discussed later in this chapter. A test to decide whether work has been done on a particular object involves determining whether the object’s energy has altered. If its energy is unaltered, no net work has been done on it, even though forces clearly may have been acting. (a)

I’m doing no work on the crate since x0

I’m doing no work on the crate since x0 W Fx 0

(b)

I’m doing work!

work is done on the load W Fx

crate speeding up

Figure 6.11 (a) No work is done on the crate since its energy is not altered. (b) The energy of the crate is changing, so work is being done on the crate.

Work done by a constant force If the net force acting on an object in a particular situation has a constant value, or if it is appropriate to utilise an average force value, then:

Figure 6.12 During the fall the force due to

gravity does work on the person and produces a displacement.

Physics file The unit for work, the joule (J), is used for all forms of energy in honour of James Prescott Joule, an English brewer and physicist, who pioneered work on energy in the 19th century.

Physics file The convention for naming units in physics is to use small letters when writing the unit in full (e.g. joule, newton, metre). A capital letter is used for the symbol only when the unit is named in recognition of a scientist’s contributions, otherwise the symbol is lower case (e.g. J for joule, N for newton, but m for metre).

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Motion

The net WORK done on an object is defined as the product of the net force on the object and its displacement in the direction of the net force. When the force and displacement are in the same direction, the work done by the stated force is given by: W = Fx where W is the work done by the stated force in joules (J) F is the magnitude of the stated force in newtons (N) x is the magnitude of the displacement in metres (m) Work is the area under a force–displacement graph.

One JOUL… of work is done on an object when the application of a net force of 1 newton moves an object through a distance of 1 metre in the direction of the net force. From the definition of work, it can be seen that if a person pushes a load a horizontal distance of 5 m by exerting a horizontal force of 30 N on the load, then the person does 150 J of work on the load. This is straightforward. A displacement in the direction of the force is achieved so work is done. If an applied force does not produce any displacement of the object (x = 0), then we say that no work is done on the object. Situations also occur in which a constant force acts at an angle θ to the direction of motion. A force acting at an angle will be less effective than the force acting solely in the direction of the displacement. The component of the force in the direction of the displacement, F cos θ, is used in calculating the work done in the required direction.

W = Fx cos θ where θ is the angle between the applied force and the direction of motion. (a)

(b)

F

(c)

F

F θ F cos θ

Direction of motion

Direction of motion

Direction of motion

Figure 6.13 (a) If a force is applied in the direction of motion of the cart, then the force is at its most effective in moving the cart. (b) When the force is applied at an angle θ to the direction of motion of the cart, the force is less effective. The component of the force in the direction of the displacement, F cos θ, is used to calculate the work. (c) When the angle at which the force is acting is increased to a right angle (θ = 90°), then the component of the force in the direction of the intended displacement is zero and it does no work on the cart—provided of course that it doesn’t lift the cart, in which case work would also be done against gravity.

Work and friction If an object is forced to move across a surface by the application of a force, its motion may be slowed by friction. In this case the applied force is doing work on the object and the frictional force can be considered to be doing ‘negative’ work on the object. In Figure 6.14 an applied force of 300 N across a displacement of 5 m does 1500 J of work on the object. If a 100 N frictional force occurs, we can state that work done by the frictional force is: −100 × 5 = −500 J Hence the net work done on the object is 1000 J. An alternative approach would involve first calculating the net force, ΣF, on the object to be 200 N. The net work done on the load (by the net force) is therefore: ΣF × x = 200 × 5 = 1000 J If work is done against a frictional force as a load continues to move, then some of the energy expended by the person pushing is converted into heat and sound energy, and transferred to the ground and the load. As the surfaces slide past one another friction would cause them to heat up slightly and make some noise. Keep in mind that on a frictionless surface the load would accelerate, increasing its energy. Figure 6.15 shows a situation in which the size of the frictional force is not large enough to prevent motion, but it is large enough to balance the applied force. As a result the object moves at a steady speed. Although the person is doing work on the object, this is opposed by friction and the net work on the object is zero. This is consistent with our earlier discussion, which stated that if the energy of the object is not altered, then no net work has been done on the object.

ΣF = 200 N Fapplied = 300 N

Net work done on crate = 1000 J

frictional force Ff = 100 N

Figure 6.14 The object slides across a displacement of 5 m. Due to friction the net work done on the object is less than the work done by the person on the object.

steady speed object’s energy is unchanged Fapplied = 100 N

frictional force Ff = 100 N

Figure 6.15 Due to friction the net work done on the object is zero since the object has no increase in kinetic energy.


193

Figure 6.16 The Great Rialto Stair Trek was one of a number of races to the tops of tall buildings around the world. The winner took less than 7 minutes to complete the 242 m (vertical) race.

Worked example 6.3A Calculate the work done against gravity by an athlete of mass 60 kg competing in the Great Rialto Stair Trek illustrated in Figure 6.16. Use g = 9.8 m s−2. Assume the athlete climbs at a constant speed.

Solution Only the weight force needs to be considered in this example as the work in the vertical direction is all that is required. m = 60 kg, g = 9.8 m s−2, x = ∆h = 242 m Force applied = weight = mg = 60 × 9.8 = 588 N W = Fx = 588 × 242 = 142 296 N m = 1.4 × 105 J

Worked example 6.3B The girl in Figure 6.13 pulls the cart by applying a force of 50 N at an angle of 30° to the horizontal. Assuming a force due to friction of 10 N is also acting on the wheels of the cart, calculate the net work done on the cart if the cart is moved 10 m along the ground in a straight line.

Solution Fapplied = 50 N, θ = 30°, Ff = 10 N, x = 10 m ΣF = Fapplied × cos 30° − Ff = 50 × 0.866 − 10 = 33.3 N Now W = ΣFx = 33.3 × 10 = 330 J

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Motion

Upward force does no work A more difficult idea to comprehend is that in the apparent absence of friction, a force can be exerted on an object yet do no work on it. For example, when a person carries an armload of books horizontally the upward force does no work on the books since the direction of the applied force (i.e. up) is at right angles to the displacement (i.e. horizontal). An examination of the definition of work—W = Fx cos θ—confirms this finding since the value of θ is 90° and, hence, the value of cos θ is 0. Similarly if a person is holding a heavy item, such as a TV, stationary, they may be exerting great effort. However, since the upward force applied to support the object does not produce any vertical (nor indeed horizontal) displacement, x = 0 and there is no work done by this upward applied force on the object.

Whenever the net force is perpendicular to the direction of motion no (net) work is done on the object.

Force–displacement graphs A graphical approach can also be used to understand the action of a force and the work expended in the direction of motion. This is particularly useful in situations in which the force is changing with displacement. The area under a graph of force against displacement always represents the work produced by the force, even in situations when the force is changing, such as during a collision. The area can be shown to be equivalent to work as follows. From Figure 6.17: the area enclosed by the graph = Fav × x Work = Fav × x When the force is changing, a good estimate of the area can be found by dividing the area into small squares and counting the number or by dividing it into thin segments. The segments can be considered to be rectangles with an area equal to the work for that small part of the displacement. The total work will be the sum of the areas of all the separate rectangles.

Physics file The area under a force–displacement graph can also be found by using calculus if the equation of the graph is known. In most instances a good estimate by counting squares or segments is sufficient.

Force (N)

Units for area = N r m =Nm =J i.e. area = work done

Displacement (m)

Figure 6.17 The area under a force–displacement graph is equivalent to the work done by a force acting in the direction of the displacement. Where the net applied force is changing, the area can be found by counting squares or by dividing the area into segments. The area of each segment then equals the work done by a constant force during that small displacement and the total area will represent the total work.


195

Worked example 6.3C The force–displacement graph on the left represents the work done on the sole of a sports shoe as it compresses against the surface of a rigid track. The displacement shown represents the amount of compression the sole undergoes. Find the work done on the shoe by the compressive forces.

100 90 80

Force (N)

70 60

Solution

50 40 30 20

Displacement (m)

0.008

0.007

0.006

0.005

0.004

0.003

0.002

0.001

10

This is a simple case of working out the area represented by each square and then counting the total number of squares to find the total work done. Be careful to consider the scale of each axis in your working. Area of one square = 10 N × 0.001 m = 0.01 J Total number of squares (part squares can be added to give whole squares) = 33 Work = 33 × 0.01 = 0.33 J

Impulse and work The concepts of impulse and work seem quite similar and, when solving problems, can easily be confused. Actually, problems focusing on forces in collisions may be solved using either concept, but it should be understood that each is derived from a different idea. Impulse comes from an understanding of the action of a force on an object over time and is equal to the change in momentum the force produces. Work is related to the action of a force on an object as it moves the object, or part of it, through some displacement. This equals the change in the object’s energy, ∆E. Summarising: • Impulse is equal to F × ∆t, is equivalent to ∆p, has the units newton seconds (N s), and can be determined from the area under a force–time graph. • Work is equal to F × x, is equivalent to ∆E, has the units joules (J), and can be determined from the area under a force–displacement graph.

6.3 summary Work • When a force does work on an object, a change occurs in the displacement and energy of the object. • The work done on an object, W in joules (J), is the product of the net applied force and its displacement in the direction of the force: W = Fx • The work done by a force acting at an angle to the displacement is given by Fx cos θ where θ is the angle between the force and the direction of the

196

Motion

displacement. When the force is at right angles to the direction of the displacement, no work is done in that direction. • The area under a force–displacement graph is equivalent to the work done. The area under the graph for a variable force can be found by counting squares or narrow segments.

6.3 questions Work 1 How much work is done on an object of 4.5 kg when it is lifted vertically at a constant speed through a displacement of 6.0 m? 2 A bushwalker climbs a hill 250 m high. If her mass is 50 kg and her pack has an additional mass of 10 kg, calculate the work she needs to do in climbing to the top of the hill.

7 The diagram shows the position of a student’s arm as the weight of a sandbag is measured using a spring balance. The balance is held still to take the reading. What net work is done on the sandbag while the measurement is being made?

3 Is the quantity you calculated in question 2 the only work that the bushwalker has done? Explain. 4 The work done by a force is: i calculated by multiplying the force by the distance moved ii measured in joules iii not affected by the angle at which the force acts. Which statement/s is/are correct? A i, ii, iii B i, ii C ii, iii D ii E iii 5 A removalist is loading five boxes onto a truck. Each has a mass of 10 kg and a height of 30 cm. The tray of the truck is 1.5 m above the ground and the removalist is placing each box on top of the previous one. a How much work does the removalist do in lifting the first box onto the truck tray? b How much energy has the removalist used in lifting this first box? c What is the total work done on the boxes in lifting all the boxes onto the truck as described? 6 If a lift of mass 500 kg is raised through a height of 15 m by an electric motor: i the weight of the lift is 4900 N ii the useful work done on the lift is 73 500 J iii the useful work done is the only energy used by the motor. Which statement/s is/are correct? A i, ii, iii B i, ii C ii, iii D ii E iii

spring balance

sandbag

8 A weightlifter raises a 100 kg mass 2.4 m above the ground in a weightlifting competition. After holding it for 3.0 s he places it back on the ground. a How much work has been done by the weightlifter in raising the mass? b How much additional work is done during the 3.0 s he holds it steady? 9 A rope that is at 35° to the horizontal is used to pull a 10.0 kg crate across a rough floor. The crate is initially at rest and is dragged for a distance of 4.00 m. The tension in the rope is 60.0 N and the frictional force opposing the motion is 10.0 N. a Draw a diagram illustrating the direction of all relevant forces. b Calculate the work done on the crate by the tension in the rope. c Find the total work done on the crate. d Determine the energy lost from the system as heat and sound due to the frictional force. 10 The graph represents the size of a variable force, F, as a rubber band is stretched from a resting length of 5 cm to 25 cm. Estimate the total work done on the rubber band by the force. 10 Force (N)

Where appropriate use g = 9.8 m s−2.

5

10 20 Extension (cm)


197

6.4 Mechanical energy Although the concept of energy is quite abstract, each of us, from an early age, will have begun to develop an understanding of its meaning. We are increasingly aware of our reliance upon the energy resources that allow our vehicles and computers to run, that keep our homes warm and fuel our bodies. Energy can take on many forms. In this section we look at the forms of energy specifically related to motion. Mechanical energy is defined as the energy that a body possesses due to its position or motion. Kinetic energy, gravitational potential energy and elastic potential energy are all forms of mechanical energy. Recall our earlier assertion that work is done when a force is applied that results in the displacement of an object in the direction of the applied force. When work is done the energy of an object will change. We will analyse situations that result in a change in the kinetic and/or gravitational potential and/or elastic potential energy of an object. A hockey puck gains energy when hit because work has been done by the stick on the puck. The amount of work done on the puck equals the puck’s change in kinetic energy. A tennis ball at the point of impact is compressed against the tennis racquet. It has gained elastic potential energy. Work has been done in compressing the tennis ball. However, the idea of work may be applied to many forms of energy. The common thread is that, regardless of the form of energy, whenever work is done there is a change in energy from one form to another. In order for any energy transformation to occur, say from motion to heat, work must be done. We observe many different forms of energy each day. We have come to take for granted the availability of light, heat, sound and electrical energy whenever we require it. We rely upon the chemical potential energies that are available when petrol, diesel and LPG are burnt to run our vehicles, and food to fuel our bodies. Whenever work is done, energy is expended.

…N…RGY is the ability to do work. Some comparative energy transformations are included in Table 6.2.

Table 6.2 Comparison of various energy transformations Energy use Household in 1 day Fan heater in 1 hour Adult food intake in 1 day

Amount of energy 150 MJ 8.6 MJ 12 MJ

Making 1 Big Mac

2.1 MJ

Climbing a flight of stairs

5 kJ

Lifting 10 kg to a height of 2 m

200 J

Kinetic energy Figure 6.18 Mechanical energy exists in

many forms.

198

Motion

An object in motion has the ability to do work and therefore is said to possess energy. This energy carried by a moving object is called kinetic energy (from the Greek word ‘kinesis’, literally meaning ‘motion’).

Figure 6.19 The kinetic energy of any object depends on its mass and the square of its speed. Doubling the velocity will increase the kinetic energy by a factor of four.

Physics file

If a moving object of mass, m, and initial velocity, u, experiences a constant net force, F, for time, t, then a uniform acceleration results. The velocity will increase to a final value, v. Work will have been done during the time the force is applied. Since work is equivalent to the change in kinetic energy of the object, there should be a relationship linking the two quantities. This can be found from the definition for work when the net applied force is in the direction of the displacement: W = ΣFx Now substituting Newton’s second law F = ma we get: W = max . . . . . . (i) Using one of the earlier equations of motion: v2 = u2 + 2ax v2 - u2 and rearranging: x = 2a v2 - u2 Substitute this for x in equation (i): W = ma( ) 2a 1 1 Rearranging gives: W = 2 mv2 − 2 mu2 but W = ∆E If it is accepted that the work done results in a change in kinetic energy, 1 then an object of mass m with a speed v has kinetic energy equal to 2 mv2.

The derivation described for kinetic energy is actually that for translational kinetic energy, the movement of a body along a path. A body can also have rotational kinetic energy, as does the Earth, if it is spinning. A different relationship is required to calculate the kinetic energy of rotation.

The KIN…TIC …N…RGY, …k, of a body of mass m and speed v is: …k = 12 mv2 Like all forms of energy, kinetic energy is a scalar quantity and is measured in joules (J). There is no direction associated with it. The kinetic energy of an object depends solely on its mass and velocity. The approximate kinetic energy of various moving objects is given in Table 6.3.

Table 6.3 Kinetic energy of moving objects Object

Mass (kg)

Average speed (m s–1)

…k (J)

Earth in orbit

6 × 10

3 × 10

2.7 × 1033

Orbiting satellite

100

8 × 103

3 × 109

Large car

1400

28

5.5 × 105

Netball player

60

8

1900

Footballer

90

8

Electron in a TV tube

9 × 10

24

−31

4

2900

7 × 10

7

2.2 × 10−15


199

Note that the relationship between work and energy, which was discussed earlier, has now been quantified for kinetic energy changes.

When an applied force results in the change in kinetic energy of an object, the work done in joules (J) can be calculated using: W = ∆…k = …k(final) − …k(initial) = 12 mv2 − 12 mu2 where m is the mass of the object in kilograms (kg) v is the final speed of the object in metres per second (m s−1) u is the initial speed of the object in metres per second (m s−1) Since the mass of the object is generally unaltered, often this can be simplified to:

W = ∆…k = 12 m(v 2 – u 2)

Interactive tutorial 4 Braking (video analysis of motion)

Therefore, if an object undergoes a known change in kinetic energy during an interaction, the work done on the object by the net force is known. Hence the average net force exerted on the object during this interaction can be calculated by assuming that ∆Ek = W = Fav x.

Worked example 6.4A Calculate the kinetic energy of an athlete of mass 60 kg running at a speed of 8.0 m s−1.

Solution m = 60 kg, v = 8.0 m s−1 1

Using …k = 2 mv2: …k = 12 × 60 × 8.02 ≈ 1900 J.

Worked example 6.4B Blood is pumped by the heart into the aorta at an average speed of 0.15 m s−1. If 100 g of blood is pumped by each beat of an adult human’s heart find: a the amount of work done by the heart during each contraction b the energy used by the heart each day in pumping blood through the aorta (use an adult’s average resting rate of 70 beats per minute). Assume that there are no other energy losses.

Solution a The work done by the heart is equal to the kinetic energy the blood gains as it is pumped into the aorta. m = 0.10 kg, v = 0.15 m s−1, u = 0 m s−1 1 Using W = ∆…k = 2 m(v2 − u2) 1 W = 2 × 0.10 × (0.152 − 02) W = 1.125 × 10−3 J = 1.1 mJ b If there are 70 beats each minute then the amount of energy transferred: …k per minute = 1.125 × 10−3 × 70 = 0.07875 J per minute …k per day = 0.07875 × 60 min per hour × 24-hour day …k = 113.4 J per day ≈ 110 J per day

200

Motion

Potential energy An object can have energy not only because of its motion, but also as a result of its shape or position. This is called potential energy. A gymnast, crouched ready to jump, has potential energy. During the jump, work is being done by the force exerted by the gymnast, and potential energy is converted into kinetic energy from the stores of chemical energy in the muscles of the gymnast’s body. There are many different forms of potential energy: chemical, gravita tional, elastic etc. Potential energy is a stored energy giving the body potential to do work or produce a force that creates motion. In this particular study we are mainly concerned with gravitational and elastic potential energy which, for the present, we will denote U.

Gravitational potential energy An athlete at the top of a high-jump has gravitational potential energy because of his position. As he falls, work is done (Figure 6.20). Recall that in this case the work done is given by: Work done = ΣFx The force acting on the body is simply the force due to gravity also called the person’s weight: Weight = mg The displacement that occurs is in a vertical direction and can be described as a change in height, ∆h. Replacing F and x with these equivalent terms gives: W = mg∆h Similarly, the work done in raising the athlete against a gravitational field is stored as gravitational potential energy; hence, the athlete has a change in potential energy: ∆Ug = mg∆h

Figure 6.20 The energy gained or lost due to a change in height within a gravitational field is called gravitational potential energy. An increase in height will require the transformation of energy from other sources. A decrease will usually increase the kinetic energy of the body.

Figure 6.21 This photograph of a pole-vaulter illustrates that elastic potential energy is stored in the pole. This energy is largely converted to kinetic energy and then the gravitational potential energy of the athlete.


201

Physics file The relationship for gravitational potential energy used here is only appropriate when the weight force due to gravity is constant. This will only be the case when the change in height is relatively small. As the distance from the Earth’s surface changes so will the strength of the gravitational field according to the relationship 1 g ∝  2 where r is the distance (in r metres) from the centre of the Earth to the body’s position. The area under a force–displacement graph can then be used to find the change in potential energy due to a change in position and the varying weight force. For the purposes of this study only relatively small changes in height close to the Earth’s surface will be considered for which the weight force can be considered constant.

The change in gravitational potential energy is due to the work done against a gravitational field and is given by: ∆Ug = mg∆h where ∆Ug is the change in gravitational potential energy measured in joules (J) m is the mass of the body (kg) Dh is the change in height (m) g is the acceleration due to gravity (m s–2) The ∆Ug of a body depends only on the vertical height of the object above some reference point, in this case the ground. It does not depend on the path taken since it is based on the direction of the gravitational field. It is the work done against or by the force of gravity that leads to changes in gravitational potential energy. Similarly, the work it can do when falling does not depend on whether the object falls vertically or by some other path, but only on the vertical change in height, ∆h. The reference level from which the height is measured does not matter as long as the same reference level is used throughout a given problem solution. It is only changes in potential energy that are important. For example, the height of the high jumper is best referenced to the ground she jumped from, and commonly it is her centre of gravity that is analysed. The height of a luggage locker in an aircraft makes a lot more sense when referenced to the floor of the aircraft than it would referenced to the ground. The need for considering a change in height in comparison to a reference level is also made apparent when considering a person standing at ground level. If the person is standing beside a hole and his centre of gravity is considered, he will have gravitational potential energy with reference to the bottom of the hole. It is also quite justifiable to suggest that even with reference to the ground he has gravitational potential energy; he could fall over! Gravity would do work on him and his gravitational potential energy would change. The change in height would be with reference to the person’s centre of mass.

Worked example 6.4C A (260 m)

B (389 m)

C (0 m)

A ranger with a mass of 60 kg, checking the surface of Uluru for erosion, walks along a path that takes her past points A, B and C. a What is her gravitational potential energy at points B and C relative to A? b What is the change in the ranger’s potential energy as she walks from B to C? c If the ranger was to walk from B to C via A would it alter your answer to part b? Explain.

Solution a In this question heights are being referenced to point A. The person would have had zero gravitational potential energy at A using this reference. m = 60 kg, g = 9.8 m s−2, hA = 260 m, hB = 389 m, hC = 0 m Potential energy change from A to B: ∆Ug = mg(hB − hA) = 60 × 9.8 × (389 − 260) = 7.6 × 104 J

202

Motion

Potential energy change from A to C: ∆Ug = mg(hC − hA) = 60 × 9.8 × (0 − 260) = −1.5 × 105 J b There is no need to calculate ∆Ug from B to C separately as the difference between the two previous results can be used. Potential energy change from B to C: ∆Ug = ∆Ug (A to C) − ∆Ug (A to B) = −1.5 × 105 − 7.6 × 104 = −2.3 × 105 J c It makes no difference what path is taken to achieve the change in height. Potential energy change throughout this example is being determined relative to an initial height. In general, if an object is originally at a height h0, then the change in potential energy as it moves to a different height, h, is: ∆Ug = mgh − mgh0 = mg(h − h0) In general terms, the change in potential energy of an object when it is moved between two heights is equal to the work needed to take it from one point to another.

Elastic materials and elastic potential energy The third aspect of mechanical energy that we will study is elastic potential energy. Like gravitational potential energy it occurs in situations where energy can be considered to be stored temporarily so that, when this energy is released, work may be done on an object. Elastic potential energy is stored when a spring is stretched, a rubber ball is squeezed, air is compressed in a tyre or a bungee jumper’s rope is extended during a fall. Since each object possesses energy due to its position or motion, these all suit our earlier definition of mechanical energy. We will see that when some materials are manipulated we can think of this as work being done to store energy. This energy is often released or utilised via work being done on another object. Materials that have the ability to store elastic potential energy when work is done on them and then release this energy are called elastic materials. Metal springs are common examples, but also realise that many materials are at least partially elastic. If their shape is manipulated, items such as our skin, metal hair clips and wooden rulers all have the ability to restore themselves to their original shape once released—within limits of course! Materials that do not return mechanical energy when their shape is distorted are referred to as plastic materials. Plasticine is an example of a very plastic material.

Ideal springs obey Hooke’s law Springs are very useful items in our everyday life due to the consistent way in which many of them respond to forces and store energy. When a spring is stretched or compressed by an applied force we say that elastic potential energy is being stored. In order to store this energy work must be done on the spring. Recall that in section 6.3 we have discussed that if a force of a constant value is applied to an object (and a displacement occurs in the direction of that force) then the quantity of work done can be calculated using W = Fx. This formula can therefore be used when a set force, F, has been applied to a spring and a given compression or extension, Δx, occurs.


203

Applied force (N)

Extension (m)

Figure 6.22 Ideal materials obey Hooke’s law:

F ∝ k∆x.

PRACTICAL ACTIVITY 26 Hooke’s law

However, we are usually interested in examining how a spring will behave in a range of conditions. Consider the situation in which a spring is stretched by the application of a steadily increasing force. As the force increases, the extension of the spring, Δx, can be graphed against the applied force, F. You can imagine hanging a spring vertically and gradually adding more and more weight to it so that is stretches. Many items, such as well-designed springs, will (at least for a small load) extend in proportion to the applied force. For example, if a 10 newton force produced an extension of 6 cm, then a 20 newton force would produce an extension of 12 cm. These items are called ideal springs. The resulting graph of applied force versus extension would be linear as in Figure 6.22. Note that the gradient of this graph tells us the force, in newton, required to produce each unit of extension. The gradient of the graph is called the spring constant, k, measured in N m−1. The gradient therefore indicates the stiffness of the spring, and for an ideal spring this gradient has a set value (i.e. the F vs Δx graph is a straight line). A very stiff spring that is difficult to stretch would have a steep gradient; that is, a large value of k. Although k is usually called the spring constant, it is sometimes called the stiffness constant or force constant of a spring. A spring constant of k = 1500 N m−1 indicates that for every metre that the spring is stretched or compressed, a force of 1500 N is required. This does not necessarily mean that the spring can be stretched by 1 m, but it tells us that the force and the change in length are in this proportion. The relationship between the applied force and the subsequent extension or compression of an ideal spring is known as Hooke’s law. Since for ideal springs F ∝ Δx, we can say F = kΔx. However, as we are often interested in using the energy stored by stretched or compressed springs we tend to refer to the force that the distorted spring is able to exert (rather than the force that was applied to it). Newton’s third law tells us that an extended or compressed spring in equilibrium is able to exert a restorative force equal in size but opposite in direction to the force that is being applied to it. Therefore Hooke’s law is often written in the form shown below.

HOOK…’S LAW states that the force applied by a spring is directly proportional, but opposite in direction, to the spring’s extension or compression. That is: F = −kDx where F is the force applied by the ideal spring (N) k is the spring constant (N m−1) (also called force constant or stiffness constant) Dx is the amount of extension or compression of the ideal spring (m)

Calculating elastic potential energy Work must be done in order to store elastic potential energy in any elastic material. Essentially the energy is stored within the atomic bonds of the material as it is compressed or stretched. The amount of elastic potential energy stored is given by the area under the force–extension graph for the item. For materials that obey Hooke’s law (such as the material shown in Figure 6.22), an expression can be derived for the area under the F–x graph.

204

Motion

The elastic (or spring) potential energy, Us, stored in an item is always given by the area below the force–extension graph for the item. The unit of Us is the joule, J. In the case of an ideal material that obeys Hooke’s law, the elastic potential energy is given by the expression: Us = 12 k∆x2 where Us is the elastic potential energy stored during compression or extension (J) k is the stiffness constant of the material (N m−1) ∆x is the extension or compression (m)

area under graph = work done

Force applied (N)

Work done = area under F–x graph = area of a triangle 1 = 2 × base × height 1 = 2 × ∆x × F But since F = k∆x: 1 Work done = 2 × ∆x × k∆x 1 W = 2 k∆x2 = the elastic potential energy stored during the extension/ compression Although many materials (at least for a small load) extend in proportion to the applied force, many materials have force–extension graphs more like that shown in Figure 6.23. For these materials the area under the F–∆x graph must be used to determine the elastic potential energy stored.

Extension (m)

Figure 6.23 Elastic potential energy is a form of mechanical energy. Work is done as elastic potential energy is stored, as indicated by the area under the F–∆x graph.

Physics file Take care! The two forms of potential energy, elastic (or spring) potential energy, Us, and gravitational potential energy, Ug, have been introduced. Take extra care when analysing situations like pole-vaulting, since at some stages in this event both forms of potential energy are present at the same time!

Worked example 6.4D 120 100 F applied (N)

Three different springs A, B and C, are exposed to a range of forces and the subsequent extension measured. The data collected for each spring has been graphed in the F–Δx graph at right. a Justify the statement that spring B is the only ideal spring shown. b Calculate the stiffness constant of spring B. c Calculate the work done in extending spring B by 25 mm. Assume the process of storing energy is 100% efficient. d Estimate the work done in extending spring C by 25 mm. e If all springs are extended such that Dx = 40 mm, which spring will have stored the most elastic potential energy? Justify your choice.

A B

80 60 40

C

20 10 20 30 40 50 60 x (mm)

Solution a Ideal springs produce an extension that is consistently proportional to the applied

force. Since spring B has an F–Dx graph which is a straight line emerging from the origin, it is behaving ideally and obeying Hooke’s law until an extension of ~30 mm is reached. Spring C does not obey Hooke’s law, that is, F is not directly proportional to Dx, since the graph is not a straight line. On close inspection it can be seen that the initial application of a small force did not produce any extension in spring A. (This is a common behaviour of real springs where a certain minimum amount of force must be applied before any extension will occur). This means that spring A has not obeyed Hooke’s law and therefore is not an ideal spring.


205

b c

k = gradient of F–Dx graph = rise/run = 90/0.030 = 3.0 × 103 N m−1 Since spring B obeys Hooke’s law, the equation Us = 12 kDx2 can be applied. W = DUs = Us[final] − Us[initial] As there is initially zero energy stored: 1 W = 2 kDx2 1 = 2 × 3.0 × 103 × 0.0252 = 0.94 J d Spring C does not obey Hooke’s law, so the work done must be calculated using the area under the F–Dx graph: 1 square of area is equal to (0.005 × 10) or 0.05 joules. There are approximately 7.5 squares of area. Therefore, W ≈ 7.5 × 0.05 ≈ 0.38 J e Elastic potential energy is given by the area under the F–Dx graph. At an extension of 40 mm spring A will have the greatest area under the graph, i.e. it will have stored the most elastic potential energy.

Physics in action

James Joule By the mid 19th century, several scientists had begun to write of the heating process as an energy change from work (mechanical energy) to heat. It was eventually realised that all forms of energy were equivalent and that when a particular form of energy seemed to disappear, the process was always associated with the appearance of the same amount of energy in other forms. This led to the development of the principle of conservation of energy. At this same time, James Joule conducted a series of experiments fundamental to our present understanding of heat. Joule noticed that stirring water could cause a rise in temperature. He designed a way of measuring the relationship between the energy used in stirring the water and the change in temperature. A metal paddle wheel was rotated by falling masses and this churned water around in an insulated can. The amount of work done was calculated by multiplying the weight of the falling masses by the distance they fell. The heat generated was calculated from the mass of the water and the temperature rise. Joule found that exactly the same quantity of heat was always produced by exactly the same amount of work. Heat was simply another form of energy, and 4.18 joules of work was equivalent to 1 calorie of heat. Joule’s work led to some unusual conclusions for his day. He stated that as a container of cold water is stirred, the mechanical energy is being transformed into thermal energy, heating the water. Theoretically, this means that a cup of water stirred long enough and fast enough will boil—a novel, if laborious, way of making a cup of coffee. Of course, the rate at which we can normally add energy by stirring is less than the transfer of energy to the surrounding environment. For the cup of water to boil Figure 6.24 James Pre scott Joule it would need to be very well insulated. . As a result of Joule’s investigations and other experiments of the time, we now interpret the process of heating or cooling as a transfer of energy. When heat ‘flows’ from a hot object to a cold one, energy is being transferred from the hot to the cold.

206

Motion

water paddle pulley

falling masses

Figure 6.25 Joule’s original apparatus for investigating the mechanical work equivalent of heat energy. The falling weights caused the paddle to turn. The friction between the wheel and the water created heat energy in the water. For the first time, heat energy could be measured and related to other forms of energy.

ne being do ical work f the falls is n a h c e o of m e bottom amount iderable of the water at th s n o c e f th perature result o .26 As a waterfall, the tem e top. Figure 6 th ater of a than at on the w C or 2°C higher ° 1 usually

6.4 summary Mechanical energy • Energy is the ability to do work. Whenever work is done, energy is transformed from one form to another. • Kinetic energy is the energy a body has because of 1 its motion. Ek = 2 mv2 where Ek is the kinetic energy in joules (J), m is the mass in kilograms (kg) and v is the speed in metres per second (m s−1). • Potential energy is stored energy with the potential to allow work to be done. It may take many forms including chemical, elastic and gravitational. • Gravitational potential energy is the energy a body has because of its position within a gravitational field: ∆Ug = mg∆h where Ug is the gravitational potential energy in joules (J), m is the mass in kilograms (kg)

and ∆h is the change in height from a reference height in metres (m). • Ideal materials extend or compress in proportion to the applied force; that is, they obey Hooke’s law: F = −k∆x • The elastic potential energy, Us, stored in an item is given by the area below the force–extension graph 1 for that item, or Us = 2 kx2 for an ideal spring that obeys Hooke’s law. • When work is done to store energy, one or more of the following may be applied: W = ∆Ug or W = ∆Ek or W = ∆Us or W = area under F–∆x graph


207

6.4 questions Mechanical energy 1 Calculate the kinetic energy of a: a 1.0 kg mechanics trolley with a velocity of 2.5 m s−1 b 5.0 g bullet travelling with a velocity of 400 m s−1 c 1200 kg car travelling at 75 km h−1.

The following information relates to questions 7 and 8. The force–extension graphs for three different springs are shown below. 300 F applied (N)

Where appropriate use g = 9.8 m s−2.

2 Calculate the gravitational potential energy relative to the ground when a: a mass of 1.0 kg is 5 m above the ground b bird of mass 105 g is 400 m above the ground c 1200 kg car has travelled a vertical height of 10 m up a slope. 3 A 100 g rubber ball falls from a height of 2.5 m onto the ground and rebounds to a height of 1.8 m. What is the gravitational potential energy of the ball relative to the ground at its: a original position? b final position? c final position relative to its original position?

A B

200

100

C

0.05 0.10 0.15 0.25 x (m)

7 Calculate the spring constant for each spring and determine which is the stiffest spring. 8 Each spring has a force of 100 N applied to it. Calculate the elastic potential energy stored by each spring.

4 Which object has the greatest amount of energy? A A spring with a spring constant k = 40 000 N m−1 compressed by 5.0 cm B A cricket ball of mass 150 g stuck on the roof of a grandstand 14 m above the ground C A cricket ball of mass 150 g travelling at 10 m s−1 at a height of 10 m above the ground

9 A piece of gymnasium equipment involves compressing a spring whose spring constant is 2500 N m−1. How much elastic potential energy is stored in the spring when it is compressed by: a 5.00 cm? b 10.0 cm? c 15.0 cm?

5 What net braking force must be applied to stop a car within a straight-line distance of 50 m, if the car has a mass of 900 kg and was initially travelling at a velocity of 100 km h−1?

10 The gymnasium equipment described in question 9 is adjusted so that its spring constant is now 3000 N m−1. If 12.5 J of energy is now stored in the spring, by what distance has the spring been compressed?

6 A small steel ball with a mass of 80 g is released from a resting height of 1.25 m above a rigid metal plate. Calculate the: a change in gravitational potential energy b kinetic energy of the ball just before impact c velocity of the ball just before impact.

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Motion

6.5 Energy transfo rm

ation and pow

er

Besides the mechanical energy discussed in section 6.4, other forms of energy exist, for example nuclear, heat, electrical, chemical and sound energy. Atomic theory has led to each of these other forms being understood as either kinetic energy or potential energy at the molecular level. Energy stored in food or fuel and oxygen can be considered as potential energy stored as a result of the electrical forces in the molecules. Despite the apparently different nature of the various forms of energy, any energy can be transformed from one form to another. The connecting factor is that all forms can do work on a body and therefore can be measured and compared in this way. A stone dropped from some height loses gravitational potential energy as its height decreases; at the same time its kinetic energy will increase as its speed increases.

Transformation of energy Energy transfers or transformations enable people and machines to do work, and processes and changes to occur. Elastic potential energy stored in a diving board must be transformed into the kinetic energy of the diver at the pool. Contracting a muscle converts chemical potential energy stored in the muscle to the kinetic energy of a person’s motion. In each example, the transformation of energy means work is being done.

Interactive tutorial 5 Kinetic and gravitational potential energy

Work is done whenever energy is transformed from one form to another. In many cases a transformation of energy produces an unwanted consequence—a substantial amount of the energy is ‘lost’ as heat energy. Of a typical adult’s daily food intake of about 12 MJ at least 80% is converted into heat energy during normal activity. Such transfers can be depicted by an energy-conversion flow diagram. (a)

(b) gravitational potential energy work done h

work done

heat

kinetic energy work done sound

work done heat

Figure 6.27 Whenever work is done, energy is transformed from one form to another. (a) As a body falls, gravitational potential energy is transformed to kinetic energy and heat, from the friction with the air. Once the body lands, further energy transformations will take place. (b) An energy-conversion flow diagram can be useful in visualising the transformations that take place.

A simple, although infinitely unlikely, example is shown in Figure 6.27. As the body falls to the ground there will be a number of energy


209

transformations. An energy flow diagram illustrates these changes. While the body falls, work will be done on the body by the gravitational field, and gravitational potential energy becomes kinetic energy, the energy of movement. There will also be some energy converted into heat by the action of air resistance. When the body hits the ground, the kinetic energy is converted into elastic potential energy by the compression of the body, and to other forms, particularly heat but also to sound and kinetic energy. Each transformation requires a force to do work on the body.

The efficiency of energy transformations The percentage of energy that is transformed to a useful form by a device is known as the efficiency of that device. All practical energy transformations ‘lose’ some energy as heat. The effectiveness of a transfer from one energy form to another is expressed as: efficiency (%) =

useful energy transferred × 100 useful output × 100 = total energy supplied total input

Table 6.4 Efficiencies of some common energy transfers

Figure 6.28 In each of these situations below an energy transformation is taking place. Can you identify the forms involved in each transformation?

Device

Desired energy transfer

Efficiency (%)

Large electric motor

Electric to kinetic

90

Gas heater or boiler

Chemical to heat in water

75

Steam turbine

Heat to kinetic

45

High-efficiency solar cell

Radiation to electric

25

Coal-fired electric generator

Chemical to electric

30

Compact low-energy fluorescent light

Electric to light

25

Human body

Chemical to kinetic

25

Car engine

Chemical to kinetic

25

Open fireplace

Chemical to heat

15

Filament lamp

Electric to light

5

In all the energy transformations included in Table 6.4, the energy lost in the transfer process is mainly converted into heat. Most losses are caused by the inefficiencies involved in the process of converting heat into motion. In the real world, energy must be constantly provided for a device to continue operating. A device operating at 45% efficiency is converting 45% of the supplied energy into the new form required. The other 55% is lost to the surroundings, mainly as heat but also some as sound.

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Motion

Physics in action

Air resistance in sport Air, like water, is a fluid and therefore there is a force between the particles of the air and the surface of any object moving through it. This force is called air resistance or drag. Drag is due to the air particles that can be thought of as obstacles in the path of a moving object. There will also be frictional forces as air particles slide past the object. At low speeds the effect of air resistance is slight. However, it has been found that air resistance is proportional to the square of the velocity (i.e. Fa ∝ v2). A doubling of speed will increase air resistance approximately four times. At the racing speeds that Olympic cyclists reach of 50 km h−1 or more, 90% of the cyclist’s energy is required just to push the bicycle and rider through the surrounding air. The remaining 10% is needed to overcome frictional forces between the wheels and the ground. Figure 6.30 Th Air resistance is affected by the frontal area of crossby the RMIT e technological adv ances in b , have led to section and the shape of the bicycle. Designers have tried ike Australian riders in re efficient designs suc design, pioneered to reduce the frontal area of racing bikes and their riders cent intern h ational com as that used by by dropping the handle bars and raising the position of the p

etition.

Drag (N)

20

10

0

5

9.0

13.5

Velocity (m s−1)

Figure 6.29 The efficiency of a cyclist is affected by the velocity of the bike. As the velocity increases, the drag or air resistance can use as much as 90% of the energy the cyclist’s input.

pedals. This allows the rider to race bent forward, reducing the area presented to the air. Streamlining the bike helps still further. Some shapes move through fluids more easily than others. Streamlined bicycles, such as those used by the Australian Cycling Team in international competition, have low-profile frames with relatively smaller front wheels. Brake and gear cables are run through the frame rather than left loose to create drag. Moulded three-spoke and solid-disk wheels help still further. The riders’ clothing and helmets have also been streamlined. Cyclists wear pointed shoes, streamlined helmets and skin-tight one-piece lycra bodysuits. Together these reduce the air resistance on a rider by as much as 10% at higher speeds. At race time riders shave their legs to reduce energy losses just that little bit more.

Conservation of energy No matter what energy transformation occurs overall, no energy is gained or lost in the process. It is a fundamental law of nature that energy is conserved. Consider the example of a diver as depicted in Figure 6.31. When the diver is at the top, his or her gravitational potential energy will be at a maximum. As the diver free falls the gravitational potential energy decreases but the kinetic energy will increase with the increased velocity. Some—a small amount—of energy will be converted into heat due to contact with the air. The moment before the diver reaches the reference level all the gravitational potential energy has been converted into other energy forms, mostly kinetic energy. The total at this point will be exactly


211

All E p

equal to the potential energy at the top. The total energy in the system remains the same. Energy has been conserved.

The TOTAL …N…RGY in an isolated system is neither increased nor decreased by any transformation. Energy can be transformed from one kind to another, but the total amount stays the same.

1 — 2 Ep 1 — 2 Ek

This applies to any situation involving energy transfer or transformations in an isolated system. In this particular case the sum of the gravitational potential energy and the kinetic energy at any point is called the total mechanical energy.

Kinetic energy + potential energy = total mechanical energy

All E k

Figure 6.31 A diver loses gravitational potential energy but gains kinetic energy during the fall.

PRACTICAL ACTIVITY 27 Conservation of energy

212

Motion

Here, the total mechanical energy remains constant. As an object falls, gravitational potential energy decreases but kinetic energy increases to compensate, so that the total remains constant. At any point during an object’s free fall: 1 total mechanical energy = 2 mv2 + mgh There are many examples of this conservation of energy. In athletics, the pole-vaulters and high-jumpers base their techniques on this principle. Throwing a ball in the air is another example. When the ball leaves the hand, its kinetic energy is at a maximum. As it rises, its velocity decreases, reducing the kinetic energy, and its potential energy increases by the same amount. At any point Ek + Ep will equal the initial kinetic energy. At the top of the throw, the ball will have a vertical velocity of zero and, in a vertical direction, the energy will be totally gravitational potential energy (any horizontal motion will be represented by a remaining amount of Ek). The transformation will reverse as the ball falls. Gravitational potential energy will decrease as the ball returns towards its original height and, with its speed increasing, the kinetic energy will increase once more. A more complex example is provided by the interactions as a gymnast repeatedly bounces on a trampoline. Figure 6.32 is a series of frames from a video of a gymnast carrying out a routine on a trampoline. Kinetic energy and gravitational potential energy changes are shown in the graph below the frames. Despite the complexity of the motion, the total energy of the gymnast remains the same during each airborne phase, as illustrated in the graph. On landing on the bed of the trampoline, the energy is transferred to elastic potential energy within the trampoline and both kinetic and gravitational potential energy fall. On take-off some of this energy will be permanently transferred to the trampoline and its surrounds, thus lowering the total available to the gymnast. This is represented by the reduced total energy for each successive jump. Were the gymnast to flex his legs then additional energy would be added to the mechanical energy available and this total could be maintained or even increased until the gymnast finally ran out of available energy himself.

81

20

43

185 122

4000

flight on bed

flight

125

194

159 174

207

147

on bed

218

flight

3000

gravitational potential energy plus kinetic energy

2000

gravitational potential energy kinetic energy

1000

Energy (J)

102

64

20

40

60

80 100 120 140 Number of frames

160

180

200

220

Figure 6.32 During each airborne stage of a gymnast’s trampoline routine (indicated on the graph by shading), mechanical energy is conserved. The graph shows the relationship between total energy and gravitational potential energy and kinetic energy. Each time the gymnast lands, energy is transferred to the trampoline. The energy returning from the springs after each landing allows the routine to continue.

Worked example 6.5A Calculate the initial velocity required for a high jumper to pass over a high bar. Assume the jumper’s centre of gravity rises through a height of 1.5 m and passes over the bar with a horizontal velocity of 1.2 m s−1, and that all of his initial horizontal kinetic energy is transferred into Ug and …k. Use g = 9.8 m s−2.

Solution As the total mechanical energy is assumed to be conserved after landing, the initial horizontal kinetic energy equals total mechanical energy at the peak height: 1 2 2 mu = …k + Ug (at peak height) = 12 mv2 + mg∆h The mass cancels out, giving an expression independent of the mass of the athlete. The same speed at take-off will be required for a light person as a heavy one. (If this doesn’t seem to make sense remember that all objects fall at the same rate regardless of their mass.) 1 2 1 2 2 u = 2 v + g∆h Substituting the values from the question: 1 2 1 2 2 u = 2 × 1.2 + 9.8 × 1.5 1 2 2 u = 15.42 and u = √2 × 15.42 = 5.6 m s−1. In reality the take-off speed will need to be a little greater since there will be some losses to friction.


213

Worked example 6.5B A climber abseiling down a cliff uses friction between the climbing rope and specialised metal fittings to slow down. If a climber of mass 75 kg abseiling down a cliff of height 45 m reaches a velocity of 3.2 m s−1 by the time the ground is reached, calculate the average frictional force applied. Use g = 9.8 m s−2.

Solution m = 75 kg, u = 0 m s−1, v = 3.2 m s−1, h = 45 m Gravitational potential energy at the top of the cliff: …p = mgh = 75 × 9.8 × 45 = 33 075 J Kinetic energy at ground level: …k = 12 mv2 = 12 × 75 × 3.22 = 384 J Total energy transformed to forms other than gravitational potential and kinetic energy: ∆… = …p − …k = 33 075 − 384 = 32 691 J This change in energy will be equivalent to the work done by the frictional force; that is: Work = ∆… = Ff × x and: D… 32 691 Ff = = = 726 N ≈ 730 N 45 x Physics file The British Imperial unit for power is the horsepower, hp, dating from the time of the Industrial Revolution when the performance of steam engines was compared with that of the horses they were replacing. 1 hp = 746 W. The SI unit for power honours the inventor of the steam engine, James Watt.

Power Why is it that running up a flight of stairs can leave you more tired than walking up if both require the same amount of energy to overcome the force of gravity? The answer lies in the rate at which the energy is used. When horses were first replaced by steam engines, the engine was rated by how fast it could perform a given task compared with a horse. An engine that could complete a task in the same time as one horse was given a rating of one horsepower. More formally, power is defined as the rate at which energy is transformed or the rate at which work is done.

POW…R =

work done energy transformed = time taken time taken

or

W D… = Dt Dt where P is the power developed in watts (W) resulting from an energy transformation Δ… occurring in time Δt. Δ… is measured in joules (J), time is measured in seconds (s). P=

Determining the power developed is fairly straightforward when mechanical work is done; but consider a situation in which a person pushes a lawnmower, say, at constant speed. Here, there is no increase in kinetic energy, but energy is being transformed to overcome the frictional forces acting against the lawnmower.

214

Motion

In the special case of an applied force opposing friction or gravity and doing work with no increase in the speed of the object, we can say: As W = Fx then: Fx P= x Dt and as v = then: Dt P = Favvav where P is power developed (W) Fav is average applied force (N) vav is average speed (m s−1) This is useful when finding the power required to produce a constant speed against a frictional or gravitational force. The rate of energy use is as much a limiting factor of the work a person can do as the total energy required. A person may be able to walk or climb a long distance before having to stop because all available energy is used. The same person will fall over exhausted after a much shorter time if the same journey is attempted at a run. Power is the limiting factor, the rate at which a person’s body can transform chemical energy into mechanical energy. Few humans can maintain one horsepower, about 750 W, for any length of time. Table 6.5 includes comparative figures for the power developed in various activities and devices.

Figure 6.33 It is not the amount of energy required that stops the rest of us from winning the 400 m sprint, but the rate at which we can effectively convert it to useful work.

Table 6.5 Average power ratings for various human activities and machines Activity or machine

Power rating (W)

Sleeping adult

100

Walking adult

300

Cycling (not racing)

500

Standard light globe

60

Television Fast-boil kettle Family car

200 2400 150 000

Worked example 6.5C The fastest woman to scale the Rialto building stairs in the Great Rialto Stair Trek in a particular year climbed the 1222 steps, which are a total of 242 m high, in 7 min 58 s. Given that her mass is 60 kg, at what rate was she using energy to overcome the gravitational force alone? Use g = 9.8 m s−2.

Solution The work is against gravity so: U mgDh P= g = Dt Dt m = 60 kg, g = 9.8 m s−2, ∆h = 242 m ∆t = (7 min × 60) + 58 s = 478 s 60 × 9.8 × 242 P= = 3.0 × 102 W 478


215

6.5 summary Energy transformation and power • Whenever work is done, energy is converted from one form into another. • The efficiency of an energy transfer from one form to the required form is: energy output efficiency (%) = × 100 energy input • Whenever energy is transformed, the total amount of energy in the system remains constant. This conserv ation of energy is a fundamental natural principle.

• The total mechanical energy will remain constant in an isolated system. That is: Ek + Us + Ug = constant • Power, P (in watts, W), is the rate at which work is done or energy transformed: W DE P= = Dt Dt • In the particular case of work being done to overcome friction, with no resulting increase in speed: P = Favvav

6.5 questions Energy transformation and power Use g = 9.8 m s−2 where required. 1 Describe the energy transformations that take place when: a a car slows to rest b a gymnast uses a springboard to propel themselves into the air c an archer draws back and then releases an arrow vertically upward d an athlete’s foot hits a track. 2 Draw an energy transformation flow chart for a swimmer diving off a diving board and into a pool of water. 3 A boy of mass 46 kg runs up a 12-m high flight of stairs in 12 s. a What is the gain in gravitational potential energy for the boy? b What is the average power he develops? 4 A coach is stacking shot-puts, from the shot-put event, onto a shelf 1.0 m high following an athletics meeting. Each shot-put has a mass of 500 g and all are being lifted from the ground. The coach stacks 15 shot-puts, at the same level, in 2.0 minutes. a How much useful work has been done in lifting all the shot-puts? b What is the total gravitational potential energy of all the shot-puts on the shelf? c What was the coach’s average power output in performing this task? d The actual power output would be considerably greater than the answer to part c. Suggest two possible reasons for this difference.

216

Motion

5 One of the shot-puts in question 4 rolls off the shelf just after the coach has finished. a What is the gravitational potential energy of the shot-put when it is halfway to the ground? b What is the kinetic energy of the shot-put when it is halfway to the ground? c What happens to the kinetic energy of the shotput when it hits the ground? 6 Tarzan is running at his fastest speed (9.2 m s−1) and grabs a vine hanging vertically from a tall tree in the jungle. a How high will he swing upwards while hanging on to the end of the vine? b What other factors that have not been considered may affect your answer? 7 In high jumping, the kinetic energy of an athlete is transformed into gravitational potential energy. With what minimum speed must the athlete leave the ground in order to lift his centre of gravity 1.80 m high with a remaining horizontal velocity of 0.50 m s−1? 8 A 100 g apple falls from a branch 5 m above the ground. a With what speed would it hit the ground if air resistance could be ignored? b If the apple actually hits the ground with a speed of 3.0 m s−1, what was the average force of air resistance exerted on it? 9 A 150 g ball is rolled onto the end of an ideal spring whose spring constant is 1000 N m−1. The spring is temporarily compressed.

a The ball compresses the spring by a maximum distance of 10 cm. How much elastic potential energy is stored in the spring at this compression? b How fast must the ball have been travelling just before it began to compress the spring? Ignore any frictional effects. c If in another trial the ball reached a speed of 5.0 m s−1 before compressing the spring, how far would the spring be compressed?

10 As a 30 kg child compressed the spring of a pogo stick, it stored 150 J of elastic potential energy. Assuming the spring is 50% efficient: a how much kinetic energy will the child be given as the spring rebounds? b with what speed will the child rebound? c ignoring air resistance, what gain in height will the child achieve?

chapter review Use g = 9.8 m s−2 where required.

Force (N)

The following information relates to questions 1–4. A ball of mass 50 g strikes a brick wall. It compresses a maximum distance of 2.0 cm. The force extension properties of the ball are shown below.

800 700 600 500 400 300 200 100 0.01 0.02 Compression (cm)

1 What work does the wall do on the ball in bringing it to a stop? 2 How much …p is stored in the ball at its point of maximum compression? 3 If the ball–wall system is 50% efficient, what is the rebound speed of the ball? 4 At the instant that the ball had only been compressed by 1.0 cm, had the wall done half of the work required to stop the ball? Explain. The following information relates to questions 5 and 6. An arrow with a mass of 80 g is travelling at 80 m s−1 when it reaches its target. It penetrates the target board a distance of 24 cm before stopping. 5 Calculate the arrow’s kinetic energy just before impact. 6 Calculate the average net force between arrow and target. 7 A 70 kg bungee jumper jumps from a platform that is 35 m above the ground. Assume that the person, the rope and the Earth form an isolated system. a Calculate the initial total mechanical energy of this system. b Write a flow chart displaying the energy transformations that are occurring during the first fall.

c The person can fall a distance of 10 m before the rope attached to her feet begins to extend. How much kinetic energy will the person have at this moment? d After a fall of a further 15 m, the person momentarily stops and the rope reaches its maximum extension. How much elastic potential energy is stored in the rope at this moment? e Since the bungee jumper bounces and eventually comes to rest, this is not a truly isolated system. Explain. 8 A stone of mass 3 kg is dropped from a height of 5 m. Neglecting air resistance, what will the kinetic energy of the stone be in joules just before the stone hits the ground? A 3 B 5 C 15 D 147 E 150 The following information relates to questions 9 and 10. An object of mass 2 kg is fired vertically upwards with an initial kinetic energy of 100 J. Assume no air resistance. 9 What is the speed of the object in m s−1 when it first leaves the ground? A 5 B 10 C 20 D 100 E 200 10 Which of A–E in question 9 is the maximum height in metres that it will reach? The following information relates to questions 11–16. After a particularly wet winter, a weir is overflowing at the rate of 800 litres of water every second (1 litre of water has a mass of 1 kg). The water takes 1.3 s to fall to the river below. 11 With what vertical velocity does the water hit the river below?


217

12 What height does the water fall through to the river below? 13 What weight of water falls over the weir every 10 s? 14 Calculate the work that has been done on this weight of water by gravitational forces by the time it reaches the river. 15 Calculate the power developed by the falling water at the instant before it hits the river. 16 Use a diagram to illustrate the energy transformations that occur as the water falls from the weir to the river below. The following information relates to questions 17–20. A roller-coaster is shown in the following diagram. Assume no friction. A C 30 m 25 m B

D 12 m

17 Calculate the speed at points B, C and D, assuming an initial speed of 4.0 m s−1 at point A. 18 Draw a graph of potential energy and kinetic energy against vertical displacement for this motion. Use separate lines for each form of energy and draw in a third line to represent the total mechanical energy, assuming no frictional losses. It is found that the roller-coaster actually just reaches point C with no remaining speed. 19 What are the energy losses due to friction and air resistance between A and C?

218

Motion

20 With what efficiency is the roller-coaster operating over this section of track? 21 Two players collide during a game of netball. Just before impact one player of mass 55 kg was running at 5.0 m s−1 while the other player, of mass 70 kg, was stationary. After the collision they fall over together. What is the velocity as they fall, assuming that momentum is conserved? 22 A 300 kg marshalling boat for a rowing event is floating at 2.0 m s−1 north. A starting cannon is fired from its bow, launching a 500 g ball, travelling at 100 m s−1 south as it leaves the gun. What is the final velocity of the marshalling boat? 23 A 150 g ice puck collides head on with a smaller 100 g ice puck, initially stationary, on a smooth, frictionless surface. The initial speed of the 150 g puck is 3 m s−1. After the collision the 100 g ice puck moves with a speed of 1.2 m s−1 in the same direction. What is the final velocity of the 150 g ice puck? 24 ‘When I jump, the Earth moves’. Is this true? Using reasonable estimates and appropriate physics relationships explain your answer. The following information relates to questions 25–27. In a horrific car crash, a car skids 85 m before striking a parked car in the rear with a velocity of 15 m s−1. The cars become locked together and skid a further 5.2 m before finally coming to rest. The mass of the first car, including its occupants, is 1350 kg. The parked car has a mass of 1520 kg. 25 What is the velocity of the two cars just after impact? 26 What is the impulse on each car during the collision? 27 What is the average size of the frictional force between road and car that finally brings them to rest?

area of study review Motion The following information applies to questions 1–3. The acceleration due to gravity may be taken as g = 9.8 m s−2 and the effects of air resistance can be ignored. An Olympic archery competitor tests a bow by firing an arrow of mass 25 g vertically into the air. The arrow leaves the bow with an initial vertical velocity of 100 m s−1. 1 At what time will the arrow reach its maximum height? 2 What is the maximum vertical distance that this arrow reaches? 3 What is the acceleration of the arrow when it reaches its maximum height? 4 Two students drop a lead weight from a tower and time its fall at 2.0 s. How far does the weight travel during the 2nd second, compared with the first second? The following information applies to questions 5–7. A car with good brakes, but smooth tyres, has a maximum retardation of 4.0 m s−2 on a wet road. The driver has a reaction time of 0.50 s. The driver is travelling at 72 km h−1 when she sees a danger and reacts by braking.

A

B

C

1.0 cm

D

3.0 cm

5.0 cm

12 Determine the average speed of the marble for the following distance intervals: a A to B b B to C c C to D 13 Determine the instantaneous speeds of the marble for the following times: a t = 0.05 s b t = 0.15 s c t = 0.25 s 14 Describe the motion of the marble.

5 How far does the car travel during the reaction time?

The following information applies to questions 15–18. A tow-truck, pulling a car of mass 1000 kg along a straight road, causes its velocity to increase from 5.00 m s−1 west to 10.0 m s−1 west in a distance of 100 m. A constant frictional force of 200 N acts on the car.

6 Assuming maximum retardation, calculate the braking time.

15 Calculate the acceleration of the car.

7 Determine the total distance travelled by the car from the time the driver realises the danger to the time the car finally stops.

16 What is the resultant force acting on the car during this 100 m?

The following information applies to questions 8–11. Two physics students, Helen and Emily, conduct the following experiment from a skyscraper. Helen drops a platinum sphere from a vertical height of 122 m while at exactly the same time Emily throws a lead sphere with an initial downward vertical velocity of 10.0 m s−1 from a vertical height of 140 m. Assume g = 9.80 m s−2 and ignore friction. 8 Determine the time taken by the platinum sphere to strike the ground. 9 Calculate the time taken by the lead sphere to strike the ground.

17 Calculate the force exerted on the car by the tow-truck. 18 What force does the car exert on the tow-truck? 19 A car that is initially at rest begins to roll down a steep road that makes an angle of 11.3° with the horizontal. Ignoring friction, determine the speed of the car in km h−1 after it has travelled a distance of 100 m (g = 9.8 m s−2). The following information applies to questions 20–23. A 100 kg trolley is being pushed up a rough 30° incline by a constant force F. The frictional force Ff between the incline and the trolley is 110 N. g = 9.8 m s–2

10 Determine the average velocity of each sphere over their respective distances. 11 In reality, will the diameters of the respective spheres affect the outcome of the experiment? The following information applies to questions 12–14. During a physics experiment a student sets a multi-flash timer at a frequency of 10 Hz. A nickel marble is rolled across a horizontal table. The diagram shows the position of the marble for the first four flashes: A, B, C and D. Assume that when flash A occurred t = 0, at which time the marble was at rest.

F

30˚

20 Determine the value of F that will move the trolley up the incline at a constant velocity of 5.0 m s−1. 21 Determine the value of F that will accelerate the trolley up the incline at a value of 2.0 m s−2. 22 Calculate the acceleration of the trolley if F = 1000 N. 23 What is the value of F if the trolley accelerates up the incline at 10 m s−2?


219 219

24 Two masses, 10 kg and 20 kg, are attached via a steel cable to a frictionless pulley, as shown in the following diagram.

a

d How much work is done on the car during the 5.0 s interval? e Determine the power output of the car’s engine during the 5.0 s interval. f How much heat energy is produced due to friction during the 5.0 s interval?

g = 9.8 m s–2 FT FT

10 kg

20 kg

a

a Determine the acceleration of each mass. b What is the magnitude of the tension in the cable?

The following information applies to questions 27–30. The following diagram shows the trajectory of a 2.0-kg shot-put recorded by a physics student during a practical investigation. The sphere is projected at a vertical height of 2.0 m above the ground with initial speed v = 10 m s−1. The maximum vertical height of the shot-put is 5.0 m. (Ignore friction and assume g = 9.8 N kg−1.) C

25 An 800 N force is applied as shown to a 20.0 kg mass, initially at rest on a horizontal surface. During its subsequent motion the mass encounters a constant frictional force of 100 N while moving through a horizontal distance of 10 m.

20.0 kg

5.0 m 3.0 m

F = 800 N 60°

B A 2.0 m

27 What is the total energy of the shot-put just after it is released at point A?

Ff = 100 N

28 What is the kinetic energy of the shot-put at point B? a Determine the resultant horizontal force acting on the 20.0 kg mass. b Calculate the work done by the horizontal component of the 800 N force. c Calculate the work done by the frictional force. d Calculate the work done by the resultant horizontal force. e Determine the change in kinetic energy of the mass. f What is the final speed of the mass? g Describe the effect of the frictional force on the moving mass.

29 What is the minimum speed of the shot-put during its flight? 30 What is the total energy of the shot-put at point C? 31 A 5.0 kg trolley approaches a spring that is fixed to a wall. During the collision, the spring undergoes a compression, ∆x, and the trolley is momentarily brought to rest, before bouncing back at 10 m s−1. Following is the force–compression graph for the spring. (Ignore friction.) v spring

50 40 30 20 10 0

0

2

t (s)

4

6

a How much kinetic energy (in MJ) does the car have at t = 5.0 s? b What is the resultant force acting on the car? c What force is provided by the car’s engine during the 5.0 s interval?

220

Motion

5.0 kg

Force (kN)

v (m s–1)

26 The figure shows the velocity–time graph for a car of mass 2000 kg. The engine of the car is providing a constant driving force. During the 5.0 s interval the car encounters a constant frictional force of 400 N. At t = 5.0 s, v = 40.0 m s−1.

12.0 10.0 8.0 6.0 4.0 2.0 0.0 0.0 1.0

2.0 3.0 4.0 5.0 Compression (cm)

6.0

a Calculate the elastic potential energy stored in the spring when its compression is equal to 2.0 cm. b What is the elastic potential energy stored in the spring when the trolley momentarily comes to rest? c At what compression will the trolley come to rest? d Explain why the trolley starts moving again.

e What property of a spring accounts for the situation described above? f Describe a situation in which the property of the spring in this example could be used in a practical situation.

Kinetic energy (J)

32 A nickel cube of mass 200 g is sliding across a horizontal surface. One section of the surface is frictionless while the other is rough. The graph shows the kinetic energy, …k, of the cube versus distance, x, along the surface. 5.0 4.0 3.0 2.0 1.0 0.0 0.0 1.0

B C

D

E

F

G

80 70 60 50

30 20

2.0

3.0 4.0 x (cm)

10

5.0 6.0

The following information applies to questions 33 and 34. The diagram is an idealised velocity–time graph for the motion of an Olympic sprinter. Velocity (m s–1)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Time (s)

0

a Which section of the surface is rough? Justify your answer. b Determine the speed of the cube during the first 2.0 cm. c How much kinetic energy is lost by the cube between x = 2.0 cm and x = 5.0 cm? d What has happened to the kinetic energy that has been lost by the cube? e Calculate the value of the average frictional force acting on the cube as it is travelling over the rough surface.

A

40

10 9 8 7 6 5 4 3 2 1 0

Position (m)

Time (s)

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80

35 During which of the section(s) (A–G) is the boy: a travelling towards the north? b stationary? c travelling towards the south? d speeding up? e slowing down? 36 For the boy’s 80 s ride, calculate: a the total distance covered b the average speed. 37 Determine the velocity of the boy: a when t = 10 s b during section B c when t = 60 s. The following information applies to questions 38–40. A mass of 0.40 kg hangs from a string 1.5 m long. The string is kept taut and the mass is drawn aside a vertical distance of 0.30 m, as shown in the diagram below. A pencil is fixed in a clamp so that when the mass is released it will swing down and break the pencil. The mass swings on but now only moves through a vertical distance of 0.14 m. (Assume: g = 9.8 m s−2.)

33 What distance was this race? 0m

1.5

34 Determine the average speed of the sprinter: a while she is racing to the finish line b for the total time that she is moving. The following information applies to questions 35–37. The diagram gives the position–time graph of the motion of a boy on a bicycle. The boy initially travels in a northerly direction.

0.14

0.30


221 221

38 Calculate the velocity of the mass the instant before it strikes the pencil. 39 Calculate the work required to break the pencil. 40 Can you account for the loss in energy?

48 Students were conducting an experiment to investigate the behaviour of springs. Increasing masses (m) were hung from a vertically suspended spring and the resulting force–extension graph was plotted as shown. 10.0 8.0 6.0 4.0 2.0 0.0

41 What is the car’s deceleration (in m s−2) during the 12 s interval?

Force (N)

The following information applies to questions 41–44. A small car is found to slow down from 90 km h−1 to 60 km h−1 in 12 seconds when the engine is switched off and the car is allowed to coast on level ground. The car has a mass of 830 kg.

The following information applies to questions 49–51. The diagram depicts a machine that can lift a mass M through a vertical height h in a time interval ∆t. The machine lifts a 1000 kg mass through a vertical distance of 8.0 m in 0.98 s.

46 a Describe the motion of your chair when you stand up and push it back from your desk. b How would the chair behave if it were on castors? c Explain how your answers to parts a and b are illustrations of Newton’s laws. 47 An engine pulls a line of train cars along a flat track with a steady force, but instead of accelerating, the whole train travels at a constant velocity. How can this be consistent with Newton’s first and second laws of motion?

40 60 80 100 120 Extension (mm)

a Estimate the value of the spring constant. b Use your answer to part a to calculate the elastic potential energy stored in the spring when the extension is 100 cm. c What other method could you have used to estimate the energy stored in the spring when the extension is 100 cm?

43 Determine the distance that the car travels during the 12 s interval. 45 A spaceship with a mass of 20 tonnes (2.0 × 104 kg) is launched from the surface of Earth, where g has a value of 9.8 N kg−1 downwards, to land on the Moon, where the value of g is l.6 N kg−1 downwards. What is the weight of the spaceship when it is on Earth and when it is on the Moon?

20

42 What was the average braking force acting on the car during the time interval?

44 Explain what happens to all the initial kinetic energy of the car.

0

g = 9.8 m s–2

M

M

49 Determine the change in gravitational potential energy (kJ) of the mass when it is lifted through a vertical distance of 8.0 m. 50 How much work is done on the mass every 0.98 s? 51 Determine the power rating of this machine in kW.

222

Motion

Unit

2

area o f stud y2

Unit

e k i l e Wav s e i t r prope t h g i l f o

outcome

area On completion of this able of study, you should be n the to describe and explai mpare wave model of light, co el of it with the particle mod served light and apply it to ob actical light phenomena in pr investigations.

chapter 7

u t a n e h T

s e v a w f o re

W

ould you know what was coming if you were sitting on a picturesque beach in Hawaii and suddenly the coastal water in front of you seemed to retract before your eyes, leaving the tethered fishing boats sitting on a bed of sand? You may have enough knowledge of coastal waves to realise that the water at the shore was being dragged back to help build a giant tsunami way out to sea. Tsunami is the Japanese term for the phenomenon that used to be called a tidal wave. Since it’s nothing to do with the tide, the name tidal wave has been dropped. Regardless of what you’d call it, you would be advised to get to high ground, and quickly! The ability to forewarn the affected coastal people of the occurrence of a tsunami anywhere in the world would undoubtedly save lives. Appropriately, it may well be our knowledge of a different category of waves—gravity waves—that may some day allow us to do this. Gravity waves are not just any ordinary type of wave. Albert Einstein in his general theory of relativity predicted their existence early last century. General relativity treats the universe as a four-dimensional surface called space–time. Gravitational waves are the curvature of space–time caused by the motion of matter. If a gravity wave arrived at Earth it would cyclically shrink and stretch the dimensions of everything around us, but by such minuscule amounts that even the strongest gravity waves are nearly impossible to detect. Einstein’s general theory of relativity was the same theory that successfully predicted the ‘bending’ of the path of light by the gravitational fields of massive objects. It was not until the 1970s that strong experimental evidence for the existence of gravitational waves in space was found, though they haven’t been detected here on Earth yet. Aside from showing us where the black holes, supernovae, etc. are located throughout the universe, the detection of gravity waves should tell us all about the big bang and break down our limits regarding how far into space we can ‘see’. Along the way Australian gravity physicists have invented a device that can accurately monitor coastal ocean waves and provide warnings of potentially lifethreatening swells. We too will focus our attention on the water as we begin our own study of the nature of waves.

by the end of this chapter you will have covered material from the study of the wave-like properties of light including: • how scientists use models to organise and explain phenomena • the differences between transverse waves and longitudinal waves • how to represent waves • how to define waves by their amplitude, wavelength, period and frequency • the speed of travel of waves • the relationship between the speed of travel, frequency, period and wavelength of a wave.

7.1 Introducing wa ve

s

Why combine the study of waves and light? As you embark on a study of the wave nature of light you will be walking in the footsteps of many famous physicists from the past who were devoted to the quest of revealing the true nature of light. In the following chapters the question as to whether light has a wave nature is addressed. Before such a discussion can begin, we must have an understanding of the nature of waves themselves! Like the physicists who preceded us, we will study the waves that can be seen on the surface of water and the waves that can be made to travel along springs and strings. Through this examination we will be able to describe how waves behave and collate a list of the properties of waves. In particular, we will be looking for the rules of behaviour that seem to be true for waves alone, and not for other mechanisms of motion. Once we have put together the rules describing the behaviour of waves, the question as to whether light has a wave nature can be addressed. You may have recognised that our quest is really just a quest to find a satisfactory model for the behaviour of light. Scientists rely heavily on models when they attempt to explain all kinds of phenomena. If an unknown or mysterious entity or observation can be linked with something with which we are familiar, then we can get closer to understanding it. For example, in the early 1900s physicists described the unknown structure of the atom by modelling it on the familiar structure of the solar system. They depicted the orbits of electrons around the nucleus as comparable to the orbits of the planets around the Sun. This was a most useful model at the time and, although not completely accurate, it set the scene for future progress regarding our knowledge of the atom. If waves are to be our chosen model for light then they must appear to behave largely in the same manner as light. That is, if a wave model for light is to be accepted then it will need to be able to explain the known behaviours of light. A very successful model would illustrate all of the behaviours of light. Perfect modelling is rare in science. Rather it is more likely that we make use of the insight that a particular model provides and, as was the case with our early models of the atom, use it as a stepping-stone to furthering our understanding.

Figure 7.1 If we can learn enough about the properties of waves we can address the question ‘Does light have a wave nature?’.

Physics file Australia has joined the quest to detect gravity waves with the commencement of construction of the Australian International Gravitational Observatory (AIGO) just north of Perth, Western Australia. This facility will use tiny changes in the path of laser light to detect the elusive gravity waves.

Waves Sometimes it is really obvious that energy is being transferred. A golf club hits a golf ball and the ball flies through the air; or the water stored in a dam is released, making a turbine spin; or a volcano erupts suddenly, spurting out hot lava and heating the surrounding region. In all of these cases energy is transferred from one location to another. Earlier in the course you looked at the concept of energy in detail and studied its various forms. For this chapter an understanding that energy allows work to be done and items to be moved around is sufficient. There is another manner in which energy can be transferred from one location to another. This mechanism does not involve a single body carrying the energy with it from its origin to its final location, but rather the energy

Chapter 7 The nature of waves

225

is carried through the particles of a substance. A dramatic example of this is a tsunami—a huge ocean wave created when there is a movement in the Earth’s crust under the sea. The energy created at the location of the shift in the crust is passed along by the particles of the ocean water at speeds of up to 800 km h−1, and can reach the coastline in the form of a towering water wave that causes devastation. None of the water particles that flow onto the shore will have been originally located near the source of the tsunami. Only the energy has been passed along.

Figure 7.2 (a) ‘Particles’ carry energy as they move. This energy can be transferred to another item as it collides with it. (b) Waves carry energy through a medium without the need for an item to have travelled from the source to the receiver.

PRACTICAL ACTIVITY 28 Disturbance and propagation of a disturbance

Energy being transferred from one location to another (by the passing of the energy from one particle to the next) within a substance is called a mechanical wave. The substance carrying the wave is called the medium. Note that, in order to pass on the wave, the particles within the medium each temporarily possess some (kinetic) energy and pass it along to the adjacent particle by physically vibrating against it. As the wave energy passes through, each individual particle of the medium will not have any overall change in its position. This is why a floating piece of driftwood will be observed to merely bob up and down as waves pass by.

All WAV…S involve the transfer of energy without a net transfer of matter. Later we will see that mechanical waves are not the only category of waves that exist. Radio waves and microwaves, for example, also transfer energy from one place to another without a net transfer of matter. There are many waves that can carry energy without requiring a medium. Some of these will be visited later in the course. Remember that our objective is to gain an understanding of the general properties of waves. We shall focus our attention on the tangible and readily observed mechanical waves that can be seen to travel in water, springs and strings.

Mechanical waves A mechanical wave involves the passing of a vibration through an elastic medium. Energy must be present at the source of the wave and this energy is described as being carried by the wave. Overall, the medium itself is not

226

Wave-like properties of light

displaced. Examples of mechanical waves include the vibrations in the Earth that we call an earthquake, the sound waves emitted by a loudspeaker, and the disturbance that travels along a guitar string when it is plucked. A model of an elastic medium is shown in Figure 7.3. Balls joined together by springs represent the particles of an elastic medium. Each ‘particle’ occupies its own mean (average) position. An initial disturbance of the first particle to the right will result in energy being passed along from particle to particle. The particles are not all disturbed at the same time; rather the disturbance gradually passes from one particle to the next. Also note that, for example, as particle 2 pushes against particle 3, particle 3 will push back on particle 2. Hence particle 2 is returned to its mean position after it has played its role in passing on the energy. Ideally all of the energy that was present initially will be passed right through the medium. In practice, the temperature of a medium will increase ever so slightly due to the movement of its particles.

1

2

3

4

5

Figure 7.3 This model of an elastic medium helps us to envisage the passage of mechanical waves through a medium.

Wave pulses and continuous waves When a single disturbance is passed through a medium in the manner discussed, we say that a wave pulse has occurred. Each particle involved in carrying the energy is displaced once as the pulse passes through, and then the particles gradually oscillate back to their mean positions. Many examples of wave motion, however, involve more than one initial disturbance or pulse at the origin. Continuous waves are created when there is a repetitive motion or oscillation at the wave source. Energy is carried away from the source in the form of a continuous wave. A vibrating loudspeaker producing sound waves in air forms a continuous wave, for example. When a medium is carrying a continuous wave, the particles of the medium will vibrate about their mean position in a regular, repetitive manner. These are also called periodic waves as the motion of the particles repeats itself after a particular period of time. (a)

one initial disturbance

PRACTICAL ACTIVITY 29 Waves in a rope

wave pulse

(b)

continuous vibration at source

Figure 7.4 (a) A single wave pulse can be sent along a slinky spring. (b) A continuously vibrating source can establish a periodic wave.


227

Transverse and longitudinal waves

Physics file

As all waves carry energy, for any wave the direction of travel of energy can be considered. There are two clearly different categories of mechanical waves. Longitudinal waves involve particles of the medium vibrating parallel to the direction of travel of the energy. An example of this is shown in Figure 7.5a. As the operator vibrates his hand in a line parallel to the axis of the spring, a longitudinal pulse is created. The particles of the medium (or the windings of the spring in this case) will vibrate in the direction shown. The vibrations are parallel to the direction of travel of the wave. Sound waves are a common example of longitudinal waves. When a speaker cone vibrates, it causes nearby air molecules to vibrate as shown in Figure 7.5b and this is parallel to the direction in which the sound energy is sent. Transverse waves are created when the direction of the vibration of the particle of the medium is 90° (perpendicular) to the direction of travel of the wave energy itself. Figure 7.4a shows an example of how this could be achieved. As the operator shakes her hand in a direction perpendicular to the axis of the spring, a transverse disturbance is created. Each particle of the medium will be moved as a pulse passes through. The particles each vibrate around their mean position, but this vibration is perpendicular to the direction in which the energy is travelling.

Water waves are often classified as transverse waves, but this is an approximation. If you looked carefully at a cork bobbing about in gentle water waves you would notice that it doesn’t move straight up and down but that it has a more elliptical motion. It moves up and down, and very slightly forward and backward as each wave passes. However, since this second aspect of the motion is so subtle, in most circumstances it is adequate to treat water waves as if they were purely transverse waves.

PRACTICAL ACTIVITY 30 Waves in a slinky

(a)

vibration of source

1. vibration of medium 2. next pulse created 3. 4.

(b)

wave energy

vibration of source wave energy

vibration of air molecule speaker

Figure 7.5 (a) When the vibratory motion and the direction of travel of the wave energy are parallel to one another, a longitudinal wave has been created. (b) Sound waves are longitudinal waves since the molecules of the medium (air molecules) vibrate in the direction of travel of the energy.

Sources of one-, two- and three-dimensional waves Another convenient classification system for waves considers the number of dimensions in which the wave energy travels. One-dimensional waves occur when longitudinal or transverse waves are sent along a spring or rope. The energy travels along the length of the conducting medium.

228


Two-dimensional waves allow energy to be spread in two dimensions. Waves travelling across surfaces are two-dimensional. A ripple travelling outward across the water’s surface when a stone is dropped into a pond is a familiar example of these (see Figure 7.6a). Earthquakes, among other effects, produce two-dimensional seismic waves that are mechanical waves travelling across the surface of the Earth. The Sun has a version of these too. Solar flares have been found to be the cause of solar quakes. These twodimensional waves travel across the surface of the Sun and, although they travel across distances equal to ten Earth diameters, they look just like ripples in a pond. When you speak you create three-dimensional waves since the soundwave energy spreads out in all three dimensions, though obviously the majority of the energy travels directly outward from the source. Designers of particular speaker systems attempt to ensure that sound waves are spread out equally in all directions. Figure 7.6b shows a three-dimensional pressure wave emitted by a bomb blast.

Physics file Seismic wave detectors don’t just pick up the vibrations from earth tremors. The demise of the space shuttle Columbia, the sinking of the Russian submarine Kursk and the collapse of the World Trade Center towers in New York all registered on different seismographs around the world.

Figure 7.6 (a) The ripples on the surface of this pond are described as two-dimensional waves since energy travels outwards in two dimensions. (b) Energy travelling outward in all directions, as in this bomb blast, forms a threedimensional wave.


229

Physics in action

Modelling a longitudinal wave position

time

If you don’t have a slinky spring handy you can still get the idea of a longitudinal wave using the handy model provided by Figure 7.7. Use two A5 pieces of paper. Place one sheet so that it covers all except the top few millimetres of the diagram. Place the other sheet so that there is a 2 mm slot created between the sheets at the top of the diagram. Now maintaining the 2 mm slot between the pages, slide the pages down the diagram, taking about 4 seconds to reach the bottom of the diagram. As you watch the slot you should be able to see ‘longitudinal waves’ travelling to the right. Try varying your sliding speed. Then figure out how it works!

Figure 7.7 Looking at these wavy lines through a slit gives the impression of longitudinal waves moving to the right.

7.1 summary Introducing waves • Scientists use models to link an unknown entity or observation to something that we are familiar with, in order to gain a better understanding of it. • Knowledge of general wave properties will allow the possible wave nature of light to be assessed. • Energy must be present at the source of any wave. • All waves involve the transfer of energy without a net transfer of matter. • A substance carrying a wave is called a medium. • A mechanical wave is the passing of energy from one particle to the next within an elastic medium. • A wave pulse occurs when a single disturbance is passed through a medium.

230


• Continuous waves are created when there is a repetitive motion or oscillation at the wave source. Energy is carried away from the source in the form of a continuous or periodic wave. • Longitudinal waves occur when particles of the medium vibrate in the same direction as the direction of travel of the energy. • Transverse waves are created when the direction of the vibration of the particle of the medium is perpendicular to the direction of travel of the wave energy itself.

7.1 questions Introducing waves 1 Describe two ways in which energy can be transmitted.

a Is the wave in the spring longitudinal or transverse? b Is the wave in the spring continuous or a pulse? c Draw an example of how the spring might look at one moment in time.

2 What is the difference between a continuous wave and a pulse? 3 Classify each of the items below as a continuous wave, a pulse or neither. a An opera singer holding a note for a long time b An explosion c A flag flapping in the wind d Dominoes standing up in a row and the first one is knocked onto the second, etc. e A tsunami that is caused by a single upward shift in a section of a seabed. 4 One end of a long spring is tied to a hook in a wall and the spring is pulled tight. The free end is then shaken up and down. a Is the resultant wave transverse or longitudinal? b Describe the motion of a particle that is part of a longitudinal wave compared with one that is part of a transverse wave. 5 A slinky spring runs from east to west across the floor of a room and is held at each end. At one end a person gives one quick shake by moving his hand in a northerly and then a southerly direction. a Is the wave in the spring longitudinal or transverse? b Is the wave in the spring continuous or a pulse? c Draw an example of how the spring might look at one moment in time. 6 A slinky spring runs from east to west across the floor of a room and is held at each end. At one end a person oscillates her hand periodically in an easterly and then a westerly direction.

7 Which of the following statements is incorrect? A Mechanical waves are made up of a series of pulses. B Mechanical waves must have a vibrating item at their source. C All waves transmit energy but don’t transmit materials. D All waves travel at right angles to the vibration of the particles in the medium. 8 A spring was initially at rest and under slight tension when a series of compressions were sent along it as shown. Z

Y

X

undisturbed spring

a How many oscillations had the hand completed at the moment shown? b In what direction are the following points about to move? i  X ii  Y iii  Z 9 Using apparatus like that shown in Figure 7.3, draw a sequence of five or six diagrams showing the passage of a transverse wave pulse along the entire length of the spring. 10 Explain the following observation: Although trans verse waves cannot travel through the middle or lower sections of a body of water, they can travel along its surface.


231

7.2

res u t a e f e v a w g Representin Displacement–distance graphs If a continuous wave was travelling across the surface of water, and we were able to freeze it instantaneously, a cross-section would look something like Figure 7.8a. If the wave then continued, a brief moment later it will have moved slightly to the right and the water particles will have taken up new positions as shown in Figure 7.8b and then Figure 7.8c. The floating cork, like the particles of the medium itself, demonstrates a vertical vibratory motion. It is displaced up then down, then up, then down. A continuous transverse wave could be sent along a piece of rope or a spring, and the particles of the medium would display a similar behaviour to the up and down motion of the water particles.

(a)

(b)

(c)

wave source

wave travels right

original water level

cork now lower

crest trough

Figure 7.8 As the wave moves to the right the displacement of the particles of the medium can be tracked using a cork. (a) The cork is on the crest of a wave. (b) The cork has moved lower as the wave moves to the right. (c) The cork is now in the trough of a wave.

A more convenient way of representing waves is to draw a graph of particle displacement against distance from the source. Keep in mind that the mean position of each water particle is the undisturbed level (flat surface) of the water. On the vertical axis we plot the displacement of each particle from its original level at a particular moment in time. The horizontal axis is used to represent the various locations across the water’s surface. Therefore the graph shows the displacement of all particles along the path of the wave, at a particular instant. In this case the chosen instant is the wave position shown in Figure 7.8c. The shape of the graph in Figure 7.9 relates directly to what we see on the surface of the water. However, these types of graphs can also be used to represent waves that are not so readily visible. Sound waves in air are

232


Particle displacement

Figure 7.9 The graph of displacement versus distance from the source of a wave is effectively freezing the wave at a moment in time, in other words taking a snapshot. Distance from the source

often represented by displacement–distance graphs, but in this case the vertical axis is used to show the forward and backward displacement of the air molecules as the sound wave passes through. Re-visit Figure 7.5a. If a longitudinal pulse was sent down a spring (by giving a quick push along its axis), then the vertical axis could be used to represent the forward and backward displacement of the particles of the medium.

The speed of waves Rather than just examining one snapshot, a sequence of graphs can be used to represent a wave that is moving across to the right (see Figure 7.10). By tracking the progress of one crest as it moves to the right, the speed at which the wave is moving can be determined. The use of a dashed line in Figure 7.10 is just to help you keep track of the initial trough and crest that were created. Note that points P and Q and all particles of the medium simply oscillate vertically, while the crests and troughs ‘move’ steadily to the right. 5.0

P

0.01

t=0 0.04

0.05

0.06

Q

5.0

P Displacement (× 10–3 m)

0.03

Q

0.02

0.07

0.08

t = 0.025 s 0.03

0.04

0.05

0.06

5.0

0.07

0.08

t = 0.050 s Q 0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

P 5.0

t = 0.075 s P 0.01

5.0

P

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

Q t = 0.100 s Q

0.03

0.04

0.05

0.06

Distance from source (m)

0.07

0.08

Figure 7.10 As each disturbance is created it will be carried away from the source by the medium.


233

Worked example 7.2A Use the series of graphs shown in Figure 7.10 to determine: a the average speed of the wave b the horizontal speed of particle P c the average vertical speed of particle P between t = 0 s and t = 0.025 s.

Solution a Since speed is the measurement of the distance an item travels in a certain time, let’s examine the progress of the first crest as it travels from d = 0.01 m to d = 0.05 m in a time period of 0.100 seconds. distance travelled Average speed = time taken (0.05 - 0.01) 0.200 0.04 = 0.100 = 0.40 m s−1 or 40 cm s−1 b Particle P is vibrating vertically. It has zero horizontal speed. c Particle P covers a vertical distance of 5 × 10−3 m (or 5 mm) in this time period. distance travelled Average speed = time taken 0.005 = 0.025 = 0.20 m s−1 In mechanical waves the speed of the wave is largely determined by the properties of the medium and, of course, by the type of disturbance that is being carried by the medium. (Sometimes the speed of a wave can also be affected by the frequency of the source; this is discussed later.) You may have observed a common example of how the properties of a medium can be altered in order to change the speed of a wave. Try sending a pulse along a slinky spring and make a mental note of how quickly it is carried away. Now stretch the spring across a greater distance, increasing the tension in the spring, and send a similar pulse along it. You should have been able to noticeably increase the speed at which the wave travels. Tension is one example of a property of an elastic medium that affects wave speed. Table 7.1 shows some common waves and typical speeds at which they are carried by their medium.

=

Table 7.1 Typical speeds of waves in some common mediums

Figure 7.11 Infrared (heat) waves travel away from the source at the same speed as light in a vacuum or in air, 3 × 108 m s–1. Different temperatures show up as different colours in an infrared photograph.

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Source of wave

Medium

Typical speed (m s–1)

Mechanical pulse

slinky spring

1200

Guitar plucking

guitar string

1300

Sound source

air at 20°C

1344

water

1450

rock

1500–3500

vacuum (no medium)

3 × 108

Infrared waves

The frequency and period of a wave Every mechanical wave must have a vibrating source. The rate at which the source vibrates directly affects the nature of the wave formed. The frequency of a source is the number of full vibrations or cycles that are completed per second. For example, a dipping rod in a ripple tank may move up and down 30 times each second. It will therefore create 30 crests and 30 troughs on the water’s surface every second. If any given point on the water’s surface were selected, then 30 complete waves would travel past this point per second. Frequency is a measurement of cycles per second (s−1), and this unit has been appropriately named after Heinrich Hertz (1857–1894), who did important work with radio waves. Hence 1 cycle per second equals 1 hertz (Hz).

The FR…QU…NCY of a wave source, f, in hertz (Hz), is the number of vibrations or cycles that are completed per second. Or, the frequency of a wave travelling in a medium is the number of complete waves that pass a given point per second. The time interval for one vibration or cycle to be completed is called the period, T, which is measured in seconds (s). This will also be the time between successive wave crests arriving at a given point. Since a decrease in the frequency of a wave will result in a longer period between waves, the relationship between frequency and period is an example of inverse variation. For example, if ten crests pass a given point in 1 second, then the frequency of the wave must be 10 Hz and the period of the wave would be one-tenth of a second or 0.1 s.

FR…QU…NCY, f = 1 T where f is the frequency of the wave in hertz (Hz) T is the period of the wave in seconds (s)

Worked example 7.2B A student lays a long heavy rope in a straight line across a smooth floor. She holds one end of the rope and shakes it sideways, to and fro, with a regular rhythm. This sends a transverse wave along the rope. Another student standing halfway along the rope notices that two crests and troughs travel past him each second. a What is the frequency of the wave in the rope? b What is the frequency of vibration of the source of the wave? c How long does it take for the student to produce each complete wave in the rope?

Solution a Frequency is defined as the number of complete waves that pass a given point per second, so f = 2 Hz. To produce a wave with a frequency of 2 Hz, the source must have the same frequency of vibration; that is, 2 Hz. c f = 1 T 1 1 ∴ T = = = 0.5 s f 2 It takes 0.5 s for each cycle to be completed.

b


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Physics file

The effects of mechanical waves can be investigated using displacement– time graphs. In these graphs the movement of one particle of the medium is monitored as a continuous wave passes through. As with the previous graphs we studied, the vertical axis may be used to represent displacements perpendicular to the wave’s direction (as in transverse waves) or parallel to the wave’s direction (as in longitudinal waves). In both cases, the displace ment is measured relative to the mean position of the particle. Since the horizontal axis indicates time values, the period of the continuous wave can be directly read from the graph. Figure 7.12 shows the displacement– time graph that would apply to the situation described in Worked example 7.2B. Note that the graph covers two complete cycles; that is, two complete waves have passed by. Particle displacement (cm)

Keep in mind that displacement–time graphs are looking at the motion of a particular particle. Recall our original definition of a wave as involving energy moving in a medium and realise that these graphs are not showing energy travelling. Therefore these diagrams are not actually graphing a ‘wave’. The familiar shape of this graph occurs because the motion of the particle is periodic; that is, it is a repeating cycle.

Displacement–time graphs

period, T

0.25

0.50

0.75

1.0

Time (s) period, T

Figure 7.12 When determining the period of a wave directly from a displacement–time graph, it does not matter at which part of the cycle you begin the period measurement.

Wavelength and amplitude

Physics file In many of the waves examined in this chapter there is no decrease in amplitude shown as the wave travels through its medium. This is an idealisation. You will have noticed that pulses sent along springs will die out eventually. Internal resistance within real springs turns some of the wave’s energy to heat. The energy of a circular wave is spread over a larger and larger wavefront as the circumference of the circular wavefront grows. As it moves outwards, each section decreases in amplitude because it carries a smaller portion of the wave’s total energy.

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Recall that earlier we examined graphs that show the displacement of all particles along the path of a continuous wave, at a particular instant. Graphs of particle displacement versus distance from the source can be used to determine the wavelength of a continuous wave. Examine Figure 7.13. Clearly there are particles within the medium that have identical displacements at the same time, such as points A and B. The wavelength of a continuous wave is the distance between successive points with the same displacement and moving in the same direction. These points are said to be in phase with one another. The symbol used for wavelength is the Greek letter lambda, λ. Like all length measurements in physics, the standard unit used is the metre (m). In Figure 7.13 the points X and Y have the same displacement and direction of movement and so they can also be described as being one wavelength apart. Note that although points P and Q have the same displacement, they will not be moving in the same direction. They are only 1 2 λ apart. In the next section we will examine how the frequency of the wave source and the velocity that the medium allows the wave combine to determine the wavelength of the wave that is produced. The amplitude, A, of a wave is the value of the maximum displacement of a particle from its mean position. The displacement of particles in a continuous wave will vary between a value of A and −A, as shown in Figure 7.13. The more energy provided by the source of the wave, the larger the amplitude of the wave. For example, in water waves the amplitude obviously corresponds directly to the height of the wave. In sound waves the amplitude determines the loudness of the sound.

Particle displacement

one wavelength, λ A

λ X

Y P

A

B Q

amplitude, A

R

–A

λ

Distance from source (m)

Figure 7.13 When determining the wavelength of a wave directly from a displacement–distance graph, it does not matter at which part of the cycle you begin the wavelength measurement.

Physics in action

Huygens’s principle

(a)

All sorts of waves, such as the circular water waves seen in Figure 7.14a, can also be represented in diagrams like that shown in Figure 7.14b. Lines are used to represent a certain part of the wave, such as the crests. If the diagram were drawn to scale the distance between the lines would represent the wavelength, λ. These diagrams are particularly useful should you want to indicate the region over which the wave energy has spread. In 1678 Christiaan Huygens suggested a model that provides an explanation for how waves are carried through a medium. His model coincides with what we see in situations like that shown in Figure 7.14a. Huygens’s principle is a method that uses geometry to predict the new position of a wavefront, if the original position of the wavefront is known. The principle states

(c)

(b)

ray

rays giving direction of propagation

ray

source

initial wavefront ray

λ

initial wavefront

new wavefront

new wavefront

Figure 7.14 (a) Circular water waves. (b) Evenly spaced lines can represent the crests of a wave travelling outward, according to Huygens’s principle and (c) every point on a wavefront is a source of secondary circular wavelets.


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that every point on a wavefront may be considered the source of small secondary circular wavelets. These wavelets spread out with exactly the same speed as the original wavefront. The new wavefront is then found by drawing a tangent to all of the secondary wavelets. This is called the envelope of the wavelets and is shown in Figure 7.14c. Figure 7.14c shows the points on a wavefront that are

sources of secondary circular wavelets. These wavelets move at speed v and so during time interval t cover a distance of vt. The speed, v, has been assumed to be the same for all wavelets. Although we have only examined the spread of a circular wave, Huygens was renowned for the use of his principle in explaining the reflection and refraction of waves at boundaries (which is discussed later in this text).

7.2 summary Representing wave features • A mechanical wave can be represented at a particular instant by a graph of particle displacement against distance from the source. • The frequency of a wave, f, is the number of vibrations or cycles that are completed per second, or the number of complete waves that pass a given point per second. Frequency is measured in hertz (Hz). • The period, T, is the time interval for one vibration or cycle to be completed. 1 • Frequency f = where f is the frequency of the wave T in hertz (Hz), and T is the period of the wave in seconds (s).

• A graph of particle displacement versus time can be drawn for the particles of a medium that is carrying a continuous wave. The period of the wave can be read directly from this graph. • Graphs of particle displacement versus distance from the source can be used to determine the wavelength of a continuous wave. • The wavelength, λ, of a continuous wave is the distance between successive points having the same displacement and moving in the same direction; that is, the distance between points that are in phase. • The amplitude, A, of a wave is the value of the maximum displacement of a particle from its mean position.

7.2 questions Representing wave features 1 Calculate the frequency and period of: a a spring that undergoes 40 vibrations in 50 seconds b a pendulum that completes 250 full swings in one and a half minutes. 2 In a ripple tank the trough of a water wave travels 70 cm in 2.5 seconds. Calculate the speed of the wave in metres per second. 3 What usually happens to the amplitude of the vibra tion of a circular water wave as it spreads out? Why? 4 A pebble is dropped into a pool and after 3.00 seconds 24 wave crests have been created and travelled out from where the pebble entered the water. What is

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the frequency and period of the water wave that was created? 5 A piston in a car engine completes 250 complete up-and-down movements every half a minute. a What is the frequency of vibration of the piston? b What is its period? c Assuming that the piston started from a central position and moved up, where will it be after: i 1 period? 1 ii 1 4 periods? 1 iii 1 2 periods?

7 Examine the wave represented in Figure 7.10. What is its wavelength? 8 A longitudinal wave enters a medium and causes its particles to vibrate periodically. Draw a displacement– time graph that could demonstrate the motion of the first affected particle of the medium for the first two cycles. Begin with a positive displacement, i.e. in the direction of travel of the wave.

Particle displacement (cm)

9 The displacement–distance graph shows a snapshot of a transverse wave as it travels along a spring towards the right. 20 10 –10 –20

P S

Q 0.20 0.40

0.60

0.80

1.00 1.20

1.40

1.60

R

b At the moment shown, state the direction in which the particles Q and S are moving. c Assuming that the wave is travelling at 12 m s−1 to the right, and no energy is lost, draw the displacement–distance graph for this wave 0.05 seconds after the moment shown. Label the points P, Q, R and S. 10 The displacement–time graph shows the motion of a single air molecule, P, as a sound wave passes by travelling to the right. Displacement of P (× 10–8 cm)

6 Which of the following statements is correct? A Period is the measurement of the length of a wave. B The amplitude of a wave is dependent upon the frequency. C The more energy put into a wave the greater the wavelength. D The more energy put into a wave the greater the amplitude.

0.5 1

2

3

4

–0.5

Time (ms)

a Use the graph to determine the amplitude, period and frequency of this sound wave. b State the displacement of the particle P at: i  t = 1 ms ii  t = 2.5 ms iii  t = 5.5 ms c Draw the displacement–time graph for particle Q, which is positioned half a wavelength to the right of particle P. Show the same 4 ms time interval. d If sound is actually a longitudinal wave, why does this graph look more like a transverse wave?

Distance along spring (m)

a Use the graph to determine the wavelength and the amplitude of this wave.


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s

ion t c a r e t n i e v a 7.3 Waves and w

wavelength = distance one wavefront moves before the next wave is created λ The frequency of a source of a mechanical wave and the velocity of that

The wave equation

Figure 7.15 The medium carrying the wave and the frequency of its source together determine the wavelength of a wave.

PRACTICAL ACTIVITY 31 Waves in a ripple tank

Interactive tutorial 6 The wave equation

wave in the medium together determine the resulting wavelength of the wave. For example, a horizontal bar vibrating at frequency f may be used as the dipping element in a laboratory ripple-tank as shown in Figure 7.15. Once one crest is created, assume that it travels away from the source at a known speed, v. Since the definition of speed is: distance travelled speed = time taken this can be rearranged to: distance travelled = speed × time taken. Consider the first period, T, of the wave’s existence. The distance that the first wave will be able to cover before the next wave is created behind it is determined by the speed at which the medium allows the wave to travel. The distance that the first wave travels during one period—by definition—is the wavelength of the wave, λ. Therefore we acknowledge that the ‘distance travelled’ = λ when the ‘time taken’ = T. Substituting into: distance travelled = speed × time taken λ=v×T The frequency and period of a wave are inversely related: 1 T= f Hence, the above relationship can also be expressed as: v λ= f

The WAV… …QUATION links the speed, frequency and wavelength of a wave: v = fλ 1 Note that substituting f = T into the wave equation gives: λ v=fλ= T where v is the speed of the wave in metres per second (m s-1) f is the frequency of the wave in hertz (Hz) λ is the wavelength of the wave in metres (m) T is the period of the wave in seconds (s) Note that for a medium of a given speed, the use of a higher frequency source would result in waves that are closer together; that is, waves of a shorter wavelength. A low frequency source would produce longer wavelength waves (see Figure 7.16). For a given wave speed: 1 λ∝ f

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(a)

(b) low frequency source λ

longer wavelength

longer wavelength

high frequency source shorter λ wavelength

tone of same frequency

slower speed shorter wavelength

shorter wavelength

Figure 7.16 (a) For a medium of a given speed, the use of a low frequency source produces

faster speed longer wavelength

waves with a long wavelength. (b) With less time between the creation of successive waves, a high frequency source produces waves with a shorter wavelength.

An implication of the wave equation that is worth noting is that a source that has a specific frequency of vibration is able to produce waves of different wavelengths, depending upon the medium that carries the wave. Consider a submarine that puts out a high frequency tone of 20 000 Hz. If this same frequency tone were sent both into the water and into the air, the waves produced in the water would have a much longer wavelength than the waves produced in the air. This is because sound waves travel about four times faster in water than in air (see Figure 7.17). For a source of a given frequency: λ∝v

Figure 7.17 Since sound waves travel much faster in water than in air, the waves produced by a tone of a given frequency have a much longer wavelength when they travel through water than when they travel through air.

PRACTICAL ACTIVITY 32 Reflections of waves in a ripple tank

Worked example 7.3A A person standing on a pier notices that every 4.0 seconds the crest of a wave travels past a certain pole that sticks out of the water. The crests are 12 metres apart. Calculate: a the frequency of the waves b the speed of the waves.

Solution a The period of the wave is 4.0 s. 1 T 1 f= 4.0 = 0.25 Hz b Since the crests are 12 m apart the wavelength is 12 m. v = f λ = 0.25 × 12 = 3.0 m s-1

Since f =

Waves meeting barriers Mechanical waves travel through a medium. Commonly a situation will occur in which the wave travels right through to a point where the medium physically ends. An example of this is the wave created as a child leaps into a pool; it travels until it reaches the pool wall. At the boundary of the medium, the energy that was being carried by the wave may undergo different processes. Some of the energy may be absorbed by or transmitted into a new medium, and some energy may be reflected.

Physics file The phase change of a wave on reflection from a fixed end can be explained in terms of Newton’s third law of motion. When the pulse arrives at the fixture, the rope exerts a force on the fixture. The fixture exerts an equal and opposite force on the rope. This produces a pulse that is in the opposite direction to the original pulse; that is, a change in phase has occurred.


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The extent to which these processes occur depends on the properties of the boundary. We shall examine the case of a transverse wave pulse travelling in a heavy rope that has one end tied to a wall. As shown in Figure 7.18a the wave travels to the boundary and we can see that it is reflected with almost no energy loss since the original amplitude is maintained. The wave, however, has been inverted; this can also be described as a reversal in phase. (The definition of phase was discussed in the previous section.) Since a crest would reflect as a trough and a trough would reflect as a crest, we 1 can say that the phase of the wave has been shifted by 2 λ.

A WAV… R…FL…CTING FROM TH… FIX…D …ND of a string will undergo a phase reversal; l that is, a phase shift of . 2 Figure 7.18 (a) The reflection of a wave at an unyielding boundary produces a phase shift of 1 λ. Note that otherwise the shape of the wave is 2 unaltered. (b) The reflection of a wave at a freeend boundary does not produce a phase shift.

Now consider the situation in which the end of the rope is free to move. As shown in Figure 7.18b, the wave travels to the end of the rope and we can see that it is reflected with no reversal in phase. Since a crest would reflect as a crest and a trough would reflect as a trough, we can say that there was no change of phase.

A WAV… R…FL…CTING FROM TH… FR…… …ND of a string will not undergo a phase reversal.

Physics in action

Reflections NOT wanted! The stealth aircraft is designed so that its body is as poor a reflector as possible. The main way in which a passing aircraft is detected by others is with the use of radar. A radar transmitter sends out pulses of radio waves or microwaves and a receiver checks for any reflections from passing aircraft. By analysing the reflections, radar systems can work out the position, speed and perhaps even the identity of the passing aircraft. Stealth aircraft are designed to create as little REPLACEMENT TO BE PROVIDED reflection of these waves as possible. The shape of the stealth aircraft is the most important factor. It does not have any large vertical panels on the fuselage that would act like mirrors, nor a large vertical tail. It has no externally mounted devices such as missiles or bombs. It does not include any surfaces that meet at right angles. These would act like the corners in a billiard table and bounce the waves right back to their source. Instead Figure 7.19 The Lockh curved surfaces on the stealth aircraft are designed to eed Martin fighter. F-35 Joint reflect waves sideways or upward wherever possible. Strike Figh ter is a lon A thick coat of special paint that absorbs radio waves g-range ste alth is used on its surface. Although not completely undetectable, with the right shape and coating a large stealth plane can produce the same amount of wave reflection as an average sized marble!

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Superposition: waves interfering with waves In the case of a continuous wave being sent towards a boundary, a situation can be created in which two waves may be travelling in the one medium, but in different directions. The incident waves will meet the waves that have already reflected from the boundary. When two waves meet they interact according to the principle of superposition.

The principle of SUP…RPOSITION states that when two or more waves travel in a medium the resulting wave, at any moment and at any point, is the sum of the displacements associated with the individual waves. Consider a spring in which a transverse pulse has been sent from each end, as shown by the sequence of events in Figure 7.20. When the pulses reach the same point in the spring, the resulting wave will be the sum of the displacement produced by the individual pulses. The principle of superposition is therefore the same as the ‘addition of ordinates’ process that is carried out on graphs. Simply sum the y-values of each of the pulses to see the resulting wave. pulse 1

y A

Physics file When a note is played on a musical instrument sound waves with many different wavelengths are produced simultaneously. The richness of a tone is largely determined by how many different wavelengths make up the sound wave. The tone with the longest wavelength determines the overall perceived pitch of the note but the number of overtones (other wavelengths present) will add to its timbre.

PRACTICAL ACTIVITY 33 Interference of water waves

pulse 2

x sum of individual pulses

y A

x y 2A 2A

A

A

y

pulse 2

x pulse 1

A x y

pulse 2

pulse 1

A x

Figure 7.20 Superposition of two pulses of the same amplitude travelling towards one another.

In Figure 7.20 the initial pulses have particle displacements in the same direction and therefore constructive interference occurs. Notice that after interacting with each other, the two pulses have continued on unaffected. This is an observed property of waves. They are able to pass through one another, momentarily interact according to the superposition principle, and then continue on as if nothing had happened.


243

Wave 1

y A x

Wave 2

In Figure 7.21 destructive interference occurs since the initial pulses have particle displacements in opposite directions. If the crest of one pulse has exactly the same dimensions as the trough of the approaching pulse then the two pulses will momentarily completely cancel each other out, as shown in Figure 7.21. If the amplitude of one of the waves is larger than the other then only partial cancellation will occur. y

y

pulse 1

2A x

pulse 2

x

y

Wave 1 + Wave 2

sum of pulses y

x 3A

y

pulse 1

x

x pulse 2

Figure 7.21 Superposition of two pulses of equal but opposite amplitudes travelling towards one another.

Figure 7.22 The superposition of continuous

waves that are in phase and travelling in the same direction will result in constructive interference. Physics file We have looked at how waves can reflect back along a string from a fixed end. Essentially this is what happens to the waves sent along a bowed violin string or a plucked guitar string. The numerous reflected waves add together according to the principle of superposition with some important effects. For each mode of vibration shown in Figure 7.23, at some spots on the string constructive interference will occur. In other spots destructive interference occurs. Since each particular mode of vibration has set locations for these spots, the wave is called a standing wave.

In the case of interference between continuous waves, the principle of superposition is still applicable. If two waves are exactly in phase and are travelling in the same direction, then constructive interference will occur along the entire length of the wave. The two waves need not have the same amplitude. In Figure 7.22 one wave is twice the amplitude of the other wave and the resultant wave is shown. Interesting effects are observed when two waves of different wavelengths are travelling in the same direction and interfere with one another. Figure 7.24 shows the addition of two waves, where one wavelength is exactly three times longer than the other. This is a relatively simple example. Imagine the complexity of the sound-wave patterns produced when instruments in an orchestra are played simultaneously. Or of the wave patterns that are produced on the surface of water in a busy harbour. y

wave 1 resultant wave

x

wave 2

Figure 7.23 One mode in whi ch

a string can vibrate involves destructive interference occ urring right at the centre point of the string.

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Figure 7.24 The addition of waves of different wavelengths results in complex wave patterns.

Physics in action

Diffraction and interference effects We have seen how a wave can spread out from a point source, but waves are also capable of bending around obstacles or spreading out after they pass through a narrow gap. This bending of the direction of travel of a wave is called . Figure 7.25 shows the diffraction of water waves as they pass through an aperture. Diffraction effects can be seen with two-dimensional waves, such as on the surface of water, and also with three-dimensional sound waves. This explains why we can hear sounds that were originally made around the corner of a building. The sound waves bend their direction of travel—that is, diffract—around the corner of the building to reach the listener’s ears. The extent to which diffraction occurs depends on the relative dimensions of the aperture or obstacle that the wave passes, and the wavelength of the wave. Most noticeable diffraction occurs if the wavelength is at least as large as the aperture is wide. When waves of a small wavelength are sent through a large aperture, as in Figure 7.26, less noticeable diffraction occurs. Spreading water waves produce interference patterns that are characteristic of waves. Consider two sources of spherical waves (Figure 7.27). In some locations constructive

Fig

ure 7.25 R the wavefr ather than only tra v onts spre ad out to elling directly forw fill the reg a ion behind rds, notice how the obsta cles.

length is at rs when the wave cu oc n tio ac ffr di ficant Figure 7.26 Signi the aperture. as e rg la as st lea interference occurs and waves of relatively large amplitude are seen. These regions have lots of contrast in the photograph; that is, alternating bright and dark bands are seen. In other regions destructive interference occurs. Troughs arriving from the other source always cancel out the crests that arrive at these locations, and the surface of the water remains relatively undisturbed. The regions of destructive interference appear grey and flat in the photograph. These regions of destructive interference appear to radiate from a point between the sources. Both diffraction and interference effects are only observed when energy is being carried by waves, not when energy is being carried by particles.

Figure 7.27 The interfe ren

in phase.

ce pattern

produced b y

two point s ou

rces

PRACTICAL ACTIVITY 34 Diffraction of continuous water waves


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Now we can look at light! Now that we have put together the rules describing the characteristics of waves, the question as to whether light has a wave nature can be addressed. Waves have numerous characteristics and they have been worth examining in their own right. We have been able to conclude the following: • Waves involve the transfer of energy without an overall transfer of matter. • Mechanical waves require a vibrating item at their source and a medium to carry them. • Waves can be categorised as longitudinal or transverse. • The wave equation, v = fλ, describes the relationship between the speed, frequency and wavelength of a wave. • Waves can reflect at boundaries and this will sometimes produce a change of phase. • Waves can be added according to the principle of superposition and this can result in constructive or destructive interference. In Chapter 8 we will go on to discuss whether it is appropriate to use waves as our chosen model for light. For this to be fitting, light must appear to behave largely in the same manner as waves do. That is, if a wave model for light is to be accepted, then it will need to explain the known behaviours of light. A very successful model would illustrate all of the behaviours of light. This is not likely. It is more likely that we will be able to make use of the insight that waves provide, and use this insight to further our understanding of the nature of light.

7.3 summary Waves and wave interactions • The frequency of the source and the speed of the wave in the medium determine the wavelength of a mechanical wave. • The wave equation states: l v = fλ = T where v = speed of the wave in metres per second (m s−1) f = frequency of the wave in hertz (Hz) λ = wavelength of the wave in metres (m) T = period of the wave in seconds (s). l • For a wave of a given speed, λ ∝ . f • For a source of a given frequency, λ ∝ v. • A wave reflecting from a fixed end of a string will l undergo a phase reversal; that is, a phase shift of . 2

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• A wave reflecting from a free end of a string will not undergo a phase reversal. • The principle of superposition states that when two or more waves travel in a medium the resulting wave, at any moment, is the sum of the displacements associated with the individual waves. • Constructive interference occurs when two waves meet that have particle displacements in the same direction. • Destructive interference occurs when two waves meet that have particle displacements in opposite directions.

7.3 questions Waves and wave interactions 1 a What happens to the wavelengths of the waves in a ripple tank if the frequency of the wave source is doubled? b What happens to the speed of the waves in a ripple tank if the frequency of the wave source is halved? 2 The source of waves in a ripple tank vibrates at a frequency of 15.0 Hz. If the wave crests are 40.0 mm apart, what is the speed of the waves in the tank?

7 Two waves are travelling in the same direction in a medium. They undergo constructive interference along the entire length of the wave. What two statements can be made about the two waves? 8 Assuming the following diagram shows the displace ment–distance graphs of two waves at a particular instant, show the addition of the two waves according to the principle of superposition.

3 A wave travels a distance of 50 times its wavelength in 10 seconds. What is its frequency? 4 A submarine’s sonar equipment sends out a signal with a frequency of 35 kHz. If the wave travels at 1400 m s−1, what is the wavelength of the wave produced? 5 Which of the following statements is incorrect? A When two pulses interact the resulting wave, at any moment, is the sum of the displacements associated with the individual waves. B After two waves interact with each other they will continue on through the medium unaffected. C For two pulses to interfere destructively they must have opposite amplitudes. D For two continuous waves to interfere construct ively they must have identical amplitudes. 6 Will a transverse wave reaching the fixed end of a string undergo a phase reversal?

9 Draw the resultant displacement versus distance graph for two superimposed continuous waves that are in phase and travelling in the same direction. Each wave has a wavelength of 4 cm and amplitude of 1 cm. Show two complete cycles. 10 Draw the resultant displacement versus distance graph for two superimposed continuous waves travelling in the same direction. Each wave has a wavelength of 4 cm and amplitude of 1 cm, but one wave is one-quarter of a wavelength behind the other.

chapter review The following information applies to questions 1–3. A pulse is travelling along a light spring. The diagram below shows the position of the pulse at t = 0 s. The pulse is moving at a speed of 40 cm s−1 to the right. Drawn to scale

t=0s

P

Q

10 cm

40 cm

90 cm

2 Draw the displacement–distance graph for the pulse 0.5 seconds later. Clearly show the location of point P. 3 Draw the displacement–time graph for the point Q for a time interval of 2.0 seconds, beginning at t = 0. 4 List an example of a one-, a two- and a three-dimensional wave. 5 A guitar string is plucked near one end. A wave moves along the string and another wave is produced in the air. State whether each wave is transverse or longitudinal.

1 Use a set of scaled axes to draw the displacement–distance graph for the pulse at the moment shown.


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The following information applies to questions 6–9. The diagram shows two successive amplitude–distance graphs for a periodic transverse wave travelling in a string. The time interval that passed between the tracings of the two graphs is 0.20 s. The graphs are drawn exactly to scale.

Amplitude

1 cm P

1 cm

Amplitude

Distance

1 cm P

1 cm Distance

6 State the amplitude of the wave. 7 State the wavelength of the wave. 8 Calculate the velocity of the wave. 9 Calculate the frequency and period of the wave. 10 Which of the following statement(s) is/are incorrect? (One or more answers are possible.) A All mechanical waves require a medium to carry the wave. B All mechanical waves transfer energy. C In wave motion some of the material is carried along with the wave. D Mechanical waves permanently affect the transmitting medium. 11 What is the period of the wave that: a involves 5.0 crests of water lapping against a breakwater each 20 seconds? b is produced by a flute playing the note middle C (512 Hz)?

12 Find the frequency of the waves that have the following periods. a 0.35 s b 4.0 × 103 s c 10−2 s 13 A wave pulse is sent simultaneously from both ends of a spring. When the pulses meet they momentarily completely cancel out one another. a What is the term that describes this occurrence? b Make statements about three features of the wave pulses. 14 A transverse wave travels along a string towards an end that is free to move. Which of the following statements is true? A The wave will reflect with no phase change. l B The wave will reflect with a phase change of . 2 C The wave will not be reflected. D The reflected wave is faster than the incident wave. 15 Waves travelling in a ripple tank have a wavelength of 7.0 mm and travel at 60 cm s−1. What is the frequency and period of the waves? 16 One end of a long spring is firmly connected to a wall fitting. Briefly explain how a transverse wave can be created and carried by the spring. The following information applies to questions 17–20. Wave A has a wavelength of 4.0 cm, a period of 2.0 seconds and an amplitude of 1.5 cm. Wave B has a wavelength of 2.0 cm, a period of 1.0 second and an amplitude of 1.5 cm. 17 Draw a scaled displacement–distance graph for wave A. Show two full waves. 18 Draw a scaled displacement–distance graph for wave B. Show four full waves. 19 If wave A and wave B were sent into the same medium and they are travelling in the same direction, draw the resultant displacement–distance graph. Show two full waves. 20 Draw a displacement–time graph for a particle in the medium that carries wave A only. Show two complete cycles.

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Mode ls

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I

t is a common trait of humans that when we seek to understand something we will intuitively attempt to link the unknown with the known. In your earlier schooling a physical representation or model was probably used to teach you about nature’s water cycle, or multiplication, or the properties of gases. Young students benefit from the use of tangible items such as models: things that can be seen and touched. As we grow, our knowledge and understanding can still benefit from the use of a modelling approach, but our models can be more sophisticated. When computer-generated pictures were used to model the complex equations of fractal geometry they had an amazing similarity to some structures found in nature. Fractal images model things such as coastlines and snowflakes and they have become popular works of art. A model is a system of some type that is well understood and that is used to build a mental picture or analogy for an observed phenomenon, in our case the behaviour of light. A good model will appear to behave in the same manner as the entity being investigated. A model for light needs to be able to explain the observations of light that have already been made and ideally it would predict new behaviours. Therefore, throughout this chapter, when deciding upon a model for light we must examine each of its known behaviours in turn and assess the effectiveness of the chosen model.

by the end of this chapter you will have covered material from the study of the wave-like properties of light including: • electromagnetic waves, particle and ray models for light • applying the ray and wave models to reflection and refraction • the speed of light • refractive index and Snell’s law • total internal reflection and critical angle • identifying visible light as a region of the electromagnetic radiation spectrum • colour components of white light and dispersion • polarisation of light waves and the transverse wave model.

es

per ti o r p t h g i l e l p lling sim

8.1 Mode A

B

C

Figure 8.1 Light from the lamp can only be seen if the pinholes lie in a straight line. This means that light must travel from the lamp to the eye along a straight line. (a)

Modelling with waves and particles

(b)

diverging rays

converging rays

parallel rays

(c)

(d)

Figure 8.2 (a) A beam of light is made up of a bundle of rays. (b) Rays can be diverging, converging or parallel to one another. (c) An idealised point source of light emits rays of light in all directions. (d) Very distant sources of light are considered to be sources of parallel rays.

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Now that we have a thorough appreciation of the properties of waves, the question that can be asked is: Is light a wave? If a wave is defined as the sum of its properties, does light exhibit all of the properties known to belong to waves? Curiosity about the nature of light has occupied the minds of physicists for centuries. The beginning of human interest in the nature of light dates back to the ancient Greek, Arabian and Chinese philosophers. In the early 19th century, evidence suggested that light could be modelled as a wave since it exhibited the same set of properties as other things that had already been defined as waves: water waves, sound waves, vibrations in springs and strings. If light exhibits sufficient properties in common with these known waves, then surely it too could be assumed to ‘be’ a wave? The story of the development of a scientific model for light is not straightforward. The discussion of light as a wave did not exist in isolation. The giants of physics became embroiled in a famous ongoing scientific debate that posed the question: Is light made up of particles or waves? In this section we look at how the very simplest behaviours of light can be readily modelled as either particles or waves.


Light streaming through trees on a misty morning, the projector’s beam in a dusty cinema, our limited view when peeping through a keyhole and the distinct shape of shadows are all evidence for the straight-line or rectilinear path of light. These examples provide evidence that light—transmitted in a uniform medium (i.e. a substance that is unchanging in its constitution)— travels in straight lines. Our awareness of the rectilinear propagation of light allows us to judge the distance to objects. The mechanism by which our eyes and brain interpret a three-dimensional world is complex, but it relies on the assumption that light in a uniform medium travels in straight lines. The following simple experiment can be performed to demonstrate that light travels a straight path in a uniform medium. Make a pinhole in each of three identical pieces of card. Place card A close to a light source, and position card B a little further away, as shown in Figure 8.1. Then, holding card C in front of your eye so that you can always see through the hole, adjust its position so that you can see the light from the lamp. This will only be possible when all three pinholes lie in the same line; that is, when the pinholes are collinear. The conclusion that can be drawn is that light must travel in straight lines. The above property of light was first modelled by considering that light was particle-like in nature. Consider a beam of light shining from a powerful torch. The direction of travel of the light ‘particles’ can be represented by rays (Figure 8.2a). The idea of a light ray is a useful concept as it can successfully model the behaviour of light in the situations illustrated. A beam of light can be thought of as a bundle of rays. A strong light source, such as the Sun, could therefore be thought of as producing a very large number of light rays.

Light sources, in conjunction with other optical elements such as lenses or mirrors, can produce rays of light that diverge, converge or travel parallel to each other (Figure 8.2b). In each case the rays are an indication of the direction of travel of the light; essentially light is being modelled as a stream of particles. The incandescent (filament) and fluorescent light globes in your home emit light in all directions. A point source of light is an idealised light source that emits light equally in all directions from a single point (Figure 8.2c). No single point source of light exists in reality, but a small filament lamp can be considered a good approximation. Lasers and special arrangements of light sources with mirrors or lenses can produce parallel rays of light in a beam. Very distant point sources of light can also be considered to be sources of parallel light rays. For example, on the Earth we treat the light rays that reach us from the Sun as though they were parallel to each other. This is because at such a large distance from the source, the angle between adjacent rays would be so tiny as to be considered negligible (Figure 8.2d). Although a particle description for light and the accompanying ray model are convenient for representing the behaviour of light in all of these cases, it has long been understood that light is not made up of ordinary particles. Light involves the transfer of energy from a source, but there are no tangible particles carrying this energy. With developing technology over the last two centuries physicists have been able to make more and more sophisticated observations of light. Later in the chapter we will see that a more refined electromagnetic wave model of light incorporates the wavelike properties of light. The ray approach is still useful. If light is considered to be a wave emanating from its source, then rays may simply be used to represent the direction of travel of the wavefronts (see Figure 8.3). The point source of light discussed above may be considered to be a point source of spherical wavefronts, like the ripples that travel out from a stone dropped into a pond (Figure 8.4).

Modelling reflection The reflection of waves was discussed in Chapter 7. Light has been observed to obey the same laws of reflection that apply to waves and so evidence is provided for the argument that light is a wave. Using a wave model, the reflection of light would be represented as a series of wavefronts striking a surface and reflecting as shown in Figure 8.5. However it is far more common to model the reflection of light using ray diagrams and the conventions associated with them. (a)

wavefronts travel outward from torch rays travel outward from torch

Figure 8.3 Rays can be used to represent the direction of travel of light waves leaving the torch.

spherical wavefronts travel outwards

point source

Figure 8.4 A point source of light may be considered to be a point source of spherical waves. Both the particle and the wave model are consistent with the observation that the intensity reduces with the square of the distance from the source.

(c)

(b)

Figure 8.5 When studying reflection, ray diagrams are the most convenient way of representing the path of light.

Chapter 8 Models for light

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Physics file In Year 12 you will study the model of light developed last century. The photon model draws on aspects of the wave and particle models. It refers to the light travelling in the form of tiny packets of energy called photons. The energy of the photon is determined by the frequency of the light.

plane (flat) mirror

inc

ide

normal

nt

ray

The LAW OF R…FL…CTION states that the angle of incidence, i, is equal to the angle of reflection, r (i = r). The incident ray, the normal and the reflected ray will all lie in the same plane.

angle of incidence i angle of reflection r ray ted c e l ref

Figure 8.6 When light reflects from a plane mirror, the angle of incidence equals the angle of reflection: i = r. incident ray

Consider the plane mirror drawn in Figure 8.6. We define a normal to the surface of the mirror as the line perpendicular (at 90°) to the mirror’s surface at the point where an incoming or incident ray strikes the surface of the mirror. The angle between an incident ray and the normal is the angle of incidence, denoted i. The ray strikes the mirror and reflects with an angle of reflection, r, which is the angle between the reflected ray and the normal. Experiment shows that whenever reflection occurs, the angle of incidence always equals the angle of reflection. In addition, the light reflects in such a way that the incident ray, the normal and the reflected ray all lie in the same plane. The law of reflection can then be re-stated using a ray model for light.

~4% ~96%

glass layer metal layer paint layer

Figure 8.7 Most of the incident light on a mirror is reflected from the silvered surface at the back of the mirror. The glass on the front and the paint on the back serve to protect the reflective surface from damage.

(a)

(b)

A normal household mirror is constructed with three separate layers: a layer of transparent glass, a thin coating of aluminium or silver deposited onto the glass to reflect the light and a backing layer of protective paint (Figure 8.7). When a beam of light strikes the surface of the mirror a tiny amount of the light energy (about 4%) is reflected from the front surface of the glass, but most of the light continues to travel through the glass and is reflected from the metal surface at the back. These reflected rays produce the image that is seen in the mirror.

Regular and diffuse reflection To some extent at least, light will reflect from all surfaces, but only some surfaces will produce a clearly defined image. If parallel rays of light are incident on a plane mirror or a flat polished metal surface, they will remain parallel to each other on reflection (Figure 8.8a). This is regular reflection (sometimes called specular reflection) and, as a result, a clear image can be produced. Common examples of regular reflection include the reflection of light from plane mirrors, glossy painted surfaces and still water such as in a lake. When light is reflected from a roughened or uneven surface, it is scattered in all directions as shown in Figure 8.8b. This is diffuse reflection. Parallel rays of incident light will be reflected in what seem to be unpredictable directions. Each ray obeys the law of reflection, but the surface is irregular so that normals drawn at adjacent points have completely different directions. Thus, light is reflected in many different directions. Most materials produce diffuse reflection. For example, when looking at this page, you can see the printing because the lighting in the room is reflected in all directions due to diffuse reflection. If the page behaved as a regular reflector, you would also see the (reflected) images of other objects in the room. Diffuse and regular reflection are the two extreme cases of how light can be reflected. In reality most surfaces display an intermediate behaviour. For Figure 8.8 (a) Regular reflection from a smooth surface occurs when parallel rays of incident light are reflected parallel to each other. (b) Diffuse reflection occurs at an irregular surface. Here, the incoming parallel rays are reflected at all angles.

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example, the pages of a glossy magazine may allow a blurry image of the reader’s face to be formed, but the printing can still be seen. The surface produces reflection that lies somewhere between pure diffuse reflection and pure regular reflection. Can you think of any other surfaces that do this? What do they all have in common? To predict the extent to which diffuse and regular reflection occur at a surface, one must examine the surface on a microscopic scale. If the irregularities in the surface are small compared with the wavelength of the incident light, then regular reflection occurs. If the irregularities are comparable in size to the wavelength of the light, then more diffuse reflection occurs. The wavelength of light is discussed more fully later in the chapter. Physics in action

Eclipses From our everyday experience, we know that a shadow is formed when a solid body obstructs the path of light from the source. This can occur on a large scale when the Moon and Sun are aligned with the Earth to produce different types of eclipses. Figure 8.9 shows how the Earth, Moon and Sun must be aligned to produce a lunar eclipse. In a lunar eclipse, the Earth’s shadow falls across the face of the Moon. However, the Earth’s shadow consists of two distinct parts. The complete shadow, the umbra (the Latin for shadow or shade), is the darkest region, and when the Moon passes through this, a total eclipse of the Moon occurs. There are also regions known as penumbra (literally ‘almost an umbra’) where the Moon is neither fully illuminated nor fully in shadow. When the Moon only passes through the penumbra of the Earth’s shadow, the Moon appears dimmer than usual. Even during a total eclipse, the Moon does not completely disappear. Rather, it appears very dim and red in colour. This is because it is still illuminated by a small amount of light that has travelled through the Earth’s atmosphere. The atmosphere acts like a prism, splitting the white sunlight into its component colours through the process of refraction, and the distances and angles involved are such that most of the light that reaches the Moon is red. Looking at the relative positions of the Earth, Moon and Sun in Figure 8.9, one might expect a lunar eclipse to occur once during each revolution of the Moon around the Earth, i.e. once per month. This does not occur because the plane of the Moon’s orbit is slightly tilted with respect to the Earth’s orbit around the Sun. This means that the Moon often travels only partially into the shadow region and so only a portion of its surface is obstructed. The Moon travels at a speed of about 1 km s-1 through the Earth’s shadow, which means that the longest time a total eclipse of the Moon can last is 1 hour 42 minutes. Between 9 and 12 total eclipses of the Moon can be seen from Earth every decade. A solar eclipse occurs when the Moon comes between the Earth and the Sun, casting its shadow onto the Earth. Figure 8.10 illustrates two types of solar eclipse.

Sun

Earth

Earth’s orbit

Moon’s orbit

Earth’s main shadow

Figure 8.9 During a lunar eclipse the Moon travels into the Earth’s

shadow. The Moon orbits the Earth in an elliptical path, so its distance from the Earth varies. The relative distances between the Earth, Moon and Sun determine whether an eclipse is total or annular (from the Latin word annulus, meaning ‘ring’). When the Moon is relatively close to the Earth, and the Sun, Moon and Earth are aligned, the Moon’s umbra reaches the Earth (Figure 8.10a). Observers in the region of the Earth’s surface covered by the umbra see a total eclipse of the Sun. Observers only just outside the main shadow but still within the penumbra see a partial eclipse of the Sun, in which only a portion of the disc of the Sun is obscured. A total eclipse of the Sun will never last more than 7 minutes at any location on the Earth and most last only 2 or 3 minutes. When the Moon is further from the Earth, the umbra does not quite reach the Earth’s surface. Viewed from the Earth, the Moon does not completely block out the Sun. Its angular size is too small to cover the whole disc of the Sun, and at mid-eclipse a thin ring of the Sun’s disc can still be seen around a dark Moon. This explains the term annular eclipse. Annular eclipses occur slightly more frequently than do total eclipses.


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(a)

Moon’s orbit

Sun

(c) Earth

Moon

(b)

Moon’s orbit

Sun

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Moon

Figure 8.10 (a) During a total solar eclipse the Sun disappears behind the disc of the Moon. Depending on the relative positions of the Earth, Moon and Sun, this can last for as long as 7 minutes. (b) and (c) When the Moon is at its furthest from the Earth, its disc is no longer large enough to cover the Sun, and an annular eclipse occurs, in which a thin ring (or annulus) of the Sun’s disc remains visible and the Moon blocks out only the central region.

Physics file When light travels past our eyes it cannot be seen. Light is invisible unless some of it is reflected into our eyes by tiny particles in the air. The particles might be dust, fog or smoke. An effective demonstration of this is to mark the path of a laser beam in a darkened room with chalk dust.

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Reflection, absorption and transmission After a beam of light strikes an object, there are three processes that can occur: some of the light may be reflected from the surface, some may be transmitted through the material, and some may be absorbed into the surface. This behaviour of light recommends a wave model, as viewing light as a particle would make it difficult to explain the light energy undergoing three different processes. Most materials are opaque to visible light; that is, they do not allow any light to pass through them. For example, brick, plaster and cardboard are impervious to light. Opaque materials will reflect some light and absorb the rest. Other materials are transparent. A transparent material will allow a significant amount of light to pass through it. It may absorb some, and some may even be reflected from the surface of the material. Clear glass, perspex, water and plastic food wrap are common examples of transparent materials. Some materials classified as transparent allow some of the incoming light to pass through but distort the path of this light so that no clear image can be seen through the material. Although the rays of light have passed through the material, the relationship between them has been altered. Such materials are called translucent, and examples include frosted glass, tissue paper and fine porcelain. Translucent materials are particularly useful if an area needs to be illuminated but privacy is required. Frosted or mottled glass is often used for bathroom windows. In other situations, a translucent material is used to deliberately scatter light. For example, the cover around a fluorescent lamp or the ‘pearl’ finish of an incandescent globe can soften household lighting by diffusing it, thereby producing less harsh shadows. Light is a form of energy, and when light is absorbed by a material, the energy it carries is converted directly into heat, warming the material up.

Some of the light energy will also be reflected, bouncing directly from the surface. Experience tells us that a shiny, smooth surface tends to reflect a greater proportion of an incoming light beam than a roughened surface. Figure 8.11 illustrates the three possibilities for the behaviour of light falling, or incident, on a transparent material. Many transparent materials will only absorb tiny amounts of the light energy falling on them. For this reason, we will choose to ignore absorption in our discussions. However, it is important to note that no material is able to allow 100% of the incident light to pass through. There are no perfectly transparent materials; some reflection and absorption of the incident light will always occur. Figure 8.12 shows the effect of the transmission and reflection of light occurring simultaneously. If you look into a shop window you can often see an image of your own face and the streetscape behind you as well as the items on display in the window. The image of your face and the streetscape are the result of reflection: the window is acting as a mirror. However, you also see the items inside the shop as a consequence of the transmission through the glass of the light reflected from objects inside the shop.

incident light

reflected light

boundary between two surfaces absorbed light transmitted light

Figure 8.11 Light incident on the surface of a transparent material is partly reflected, partly transmitted and partly absorbed by the material. The relative amounts of the light experiencing these processes will depend on the nature of the material in question.

Physics file

Figure 8.12 Multiple images are formed by the window, which is simultaneously reflecting and transmitting light from outside and inside the shop respectively.

Euclid, a philosopher and mathematician (330–260 bc) described the law of reflection in his book Catoptrics. However, Euclid also upheld Plato and Ptolemy in their misguided belief in extramission. Euclid claimed that vision was possible because rays from our eyes spread out in all directions and fell on the objects that we see. He proposed that more rays fell on closer objects and so they were seen more clearly. Very small or distant objects were supposed to be difficult to see because they would lie between adjacent rays. Echoes of this idea continued well into the 14th century when vision was still described in terms of ‘extramitted’ visual rays emanating from the eye. Roger Bacon in the late 16th century, however, proposed that light actually travelled from the object to the observer’s eyes.


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Physics in action

The pinhole camera The operation of a pinhole camera provides further evidence that light travels in straight lines. The operation of a camera is more easily explained using a ray model for light. Rays are used to represent the direction of travel of light through the camera. A pinhole camera consists of a sealed box with a small pinhole in the centre of one side. When open, the pinhole allows a limited amount of light into the camera and forms an image on the opposite inside wall of the box. If the opposite wall of the box is lined with photographic film, a permanent image can be developed. The amount of light that can enter the camera is determined by the size of the pinhole, and because this is small, the object to be photographed must either be well illuminated or be a luminous object itself. Typical exposure times reach several minutes, so it is only practical to use a pinhole camera to photograph stationary objects. Ray tracing can be used to determine the size and nature of the image that will be produced on the film. First, consider only the uppermost tip of the object being photographed. An infinite number of rays can be considered to emanate from this point. Only a few of these rays will pass into the camera because of its tiny aperture. These rays continue on to strike the photographic film lining the back wall of the camera. These rays will not all fall on exactly the same point on the photographic plate, but if the pinhole is small, they lie sufficiently close together for an image to be formed. The geometry of the camera dictates that rays leaving the top of the object strike the bottom of the film. Similarly, the rays from the bottom of the object strike the top of the film. This means that the image on the film is inverted relative to the object (Figure 8.13). The image in a pinhole camera is faint because only a little light has been allowed to reach the photographic film. To make a brighter image, a larger diameter pinhole might be used, but this will not produce satisfactory results. The image will be brighter but it will be blurred. This is because the larger hole will allow rays from one point on the object to strike different points on the film. We say that the rays are not focused. On examining the geometry in the ray diagram for a pinhole camera, it is clear that a relationship exists between

screen, film or photographic plate

pinhole distant object

real inverted image

Figure 8.13 The pinhole camera. If the object is at a distance that is 10 times the distance from the pinhole to the film, then the size of the image will be one-tenth of the size of the object. the distances from the pinhole to the object and image and the relative heights of the object and image. It can be seen by using similar triangles that if an object is at a distance equal to 10 times the distance from the pinhole to the film, the height of the image will be one-tenth of that of the object. You can build your own pinhole camera using any container (card or metal) that can be sufficiently sealed to block out all light except that falling on the pinhole. It may help to paint the inside of the box matt black to prevent scattered light from reflecting off the walls and back on to the film. To make the pinhole, punch a nail hole in one wall and cover it with aluminium foil. A pinhole in the foil of about 1 mm diameter will produce good results. You will need to load the film and seal the box in darkness; it is a good idea to practise this a few times first. Alternatively, you can replace the wall opposite the pinhole with tracing paper or another translucent material to act as a viewing screen. This also needs to be shielded from exterior light so that the image is not flooded out. This can be done by surrounding this end of the camera with a cardboard tube. If a photograph is to be taken, mounting the camera on a stand is a good idea. The camera will produce best results with bright, distant objects. Outdoor scenery works well.

8.1 summary Modelling simple light properties • Light travels in a straight path in a uniform medium. • A straight ray model of light implies its particle nature, but rays can also be used to represent the direction of travel of light waves. • When describing the reflection of light, light can be readily modelled using rays. • The law of reflection states that the angle of incidence is equal to the angle of reflection (i = r), and the

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incident ray, the normal and the reflected ray lie in the same plane. • Smooth reflective surfaces produce regular (specular) reflection, whereas rough surfaces produce diffuse reflection. • Light can be reflected, transmitted and/or absorbed at the surface of a material. • Materials can be classified as transparent, translucent or opaque to the passage of light.

8.1 questions Modelling simple light properties 1 Describe three situations or phenomena that provide evidence for the statement ‘light travels in straight lines’. 2 On a particular day in Melbourne at noon, the Sun was at an angle of elevation of 70° above the horizon. Find the length of the shadow of: a a 10 m flagpole b a 1.8 m person c a 50 m building.

8 An observer stands at position P, near a plane mirror as shown. Which of the objects A, B, C and D can be seen in the mirror?

3 A child has glow-in-the-dark stars on her bedroom ceiling. The reason they can be seen in a darkened room at night is because they: A reflect light. B emit light. C transmit light. D absorb light. 4 a Describe the construction of an ordinary plane mirror. b Under certain conditions, a double image can be seen in a mirror. Why? 5 Use the law of reflection to trace the path of the rays of light shown in the diagram. Calculate the angle of incidence and the angle of reflection at each surface. a

b 40n

iii a pane of glass iv aluminium foil v matt paint on a wall vi frosted glass. b Why is it impossible to see an image of yourself in a sheet of paper?

B

P

A

D C

9 Oceanographers refer to the region in the oceans in which some light from the Sun is able to penetrate as the ‘photic’ zone. If seawater is transparent, why doesn’t the photic zone extend to the ocean floor? 10 Two mirrors are placed at right angles as shown in the diagram, and a small object is viewed in the mirrors. Draw the path for rays travelling from the object to the observer as they reflect from the mirrors. (Hint: there are three possible paths.)

60n

eye 90n

90n

6 Describe a situation in which both the partial reflec tion and partial transmission of light occur. How can you tell that both phenomena are occurring at the same time?

7 a Classify the following surfaces as producing dif fuse reflection or regular reflection: i the duco of a new car ii the surface of calm water


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8.2 Refraction of light Refraction Interactive tutorial 7 Refraction

Light travels in a straight path if it is travelling in a uniform medium, but as soon as light enters a different medium its path may be bent. Evidence of the bending of light is shown in Figure 8.14 in which a person’s face can be seen through a glass of water. Some of the person’s face can be seen directly. Light must be travelling along a straight path from the person’s face to the observer’s eyes. However, notice that parts of the person’s face can also be seen through the glass of water. The light rays from the person’s face passing through the glass of water have been re-directed or bent by the water towards the observer’s eyes. The bending or change of direction of light as it passes from one medium to another is called refraction.

Figure 8.14 Refraction occurs because the light changes speed as it enters a medium of different optical density. In this case the light reflected from the person’s face is bent as it enters and leaves the glass of water. As a result the face is seen ‘inside’ the glass of water.

Various common phenomena are caused by refraction. Examples include the bend which appears in a straw that is standing in a glass of water (Figure 8.15), the strangely shortened appearance of your legs as you stand in a waist-deep swimming pool, and the ‘puddles of water’ that you see on the road ahead on a warm day.

R…FRACTION is the bending of the path of light due to a change in speed as it enters a medium of different optical density. Figure 8.15 The refraction of light makes the straw appear to have a bend in it. The appearance of the straw is explained in Figure 8.20.

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To fully understand these phenomena, the refraction of light can be investigated by using a block of glass and a narrow beam of light. Figure 8.16 shows a light beam travelling through air and entering a semicircular glass block. When light strikes the surface of a material some of the incident light is reflected, some is transmitted and some is absorbed by the material. The transmitted ray deviates from its original direction of travel. This change in direction occurs at the boundary between the air and the glass, and the ray

is said to have been refracted. Refraction occurs because the light changes speed as it enters a medium of different optical density. Later discussion will examine the speed of light in different media and how greater changes in speed cause more significant deviation of the beam.

Refraction and a ray/particle approach The refraction of light can be represented using a ray/particle approach. The ray of light that strikes the boundary between two media is called the incident ray. A normal to the boundary is drawn at the point where the incident ray strikes. The angle between the normal and the incident ray is called the angle of incidence, i. The angle between the normal and the transmitted or refracted ray is called the angle of refraction, r. The incident ray, the normal and the refracted ray all lie in the same plane (Figure 8.17). The angle of deviation, D, is the angle through which the ray has been deviated; hence D = (i − r). Refraction is only noticeable if the angle of incidence is other than 0°. If the incident ray is perpendicular to the boundary, i.e. i = 0°, the direction of travel of the transmitted ray will not deviate even though the speed of light has altered. An example of this can be seen in Figure 8.17 as the ray leaves the prism and continues in a straight path. When light travelling through air enters a more optically dense medium such as glass (in which it must travel more slowly), it will be refracted so that the angle of refraction is smaller than the angle of incidence. We say that the path of light has been deviated ‘towards the normal’. When light passes from glass to air it speeds up, as it has entered a less optically dense medium. The angle of refraction will be larger than the angle of incidence. The path of light is described as being refracted ‘away from the normal’.

The behaviour of light undergoing refraction can be summarised by two statements. • When a light ray passes into a medium in which it travels more slowly (a more optically dense medium), it is refracted towards the normal. • When a light ray passes into a medium in which it travels faster (a less optically dense medium), it is refracted away from the normal. The path of refracted light is ‘reversible’. Figure 8.18 shows a ray of light incident on the left-hand side of a rectangular prism. It undergoes refraction towards the normal at the air–glass boundary. It then continues in a straight path through the glass until it strikes the glass–air boundary where it is refracted away from the normal. At each boundary the ray’s path deviates through the same sized angle; hence, the ray that finally emerges from the prism is parallel to the original incident ray. If the light ray was sent in the opposite direction through the prism, i.e. if the starting and finishing points of the light ray were swapped, the light ray would trace out this same path—but in reverse.

Figure 8.16 When a light ray strikes the surface of a glass prism the transmitted ray is refracted because of a change in speed of the light. The bending occurs at the boundary of the two media.

angle of incidence i r

D

angle of deviation angle of refraction

Figure 8.17 The angles of incidence, refraction and deviation are defined as shown. If an incident ray is perpendicular to the boundary between two media, i.e. i = 0°, the direction of travel of the ray does not deviate. An example of this can be seen as light leaves the glass prism.

Figure 8.18 The path of light through any optical element is reversible since the amount of deviation at any boundary is determined by the change in speed of the light.

Refraction and the wave approach Although the way in which light is reflected is modelled equally effectively using either a particle or wave approach, this is not the case for refraction. The wave model is better able to explain the change in direction that is observed as light enters a medium in which its speed is altered.


259

Consider a light wave to be travelling at an angle towards a boundary between two media, as shown in Figure 8.19a. For example, the light may be travelling from air into water. As soon as the light wave enters the water it will slow down. At the moment shown in the diagram the section of wavefront AB that first enters the water will be travelling at a slower speed than the section of the same wavefront BC that has not yet entered the water. This first section of the wavefront then effectively lags behind the position that it would have held had it been able to continue at its initial faster speed. The overall result of this delay is that the direction of travel of the overall wavefront is altered. Figure 8.19b shows how, once a number of wavefronts have passed into the second medium, the direction of travel of the overall wave has been deviated from its original course. A similar arrangement can be constructed for the refraction of light as it speeds up.

PRACTICAL ACTIVITY 35 Refraction of continuous water waves

(a)

S

original direction of travel

(b) C

S

air

wavefront air (fast medium) R

water (slow medium)

B

C

R water

A

B A

new direction of travel

Figure 8.19 The change in the direction of light that is associated with a change of speed is called refraction. Refraction can be modelled by treating light as a wave.

The bent straw

O

Figure 8.20 The immersed portion of the straw is apparently shifted upwards due to refraction. This is because the rays appear to have come from a raised position in the glass of water.

Objects partially immersed in water will be distorted because of refraction; that is, they will appear to have a kink in them. The photograph of the ‘bent’ straw (Figure 8.15) illustrates this. Figure 8.20 shows the path of the rays which produce this illusion. As each ray of light emitted from the base of the straw encounters the water–air boundary it is refracted away from the normal, since the ray enters a less optically dense medium. The observer perceives these rays from the base of the straw to have come from a position higher up in the glass of water. Ray tracing can be carried out for each point along the straw, resulting in the image shown. The straw appears to have a bend in it because the part of the straw that is submerged appears closer to the air–water boundary.

Apparent depth Just as the position of the submerged portion of the straw is apparently shifted, the actual depth of a body of water, or any transparent substance, cannot be judged accurately by an external observer because of refraction effects. Young children often jump into a pool believing it to be much shallower than it really is. Have you ever reached for an object at the bottom of a body of water and been surprised to find that you can’t reach it? Consider an object O at the bottom of a pool as shown in Figure 8.21. Rays from the object are refracted away from the normal at the water–air boundary. If the observer looks down into the water from directly above, he or she will perceive these rays to have come from a closer position as

260


shown. Since the floor of the pool will appear to be much closer than it really is, the water seems safely shallow. The extent to which the depth of the water is altered is affected by the angle from which it is viewed.

observer

Physics file Although ray diagrams are commonly used to represent the refraction of light, keep in mind that these describe only the direction of travel of the refracted light and do not tell the whole story. Although wave diagrams of refraction are more complicated, they are more convincing since they reveal the reasons for the change in direction of travel of the light.

air water apparent depth of pool

O

O

Figure 8.21 The apparent depth of a body of water appears less than its actual depth because of refraction. Light rays from a point at the bottom of the pool are refracted at the water’s surface. The observer perceives these rays to have come from an elevated location and interprets this as an indication of shallow water.

Worked example 8.2A Predict the approximate path of light through the following prisms. In each case identify the normal, the angle of incidence and the angle of refraction.

a

b

air

air

glass

perspex

c

d air air

glass

glass water


261

Solution a

b

r

i

r

i air

glass

perspex

c

d i

air

air

glass

glass

i

water

r

Physics file Willebrord Snell (1591–1626) is commonly accredited with the discovery of the law of refraction in about 1621. He did not immediately publish his findings and meanwhile the French scientist René Descartes (1596–1650) published his own derivation of the law of refraction. This caused a dispute within the scientific community at the time, with some claiming that Descartes had seen Snell’s work. It is worth noting that in France the law of refraction known elsewhere as Snell’s law is called Descartes’ law.

At each boundary between media the light must refract either towards or away from the normal to the boundary. The light ray is refracted towards the normal on entering each prism and away from the normal when leaving the prism. When an incident ray meets a boundary with an incident angle of 0°, no deviation occurs.

The law of refraction Scientists spent many years trying to find the relationship between the angle of incidence and the angle of refraction produced at the boundary between a given pair of media. At very small angles, doubling the angle of incidence appeared to double the angle of refraction, but this relationship does not hold for larger angles of incidence. In about 1621 Willebrord Snell, a Dutch scientist, found that for a given pair of media the sine of the angle of incidence was directly proportional to the sine of the angle of refraction, i.e. sin i ∝ sin r. This relationship is now known as Snell’s law:

Figure 8.2 2

262

r

sin i = constant sin r

. . . . . . (i)

Each combination of a pair of materials has a different constant. For example, the constant for the air–water interface is different from that for the glass–water interface. The constant is called the relative refractive index, denoted n*. Literally, an index (or listing) of numbers was created to describe the amount of refraction or bending occurring at the boundary of numerous pairs of transparent media. A higher relative refractive index for a given pair of media indicated that more bending occurred. This was a cumbersome system which was refined, as will be explained later in this chapter. Willebrord

Snell.


sin i sin r Each pair of media will have a specific relative refractive index. R…LATIV… R…FRACTIV… IND…X: n* =

Worked example 8.2B A student shines a thin beam of light on to the side of a glass block. She notes that when the angle of incidence is 40°, the light passes into the block with an angle of refraction of 25°. When the angle of incidence is 70° the angle of refraction is 39°. Determine the relative refractive index, n*, of the air–glass boundary.

PRACTICAL ACTIVITY 36 Investigating refraction: Snell’s laws

Solution The first pair of data gives: sin i sin 40° = = 1.5 sin r sin 25° The second pair of data gives: n* =

sin i sin 70° = = 1.5 sin r sin 39° i.e. the same relative refractive index, as expected. n* =

Optical density and the speed of light Different transparent media allow light to travel at different speeds. Light travels fastest in a vacuum, more slowly in water and even more slowly in glass. We say that glass is more optically dense than water. Table 8.1 shows the speed of light in various media. The amount of refraction occurring at any boundary depends upon the extent to which the speed of light has been altered, i.e. the ratio of the two speeds of light in the two different media. Figure 8.23 represents light travelling in air and meeting three substances of different optical density. Light bends most when its speed is most significantly altered. The medium carrying the incident ray is identified as medium 1 and the medium carrying the refracted ray is medium 2. The angle of refraction depends on the speed of light in the two media and the angle of incidence. sin i v1 = . . . . . . (ii) sin r v2

Table 8.1 The speed of light in different media (quoted for yellow light λ = 589 nm) Medium

Speed of light (m s-1)

Vacuum

3.00 × 108

Air

3.00 × 108

Water

2.25 × 108

Glass

2.00 × 108

air water

air glass

air diamond

where v1 is the speed of light in medium 1 and v2 is the speed of light in medium 2.

The index of refraction Since it is only the ratio of the speeds of light in the two different media that determines the degree of refraction, each medium can be allocated an absolute refractive index, n. This is obtained by comparing the speed of light in the medium in question with the speed of light in a vacuum: n=

speed of light in a vacuum speed of light in the medium

least refraction

most refraction

Figure 8.23 In each case light is entering a medium of greater optical density. The bending of the path of the light depends on the ratio of the speeds in the two different media. The path of light deviates most when the change in speed is greatest.

c = speed of light in a vacuum = 3.0 × 108 m s−1 c n= v medium


263

Table 8.2 Absolute index of refraction, n (quoted for yellow light λ = 589 nm) Medium

Index (n)

Vacuum

1.0000

Air

1.00029

Ice

1.31

Water

1.33

Quartz

1.46

Light crown glass

1.51

Heavy flint glass

1.65

Diamond

2.42

By definition the refractive index of a vacuum would be exactly 1. Light travels only marginally more slowly in air and so the refractive index of air is 1.0003, but in most cases a value of 1.00 is sufficiently accurate. Materials in which light travels slowest will have the highest indices of refraction. For example, if a particular medium allowed light to travel at half the speed it does in a vacuum, then the refractive index of the medium would be 2. The refractive index can therefore be considered an indication of the ‘bending power’ of a material. Table 8.2 lists the absolute refractive indices for various media. By definition of the absolute refractive index: v1 c n2 × = c v2 n1 Hence, Snell’s law can now be expressed as: sin i n2 = sin r n1

Physics file The refractive index is also dependent on the colour of the light travelling through the medium. Since white light is made up of light of different colours, the refractive index must be quoted for a specific wavelength of light. Typically yellow light of wavelength 589 nm is used since this can be considered an average wavelength of white light. If the refractive index for a particular sample of crystal quartz was quoted as 1.55, red light would have a slightly higher refractive index of 1.54 and violet light would have a slightly lower refractive index of 1.57. This observation produces dispersion, discussed later in the chapter.

N i = 40°

Snell lived nearly 200 years before scientists were able to measure the speed of light in air with any degree of accuracy, so he was not aware that the refractive index was linked to the speed of light in a particular medium. We can now combine the equations (i), (ii) and (iii) developed above so that Snell’s law is fully expressed.

SN…lL’s law:

sin i v1 n2 = = = n* sin r v2 n1

Worked example 8.2C A ray of light passes from air into quartz, which has an absolute refractive index of 1.46. If the angle of incidence of the light is 40°, calculate: a the angle of refraction b the angle of deviation of the ray c the speed of light in the quartz. Assume the index of refraction of air is 1.00 and the speed of light in air is 3.0 × 108 m s−1.

Solution

medium 1 = air medium 2 = quartz

boundary

r angle of deviation

Draw a diagram to model the situation. As the light is slowed down the rays should bend towards the normal. a List the data: i = 40°, n1 = 1.00, n2 = 1.46 and r = ? sin i n2 Then: = sin r n1

sin 40° 1.46 = 1.00 sin r

1.00 × sin 40° 1.46 ∴ r = 26° Note that r is smaller than i, as expected. b The angle of deviation is equal to the difference between i and r. D = (i − r) = 40° − 26° = 14° Hence sin r =

264

. . . . . . (iii)


c

n1 = 1.00, n2 = 1.46, v1 = 3.0 × 108 m s−1 v1 n2 = v2 n1 3.0 × 108 1.46 = v2 1.00 v2 = 2.1 × 108 m s−1

Physics in action

Huygen’s wavelets and refraction In 1678 the Dutch mathematician Christiaan Huygens published his ideas on the nature and propagation of light. (At about the same time Newton developed his corpuscular theory.) Huygens’s idea was that light acted like a wave. In his model he suggested that each point along a wavefront of light could be considered to be a point source for small, secondary wavelets. Each wavelet was spherical, and the wavelets radiated from their point source in the general direction of the wave propagation (i.e. the light beam). The envelope or common tangent of the wavelets then became the new wavefront as shown in Figure 8.24.

incident light

λ1

λ1 A λ2

θ1

θ2

B

boundary

λ2 refracted light

ray

Figure 8.25 Huygens’s approach allowed the refraction of light—as quantified by Snell’s law—to be accurately modelled. ray

source

initial wavefront ray new wavefront

Figure 8.24 The envelope of the wavelets caused the formation of the new wavefront. Figure 8.25 shows how Huygens’s principle can be used to explain refraction. In the initial medium the spacing between wavefronts is λ1. In the second (slower) medium the wavelength, λ2, will be reduced in correspondence with the ratio of the velocity of light in each medium such that:

Consider the wavefront that is just approaching the boundary labelled AB and the wavefront just leaving it. Using the right-angled triangle drawn above the boundary (with hypotenuse AB), we can state: l sin θ1 = 1 AB Using the right-angled triangle below the boundary with hypotenuse AB, we can state: l sin θ2 = 2 AB Therefore: AB l1 sin θ1 l = 1 × = sin θ2 AB l2 l2 Hence Snell’s law, which states that: sin θ1 l1 v1 = = sin θ2 l2 v2 can be derived from a wave model of light.

l1 v1 = l2 v2


265

apparent position of star

light from star is refracted as it travels through the atmosphere

real position of star

layers of increasing optical density

Figure 8.26 The refractive index of the atmosphere is not uniform; hence, the path of light from a star is refracted, apparently altering its position.

Refraction in the atmosphere Although air has previously been considered to be a uniform medium, there are circumstances when in fact it is not uniform. Consider the envelope of air surrounding the Earth. The atmosphere is not uniform: the optical density of air increases closer to the Earth’s surface. Although this variation occurs gradually, the situation can be represented by a series of horizontal layers of increasing refractive index as shown in Figure 8.26. As light from an object such as a star travels through each boundary between layers it is refracted towards the normal. The observer therefore believes the light to have come from a position higher in the sky. In reality the bending must occur gradually rather than at distinct intervals, and the amount of bending has been greatly exaggerated in the diagram. A maximum amount of refraction occurs when objects are quite low in the sky. This is because the angle of incidence on the atmosphere is greatest and the light must travel through a wider atmospheric band. The amount of refraction is not noticeable to the human eye as when a maximum amount of refraction occurs stars would only be shifted in position by less than 1°. Refraction by the atmosphere also extends the length of the day. Whenever you watch a sunset you see the Sun for a few minutes after it has actually passed below the horizon. This is because light from the Sun is refracted as it enters and travels through the Earth’s atmosphere, as shown in Figure 8.27. This effect is greatest at sunset and sunrise when the angle of incidence of the Sun’s rays on the atmosphere is greatest. The atmosphere consists of moving layers of air and so the optical density of the layers is continually changing. Since light rays must travel through this varying medium, light from the one object will follow slightly different paths at different times. This is one reason for the twinkling of stars and the apparent wriggling of distant objects on a warm day. Sun appears to be on the horizon atmosphere position of observer

actual path of sunlight

Figure 8.27 The Sun is still visible even though it is actually below the horizon.

266


Physics in action

Mirages light from sky

cool air hot air mirage

Mirages are often believed to be the insane illusions of thirsty desert wanderers, but a mirage is the image of a real object and can be explained in terms of the refraction of light. A mirage is a displaced and often distorted image occurring when layers of air of different temperature cause the path of light to bend. The severity and consistency of this temperature gradient determines many features of the observed apparitions. An inferior mirage refers to the downward displacement of an image. A superior mirage means the image is displaced upwards. The following discussion examines just a few of the many different types of mirages that occur.

Inferior mirages An inferior mirage occurs when the air at ground level is warmer than the air immediately above, i.e. the air is being heated from below. This situation often arises in the afternoon of a hot, sunny day above a black bitumen road or above the sands of the desert. Air of higher temperature is less optically dense and hence has a lower refractive index. Light from the sky is refracted as it travels through the layers of air of

Figure 8.28 Desert mirage. When air is heated from below, an inferior mirage can occur in which an image is displaced downwards. This diagram is simplified. The air layers of different temperature will not be uniform nor parallel to the plane of the ground. Light from the sky is gradually refracted by the air layers so that it is travelling slightly upwards when it enters the eye. The diagram exaggerates this bending. The observer sees an image of the sky on the distant road ahead and interprets this as a body of water.

different optical density as shown in Figure 8.28. As a result the light ray is travelling upwards as it enters the observer’s eyes and the image is then seen on the ground ahead. Driving on a warm day, you often see an image of the sky on the road ahead; this is interpreted as a body of water.

Floating on water A fascinating inferior mirage often occurs above shallow bodies of water in the early morning. The water retains its heat overnight but the surrounding land does not. Cool air from above the land flows over the warmer water and is heated from below. Thus air temperature decreases with height. However, the temperature gradient is not uniform. The temperature drops quickly near the water’s surface, but at greater heights the decrease in temperature is more gradual. If an observer looks at a person in the distance the image of the person is displaced downwards but the bottom of the object is displaced more than the top of the object since these lower rays travel through a stronger temperature gradient. Thus the person is irregularly enlarged. This phenomenon is called towering.

Figure 8.29 The picture on the left shows a mirage on a road. In the picture on the right, the horizon shows the effect of towering.


267

8.2 summary Refraction of light • Refraction is the bending of light due to a change in its speed as it enters a medium of different optical density. The greater the change in the speed of light the greater the bending. • The angle between the normal and the incident ray is called the angle of incidence. The angle between the normal and the transmitted or refracted ray is called the angle of refraction. • When a light ray passes into a medium in which it travels more slowly (a more optically dense medium) it is refracted towards the normal. • When a light ray passes into a medium in which it travels faster (less optically dense) it is refracted away from the normal.

• Each transparent medium is allocated an absolute refractive index, n, determined by the speed at which light can travel in the medium compared with the speed of light in a vacuum, c: c n= v medium • Snell’s law describes the relationship between the angle of incidence, i, and the angle of refraction, r, for a given pair of media: sin i v1 n2 = = = n* sin r v2 n1

8.2 questions Refraction of light 1 a The figure below represents a situation involving the refraction of light. Which of the lines labelled A–E is: i the boundary between two media? ii the normal? iii the incident ray? iv the refracted ray? v the reflected ray? b Explain what happens to the speed of light as it crosses the boundary between medium 1 and medium 2. How do you know? A

c Which medium has the highest refractive index? d Which medium has a refractive index very close to the refractive index of this sample of glass? A

B i

i glass medium A

B C C

D i

i medium 1 D medium 2

E 2 The following diagrams show light passing from glass into different media labelled A, B, C and D. a Which media are more optically dense than the glass? b Which media have a refractive index less than the refractive index of glass?

268

glass medium B


glass medium C

glass medium D

3 Using Figure 8.25 as your reference, show how a wave model can be used to explain the refraction of light as it passes through the boundary into a medium in which its speed is increased. 4 Explain the following observations. a When you are standing in a shallow pool you appear shorter than usual.

b On a warm day a person sees a ‘puddle’ on the road ahead. 5 When light passes through a pane of glass it is refracted. This does not cause the distortion of an image seen through the glass because: A the emerging rays are perpendicular to the inci dent rays. B the index of refraction of glass is too small to cause distortion. C the displacement of light rays is too small to be noticed unless the glass is very thick. D most of the light is reflected, not refracted. 6 a The speed of light in a particular transparent plastic is 2.00 × 108 m s−1. Calculate the refractive index of the plastic. The speed of light in a vacuum is 3.00 × 108 m s−1. b What is the speed of light in water (n = 1.33)? 7 A student wishes to determine the refractive index of a particular sample of glass by experiment. By passing a narrow beam of light from air into the glass, she measures the angles of refraction, r, produced using a range of incident angles, i. Her results are shown.

8 a What is the relative refractive index for light passing from water into diamond if an incident angle of 30° produces an angle of refraction of 16°? b Light travels from water (n = 1.33) into glass (n = 1.60). The incident angle is 44°. Calculate the angle of refraction. 9 A ray of light is incident on the surface of water in a fish tank. The incident ray makes an angle of 32.0° with the surface of the water. The light that is transmitted makes an angle of 50.4° with the surface. Calculate: a the angle of incidence b the angle of refraction of the transmitted light c the angle of reflection d the angle of deviation of the transmitted ray. 10 A ray of light travels from air, through a layer of glass and then into water as shown. Calculate angles a, b and c. air (n = 1.00)

glass (n = 1.50)

water (n = 1.33)

40n a

i (degrees) 25

30 35

r (degrees) 16

19

40 45

22 25

28

50 55

60 65

31

35

33

37

a Plot a graph of sin i versus sin r. b Determine the gradient of the graph, i.e. the relative refractive index of light passing from air into glass. c Assuming the refractive index of air is 1.00, what is the refractive index of the glass sample used? d Calculate the velocity of light in the glass.

b c


269

R

8.3 Critical angle, TIR and EM Physics file

Critical angle and total internal reflection

When light refracts at a surface the transmitted ray becomes less intense as the angle of incidence increases. Place any small flat piece of glass on this page; a microscope slide will do. Look onto the slide from directly above and you will easily observe the writing below. Now move your head so that you are looking through the slide from a gradually decreasing angle of elevation. The page beneath the glass should become gradually darker. The amount of light transmitted through the glass and towards your eyes is becoming less.

When light is incident upon the boundary between two media, reflection, transmission and absorption may occur. As the angle of incidence increases, the intensity of the reflected beam increases and less light is transmitted. Consider the case of light travelling from water into air. Since the transmitted light enters a less optically dense medium it travels faster and is refracted away from the normal. The series of diagrams in Figure 8.30 shows the effect that increasing the angle of incidence has on the transmitted light. As the angle of incidence is increased, the angle of refraction also increases. This continues until, at a certain angle of incidence called the critical angle, ic , the angle of refraction will be almost 90° and the transmitted ray travels just along the surface of the water. The angle of refraction can increase no further. If the angle of incidence is then increased beyond the critical angle no ray is transmitted and total internal reflection will occur. This is appropriately named, as none of the incident light energy is able to be transmitted into the next medium; it is totally reflected into the medium carrying the incident ray. In effect, as the angle of incidence increases, the intensity of the reflected beam gradually becomes stronger, until at an angle of refraction of more than 90° all of the light is reflected and no light is transmitted at all. The critical angle can be found for any boundary between two media by using Snell’s law. If the refractive indices of the two media are known, a presumption of an angle of refraction of 90° allows the critical (incident) angle to be calculated: sin i n2 = sin r n1

i = 45° water air r = 70°

i = 49° water air

r< ∼ 90°

i = 50° water air

Figure 8.30 The critical angle for light travelling from water into air is approximately 49°. If the incident angle is greater than 49° total internal reflection occurs.

If the incident angle is equal to the critical angle, i.e. i = ic , then r = 90°. The above equation becomes: sin ic n = 2 sin 90° n1 Now sin 90° = 1; hence the critical incident angle is given by: n sin ic = 2 n1

The CRITICAL ANGL…, ic, is the angle of incidence that produces an angle of refraction of 90° as light is transmitted into a medium in which it travels at a higher speed. n sin ic = 2 n1

Worked example 8.3A An underwater light shines upwards from the centre of a swimming pool that is 1.50 m deep. Determine the radius of the circle of light that is seen from above. (nair = 1.00, nwater = 1.33)

270


Solution

Physics file

Step 1. Determine the critical angle for the water–air boundary. n sin ic = air nwater 1.00 sin ic = 1.33 1.00 ic = sin−1 1.33 ic = 48.8° Step 2. Draw a diagram to represent R air P the situation. Considering one-half of the cone of water light produced, since the critical angle 1.50 m is 48.8°, the angle QOP is 48.8°. Step 3. Using trigonometry: 48.8° R tan 48.8° = 1.50 O hence R = 1.50 tan 48.8° light source = 1.71 m The radius of the circle of light is 1.71 m.

( )

Q

48.8°

Total internal reflection is used in many optical instruments, including cameras, periscopes and binoculars. When light reflects from a mirror there is always some loss in the intensity of the incident light, and reflection can occur from both the front and rear surfaces of the mirror, causing problems. By contrast, almost no loss of intensity occurs with total internal reflection. Since the refractive index for glass is about 1.5, the critical angle for light travelling from glass to air is approximately 42°, and a glass prism with internal angles of 45° can be used as a mirror in the applications discussed below. Figure 8.31a shows the construction of a simple periscope. Light enters the top glass prism perpendicular to the glass surface and so no refraction (deviation) occurs at this stage. The light passes through the prism and strikes the back surface at an angle of 45°. The angle of incidence is greater than the critical angle and so the light can only be totally internally reflected. The light travels down the tube of the periscope, enters the lower prism and is again reflected. The surfaces of the prism do not need to be silvered for reflection to occur. Binoculars use a compound prism that is constructed of four 45° prisms (Figure 8.31b). The light actually undergoes four reflections on its passage from the objective lens to the eyepiece. This lengthens the path that the light must travel and hence more compact binoculars can be made for a given magnification. Today’s binoculars are as effective as the telescopes of the past, which had to be many times longer.

Since the refractive index of any given medium depends on the colour of the light travelling through the medium, each colour of light will have a slightly different critical angle. For example, if the critical angle for red light travelling from glass to air was 40.5°, then yellow light would have a slightly smaller critical angle of 40.2°. Violet light would have an even smaller critical angle of 39.6°. Although these values are very similar, they are sufficiently different to cause the dispersion of white light into its component colours—the colours of the rainbow!

(a)

(b)

eyepiece

objective

Figure 8.31 Good quality periscopes and binoculars use 45° glass prisms for the total internal reflection of light rather than mirrors. This means reduced loss in the intensity of light and eradicates the problems caused by reflection occurring at both the front and rear surfaces of mirrors.

PRACTICAL ACTIVITY 37 Total internal reflection in prisms


271

Physics in action

Optical fibres An optical fibre uses total internal reflection to carry light, with very little energy loss. Since the 1950s fibre optics has been used in the flexible fibrescope, a device that allows doctors to see inside the human body. By 1970 the production of very pure fibres reduced energy losses, making fibre optics feasible for use in communication. Since then this technology has made it possible to transfer large amounts of data at remarkable speeds, enabling the dream of a worldwide computer network to be realised.

The transmission process Essentially optical fibres carry information in a digital format; that is, light signals turn on and off at very fast rates. Optical communication systems take information from a device, convert it into a digital form if needed, and impress this digital signal onto a carrier frequency that has been produced by a laser or LED. This is called modulation. The signal is then fed into an optical fibre for transmission (coupling). A critical factor is the range of frequencies of light that the optical fibre is able to carry efficiently. Each fibre has a limiting bandwidth (frequency range) and a set number of different signal wavelengths that are allowable in this bandwidth. For example, the different signal wavelengths employed may be not allowed to be any closer than about 0.3 nm in spacing. (See later discussion.) During transmission attenuation (energy losses along the way) will occur. Therefore repeater stations are used to receive the weak incoming signal and boost it before sending it further along the fibre. Regenerators may also be used to remove noise and distortion in the signal at this stage. Depending on the quality of the fibre, repeater stations are inserted about every 100 km along a long-distance optical cable. When a signal reaches its destination it must be de-modulated (removed from its carrier wave) and the signal processed as required by the end user. The signal must arrive with only as much distortion as can be compensated for at this end.

(a)

Figure 8.32 A magnifie

d optical fib re torch.

Different types of fibre Optical fibres can be divided into two categories depending on the manner in which information is carried: single-mode (thin core) fibres and multimode (large core) fibres. The term ‘mode’ is synonymous with pathway. A single-mode fibre allows only one path on which light can travel (Figure 8.33a). Note the tiny dimensions of the fibre. A micron is one-millionth of a millimetre, and a human hair is typically about 70 microns in diameter. Single-mode fibres are used in high-speed, long-distance telecommunications. For example, Melbourne and Sydney are linked by optical cable containing numerous single-mode fibres. (b)

core 8.3 micron

cladding 125 micron

core 50 100 micron

coating 250, 500 or 900 micron

cladding 125 or 140 micron

coating 250–900 micron

Figure 8.33 Compare the dimensions and structure of the different fibres. (a) Single-mode fibres, such as those used in high-speed telecommunications, result in much less distortion of the optical signal. (b) Multimode fibres, such as those used in local area networks, are cheaper to produce but have more distortion problems. These are adequate for use over shorter distances.

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A multimode fibre can have two different forms: the stepindex multimode fibre or the graded-index multimode fibre. A step-index multimode fibre has the same structure as the single-mode fibre described above, but it has a much larger core made of uniform glass (Figure 8.33b). A graded index fibre also has a large core, but its refractive index gradually decreases from the centre to the outer diameter of the fibre.

The path of light in a step-index multimode fibre Light rays are sent down the central core fibre. If the fibre is straight, most of the rays will travel along the axis of the fibre. Some light will strike the boundary between the core and the cladding, particularly if the fibre is bent. Any ray striking the boundary at an angle greater than the critical angle is totally internally reflected. The size of the critical angle is determined by the refractive indices of the core and cladding (see Figure 8.34).

Worked example 8.3B A particular step-index multimode fibre has a core of refractive index 1.460 and cladding of refractive index 1.440. Calculate the critical angle of the core–cladding boundary of this optical fibre.

Solution List the data: ncore = 1.460, ncladding = 1.440

( )

1.440 1.460 ic = 80.51°

sin ic = sin−1

The core and cladding are designed so that the critical angle for an optical fibre is typically greater than 80°. Hence, only light rays undergoing glancing collisions with the core– cladding boundary are totally internally reflected. Although this causes more light energy to be lost at coupling, it means that all of the light rays emerging from the end of the fibre have travelled a path of approximately the same length.

cladding (n = 1.440)

81°

rays striking here are lost

86°

9°

cone of acceptance

light source

13° core (n = 1.460) 13° 9°

rays striking here are lost

81°

86°

cladding (n = 1.440)

Figure 8.34 Outer rays from the light source that enter the core at an angle of incidence greater than 13° will continue on to strike the core–cladding boundary at an angle of incidence less than the critical angle of 81°. These rays will therefore be transmitted into the cladding and be lost. Rays originally within the cone of acceptance strike the cladding at an angle greater than 81° and are therefore totally internally reflected. These reflected rays carry the signal along the fibre.

Electromagnetic waves What is light? In the late 1600s it was known to involve the transfer of energy from one place to another. In Isaac Newton’s time a corpuscular (particle) model and a wave model for light had seemed equally valid. We have discussed these two proposed models of light along with their respective explanations of the reflection and refraction of light. In spite of considerable endeavour by scientists it was not until the early 1800s that one model prevailed. Thomas Young discovered that sources of light were able to interfere with each other just like sound waves and water waves do. This finding led to a universally accepted wave theory for light. Furthermore, the speed of light could be measured for the first time, and the wave model of


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refraction (discussed in section 8.2) was validated. Meanwhile another area of physics had been developing. By the 1860s investigations being carried out on different forms of electromagnetic radiation led to the finding that visible light itself is just one of the many forms of electromagnetic radiation (EMR). Electricity and magnetism were once considered to be separate subjects. However, moving charges create magnetic fields. Similarly a changing magnetic field can be used to create electricity. In 1864 James Clerk Maxwell used mathematical equations to describe how charges moving periodically in a conductor would set up alternating electric fields and magnetic fields in the nearby region. Maxwell knew that the magnetic and electric fields travelled through space. He calculated their speed and found it to be 300 000 km s−1, exactly the same as the speed of light! Also, he devised mathematical expressions to describe the magnetic and electric fields. The solution to these expressions was found to be the equation of a wave. Maxwell had shown that light is an electromagnetic wave. Today we know that the electromagnetic spectrum includes a wide range of frequencies (or wavelengths). All electromagnetic waves are created by accelerating charges which result in a rapidly changing magnetic field and electric field travelling out from the source at the speed of light, as shown in Figure 8.35. Note that the electric field component and the magnetic field component are at right angles to each other and to their direction of travel. Electromagnetic radiation meets the description of a transverse wave as discussed in Chapter 7. x electric field

direction of motion of wave velocity c = 3 × 108 m s–1

y

magnetic field

z wavelength λ

Figure 8.35 Since all electromagnetic waves travel with the same velocity the only thing that differentiates one form of EMR from another is the frequency (and, therefore, the wavelength).

The many forms of EMR are essentially the same, differing only in their frequency and, therefore, their wavelength. The electromagnetic spectrum is roughly divided into seven categories depending on how the radiation is produced and the frequency. The energy carried by the electromagnetic radiation is proportional to the frequency. High-frequency short-wavelength gamma rays are at the high-energy end of the spectrum. Low-frequency long-wavelength radio waves carry the least energy. Humans have cells in their eyes which can respond to EMR of frequencies between approximately 400 THz and 800 THz; these frequencies make up the visible light section of the electromagnetic spectrum.

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Recall from Chapter 7 that for any wave the relationship between its frequency and its wavelength is given by v = fλ. All electromagnetic radiation travels at a speed of 3.00 × 108 m s−1 in a vacuum and so this significant speed has been allocated the symbol c.

c For all …L…CTROMAGN…TIC RADIATION f = l where f is the frequency of the EMR (Hz) c is the speed of the EMR = 3.0 × 108 m s−1 λ is the wavelength of the EMR (m) Figure 8.36 shows the different categories of EMR. Note the range of frequencies and wavelengths is enormous. The range of frequencies (or wavelengths) constituting visible light occurs near the middle of the spectrum. Our eyes cannot perceive any wavelengths of EMR outside of this range. Frequency

Wavelength

1022 Hz

10–14 m

Ultraviolet

1016 Hz

10–8 m

Visible spectrum

800–400 THz

400–800 nm

Infrared

1012 Hz

10–4 m

Microwaves

1010 Hz

10–2 m

TV

108 Hz

10 m

Radio

106 Hz

102 m

Gamma rays

X-rays

Figure 8.36 The electromagnetic spectrum.

Worked example 8.3C The EMR given off by a sample of sodium as it is burned has a wavelength of 589 nm. What is the frequency of this radiation? How would we detect the radiation?


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Solution c l 3.0 × 108 = 589 × 10-9 = 5.09 × 1014 Hz = 510 THz This frequency of EMR lies in the visible light section of the electromagnetic spectrum, therefore we would see it! It is actually yellow light. f =

Physics in action

Other forms of EMR by sensing infrared radiation and converting it into a visible picture.

Radio waves Accelerating a positive or negative charge can produce EMR. Electrons oscillating in a conducting wire, such as an antenna, produce the radio waves that bring music to your home. The long-wavelength low-energy electromagnetic waves blanket the surrounding region, and aerials can receive the signal many kilometres from the source. As a result of the radio waves, electrons in the receiving aerial wire will oscillate, producing a current that can be amplified. Radio waves can be transmitted over very long distances, including around the Earth’s surface, by reflection from layers in the atmosphere.

Microwaves Microwaves are EMR of wavelengths ranging from about 1 mm to about 10 cm. The microwaves that cook your dinner are produced by the spin of an electron or nucleus. Microwave links are used to allow computer systems to communicate remotely, and radar equipment uses microwave frequencies of centimetre wavelengths.

Infrared waves Infrared or heat radiation includes the wavelengths that our skin responds to. When you feel the warmth from the Sun or an electric bar heater you are actually detecting infrared radiation. All objects that are not at a temperature of absolute zero radiate EMR. The hotter the object the more radiation is emitted, and the further along the spectrum the radiation is. Night scopes and infrared spy satellites create an image

PRACTICAL ACTIVITY 38 Light and a continuous spectrum

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Ultraviolet waves Ultraviolet waves have wavelengths shorter than violet light— so our eyes cannot detect them—but no greater than about 10 nm. Many insects can detect the ultraviolet light that is commonly reflected from flowers. Although ultraviolet light is less energetic than gamma- or X-rays, it is known to cause skin cancer, particularly with increased exposure. Silicon atoms are able to absorb some frequencies in the ultraviolet region of the spectrum, reducing your chances of getting sunburnt through glass.

X-ray waves X-rays are produced when fast-moving electrons are fired into an atom. The name is a result of scientists not knowing what they were when they were first detected, hence the letter ‘X’. X-rays can pass through body tissue and be detected by photographic film, and so are used in medical diagnosis. They have extensive safety testing, security and quality control applications in industry.

Gamma-ray waves The highest energy, smallest wavelength radiation is the gamma ray, which is produced within the nucleus of an atom. Gamma rays are one of the three types of emissions that come from radioactive (unstable) atoms. Gamma rays are extremely penetrating and require dense material to absorb them.

Coloured light, different wavelengths Our eyes are responsive to many different colours of light from the deepest red through to the brightest violet, the visible spectrum (Figure 8.37). Each variation in colour or shade is caused by light of a different wavelength. Traditionally the colours quoted as making up the visible spectrum are red, orange, yellow, green, blue and violet. However, as shown in Figure 8.37, the actual allocation of separate names for the colours is difficult since they merge into one another. The wavelengths associated with visible light are very small: they range from approximately 390 nanometres (or 3.9 × 10−7 m) for violet light to around 780 nanometres for red light.

Physics file

red wavelength y 780 nm

You wouldn’t expect a person renowned as a great scientist to be superstitious. For many years the spectrum of colour was listed as being made up of seven separate colours rather than the six colours listed today. Isaac Newton carried out famous experiments producing the spectrum of colour and recombining it into white light. In his writings indigo (a very dark blue) was stated as lying between blue and violet. The separate identification of indigo light is strange as it really does not appear as prominently as the other six main colours. Newton was rather mystical in his religious beliefs and seven was considered to be a ‘perfect’ number somehow related to the natural laws governing the universe, and so he deliberately identified seven colours in the visible spectrum.

violet wavelength y 390 nm

Figure 8.37 Visible light is one category of EMR. The spectrum of visible light contains a myriad of colours. Each different colour or hue is light of a different wavelength.

Colour addition or mixing light sources In 1807 Thomas Young discovered that combining red, green and blue light on a screen produced white light. In fact various combinations of these three colours of light could create all of the other colours of the spectrum. Red, green and blue are therefore called the primary colours of light. None of the primary colours can be produced by a combination of the other primary colours.

Relative intensity

PRACTICAL ACTIVITY 39 Colour addition and subtraction

Relative intensity

Sunlight

Incandescent light

Relative intensity

The colour of an object that we see is actually a physiological response to the particular wavelength(s) of light entering our eyes. Our coloursensing system, consisting of the eye, nerve conductors and the brain, can discriminate between hundreds of thousands of different colours. However, different combinations of wavelengths of light can evoke the same response from our brain. In other words, there are a number of different ways in which to make an object appear a particular shade of yellow, for example. We perceive light as white light if it contains roughly equal amounts of each of the colours of the visible spectrum. The page of this book appears white because it is reflecting all of the colours (wavelengths) of visible light in roughly equal proportions. Sunlight, incandescent light and fluorescent light all produce the same general sensation of white light. Figure 8.38 shows their component colours. The light from incandescent and fluorescent globes does not appear to be quite as white as sunlight. This is because sunlight is very evenly distributed across the spectrum, but an incandescent source radiates considerably more red light than blue light, and a fluorescent source favours blue wavelengths of light. When incident light strikes the surface of an object, it may be absorbed, transmitted and/or reflected. If all of the white light falling on a surface is absorbed, the object will appear black as no light is reflected. The colour of an object is often determined by which colours of light it reflects and absorbs when white light is shone upon it.

Fluorescent light

Figure 8.38 The colour components of sunlight, incandescent light and fluorescent light. All are referred to as sources of white light, but their spectral compositions vary, affecting the colour of an illuminated object.


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(a)

(b)

Figure 8.39 Pairs of the primary colours of light overlap to produce the secondary colours yellow, cyan and magenta. When all three primary colours of light overlap white light is produced.

(c)

Blue

Green

Red

Mixture white cyan yellow magenta

Figure 8.40 The yellow seen on the television screen is, like all other colours, composed of only red, green and blue dots. The tiny dots on the screen are so close together that they cannot be recognised as separate dots, but blend together to form a continuous picture. Different areas of the television screen are made to produce red, green and blue dots in varying proportions, thus creating the various colours seen on the television screen.

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Figure 8.39 shows the three primary colours of light overlapping to produce other colours. The particular colours formed by the overlapping of pairs of primary colours are called cyan, magenta and yellow. Any group of colours which combine to form white light are called complementary colours. All three primary colours when combined form white light. Combining any two primary colours forms the complementary colour of the remaining primary colour. So, for example, when red and green are combined they form the complement of blue, which is yellow. Yellow is the complement of blue. Cyan is the complement of red, and magenta is the complement of green.

Colour television Television screens produce coloured pictures (Figure 8.40a) yet only utilise three different colours: red, green and blue. These colours are produced when electron beams strike the tiny phosphor dots lining the screen. Figure 8.40b shows a greatly magnified picture of a screen. It is actually made up of thousands of tiny coloured dots called pixels. Different parts of the television screen appear to be different colours because of the relative abundance of the three primary colours. Figure 8.40c shows how the different colours are created. If white is required all three colours will be produced. Because these dots are so close, our eye interprets this as a uniform area of white light. If magenta light is required only the blue and red dots will be stimulated. In fact all colours are produced by altering the proportions of red, green and blue dots stimulated in any particular area of the screen.

8.3 summary Critical angle, TIR and EMR • As the angle of incidence of light onto a transparent surface is increased, proportionally more light is reflected and less light is refracted. • When light enters a less optically dense medium it is refracted away from the normal. At the critical incident angle, ic , the angle of refraction is 90°. • If the incident angle is greater than the critical angle, ic , total internal reflection occurs. • The critical angle is given by: n2 sin ic = n 1

• Visible light is only a small part of the electromagnetic spectrum. The many forms of EMR are essentially the same, differing only in their frequency and, therefore, their wavelength. • All EMR travels at a speed of 3.0 × 108 m s−1 in a vacuum. • White light contains approximately equal proportions of red, orange, yellow, green, blue and violet light. • The primary colours of light are red, green and blue. Combining the three primary colours of light produces white light.

8.3 questions Critical angle, TIR and EMR 1 Can total internal reflection occur as light strikes the boundary from: a air (n = 1.00) to glass (n = 1.55)? b glass to air? c glass to water (n = 1.33)? d glass (n = 1.55) to glass (n = 1.58)?

4 The critical angle for light passing from oleic acid into air is 43.2°. Calculate the index of refraction of oleic acid.

2 The diagram shows four rays incident on the boundary between glass and air. Ray 2 meets the boundary at the critical incident angle. For each of the rays 1–4 choose the option that best describes what happens as it strikes the boundary.

6 Calculate the wavelength of: a microwaves of frequency 3 × 1010 Hz b ultraviolet radiation of frequency 1015 Hz.

normal air glass ic

1

2 3

4

A The ray is reflected only. B The ray is refracted only. C The ray is reflected and refracted. D The ray is reflected and transmitted. 3 Determine the critical angle for light travelling from: a diamond (n = 2.42) into air b flint glass (n = 1.60) into air c water (n = 1.33) into air d glass (n = 1.50) into water (n = 1.33).

5 The speed of light in a particular sample of clear plastic is 1.80 × 108 m s−1. Determine the critical angle for light passing from this plastic into air.

7 a List three different types of electromagnetic radia tion and describe a use for each. b List two properties common to all forms of electro magnetic radiation. 8 The primary colours for light are: A red, green and yellow. B red, blue and yellow. C red, green and blue. D yellow, cyan and magenta. 9 Students are experimenting with the lighting for their school play. They want to produce some dramatic lighting effects. Determine the colour formed from a mixture of: a red and blue light b red, blue and green light c blue and yellow light d green and magenta light. 10 A spinning top is decorated with all the colours of the rainbow, yet when spun it appears almost white. Why?


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waves t h ig l f o n o i t a ris 8.4 Dispersion and pola Dispersion We have examined how the recognition of the wave nature of light allowed the development of a full explanation of the refraction of light as it changes speed. For example, the change in direction of travel of the light wave as it entered an optically denser medium occurred because a section of the wavefront entered a slower medium. The slowing down of this section of the wavefront, but not the section still travelling in the original medium, causes the overall wave to veer from its original direction of travel. Recall that white light is made up of many different frequencies (colours) of light. For some materials the speed at which light is transmitted is actually slightly different for different frequencies (colours) of light. This means that on refraction different colours of light will take slightly different paths. This results in the spreading out of the white light into its component colours. This is called the dispersion of white light. Prisms split white light into its component colours. It took scientists many years to be able to explain this phenomenon. Prior to Isaac Newton it was thought that glass prisms altered the incoming white light by varying degrees to produce the spectrum of colour. Newton carried out his investigations into dispersion and was the first to conclude that white light is actually made up of the colours of the spectrum and therefore recombining these colours would produce white light. Figure 8.41 shows the dispersion of white light as it passes through a triangular prism. The light is dispersed both on entering and leaving the prism, so that as the light emerges the range of colours spreads over quite a wide angle. There are no distinct boundaries at which one colour finishes and another begins.

Figure 8.41 Dispersion of white light by a triangular glass prism. On entering and leaving the prism, the violet light is most significantly altered in speed and so it is refracted through the greatest angle. Red light is slowed less and so is refracted the least.

As light enters a prism, it refracts due to a change in speed. Why does light slow down when it enters a more optically dense medium? The light energy is being momentarily absorbed and then re-radiated by the atoms that make up the medium. Different colours of light interact differently with these atoms. As a result they travel at different speeds within the

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medium and so are refracted through different angles. Of the colours that constitute the visible spectrum, violet light is slowed down the most and so is refracted through the greatest angle. Red light is slowed least and so is refracted the least. A similar situation occurs when light speeds up on entering a new medium. Different colours are refracted through different angles. Effectively, a particular medium, glass for example, has a different refractive index for each colour of light. Light flint glass has a refractive index of 1.62 for red light and 1.67 for violet light. Quartz has a refractive index of 1.45 for red light and 1.47 for violet light. In a vacuum, however, all colours of light travel at the same speed of 3.00 × 108 m s−1.

Worked example 8.4A A narrow beam of white light enters a crystal quartz prism with an angle of incidence of 35°. In air, the white light travels at a speed of 3.00 × 108 m s−1. In the prism the different colours of light are slowed to varying degrees. The refractive index for red light in crystal quartz is 1.54 and for violet light the refractive index is 1.57. Calculate: a the angle of refraction for the red light b the angle of refraction for the violet light c the angle through which the spectrum is dispersed d the speed of the red light in the crystal quartz.

Physics file Diamond has a relatively high refractive index of 2.42; hence, the critical angle for diamond is a relatively small 24°. White light is slightly dispersed on entering a diamond. Because of the shape of a diamond, once light has entered the diamond any ray striking the diamond–air boundary is likely to have an incident angle greater than 24° and it is therefore totally internally reflected. The special shape of a diamond means that the dispersed beam of light is likely to undergo a number of internal reflections before it meets a boundary at an incident angle of less than 24°, each reflection spreading the beam a little wider. After a number of internal reflections the light leaves the diamond. If the diamond is appropriately shaped, single colours of light are seen to be scattered by the diamond.

Solution sin i n2 = sin r n1 sin 35° 1.54 = 1.00 sin r r = 21.9° n b sin i = n2 sin r 1 sin 35° 1.57 = 1.00 sin r r = 21.4° c Angle of dispersion = 21.9° − 21.4° = 0.5° n v d n2 = v1 1 2 1.54 3.00 × 108 = 1.00 v2 v2 = 1.95 × 108 m s-1

a

Figure 8

.42 light can Flashes of colou red be a diamon seen emerging fr d as it is viewed fr om different om angles.

Physics in action

Rainbows Water droplets in the air disperse white light into colours to produce a rainbow just as a glass prism does. Rainbows are seen only when you have your back to the Sun, and many water droplets form a cloud in front of you. White light enters the water droplet, reflects from the back of the droplet (due to total internal reflection) and then leaves the droplet. On both entering and leaving the droplet the white light is slightly dispersed since water has a slightly different refractive

index for each colour of light (Figure 8.43). To see what will happen, we will examine the path of the two extremes of the spectrum, red and violet light. The paths of all of the other colours of light will lie between these. Because of dispersion, the red and violet rays leave the drop in different directions. Therefore an observer cannot see both the red and violet light emitted from the one droplet. If the violet light from a particular raindrop is entering your


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eye then all of the other colours reflected by that droplet must miss your eye. The different colours observed in a rainbow must come from different raindrops. Those raindrops sending the red light to your eye must be higher in the sky since the red light emerges more downward than the violet light. In fact all of the droplets which send red light to your eye lie on an arc of about 42° as measured from the original direction of travel of the Sun’s rays (Figure 8.44). The droplets reflecting violet light lie on an arc of about 40°, causing the shape of the rainbow!

white light from the Sun

direction of Sun’s rays all drops on this arc appear red to viewer’s eye

total internal reflection

red violet

let

vio

let

vio

re d

dispersion

No two people can actually see the same rainbow because to view a rainbow the light reflected and dispersed by a particular set of raindrops is directed towards your eye. Those same drops cannot send the same colours of light to your neighbour. A portion of the particular drops producing red light for you may be forming the green section of your neighbour’s rainbow.

dispersion violet

re d

42n 40n

all drops on this arc appear violet to viewer’s eye

red

Figure 8.43 The rainbow is commonly seen because of the dispersion and reflection of light by water droplets. Less intense rainbows sometimes accompany a brighter rainbow. These are due to light reflecting inside the droplet more than once before emerging.

Figure 8.44 All drops lying on the outer arc reflect red light in the direction of the observer. All droplets on the lower arc reflect violet light to the observer’s eye.

Polarisation (a)

direction of travel (b)

direction of travel

Figure 8.45 (a) Unpolarised light, such as that emitted by the Sun or a globe, has electric field variations that are not in alignment with one another. (b) Polarised light waves, such as laser light waves, have the electric fields that vary in the same plane as one another.

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Further evidence for the wave nature of light is the finding that light can be polarised. Consider that light is an electromagnetic wave with associated electric and magnetic fields that vary. Each of these fields varies at a right angle to the direction in which the wave travels. For this discussion we need only think about the varying electric field associated with light, as it is this that largely determines how light interacts with materials. We can therefore represent light as shown in Figure 8.45. Think of how light is produced by a normal light globe inside a torch. Light is emitted from many different atoms in the filament and lots of light waves may be sent in a particular direction. However, the electric fields of these light waves will not be aligned. This is shown in Figure 8.45a. This is called unpolarised light. Most light sources, including our Sun, produce unpolarised light. If the light waves did have their electric fields aligned with one another as shown in Figure 8.45b, we would call this polarised light. The fact that unpolarised light can be converted into polarised light provides strong evidence that light is actually a wave.

Techniques for polarising light The most familiar way in which unpolarised (non-aligned) light can be converted to polarised (aligned) light is by using polarising filters. These

filters have molecules that will block all electric-field-wave components except those whose plane is aligned in a particular direction. Figure 8.46 demonstrates this process using a ‘slit’ to represent the filter. Keep in mind that filters are actually solid materials, usually special plastics. A light wave that has its electric field varying in a plane aligned with the filter will pass straight through the filter, maintaining its original amplitude. All of the light energy passes through. Figure 8.46a shows a vertical polarising filter. It allows light with vertically orientated electricfield variation to pass through. A light wave that has its electric-field plane completely out of alignment with the filter will be blocked. That is, the light will be absorbed by the filter, as shown in Figure 8.46b. If a light wave has only a component of its electric-field plane corresponding to that of the filter, then only this component of the wave will be transmitted. The emerging wave has significantly reduced amplitude. Therefore a portion of the light energy does not pass through the filter, as indicated in Figure 8.46c. The emerging light is described as ‘vertically polarised’. Should unpolarised light be incident on the filter, only vertically polarised light would emerge. A pair of polarising filters can therefore be placed at right angles to one another to prevent all light from passing through. One filter may block all of the horizontal electricfield components and the other filter may block the vertical components, as shown in Figure 8.47. (a)

filter

(b)

filter

Physics file The fact that light can be polarised provides strong evidence that light is actually a transverse wave since longitudinal waves cannot be polarised.

PRACTICAL ACTIVITY 40 Polarisation effects with light

(c)

filter

amplitude is reduced

wave is blocked

Figure 8.46 (a) This filter allows vertically polarised light to pass through. (b) All horizontally polarised light is blocked. (c) The horizontal component of the light is suppressed, resulting in vertically polarised light of reduced amplitude. The emerging light will be less bright than the incident light.

Why Polaroid sunglasses work When outdoors on a bright, sunny day, the smooth, highly reflective, horizontal surfaces around you are a significant contributor to the amount of light entering your eyes. Bring to mind the glare that can occur from the surface of water or snow. Fortunately light that is reflected from smooth, horizontal surfaces tends to be polarised (aligned) in a horizontal direction. An appropriately orientated polarising filter can be used. Lenses in a pair of Polaroid sunglasses are polarising filters orientated to block the horizontal wave components, allowing only vertical components through. Hence the intensity of light—that is, the glare—is markedly reduced.

Figure 8.47 Polarising materials crossing over at right angles to one another will prevent any light from passing through.


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8.4 summary Dispersion and polarisation of light waves • Dispersion is the spreading of white light into its component colours in a spectrum. • Dispersion occurs as white light enters or leaves a prism because different colours of light are refracted by different amounts. • Violet light has the greatest refractive index and is refracted through the greatest angle. Red light has the smallest refractive index and is refracted through the smallest angle.

• Unpolarised light waves have electric-field variations that are not in alignment with one another. Polarised light waves have electric fields that vary in the same plane as one another. • Polarising filters are able to convert unpolarised light into polarised light, providing strong evidence that light is actually a wave.

8.4 questions Dispersion and polarisation of light waves 1 a Which colour of light travels fastest in perspex: red, green or blue? b If red, green and blue light passes from air into a perspex block, which colour of light would be slowed down the most? c Which colour of light—red, green or blue—would be refracted the most as it passes into the perspex block? 2 How does polarisation support a wave model for light? 3 With the use of a diagram, show how a narrow beam of white light will be dispersed on both entering and leaving a triangular glass prism. 4 You have two identical pairs of sunglasses. How could you find out whether the sunglasses were polarising or not polarising? 5 A piece of glass and a diamond are cut to exactly the same size and shape. They are illuminated by the same white light. Why does the diamond appear to sparkle more colourfully than the glass? 6 Explain the effect that a polarising filter has on unpolarised light.

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7 A particular prism of glass has a refractive index of 1.55 for violet light and 1.50 for red light. A beam of white light is incident on the prism at an angle of 40.0° and is dispersed. a Which colour of light will have the slower speed in glass: red or violet? b Which colour is refracted more: red or violet? c What is the angle of refraction for the red light at the air–glass boundary? d What is the angle of refraction for the violet light at the air–glass boundary? e Over what angle will the spectrum be spread? 8 A particular plastic has a refractive index of 1.455 for red light and 1.650 for violet light. For light passing from this plastic into air, which colour of light would have the greatest critical angle? 9 Explain why dispersion provides evidence for a wave nature of light. 10 A polarising filter is positioned so that it produces vertically polarised light. If another filter is orientated at an angle of 30° to this filter, what happens to the intensity of the light?

chapter review 1 As light travels from quartz to water does it: a speed up or slow down? b refract towards or away from the normal? 2 Describe the practical work which you would carry out if you were given a glass prism and asked to determine its refractive index and critical angle. 3 a List seven different categories of electromagnetic radiation and give an application of each. b List two features common to all types of electromagnetic radiation. 4 Monochromatic light with a wavelength of 590 nm is passing through a medium. Its frequency is measured as 3.81 × 1014 Hz. Determine whether the medium through which the light travels is air. Justify your decision. 5 At the boundary between medium 1 and medium 2, the angle of refraction is smaller than the angle of incidence. a Which medium allows light to travel faster? b Which medium has the higher index of refraction? 6 A ray of light exits a glass block. On striking the inside wall of the glass block, the ray makes an angle of 58.0° with the glass–air boundary. The index of refraction of the glass is 1.52. a What is the angle of incidence? b Assume nair = 1.00. What is the angle of refraction? c What is the angle of deviation of the ray? d What is the speed of light in the glass? 7 List these forms of electromagnetic radiation—X-rays, microwaves, visible light, radio waves—in order of increasing: a frequency b wavelength c energy 8 Scuba divers notice that when they are only submerged a few metres the ocean waters appear blue. Why? 9 Which of the following boundaries between two media will result in the greatest refraction of any light ray that crosses the boundary? A Water into air B Water into diamond C Air into diamond D Glass into air 10 Explain why snowboarders and fishermen are likely to wear polarising sunglasses. 11 Explain the following observations. Use a diagram where appropriate. a A star can still be seen even though it is actually positioned below the horizon.

b A stone at the bottom of a shallow pond seems closer to you than it really is. 12 State three behaviours of light for which the ray model is useful when explaining the behaviour. 13 Explain how the spectrum of colour reflected from a black object is different from the spectrum of light reflected by a grey object. 14 In mid-afternoon a diver looks up from below the surface of the water. His judgement of the position of the Sun will be: A higher than it really is. B lower than it really is. C the same as viewed from above the water. 15 A narrow beam of white light enters a crown glass prism with an angle of incidence of 30°. In air, the white light travels at a speed of 3.00 × 108 m s−1. In the prism the different colours of light are slowed to varying degrees. The refractive index for red light in crown glass is 1.50 and for violet light the refractive index is 1.53. Calculate: a the angle of refraction for the red light b the angle of refraction for the violet light c the angle through which the spectrum is dispersed d the speed of the violet light in the crown glass. 16 The speed of light in air is 3.00 × 108 m s−1. As light strikes an air–perspex boundary, the angle of incidence is 43.0° and the angle of refraction is 28.5°. Calculate the speed of light in the perspex. 17 Which one or more of the following models (particle model, electromagnetic wave model, ray model) are used by scientists to explain the following behaviours of light. a Reflection b Dispersion c Total internal reflection d Polarisation 18 What is the relative refractive index for light passing from perspex into water if an incident angle of 17.0° produces an angle of refraction of 14.5°? 19 a What is the critical angle for light passing from: i diamond (n = 2.42) to air? ii glass (n = 1.50) to air? iii water (n = 1.33) to air? b What effect does the relative size of the refractive index have on the size of the critical angle? 20 Why does a ray of light that passes through plate glass emerge parallel to its original direction of travel? Has the ray of light been refracted?


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chapter 9

l a c i t p o d n a s e s s n m e l e , t s s r y o s r r i M

W

ithout question the most useful optical device in existence is the lens. Our entire view of the world relies on a pair of lenses situated at the front of our eyes. Many everyday and scientific devices utilise lenses. Magnifying glasses, movie projectors, telescopes, binoculars and microscopes rely on specially designed systems of lenses. All of these devices manipulate the path of light to produce useful images that would otherwise be more difficult, or indeed impossible, to see. Telescopes were first pointed at the night sky in the early 17th century. Since then technology has greatly advanced the capabilities of these devices. The clearest optical images of our night skies are obtained by enormous reflecting telescopes. These are placed on mountain tops to reduce the degrading effects of the shimmering atmosphere. The Hubble Space Telescope, launched in 1990, orbits the Earth above the atmosphere so that clearer images can be produced. It has become infamous because once launched it was discovered that the images did not provide the dramatic improvement in clarity that was expected. An error in design meant that the clarity of the images was hardly better than that produced by the largest telescopes on the ground! This was an error that proved very expensive to fix. Whole new technologies had to be developed to be able to send a repair team into orbit! In this chapter we use a ray model of light to examine how mirrors and lenses alter the path of light to produce images. We look at the design of optical systems and the function of the eye.

by the end of this chapter you will have covered material from the study of the wave-like properties of light including: • describing the use of models to explain observed phenomena • application of a ray model of light to behaviours of light, including reflection and refraction • optical devices, including lenses and mirrors • colour dispersion in lenses • application of mathematical modelling to light phenomena.

9.1 Geometrical op ti

cs and plane

mirrors

The wave model of light acknowledged In Chapter 8 we have examined how the wave model of light is paramount in explaining numerous properties of light. We have seen that by considering light to be a transverse electromagnetic wave an understanding of light has been reached in relation to: • the linear propagation of light waves in a uniform medium • the regular reflection of light waves from smooth surfaces and diffuse reflection from irregular surfaces • the refraction of light waves as they change speed at the boundary between two media • the dispersion of white light into its component colours (wavelengths) • the existence of light as a part of the continuous electromagnetic spectrum • the polarisation of light waves. In addition to these wave ideas that we have already studied, the diffraction and interference properties of light were discovered in 1803 by Thomas Young. Although not part of this study, these discoveries were crucial in demonstrating the wave-like nature of light. Young showed that when light passes through a narrow slit, bright bands are formed in regions on a screen that a particle model would predict to be in shadow. Light was observed to bend its path as it passed through the slit; that is, diffract. When a pair of slits is used, alternating bright and dark bands form, imitating the interference that occurs between two sets of water waves. These important wave-like behaviours of light are discussed in Heinemann Physics 12. In the early 19th century scientists were satisfied that light truly had a wave-like nature. In 1819 Augustin Fresnel presented a wave theory of light that explained diffraction and interference effects, and it appeared that the matter was settled. It was not until the next century that observations were made that once again questioned the pure wave-like nature of light. As your later studies may illustrate, there are some behaviours of light that a pure wave model simply cannot explain.

Figure 9.1 In 1803, Thomas Young discovered the diffraction and interference properties of light.

Geometrical optics When studying the path of light through many common optical devices, maintaining a wave model for light is unnecessarily complex. In this chapter, rays are used extensively to describe the path of light in optical systems. Note that this approach is justified only in the absence of diffraction. Diffraction is the bending of the direction of travel of light as it passes through an aperture; light will ‘flare out’ as it goes through a narrow gap and rays cannot adequately represent its path. However, significant diffraction only occurs if the size of the aperture is approximately as small as the wavelength of the light itself. In this chapter the dimensions of the mirrors, prisms and lenses discussed are much greater than the wavelength of light and hence diffraction effects can be ignored. In these circumstances light travels in straight lines according to the laws of reflection and refraction, and so pathways are accurately represented with rays. These conditions are called the conditions of geometrical optics.

Chapter 9 Mirrors, lenses and optical systems

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(a)

(b)

incident light waves

ray path

incident light waves ray paths

lens

reflected light waves mirror

Figure 9.2 The path of light through many optical systems involving (a) lenses and (b) mirrors can be accurately represented using rays. The modelling of image formation in simple optical systems is far simpler using geometrical optics rather than a wave-optics approach.

Ray tracing to locate and describe images

PRACTICAL ACTIVITY 41 Reflections in a plane mirror

To investigate the images produced by mirrors and lenses ray tracing can be carried out. Using rays, known pathways of light are modelled on a scaled, two-dimensional diagram and the characteristics of the resulting image can be identified. The image can then be fully described by its: • nature—Is the image real or virtual? (discussed later) • orientation—Is the image upright or inverted? • position—Where is the image in relation to the mirror? • size (including magnification)—What is the height of the image? By what factor has the size of the image changed?

Images in a plane mirror

(a)

A you

your friend 2m

(b) real rays

virtual rays

you

your image

mirror 1m

1m

Figure 9.3 (a) Diverging rays from your friend’s foot are focused by the eye in order that you see them. (b) Diverging reflected rays enter the eye in the same way and appear to have come from the same point.

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To understand how an image forms in a plane mirror, it is helpful to distinguish between how we see an object when we are looking at it directly and the image of the object in a plane mirror. To begin, consider an observer looking directly at another person’s feet. The observer will see a foot because light from some external source is reflected from a point on the foot and this enters the observer’s eye. Specifically, the paths of two slightly diverging rays can be traced from a point on the foot to the observer’s eyes (Figure 9.3a). The pupil of the eye is not just a point aperture. It collects light over an area and the eye will focus diverging rays onto the retina at the back of the eye. This is the mechanism by which we see and hence the foot is judged to be at A. When a person stands in front of a plane mirror, light from a point on his or her foot will travel to the mirror and be reflected. All rays obey the law of reflection, and a pair of slightly diverging rays can be traced as they leave the foot, reflect from the mirror and enter the observer’s eyes (Figure 9.3b). To form an image, the observer’s eyes focus the diverging light rays just as they did in the previous situation. Knowing that light travels in straight lines, the eye–brain system interprets the rays as having come from a single point behind the mirror. This is where the image is seen in the mirror. Figure 9.3b illustrates geometrically how the eye–brain system subconsciously extrapolates diverging rays of light until they meet and are interpreted as an image. It is important to remember that the rays behind the mirror are not actually there. They are virtual rays, and by convention they are shown as dashed lines. When an image forms behind an optical element such as a mirror, it is classified as a virtual image. Other optical systems form a

real image where the rays converge to meet in reality. A real image can be shown on a screen—like the image from a slide projector. Virtual images cannot be projected on to a screen. This ray-tracing exercise can be repeated for all parts of the body to locate the complete image of the whole person. Every point on the object is matched by a corresponding image point, which is the same distance behind the mirror surface as the object is in front. The object and image are exactly the same size. All images formed by plane mirrors have the following characteristics: • The image is always upright. • The image is the same distance behind the mirror as the object is in front. • The image is the same size as the object.

reflecting surface of plane mirror 50 cm

reflecting surface of plane mirror 50 cm

50 cm

Worked example 9.1A Use ray tracing to locate the image of the pom-pom on the top of this girl’s hat, which is 50 cm from the mirror as shown.

i r

Solution To locate an image, the paths of at least two rays coming from a point on the object must be followed. Each ray obeys the law of reflection, i = r, and the image will lie at the point where they meet after extrapolating the rays back. This point lies 50 cm behind the mirror.

Multiple mirrors If two plane mirrors are arranged at 90° to one another, incident light from an object will reflect from one mirror and then the other as shown in Figure 9.4. The ray will finally emerge parallel to the incoming light ray. If the angle of incidence for the incoming light ray is α, then the angle of reflection is also α. This means the angle of incidence as it strikes the second mirror is now 90° − α, and it leaves with an angle of reflection 90° − α. The beam therefore has been deflected through a total angle of α + α + (90° − α) + (90° − α) = 180°. Being deflected through 180° means that it reflects from the mirrors along a path that is parallel to its incoming path. If three mirrors are set up at 90° to each other—like the corner of a cube—then again any incident ray will reflect and emerge parallel to the incident ray in three dimensions. This situation is exploited in reflectors used on cars and bicycles and seen on roadside posts. If you look carefully at the surface of a bicycle reflector it appears to be made up of rows and rows of tiny little box corners. Light from the headlights of an approaching car is reflected back towards the car so that a driver can see the cyclist. A three-dimensional corner reflector placed on the Moon by astronauts reflects laser light from Earth directly back to its source. A measurement of the time this journey takes allows the distance to the Moon to be calculated to within centimetres. Finally, if two mirrors are placed at an angle θ to each other, light from an object will reflect to produce a number of images. Mirrors at 90° to each other will produce three images. As the angle is reduced, the number of images increases. The number of images seen in a pair of mirrors placed at an angle θ to each other is given by the following relationship: 360° number of images = − 1 θ Use a pair of mirrors to verify this.

Physics file When looking into a mirror, you do not appear as others see you. For example, a careful inspection of the image of your left hand in a mirror reveals that it appears as a right hand. This is an example of the back-to-front effect in a plane mirror, which results in a change of ‘handedness’. The image is neither vertically nor laterally inverted (though this is often stated). Something at the top left of the object is also at the top left of the image as we see it. However, the image of your left eye in a mirror will be directly in front of your left eye as you face the mirror, but this eye would appear to be the right eye of the ‘person’ standing in the mirror.

90n A A

A

90n A

A

90n A

Figure 9.4 Two mirrors are placed at right angles to one another. Light is reflected from each mirror and emerges parallel to its original direction of travel.


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9.1 summary Geometrical optics and plane mirrors • In the 19th century the wave nature of light was accepted. • Geometrical optics is the branch of optics that models the path of light using straight rays. This is accurate as long as diffraction (spreading) effects are negligible. • Ray-tracing procedures can be used to determine the image formed by a mirror or lens. An image of a point exists when rays from the point converge (or appear to converge) at another point.

• When using a ray diagram at least two rays must be traced in order to locate an image. • The complete description of an image requires that the position, nature, orientation and size of the image be stated. The nature of an image is either ‘real’ or ‘virtual’. The orientation of an image is either ‘upright’ or ‘inverted’ relative to the object. • Diverging rays of light entering our eyes are interpreted as having come from a single point behind a plane mirror, where a virtual image is seen. • A plane mirror produces an image that is always virtual, upright and unchanged in size.

9.1 questions Geometrical optics and plane mirrors 1 What are two behaviours of light that can be modelled effectively using both a particle and a wave model for light? 2 In what circumstances is it acceptable to model the path of light with rays rather than waves? 3 When light passes through tiny apertures (such as a pinhole) the image produced has blurred edges. Why does this suggest that the particle model of light is inadequate? 4 Is the image formed in a plane mirror always: a upright or inverted? b enlarged, diminished or the same size as the object? c real or virtual?

b How high above the ground should the bottom of the mirror be positioned? 9 Complete the ray paths in the following diagrams to locate the image of the object in the plane mirror. a object

mirror

b

5 In order to fully describe an image what four characteristics should be commented upon? 6 a List the features common to all images produced by plane mirrors. b What is a virtual image? Sketch a ray diagram to illustrate your answer. 7 A girl stands 7 m away from a full-length vertical plane mirror in a boutique. a How far must she walk to appear to be 3 m from her image? b She walks towards the mirror at 1.5 m s-1. With what speed does her image appear to walk towards her? 8 A person 1.6 m tall stands in front of a vertical dressing mirror. Her eyes are at a height of 1.5 m above the ground. a What length mirror is needed for the person to be able to see a full-length image of herself?

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observer’s eye

object observer’s eye

mirror

10 a At the football, short people are often stuck behind others who are up to 20 cm taller. Design yourself a periscope that could assist in such a situation. b Your idea for a ‘footy-scope’ is taken up by the AFL and manufactured. However, at the football, the bottom mirror falls out of a poorly constructed periscope. If the user were to look directly up the tube, how would the image differ from that seen in the two-mirror periscope?

Applications 9.2 concave m of cur ved mirrors: irrors The applications of plane mirrors in everyday life are limited due to the features of the image that can be produced. For example, a plane mirror will be useless where magnification is required—the image in a plane mirror is always the same size as the reflected object. Curved mirrors provide a greater variety of options, depending on the extent and sense of the curvature of the mirror and the placement of the object (Figure 9.5). It’s simply a matter of choosing the right mirror for the task to be undertaken. Shop-security mirrors, shaving and make-up mirrors and dentist’s mirrors all use a curved mirror to produce an image that is appropriate for the situation in which it is used.

(a)

(b)

Figure 9.5 In one mirror the image is upright and significantly reduced in size. The image in the other mirror has been magnified many times.

All curved mirrors are either concave or convex. A concave mirror is curved like the inside of the bowl of a spoon, whereas a convex mirror is shaped like the back of a spoon. (One way to remember which is which is to recall that a concave mirror forms a small cave or may cave in.) Curved mirrors are usually spherical in shape—as if the mirror has been made from a portion of a sphere—because these are cheap and easy to manufacture. More specialised applications require a mirror that is parabolic in shape. As you will see, the image of an object seen in each mirror type can be analysed in the same way. To understand how a curved mirror produces an image, it is important to understand how the curved mirrors reflect light. Any curved mirror can be considered to be made up of a number of tiny plane mirrors. A twodimensional representation of this is shown in Figure 9.6, although a real mirror will of course be three-dimensional. If parallel rays of light shine directly onto the surface of each mirror, the rays reflect, thus obeying the law of reflection for the position of the (plane) mirror at which it strikes. Parallel rays will converge to a single point on reflection from a concave mirror. This point is called the focal point or focus of the mirror. Parallel rays striking a convex mirror will diverge from an imaginary focal point located behind the mirror. Unfortunately, most spherical concave mirrors are only capable of bringing parallel light rays to an approximate focus, as the reflected rays do not intersect at precisely the one single point (Figure 9.7a). This causes blurring of an image near the edges of the mirror. This distortion is called spherical aberration, and is particularly noticeable if the mirror is large compared to its radius of curvature. If an undistorted image, free of spherical aberration, is required, a parabolic mirror is needed. A parallel beam of light will reflect from a parabolic mirror to a sharp focus (Figure 9.7b).

Figure 9.6 Each ray obeys the law of reflection, resulting in (a) converging rays or (b) diverging rays. In each case, a focal point can be defined. A concave mirror has a real focus, and the focal length is positive. The focus for the convex mirror is virtual, since its position is determined by extrapolating the reflected rays behind the mirror.

Physics file In reality, parallel rays of light are not brought to a perfect focus by a spherical mirror. The rays directed near to the centre of the mirror will meet in a region near an approximate focus, but rays farther from the centre miss this region completely. As a consequence, any image formed in a spherical mirror will be distorted. This is called spherical aberration, and, literally, an aberration is a distortion. The farther from the centre of a mirror one looks, the more distorted is the image.


291

(a)

spherical mirror

(b)

parabolic mirror

Spherical mirrors are often used in inexpensive torches and in security and dental mirrors where the accuracy of the focal point is not a great concern. Parabolic mirrors are more difficult to produce and therefore are more expensive. Parabolic mirrors are usually reserved for situations in which a precise focus and undistorted images are required, such as in reflecting telescopes. The focus of a concave mirror can be used in reverse. It is possible to place a point light source at the focus and have the light reflect from the mirror, creating a beam of parallel rays. Car headlights, halogen spotlights and the torches used by security guards all employ this design. Because the rays of light emitted from the light source strike the mirror and emerge parallel to one another, the beam is strong even at a large distance (Figure 9.8). parabolic mirror

F

i r

parallel beam of light

F, focal point

Figure 9.7 (a) Spherical mirrors do not have a precisely defined principal focus, and the further a ray of light is from the principal axis of the mirror, the greater the inaccuracy. The effect is most noticeable if the mirror is large compared to its radius of curvature. (b) A parabolic concave mirror will bring parallel rays to a perfect focus.

globe placed at focal point produces a parallel beam

Figure 9.8 A bright light globe placed at the focal point of a parabolic mirror produces a parallel beam of light. This arrangement is used in spotlights where a penetrating beam is wanted, e.g. car headlights, rescue searchlights.

A dentist uses a concave mirror placed close to the back of a tooth to see a magnified upright image. Shaving and make-up mirrors also produce magnified upright images but, again, you have to stand close to the mirror. Concave mirrors are also capable of forming small, inverted images (Figure 9.11). The type of image produced by a particular concave mirror depends on the placement of the object relative to the mirror.

Image formation in a concave mirror Figure 9.9 A concave mirror forms an enlarged image of a close object.

PRACTICAL ACTIVITY 42 Concave mirrors

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Before explaining how the image is found, there are some terms used to describe a spherical mirror which need to be understood (Figure 9.10). The centre of the mirror is called the pole, denoted P. The principal axis is a straight line perpendicular to the surface of the mirror at the pole. Rays parallel to the principal axis will meet at the principal focus, denoted F. The distance between the pole and the principal focus is the focal length of the mirror, denoted f. The radius of the sphere from which the mirror is assumed to have been made is the radius of curvature of the mirror, R. Finally, the centre of curvature of the mirror is the centre of the sphere of which the mirror forms a part and is denoted C. Just as for the plane mirror, the path of some rays must be determined as they are reflected from the mirror. First, it is assumed that one of the rays leaves the base of the object and travels along the principal axis to the pole of the mirror. This ray reflects along itself and will form the base of the image.

A concave mirror is a portion of a sphere.

principal axis

R, radius of curvature P, pole

(a) Ray 1 principal axis object

F

(b) Ray 2

C, centre of curvature F, principal focus

C

object

f, focal length

(c) Ray 3

Figure 9.10 Terminology used to describe spherical mirrors.

i r

object

The position of the top of the image is determined by constructing at least two rays emanating from the top of the object. The image of this point forms where these rays intersect. There are a number of specific rays that strike the surface of the mirror whose path can be easily predicted. Naturally, these rays obey the law of reflection.

Graphical ray tracing: concave mirrors

P

(d) Ray 4 object

F

(e)

To determine the image of an object in a curved mirror, the object is positioned on the principal axis to the left of the mirror. Between two and four rays are traced from the top of the object; the top of the image is found at the intersection of these rays. The rest of the image fills in the space from this point to the principal axis, and is perpendicular to the principal axis. The four rays whose path can be determined with ease in a concave mirror are described below and shown in Figure 9.11. • Ray 1: A ray of light which travels parallel to the principal axis will be reflected through the principal focus of the mirror. See Figure 9.11a. • Ray 2: A ray of light emanating from the object and travelling through the centre of curvature of the mirror will strike the surface at an angle of 90°, i.e. along a normal. This ray will therefore be reflected back along the path from which it came. See Figure 9.11b. • Ray 3: The ray that strikes the pole of the mirror will reflect at an angle equal to the angle of incidence. At the pole, the mirror acts like a plane mirror and the principal axis becomes a normal. See Figure 9.11c. • Ray 4: The ray emanating from the object and travelling through the principal focus before striking the mirror will be reflected from the mirror parallel to the principal axis. See Figure 9.11d. If any two of these four rays are traced, the point at which they intersect determines the location of the image. See Figure 9.11e. Once the position of the image is found, we know that any ray emanating from the top of the object and striking the mirror must reflect through this point, and a clear image is produced.

object image

Figure 9.11 The path of four rays leaving the top of an object reflect from the mirror to produce an image. (a) This ray travels parallel to the principal axis and reflects through F. (b) Ray 2 passes through C and is reflected back along its incident path. (c) The ray striking the pole will reflect as if it has struck a small plane mirror, i = r. (d) The ray passing through F reflects parallel to the principal axis. (e) Intersecting rays locate the image.

Physics file The quickest method to measure the focal length of a concave mirror is to stand near a window, holding the mirror so that an image of a distant object forms on a small white card or screen that you hold in your other hand. Light rays from a distant source are considered to be parallel to one another, so you are effectively focusing parallel rays of light. The distance from the mirror to the screen approximates the focal length of the mirror.


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Ray diagrams

Physics file Any ray of light, parallel and close to the principal axis, reflecting from a spherical mirror, will travel to the focal point, F. The distance from the focus to the pole of the mirror is called the focal length of the mirror, f, and this is half the distance from the pole to the centre of curvature for the mirror, R. Concave mirrors have a real focal point as the rays intersect at this point. Convex mirrors have a virtual focal point as the rays only appear to have come from this point: R f= 2

Particular conventions apply to the construction of a ray diagram (Figure 9.12). • A vertical line called the optical axis represents the reflecting surface, i.e. the back of the mirror. Although the mirror surface is curved, the optical axis is a straight vertical line. A ray diagram tends to use a much larger vertical scale than horizontal scale, thus allowing the mirror to be represented by a straight line with little loss of accuracy. • The optical axis is perpendicular to the principal axis, and the pole, P, is placed at the intersection between the optical axis and the principal axis. A small curved mirror symbol is placed here to indicate the type of mirror being used. • The principal focus of the mirror, F, and its centre of curvature, C, are located on the principal axis to scale. • Traditionally, the object is on the left of the optical axis, along with the eye that will view the image. The object is usually represented as a small vertical arrow. optical axis

f

u

C

F

principal axis object R

C

image

P

F v

Figure 9.12 Layout for a ray diagram to find the image of an object in a concave mirror.

When constructing a ray diagram, the vertical scale for the diagram does not have to be the same as the horizontal scale. This is particularly useful if a tiny object such as a small whisker is to be viewed in a mirror with a focal length of 20 cm. The distance from the object to the mirror (or optical axis) is called the object distance, denoted u, and the distance from the optical axis to the image is the image distance, v. As already stated, at least two rays from the top of the object must be drawn to locate the position of the image. It is a good idea to use a third ray to check that sufficient care has been taken. All three rays should meet at a single image point. Remember that four characteristics of the resulting image can be com mented on: the nature, orientation, position and size of the image. Figure 9.13 shows the images produced when the same object is placed various distances from a concave spherical mirror. When an object is placed at a relatively large distance (u > R) from a concave mirror, the image formed is real, diminished and inverted (Figure 9.13a). Figure 9.13b shows that when the object is placed at the centre of curvature of the mirror (u = R), the image formed is also real and inverted but is exactly the same size and distance from the mirror as the object. When the object is placed between the centre of curvature and the focal point (f < u < R), as depicted in Figure 9.13c, the image is again real and inverted, but now it is magnified. When an object is placed at the principal focus of a concave mirror (u = f ) the reflected rays will emerge parallel to one another. As discussed earlier,

294


(a)

(b)

object

C image

object

Image real diminished inverted

F

C

F

Image real same size inverted

image object

(c)

Image real magnified inverted

F

C

image

object

(d)

C

F

object

(e) Image formed at infinity

F

C

Image virtual magnified upright

Figure 9.13 (a)–(c) A concave mirror will produce a real image when the object is placed farther than the focal point, and (d), (e) a virtual image when at or closer than the focal length.

this is the arrangement employed in car headlamps, torches and the like, in which a globe is placed at the focal point of a concave mirror. The rays leaving the top of the object strike the mirror and emerge parallel to one another so the image can be considered to be at infinity (see Figure 9.13d). Figure 9.13e illustrates what happens when an object is placed inside the focal length of the mirror (u < f ). The reflected rays diverge like those from a plane mirror. This means that the rays cannot intersect to create a real image. Rather, our eyes interpret the diverging rays as coming from an image behind the mirror—and a virtual, upright and magnified image results. The location of the image is found by extrapolating the rays back to their intersection point as shown. Since it is a virtual image, it cannot be projected onto a screen. If one is available, mount a concave mirror on a wall and check that the predictions from the ray diagrams in Figure 9.13 are valid. Stand a few metres back from the mirror and then slowly walk towards the mirror. As you walk towards the mirror you should notice that your image alters. A concave mirror can produce different images depending on how close the object is to the mirror. As mentioned earlier, the complete description of an image requires its nature, orientation, position and magnification to be stated. Magnification refers to the size of an image relative to the object. If the image is larger than the object, it can be described as enlarged or magnified. A smaller image is diminished or reduced. Sometimes the actual height of an image has to be found—a carefully drawn scale diagram is required here.


295

The magnification, M, of an image is the factor by which the size of the object has been multiplied; this is the ratio of the height of the image to the height of the object. A positive magnification value is used to indicate that the image is upright relative to the object. A negative magnification value is used to indicate that the image is inverted relative to the object. If the magnification factor is greater than one, |M| > 1, this indicates a magnified image. A diminished image has a magnification factor of less than one, |M| < 1. If |M| = 1, the image and object are the same size. For example, a mirror producing a magnification of −1.2 means that the image is inverted and 20% larger than the object. The magnification of an image is also given by the ratio of the image v distance to the object distance (M = - ). Notice from Figure 9.12 that the u ratio of the heights of the image and object is exactly the same as the ratio of the image and object distances.

The MAGNIFICATION FACTOR, M, in optics is given as the ratio of the image height to the object height. image height image distance = M= object height object distance H v M = H i = - u o

Summary of images formed by concave spherical mirrors Table 9.1 summarises the images formed by concave mirrors. Note that all real images are inverted and all virtual images are upright.

Table 9.1 Summary of images formed by concave spherical mirrors Position of object

optical axis

object

(1.5 m)

F (0.5 m)

Position of image

Description of image Real or virtual Upright or inverted

Enlarged or diminished

Beyond C

between C and F

real

inverted

diminished

At C

at C

real

inverted

same size

Between C and F

beyond C

real

inverted

enlarged

At F

image at infinity

—

—

—

Between F and P

behind mirror

virtual

upright

enlarged

Worked example 9.2A A man stands in front of his shaving mirror and is disappointed with the image he sees. The focal length of the concave mirror is 50 cm, and he is standing 1.5 m from its pole. Use a ray diagram to explain what he sees in the mirror.

0.75 m object

image

Solution

F 0.25 m

0.5 m optical axis

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Draw a scale diagram representing the situation. Choose two rays to trace. In this case rays 1 and 4 have been drawn. The intersection between the reflected rays lies below the principal axis. The image is therefore real and inverted, but half the size of the original whisker—not much use when shaving!

Worked example 9.2B

optical axis

A dentist wishing to view a cavity in a tooth holds a concave mirror of focal length 12 mm at a distance of 8 mm from the tooth. Describe the image produced. F object (12 mm) (8 mm)

Solution Draw a graphical scale diagram representing the situation. Choose two rays to trace. In this case rays 1 and 3 have been drawn. The reflected rays do not intersect, and so they must be traced back to locate the virtual image. The dentist sees a virtual, upright image of the tooth that is magnified approximately three times.

optical axis

optical axis F

object

virtual image

F

object

These two worked examples have illustrated problems in which the characteristics of the mirror and image are given. It is the image that has to be found. The following worked example shows a more difficult problem in which the position of the object and image are given, and the focal length has to be found.

Worked example 9.2C A person stands 3.0 m from a concave mirror that produces an upright image magnified by a factor of 1.5. Determine the focal length of the mirror.

Solution

optical axis

Draw a graphical scale diagram representing the situation. Because the height of the person is not given, make the object 2 cm high. Choose ray 3 to draw. If the image is upright, it must also be virtual. Being magnified by 1.5, the image must be 3 cm high at some point behind the mirror. Now we know the position of the top of the image, so we work from ray 1 to find the focal length. A ruler is placed aligning the top of the image to the point where the ray from the top of the object meets the optical axis. This ray will reflect through the focus. Measuring from the following diagram, and using the scale, f = 9 m.

ray 3

3 cm

principal axis object

position of image

Horizontal scale: 1 cm = 1 metre

optical axis ray 1

principal axis object

F

3m

image

9m


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9.2 summary Applications of curved mirrors: concave mirrors • A concave mirror is curved like the inside of the bowl of a spoon. A convex mirror is curved like the back of a spoon. • A curved mirror is usually spherical or parabolic in shape. Parabolic mirrors create more precise images but are more difficult and costly to produce. Spherical mirrors are used in everyday situations. • A concave mirror can produce different images depending on the position of the object. • There are four rays, incident on a concave mirror, whose path can be predicted to determine the

position, nature, orientation, size and magnification of the image. At least two rays must be traced to locate an image when using a ray diagram. • The nature of an image is either ‘real’ or ‘virtual’. The orientation of an image is either ‘upright’ or ‘inverted’ relative to the object. image height image distance • Magnification = = object height object distance v Hi M = = − u Ho

9.2 questions Applications of curved mirrors: concave mirrors 1 a Distinguish between a concave and a convex mirror. Give one application for each. b What is meant by the term ‘radius of curvature’ for a spherical mirror? c The radius of curvature of a spherical concave mirror is 50 cm. What is its focal length? d Use a diagram to define the following terms associated with a spherical concave mirror: centre of curvature, principal axis, principal focus, pole. 2 a A small light bulb is placed at the focus of a para bolic concave mirror. Use a diagram to show the path of the light after it reflects from the mirror. b State one use for such an arrangement of a globe and mirror. c What is spherical aberration? 3 While shaving, a man stands 24 cm from a concave mirror that has a focal length of 36 cm. His aim is to shave off his 2 mm long whiskers. Use a ray diagram to answer the following. Different horizontal and vertical scales will aid accuracy. a What is the nature of the image? b How far will the image of each whisker be from the mirror? c What will be the size of the image of each whisker? 4 A pen is held 60 cm from a concave mirror of focal length 20 cm. Describe the nature of its image. 5 A person looks into the bowl of a soup spoon and sees an inverted image of themselves. Use a ray diagram

298


to model this situation, estimating the focal length of the spoon and any other distances you need. 6 A concave mirror of focal length 20 cm is used to produce an image of an object. How far from the mirror must the object be placed in order to form an image that is: a inverted and the same size as the object? b upright and twice the size of the object? 7 A cavity in a tooth is actually 2 mm long. A dentist uses a concave mirror of focal length 15 mm to view the tooth. The mirror is held 8 mm from the tooth. Approximately how big will the cavity appear? Describe the nature of the image formed. 8 A person wants to use a make-up mirror with a focal length of 40 cm to insert contact lenses. She stands too far back and sees an image of her face that is true to size but inverted. How far is her face from the mirror? Where can she stand to see an upright, magnified image of herself? 9 A student wants to use a concave mirror to magnify a specimen by a factor of 3. A concave mirror with a focal length of 40 cm is chosen. Use a ray diagram to determine the position of the specimen and the characteristics of the image. 10 The image of an object in a concave mirror is 20 cm behind the mirror when the object is placed 5 cm in front of the mirror. What is the focal length of the mirror?

9.3 Convex mirrors Convex mirrors are very common in everyday life, often seen as large mirrors used for security in the ceiling corners of shops or for traffic safety at ‘blind corners’, and as rear-view mirrors for particularly long vehicles. Convex mirrors can be spherical, parabolic, ellipsoidal or hyperbolic in shape, but the last three are only used in telescopes and other specialised equipment.

parallel incident rays C

Figure 9.14 Convex mirrors are used on the road to help drivers view traffic around sharp corners.

The reflecting surface of a spherical convex mirror is again a portion of a sphere, but in this case the reflective surface is on the outside of the sphere. Incident light rays parallel to the principal axis of a convex mirror will diverge on reflection. If the rays are drawn back behind the mirror, they appear to come from one single point—the virtual principal focus of the mirror (Figure 9.15). The distance from the pole of the mirror to this point is

F virtual focal point

principal axis

Figure 9.15 The parallel rays reflected from a convex mirror will diverge in such a way that they appear to come from a single point— a virtual focus.


299

object

F

principal C axis

(b) Ray 2

F

C

(c) Ray 3

P

F

C

F

C

(d) Ray 4

object

(e) image object

F

C

Figure 9.17 Four rays reflected by a convex mirror are chosen to locate and describe the position of the image. The position of the image can be found by using only two of the rays, and any of the others can be a check. These rays are chosen because their path can be predicted with accuracy.

300

convex mirror C

r

i

i r

plane mirror

r i

Figure 9.16 Convex mirrors allow a very wide field of view, and so are used when the image of a large region is required.

object

object

i r

narrow field of view

(a) Ray 1

wide field of view

the focal length of the mirror and, as with a concave mirror, the focal length is half the radius of curvature. Convex mirrors (like plane mirrors) can only produce a virtual image. A convex mirror always provides a wide field of view and an image that is always upright. Examine Figure 9.16. Two observers are looking into mirrors of the same size. The observer looking into the convex mirror is able to see rays coming from a much wider field of view than the observer looking into the plane mirror.


Graphical ray tracing To understand how an image is formed in a convex mirror, we must again trace the path of rays from a common point on the object as they reflect from the mirror. This is exactly the same process that was followed with plane and concave mirrors. Again, the object is placed to the left of the mirror with its base on the principal axis. Any two of four rays can be drawn from the top of the object to find the image. These rays are chosen because we are able to predict exactly the path that they follow. Either of the other two rays may be used as a ‘check’. • Ray 1: This ray travels parallel to the principal axis and reflects as if it came from the virtual principal focus of the mirror. See Figure 9.17a. • Ray 2: This ray travels towards the centre of curvature of the mirror and strikes the mirror surface along the normal at the point, i.e. the angle of incidence is 0°, and so it will reflect along the path from whence it came. See Figure 9.17b. • Ray 3: This ray strikes the pole of the mirror and reflects at an angle equal to the angle of incidence. The principal axis acts as the normal at this point and so the ray passes below the principal axis at the distance of the object through a point which is the same height as the object. See Figure 9.17c. • Ray 4: The ray of light directed towards the virtual principal focal point reflects parallel to the principal axis. See Figure 9.17d. The path of the light described in these four rays emphasises that light reflecting from a convex mirror will always diverge in front of the mirror. As these rays never meet, a real image cannot be formed and so no image from a convex mirror may be projected onto a screen. Rather, the image will be

virtual, forming where the rays appear to ‘meet’ behind the mirror. Once the tip of the image has been located, the rest of the image will lie between this point and the principal axis. When constructing a ray diagram for a convex mirror, the same conventions apply as for concave mirrors. A vertical line, the optical axis, representing the reflecting surface, is drawn perpendicular to the principal axis for the mirror. A small convex mirror symbol, whose pole lies at the origin for the axes, indicates the type of mirror used. The object is always placed to the left of the mirror and the principal focal point will lie to the right of the mirror (behind it). The horizontal and vertical scales need not be the same, and, in many situations, should be different in order to aid accuracy. Figure 9.18 shows the effect of placing the same object at various distances from a convex mirror. All the images lie behind the mirror between the pole and the focal point, and all are upright and diminished. Each image, however, has a different magnification that depends on the position of the object from the mirror. As the object is moved closer to the mirror, the image size increases. Because of the curvature of the mirror the image can never be quite as large as the object; hence the magnification is always less than one (|M| < 1). (a)

image F

C

image F

C

F

C

object

(b)

object

(c)

object

image

Figure 9.18 Ray tracing for a convex mirror. All images are upright, virtual and diminished. As the object is brought closer to the mirror, the image increases in size, but it will never be the same size as the object.


301

Worked example 9.3A A shop uses a convex mirror of focal length 2.0 m for security purposes. A person 1.5 m tall is standing 4.0 m from the mirror, describe the nature of the image seen. What is the magnification of the image?

Solution Draw a scale diagram to represent the situation. optical axis

object (4 m)

ray 1

F (2 m)

C

ray 2

principal axis

image

object

F

C

Select two rays to trace. In this case rays 1 and 2 have been drawn. The real rays diverge after reflection. Tracing these rays behind the mirror shows that they intersect above the principal axis, forming a virtual, upright image. The person’s image is H diminished and, from the scale diagram, the magnification M = i = 0.4. Ho

Worked example 9.3B An image in a convex mirror of focal length 15 cm is 10 cm behind the mirror. Determine the position of the object and the magnification of the image.

Solution Draw a scale diagram to represent the situation. The size of the image is not given, so let us make it 1 cm high. Choose ray 1—this will reflect from the mirror at a point where the line joining the focus and the top of the image meets the optical axis. Hence the incident ray 1 can be drawn.

optical axis ray 1 ray 1

principal axis image Horizontal scale: 1 cm = 5 cm

302


F

This immediately gives Ho = 3 cm. A second ray is chosen—ray 3. The reflected ray is drawn first. Here, the ruler must be placed across the top of the image and the pole of the mirror. This gives the virtual and the real parts of the reflected ray. The real incident ray is then drawn striking the pole so that i = r. The object is placed where the two incident rays intersect. The position is now measured and, using the scale, u = 30 cm. From the scale diagram, H 1 M  = i = . Ho 3 optical axis

ray 1 ray 3

object

image

F

principal axis

ray 3

Physics in action

Parallax helps locate virtual images Draw a large dot on a window. Step back a metre or so from the window and as you look out, move your head slightly from side to side—a couple of centimetres is sufficient. The dot will appear to alter its position relative to the background scenery. This is a parallax effect and it is one of the visual clues that we use when judging the distance to an object. If the relative position of two objects alters as an observer moves, then the two objects must be at different distances from the observer. To verify this, draw another large dot on the window directly below the first dot. As you move your head from side to side, the dots do not move relative to each other since they are both the same distance from you. Plane, concave and convex mirrors can form virtual images. There aren’t really any rays behind the mirror and so we cannot use a screen to locate the actual position of the virtual image. Instead parallax helps to locate the position of the image. If an object such as a tall candle is placed in front of a small plane mirror, and an identical object is placed behind the mirror, this second candle can be moved to a position so that when viewed from in front of the mirror it appears to be an extension of the image seen in the mirror, as in Figure 9.19. If the sideways motion of the observer’s head shows an absence of parallax, then the second candle must be the same distance from the Figure 9.1 observer as the image produced by the mirror. Thus the 9A virtual im n absence of para location of the virtual image has been found. ll

age to be de

termined.

ax allows

the locatio n

of a


303

Physics in action

The mirror formula u O ray 3

Ho

P

O` I`

F

Hi ray 4

P`

I

Figure 9.20 When using the mirror formula, keep in mind the sign conventions that must be followed.

f v The mirror formula provides the relationship between the focal length of a mirror, f, the distance of the object from the pole of the mirror, u, and the distance of the image from the pole of the mirror, v. The relationship is: 1 1 1 = + f u v In order to use this relationship with both convex and concave mirrors, various sign conventions must be applied: • ‘Real’ distances are measured on the same side of the mirror as the object, and are positive. • A ‘virtual’ distance is behind the mirror and is negative. This means that u, the distance from the object to the pole of the mirror, is always positive, as the object is always placed in front of the mirror. v, on the other hand, is the distance from the image to the mirror, and may be positive for a real image or negative for a virtual image. • f is the focal length of the mirror and will be positive for a concave mirror (as it has a real focal length) and negative for a convex mirror (virtual focal point).

Worked example A person stands 30 cm from the pole of a concave mirror and an inverted image is formed 60 cm from the mirror. What is the focal length of the mirror?

Solution List the data and assign positive and negative values as appropriate: u = +30 cm, v = +60 cm (v is positive because inverted images in a concave mirror have to be real.) To find f, the mirror formula is used: 1 1 1 1 1 = + = + f u v 30 60 2 1 = + 60 60

304


1 3 So, = f 60 60 f = = 20 cm 3 (A positive focal length was expected as only a concave mirror can produce a real, inverted image.) In these problems, it is often easier to express all numbers as fractions until the last step of the calculation. This is quicker and rounding errors are reduced.

Worked example A car is 40 m from a convex mirror of focal length 10 m. Determine the position of the image, and describe the image produced.

Solution List the data and assign positive and negative signs: u = +40 m, f = −10 m (This is negative because a convex mirror has a principal focus behind the mirror.) To find v, use the mirror relationship: 1 1 1 = + f u v So, 1 1 1 = − v f u 1 1 1 = − − v 10 40 4 1 = − − 40 40 5 = − 40 Hence, v = −8 m. The image distance is negative, hence the image is virtual. The image must also be upright since virtual images are always upright.

9.3 summary Convex mirrors • Rays parallel to the principal axis will diverge on reflection from a convex mirror. They reflect as if they come from a single point behind the mirror— the principal focus. • A convex mirror is useful because it has a wide field of view and always produces upright images. • The focal length of a spherical convex mirror is half its radius of curvature: R f = 2

• A ray diagram drawn to scale is used to trace the path of known rays that are used to locate the image produced in a convex mirror. • Convex mirrors always produce images that are virtual, upright and diminished. The size of the image depends on how close the object is to the mirror, but the magnification is always less than 1.

9.3 questions Convex mirrors 1 The radius of curvature of a convex mirror is 70 cm, what is its focal length? 2 Use a diagram to help define what is meant by the terms virtual focal point, centre of curvature and focal length of a convex mirror. 3 Under what conditions would a convex mirror be used in preference to a plane mirror? 4 a An ant crawls over the surface of a convex mirror. Describe the nature of the image of the ant. b Explain why a convex mirror can never produce a real image. 5 The image of an electric light filament be projected onto a screen by: a a plane mirror. b a concave mirror. c a convex mirror. 6 Curved mirrors are used for different purposes. Three examples are shaving and make-up, shop security, and a rear-view mirror for a truck. Use ray

tracing to decide on the shape and approximate focal length suitable for the given applications, taking the position of the object and required features of the image into consideration. 7 Locate the images formed when a match 3 cm long is placed 10, 5 and 2 cm from a convex mirror of focal length 5 cm. 8 A convex mirror of focal length 3 m is placed at a ‘blind corner’. A person looking at the mirror sees a truck coming; the truck is actually 30 m from the mirror. Describe the nature of the image. What is the magnification of the truck? 9 A person 1.25 m tall stands 5 m from a shop-security mirror of focal length 2.5 m. How tall is the person’s image? 10 A child stands in front of a convex mirror with a focal length 2.0 m at Luna Park. If her image is half her real height, determine how far she is in front of the mirror.


305

es

9.4 Refraction and lens converging lenses

plano-convex

bi-convex

diverging lenses

plano-concave

bi-concave

Figure 9.21 Spherical convex and concave lenses. Convex lenses are thicker in the middle. Concave lenses are thinner in the middle. The curved surfaces form part of the surface of a sphere.

(a)

(b)

Figure 9.22 Lenses cause light to (a) converge or

(b) diverge. (a)

In the preceding sections we have seen how ray tracing can be used to demonstrate the formation of images by plane and curved mirrors. A similar approach can be taken regarding image formation by lenses. Keep in mind that our geometrical optics approach is justified since diffraction (spreading) effects can be ignored. We assume that since we are dealing with lenses that are much larger than the wavelength of light, the light will not diffract appreciably. So it is quite appropriate to represent the path of light with straight rays. The function of a lens can be demonstrated by examining the refraction of light rays as they change speed on entering and exiting a lens. A lens is a transparent optical element that is designed to cause light to converge or diverge. Lenses are usually made of glass or plastic. Since refraction must occur as the light both enters and exits the lens, the curvature of both surfaces of the lens must be chosen specifically for each application. The lenses we will be studying fit into two categories: convex and concave lenses. Convex lenses have one or both sides curved outwardly, like the back of a spoon. These are called plano-convex and bi-convex respectively (Figure 9.21). Concave lenses have one or both surfaces curved inwardly like the bowl of a spoon. These are called plano-concave and bi-concave respectively. The curvature of the surface of most lenses is spherical, i.e. the surface of the lens is actually a portion of a sphere. Spherical lenses do have some problems related to accurate image formation (discussed later in the chapter) but are very commonly used. Since spherical lenses are much easier and cheaper to produce than lenses of alternative shapes, non-spherical lenses are only used in very high precision optical equipment. This study involves only the more commonly used thin bi-convex and thin bi-concave spherical lenses, and they will simply be referred to as convex and concave lenses. Figure 9.22 demonstrates the effect that convex and concave lenses have on a beam of light. Through a convex lens the beam converges and through a concave lens the beam diverges. The law of refraction explains why this occurs. Figure 9.23 shows the path of light rays as they enter small glass prisms. The path of light is bent at the front and back surfaces of the prism according to the law of refraction. The shapes of these prisms have been chosen and arranged so that the separate light rays converge to a single point (Figure 9.23a) or diverge from a single point (Figure 9.23b).

F

Spherical lenses—terminology (b)

F

Figure 9.23 A beam of light can be considered as a bundle of separate rays. Light will (a) converge or (b) diverge after passing through a lens.

306


In order to understand how a lens will produce an image, a number of terms must be defined. Many of these are similar to the definitions used when describing curved mirrors (Figure 9.24). The optical centre of the lens is a point (usually inside the lens) through which a ray may pass undeviated in its direction of travel. Since our study only involves thin bi-convex and bi-concave lenses, the optical centre is located at the physical centre of the lens. (If thick lenses were to be considered this might not be the case.) The principal axis of the lens is the line drawn through both the optical centre and the centre of curvature of a lens. It will therefore intersect the centre of the lens surface at an angle of 90°.

If rays of light are directed parallel to the principal axis of a convex lens they will converge to a point called the principal focus or the focal point, F. This is called a real focal point since rays really do intersect at this point. Each ray is bent towards the principal axis at both the front and rear surfaces of the convex lens. For a concave lens, rays parallel to the principal axis are diverged as if they came from a single point called the virtual focus, F. All lenses are reversible and have a principal focus on either side. The distance between the optical centre and the (real or virtual) focal point of a convex or concave lens is called the focal length, f. The focal length of a lens is determined by the index of refraction of the lens and its shape. A thick lens, with a small radius of curvature, has a shorter focal length. Since a thick lens bends light by a greater amount, it is referred to as a strong lens. If the same glass was used to make a thinner lens, with a longer radius of curvature, it would be a weaker lens. (a)

Physics file The principal focal point of a convex lens is the position at which rays parallel to the principal axis are brought to a focus. Therefore, to determine the focal length of a convex lens simply choose a distant object such as the Sun or a distant tree. Hold a white card on the opposite side of the lens to the object and adjust the position of the card until a sharp image is formed on it. The distance from the lens to the card is the focal length of the lens.

(b) principal focus principal focus principal axis

f

f

principal axis

(a)

optical centre optical centre

Figure 9.24 Terminology associated with spherical lenses. (a) If rays parallel to the principal axis enter a convex lens they are converged to the principal focal point, F. (b) For a concave lens, rays parallel to the principal axis diverge as if they came from a single point called the virtual focus.

Ray paths through spherical lenses

(b)

Many optical systems incorporate spherical lenses, but in each application the nature and size of the image required will be different. For example, microscopes and telescopes are designed to produce enlarged images that may be real or virtual, but cameras must produce diminished, real images. As for mirrors, ray-tracing techniques can be used to determine the nature of the image formed by either a concave or a convex spherical lens. Although an infinite number of rays of light are emanating in every direction from every point on the object in question, it is only necessary to trace the path of two known rays through the lens to determine the position of the image. Each ray of light passing through a lens is bent at both the front and rear surfaces of the convex lens, but for simplicity optical diagrams show one change in direction occurring in the middle of the lens. For spherical lenses there are three particular rays whose path can be predicted, and any two of these rays can be used to locate the image (Figure 9.25). • Ray 1: Any ray passing through the optical centre of a spherical lens is undeviated (Figure 9.25a). • Ray 2: Any ray that travels parallel to the principal axis will be deviated through the principal focus of a convex lens, or deviated so that it appears to have come from the principal focus of a concave lens (Figure 9.25b). • Ray 3: Any ray that passes through the principal focus of a convex lens or is directed towards the principal focus of a concave lens will be deviated so that it emerges parallel to the principal axis (Figure 9.25c).

F

F

(c) F

F

Figure 9.25 (a) A ray passing through the optical centre is undeviated. (b) A ray parallel to the principal axis passes (or appears to pass) through the principal focal point. (c) A ray which passes through the focal point emerges parallel to the principal axis.


307

Graphical ray tracing for convex lenses Just as for spherical mirrors, graphical ray tracing can be used to locate the image formed by a spherical lens. For accurate results the ray diagram must be drawn carefully to scale, although generally a much larger horizontal than vertical scale is useful. A horizontal line represents the principal axis for the lens. A vertical line called the optical axis or vertex represents the lens. To indicate the type of lens in use a small symbol is drawn at the intersection of the principal and optical axes. The principal foci of the lens should be labelled. All refraction is shown as occurring at the optical axis. By convention, the object is positioned on the principal axis and to the left of the lens. A small upright arrow is often used to symbolise the object. The distance from the object to the optical axis is called the object distance, denoted u. The ray paths are usually traced from the uppermost point of the object to the corresponding image point. The orientation of the entire image is inferred from this one image point. The distance from the optical axis to the image is called the image distance, denoted v. Although only two ray paths need to be traced to locate the image, it is often advisable to check the accuracy of your image location by using a third ray. When these diagrams are drawn accurately the image formed can be fully described. Recall that there are four aspects of an image which may be determined: • nature—whether the image is real or virtual • orientation—whether the image is upright or inverted • position—the distance of the image from the optical axis • size—the image is described as enlarged, reduced or the same size as the object. This is often described in terms of magnification. Figure 9.26 shows the images produced when the same object is placed at various distances from a convex lens. In each diagram the paths of rays 1, 2 and/or 3 have been traced. Note that real rays are represented by solid lines and any virtual rays, which are merely a backwards extrapolation, are represented by dashed lines. Figure 9.26a demonstrates that when an object is placed a relatively long way from a convex lens (i.e. u > 2f ) a diminished real image is formed below the principal axis. It is inverted. Recall that real images are formed when ray paths actually do intersect. Real images can be projected onto a screen. The image is between F and 2F on the opposite side of the lens to the object and is smaller than the object. This sort of arrangement would be useful in a camera where a reduced real image is needed. Figure 9.26b shows that when the object is placed at a distance of twice the focal length of the lens (i.e. u = 2f ), the real image formed is below the principal axis, inverted, and the same size and distance from the lens as the object. This would be useful if an image that was identical to the object was required. Figure 9.26c shows that when the object is placed between 2F and the focal point of the lens (i.e. f < u < 2f ), the real image is below the principal axis, inverted and enlarged. In this region, if the object is moved closer to the lens the image becomes larger and is further from the lens. A slide projector uses such an arrangement.

308


Figure 9.26d demonstrates that when an object is placed at the focal point of the lens (i.e. u = f ) the emerging rays are parallel to one another, so the image can be considered to be at infinity. Figure 9.26e demonstrates that when an object is placed inside the focal length of the lens (i.e. u < f ) the rays diverge after refraction by the lens. An observer would presume that these rays have come from an image point on the left side of the lens. The rays can be extrapolated backwards to locate the virtual image as shown. The image is upright and enlarged. This is how a simple magnifying glass works. The object must be inside the focal length of the lens. (a)

Description of image

object F F

2F

(b)

2F

real inverted diminished

2F

real inverted same size

image

object F F

2F

image (c)

object F

2F

real inverted enlarged

F

2F

image (d)

object F

(e)

2F image at infinity

F

2F

image virtual upright enlarged

object 2F

F

F

2F

Figure 9.26 (a)–(c) When the object is placed beyond the focal point of a convex lens, a real image is formed. As the object is brought closer, a larger image is created. (d) The image of an object at the focal point of a convex lens is considered to be at infinity. (e) A convex lens forms an upright, virtual image of an object placed inside the focal point.


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Worked example 9.4A A child examines a beetle by using a magnifying glass of focal length 7 cm. He holds the magnifying glass 14 cm from the beetle and is disappointed with the image. Use graphical ray tracing to explain: a why he is disappointed b where he should hold the magnifying glass to see an enlarged image. PRACTICAL ACTIVITY 43 Convex lenses

Solution a Select suitable horizontal and vertical scales, and place the object 14 cm from the optical axis.

F

F

f = 7 cm

u = 14 cm

Trace the path of two rays. A real image is produced which is the same size as the object. The boy is disappointed because the image is not enlarged.

ray 2 ray 1 object

image F

F

f 7 cm

u 14 cm

b To produce a magnified, upright image the object should be positioned inside the focal point of the lens as shown. The closer the object is to the focal point the larger the image. image

ray 2

F

object

F ray 1

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9.4 summary Refraction and lenses • The path of light is refracted as light both enters and exits a lens. A convex or converging lens brings parallel rays to a focus. A concave or diverging lens spreads parallel rays further apart. • The focal length of a spherical lens is determined by its index of refraction and radius of curvature. Thinner lenses, with less severe curvature, have longer focal lengths. • For spherical convex and concave lenses there are three rays whose paths can be predicted easily, and any two of these rays can be used to locate an image.

Ray 1: A ray through the optical centre is undeviated. Ray 2: A ray parallel to the principal axis passes (or appears to pass) through the principal focus. Ray 3: A ray through the principal focus emerges parallel to the principal axis. • Graphical ray tracing can be used to locate the image formed by a spherical convex lens. • The nature, orientation, position and size of the image depends on the position of the object relative to the focal length of the lens. • In general, convex lenses form real images of distant objects and virtual images of very close objects.

9.4 questions Refraction and lenses 1 a Draw a scale diagram of a convex lens of focal length 10 cm and label the principal axis, optical axis, optical centre and principal focus. b Describe the path of three different rays as they pass through a spherical convex lens. 2 Why can an upright, real image never be formed by a single convex lens? 3 Two convex lenses are exactly the same size and shape but made of different plastics. Lens A is made of plastic of a higher refractive index than lens B. a Which lens has the longer focal length? b Which is the stronger lens? 4 Can two convex lenses of different refractive indices have the same focal length? 5 a A certain spherical convex lens has a focal length of 20 cm. If that lens is immersed in water will the focal length increase or decrease? Use a diagram to explain your answer. b A bubble of air in a glass of water could act like a spherical lens. Will it cause a parallel beam of light to converge or diverge? Explain. 6 A convex lens of focal length 20 cm is used to form an image of the full Moon on a screen. The Moon is a distance of 3.9 × 108 m from the Earth and has a diameter of 3.5 × 106 m. How far must the screen be placed from the lens?

7 A real image is formed 36 cm from a lens when the object is also 36 cm from the lens. The focal length of the lens is: A 36 cm B 18 cm C 72 cm D 24 cm 8 An object is 10 cm from a convex lens of focal length 15 cm. The image formed is: A real, inverted and enlarged. B real, inverted and the same size as the object. C virtual, upright and reduced. D virtual, upright and enlarged. 9 A convex lens has a focal length of 20 cm. Use graphical ray tracing to determine the nature, size and position of the image formed when: a a match 3 cm high is placed upright 30 cm from the lens b a coin of diameter 2.5 cm is stood on edge 45 cm from the lens c a coin of diameter 2.5 cm is stood on edge 5 cm from the lens. 10 A real image is formed 1.2 m from a convex lens of focal length 0.4 m. The object is 5 cm tall. Use graphical ray tracing to determine: a the distance of the object from the lens b the height of the image.


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9.5 Concave lenses Individual concave spherical lenses are not as commonly used as convex lenses, but they are still important in many optical systems, some of which are discussed at the end of this chapter. As has been demonstrated, concave lenses are essentially diverging lenses, i.e. light is refracted away from the principal axis. As a result the images formed by a single concave lens are always reduced in size. Artists sometimes use a concave lens to see how a particular landscape may look if painted in miniature. Through a concave lens the landscape is seen as an upright, diminished image.

Figure 9.27 A concave lens forms a reduced upright image.

Graphical ray tracing and concave lenses Graphical ray tracing techniques can again be used to determine the nature, position and size of the image formed by any concave lens. Recall that for concave lenses there are three rays whose paths through the lens can be predicted (see Figure 9.25). The horizontal principal axis and the vertical optical axis are drawn using suitable scales, as shown in Figure 9.28. The foci are labelled. A small symbol indicates that a concave lens is in use. Conventional ray diagrams require that the object is placed on the principal axis to the left of the optical axis. All refraction is shown as occurring at the optical axis which represents the lens. The ray paths are usually traced from the upper-most point of the object to the corresponding image point. Although only two ray paths need to be traced to locate the image, it is again advisable to check the accuracy of the image location using a third ray if possible. Figure 9.28a shows an object placed on one side of a concave lens. Ray 1 is shown passing through the optical centre of the lens undeviated. Ray 2 travels parallel to the principal axis and is deviated so that it appears to have come from the principal focus of the concave lens. Hence, the rays are diverging as they emerge from the lens. An observer perceives them to have come from a point behind the lens. To find the image, the diverging rays are extrapolated backwards to an intersection point. Therefore a virtual image is formed at this point.

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Figure 9.28 shows the image produced when the same object is placed at various distances from a concave lens. In each diagram, the paths of rays 1 (a) and 2 have been traced. Note that, regardless of the position of the object, a virtual, upright, diminished image is always formed. The image produced object 2f is always on the same side of the lens as the object. When the object is brought closer to the lens the image becomes larger, but it can never be quite as large as the object.

F image

(b)

Worked example 9.5A A person uses a concave lens of focal length 10 cm to ‘shrink’ a sketch of a tree to see what it would look like in miniature. The sketch is of a 6 cm tall tree and the lens is held 15 cm above the page. Use graphical ray tracing to determine: a whether the image formed is real or virtual b the image distance c the size of the image.

object 2f

F image

2f

object F image

(c)

Solution a Select suitable horizontal and vertical scales and place the object 15 cm from the optical axis. Label the focal points.

(d) object 6 cm

2f

F

image

(e)

F

object

f 10 cm

2f

F

image

u 15 cm

Figure 9.28 A concave lens causes rays to diverge and so a virtual, reduced image is always formed.

ray 2

F

v = 6 cm

ray 1


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Trace the path of two rays through the lens. The rays would be diverging as they emerge from the lens and so a virtual image is formed. b According to the horizontal scale used, the image distance is 6 cm. c According to the vertical scale used the image height is approximately 0.4 times the height of the object, or 2.4 cm.

The lens equation and magnification Although ray tracing indicates the nature of the images formed by concave and convex lenses, it can be time-consuming and only gives approximate values of the image distance and height because of construction errors. A more accurate analysis can be carried out algebraically, as was the case with spherical mirrors. The lens equation relates the focal length of a spherical lens, f, to the object distance, u, and image distance, v.

1 1 1 The L…NS …QUATION states that: = + f u v where u is the object distance v is the image distance f is the focal length of the convex or concave lens. In order to use the lens equation, sign conventions must be adopted for various quantities: • f is the focal length of the lens. f is positive for a convex lens (real focal point) and negative for a concave lens (virtual focal point). • u is the distance from the object to the lens (object distance). This is always positive as the object is always placed to the left of the lens. • v is the distance from the lens to the image (image distance). This is positive if the image is real: a real image is always formed on the right side of the lens; v is negative if the image is virtual: a virtual image is always formed to the left of the lens. Earlier in this chapter we examined the magnification produced by spherical mirrors. The definition of magnification remains the same. Magnification, M, is the ratio of the image height, Hi , and object height, Ho. Magnification can also be determined from the ratio of the object and image distances.

H v M = i = - Ho u When the same sign conventions previously applied to the lens equation are used, the polarity of the magnification (or image height) indicates the nature of the image formed. A positive magnification value indicates that an upright and therefore virtual image has been formed. Hi will be positive as the image is formed above the principal axis. A negative magnification value indicates that an inverted, and therefore real, image has been formed. Hi will be negative as the image is formed below the principal axis.

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Worked example 9.5B A slide transparency is placed 11 cm from a convex lens with focal length 10 cm, in a slide projector. A sharp image is projected onto the screen. The slide is square with side length 3 cm. a How far should the screen be placed from the lens of the projector? b How large is the image on the screen? c Must the slide be placed upright or inverted inside the projector?

Solution a List the data: u = 11 cm, f = +10 cm Substitute into the lens equation: 1 1 1 = + f u v 1 1 1 + = 11 v 10 1 1 1 11 − 10 1 = − = = v 10 11 110 110 Therefore v = 110 cm (a positive answer indicates a real image is formed, as required). The screen should be placed 110 cm from the lens of the projector. H b Magnification: M = H i = − v u o Hi v Hence = − Ho u Hi 110 = − 3 11 Hi = − 30 cm The image on the screen is a square of side length 30 cm. c The negative value of Hi indicates that the image is inverted relative to the slide. The slide must be placed upside down in the projector to see an upright image.

Worked example 9.5C A concave lens is held horizontally 30 mm above the page of a book. The text on the page has a maximum height of 6.0 mm. The image of the text appears to be 10 mm beneath the lens and is reduced in size. a Is the image formed real or virtual? b Calculate the focal length of the lens. c What magnification is produced by the lens?

Solution a The image of the text is on the same side of the lens as the text and so it must be virtual. Concave lenses can only form virtual images. b List data: u = 30 mm, v = −10 mm (This is negative since a virtual image is formed.) 1 1 1 + = u v f 1 1 1 + = 30 −10 f 1 1 − 3 −2 = = f 30 30 Therefore f = −15 mm c M = − v = − (−10) = 1 (A positive answer was expected since the image is virtual u 3 30 and, therefore, upright.)


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9.5 summary Concave lenses • Regardless of the object position a concave lens always forms a virtual, upright and diminished image. The image is always on the same side of the lens as the object. When the object is brought closer to the lens the image becomes larger, but it can never equal the size of the object. • The lens equation shows the relationship between the focal length of a spherical lens and the object and image distances: 1 1 1 = + f u v

where f is the focal length of the convex or concave lens u is the object distance v is the image distance. • Magnification is the ratio of the heights of the image and object. The magnification produced by a particular arrangement of a lens and object is given by: v H M = i = − u Ho

9.5 questions Concave lenses 1 Describe the paths of three known rays through a spherical concave lens. 2 Why can a concave lens never produce a real image? 3 An object is placed 20 cm from a concave lens of focal length 10 cm. Is the image formed: A virtual, upright and enlarged? B virtual, upright and the same size as the object? C virtual, upright and reduced? 4 A concave lens of focal length 15 cm is held 30 cm above a page. a Use graphical ray tracing to model the arrange ment. b Is the image real or virtual? c How far from the lens is the image? d An illustration on the page is 3 cm high. How large would it appear if viewed through the lens? 5 A concave lens is held 15 cm above a page and the text appears to be one-third its original size. a Use graphical ray tracing to model the arrange ment. b Is the image real or virtual? c What is the focal length of the concave lens? d Approximately how far from the lens is the image formed? Use the lens equation and magnification formula to help you answer the following questions.

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6 A child examines an insect with a magnifying glass of focal length 15 cm. The insect is 5 mm in length and 10 cm from the lens. a How far is the image from the lens? b What magnification occurs? c How large is the image of the insect? 7 Use the lens equation to check your answers to questions 4 and 5. How accurate was your ray tracing? 8 A convex lens of focal length 20 cm is used to form an image on a screen which is eight times larger than the original object. a How far is the object from the lens? b How far is the object from the image? 9 A 3.0 cm high object is placed 12 cm from a lens of focal length 10 cm. A real image is formed on a screen. The lens is then replaced by a lens of twice the focal length. Comment on the effect that changing the lens has on: a the nature of the image b the position of the image c the size of the image. 10 A person is arranging a slide projector and screen. The projector is arranged so that the slide is 4.0 cm from the lens. The image needs to be 30 times larger than the original slide. What focal length lens does the projector require?

9.6 Optical system s We now examine a number of optical systems to discover how the principles of optics can be put to use in telescopes and microscopes. The engineering involved in the design of optical systems is complex. Much work has gone into discovering how to form the clearest possible images, and this has been based on trial and error. Engineers design a system, build it, test it and refine it! The examples we look at contain only two or three optical elements. Sophisticated instruments contain many more elements in the desire to produce clearer aberration-free images.

Aberration When discussing the path of light through spherical lenses we have assumed that sharp, perfect images are formed. You may have noticed that in fact the images formed have been blurred and distorted, particularly near the outer edge, and the image sometimes has had coloured edges. Deviations from the perfect image are an aberration. When scientists are designing optical systems to carry out particular functions they must attempt to eradicate as much aberration as possible. This is one reason why good quality optical equipment never has just one or two lenses; often combinations of numerous lenses are required.

Chromatic aberration In the previous chapter we found that different colours of light are refracted different amounts by a particular medium. We examined how a glass prism, having a slightly different refractive index for each colour of light, dispersed white light into its separate colours. Similarly, lenses will bend the path of different colours by slightly varying amounts. Using ray modelling Figure 9.30a shows (an exaggeration of) how red light is bent less than violet light in a convex lens. Thus the focal point for red light is further from the lens than the focal point for violet light. The other colours of the spectrum will be focused between these points. This spreading of colour is called chromatic aberration. An achromatic doublet, often made from a crown glass convex lens and a flint glass concave lens, is used to correct for chromatic aberration as shown in Figure 9.30b. By cementing a concave lens of a different glass type behind the convex lens the focal length for the red light is slightly extended, but the focal length for the blue light is considerably extended. Therefore, the focal points for the different colours can be positioned at the same place, eliminating chromatic aberration.

Figure 9.29 Sophisticated optical instruments contain a number of elements to correct for aberration and allow versatility.


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(a) white light

red violet

F'V

F'R

violet white light

red fV fR

(b) white light

red violet violet

white light

red

Figure 9.30 (a) Lenses do not refract all colours equally, which causes chromatic aberration. (b) An achromatic doublet eliminates chromatic aberration.

Spherical aberration

Figure 9.31 Spherical lenses are subject to spherical aberration; the outer rays have the shortest focal length.

We have previously discussed the paths of rays through spherical concave and convex lenses. However, when rays enter at a significant distance from the principal axis, they do not follow these same rules. Figure 9.31 shows how rays passing through the outer edges of the lens do not converge at the principal focal point. They are brought to a focus closer to the lens than rays passing through the centre area of the lens, resulting in a blurred image. This is called spherical aberration and it is most noticeable in high power lenses. Spherical aberration can be eliminated by using specially ground nonspherical lenses, but these are very expensive. Often combinations of lenses can be used to reduce spherical aberration. Spherical aberration in the eye is reduced because the surface of the cornea is not perfectly spherical but slightly flattened at the outer edges; the lens in the eye also has a lower refractive index near the edges, and the retina on which the image is formed is a curved surface.

Simple camera A camera is an instrument that uses a converging lens system to create a real image on a light-sensitive film. When the film is processed it can be used to print a photograph. Cameras come in many forms, each offering different features depending on the type of photographs taken and the quality of the images required. Apart from the very cheap fixed-focus cameras, there are three main adjustments which are carried out when taking a photograph. Automatic cameras will do these adjustments for you.

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Focusing The lens system in a simple camera has a fixed focal length and so the distance from the lens to the film, the image distance, must be adjusted depending on the proximity of the object being photographed. Recall that the relationship between object and image distances and focal length is given by the lens equation: 1 1 1 = + f u v When photographing closer objects (smaller u) the lens is moved further from the film (larger v). This is why a lens moves out from the camera to focus on a close object.

Worked example 9.6A A camera with a focal length of 50 mm is focused on a distant object. The lens position is now adjusted so that an object at a distance of 1.0 m is clearly in focus. a Was the lens moved towards or away from the film to focus on the nearer object? b How far was the lens moved?

Solution a When the camera is focused on a distant object, the image distance must be the same as the focal length of the lens, since effectively parallel rays converge onto the film. The film is therefore 50 mm from the lens. To focus on a closer object, should v increase or decrease? v is given by a rearrangement of the lens equation. 1 1 1 = − v f u In this example f = 50 mm. 1 1 1 = − v 50 u Since in moving the object closer, u is reduced, v must increase (i.e. lens is moved away from the film). b In this case u = 1.0 m = 1000 mm. Therefore: 1 1 1 = − v f u 1 1 = − 50 1000 19 20 − 1 = = 1000 1000 ∴ v = 52.6 mm (i.e. the lens is moved further from the film by a distance of 2.6 mm).

Shutter speed In order to produce a photograph, a film must be exposed to a certain amount of light. The film speed quoted for each type of film indicates how much light is required. The amount of light to which the film is exposed can be changed by altering the camera aperture and/or the shutter speed. The exposure time of the film is determined by the shutter speed. Exposure times may be as short as a thousandth of a second or as long as 1 second. High-speed action requires fast shutter speeds so that the scene does not appear blurred. Stationary objects can be exposed for longer


319

times. Fast shutter speeds can only be used successfully in well-illuminated situations, with very sensitive film, otherwise the film will be underexposed and the photograph will appear dark.

Aperture or f-stop Variable apertures can be situated at a number of different locations in the lens system of a camera, but they are commonly found behind the first few lens elements. A diaphragm of variable radius determines the effective diameter of the lens. The effective diameter of the lens is then quoted as an ‘f-number’, which is the ratio of the focal length to the lens diameter. For example, an f-number of f/8 means that the effective diameter of the lens is one-eighth of its focal length. Reducing the aperture to f/16 means that you have halved the effective diameter of the lens; therefore much less light will reach the film. A careful balance must be reached between shutter speeds and f-stop values. The faster the shutter speed the greater the lens diameter required to give a proper exposure.

The SLR camera Unlike the simple camera, the single lens reflex (SLR) camera allows you to see the exact view that you are about to photograph rather than viewing through an alternative chamber. This is particularly important if taking close-up photographs, as the boundaries of your view may differ from those reaching the sensor. This exact view is possible because of a small mirror and a pentaprism which reflect the rays after they have passed through the lens, as shown in Figure 9.32. Once you have focused your shot, pressing a button lifts the mirror momentarily out of the way. The shutter covering the sensor opens and the sensor is exposed. pentaprism

series of lenses

focusing screen

main reflex mirror sensor

secondary mirror light sensor

Figure 9.32 The SLR camera allows the same lens to be used for view-finding and taking the photograph.

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Optical instruments

large visual angle

When an object is viewed by the eye its apparent size depends on the angle which is subtended at the entrance to the eye. This is called the visual angle (Figure 9.33). Any object that is moved closer to the eye will subtend a larger visual angle and consequently a larger image forms on the retina. However, the object cannot be clearly viewed at distances less than the near point. The magnification produced by optical instruments such as microscopes and terrestrial telescopes occurs because the path of light from an object to the eye is altered to increase the visual angle of the object.

near point

The refracting telescope

smaller visual angle

The refracting telescope is constructed of two convex lenses placed a specific distance apart (Figure 9.34). The lens closest to the object being viewed is called the objective lens, the other is the eyepiece lens. The objective lens has a relatively long focal length. Light rays from a distant object are parallel and are brought to a focus, forming a real image in the focal plane of the objective lens. The eyepiece lens then magnifies this image. The eyepiece lens is positioned so that its focal point is concurrent with that of the objective lens. Rays from the real image then pass through the eyepiece lens and emerge parallel. The eye must still carry out its normal function of focusing parallel rays from a distant object; however, the visual angle has been increased, resulting in magnification. The parallel rays originally travelling from the object would form a very small visual angle at the eye. This is the same as the angle denoted α in Figure 9.34. Yet after refraction by the two lenses the visual angle has been increased to β. Note that the image formed is upside down, which is fine if you are viewing the stars but not if you are watching a yacht race! Terrestrial telescopes therefore also contain a third lens to re‑invert the image.

object appears large

object appears small

Figure 9.33 The apparent size of an object is due to the size of the image produced on the retina. A large, distant object and a small, close object can subtend the same visual angle.

Physics file Galileo Galilei is often mistakenly accredited with the invention of the telescope. He did not invent it but he did construct telescopes of vastly improved performance. With his telescopes Galileo made incredible, previously undiscovered observations such as the details of the surface of the Moon and the phases of Venus.

The reflecting telescope To collect light from very distant and faint objects large-diameter telescopes are needed. Larger telescopes form brighter images, but their size causes practical problems. The largest optical telescopes in the world are all reflecting telescopes: they use a mirror rather than a lens as the main lightcollecting element. These telescopes still employ eyepiece lenses. The long focal length objective lens

focal plane of both objective and eyepiece lenses

short focal length eyepiece lens

parallel rays of light β

α top of image formed by objective lens focal length of objective lens

line from the top of the objective lens image through the centre of the eyepiece gives the direction of parallel rays

parallel rays forming an image at infinity focal length of eyepiece lens

Figure 9.34 A simple refracting telescope magnifies by increasing the visual angle; the resulting image is inverted.


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Physics file Resolution is the ability of the telescope to separate the parts of an image that are so close together that they would otherwise appear as a single point. In the design of astronomical telescopes it is the resolution that is important and not necessarily the magnification. After all, the image of a point source of light such as a star is just another point source of light! Telescopes with a high resolution allow us to see individual stars in what may have looked like a fuzzy blob.

Physics file Even when using mirrors there are significant practical problems with very large telescopes. Mirrors that are more than about 4 m in diameter are actually so heavy that the mirror’s shape changes slightly as the telescope’s position alters—the glass sags under its own weight. The largest telescopes are of the order of 6.0 m in diameter.

mirrors used have their reflecting surface at the front, unlike the backsilvered mirrors we studied earlier. The advantages of using mirrors rather than lenses are numerous. Unlike lenses, large mirrors can be supported from behind and can be more easily shaped into parabolic surfaces to reduce spherical aberration. Large flawless mirrors are much cheaper to produce than large flawless lenses. Mirrors are not subject to chromatic aberration. There are two types of optical reflecting telescopes in common use, the Newtonian reflector and the Cassegrain reflector. The principle behind their operation is the same as for the refracting telescope: to produce an image at the focal plane of the eyepiece lens. The eyepiece lens then acts as a magnifier and enlarges the real image formed by the collecting mirror. The light-collecting mirror and eyepiece lens need to be arranged so that their focal planes coincide. To do this a second smaller mirror is used to redirect the rays, otherwise the observer’s head would need to be inside the telescope itself! The Newtonian reflector (Figure 9.35a) simply uses a flat mirror to redirect rays to the side of the telescope. This is simply for more convenient viewing when astronomers are manipulating the position of these enormous telescopes. In the Cassegrain reflector (Figure 9.35b) a small hyperbolic mirror reflects the rays back through an opening in the main mirror. (a) F1 plane mirror at 45°

Physics file Although the use of a parabolic reflector solves the problem of spherical aberration, parabolic reflectors have a significantly reduced field of view and suffer from an aberration defect called a coma. Stars imaged away from the centre of the field of view appear elongated and look like tiny water droplets.

F2

concave mirror (objective ‘lens’)

eyepiece lens

(b) parallel rays of light forming an image at infinity

focal point of eyepiece lens and mirror combination

diverging mirror to lengthen the focal length of the objective mirror parallel rays of light from a point on a distant object focal point of objective mirror

eyepiece lens b

a

objective mirror c focal length of objective mirror = c + a effective focal length of objective mirror = c + b

Figure 9.35 (a) The Newtonian reflector and (b) the Cassegrain reflector.

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The compound microscope The compound microscope is constructed from two convex lenses. The objective lens is shorter in focal length than the eyepiece lens (Figure 9.36). When an object to be examined is placed outside the focal point of the objective lens, a real, enlarged image is formed between the two lenses. The lenses must be arranged so that this first image, I1, lies inside the focal length of the eyepiece lens. This real image can then be considered an object for the eyepiece lens, which acts as a magnifying glass. As it is inside the focal length of a convex lens, a virtual, enlarged image is formed, I2. This is inverted relative to the original object and is many times larger. Spend a minute looking at the ray paths through the microscope. All of the rays shown follow the three paths discussed earlier in the chapter. eyepiece lens

object

objective lens Fo

Fo

Fe

Fe I1

I2

final image

Figure 9.36 In the compound microscope the objective lens can be considered to create an image which the eyepiece lens then magnifies further.

Physics in action

Structure of the human eye The human eye behaves as a converging lens, casting a real image on the light-sensitive screen, the retina. The structure of the eye is shown in Figure 9.37. The eye approximates a sphere of jelly-like substance contained in a tough coating called the sclera. The sclera is white and opaque, except at the front where the cornea is located. Light enters the eye through the cornea. The majority of the bending of light occurs at the air–cornea interface since this is where the most dramatic change in refractive index occurs (ncornea = 1.376). Light continues into the aqueous humour with negligible refraction since its refractive index is very close to that of the cornea (naqueous humour = 1.336). The iris is located behind the aqueous humour. It behaves like the aperture of a camera, dilating under dim conditions to allow more light through the pupil, and contracting in bright light. Interestingly, the iris will also contract when a person is viewing something at a very close distance. This is linked to the focusing behaviour of the eye: contracting increases the sharpness of the image. The iris can contract the pupil to 2 mm in diameter in bright light and expand it to nearly 8 mm in diameter in complete darkness.

aqueous humour pupil

cornea

iris

crystalline lens vitreous humour blind spot choroid

retina

macula

sclera

optic nerve

Figure 9.37 The structure of the human eye.


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Accommodation Light continues from the aqueous humour into the lens, which is a pliable, fibrous mass with an elastic membrane. It is about 9 mm in diameter and 4 mm thick. The lens is not of a uniform refractive index but it varies from 1.386 near the edge to 1.406 at the centre. The lens carries out the fine focusing of the system, forming a sharp inverted image at the back of the eye, through changes in its shape controlled by the ciliary muscles. This process is called accommodation. When these muscles relax, the lens, which is suspended by ligaments, becomes thinner and has a longer focal length for viewing distant objects. This is shown in Figure 9.38a. When the ciliary muscles contract, the lens bulges, resulting in a shorter focal length for viewing closer objects (Figure 9.38b). Remember, most of the work of bending the path of the light is done by the cornea, the lens just takes care of the fine tuning! (a)

relaxed muscle

taut ligaments

distant object

contracted muscle

(b)

slack ligaments

near object

Figure 9.38 (a) When the ciliary muscles are relaxed, images of distant objects can be focused. (b) When the ciliary muscles are tense the lens bulges, focusing images of close objects. The aqueous humour and vitreous humour provide nourishment to the lens and cornea; there are no blood vessels traversing them which would block the path of light. The lack of blood vessels means that these elements can be successfully transplanted with less likelihood of rejection by the immune system. Corneal transplants are now quite common.

The retina, rods and cones An inner shell called the choroid lines the sclera. The choroid carries major blood vessels to nourish the retina and absorbs stray light, just like the matt black lining inside a camera. A thin layer of more than 100 million photoreceptor cells lines much of the choroid. This is called the retina. It contains two types of photoreceptors—about 120 million rods and 7 million cones. These names relate to their shape, as Figure 9.39 illustrates. Rods are responsible for vision in dim light (scotopic vision). They function like high-speed black and white film. The images relayed are not well defined and lack colour. The cones respond to brighter levels of illumination (photopic vision), giving well-defined colour images. The relative numbers of rods and cones are different in different animals, depending on the animal’s vision requirements. Nocturnal animals have all or mostly rods in their retina, allowing them to see in very low light levels. The macula is a small depression less than 3 mm in

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Figure 9.3

9 The rod

s and con es in a hu

man eye .

diameter, at the centre of the retina. At its centre is a region called the fovea where no rods, only cones occur. The cones here are thinner and more densely packed than anywhere else on the retina. As a result the fovea can provide the sharpest and most detailed images under good illumination. The fovea is located directly opposite the pupil. Humans continually alter the direction in which their eyes are oriented, so that the image of the object of interest is formed on the fovea. However, to view a very dim object, such as an elusive comet, astronomers train themselves not to look directly at the comet, so as to use the non-foveal area of the retina. The relative abundance of cones diminishes with distance from the fovea. The further away from the fovea, the more rods predominate. As you gaze directly ahead, how much colour is apparent at the periphery of your field of view? Many birds have two foveal areas in each eye, allowing them sharp forward and sideways vision. The next time you talk to a budgie in a cage determine which foveal area he is using to examine you. Rabbits have a horizontal foveal strip, allowing them to keep a sharp eye on the whole horizon.

The blind spot The rods and cones in the retina are connected to the optic nerve by a network of nerve cells. At the point where the optic nerve connects to the retina, there can be no rods or cones and so a blind spot exists for each eye. This is normally not noticed as the brain is capable of ‘filling in’ the missing information from the surroundings. You can detect the blind spot in each eye, one eye at a time. Cover your right eye and hold this book upright about 30 cm in front of your face. Stare directly at the dot on the right. Now slowly move the book towards you. At some point you should become aware that the cross has disappeared. Its image falls on your blind spot. Continue to move the book towards you and the cross should reappear. Extending the size of the cross could help you to determine the size of the blind spot.

Correcting eye defects

(a)

A normal eye is described by optometrists as one that is able to clearly focus on any object that is between 25 cm and an ‘infinite’ distance from the eye. These object locations are called the near point and far point respectively. To locate your own near point, close one eye and hold this book a few centimetres from your nose. The text should appear blurred. Then slowly move the book away from you until the words come into focus. The page is now at the near point for that eye. You may find that your near point is as close as 10 cm from your eye, and that each eye has a different near point.

(b)

Myopia or short-sightedness Myopia is a condition in which parallel rays from a distant object are brought to a focus in front of the retina, rather than on it, as shown in Figure 9.40a. The lens is too powerful for the length of the eye, or the eye’s shape is too oblique so that the retina is too far from the lens. Images of distant objects are not clearly focused. The person can only clearly focus images of close objects and so this condition is called short-sightedness. The far point of a severely myopic eye may be as close as 50 cm as shown in Figure 9.40b. The near point would also be closer than usual to the eye. To correct this condition, the rays entering the eye must diverge through a concave lens, thus extending the focal length of the visual system and placing the focus correctly on the retina (Figure 9.40c). Mildly myopic people often have glasses that they only wear when driving or visiting the cinema. These people function quite easily in everyday life without their glasses.

Hypermetropia or long-sightedness Hypermetropia is a condition in which a person’s lens is too weak to bend rays sufficiently to bring them to a focus at the retina. Alternatively, the retina is too close to the lens because of an eyeball that is too short in length. Here the point of focus lies beyond the retina, as shown in Figure 9.41a. Since closer objects require the maximum deviation of rays by the eye, hypermetropic eyes are unable to focus on close objects. These people only see distant objects clearly and so this condition is called long-sightedness. The near point for a hypermetropic eye may be as far as 1 metre (Figure 9.41b). Hypermetropic people often read by holding the page at arm’s length. To correct this condition, a converging lens is used, which increases the bending ability of the visual system. In fact the glasses effectively move the image of the close object out beyond the far point where the person can see it (Figure 9.41c). The eye must then accommodate to see distant objects clearly, indeed some very hypermetropic people must remove their glasses to look at distant objects.

distance to far point (c)

distance to far point = focal length

Figure 9.40 Myopia. (a) Rays passing through the lens are bent so that they intersect before reaching the retina, resulting in a blurred image on the retina. (b) As the lens can only carry out severe bending of light rays, only relatively close objects can be focused on the retina. (c) A concave lens compensates for the overconvergence of light rays by the eye lens, so that the rays intersect at the retina, forming a clear image of distant objects.

(a)

25 cm (b)

distance to near point (c)

Figure 9.41 Hypermetropia. (a) Rays from a nearby object do not converge sufficiently to focus on the retina. (b) The near point of a hypermetropic person is more than 25 cm from the eye. (c) A convex lens compensates for the inadequate convergence of rays by the eye lens. A virtual image is then created beyond the person’s near point.

image

object 25 cm

distance to near point


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The power of lenses In optometry the focal length of a lens is not considered; rather the dioptric power of the lens is quoted. The dioptric power of a lens is the reciprocal of its focal length: D=

1 f

The unit for power is m−1 or dioptre, with the symbol D. A lens with a focal length +20 cm or +0.2 m has a dioptric power of +1/0.2 or +5 D. (Obviously this use of the term power is completely unrelated to power as a measure of the rate of energy transfer!) We previously examined how optometrists prescribe glasses which will bend light by a specific amount to correct the focusing flaws of the eye. We often refer to glasses as being strong or weak. It is logical that a lens with a short focal length, and therefore a stronger bending ability, has a higher power value. Dealing with powers rather than focal lengths saves a considerable amount of arithmetic. If two thin lenses are in contact, the overall focal length of the lens system is given by: 1 1 1 = + f f1 f2 But the combined power is simply D = D1 + D2. This is why when you visit the optometrist he/she aligns a number of lenses in front of your eye and asks you how clearly you can view a particular item. If a combination of a 1.0 dioptre lens and a 0.3 dioptre lens allows you to see a clear image, then the optometrist will prescribe glasses of power 1.3 dioptre.

9.6 summary Optical systems • Chromatic aberration is the spreading of colour by a lens. This occurs because the lens has a different index of refraction and therefore focal length for each colour of light. An achromatic doublet corrects for chromatic aberration. • Spherical aberration occurs because rays passing through the outer region of the lens are brought to a focus closer to the lens than rays passing through the centre area of the lens. This results in a blurred image. • A camera is an instrument that uses a converging lens system to create a real image on a light-sensitive sensor. The three main adjustments which are carried out when taking a photograph are shutter speed, f‑stop and focusing. • The refracting telescope uses two converging lenses. The image formed by the objective lens is positioned in the focal plane of the eyepiece lens, which acts as a

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•

•

• • •

magnifying glass. The final image is inverted relative to the object. Reflecting telescopes use a mirror as the main light-collecting element. These are produced free of spherical and chromatic aberrations. The compound microscope is constructed from two convex lenses. A real, enlarged image is formed between the two lenses; this forms inside the focal length of the eyepiece lens, and hence a virtual, enlarged image is formed. Short-sightedness (myopia) can be corrected with a concave lens. Long-sightedness (hypermetropia) can be corrected with a convex lens. The dioptric power of a lens, D, is the inverse of its focal length: 1 D= f

9.6 questions Optical systems 1 Describe two different types of aberration that can occur in spherical lenses and explain why they occur.

6 You want to buy a large telescope to carry out astro nomical observations. What are the advantages of the reflecting telescope over the refracting telescope?

2 In a camera that is focused on an object on the horizon, is the distance from the lens to the film or sensor smaller than, equal to, or greater than the focal length of the lens?

7 A student has four lenses available of focal lengths 10 cm, 20 cm, 50 cm and 3.0 m. The lenses are all of the same diameter. She wishes to construct a refracting telescope using only two lenses. a Which two lenses should she use? b Which will be the objective and which will be the eyepiece lens? c How far apart should the lenses be placed? d Will the image formed be upright or inverted?

3 A camera is mounted on a stand and used to photo graph a building which is at a distance of 300 m. The building is 30 m high. The lens system of the camera has an effective focal length of 4.0 cm. a How far is the lens from the film or sensor? b How large is the image of the building on the film or sensor? 4 The convex lens of a simple camera has a focal length of 5.0 cm. a If the object being photographed is 1.5 m from the lens, how far should the film or sensor be from the lens? b If the object being photographed is moved closer to the camera, should the lens be moved towards or away from the film or sensor to produce a clear image on the film? 5 If you were photographing a a dimly lit sculpture b a speeding racing car c a distant castle on a clear day would you choose a relatively high or low: i shutter speed? ii sensor speed? iii f-number?

8 A student proposes doing some astronomical photo graphy by connecting a camera to a Newtonian reflecting telescope. Another student claims that this is impossible since mirrors only produce virtual images and photography requires a real image to be formed. Which student is correct? Why? 9 A compound microscope has two lenses placed 25 cm apart. The objective lens has a focal length of 4 cm and the eyepiece lens has a focal length of 6 cm. An object is placed 5 cm below the objective lens. Calculate the position of the image formed by: a the objective lens b the eyepiece lens.

chapter review 1 Which of the following statements regarding a virtual image is correct? A A virtual image can be projected onto a screen. B A virtual image really exists. C A virtual image is produced by intersecting rays. D A virtual image is located at a point from which rays of light seem to diverge. 2 Which of the following is correct? A C represents the centre of a curved mirror. B F represents the point where parallel rays are converged by a concave mirror.

C C represents the point where parallel rays are converged by a concave mirror. D P represents the principal focus of a curved mirror. 3 a If you stand 3 m in front of a plane mirror, how far from you will your image appear to be? b Use ray tracing to show how a plane mirror produces an image of your face. 4 Explain the meaning of spherical aberration as it applies to a spherical mirror.


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5 A shop owner installs a large convex mirror for security purposes. Describe the nature of the images formed in the mirror. Why is such a mirror useful? Use ray tracing to solve questions 6–8. 6 A child 1 m tall is standing 6 m from a concave mirror whose radius of curvature is 4.0 m. a How far from the mirror is the image? b Calculate the height of the image. c Describe the nature of the image of the child. 7 A man wishes to shave a 2 mm long whisker. He stands so that the whisker is 30 cm from a spherical concave mirror of focal length 40 cm. a Describe the nature of the image. b How far from the mirror is the image? c What is the magnification of the image? d How large is the image of the whisker? 8 An object is placed 20 cm in front of a convex mirror with a radius of curvature of 30 cm. a How far is the image from the mirror? b Describe the nature of the image. c Determine the magnification of the image. 9 State which of the following use a convex mirror: make-up mirror, car headlight, searchlight, rear-view mirror, torch mirror, dentist’s mirror, shop-security mirror, solar collector. 10 Use ray tracing to illustrate two different situations in which a concave mirror produces an image that is twice the size of the object. 11 What properties of a spherical lens determine its focal length? 12 A concave lens is held 20 cm above a page and the text appears to be one-quarter its original size. a Use graphical ray tracing to model the arrangement. b Is the image real or virtual? c What is the focal length of the concave lens? d Approximately how far from the lens is the image formed? 13 A person accurately determines that her glasses have a focal length of +30 cm. Is she myopic or hypermetropic?

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14 A lens forms a real image 6.0 cm from the lens when an object is placed 18.0 cm from the lens. What is the focal length of the lens? A 6.0 cm B 18.0 cm C 4.5 cm D −4.5 cm 15 A convex lens has a focal length of 40 cm. Use graphical ray tracing to determine the nature, size and position of the image formed in each case. a A match 3 cm high is placed upright 60 cm from the lens. b A chess piece 2.5 cm tall is stood 90 cm from the lens. c A coin of diameter 2.5 cm is stood on edge 10 cm from the lens. 16 A convex lens is able to form (one or more correct answers): A real, diminished images. B virtual, diminished images. C real, enlarged images. D virtual, enlarged images. 17 The refracting telescope and the compound microscope both use two convex lenses. a Which device uses objective lenses of a shorter focal length? Why? b What differences occur in the nature of the images produced by each device? 18 Where must the object be located if the image distance is equal to the focal length of a convex lens? 19 A person looks through a concave lens and notices that a building that is 4.0 m tall appears only 4.0 cm tall. The building is 10 m from the person. a Is the image upright or inverted? b What magnification has the lens produced? c How far does the image appear to be from the lens? d What is the focal length of the lens? 20 A candle sits on top of a birthday cake. Its flame is approximately 2 cm high. a Where should a child hold a magnifying glass of focal length 8 cm if the candle is to cast an image on the wall which is 0.50 m from the candle? b How tall is the image of the candle?

area of study review Wave-like properties of light 1 Briefly discuss your understanding of what constitutes a wave. 2 Which one of the following is a correct statement concerning light? A Light travels through glass at a speed of 3.00 × 108 m s−1. B Red light has a smaller wavelength than blue light. C In a vacuum red and blue light travel at the same speed. D Increasing the intensity of a light source increases the speed of the light emitted. 3 The crest of a transverse wave approaching a non-yielding boundary or fixed end will: A be reflected as a crest. B be reflected as a trough. C have all its energy absorbed by the boundary and not be reflected. D will be reflected as a longitudinal wave. 4 The sound energy produced by speaking is propagated through the surrounding medium in the form of: A transverse waves. B longitudinal waves. C randomly vibrating air molecules. D electromagnetic radiation.

7 The destructive interference of two light waves occurring at a point: A clearly supports the wave theory of light. B clearly supports the particle theory of light. C clearly supports the theory of wave–particle duality. D does not support any particular theoretical model of light. 8 Briefly explain the phenomenon of refraction of light. 9 The speed of light in diamond is 1.24 × 108 m s−1. Determine the refractive index of diamond. 10 A ray of light passing from air into glass has an angle of refraction of 19°. a If the absolute refractive index of glass is equal to 1.50, determine the angle of incidence of the light ray. b Calculate the critical angle for light travelling from glass (n = 1.50) to air. 11 The following diagram shows a ray of light travelling from medium 1 into medium 2, with absolute refractive indices n1 and n2 respectively. Assume θ1 = 19° and θ2 = 29° and medium 2 is air (n2 = 1.0).

(cm)

1.5 1 0.5 0 –0.5 –1 –1.5

θ2

n2

5 Consider the transverse wave illustrated in the following diagram. The frequency of the wave is 50 Hz.

medium 2 medium 1 n1 θ1

50

150

250

350

a Calculate the value of n1. b Assuming that the speed of light in air = 3.0 × 108 m s−1, calculate the speed of light in medium 1. c Determine the critical angle for light passing from medium 1 into medium 2. d Explain what happens to the light ray if the critical angle is exceeded. e Which medium is the more optically dense? Justify your answer.

(cm)

a Determine the period, T, of the wave. b What is the amplitude, A, of the wave? c Determine the wavelength, λ, of the wave. d Calculate the speed, v, of the wave. 6 A treasure-seeking ship, equipped with sonar, uses an 8.0 kHz signal to locate a sunken wreck at a distance 1.0 km from the wave emitter. The speed of sound in the water is 1.5 km s−1. a What is the wavelength of the signal in the water? b How long after emitting the signal was the echo detected? c Would increasing the frequency of the signal increase its speed through the water? d What factor(s) would influence the speed of the signal through the water?

12 Determine the value of the refractive index for yellow light of the glass prism shown in the following figure. F 40°

E D C

A B yellow light


329 329

13 A glass plate is 6.00 mm thick and has a refractive index of n = 1.55. (Use c = 3.00 × 108 m s−1.) a Calculate the speed of light in the glass. b Determine the time it takes for a pulse of yellow light (λ = 590 nm) to pass through the plate. c What is the frequency of yellow light in air? 14 A student detects electromagnetic radiation of wavelength 10−2 m. What would be the most likely source of this electro magnetic radiation? A A television transmitter B A radio transmitter C A remote control unit D A microwave oven 15 Which of the following does not require the presence of a medium for transmission? A Radio waves B Infrared radiation C X-rays D All of the above. 16 Electromagnetic radiation is produced by: A stationary charges. B charges moving with constant velocity. C accelerating positive or negative charges. D high velocity neutrons. 17 The most penetrating form of electromagnetic radiation is: A radio waves. B infrared radiation. C X-rays. D gamma rays. 18 Classify each of the following waves as transverse or longitud inal: sound waves, microwaves, ripples on a pond, radio waves. 19 In an optical fibre, the optical density of the central glass core is greater than that of the cladding, so that total internal reflection occurs along the core. Explain the meaning of the term ‘total internal reflection’. 20 Explain the following statement and provide an example: ‘For 1 waves travelling in a given medium, f ∝ ’. l 21 An underground explosion occurs and 4.0 seconds later it is detected by a seismograph that is 10.0 km from the blast. Determine the speed of travel of the seismic wave. 22 Determine the wavelength of a sample of red light that is travelling in air. The period of the electromagnetic wave is 2.63 × 10-15 s.

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23 A transverse wave is sent along a rope that is fixed at one end. The phase change that occurs at the fixed end is: λ λ A λ B C D 0λ 2 4 24 An object would be described as yellow: A if the object absorbs all the yellow light incident on it. B if the object reflects all incident colours except yellow. C if the object reflects mostly yellow light and absorbs all other colours. D only if it can be seen when illuminated by yellow light. 25 At the boundary between two transparent media, increasing the angle of incidence of a beam of light will increase the proportion of the incident light which is: A transmitted. B absorbed. C reflected. D refracted. 26 Explain why a beam of white light passing through a glass prism can emerge as coloured light. 27 Discuss the differences in the images produced by concave and convex mirrors. 28 An object of height 1.0 cm is placed at various distances from a concave mirror of focal length f = 10 cm. The distance of the object from the mirror is denoted by u. For each given value of u, determine the corresponding image distance from the mirror, v, the image height Hi, and the nature of the image. a u = 20 cm b u = 15 cm c u = 10 cm d u = 5.0 cm 29 A concave mirror: A always produces a real image. B always produces an inverted image. C always produces a magnified image. D always produces a virtual image. E can produce both real and virtual images. 30 Determine the focal length of the concave mirror that will project the image of a lamp that is magnified 5 times onto a screen located 9.0 m from the lamp. 31 A physics student is asked to design a system that will project an image of a lamp, magnified 5 times, onto a screen 12 m distant from the lamp. Determine the type of mirror required, its focal length and its position in relation to the lamp.

32 A convex rear-view mirror of a car has a focal length of 20 cm. If a car 1.5 m high is 20 m behind the mirror, what will be the height of the image and where will it be formed? 33 Describe the factors that determine the focal length of a convex lens. 34 Which of the following statements concerning a convex lens is true? A Only a real image can be formed using a convex lens. B Two convex lenses of different refractive indices must have the same focal length. C The shorter the focal length, the more powerful the lens. D All images produced by a convex lens are magnified. 35 a An object of height 10 cm is placed on the principal axis at a distance of 8.0 cm from a converging lens of focal length 4.0 cm. i What is the distance of the image from the lens? ii Describe the image. b The object is moved to a new position at a distance of 2.0 cm from the lens. Describe the image. 36 The diagram shows a customer standing 3.0 m from a shopsecurity mirror of focal length 1.0 m.

F 3.0 m

1.0 m

a Use a ray diagram to model the formation of an image by the mirror. b Describe the image. 37 The following diagram shows an object of height 1.0 cm located on the principal axis at a distance of 2f from a concave mirror of focal length f. Which one or more of the rays A, B and C are incorrectly drawn? A O 2f

C

B f

38 The convex lens of a projector, focal length 1.25 m, is damaged. The only other lens available is a concave lens with focal length of equal magnitude. Replacing the convex lens with the concave lens will result in: A an image being projected onto the screen identical to that produced by the convex lens. B an image being projected onto the screen of identical size and clarity but inverted. C a greatly reduced image being projected onto the screen. D no image being projected onto the screen. 39 Which of the four rays A, B, C and D, shown in the following diagram, is incorrect? A B

f

C D

40 A camera uses a convex lens of focal length 8.40 cm, located at a distance of 8.60 cm from the film or sensor. a At what distance from the lens should an object be located in order to be most clearly photographed? b What magnification will be produced when the object is placed in the optimum position? c Describe the nature of the image produced on the sensor when the object is located at the optimum position. d If the photographer wants to take a picture of a distant mountain range, describe the focusing adjustment required for the lens. 41 Briefly explain the term ‘chromatic aberration’. 42 What is a ‘polarised wave’? 43 The phenomenon of polarisation demonstrates that: A light consists of transverse electromagnetic waves with their field vectors pointing in random directions. B light consists of longitudinal electromagnetic waves with their field vectors pointing in random directions. C light consists of a continuous stream of particles. D light consists of a mixture of longitudinal and transverse waves. 44 Explain colour dispersion in lenses. 45 Describe the polarisation of visible light and its relation to the transverse wave model. 46 List three assumptions of the ray model of light.


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47 The ray or particle model of light successfully explains: A diffraction. B the existence of light pressure. C that light travels faster in a vacuum than in a refractive medium. D the simultaneous reflection and refraction of light. 48 The phenomenon of dispersion occurs because: A the component colours of white light have slightly different velocities in any medium whose optical density is greater than that of air or a vacuum. B the absolute refractive index of any transparent medium is independent of the frequency of light travelling through the medium. C the speed of light in a medium is a function of the temperature of the medium. D the amplitude of red light is greater than that of blue light. 49 The wave model of light successfully explains: A interference and diffraction. B the inverse square relationship of light intensity with distance.

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C the photoelectric effect. D the spectrum of light emitted by a hot object. 50 According to the ray model, light energy is delivered in tiny discrete bundles called photons. Which of the photons listed below would have the highest energy? A red light B blue light C yellow light D X-rays 51 The ray model of light: A correctly predicts that light should travel faster in a more refractive medium. B correctly predicts that light should travel faster in a less refractive medium. C incorrectly predicts that light should travel faster in a more refractive medium. D correctly predicts the simultaneous reflection and refraction of light.

Unit

1&2 area o f stud y3

s e i d u t s d e l i a t e D detailed studies e detailed Chapters 10–15 are th rtake one detailed studies. You will unde e detailed study study in each unit. Th t be different from chosen for Unit 1 mus osen for Unit 2. the detailed study ch

y Chapter 10 Astronom sics Chapter 11 Astrophy om the nucleus Chapter 12 Energy fr ions: Flight Chapter 13 Investigat ions: Sustainable Chapter 14 Investigat energy sources ysics Chapter 15 Medical ph

chapter 10

y m o n o r t As outcome On completion of this chapter, you should be able to use observations to explain the motions of stars and planets, and describe models of planetary motion.

A

bout ten billion years ago, in what we now call our corner of the galaxy, there was a huge explosion—a fireball of truly astronomical proportions. A giant, fast-burning star literally blew itself to pieces as a ‘supernova’. In the short time before the cosmic dust was scattered, the energy released created temperatures so high that some of the star’s hydrogen and helium atoms were fused to form much larger atoms—atoms such as carbon, oxygen, iron and uranium. Much later, some of that dust, interspersed with the primordial hydrogen and pulled by the slow but relentless force of its own gravity, started to coalesce into a great whirling cloud. Most of the material eventually collapsed into the centre with such force that the temperatures created ignited a new nuclear fire. Our Sun was formed. Some of the heavier material in the cloud—in a dance played out between the gravitational pull of the new star and its own fast whirling motion—settled into smaller clumps spinning around the Sun. The planets were born.

by the end of this chapter you will have covered material from the study of astronomy including: • the motion of the stars, Sun, Moon and planets in our sky • the development of models of the solar system • the Copernican revolution and Galileo’s contribution • the telescope—types, uses and discoveries.

The story con tinues ... The heat that was released as the dust cloud collapsed ensured that everything was molten. Collisions between huge masses of molten rock created smaller missiles which flew around the new solar system pounding the surfaces of the new planets. Gradually the heat radiated away into space and the planets cooled. Some became so cold that even gases such as methane froze into still, silent seas. The closer planets, warmed by the Sun’s radiation, retained liquid water and an atmosphere of gases such as ammonia and methane.

Figure 10.1 ‘Starbirth’. This famous photograph shows huge gaseous pillars several light-years high. The ‘fingers’ are evaporating gaseous globules (EGGs) and are thought to be denser regions in which stars are forming. The less dense gas around the EGGs has been ‘blown away’ by strong ultraviolet light from nearby stars. This nebula actually has the rather unromantic name of M16 and is near the constellation of Sagittarius.

The chemistry going on in this rich soup of water, dissolved gases, and minerals was amazing. Some molecules grew larger and more complex. Perhaps the mixture was seeded with materials from outer space, perhaps they were driven by forces we don’t understand fully, but eventually the molecules started to form self-replicating structures. Life had arrived on Earth, if not on some of the other planets. The Earth was a very special planet. It contained a rich mix of the elements created in that cosmic fireball, notably elements such as carbon and oxygen that were essential for life. At just the right distance from the Sun it received enough warmth to keep the water liquid, but not so much that the water was boiled off. The 24-hour rotation of the Earth kept the night and day variation in temperature within reasonable bounds. The 1 23 2 ° tilt of the axis produced ice at the poles that also helped to stabilise the temperatures. Eventually life emerged from the ocean and began to colonise the land. Hundreds of millions of years later here we stand, looking back at, and wondering about, the processes that have created us. This is our modern creation myth. All human cultures on Earth have a creation myth, a story that gives them a feel for their origins, a sense of

Chapter 10 Astronomy

335

Figure 10.2 Life emerged from the ocean and colonised the land.

Figure 10.3 From the beginning of history humans have wondered about the origins and meaning of the universe.

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Detailed studies

meaning. Earlier stories speak of Gods or of timeless beings that brought the world into existence. Aboriginal Australians speak of the ‘Dreamtime’ in which great animals brought the land and sky into being. Our modern creation story begins with the early Greeks. It was Pythagoras (c. 540 bc) who declared that the world could be understood in mathematical terms, and Aristotle (c. 340 bc) who tried to make sense of the world with his picture of an Earth-centred universe. Ever since, natural philosophers (more recently referred to as physicists) have used whatever tools they could find or make, along with their imagination and power of reason, to find out more about the universe. Today, this story has flowered into a wonderfully detailed and truly awe-inspiring picture of our world. This is the story of astronomy (measuring the stars) and cosmology (understanding the universe). Clearly our modern creation story is not just of interest to scientists. Almost all of us wonder about our origins, and our destiny. Indeed, seeking answers to the big questions of ‘life, the universe and everything’ seems to be an essential part of being human. Astronomy can help us in this search. It certainly will never tell us all the answers. Other sciences, as well as the arts, the humanities and religions, all contribute to our picture of what it is to be human. In a real sense, however, astronomy and cosmology paint the background to which these other endeavours add details. It is a fascinating picture. In this detailed study of astronomy we are concerned mainly with how we have come to understand the night sky as we see it with the naked eye and through telescopes.

10.1 Motion in the h ea

vens

There is probably no more spectacular natural display than the heavens as seen from a dark setting away from city lights. Imagine yourself lying awake (in a warm sleeping bag) on a mountain top on a clear moonless night. The Milky Way, our galaxy, is particularly beautiful. In the southern hemisphere we are privileged to be able to see into its heart. It may look like a milky band of cloud, but if we look with binoculars we see that it is actually composed of myriad stars and nebulae (seen as small fuzzy patches). In late summer, early evening, the two brightest stars in the night sky—Sirius and Canopus—are high overhead: Sirius a little north of the zenith (directly overhead) and Canopus a little to the south. Just to the north-east of Sirius we find the constellation of Orion the Hunter with his distinctive belt and sword—well, distinctive when seen from the northern hemisphere. From our part of the world Orion appears upside down and we often refer to this group of stars as the ‘Saucepan’. (a)

(b)

PRACTICAL ACTIVITY 44 Watching the night sky

(c)

Figure 10.4 (a) Orion as seen from the southern hemisphere. Note the colours of the stars. (b) The Great Nebula in Orion can be seen with binoculars. (c) Orion the Hunter is upside down when viewed from the southern hemisphere.

Look a little more closely at the two bright stars that make up Orion’s shoulders (Betelgeuse and Bellatrix). They are the two stars a little below the Saucepan. You will notice that Bellatrix is distinctly bluish while Betelgeuse is quite red. Compare them with Sirius (a little to the south and east of Orion), which is quite white. The colours of the stars are of great importance to astronomers. We can only know the stars by the light we receive, and the colour of this light is an important clue to their nature. In the past century many techniques have been developed to analyse this light and to extend our vision into the invisible ‘light’ well beyond the visible spectrum. You may also be able to see some planets. Planets tend to be brighter than most stars and do not ‘twinkle’ (shimmer due to atmospheric effects). Venus, the so-called ‘evening star’ (or ‘morning star’) can be particularly stunning. It is not really a star of course, but it is the brightest object in the night sky, apart from the Moon. Mars, Jupiter and Saturn can be quite spectacular also and are often brighter than most stars. Through binoculars we can make out Jupiter’s moons, and, with a small telescope, Saturn’s


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Figure 10.5 The Southern Cross as seen when it is high in the sky.

rings. As we watch the stars from our mountain top we notice that they are gradually moving, rising from the east and setting in the west, as do the Sun and the Moon. In fact, when we wake at 4 a.m. the next morning the starscape is quite different. Orion has set and Scorpius is now high above us. The Southern Cross, which in the evening was low in the south-east, is now quite high in the southern sky. It is sometimes a surprise to those of us who live in cities and rarely notice the night sky to discover what the ancients knew as common knowledge. If we were to stay awake all night and carefully watch the sky, we would find that the stars rotate a full circle each day about a point or ‘pole’ in the sky. This point is the south celestial pole (SCP) in the southern hemisphere and the north celestial pole (NCP) in the northern hemisphere. From Australia we can see the SCP (see Figure 10.6) but not the NCP. In fact, as we shall see, the altitude of the SCP above our southern horizon is equal to our latitude.

Figure 10.6 The diurnal (daily) motion of the stars. In this 10-hour exposure from the AngloAustralian Observatory at Siding Springs the stars have rotated 150° around the south celestial pole.

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Physics in action

A strong recommendation Although it would be possible to complete this study entirely from within the classroom, we strongly suggest that at some point you try to take a trip to a country location where you will have a clear view of the night sky away from the light pollution of cities and towns. The ideal would be a trip to the high country, but even a 50-km trip away from the city lights can provide a much better view of the night sky than is possible from the outer suburbs. Moonlight will also overwhelm many of the stars, so choose a night when the moon will not be up in the late evening. However, the crescent moon, just after new moon, can be a wonderful sight in the early evening. Take along a star chart of some sort (and a dim red torch to read it by) as well as the ‘planet rise and set’ table published in the daily press. The latter will enable you to identify any planets, as they cannot be shown on a star chart. Even better, take an astronomy yearbook which lists all sorts of information and suggestions for viewing for the current year. (See the list included in ‘To help you in your sky gazing’ near the end of this Physics in action.) All you really need are your own two eyes, but a pair of binoculars can significantly enhance the experience. The lens diameter of the binoculars is more important than the magnification. A pair of 7 × 50 (magnification × diameter) binoculars will give a brighter image than an 8 × 30 pair, for

example. It is well worth trying to find a way to keep them steady by mounting them on a tripod or other firm support. Another useful addition may be short black cardboard tubes on the front of the lens to protect them from stray light, as well as from dew on cold nights. Also, make sure the lenses are clean, the slightest smear from a finger on the lens will tend to put halos around the stars. If you are appropriately equipped it can be a wonderful experience to sleep outside ‘under the stars’. Try to wake up several times through the night to see the changes that have occurred during the night. The early morning sky will be quite different from the evening sky. The following table lists some of the objects particularly worth looking for, but don’t neglect to simply enjoy the experience as a whole. Then spend some time identifying the brightest stars and the more obvious constellations. Notice the different colours of the stars (binoculars are helpful here), and the planets if they are visible. The Milky Way is our ‘home galaxy’, and on a clear dark night can be a wonderful sight indeed. Because we are looking at it from within, it stretches right around our sky. We are lucky that the centre of our galaxy, the brightest part of the Milky Way, is actually in the southern sky.

Table 10.1 Some astronomical sights With the naked eye

• Use a star chart to identify the major constellations: Start with the Southern Cross and Pointers (part of Centaurus), Orion (with the ‘Saucepan’) and Scorpius. Then look for some key stars such as Sirius, Canopus, Aldebaran, Achenar and Formalhaut. • The Milky Way—our galaxy: It is particularly brilliant in winter and spring evenings (or the early morning in autumn). The centre of the galaxy is in the region near Sagittarius and Scorpius, so try to observe it when that area is high in the sky (i.e. in the evenings from July to September). • The Large and Small Magellanic Clouds: The LMC and SMC are patches of light near the south celestial pole. They are quite easily seen in a dark sky and look like pieces of the Milky Way that have ‘broken off’. Indeed they are actually satellite galaxies of the Milky Way. • The colours of stars and planets: Look at the colours of the stars in Orion. Betelgeuse is one of the reddest stars while Bellatrix is one of the bluest. Because of the brilliance of stars it is not easy to see the colour with the naked eye; but if you have the chance to take a photograph, the colour will be more obvious. • The wandering planets: In one evening you will not see the planets moving among the stars, but if you can spread your observations over weeks or months you will see that the planets appear to move among the stars. If you are lucky you may notice that the direction of a planet’s motion, with respect to the stars, changes and starts moving ‘backwards’ for a while. This is the ‘retrograde motion’ which made it so difficult for early astronomers to explain the motion of planets. • Conjunctions between stars, planets and the Moon: Astronomy yearbooks list various conjunctions of planets with each other, stars or the Moon when the objects involved appear to come close. These are fun to watch and, except for those involving the Moon, occur over a period of days or weeks.


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With a pair of binoculars

• In good conditions, four of Jupiter’s moons may be visible through binoculars. Look for the small ‘stars’ in a line nearby and watch them over the course of a few days. • The colours of stars and planets are more obvious through binoculars. Look particularly at the colours of stars such as Aldebaran, Betelgeuse and those of the Southern Cross. • The larger craters on the Moon will be visible, particularly where the shadows are long near the edge of the dark region (the ‘terminator’), so look for them at phases other than full. Also look for the mare (seas) on the Moon. • There are many star clusters, some quite open such as Coma Berenicis or the Pleiades (the Seven Sisters), which is the most famous star cluster in the sky, and others small and dense, such as the Jewel Box near the Southern Cross. They are a wonderful sight. • Look for the Great Nebula in Orion—the middle star in the sword (or Saucepan handle). • One of Galileo’s great discoveries was the fact that Venus showed phases that changed with its apparent size. The crescent phases may just be visible in good conditions.

With a telescope

• If a small telescope is available this may enhance your view of any of the observations suggested above, but many people find observation with binoculars just as satisfying.

To help in your sky gazing

• Find the rise and set times of the planets, Moon and Sun in the daily paper. • Obtain a ‘starfinder’ or ‘planisphere’: a cardboard or plastic disk that rotates showing the stars that will be in the sky at any particular time. (These are available from museum shops, Australian Geographic stores, telescope stores and some bookshops.) • A year guide such as Astronomy 20XX Australia [where XX is the current year], by Dawes et al. (Quasar Publishing) is helpful. • There are a number of publications which list sky objects to observe through binoculars or small telescopes. A good one is Space Watching published by Australian Geographic, but there are a number of others.

(a)

(b)

Figure 10.7 (a) The Pleiades (the Seven Sisters) is the best-known open star cluster and is a lovely sight through binoculars. (b) The Aborigines incorporated the Seven Sisters and Orion’s belt into their Dreamtime legends. One story has Jarn the hunter chasing Marigu (which we know as the Seven Sisters) to try to catch a wife.

The celestial sphere As we watch the nightly, or diurnal, rotation of the stars it is easy to believe that the stars are all attached to a huge celestial sphere with the Earth at its centre, and around which the sphere rotates once a day. Indeed this is what most people down through the ages have believed. Even though we now know that the stars are spread throughout a vast universe, and that

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it is the Earth that is rotating, for many purposes astronomers still find it convenient to think of a rotating celestial sphere. Indeed, the equations used by astronomers to calculate the positions of the stars make no distinction between the Earth’s rotation and the rotation of the celestial sphere. The SCP is the point directly above the Earth’s South Pole and the NCP is directly above the North Pole. The celestial equator is the projection of the Earth’s equator onto the celestial sphere. In looking at Figure 10.8 it is easy to see that our view from the Earth will be the same whether we regard the Earth or the celestial sphere as rotating. Because, as we watch at night, it looks as though it is the sphere that is rotating, this is the way astronomers normally talk about it when describing the (apparent) motion or position of stars. (It is worth pointing out here that the ‘tilt’ of the Earth’s axis has no relevance to the stars—it only concerns the Sun, which we will look at later.) Celestial navigation depends on an understanding of the positions and motion of the stars. That the SCP is at an altitude (angle above the horizon) equal to our latitude can be seen from a little geometry. In Figure 10.9, angle b is the latitude of Melbourne, 38° south of the equator. As the zenith is perpendicular to the line to the northern horizon, angle a, the altitude of the celestial equator above the northern horizon, must be 52°. As the latitude of the South Pole is 90° south, the angle from the zenith down to the SCP (angle c) must also be 52°, and thus the altitude of the SCP above the southern horizon (angle d) will be 38°. As you can see, this will be true for any latitude: the altitude of the celestial pole (north or south) above the horizon enables one to determine the latitude of the viewing point.

NCP

N Earth S celestial equator

Orion

Southern Cross SCP

Figure 10.8 To describe the diurnal motion of the stars, astronomers picture them as attached to a huge celestial sphere which rotates daily around the Earth about axes which extend from the Earth’s North and South Poles. While we have drawn some stars on the sphere as they appear to us from inside the sphere, in reality their distances from Earth may be hugely different.

no

rth

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riz

on

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izo

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ato r

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Figure 10.9 On any diagram showing the Earth and the celestial sphere to scale, the Earth would be an invisible tiny dot in the centre. This is the reason the horizon line is shown as though it passes through the centre of the Earth rather than as a tangent to the Earth’s surface. The altitude of the SCP above the southern horizon is equal to the latitude of the viewing position.


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PRACTICAL ACTIVITY 45 Measuring the night sky: altitude–azimuth

Physics file Actually the horizon as seen from the Earth’s South Pole includes just a little of the northern sky and vice versa. This is the result of the refraction that occurs as the light from the stars on the celestial equator enters the Earth’s atmosphere. The same effect also occurs everywhere else. For this reason the Sun rises a few minutes before, and sets a few minutes after, it would if the Earth had no atmosphere. Can you see why the actual amount of extra time increases with latitude?

Before considering the sky as we see it, first let’s take an imaginary trip to the South Pole, as the sky seen from the poles is simple to understand. As you can see from Figure 10.8, an observer at the North or South Poles will see the stars rotating around in horizontal circles clockwise around the celestial pole directly overhead. No stars will ever rise or set at all. The half of the sky that is seen from the South Pole is referred to as the southern sky and that seen from the North Pole is called the northern sky. Remember also that the size of the celestial sphere is virtually infinite in comparison with the size of the Earth and so the horizon from both poles actually meets the celestial sphere at the same place. As we return from the South Pole and head north, the SCP gradually drops from the zenith down toward the south horizon (as we saw in Figure 10.9). Let’s go all the way to the equator. Once at the equator, the SCP will have dropped right down to the south horizon and we will just begin to see the NCP on the northern horizon. As we watch the stars from a point on the equator they will all rotate around the two celestial poles on our northern and southern horizons. This time we will see all of the stars rising, and then setting 12 hours later—except for the fact that the Sun will come up and obscure our view! As Orion is on the celestial equator it will rise in the east, go right overhead, and set in the west. On the other hand, the Southern Cross, which is low in the southern sky, will describe a smaller half circle around the SCP. (a)

(b)

Southern Cross

zenith 90°

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Melbourne

Saucepan

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at

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SP

our horizon

0°

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NCP the sky we never see

the sky we can't see now

Figure 10.11 The celestial sphere as seen from

Melbourne. As the SCP is 38° above the horizon, all stars within 38° of the SCP are always in our sky. All the stars within 38° of the NCP (that is, those more than 52° north of the celestial equator) are never seen in our sky. The stars further than 38° from the SCP rise and set, with those closer to the pole spending more time in our sky than those near the NCP.

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celestial equator

the sky we always see

52°

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Saucepan

SCP

Detailed studies

SCP

east

equator

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Southern Cross

Figure 10.10 (a) The southern sky as seen from the South Pole. No stars rise or set. Orion (the Saucepan), which is on the celestial equator, sweeps around the horizon each day. The Southern Cross describes a smaller, horizontal circle high in the sky. No directions are shown as all directions are north, but the rotation of the sky is clockwise, which can be seen as a westerly direction. (b) The sky as seen from the equator. All the stars rise from the eastern horizon and set 12 hours later.

Returning to Australia we will see the SCP again move upwards from the southern horizon to an altitude equal to our latitude. The stars in the north will still be appearing to rotate around the NCP, but the NCP itself will have sunk below our horizon as soon as we left the equator. Most of the stars in our sky will rise and set, but those close to the SCP will always be in the sky (although sunlight may prevent us seeing them). We can therefore divide the celestial sphere up into three parts: the stars always in our sky, those which rise and set, and those we never see. Figure 10.11 shows the celestial sphere as we see it from our point of view where we are at the ‘top’

of the Earth. Again, remember that the horizon is shown through the centre of the Earth because the celestial sphere is actually huge compared with the size of the Earth. The stars that are always in our sky are referred to as circumpolar stars. Trails from circumpolar stars are clearly visible in Figure 10.6. The Southern Cross, being within about 33° of the SCP, is circumpolar from all latitudes below about 33° south, which includes all the southern capital cities of Australia. Stars beyond 33° from the SCP will spend decreasing amounts of time in our sky. Those on the celestial equator (such as Orion) will rise due east, reach an altitude (from Melbourne) of 52° (that is, 90° minus the latitude) in the northern sky and set due west after spending 12 hours in the sky. Stars north of the celestial equator will spend less than 12 hours in our sky, while stars within 38° of the NCP will never be seen in the Melbourne sky. Physics in action

Where are the celestial poles? The position of the celestial poles can be found quite simply on a time exposure of the sky such as that shown in Figure 10.6. However, it is not so easy to find the SCP when looking at the night sky. While those of us in the southern hemisphere enjoy a more spectacular view of the Milky Way, northern sky watchers do have one advantage; a star—not surprisingly called the North Star or, more correctly, Polaris—closely marks the NCP. However, there is no ‘south star’ and we have to estimate the position of the SCP from the positions (a)

(b) Southern Cross

of the stars around it, notably the Southern Cross. There are several well-known techniques for doing this. The simplest is 1 probably to extend the long axis of the cross about 4 2 times. Another is to extend the long axis of the cross and take the perpendicular bisector of the two pointers. The SCP is where those two lines meet. A third is to find Achernar. The SCP is then half way between the top star of the cross, Gacrux (γ Crux), and Achernar (Figure 10.12). Finding the SCP was very important to the early navigators as its altitude above the horizon determines the latitude of the viewer and, of course, its direction is true (not magnetic) south. Various complicated, but more accurate, schemes involving the positions of the stars were used to determine the position of the SCP.

Cassiopeia Pointers Cepheus

SCP

Polaris NCP

Achernar Draco

Ursa Minor

Figure

Figure 10.12 (a) Finding the SCP from the Southern Cross, the Pointers and Achernar. (b) The North Star—Polaris—closely marks the position of the NCP.

10.13 Sextants such as this w stars, which, in ere used to fin conjunction w d the positions ith various tabl of a ship to be of es, enabled th determined. e latitude


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tio n +60n +30n

16 18

Cele

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14

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stial equato r

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8 6

0n

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SCP –90n

Figure 10.14 The coordinate system used to identify the positions of celestial objects uses declination, which is the equivalent of latitude, and right ascension (RA), which is the equivalent of longitude. The star shown has coordinates RA 2 h 30 min, dec. –20º.

Physics file The point of zero right ascension is the vernal equinox. This is the point at which the Sun moves from the southern sky into the northern sky on about March 21 each year. The word ‘vernal’ refers to the fact that this is the start of spring in the northern hemisphere. An older expression for this point in the sky was ‘The first point of Aries’. In ancient astrology the vernal equinox was actually in Aries, but due to the precession of the Earth’s axis (which we discuss later) it has moved into Pisces, and will move into Aquarius in about 2600 ad. Hence the term, ‘the Age of Aquarius’, meaning the start of a new era.

As the stars appear ‘fixed’ to the celestial sphere it is useful to be able to describe their positions accurately. To describe the position of any place on Earth we use latitude and longitude. The system for describing the positions of the stars on the celestial sphere is very similar. The latitude equivalent is called the star’s declination (dec.) and is given in just the same way as we describe latitude. Indeed the declination scale in Figure 10.14 can be seen as simply the projection of the Earth’s latitude scale onto the celestial sphere. The celestial equator therefore has a declination of 0°, the NCP a declination of +90° and the SCP a declination of –90°. It is useful to remember that the Saucepan (in Orion) sits on the celestial equator, and so extends from declination 0° to about –6°. The Southern Cross spans about 3° either side of declination –60°. It follows that at any place of a certain latitude, all the stars passing directly overhead at some stage during the day will have a declination equal to that latitude. It is easy to think of this at the poles and the equator. The stars directly overhead at the North and South Poles must have declinations of +90° and –90° respectively; that is, they are at the NCP or SCP. On the other hand, the stars passing over any point on the Earth’s equator (latitude 0°) must be on the celestial equator (declination 0°). The same is true at any other latitude: the stars passing directly overhead will have a declination equal to the observer’s latitude. The celestial equivalent of longitude is called right ascension (RA). As the celestial sphere rotates, the stars will rise (ascend) from the east (on your right if you are looking north). To determine right ascension, the time at which a star passes the meridian (an imaginary line running up from the north, through the zenith and down to the south) is measured and then compared with the time at which the zero of RA passed the meridian. Because of this way of determining the RA, astronomers normally measure it in hours, minutes and seconds, rather than degrees. Just as the zero of longitude has to be chosen arbitrarily (at Greenwich, London), the zero of RA is chosen at a particular point on the celestial equator. This point is called the vernal equinox. The RA of a star is close to, but not exactly equal to, the number of hours and minutes after the vernal equinox has crossed the north–south meridian that the star crosses the same line. We will see why shortly.

Star time—the sidereal day PRACTICAL ACTIVITY 46 Measuring the night sky: equatorial coordinates

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We said earlier that the stars rotate a full circle each day. This, as you may know, is not quite correct. In fact they take a little less than 24 hours to return to their position of the night before. If, for example, you see the Saucepan due north at 9:00 p.m. one night, the next night it will be due north at 8:56 p.m. The reason for this is the Earth’s motion around the Sun. Consider the two people shown on the equator in Figure 10.15. One is at midday and the other is at midnight. The first can see the Sun directly overhead and the second can see a particular star, say Rigel, directly overhead. After one full 360° rotation of the Earth, the midnight observer will see Rigel overhead again (the stars being so far away that the direction is still the same). However, at this stage the midday observer will not see the Sun exactly overhead.

Physics file

distant star distant star next sidereal day

Sun

distant star day one Earth’s orbit

Figure 10.15 One sidereal day is the time taken for the Earth to do a full 360° rotation. However, this is slightly shorter than one solar day, which is the time between midday one day and midday the next. (Note that the angles in this diagram have been exaggerated for clarity.)

Each day we move about one degree in our orbit around the Sun (as there are 365 days in the year). This means that the Earth has to rotate a little more than a full 360º before the midday observer will see the Sun directly overhead again. As the time between midday one day and midday the next is defined as 24 hours, the time for one complete 360º rotation of the Earth on its axis is a little less than 24 hours. In fact, as this small difference has to make up one day in a year, it is less by 1/365th of a day, which you can confirm is close to 4 minutes. The time from Sun due north one day to Sun due north the next day is referred to as a solar day, and so the 23 h-56 minday is a sidereal day (literally a ‘star day’).

The average length of a solar day is 24 hours; a so-called . Because the Earth’s orbit is not circular, but slightly elliptical, the time between the Sun crossing the north–south meridian varies a little. When the Earth is further from the Sun it travels a little more slowly around its orbit and, therefore, it takes a little less time to return to the midday position. The tilt of the Earth’s axis relative to the plane of its orbit also leads to a slightly shorter solar day at the equinoxes compared with the solstices. Although these differences each day are small, they add up, and so the Sun actually gets about a quarter of an hour ahead of the clock in November but a quarter of an hour behind in February. +15 +10 Minutes

next solar day

+5 0 –5

J F M A M J J A S O N D

–10 –15

Figure 10.16 The difference between clock and solar time through the year.

10.1 summary Motion in the heavens • The stars appear to rotate as if fixed to a huge celestial sphere that is centred on the Earth. The sphere appears to rotate on an axis that passes through the Earth’s North and South Poles. The axis meets the celestial sphere at the south and north celestial poles (SCP and NCP). • The height (altitude) of the SCP in our sky is equal to our latitude. The only place from which both the NCP and SCP are visible (just) is the equator. • The stars visible from the North and South Poles are referred to as the northern and southern stars respectively. The celestial equator is the line between

them. It is the projection of the Earth’s equator onto the celestial sphere. • The position of a star on the celestial sphere is described by its right ascension (RA) and declination (dec.) coordinates. RA is the equivalent of longitude and is measured from the vernal equinox in hours, minutes and seconds. Declination is the equivalent of latitude and is measured in degrees north (+) or south (–) from the celestial equator. • A sidereal day is the time for the Earth to rotate through one full circle; that is, the time for the stars to appear to rotate exactly once around the Earth. It is about 4 minutes less than 24 hours.


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10.1 questions Motion in the heavens Unless otherwise stated, assume that the stars are observed from within Australia, in answering these questions. The latitude of Melbourne is 38°. 1 As we watch the stars through the night we find that they: A stay in exactly the same place in the sky all night. B all rise from the east and set in the west. C mostly rise from the east and set in the west. D rotate around the SCP directly overhead. 2 As we watch the Southern Cross at night we will find that: A it rises in the east and sets in the west. B it rises in the west and sets in the east. C it rises in the south and sets in the north. D it never rises or sets. 3 Orion is on the celestial equator. In the night sky Orion: A rises in the east and sets in the west. B rises in the west and sets in the east. C rises in the south and sets in the north. D never rises or sets. 4 This is a long-exposure photograph of part of the sky at night. In what direction is the camera facing and are these stars rising or setting?

5 Briefly explain the difference, if any, you would see in the location of the SCP in the sky in Brisbane (latitude 27°) compared with its location in the Melbourne sky. 6 Two stars are observed from a location with latitude –30°. Star A has a declination of –30° and star B a declination of –40°. a Which of the statements below about the path of star A will be true? b Which of the statements below about the path of star B will be true? A The star will remain overhead permanently. B The star will pass directly overhead at 12 noon each day. C The star will pass directly overhead at some time during the day. D The star can never be seen directly overhead from that location. 7 Use a starfinder or map to find the bright stars which have the following celestial coordinates: a RA 7 h 42 min, dec. +28° b RA 22 h 55 min, dec. –30° 8 Use a starfinder or map to find the approximate coordinates of: a Canopus b Antares (in Scorpio). 9 At 9:00 p.m. one night we see Orion due north. Explain why, when we look one week later at 9:00 p.m., it will not be due north. Approximately where will it be? 10 The right ascension of Orion is approximately 6 h and the right ascension of Aquarius approximately 23 h. Both are close to the celestial equator. If they can both be seen in the night sky, which one rose first and by how much?

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10.2 The Sun, the Mo o

n and the pla

nets

The most significant star in our sky is, of course, our Sun. Like the stars, it appears to rotate around us once a day. Unlike the stars, however, it does not stay in the same place on the celestial sphere all through the year. While the stars all remain ‘fixed’ on the celestial sphere and describe the same perfect circles every day, the Sun gradually moves through the stars. Now we all know that the Sun does not really ‘move through the stars’; for a start, the stars are billions of times further away! The apparent motion of the Sun against the background of the ‘fixed’ stars is actually due to the Earth’s revolution around the Sun. Of course, we cannot see the stars behind the Sun, but it is a relatively easy job to plot the Sun’s position on the celestial sphere and in that way determine its path through the stars. Figure 10.17 illustrates one way of visualising the motion of the Sun through the celestial sphere. (a)

(b)

Physics file We refer to the Sun and Moon as being quite different from the planets. However, because they move with respect to the background stars, astronomers before the time of Galileo referred to them as planets (‘wanderers’). Hence you will see references to the ‘seven known planets: Sun, Moon, Mercury, Venus, Mars, Jupiter and Saturn.’

(c)

Figure 10.17 One way to get a feel for the Sun’s motion through the celestial sphere is to use a computer program. These three pictures from the Starry Night™ planetarium software show the sky at sunset in April, May and June. In each picture the Sun is just on the horizon and so it is clear that the Sun has moved eastward relative to the stars during this time. The Saucepan (Orion’s Belt) has been circled in each picture.

The Sun through the year Figure 10.18 shows the Earth in its orbit around the Sun. The Earth’s axis 1 remains on the same 23 2 ° tilt, relative to the plane of its orbit, throughout the yearly revolution. It is the tilt of the axis that gives rise to the seasons. In December, when the South Pole is tilted towards the Sun, the southern hemisphere has summer. The Sun is higher in the sky, rises and sets further south, and spends longer above the horizon. The date when the Sun is highest is known as the summer solstice (about December 21). The winter solstice (about June 21) is the date when the Sun is lowest in the southern hemisphere sky. In the northern hemisphere the solstices are the other way around. At the March and September equinoxes the tilt is such that the boundary between day and night runs right through the poles. This means that everywhere on the Earth has 12 hours of daylight and 12 hours of night-time; hence the term ‘equinox’ (equal day and night). After the March equinox the South Pole will move into 24-hour night and the North Pole into 24‑hour day. Only the equator will experience equal day and night all year round.


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Vernal equinox PISCES

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Figure 10.18 As the Earth moves around the Sun during the year we see it move against the background of stars. The path along which it moves is called the ecliptic and the twelve constellations along the ecliptic together make up the Zodiac.

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Figure 10.19 The Sun through the year as seen from Melbourne. In summer it rises and sets about 30º south of east and west, spends longer in the sky and reaches a higher altitude.

The changing temperatures of the seasons is the result of the changing angle of the Sun’s rays as well as the length of time that the Sun is in the sky. Figure 10.20 shows equivalent ‘sunbeams’ reaching the Earth’s surface in summer and winter. Clearly the same amount of solar energy spreads over a larger area of surface in winter than in summer, leading to less heating of the surface and, hence, cooler weather. It is worth noting that the peak of summer or winter is somewhat later than the solstice due to the time lag between the maximum or minimum energy input and the total heat content of the land and atmosphere.

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Arctic Circle

Tropic of Cancer Equator

Sun’s rays

Tropic of Capricon

Antarctic Circle

Figure 10.20 The same amount of sunlight is concentrated in a smaller area in summer than in winter, leading to greater heating.

The Sun moves along the ecliptic The apparent motion of the Sun through the celestial sphere can be visualised more easily if we realise that whether it is the Earth or the Sun that is moving makes no difference to the way we see the Sun against the background stars. Imagine yourself and a friend on a merry-go-round at a fair. If you go to the centre of the merry-go-round and watch your friend, who is standing at the outside edge, you will see him or her moving in a circle against the background of the fair ground. But your friend will also see you moving against the background in just the same way. As we watch the Sun from the moving Earth we see the Sun describe a circular path against the stars once a year. (Remember that the Sun’s daily path through our sky is due to the Earth’s daily rotation, not our motion around the Sun. The Sun only moves about 1° through the celestial sphere each day.) 1 Due to the 23 2 ° tilt of the Earth’s axis, the path the Sun follows through 1 the stars is also on a 23 2 ° tilt relative to the celestial equator. The path is referred to as the ecliptic. (The connection between this path and eclipses will become obvious a little later.) The relationship between the ecliptic and the celestial sphere is shown in Figure 10.21. The Sun spends the halfyear from September to March in the southern sky and the other half in the northern sky. The two points where the Sun crosses from one hemisphere to the other are known as the vernal (March) and autumnal (September) equinoxes. Vernal refers to spring in the northern hemisphere. To avoid confusion, southern hemisphere astronomers use the same terms as their northern counterparts. As we saw earlier, the vernal equinox is taken as the zero for right ascension.

PRACTICAL ACTIVITY 47 The Sun in the day sky


349

north celestial pole autumnal equinox

ecliptic solstice Jun Sep

Mar

1

23 2 n

Dec solstice vernal equinox celestial equator south celestial pole

Figure 10.21 The ecliptic is the plane of the Earth’s orbit projected onto the celestial sphere and 1 is, therefore, inclined at 23 2 ° from the celestial equator. The two points where the ecliptic meets the equator are the equinoxes. The points furthest from the equator are the summer and winter solstices. The dates show the position of the Sun as it moves around the ecliptic once each year.

The band of constellations through which the Sun passes each year is referred to as the Zodiac. These are the constellations that astrologers refer to and so have familiar names such as Pisces, Aquarius, and so on (see Figure 10.22). Your ‘astrological sign’ is supposed to be the constellation the Sun is in at your birth; however, you will see that the signs are actually about a month out! The reason for this is that since astrologers developed their art (more than two thousand years ago) the ecliptic has moved. The 1 23 2 ° angle has remained constant, but its orientation relative to the celestial sphere has slowly changed. This has resulted in the equinoxes rotating about 30° around the celestial equator, the equivalent of almost a month of the Sun’s travel along the ecliptic. This slow rotation is called the precession of the equinoxes.

Figure 10.22 The Zodiac is the band of constellations through which the Sun passes on its annual path along the ecliptic.

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Physics in action

The precession of the equinoxes The reason for the precession of the equinoxes is that the Earth is spinning like a top and, like a top, the axis slowly ‘precesses’. If the spinning top is not exactly vertical, the axis of the top describes a circle that gradually gets larger as the top slows down. The reason for this is the pull of gravity on the spinning ‘bulge’ of the top. The Earth precesses for exactly the same reason: the pull of the Sun and Moon’s gravity on the equatorial bulge of the Earth. (The Earth’s diameter is about 40 km larger at the equator than at the

poles.) The precession of the Earth’s axis also results in the gradual apparent movement of the celestial poles. Each celestial pole is describing a slow circle through the stars, which will take 26 000 years to complete. This means that the northern hemisphere will lose its ‘pole star’. Actually, in about 4000 years time, the southern hemisphere will have its own pole star—Omega Carina. (It will be rather faint, however; about the same as the fifth star in the Southern Cross.)

Figure 10.23 Just as gravity causes a top to precess, the gravitational pull of the Sun and Moon on the Earth’s equatorial bulge causes a slow precession of its axis. This means that the celestial poles gradually trace out a circle in the sky and the equinoxes move around the celestial equator. However, it takes 26 000 years to complete the circle.

The Moon—its phases and eclipses Because the plane of the Moon’s orbit around the Earth is only about 5° different from the plane of the Earth’s orbit around the Sun, the Moon appears to follow a very similar path through the celestial sphere to that followed by the Sun; that is, the ecliptic. The big difference is that while the Sun takes a year to complete its trip around the ecliptic, the Moon only takes about a month (a ‘moonth’). The other obvious difference is that in each trip around the ecliptic the Moon goes through its cycle of different phases. Actually, while its trip around the ecliptic takes a little over 27 days, 1 the Moon’s cycle of phases takes about 292 days. The phases of the Moon result from the relative positions of the Earth, the Moon and the Sun (which illuminates the Moon). If the Moon is between the Earth and the Sun we only see the dark side of the Moon; this is called a new moon. When the Earth is between the Moon and the Sun we see the light side of the Moon; this is called a full moon. Figure 10.24 illustrates the relationship between the position of the Moon and its phases. As can be seen from the diagram, there is a very definite relationship between the phase of the Moon and the time of day at which it can be seen. A full moon must rise at sunset and will always be high overhead at midnight, for example. As well as being dark, a new moon is almost in line with the Sun and therefore very hard to see. It is worth noting that a new moon is actually illuminated by ‘earthshine’ and so it is not totally dark.

PRACTICAL ACTIVITY 48 The phases of the Moon


351

third quarter gibbous (waning)

sunlight

crescent (waning) 6 a.m. midday

midnight

N

new moon 6 p.m.

crescent (waxing)

full moon

Earth

gibbous (waxing) first quarter 1

Figure 10.24 The Moon goes through a complete cycle of phases every 292 days. This view

is looking down on the Earth’s North Pole, from which vantage both the 24-h rotation of the Earth on its axis and the 27-day revolution of the Moon in its orbit around the Earth are anticlockwise. The phases are shown as they would be seen from the Earth. The times are shown at appropriate longitudes. It can be seen, for example, that a full moon will rise as the Sun sets and that while a first-quarter moon will be high in the sky at sunset, a last-quarter moon is high in the sky at sunrise. Remember that diagrams such as this cannot be drawn to scale as both the Earth and Moon would hardly be visible. Physics file In thinking about the time of day at which we see the various phases of the Moon we need to picture the horizon as seen from any particular place on the Earth. It is important to remember that Figure 10.24 cannot be drawn to scale as both the Earth and the Moon would be very small. (The Moon is actually about sixty Earth radii from the Earth and one-sixth the size of the Earth.) So in order to picture the horizon and the visible sky at any time, draw a tangent to the Earth’s surface, but move it back to the centre of the Earth and then imagine the Moon as a very small dot rather than the large circle shown. For example, a horizon drawn at 6 p.m. on the diagram would have the Sun setting in the west, the full moon rising in the east, and a first-quarter moon high overhead.

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Detailed studies

Figure 10.24 shows the cycle of the Moon’s phases. This cycle takes about 1 29 2 days; however, during this time the Earth also moves through almost one-twelfth of its orbit around the Sun. This, in terms of the diagram, means that the direction of the Sun’s rays will have changed by an angle of almost 30°. This is why the phase cycle does not correspond to the time taken for one full revolution of the Moon around its orbit. This can be seen more clearly in Figure 10.25. The time for one full revolution (27.32 days on average) around the orbit is known as a sidereal month because this is measured against the background of stars (in much the same way that a sidereal day is measured). In one sidereal month from new moon, the Moon has revolved 360° around its orbit. However, because of the movement of the Earth (and the Moon) around the Sun, the Moon has to revolve almost another 30° in order to line up between the Earth and the Sun again for a new moon. This time—averaging 29.53 days—is referred to as a synodic month.

The dark side of the Moon The dark side and the far side of the Moon are often confused. The dark side is the side presently experiencing night, but the far side is the side we never see, not because it is dark, but because the Moon rotates on its axis in such a way that the same side is always toward us. If you look at any photograph of the Moon (from Earth) you always see the same set of craters and ‘seas’ in the same position. This is not due to a lack of rotation but because it is rotating with a period which is the same as its period of revolution

synodic month

new moon

sidereal month

Earth

Sun new moon

Earth’s orbit 1

Figure 10.25 A sidereal month (273 days) is the time for the Moon to revolve once around

the Earth relative to the background stars. The time between one new moon and the 1 next—a synodic month—is a little longer (292 days). 1

around the Earth (both 27 3 days). This is called synchronous rotation. It is not simply a coincidence that the two are exactly the same, in fact many of the moons in the solar system are in synchronous rotation around their planet. The reason is that when a moon formed, tidal forces pulled it so that it is a slightly bulging sphere rather than being spherical in shape. Gravity then pulls on this bulge and tends to hold the moon in line with its planet.

Eclipses A casual glance at Figure 10.24 would suggest that the Earth would fall into the Moon’s shadow every new moon (a solar eclipse) and the Moon would be in the Earth’s shadow every full moon (a lunar eclipse). But, if drawn to scale, the diagram would show the Earth and Moon as very small dots casting very thin shadows and, more particularly, the Moon’s orbit tilted at about 5° to the plane of the ecliptic (the Earth’s orbit around the Sun). An eclipse only occurs when the Sun, Earth and Moon are almost exactly in line, and as you will realise this does not occur at every full or new moon. Figure 10.26 shows that an eclipse will occur only when a full or new moon coincides with the Moon being on a node; that is, a point where the Moon’s orbit intersects the plane of the ecliptic. (This, indeed, is the reason for the name ‘ecliptic’.) As you can see from the diagram, this can occur about twice a year. Generally speaking, because the Sun is near enough to the line of nodes for a few weeks, at each node there will be both a lunar and a solar eclipse. This, however, does not mean that we can see four eclipses each year! Most eclipses are not total; that is, the Earth or Moon do not go into the umbra (region of complete darkness) of the shadow and so we see only a partial eclipse. Furthermore, the eclipse may occur when we are on the other side of the Earth or, in the case of a solar eclipse, the shadow may not fall on our region of the Earth.


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Physics file Astronomers use the term ‘ecliptic’ in three related, but slightly different ways: • the path of the Sun through the sky each day (diurnal motion) • the path of the Sun on the celestial sphere (annual motion) • the plane of the Earth’s orbit around the Sun. They are just different ways of describing the same basic concept and the context will generally define which meaning is appropriate.

Sun line of nodes New

(a)

(b) New Full

5˚ Full eclipse can occur

line of nodes no eclipse possible

Figure 10.26 Eclipses can only occur when the Sun and a full or new moon are on the line of nodes. (a)This occurs twice each year—six months apart. (b) At most full or new moons, the shadows are not in line.

As the Sun and Moon appear to be about the same angular size in our sky, the umbra of the Moon’s shadow in a solar eclipse only just reaches the Earth and is only a few tens of kilometres wide at most. As a result, only a very small area of the Earth’s surface will experience a total solar eclipse at any one time, although a much larger area will experience a partial eclipse of the Sun. Because the Moon has a slightly elliptical orbit, sometimes the umbra does not reach the Earth at all and the best we see is an annular eclipse. This means that those in a direct line with the Moon and Sun will see a thin ring of the Sun around the dark Moon. penumbra Moon

Earth

Sun umbra total eclipse partial eclipse

Figure 10.27 The umbra of the Moon’s shadow only just reaches the Earth and so very few places experience a total eclipse. A much wider region will experience a partial eclipse. This diagram is not to scale!

A lunar eclipse is much more common because everyone on the night side of the Earth can observe the Moon going into the Earth’s shadow. Also, because of the relative sizes of the two, the umbra of the Earth’s shadow is three or four times larger than that of the Moon and so ‘totality’ can last up to 1 hour 40 minutes if the Moon goes through the centre of the umbra. The Moon is a beautiful sight during a total lunar eclipse. It is not completely dark because, although it is in the Earth’s shadow, some light that has refracted through the Earth’s atmosphere does reach it. Just as light which has travelled through much atmosphere makes a sunset red, so the Moon can appear orange to deep red during a total lunar eclipse.

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Detailed studies

Figure 10.28 As the Moon enters the Earth’s shadow it is illuminated by light refracted through the atmosphere and so appears red or orange.

The wandering planets The stars are extremely well behaved. They follow exactly the same perfectly circular paths every day and every year. The Sun and Moon move easterly through the celestial sphere along the ecliptic at fairly constant rates, repeating their paths every year and every month. By comparison, the planets are quite untidy in their wanderings through the celestial sphere. Indeed the term ‘planet’ means ‘wanderer’ in ancient Greek. Although they stay reasonably close to the ecliptic, their motion along it is quite variable— they speed up and slow down, sometimes even stopping their generally eastward motion and going westward for a while. This ‘backwards’, or retrograde motion was very difficult for the early Greek philosophers to explain. The various attempts to explain it, however, were associated with many of the great discoveries in physics. It is the story of some of these discoveries that is the subject of our next section.

Figure 10.29 The planets normally appear to move eastwards through the stars, but here, Mars is seen to move westwards for two months in 2003 when it was the closest to Earth it had been for nearly 60 000 years.


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10.2 summary The Sun, the Moon and the planets • A year is the time taken for the Earth to revolve once around the Sun. Because of the tilt of the Earth’s axis we experience the different seasons as the Sun’s height (altitude) in our sky changes. • During the year the Sun travels through the stars along a path called the ‘ecliptic’ against the back ground of the Zodiac constellations. • The ecliptic intersects the celestial equator on March 21 (vernal equinox) and September 21 (autumnal equinox). These are the dates when night and day are exactly 12 hours each all over the Earth.

• The solstices are the dates (December 21 and June 21) at which the Sun is furthest south or north from the celestial equator, and hence highest or lowest in our sky. (This is the opposite to the northern hemisphere.) • The phases of the Moon are related to its position with respect to the Sun and Earth. • Eclipses of the Sun or Moon occur when the Sun, Moon and Earth are all in a direct line. • The planets wander through the stars, sometimes undergoing retrograde motion.

10.2 questions The Sun, the Moon and the planets 1 The word ‘solstice’ is from the Latin for ‘solar standstill’. In what sense does the Sun ‘stand still’ at the solstices? 2 Are the dates for the summer and winter solstices the same in the northern and southern hemispheres? What about the dates of the autumn and spring equinoxes?

6 The terms ‘ecliptic’ and ‘Zodiac’ are used to describe the annual path of the Sun through the background stars. What is the difference between these two terms?

3 Is it true that summer is when the Earth is closer to the Sun? What is it that is closer to the Sun in summer?

7 Using the facts that the Moon is actually about sixty Earth radii from the Earth and one-sixth the size of the Earth, draw a scale diagram of the Earth and Moon to get a feel for the situation shown in Figure 10.24.

4 What is the altitude of the Sun at midday in Albury (latitude –36°) at the summer and winter solstices?

8 At what time of day do a full moon and new moon rise? Why is this?

5 Imagine a ‘sunbeam’ of 1 m2 cross-section falling on a flat surface of the Earth (as shown in Figure 10.20). Unless the Sun is directly overhead, the 1 m2 sunbeam is spread over a larger area on the Earth. Use the different altitudes of the Sun in winter and summer in Melbourne to show that the amount of sunlight per square metre is about twice as much at the summer solstice as at the winter solstice.

9 At what time of day would you see: a a full moon high in the north sky? b a first-quarter moon setting? c a third-quarter moon rising?

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10 The Moon is in synchronous rotation with its revolution around the Earth. As a result, Moondwellers would see the Earth in a rather different way from the way we see the Moon move through our sky. Describe the differences.

10.3 Understanding ou

r world

Early models of the universe Most of what we have said so far would make perfectly good sense to the ancient Greek astronomers, such as Pythagoras, who studied the heavens in great detail. Most of them accepted, however, that the Earth was a solid, immovable sphere at the centre of the great celestial sphere, which they believed to be truly heavenly, the realm of the gods. They argued that the Earth was spherical on several grounds. For example, it was well known that the sailor in the ‘crow’s nest’ at the top of the mast could see further than those on the deck of the ship. By the same token, those on shore saw the hull of the ship disappear before the mast and sails. It was also known that as one travelled further north the Sun became lower in the south sky and the North Star rose higher. These facts were easily explained by assuming a spherical Earth. Perhaps their best proof was the curved shadow of the Earth on the Moon during a lunar eclipse. It was very easy to imagine the stars rotating on their huge celestial sphere which moved in a very stately and regular fashion. Inside the celestial sphere were more spheres for the other objects in the sky. The spheres containing the Sun and the Moon rotated regularly, but changed

Physics file

Figure 10.30 Aristotle taught that the Earth was at the centre of the universe. It was surrounded by water, air and fire. Beyond that was the heavenly realm of the Moon, the Sun and the planets which described perfect circles around the Earth. The celestial sphere, which surrounded all of this, was the region of the stars and the home of the gods.

In about 200 BC the Greek philosopher Eratosthenes found a way to estimate the Earth’s size. He knew that at the summer solstice in the town of Syene (now under the present day Aswan Dam in Egypt) the Sun was directly overhead—it cast no shadow of a vertical pole. At the same time in Alexandria, 5000 stades (a distance measure) to the north, a vertical pole did cast a shadow at midday. Measurements of this shadow and the height of the pole, enabled Eratosthenes to calculate that the Sun was 7° away from being directly overhead. He therefore concluded that the distance between the towns represented 7/360 or close to 1/50th of the distance around the Earth’s circumference. Thus the total circumference would be 50 × 5000 or 250 000 stades, which in today’s units is roughly 42 000 km. Not a bad estimate when you compare it with today’s value of close to 40 000 km!

Sun’s rays

Alexandria 7° Centre of Earth

shadow 7° es 5000 stad ow no shad

Figure 10.31 As the Moon progresses into the Earth’s shadow during a lunar eclipse, the shadow always appears curved. The Greeks realised that this meant that the Earth must be spherical.

Syene

Figure 10.32 Eratosthenes measured the Earth’s circumference in 200 bc by noting that while the midday Sun was overhead in Syene it cast a 7° shadow in Alexandria, 5000 stades to the north.


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their orientation somewhat over the period of a year or a month, albeit in a very predictable fashion. The planets were something of a problem however. How could their rather irregular motion be reconciled with their heavenly and, therefore, ‘perfect’ nature? Circular motion, such as that of the stars, was seen as perfect and so it was thought that in some way the motion of the planets must be made up of some combination of different circular motions. In the second century bc, Hipparchus suggested that the motion of the planets could be explained by a system of circles upon circles. Each planet was thought to move in a small circle called an epicycle, the centre of which moved in a larger circle called the deferent, which was centred on the Earth. In the second century ad, Ptolemy, the last of the ancient astronomers, improved the accuracy of Hipparchus’s system by developing a method for locating the centre of the deferent. He located the centre a little away from the Earth, and had the planet move around the circle at slightly varying speeds. Although a rather complicated system, it was based on the ‘perfect circle’ and was able to predict the future positions of the planets with quite a high degree of accuracy. In the third century bc Aristarchus had suggested a model based on the Earth rotating around the Sun; however, the Ptolemaic system was seen as philosophically more satisfactory and thus lasted for over a millennium.

deferent

epicycle Earth

planet

Figure 10.33 Ptolemy’s system of planetary motion based on epicycles, the centre of which moved on a circle called the deferent centred near, but not on, the Earth. When on the inside of the deferent the planet could exhibit retrograde motion as shown.

The Copernican revolution In the Middle Ages natural philosophers began to question the Ptolemaic system on the grounds that although it worked well, it was not based on any underlying principle that could be used to work out the motion of any planet. The details had to be worked out to suit the observed motion of each planet separately. By this time scientists, or ‘natural philosophers’ as they were referred to then, had begun to realise that there seemed to be a principle in nature that favoured simpler explanations over complex ones.

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Aristarchus had actually suggested a heliocentric (Sun-centred) model of the world back in ancient Greek times. He reasoned that it provided a simpler explanation for retrograde motion. However, to most people the idea of the Earth being in motion was an idea that seemed far harder to accept. In the early 16th century Nicolaus Copernicus, a gifted Polish mathematician, suggested that a Sun-centred model had some advantages over the Ptolemaic system. Not only did the heliocentric model explain retrograde motion more simply, it also explained why, for example, Mercury and Venus never appeared in the midnight sky while the other planets did. The heliocentric explanation for retrograde motion can be seen in Figure 10.34. In the Copernican system the further away the planet is from the Sun the more slowly it moves around its orbit. As a result, the Earth ‘overtakes’ the outer planets at some stage in its orbit. As this happens, the outer planet will appear to travel backwards relative to the stars in the far background. 9

8

9 7

8

6 5

Sun

4

Earth 1 Mars

2

3 1

2

9 7 6 5

4 3

Physics file The 14th century English philosopher William of Occam first expressed the principle that nature seems to favour simple, elegant principles over complex ones. This became known as Occam’s razor in reference to the fact that he suggested trying to shave unnecessary details from any explanation for a natural phenomenon. Ever since, scientists have found it a useful, although not infallible principle. Albert Einstein actually based much of his thinking on Occam’s razor, continually searching for simpler explanations for phenomena to do with light. Although the mathematics of relativity can be complex, its underlying principles are very simple.

8 4 3 5 7 6 2

1

Figure 10.34 Using the Copernican explanation of retrograde motion, Mars appears to move backwards for a couple of months each year when the Earth ‘overtakes’ the more slowly moving planet.

It was well known that Mercury and Venus never moved more than a certain angular distance from the Sun. Venus is often referred to as the morning or evening ‘star’ because it is only seen within about three hours of sunrise or sunset. The Ptolemaic system had no explanation for this, but the Copernican heliocentric system had a very simple explanation: these two planets are closer to the Sun than is the Earth. They are said to be inferior planets. This was not a value judgement, simply a reference to the fact that the radius of their orbits was less than that of the Earth. All the other planets can, at some stage of their cycle, be seen high in the sky at midnight. This means that they must be further from the Sun than the Earth is, so they are superior planets. This is illustrated in Figure 10.35. Copernicus was able to use trigonometry to calculate the radius of the orbits of the planets in terms of the radius of the Earth’s orbit; that is, the distance to the Sun. This was simple for the inferior planets, as can be seen in Figure 10.35. At its maximum elongation from the Sun (46°) we can see that: R sin 46° = V so RV = 0.72 RE RE Copernicus had no good way of determining the distance of the Earth from the Sun and so simply expressed the radii of the planet orbits in terms of astronomical units, or AU for short. 1 AU is the average distance of the Earth from the Sun. We now know it to be 150 million km. A similar,


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t rbi t orbi th ar bit r so nu inferior conjunction

Ve

E

M

so ar

46° opposition

Earth

Sun greatest elongation

conjunction

superior conjunction

Figure 10.35 Inferior and superior orbits are illustrated by the orbits of Venus and Mars. Venus is never seen to be more than 46° from the Sun, but Mars can be seen in the midnight sky. When the inferior planets are in a line with the Sun and Earth they are said to be at inferior (same side as the Earth) or superior (behind the Sun) conjunctions. The equivalent points for the superior planets are called simply ‘conjunction’ when the planet is behind the Sun (and so not visible at night) and ‘opposition’ when the planet will be seen high in the sky at midnight.

Table 10.2 Average distances of the planets from the Sun in AU Planet

Copernicus’s value

Modern value

Mercury

0.38

0.39

Venus

0.72

0.72

Earth

1.00

1.00

Mars

1.52

1.52

Jupiter

5.22

5.20

Saturn

9.07

9.54

Uranus

–

19.19

Neptune

–

30.06

Pluto*

–

39.53

* In 2006 the International Astronomical Union decided that Pluto should be downgraded from planet to dwarf planet. See the Physics file on page 379.

PRACTICAL ACTIVITY 49 Computer simulation of the night sky

but somewhat more involved, process enabled Copernicus to determine the distances to the superior planets. His values were quite close to the modern-day values, as shown in Table 10.2. Copernicus did not know about the three outer planets; their discovery had to await the invention of the telescope. While the Copernican system had a philosophical advantage in that it seemed conceptually simpler to some astronomers, it was no better at predicting astronomical phenomena than was Ptolemy’s. In fact Copernicus initially found that his system gave worse results, and he had to resort to adding small epicycles to his orbits in order to achieve similar accuracy. His epicycles were not nearly as large as Ptolemy’s however, and were not needed to produce the retrograde motion. They simply had the effect of speeding up and slowing down the planet a little to make the orbits fit the known data. Perhaps the most important aspect of Copernicus’s achievement was that he had made a break from the past. He showed that despite the success of the Ptolemaic system it might not be the best way to understand the way the world works. In fact Copernicus started a revolution. The publication in 1543 of his De revolutionibus orbium coelestium (‘On the revolutions of the celestial spheres’) is now generally seen as just the start of the so-called ‘Copernican Revolution’ and is regarded as one of the great turning points, not just of science, but of history. It was a revolution that was to include some of the great names in physics, such as Galileo and Kepler. It also pitted the new science against the political authorities of the day, including the Church.

Observation and theory Tycho Brahe was born in 1546, just a few years after Copernicus’s death in 1543. He grew up in a wealthy Danish family and became a nobleman. His wealth allowed him to pursue his passion for astronomy. In particular he wanted to catalogue the stars and the motion of the planets. In 1572 he

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experienced a truly amazing sight: a new bright star, even brighter than Venus, suddenly appeared in the constellation of Cassiopeia. Today we recognise that event as a supernova, the explosion and death of a massive star. At that time, however, the celestial realm was seen as permanent and unchangeable, the home of the Gods, and so most people argued that the new ‘star’ must have been something much closer to Earth. Brahe was determined to investigate. He figured that if the new object really was closer than the ‘real’ stars then it should be possible to see it move against the background stars, as do the planets. This is known as parallax and is a familiar phenomenon in everyday experience. If you look out the window at something in the distance and then move your head from side to side, the distant object appears to move with you, relative to the closer objects, such as the window frame, which moves against you. You may have seen a rising Moon skipping along the treetops as you drive along a country road at night. The Moon is not moving at all (well not much), it is your motion that makes the trees appear to move backward relative to the Moon (or anything that is further away than the trees). With this in mind Brahe observed the new star very carefully during the course of a single night and during the course of the next months as it faded from view again. He found no parallax between the star and the celestial sphere and therefore concluded that the star was indeed very far from the Earth; perhaps in the ‘celestial realm’ itself. Cracks were beginning to appear in the 2000-year-old view of a ‘permanent and unchangeable realm of the Gods’. As a result of his careful work on the new star of 1572, Brahe was awarded a grant by the King of Denmark to build two state-of-the-art observatories. There were no telescopes yet, but there were very carefully built instruments with which to measure the positions of the stars and planets. Brahe decided that with careful observation he could test the Copernican theory of a moving Earth. If the Earth was indeed circling the Sun, surely we should notice some parallax movement in the positions of the stars, he reasoned. There was plenty of parallax movement between the planets and the stars, but this was the reason for believing that the planets were indeed much closer to the Earth than were the stars. Earth in December

Sun

Figure 10.36 This contemporary drawing shows Brahe with some of his apparatus and assistants at his Uraniborg observatory.

to distant star

closer star

to same distant star Earth in June

Figure 10.38 Brahe argued that if the Earth was revolving around the Sun, changing parallax should alter our view of the stars through the year. Because he found no parallax he concluded that the Earth was at rest.

Figure 10.37 Brahe’s ‘Great Armillary’, one of the instruments with which he measured the positions of the stars and planets.


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After years of painstaking observation Brahe found no evidence of any star parallax and decided that Copernicus must have been wrong about the Earth moving. He did, however, decide that Copernicus seemed to be correct in suggesting that the other planets were revolving around the Sun rather than the Earth. It was, he said, that the Sun, and all the planets with it, were revolving around the Earth. This was a perfectly reasonable conclusion based on his observations. Remember our merry-go-round observations: just by watching the motion against the background it is not possible to tell which of two people rotating around each other is ‘really’ moving. So it is with the Earth and the Sun.

Figure 10.39 In this colourful 1660 drawing of Tycho Brahe’s system the Earth is in the centre with the Moon and the Sun circling around it, but all the planets are in orbit around the Sun. There was no way to tell the difference between this system and the Sun-centred system by observation alone.

Brahe’s legacy to astronomy was the huge amount of data he had collected over the many years of his endeavours. The accuracy of his work was right at the limit of what can be achieved with observation by the naked eye and is one of the most remarkable feats of scientific observation. The real analysis of this data, however, was left to an assistant he employed in 1600, just one year before he died. That assistant was the German mathematician and keen astronomer Johannes Kepler. Brahe and Kepler were a great team. Brahe was the painstaking observer who recorded masses of data faithfully and accurately. Kepler loved analysing that data and trying to make sense of it. Although they only worked together for the year before Brahe’s death, Kepler spent much of the rest of his own lifetime analysing Brahe’s data. In the meantime another great observer was using the newly invented telescope to observe the heavens. He was Galileo Galilei.

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Copernicus and Brahe had opened cracks in the supposedly unchanging and perfect world of the heavens. Kepler went further by daring to question the notion that the orbits of the planets must be combinations of perfect circles. It is said that he could almost fit the orbit of Mars to a circle, but there remained a very small discrepancy with the data. However, so confident was he in the accuracy of Brahe’s data that he decided the orbit could not be circular and so went on to investigate other possibilities. He finally decided that the best shape to describe the orbits of the planets was an ellipse. An ellipse can be drawn with a pencil in a loop of string around two drawing pins as shown in Figure 10.40. The positions of the two pins are the foci of the ellipse. Clearly, if the pins are moved together the ellipse becomes a circle, and so a circle is actually a special type of ellipse. In reality most of the planet orbits look almost circular, but Kepler found that Brahe’s data could only be satisfied if they were actually elliptical. We can get an idea of how close the Earth’s orbit is to circular from the fact that its closest distance to the Sun (in January) is 149 million km while the furthest (in July) is 151 million km.

F1

F2

Figure 10.40 An ellipse can be drawn with a loop of string and two pins. The pins are at the foci of the ellipse.

K…PL…R’S FIRST LAW: The orbit of a planet is an ellipse with the Sun at one focus. Kepler found that Brahe’s data fitted exactly if he assumed the orbits of the planets to be elliptical with the Sun at one focus. Each planet did not go around its ellipse at a steady speed however: it sped up as it came nearer the Sun and slowed down as it went further away. Again, Kepler found a simple pattern in this changing speed. If a radius was drawn from the planet to the Sun, this line swept out area at a constant rate. In other words, if it swept out an area A in 30 days in January, it swept out an equal area in 30 days in July. For this to occur, the planet has to move fastest when closest to the Sun. In fact, this means that its speed along the orbit is inversely proportional to its distance from the Sun.

K…PL…R’S S…COND LAW: The radius from the Sun to a planet sweeps out equal areas in equal times.

B A C Sun aphelion

perihelion

D

Figure 10.41 Kepler’s second law says that a planet sweeps out equal areas in equal times. This means that it is moving faster at perihelion (closest to the Sun) and slowest at aphelion (furthest from the Sun). This ellipse is exaggerated; the orbits of most planet are quite close to circular.


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Physics file As the orbits are not circular, the radius of the planet’s orbit varies a little. In fact the radius that is used in Kepler’s third law is the semi-major axis of the ellipse. It is half the long (major) axis of the ellipse. It is close to, but not the same as, an average radius, and so is often written as ‘a’ rather than R.

Having found the way to describe each planet’s orbit, Kepler went on to look at the relationships between the planets. He knew that the further planets took longer to complete their orbits; that is, they had a longer ‘year’. In fact a planet twice as far from the Sun as the Earth took more than twice the time of the Earth to complete its orbit, which meant that its actual speed along its orbit must be less. After considerable work (and nine years after the first two laws) Kepler discovered the relationship he was looking for.

K…PL…R’S THIRD LAW: The square of the period of the planet (T2) is proportional to the cube of the radius of the orbit (R3). That is: R3 = K where K is Kepler’s constant. T2 Table 10.3 illustrates Kepler’s third law. If the period is measured in earth years and the (average) radius of orbit is measured in AU, the value of K for the Earth must be 1.00 (as both R and T are 1 by definition). As you can see from the table, the values of R3 and T2 are equal (within experimental limits), making their ratio—Kepler’s constant—equal to 1.00 (when measured in AU and years) for all planets.

Table 10.3 Kepler’s third law for all planets using modern values Planet

T (Earth years)

R (AU)

T2

R3

Mercury

0.24

0.39

0.06

0.06

Venus

0.61

0.72

0.37

0.37

Earth

1.00

1.00

1.00

1.00

Mars

1.88

1.52

3.53

3.51

Jupiter

11.86

5.20

140.7

140.6

Saturn

29.46

9.54

Uranus

84.01

19.19

7 058

7 067

Neptune

64.79

30.06

27 160

27 160

Pluto

248.5

39.53

61 770

61 770

867.9

868.3

Not only did Kepler find a more accurate way of predicting the future positions of planets (something very important to a society still steeped in astrology) but, unlike Ptolemy, he found general patterns that all the planets obeyed. This suggested that some underlying physical principles must govern the motion of the planets. Kepler hypothesised a fairly unsatisfactory system based on magnetic forces, but it was left to two of the most well-known names in physics to discover those principles: Galileo and Newton.

Galileo and the telescope Galileo Galilei was born in Pisa, Italy, in 1564, about 20 years after Copernicus’s death and while Tycho Brahe was still a teenager. He became fascinated with mathematics and the possibility of using it to describe the workings of the world. In 1589 he was appointed a professor of mathematics at Pisa University where he developed many of his most important ideas about the motion of falling objects (maybe even doing experiments from the famous Leaning Tower). In 1609, however, he was

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introduced to a fascinating piece of new technology—the telescope. As often happens in science, new technology can lead to new discoveries and hence new theories. The next few decades were to be one of the prime examples of this. The principle of the telescope was developed by Dutch opticians, but Galileo appears to have been the first to turn it toward the heavens. He also refined and improved its optics. The use of the telescope to observe the sky was arguably one of the greatest single advances in all of science. It had a profound effect, not just in natural philosophy but across the whole spectrum of human intellectual endeavour. Galileo’s observations with the telescope caused a lot of trouble for the Church, which was the dominant political institution of that time. So what did Galileo see that caused such upheaval? There were four main discoveries and we shall look at each in turn.

Figure 10.42 Galileo tried to convince the Church fathers of his new discoveries.

Physics file One of Galileo’s first discoveries with his telescope was that the Milky Way was actually composed of myriads of stars. Its apparent cloudiness was due to the fact that our eyes could not resolve the individual stars. This discovery, however, did not seriously threaten the established view.

Figure 10.43 Galileo sketched the mountains on the Moon as he saw them.

Mountains on the Moon Not surprisingly, one of the first objects Galileo turned his telescope to was the Moon. What he saw convinced him that the Moon was not a ‘heavenly’ object, but an Earth-like object. It seemed, to him, to have mountains and seas, just like the Earth. He even made rough measurements of the heights of the mountains from the shadows they cast.

Spots on the Sun Even more devastating to the ‘heavenly perfection’ theories were spots on the Sun. Because of its importance to our lives, the Sun was regarded as a particularly perfect object, and so when Galileo discovered sunspots many people could not accept the thought that the Sun had blemishes and concluded that it was a problem with his telescope. Galileo persisted however, and found that the sunspots rotated around the Sun, gradually changed in size and disappeared. Soon the evidence was undeniable.


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Physics file Warning: Never look at the Sun through a telescope or binoculars! Looking at the Sun with the naked eye is dangerous; imagine how much more dangerous it is with the concentrated light in a telescope. One way to view the Sun is by putting a screen behind the telescope (or binoculars) as shown in Figure 10.44; however even this can be dangerous with a larger diameter telescope, as the lens inside the telescope may become too hot. Carefully read the information that came with the telescope before attempting to use it in this way; and don’t use the finder scope either! It may be possible to purchase a filter to put over the objective lens before projecting the image onto a screen, but do not be tempted to look through the telescope even with a filter.

Figure 10.44 Some Jesuit priests—contemporaries of Galileo—observed sunspots using a telescope to project the image onto a white surface. This picture shows how to use a telescope for this purpose.

Figure 10.45 These two drawings by Galileo show the changes in sunspots observed on two consecutive days in 1612.

Moons around Jupiter The old view of the cosmos held that everything revolved around the Earth, so when Galileo saw tiny ‘stars’ revolving around Jupiter it was yet another threat to the established ideas. After careful observation over weeks, Galileo was convinced that Jupiter had four Moons that revolved at fixed distances and times that increased with the distance from Jupiter.

The phases of Venus

Figure 10.46 Galileo’s record of the moons of Jupiter as he watched them over several weeks.

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Detailed studies

Perhaps the most convincing of Galileo’s observations were his studies of the changes in the appearance of Venus through the year. He found that Venus had phases like the Moon, but, more importantly, the apparent size of the planet changed with the phases. The phases themselves were not particularly surprising as it had been accepted that the planets were illuminated by the Sun. However, the changing size in conjunction with the phases could only be explained in terms of its motion around the Sun,

inside the orbit of the Earth. Venus was only ‘full’ when on the opposite side of its orbit, furthest from the Earth, and the thinner the crescent the larger the diameter of Venus appeared to be. These observations were completely at odds with the Ptolemaic model in which Venus was always in between the Sun and the Earth, and so could never be seen as more than a crescent. Venus

Sun

Earth

Figure 10.47 Venus appears small when full and much larger when crescent. Galileo said this must mean that Venus is going around the Sun, not around the Earth.

Unfortunately for Galileo, none of these observations could really distinguish between the Copernican heliocentric system and Tycho Brahe’s geocentric system, as both would lead to exactly the same observations— just as the view from the centre of the merry-go-round is the same as that from the edge. Nevertheless Galileo was convinced that the heliocentric system simply made more sense. Why would all the other planets go around the Sun while the Earth stayed motionless? By insisting on this interpretation Galileo got himself into a lot of strife with the Church, which had taken on most of Aristotle’s ideas and believed that the Bible said that the Earth was still. Because of his fame and popularity, the Church did not want to deal with Galileo too harshly. In 1616 they told him to stop teaching the Copernican model and when he did not, eventually confined him to house arrest in 1633. But this was mild in comparison to the fate of Giordano Bruno who was burnt at the stake in 1600 for claiming that the Earth moved and that the stars were distant suns. Without doubt, Galileo’s discoveries were a huge and important step in the transition from a society based on the authority of the Church to one in which exploration, experimentation and discovery were to become the accepted ways of finding truth, whether in matters of the mind or nature. In the same year that Galileo died in Italy (1642) Isaac Newton was born in England. It was Newton who was eventually to show that the centre of the merry-go-round is different from the outside, that the Earth was on the outside and the Sun at the centre.


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10.3 summary Understanding our world • The early Greek astronomers knew that the Earth was round, but they believed that it was fixed and in the centre of the universe, and that everything else revolved around it. • Ptolemy pictured the planets as undergoing epi cyclic motion in which their orbits were smaller circles centred on a larger circle. This explained the retrograde motion of the planets quite well. • Copernicus put forward an alternative Sun-centred (heliocentric) system based on the assumption that the Sun was at rest and the planets, including the Earth, circled it.

• Kepler, after analysing Tycho Brahe’s data, showed that the orbits of the planets were ellipses with the Sun at one focus. He also found that the planets swept out equal areas over equal times, and that the square of the period of the planet was proportional to the cube of its radius of orbit. • Galileo used a telescope to examine the heavens and found mountains on the Moon, spots on the Sun, moons around Jupiter, and that Venus changed size with its phases. These were contrary to the old views of a truly perfect heavenly realm and supported the Copernican heliocentric view.

10.3 questions Understanding our world 1 You have arrived on a new planet and are asked to find its diameter. You find that at one location the planet’s star is directly overhead at noon, but at the same time in a location 200 km south it makes an angle of 10°. What is the diameter of the planet? 2 What was Copernicus’s explanation for the fact that Venus is always an evening or morning ‘star’, and not a midnight one? What do you think Ptolemy’s explanation was? 3 In astronomical terms a superior planet is one that: A is larger than the Earth. B is heavier than the Earth. C is further from the Sun than the Earth. D can support life. 4 a If a planet is at ‘conjunction’ where will it be seen in the sky? Why? b If a planet can be seen high in the north sky at midnight, where is it in its orbit in relation to the Earth in its orbit? Can all planets be seen in this position? 5 As you walk along a track you notice that a certain large tree off to the side at right angles to the track is in line with a distant mountain. Another 50 m down the track the tree makes an angle of 12° with the distant mountain. How far away was the tree from the track originally? 6 What was Tycho Brahe’s main argument for retaining the geocentric model? What was the difference between his system and that of Ptolemy?

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7 Kepler discovered that the orbits of the planets were ellipses rather than circles; however, the ellipses were very close to circles. The length (long axis) and width (short axis) of the orbits for three planets are given below in AU. Planet

Length

Width

Mercury

0.387

0.379

Earth

1.000

0.999

Mars

1.524

1.517

a What is the length expressed as a percentage of the width in each case? b If you drew a circle of 10 cm radius to represent the short axis of the orbit, what would be the difference between this and the correct orbit? 8 a If an asteroid was found at twice the distance from the Sun as the Earth is—that is at 2.0 AU—what would be the length of its year (in terms of an Earth year)? b If an asteroid was found which had a period of 8 years, how far would it be from the Sun? 9 What was the significance of Galileo’s discovery of mountains on the Moon and spots on the Sun? 10 The phases of Venus had not been seen before the advent of the telescope; however, had it been known that Venus had phases it would not have caused a stir. What was it about Galileo’s observations of Venus that caused concern among the authorities?

10.4 The telescope: fr

om Galileo to

Hubble

Many of the turning points in the history of humankind have been due to great leaps in our ability to use new technology. Imagine the difference the ability to light and use fire made to the lives of early humans. Recent history has been profoundly influenced by the discovery and use of radio waves and the ability to control tiny electric currents, and hence information, with microscopic transistors—to name just two obvious examples. The telescope was another piece of technology that had a profound effect on the development of human culture. Galileo’s use of the telescope was just the start of the story. A telescope equivalent to his can now be bought for a few tens of dollars. At the other end of the spectrum there is a huge array of modern devices costing billions of dollars. These new telescopes have produced spectacular images and, more importantly, have enabled us to see not only further into our universe than ever before, but also right back in time, almost to the origin of the universe. As a result, our knowledge and understanding of the universe has exploded in the last century. In this section we look at the development and evolution of the telescope.

Galileo’s telescopes

Figure 10.48 One of the many spectacular images from the Hubble Space Telescope. This is the Cone Nebula. It is a huge pillar of gas and dust. The entire nebula is 7 light-years long and resides 2500 light-years away in the constellation Monoceros.

The basic principle of Galileo’s original telescope is still the basis of a simple refracting telescope. A large convex lens, the objective, with a long focal length is used to form a real image. This image is then observed with a smaller, convex lens with a short focal length, the eyepiece. The longer the focal length of the objective lens, the larger the image, and the shorter the focal length of the eyepiece lens the greater its magnifying power. In fact the magnification of a telescope is given by the ratio of the two focal lengths. The reason for this can be seen in Figure 10.49. objective

eyepiece fo

θo

fe

F

θo

h

fe

θe

real image virtual image

Figure 10.49 In a simple astronomical telescope the objective produces a real image at a distance of fo behind the lens. This image is magnified by the eyepiece lens which is placed a distance fe further back. This means that the observer sees a virtual image that appears to be at infinity.

When dealing with astronomical images, the magnification is defined as the ratio of the angle the object appears to subtend when seen through the telescope to the angle as seen by the naked eye. For example, the Moon 1 makes an angle of about 2 ° to the eye, so if it appears to be 10° in the telescope the magnification is 20 times (often written 20×). In the diagram the angles are greatly exaggerated for clarity. In fact they are really quite


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small, and that enables us to make the simplifying assumption that the ratio of the angles is equal to the ratio of the tangents of the angles. As the object in this case is effectively at an infinite distance, the image will be in the focal plane of the lens; that is, at a distance of fo behind the objective. As can be seen in the diagram: h h tanθo = and tanθe = fo fe where h is the height of the image. Making the assumption mentioned above, we can see that the ratio of the angles is equal to the inverse ratio of the focal lengths.

MAGNIFICATION OF A T…L…SCOP… =

qe fo = qo fe

Most telescopes have an objective lens of fixed focal length, often around 1 metre, but may have several eyepiece lenses of different focal lengths in order to provide different magnification for various purposes. For example, in looking at the Moon or planets, which are quite bright, we will see more detail with high magnification. However, in looking at groups of stars, or nebulae, we may want less magnification in order to obtain a wider field of view and a brighter image.

Worked example 10.4A One of Galileo’s original telescopes had an objective with a diameter of 16 mm and a focal length of 960 mm. The convex eyepiece had a focal length of 48 mm. a What was the magnification of the telescope? b The angle subtended by the three stars in Orion’s belt is 3°. What angle would Galileo have seen them subtend through his telescope? Do you think he would have been able to see them all together?

Solution a The magnification is the inverse ratio of the focal lengths: 960 = 20× 48 b Theoretically Galileo would have seen them subtend a total angle of 20 × 3° = 60°. But this would require a very wide field of view and, as can be seen from the photograph of his telescope, his telescope was very long and narrow. The objective diameter, in conjunction with the focal length, suggests a field of view less than 1° (from tan θ = 16/960). As the stars are 3° apart, he would only have been able to see one at a time.

Figure 10.50 Two of Galileo’s telescopes. The larger telescope has an objective lens of diameter 26 mm and focal length 1.33 m with a concave eyepiece giving a magnification of 14×. The smaller telescope has an objective lens of diameter 16 mm and focal length 0.96 m with a concave eyepiece giving a magnification of 20×.

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Detailed studies

Curiously enough, this type of telescope is called a Keplerian telescope as it was Kepler who first described the optics of the system. The Galilean telescope was a modification, by Galileo, of this simple telescope. He found that by using a concave lens (a diverging lens) for the eyepiece he could make the telescope shorter. It also produced an upright image; this was important for terrestrial use, but not for astronomy. Figure 10.51 shows the basic optical principle of a Galilean telescope. Although it looks somewhat different, the geometry of the ray diagram is actually similar to that for the Keplerian telescope and so the magnification is still given by the same formula.

eyepiece

objective

fe

virtual image

F (both lens)

fo

Figure 10.51 The Galilean telescope is a modification of the simple telescope shown in Figure 10.49. Instead of a convex eyepiece lens which magnifies the real image formed by the objective, the concave eyepiece intercepts the rays before they focus and produces an upright virtual image.

While the magnification of a telescope is important, the more crucial factor is actually its light-gathering power. The biggest difficulty in seeing detail in the heavens is the faint light that reaches Earth from the vast distances involved. It is quite easy to increase the magnification with a more powerful eyepiece, but it is much harder to increase the brightness of the image. There is little point in having a highly magnified image if it is so dull no detail can be seen! The light-gathering power is purely dependent on the diameter of the lens—the bigger the diameter of the lens, the more light it focuses onto the image. As the amount of light captured by the lens depends on its area, the light-gathering power is proportional to the square of the diameter (as the area is equal to πr2). Thus an objective lens that is twice as wide as another will have four times the light-gathering power.

Worked example 10.4B Figure 10.50 shows two of Galileo’s telescopes. Using the information given in the caption, which had the greater light-gathering power, and by how much?

Solution The only figure that matters is the diameter of the objective lens. The larger telescope had an objective lens of 26 mm diameter, compared with 16 mm for the smaller telescope, and thus had the greater light-gathering power. In order to compare the two, we find the ratio of the squares of the diameters: 26 2 = 2.6 16 Hence, the larger telescope had more than two and a half times the light-gathering power of the smaller telescope. All refracting telescopes suffer from two significant problems. The first is that it is difficult to make the large diameter lenses needed to gather more light. It is the curvature of a lens that determines the focal length, and in order to retain the same curvature a larger lens will become very thick at the centre. This introduces distortion due to the fact that the light going through the centre is refracted at two points (on either side of the lens) which become further apart in a thicker lens. In our ray diagrams for lenses


371

convex lens focal point for blue light

focal point for red light

two lenses

we have always assumed the lens is thin and the rays only bend in the centre of the lens. This assumption does not hold true for thicker lenses. The second problem for a refracting telescope is that the refractive index of light depends slightly on the colour of the light. The result is that the blue light in the image is focused at a slightly different place to the red light. This effect is called chromatic aberration. It can be significantly corrected by the use of an additional lens made of glass that has different dispersion characteristics, but it is hard to eliminate it entirely. These problems are just two of the many difficulties in obtaining a clear, undistorted image from a large objective lens. Some master opticians of the 19th century succeeded in creating a telescope with a 40-inch (102 cm) lens. It has a huge focal length of 19.5 m. This was the largest lens ever built and it was installed at the Yerkes Observatory near Chicago in 1897. It is still in use today.

Newton’s reflecting telescope—and others focal points for both colours

Figure 10.52 A convex lens always focuses red light a little behind blue light. This leads to coloured edges around the images called chromatic aberration. The problem can be largely overcome by using a concave lens of a different type of glass in conjunction with the convex one.

Because of the difficulties in making large lenses, almost all modern telescopes have replaced the objective lens with a curved (parabolic) mirror, called the objective or primary mirror. The great advantage of a reflecting telescope is that light does not travel through the glass, it only reflects from the silver or aluminium coated surface. This means that any defects within the glass do not affect the light and there is no chromatic aberration as all colours reflect in exactly the same way. Because the mirror can be supported from the back it is also easier to make very large rigid mirrors—this is the key to better light-gathering power. One difficulty with the use of a parabolic mirror was viewing the image, which was in front of the mirror, without blocking the light. It was Sir Isaac Newton who solved this problem and first built an effective reflecting telescope. He placed a small plane secondary mirror at 45° to the light path just in front of the principal focus. The image was then viewed through an eyepiece lens mounted on the side of the supporting tube as shown in Figure 10.53. The ray optics of a curved mirror are virtually the same as for a lens, the difference simply being that the image is formed on the other side. The magnification of a reflecting telescope is also given by the ratio of the focal

Newtonian

Cassegrain

Figure 10.53 The Newtonian reflector has a small secondary mirror which reflects the light out at the side of the telescope. The Cassegrain reflects the light back down the tube and out through a hole in the main mirror, and thus it is shorter for a given focal length.

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Detailed studies

lengths of the primary mirror and that of the eyepiece used to view the image. Similarly, the light-gathering power depends on the diameter of the mirror, although the small secondary mirror will block a little light. (The secondary mirror does not cause a ‘hole’ in the image; can you see why?) As with a refracting telescope, the long focal length of the objective mirror means that the supporting structure must also be long and unwieldy. To avoid this problem other designs have been invented. The Cassegrain telescope uses a secondary mirror that reflects the light straight back down the tube and out through a hole in the centre of the primary mirror. This almost halves the length of the telescope (for a given focal length) and makes it much more portable and stable when mounted on a tripod. For this reason many amateur astronomers use this type of telescope. Large observatory-based telescopes use other systems as well as these. In very large telescopes the observer may actually sit at the prime focus right inside the tube! More particularly, in modern telescopes various electronic devices will be positioned at the prime focus to collect the light and send the image to computers and other devices for analysis. In other telescopes the light may travel a more complex route to a focus outside the tube in such a way that the position of the image remains stationary while the telescope moves. This enables heavy or bulky equipment to be used to analyse the image.

(a)

Follow that star There is one difficulty that all telescopes have to contend with—the stars move through the night. Unless one is content with a quick glimpse of the celestial object it will move out of the field of view of the telescope in a matter of minutes, or less. Most astronomers need to keep their telescope focused on one object, or field of stars, for some time while they photograph it or analyse it with their spectrometers and other devices. For this reason telescopes must be mounted in such a way as to ‘follow that star’ (or planet or whatever). The simplest type of mount for a telescope is the altitude–azimuth system. The altitude of a star is its angle above the horizon (0° being on the horizon, 90° straight overhead). To change altitude the telescope needs to rotate in a vertical plane; that is, around a horizontal axis. In Figure 10.54a this axis is the one holding the telescope between the two forks. Azimuth is effectively the compass bearing (north being 0°, east 90° and so on) and so to change azimuth the telescope needs to rotate in a horizontal plane; that is, around a vertical axis. The whole fork structure holding the telescope in Figure 10.54a rotates on the base plate mounted on the tripod. The problem, of course, is that as the stars move across the sky, both their azimuth and altitude change at varying rates and so both controls need to be adjusted continuously to follow a star. The telescope shown in Figure 10.54a has two computer-controlled motors that can accomplish this task automatically. Most modern large professional telescopes use a similar mounting system. A more common mounting system for amateur telescopes is the equatorial mount, as shown in Figure 10.54b. This system is based on the fact that, as the Earth rotates, the stars move along lines of constant declination. You will remember that the positions of stars are given in terms of their declination and right ascension. As the Earth rotates, it is the lines of right ascension that move across the sky; the lines of declination remain fixed. The equatorial

(b)

Figure 10.54 These two telescopes illustrate the two ways of mounting a telescope. (a) This modern Cassegrain telescope has an altitude– azimuth mount with two computer-controlled motors that enable it to follow the celestial sphere. (b) This telescope has an equatorial mount and is rotated about a polar axis—which points to a celestial pole (north or south)—and a declination axis, which is at right angles to the polar axis. Can you see which axis is which in this photo?


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PRACTICAL ACTIVITY 50 Night sky exercises in astronomy

mount is set up so that one axis—the polar axis—points directly to the south celestial pole (or the north celestial pole for northern observers). As the telescope moves around this axis it will maintain a constant declination while sweeping through the right ascension coordinates. As it is moved around the declination axis (at right angles to the polar axis) the telescope will move through the declination coordinates while maintaining a constant right ascension. Once a telescope on an equatorial mount is focused on a particular star, then it is only necessary to rotate it around the polar axis to keep it focused on the same star as time goes by. Furthermore, provided the telescope is rotated around this axis at the constant rate of one full turn (360°) every 23 hours and 56 minutes (a sidereal day) it will follow the star’s movement across the sky. This has the effect of counteracting the Earth’s rotation: the polar axis is parallel to the Earth’s axis and the telescope is rotated in the opposite direction at exactly the same rate. Many amateur telescopes can be driven around this axis at the required rate by a small electric motor. This avoids the need for constant adjustments by the observer or more expensive computer-controlled motors. SCP

SCP SCP to star S

to star

N

Figure 10.55 An equatorial mount is designed to rotate the telescope around the polar axis in the opposite direction to, but at exactly the same rate as, the Earth’s rotation.

Seeing stars Physics file Looking at the stars through the atmosphere could be likened to looking up from the bottom of a swimming pool. Objects above the pool will shimmer due to the currents in the pool and will appear on greater angles because of the refraction at the surface. The effect of the air is not as great, however, as the refractive index of air is only 1.003 compared with 1.33 for water.

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Detailed studies

Having got our telescope fixed on some interesting object in the heavens, and having set up our mount so that the object stays in the field of view, we will want to examine the object carefully or even take a photograph of it. At this stage we are interested in the resolution of the telescope. This is its ability to distinguish between two very close stars, or perhaps the details on the surface of a planet. It is easy to increase the magnification of our telescope with a stronger eyepiece, but unfortunately resolution is a different matter. It is the resolution that ultimately limits the useful magnification of a telescope—there is little point in increasing the size of a blurry image! There are a number of factors that limit the resolution of the telescope; some we can do something about, others we can not.

Probably the most frustrating problem for the amateur astronomer is the shimmering effect of the atmosphere. This is the result of warmer and cooler patches in the air swirling around and causing differing amounts of refraction of the light from the stars. This is why stars ‘twinkle’. It is also one reason why observation from a suburban area is difficult as there are many different sources of heat to stir the air. To avoid these problems we need to take our telescope to the top of a high mountain in an otherwise desert area where the air is thinner and more stable. This also avoids the ‘light pollution’ of the city which makes it hard to see dim objects in the sky. Having done this we still have some other difficulties, however. One is that our mounting system has not only to keep the telescope moving at just the right rate, it also has to keep it rock steady as we look through it and move around it. This can be a problem if we have to carry the mount and telescope up a mountain on our backs. Given a rock-steady mount, still air and no light pollution, there is one limit that only money can improve. That is the size of our objective, whether it is a mirror or lens. Ultimately, what is formed on our eye, film, or electronic sensor is a set of diffraction patterns formed by the stars. Stars may be large, but even the nearest ones are so far away that even through the largest telescopes they are still only pinpoints of light. We cannot see their surface. As we have learnt earlier, light is a wave, and like any wave it forms a diffraction pattern when it passes through a hole, and that is basically what our lens or mirror is. So each star appears as a small diffraction pattern with a diameter determined by the size of the objective. The larger the objective the smaller the diffraction pattern. So there are now two reasons for investing in ever-larger mirrors: greater light-gathering power and smaller diffraction patterns. Figure 10.56 shows the images of three point sources of light produced by lenses of increasing diameter (top to bottom). The diffraction patterns in the first two photographs result in the two points on the left being hard to distinguish; that is, they are not resolved. If we have near perfect vision our eyes are able to resolve objects which have an angular separation of about 1 arc minute (arcmin). This corresponds to the angle subtended by a 20-cent coin at a distance of 100 metres. All the planets, as seen from Earth, have angular sizes less than this, which is why we can only see them as points of light with the naked eye. This is why no one had seen the phases of Venus before Galileo looked at the planet with his telescope. A good, 20-cm diameter telescope can resolve about 1 arc second (arcsec) in good conditions—that is, about 60 times better than the unaided eye. It is like seeing separate grains of sand at 100 m. For comparison, the major planets subtend angles which vary (with their distance from us) from a few arc seconds to almost 1 arcmin. Although a telescope with a diameter measured in metres should be able to resolve objects only 0.1 arcsec apart, in practice atmospheric effects limit this to around 1 arcsec or 0.5 arcsec at very best. Astronomers, not to be put off, have found two ways around this problem. One is to put the telescope outside the Earth’s atmosphere and the other is to compensate for the effects of the atmosphere. Both these solutions are very expensive!

Figure 10.56 Photographs of three pinpoints of light taken with lenses of increasing diameter (top to bottom). The resolution of the images is greatest with the largest lens.


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Physics file Astronomers speak of the resolution of a telescope in terms of angular sizes. For example, the angular size of a 20-cent coin is about 1° at a distance of 1.7 m, 1 arcmin at 100 m and 1 arcsec at 6 km. (An arcmin is 1/60th of a degree and an arcsec is a 1/60th of an arcmin.) The diffraction-limited angular resolution of a telescope is given by the formula: 2.5l q= D where θ is the resolution in arc seconds λ the wavelength of light in metres D the diameter of the mirror in metres. You can use this formula to show, for example, that the minimum size telescope needed to see a 20-cent coin at 6 km is about 13 cm. Remember that this is only the diffraction-limited resolution; it assumes everything else is perfect!

The Hubble Space Telescope Launched in 1990 by the space shuttle Discovery, the Hubble Space Telescope (HST) was designed to overcome the problems of looking through the Earth’s atmosphere by getting above it. Its 2.4-metre mirror, combined with the lack of atmospheric problems, should have given it a resolving power of only 0.1 arcsec, much better than any Earth-based telescope. The initial pictures were a great disappointment, however; they were not even as good as many Earth-based telescopes. It was eventually discovered that due to an error in manufacture the main mirror suffered from spherical aberration; it was not a perfect parabola. This led to only about 20% of its light being focused to 0.1 arcsec. As a result astronomers could not photograph the very dim objects they had hoped to see in high detail. But the whole point of the HST was to look into the detail of the very distant and, therefore, very faint galaxies, as well as the faint nebulae around stars in our own galaxy. After an extraordinary effort by all concerned, a team of astronauts in another space shuttle mission fixed the problem in 1993. They installed a set of small secondary mirrors whose curvature was to exactly compensate for the fault in the primary mirror. The result was cause for jubilation among astronomers worldwide. The correction was perfect and the telescope lived up to its best expectations. The dramatic improvement can be seen in the photographs in Figure 10.58.

Adaptive optics The other approach to overcoming the problems caused by the Earth’s atmosphere is to compensate for them. Most new telescopes are now built with adaptive optics. This means that the shape of the primary mirror (or sometimes a secondary mirror) can be very slightly altered to adapt to changes either in the mirror (due to small temperature fluctuations or its mounting) or as a result of atmospheric turbulence. The image is constantly monitored by a powerful computer that notes any changes occurring and sends messages to appropriate fast mechanical actuators, which slightly deform the mirror to correct for the changes. This operation all happens at a rate of up to 100 times per second and has enabled Earth-bound telescopes to reach, almost, their theoretical diffraction limits. (a)

Figure 10.57 Probably the best known telescope today is the Hubble Space Telescope. It weighs 13 tonnes and is 13.1 m long, and has brought us magnificent pictures of astronomical sights.

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(b)

Figure 10.58 These photographs of galaxy M100 in Coma Berenices were taken (a) before and (b) after the repair of the HST in 1993. They show a dramatic improvement in resolution. The width of both pictures is about 25 arcsec.

Physics file In order to measure the effect of the atmosphere, astronomers may shine powerful laser beams into the sky. These form artificial ‘stars’ above the turbulent regions in the atmosphere. Since the shape and intensity of the laser beam is known, any distortions caused by the atmosphere can be measured and compensated for by the adaptive optics in the telescope in ‘real time’.

Physics in action

Very big telescopes Because of the difficulty of making very large mirrors, many new telescopes actually have a number of smaller hexagonal mirrors fitted together to create one larger mirror. The orientation of each of the mirrors is controlled by a computer to bring them all into perfect focus at the same place. The Keck I telescope on the 4200 m summit of an extinct volcano in Hawaii uses 36 hexagonal mirrors, each 1.8 m wide, to make up a 10 m mirror. Sensors and actuators on the back of each mirror keep the whole system in perfect alignment.

The new Very Large Telescope (VLT) project being conducted by the European Southern Observatory is aiming to achieve an incredible 0.001 arcsec resolution by combining the light from four separate 8.2 m telescopes. Each telescope is fitted with adaptive optics and they are spread over 200 m on the top of a mountain in Chile. Complicated electronics and computing will bring together the images from the four telescopes in such a way that the resolution achieved will, in effect, be that of a single, 200-m wide telescope. This resolution would be sufficient to see our 20-cent coin at 6000 km, or to resolve the two headlights of a car on the Moon!

(a) (b)

Figure 10.59 (a) One of the largest telescopes in the world can be found on Mauna Kea, in Hawaii; the Keck I uses 36 hexagonal mirrors. The secondary mirror is in the housing at the top of the picture and reflects the light back through the hole in the centre of the large mirror. (b) The images from these four 8 m adaptive optics telescopes in Chile will be combined electronically to create an image with the effective light-gathering power of a 16 m telescope and the resolution of a 200 m telescope.


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10.4 summary The telescope: from Galileo to Hubble • A telescope magnifies a real image produced by an objective (lens or mirror) with a long focal length together with an eyepiece lens that has a short focal length. The angular magnification is given by the ratio of the focal lengths: f M = o fe • A reflecting telescope uses a curved mirror to produce the image. This does not suffer from chromatic aberration or other problems that afflict a refracting telescope.

• The power of a telescope is directly proportional to the area of the objective lens or mirror, and thus proportional to the square of the diameter. • The equatorial mount has one axis parallel to the Earth’s. The telescope can be uniformly rotated around this axis in order to follow a star. • Atmospheric effects cause stars to shimmer. This can be overcome by putting the telescope in space (HST) or by using adaptive optics. • The resolution of a telescope is ultimately limited by the diffraction pattern it produces. This is inversely proportional to the diameter of the objective.

10.4 questions The telescope: from Galileo to Hubble 1 A simple astronomical refracting telescope consists of two convex lenses: the objective lens and the eyepiece lens. a The purpose of the objective lens is to: A create an inverted virtual image. B create an inverted real image. C create an upright real image. D gather light so that the eyepiece lens can create an image. b The purpose of the eyepiece lens is to: A invert the image created by the objective. B convert the image created by the objective into a real image. C magnify the image created by the objective lens. D increase the focal length of the objective lens. 2 A telescope has an objective lens with a focal length of 50 cm and an eyepiece with focal length 2 cm. From the Earth, the Moon subtends an angle of 0.5° in the sky. What will be the apparent angle the image of the Moon makes in the telescope? 3 A student goes to buy a telescope and finds that one is available with a 10 cm diameter mirror and another with a 15 cm mirror. However, the 15 cm telescope is twice the price. Other things being equal, is the 15 cm telescope really twice as good as the 10 cm one? 4 Newton solved the problem of having one’s head in the way of the light to a reflecting telescope by putting a small mirror at 45° in the reflected light beam, before the principal focus, but this blocked some of the light to the main mirror. Did this create a ‘black hole’ in the centre of the image? Did it have any other effect?

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5 To follow the stars, a telescope needs to be rotated around its polar axis at a rate of a little more than 360° in 24 hours. Why is this, and at what rate should it be rotated? 6 a Why was it that before Galileo saw the planets through his telescope, no one had realised that the planets were disks rather than just points of light? b Why is it that a dark-adapted eye is actually better able to resolve small objects at a distance than one adapted to bright light? 7 It has been suggested that a large telescope could be used to confirm the presence of the Moon lander base left on the Moon by the astronauts in the 1970s. If the lander is about 5 m wide, what angle does it subtend from the Earth? The distance to the Moon is 380 000 km. How large would a telescope mirror need to be in order to resolve it? Do you think that the idea is practicable? 8 The two photographs in Figure 10.58 are 25 arcsec wide. How far away would a 20-cent coin have to be to subtend the same angle? 9 A space-based telescope has the big advantage that it does not suffer from atmospheric shimmer. Can you think of at least two more advantages of a space telescope? 10 The VLT project electronically combines the images of four 8.2 m telescopes spread over a 200 m distance. Explain why it is that in some ways this is like a 16 m telescope, but in others it is like a 200 m telescope.

10.5 New ways of se e

ing

Before Galileo, the heavens were seen as the unchanging realm of the Gods. Galileo showed that the heavens seemed a little less than perfect: mountains on the Moon, spots on the Sun. Some years earlier Tycho Brahe had even witnessed a new star come and go. Ever since, astronomers have been discovering a new and dynamic universe.

New planets In 1781 William Herschel, a German-born musician who had immigrated to England and become fascinated by astronomy, discovered a new planet. He was systematically surveying the sky and noticed a faint object which he thought may have been a comet. However, after tracking it for almost a year he decided that it was in an almost circular orbit beyond Saturn and that it must therefore be another planet. He had discovered a seventh to add to the six that had been known since antiquity. Herschel originally named the new planet after King George III, but after several decades the name Uranus, meaning ‘heavenly’ in Greek mythology, became more commonly used. Actually, although Herschel was the first to recognise it as a planet, Uranus had been plotted on some early star maps and is sometimes just visible to the naked eye. It had been mistaken for a star because being so far from the Sun it moves through the stars at a very slow rate, only 4° yearly compared with 12° yearly for Saturn. As astronomers carefully tracked the motion of the new planet they eventually became aware that it was not behaving just as it should according to Newton’s laws, which by now had been very convincingly applied to the planets. In 1843 young Cambridge University graduate John Couch Adams suggested that the reason for the discrepancies in Uranus’s orbit could be the gravitational pull of an eighth planet in an even more distant orbit. He even calculated the approximate position of the new planet. At the time English astronomers did not take him very seriously. In 1846 a German astronomer, Johann Gotfried Galle, was told of a similar prediction by a French astronomer, Urbain Le Verrier. He looked in the predicted position that same night and discovered the new planet. It was Le Verrier who proposed that the new planet should be called Neptune. Eventually, after much English–French wrangling Le Verrier and Adams were credited equally with the discovery. The search for a ninth planet started shortly after the discovery of Neptune. Despite various predictions about small variations in Neptune’s orbit none turned out to be genuine. Nevertheless, Percival Lowell, a prosperous Boston gentleman who was convinced that there were canals (and hence life) on Mars, funded a special wide-field camera to be installed at the Lowell Observatory in Arizona. Unfortunately for Lowell, it was not until 1930, 14 years after his death, that Clyde Tombaugh finally discovered the very faint and relatively tiny planet. It was named Pluto after the mythological god of the underworld. Pluto also starts with Percival Lowell’s initials. Despite its history, however, Pluto has been demoted and is no longer an official planet (see the Physics file). Perhaps by way of compensation, however, it will always retain a small place in history as the element plutonium was named after it. (Note that the elements uranium, neptunium and plutonium were named after the planets, not the other way around.)

Figure 10.60 A false colour photo taken by the Hubble Space Telescope of Uranus in infrared light. The planet is unusual in that the tilt of its axis is almost at a right angle to the plane of its orbit.

Physics file In August 2006 the International Astronomical Union (IAU) decided that Pluto was not a ‘planet’. It was reclassified as a dwarf planet (officially ‘Minor planet number 134340’). The problem was that more Pluto-sized objects (such as Eris, Sedna and Quaoar) were being found in the Kuiper Belt. More particularly, it was decided that a ‘planet’ sweeps its orbit clean (by its gravitation) of other objects—something poor old Pluto has not achieved. Despite the IAU, however, Pluto is likely to live on as an unofficial planet for some time yet!


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Saturn

Jupiter

Mars

Earth Venus Mercury Sun

Figure 10.61 The orbits of the visible planets drawn to scale. The dotted section is where the planet’s orbit is below the plane of the ecliptic (the Earth’s orbit). There appears to be a surprisingly large gap between Mars and Jupiter. Is there a missing planet?

A scale drawing of the orbits of the planets shows a surprisingly large gap between Mars and Jupiter (in Figure 10.61). As well, two astronomers had suggested a law, now called the Titius–Bode law, which suggested that another planet may lie between Mars and Jupiter. Six German astronomers, including Bode, decided to organise a search for the missing planet, and called themselves the ‘Celestial Police’. At the start of 1801, almost before they had got their search underway, a Sicilian astronomer, Giuseppe Piazzi, noticed a dim uncharted star that seemed to be moving slowly through Taurus. However, shortly afterward the Sun moved into Taurus and the object was no longer visible. Astronomers feared it may have been lost. A young German mathematician—Karl Friedrich Gauss—came to the rescue and, using Piazzi’s data, calculated that the object should reappear in the early morning sky in the constellation of Virgo. Sure enough, near the end of 1801 it was found just where he predicted. Piazzi named the object Ceres after the patron mythological goddess of Sicily. Its orbit was found to be at 2.77 AU, in close agreement with the prediction of Bode’s law. However it was very dim and didn’t appear to be a full-sized planet at all and so was referred to as a minor planet or asteroid. Within a few years several more small asteroids had been discovered, but the search was painstakingly slow as the positions of suspect objects had to be recorded night after night to see if they were moving through the stars. By the mid-1800s, hundreds of asteroids had been discovered in what is called the asteroid belt, which extends from about 2 AU to 3.5 AU. So are the asteroids a planet that broke up, or one that never formed? It has now been found that the combined mass of the asteroids is not enough to make

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up a planet and it is thought that perhaps Jupiter’s huge gravitational pull prevented the formation of a planet in this region. Don’t confuse asteroids with comets. The latter have been known since antiquity as occasional visitors to the skies. They have very highly elliptical orbits and spend most of their time well out beyond the orbits of the outer planets. While asteroids are rock-like, comets have been called ‘dirty snowballs’. When they come in closer to the Sun, some of their frozen ice evaporates producing the characteristic tail.

Physics in action

Curious laws for the planets? Back at the start of the 17th century, Kepler had suggested a law which appeared to relate the size of the orbits of the planets to the ratio of the spheres that could be placed around and within the five regular platonic solids: tetrahedron, cube, octahedron, dodecahedron and icosahedron. Kepler had to change the natural order, but was convinced that God had set out the planets according to geometric patterns. Although it

e orbits the size of th tside n ee tw e b ip ou d inside and d a relationsh ler discovere heres that can be place cidental. ep K 2 .6 10 Figure d the sp ply coin n planets an ems to have been sim of the know se It s. lid so nic the five plato

was a reasonably good fit for the six known planets, it was relegated to history once the outer planets were discovered.

Table 10.4 Kepler’s planetary orbits compared with the scaled radii of the spheres around the platonic solids Planet

Orbit (AU)

Solid and sphere

Scaled radius (AU)

Saturn

9.16

Cube circumsphere

9.16

Jupiter

5.25

Cube insphere

5.26

Mars

1.52

Tetrahedron insphere

1.65

Earth

1.00

Dodecahedron insphere

1.10

Venus

0.72

Icosahedron insphere

0.76

Mercury

0.36

Octahedron insphere

0.43

The cube’s circumsphere is taken as equal to the radius of Saturn’s orbit and the other radii are scaled accordingly. Another curious relationship discovered in the late 18th century was the Titius–Bode Law. Titius (in 1772) found a simple number sequence which appeared to be related to the orbits of the planets. Later, Bode brought it to prominence. The sequence is: 0 + 4, 3 + 4, 6 + 4, 12 + 4, 24 + 4, 48 + 4, 96 + 4, 192 + 4, etc. That is, apart from the first term, each term is obtained by doubling the previous multiple of three and adding four. If the addition is done and then the result divided by 10 we obtain the series: 0.4, 0.7, 1.0, 1.6, 2.8, 5.2, 10.0, 19.6, etc. Where is all this going? Well now look back at Table 10.2 (page 360) and compare these values with those of the radius of orbit of the planets in AU. Not a perfect fit, but curiously similar, except for two things. First, at that time the outer planets had not been discovered, and so what did 19.6 and the numbers beyond it mean? Second, where was the planet for the 2.8 value in the sequence? A number of astronomers started looking for these ‘missing’ planets and, as we know, Uranus at 19.2 AU was discovered less than ten years after Titius suggested the pattern.


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Physics file Another way to find asteroids is to use a ‘blink comparator’. Photographs of the same part of the sky taken at different times are compared in a device that gives the operator rapidly alternating views of the two images. Any object that has moved in the interval between photographs will appear to blink on and off and, hence, will be more noticeable.

New techniques In 1891 a significant breakthrough occurred. German astronomer Max Wolf developed an important new technique to search for asteroids. First, he motorised the drive of the right ascension axis of his telescope so that it followed the stars across the sky automatically. Then he used photographic film to record a long exposure of the view. The stars appeared as dots, but if an asteroid was in the field of view, its motion through the stars would leave a slightly blurred image. Using this method, Wolf alone discovered 228 more asteroids! Today tens of thousands of asteroids have been identified and more are discovered every month. Indeed it is quite a sport among amateur astronomers, who get the privilege of naming their own asteroid! The technique of using a motorised drive on the telescope in conjunction with long exposure photographs had become standard by the beginning of the 20th century. Apart from motion against the background stars, the photographs showed detail too faint to be seen by the naked eye. With better telescopes, drives and photographic film an explosion of new astronomical discoveries started in the early 20th century. With the development, later in the century, of new electronic techniques for recording and processing the images, this explosion continues into the present century!

New objects Before Galileo, almost all astronomers believed that all celestial objects, apart from those in the solar system, were stars. Remarkably, there is a record from a 10th century Persian astronomer showing the Andromeda Nebula, which is visible to the naked eye only under the very best conditions. Only after the use of telescopes became common in the 18th century did astronomers begin to turn their attention to certain ‘fuzzy

Figure 10.63 Binoculars will reveal a fuzzy patch around one of the stars in Orion’s sword (in the Saucepan handle). It is the ‘Great Nebula in Orion’.

Figure 10.64 The three different types of nebulae are all seen in this photograph of the area around Alnitak in Orion’s belt. NGC 2024 (emission)

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Dark nebula

Alnitak

IC 434 (emission)

NGC 2023 (reflection)

Horsehead nebula (dark)

patches’ in the sky. The Orion nebula had been noted earlier, but it was William Herschel (of Neptune fame) who started to realise the significance of studying nebulae. He compared his view of the Orion nebula with the earlier record and decided that it had changed in the intervening period and thus ‘...we may infer that there are undoubtedly changes among the fixt Stars, and perhaps from a careful observation of this Spot something might be concluded concerning the Nature of it’. Ever since, astronomers have been learning much from their study of these ‘fuzzy spots’ in the sky. Binoculars or a small telescope will reveal many nebulae in a clear dark sky. Nebulae are actually huge clouds of gas or ‘dust’ and most that we see are within our galaxy. There are three basic types of nebulae: those that emit light, emission nebulae; those that reflect light from other sources, reflection nebulae; and those that block light from stars beyond them, dark nebulae. All three types are visible near Alnitak, the eastern star in Orion’s belt (see Figure 10.64) They are not really ‘near’ Alnitak as it is a relatively close star (820 light-years) and the nebulae are about twice as far away. The light from nebulae tells us a lot about the nature of the material and the processes going on in the nebula itself. Spectral lines tell us what atoms are present, small shifts in the lines tell us about its movement, missing parts of the spectrum tell us what is absorbing the light, and so on. As we will see in the next section, astronomers like to analyse as much of the light, both visible and invisible, coming from nebulae as they can collect. Some objects which appear to be nebulae are actually not dust clouds in our own galaxy, but other galaxies right outside the Milky Way. In fact the Andromeda ‘nebula’ mentioned at the start of this section is one of the most famous. It was not until the start of the 20th century that astronomers began to realise that our own Milky Way Galaxy was much larger than they had thought, and furthermore, other complete galaxies lay at vast distances from it. Some galaxies like Andromeda and the Milky Way are huge spiralling collections of billions of stars; others are egg-shaped, or ‘globular’ conglomerations of stars, and others quite irregular. Some are only one-hundredth the size of the Milky Way, others are fifty times as massive. The two closest galaxies to the Milky Way are actually quite easily visible to the naked eye: the Large and Small Magellanic Clouds. They are actually smaller, irregular galaxies and are referred to as satellite galaxies of the Milky Way.

Figure 10.65 The M31 Galaxy in Andromeda is a spiral galaxy like the Milky Way. The two smaller galaxies above and below M31 are M32 and M110. They are actually satellite galaxies of the much larger M31.

New light ... As we look through a telescope we gather more light to create a brighter image of the object we are looking at. The light, however, is just the same as we see with the naked eye. The first indication that there was more to light than what met the eye, so to speak, was around 1800 when William Herschel placed a thermometer in the spectrum produced as sunlight streamed through a prism. He found that a higher temperature was recorded even when the thermometer was placed beyond the red end of the rainbow spectrum. The invisible radiation ‘below’ the red end of the spectrum is now called infrared radiation. In the later 1800s Maxwell showed that light was an electromagnetic wave. He also realised that there was no reason why electromagnetic waves of lower or higher frequency should not also exist. Herschel’s discovery had shown the presence of waves of slightly lower frequencies, but what about other frequencies?


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Wavelength (nm) Blue

400 Violet

500 Green

Yellow

Red

600 Orange

700

Visible spectrum Wavelength (m) 108 107 106 105 104 103 102 10 Long waves

1

10–1 10–2 10–3 10–4 10–5 10–6 10–7 10–8 10–9 10–10 10–11 10–12 10–13 10–14 10–15 10–16

Radio waves

Infrared

Ultraviolet

X rays

Gamma rays

10 102 103 104 105 106 107 108 109 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1023 1024

Frequency (Hz)

Figure 10.66 The spectrum of electromagnetic waves extends from radio waves, with wavelengths of kilometres, all the way down to gamma radiation, with wavelengths smaller than a nucleus. Astronomers like to look at the universe in ‘light’ of all wavelengths.

Figure 10.67 The Parkes Observatory radio telescope (‘The Dish’) in NSW has a diameter of 64 m. It has played a key role in the development of radio astronomy and is well worth a visit if you get the chance.

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Soon after Maxwell’s prediction, Heinrich Hertz produced electro magnetic waves of much lower frequencies, with wavelengths of metres instead of micrometres. These we now call radio waves. Not long after that Wilhelm Röntgen produced higher frequency, short wavelength (only onehundredth that of light) X-rays. Needless to say, the rest of the spectrum was rapidly filled in and we are now familiar with the spectrum as shown in Figure 10.66. Traditionally, astronomy had always used visible light to look at the universe. There actually appeared to be little option as it was soon discovered that the atmosphere was opaque to ultraviolet (UV) and infrared (IR) radiation much beyond the visible spectrum. However, a surprising new tool was discovered, more or less by accident in the early 1930s. As radio developed as a communication link, engineers became aware of a form of interference which seemed to fluctuate with the time of day. Karl Jansky, a young engineer at Bell Telephone Laboratories, eventually realised that the interference was strongest when the constellation of Sagittarius was overhead and that indeed he appeared to be detecting radio waves from the centre of our galaxy, which is in that region of the sky. Initially few astronomers took any interest in these waves, but Grote Reber, a radio engineer in Illinois, decided to take up the challenge and built the first radio telescope in his backyard. It was a 10 m metal ‘dish’ which focused the waves onto a radio receiver at the principal focus. Over the next few years he mapped the radio emissions from the entire Milky Way at two wavelengths, both of which were around 1 m. During and after World War II, radio technology advanced very quickly and astronomers began to realise they had a powerful new way of looking at the universe. Radio telescopes proliferated and soon a vast array of new data was being generated. There was one very significant problem however. Because radio waves are very much larger than light waves the angular resolution of the new devices was very poor. As we saw in section 10.3, the resolution is dependent on the ratio of the wavelength to the size of the telescope, in this case the diameter of the dish. For a large optical telescope this ratio is about 107, but for the radio telescopes it was about 10–100. This made identification of a radio source with a particular star or other object very difficult indeed. So radio telescopes were made as large as possible. This was also important in terms of their light-gathering power

(‘light’ being used in a general sense of electromagnetic waves) as the radio emissions from space were rather faint compared with optical emissions. The largest single radio telescope built is the Aricebo dish in Puerto Rico. It was actually constructed in an extinct volcano using a steel superstructure to form the concave dish that is 305 m in diameter. This still only gives it a diameter-to-wavelength ratio of less than 1000 when mapping 5 cm radio waves; this corresponds to a resolution of about 1 arcmin—compared with the 1 arcsec of a good optical telescope. Fortunately, radio astronomers have found a way around the resolution problem! Radio signals can be fed along cables and the output from several dishes combined by clever electronics. This enables a radio telescope with an effective diameter (for resolution purposes, not light-gathering) of many kilometres to be built. The Very Large Array (VLA) in New Mexico consists of 27 dishes, each 25 m in diameter, arranged in a Y shape. This array has an effective diameter of 27 km, giving it better resolution than most optical telescopes. Even larger resolution can be achieved by combining radio telescopes right across the Earth. There is even a space-based radio telescope that can be used for this purpose, giving resolutions down to better than 0.001 arcsec. Radio astronomy enables information to be obtained that would be impossible to obtain at visible wavelengths. For example, electrically charged particles moving in the magnetic field of a planet (such as those which cause auroras on Earth) give off radio waves that, in effect, make the magnetic field of the planet ‘visible’ to the radio telescope. On the larger scale, stars can produce radio waves by similar means and this tells us much about the type of star. As their name implies, so-called ‘black holes’ are not visible. However, as they suck in matter around them, it spirals in at ever-increasing speeds giving off a characteristic longer wavelength electromagnetic radiation which was first detected by radio telescopes.

... and new ways of seeing it For well over a century many wonderful images of the sights of the universe have been produced on photographic film. Unfortunately, however, only about 2% of the light that falls on a photographic film triggers the chemical reactions which set the image in place. Since it was discovered that various silicon devices could be made light-sensitive (‘solar cells’ that produce electricity, for example), many ways to exploit this technology to produce images have been found. The most sensitive device is known simply as a charge-coupled device or CCD. The term hides the complexity involved, but these are the basis of most modern video and digital cameras. CCDs can be made that utilise up to 70% of the light falling on them, so they are 35 times more sensitive than photographic film. They are now used almost universally in new telescopes. The Hubble Space Telescope would have been impossible without CCDs to collect the pictures electronically and radio them back to Earth. Indeed the development of the HST was the reason for much of the development of CCDs. Basically a CCD is a square ‘chip’ of silicon which is divided into an array of tiny light-sensitive ‘picture elements’ or pixels. A typical CCD may have about 16 million pixels arranged in 4000 rows and columns on a chip only a few centimetres square. When the image from the telescope is focused onto

Figure 10.68 Charge-coupled devices such as this are used to record the images in modern telescopes. They are much more sensitive than film and directly produce an image in electronic form for further analysis or display.


385

the CCD, each pixel builds up an electric charge; the size of the charge is related to the amount of light that falls on the pixel. The amount of charge on each pixel is then recorded electronically and processed to produce an image that can be stored in a computer. From there the image can be displayed, printed, transmitted around the world, or further processed by powerful computer programs which may, for example, compare many sets of similar photographs to look for minute changes that would indicate variable stars, asteroids, supernova and much more. Because of their great sensitivity and means of operation, CCDs are much more useful than film. They can not only record an image, but they can be used to measure the light intensity of a certain star and produce a graph of brightness versus time, for example. Or they can analyse the light in terms of the wavelengths present and produce a spectrum. A large amount of what we know about stars has been obtained from their spectra.

New platforms

Figure 10.69 The International Space Station

is an ideal platform from which to observe the universe in all its many colours: from radio waves to gamma radiation.

Transparency

Earlier we said that the Hubble Space Telescope had the huge advantage of not having to look through the Earth’s atmosphere and hence avoided the problem of ‘twinkling’ stars. There is another very important reason for getting above the Earth’s atmosphere: it absorbs much of the electromagnetic radiation coming from space. This is very important to us as it means that we are protected from dangerous ultraviolet radiation from the Sun. On the other hand, astronomers want to get all the information they can about the light, both visible and invisible, emitted by stars. Figure 10.70 shows the electromagnetic radiation that can penetrate the atmosphere. As you can see, there are effectively just two ‘windows’ that can be used by Earth-bound astronomers: from visible light into a little of the infrared radiation, as well as 1 cm to 10 m radio waves. In order to see the universe in X-ray light, strong ultraviolet radiation, deep infrared radiation or short-wavelength microwaves, we need to get above most of the atmosphere. Some of this can be done by sending equipment up on high-altitude balloons or very high-flying aircraft, but clearly the best solution is to go into orbit around the Earth, right outside the atmosphere. The Hubble Space Telescope is designed to use light of wavelengths from about 0.1 µm to 1.0 µm; that is, from strong ultraviolet through visible and just into the infrared radiation. Other space-based telescopes are specifically designed to use microwave, infrared, ultraviolet, X-ray or gamma radiation. optical window

radio window

100%

50%

atmosphere is opaque



0%

0.1 nm 1 nm 10 nm 100 nm 1 µm 10 µm 100 µm 1 mm

1 cm

10 cm

1m

10 m

100 m

Wavelength

X-rays

ultraviolet visible

infrared

microwaves

radio waves

Figure 10.70 A graph showing the transparency of the atmosphere to electromagnetic waves. A space-based telescope is able to receive waves from all parts of the spectrum.

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(a)

(b)

(d)

(e)

(c)

Figure 10.71 These five views show the entire celestial sphere as seen in (a) visible, (b) radio, (c) infrared, (d) X-ray and (e) gamma-ray radiation. The Milky Way stretches across the centre of each image.

10.5 summary New ways of seeing • New planets were discovered by a combination of luck, good guesses and prediction based on the known planets—and, of course, the telescope. • Techniques for finding new planets or asteroids rely on finding small differences in the position of a celestial object over time. Once photographic techniques were combined with motorised telescope drives, thousands of new asteroids were found. • Nebulae can be dust clouds in our own galaxy (emission, reflection or dark) or in other galaxies.

• As well as visible light, astronomers use ‘invisible’ electromagnetic radiation such as ultraviolet and infrared radiation, as well as much longer radio waves to reveal the nature of the universe. • Electronic image-recording devices have revolution ised astronomy, enabling astronomers to analyse their pictures of the universe in ways that were impossible not that long ago. • Space-based telescopes enable the universe to be seen in all wavelengths of electromagnetic radiation from radio and microwave, through visible to X-rays and gamma radiation.

10.5 questions New ways of seeing 1 It was quite some time after the invention of the telescope before the outer planets were discovered. What were some of the difficulties in finding new planets?

6 Brave astronomers who do peer through a telescope at night have to endure cold conditions in the observatory. Is this because the observatory needs the money and can’t afford heating?

2 In what way was the discovery of Neptune more of a scientific achievement than the discovery of Uranus?

7 Why do radio astronomers never have to endure cold dark nights in unheated observatories?

3 How did photography revolutionise the discovery of asteroids?

8 What are some of the advantages and disadvantages of a space-based telescope?

4 What is ‘invisible light’? What different types are there and what are their characteristics?

9 X-ray telescopes can be used in space or in highflying aircraft or balloons. Why can’t they be used in ground-based telescopes? What is the difference between an X-ray photograph of stars and the one you may have at a hospital?

5 One hundred years ago it was known that photo graphic film was sensitive to infrared and ultraviolet light. Why didn’t astronomers use these films to take photographs of the stars at these wavelengths?

10 What are the similarities and the differences between a radio telescope and a reflecting optical telescope?


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chapter review 1 As you look into a dark night sky, what are some of the thoughts that come into your mind? 2 Even though we know that there is really no huge ‘celestial sphere’ rotating around the Earth, astronomers still speak of one. Why is this? 3 The stars are said to have a diurnal and an annual rotation. What is the difference between these two expressions and what is the reason for the difference? 4 What is the altitude of the celestial equator above the north horizon in Melbourne? Would it be different from Brisbane or Hobart? If so, in what way? 5 Where does the celestial equator meet the horizon as seen from Melbourne? Would it be different from Brisbane or Hobart? If so, in what way? 6 At the South Pole no stars are visible in the middle of summer. Why not? If there was a sudden eclipse of the Sun and the stars did become visible, how would the sky differ, or not differ, from that seen in the middle of winter? 7 If you observed the stars from a point on the equator at midnight on 21 March and then looked again at midnight on 21 September how would the two views differ? 8 Because of the Earth’s atmosphere the Sun rises a little earlier and sets a little later than it would otherwise. Assuming that the atmosphere is uniform, 100 km thick and has a refractive index of 1.003, use Snell’s law to determine the amount of refraction at sunrise and sunset and, hence, the extra time that the Sun is visible. (Use Earth’s radius = 6400 km.) How closely does your answer agree with the actual extra time? 9 From Melbourne, latitude 38°, some of the stars are always in the sky (A), some spend part of the day in the sky (P) and some never appear (N). a Classify each of the following stars as either A, P or N. i A star within 20º of the SCP ii A star within 20º of the NCP iii The Southern Cross iv Orion v A star 50º north of the celestial equator vi A star 50º south of the celestial equator b For those stars you classified as P, give a rough estimate of the time they will spend in the sky. 10 Use a starfinder or chart to find the stars closest to the positions given by the following celestial coordinates: a RA 14 h 13 min, dec. +19° b RA 5 h 50 min, dec. +7° c RA 14 h 40 min, dec. –60° d RA 4 h 30 min, dec. +15°

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11 What are the coordinates of the following stars? a Sirius b Achernar c Vega d Rigel 12 Is the Sun due north at midday each day? 13 After one full sidereal day, compared with the previous day, a star on the celestial equator will: A be in exactly the same position. B be a little east. C be a little west. D have set. 14 In Melbourne, the Sun has a maximum altitude of 75° at the summer solstice and 29° at the winter solstice. What is the maximum altitude at the equinoxes and why is it different in summer and winter? 15 How is the ecliptic related to the celestial sphere? Is it fixed in place on the celestial sphere or does it move? 16 You are watching the sunset with a crescent moon still in the sky. Which of these pictures best represents the Moon as you will see it? A

B

C

D None of the above.

17 A full moon is seen high in the sky at midnight during winter whereas a full moon in summer is always lower in the sky. Why is this? 18 Jupiter’s moon Io is 3642 km in diameter and 350 000 km from the surface of Jupiter. The comparable figures for our Moon are 3476 km diameter and 378 000 km from Earth. Thus the two moons would look quite similar in size when viewed from the surface of each planet. a When the two moons are overhead, which one will appear larger in the sky? b Given that Jupiter’s diameter is about 11 times that of the Earth, and its orbit is 5 times further away, what con clusions can you draw about the likelihood of seeing, and the nature of: i a lunar eclipse from Jupiter? ii a solar eclipse from Jupiter? 19 The orbit of Mercury is 0.39 AU. What is its maximum elongation angle from the Sun? How long before or after sunrise or sunset can it be seen? 20 You can see a tree in a direct line with the top of a hill that is 5.0 km away from you, so there is no angular separation between the two. The tree is 200 m from you. You now walk a distance of 20 m at right angles to the line from the hill and tree. What is the angular separation between the two now?

21 When furthest from the Sun, the Earth is at a distance of 152.1 million km and speeding around its orbit at 29.3 km s‑1. At closest approach the distance is 147.1 million km. Use Kepler’s second law to find the speed at this distance. 22 Of the two telescopes Galileo built, shown in Figure 10.50, which would have given the larger image and which would have given the brighter image? 23 The largest refracting telescope built has a lens with a diameter of 40 inches and a focal length of 19.5 m. a How much greater is its light-gathering power than a standard 6-inch diameter lens? b What is its magnification when using an eyepiece with a focal length of 20 mm? 24 Commonly sold binoculars are rated as 8 × 30 or 7 × 50. The first number is the magnification of the eyepiece lens and the second is the diameter of the objective. a Which pair of binoculars would magnify the most? b Which would have the brightest image, and how much brighter would the image be? 25 Most astronomers don’t spend much time peering through a telescope on cold dark nights any more. What do they do instead? 26 What is the connection between radio waves and light waves?


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chapter 11

s c i s phy

o r t s A

outcome On completion of this chapter, you should be able to describe and explain methods used to gather information about stars and other astronomical objects and apply this information to models of the nature and origin of the universe.

H by the end of this chapter you will have covered material from the study of movement including: • the properties of stars: distance, apparent and intrinsic brightness, and spectral type • the Sun, our star: characteristics and the source of its energy • the Hertzsprung–Russell diagram: the nature and evolution of stars • galaxies: types, distances, Hubble and red shifts • theories of the formation of the universe, galaxies, stars and planets.

ow old is the universe? Will it last forever? Are there other universes? How was the Earth formed? Is there other intelligent life in the galaxy? Will we ever travel to the stars? These and many more such questions interest not just professional astronomers and astrophysicists, but almost anyone who has looked up in the night sky and wondered about our origins and destiny. In this study we will find fairly good answers to some of our questions, we will gain tantalising glimpses of possible answers to others, and we will discover more questions we hadn’t even thought of! Physics is the fascinating story of a great human adventure—the search for ways in which we can understand our world. It has been spectacularly successful in many areas. When we casually flip a switch to watch the TV news we don’t often think of all the physicists who made this possible— Newton, Faraday, Maxwell, Hertz and Einstein, to name just a few. None of these people had television in mind when they explored the natural world—that was just one of the less important by-products of their work. What was important to them was understanding some of the basic principles on which the universe works. As students of physics we are privileged to be able to share a little of the excitement of that achievement. Astrophysics, the attempt to answer some of the very biggest questions, is a fascinating mix of basic physics, the very latest theories and new technology. The development of ‘silicon electronics’ along with space travel has given us methods of seeing our universe in ways never before even dreamed of. The photograph of the Omega Nebula shows just one example. The Omega Nebula ‘star factory’ is a huge cloud of dust and molecular gases in which massive stars are forming and gradually boiling off the dark shroud of carbon-based smoke-sized dust. We know this as the result of the work of many astrophysicists down through the ages. Hopefully, in this short study you will gain some glimpses of this exciting area of physics—and become part of this great adventure of discovery.

11.1 The stars—how f

ar, how brigh

t?

Astrophysics is about some of the biggest questions we can ask: What is the universe? How and why did the conditions for life to evolve occur? The early astronomers thought the stars all lay in the ‘realm of the gods’, just beyond the planets that were circling the Earth—which was stationary, the centre of the universe. Galileo realised both that the Earth circled the Sun and that the stars must be much further away than the planets, or they would show apparent movement as the Earth revolved around the Sun. By Newton’s time it was realised that the stars were possibly other ‘suns’ and therefore must be huge distances away or they would appear much brighter. Newton calculated that if Sirius, the brightest star in the sky, was about the same intrinsic (actual) brightness as the Sun then it must be about a million times further away. Many people, however, refused to believe that stars could be suns. For thousands of years it was believed that they were ‘heavenly’ objects quite unlike anything in the solar system, and located just beyond the orbit of Saturn. Distances such as Galileo and Newton were suggesting were simply incomprehensible. So where are the stars, and what are they?

Although it is not necessary to have taken the Astronomy detailed study in order to take this one, some of the ideas from that study are assumed and expanded upon here. If you have done some astronomy in your junior science, or have simply taken an interest in the subject you will have no trouble with this topic. However, we do suggest that a browse through the content of the Astronomy study could help to set the scene for the ideas covered here.

Far far away The only way the astronomers who followed Galileo could measure the distance to the stars was to look for parallax in their positions as the Earth moved around the Sun. The idea is shown in Figure 11.1. As the Earth moves around its orbit the closer star should appear to move in relation to those further away. The same idea can be seen if you simply look out the window and move your head from side to side: the window frame moves ‘against’ you while the background moves ‘with’ you. background stars

Earth in June

Sun

1 AU

PRACTICAL ACTIVITY 51 Distances by parallax measurement

d

p

closer star

Earth in January

view in January

view in June

Figure 11.1 The parallax movement of a close star relative to the background ‘fixed’ stars. The parallax angle p is half the total apparent shift in angle.

Various 18th century astronomers attempted to measure the annual parallax of some bright stars (which they thought more likely to be closer). In 1729 James Bradley announced that if there was any parallax in the positions of stars it was less than 1 arc second (see Physics file on page 392). He calculated that they must be at least 400 000 times further away than the Sun (i.e. 400 000 AU). Along with Newton’s calculation for Sirius (1 million AU) this meant that attempting to measure the distances to stars seemed futile and so there were few attempts for the rest of the 18th century.

Chapter 11 Astrophysics

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Physics file To measure the small angles between stars 1° is divided into 60 ‘arc minutes’, and the arc minute (arcmin) divided again into 60 ‘arc seconds’. So the arc second (arcsec) is only 1/3600 of a degree. For example, the angular size of a 20-cent coin is about 1° at a distance of 1.7 m, 1 arcmin at 100 m and 1 arcsec at 6 km. Distances in astronomy are often given as multiples of the distance to the Sun, which is close to 1.5 × 108 km. This distance is referred to as 1 astronomical unit, or 1 AU.

Figure 11.2 Wilhelm Struve used the ‘Great Dorpat Refractor’ to look for double stars. By 1837 he had found over 3000.

Physics file What is a parsec? Astronomers define the parallax angle as the angle subtended by the radius of the Earth’s orbit (1 AU) as it would be seen at the distance of the star—as shown in Figure 11.1. The total circumference of a circle of radius d would be 2πd. As 1 arcsec is 1/3600 of a degree, and a degree 1/360 of a full circle, we can see that 1 AU must be 1/(3600 × 360) of the full circumference; that is: 2πd 1 AU = (360 × 3600) You can show that this leads to d = 206 265 AU, which is indeed the conversion factor from pc to AU.

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By the beginning of the 1800s instrument makers had developed special telescopes capable of measuring angles of only fractions of an arc second and so attempts were renewed to look for stars showing parallax. By this time it had been realised that some stars move, relative to the background stars, at a steady rate. This was not due to parallax; the star was actually moving (very slowly!) through the sky. For example, the star 61 Cygni had been observed to move 5 arcsec every year (at that rate it would take over 700 years to move 1°). This motion is known as proper motion and is very interesting to astronomers as it tells them about the motion of the stars within the galaxy. The early 19th century astronomers realised that stars with a large proper motion were more likely to be closer to us than those with none. Clearly it was also likely that brighter stars were closer. There was one other indication they used to select stars as possible candidates for parallax measurement. The development of better telescopes had led to the discovery that many stars were actually binary stars. That is, they appeared to be two stars in orbit around each other. The further apart the two stars appeared to be (in relation to the time they took to orbit each other), the closer to the Earth the stars were thought to be. So a systematic examination was made of stars that fulfilled at least two of these three criteria. In 1835 Germanborn Wilhelm Struve, working in Russia, decided to look at Vega, a single star, but a bright one with a large proper motion. He looked for parallax against nearby stars, just as Galileo had suggested many years before. Two years later he announced that Vega showed a parallax of just one-eighth of an arc second (quite close to the modern value)—a tiny value, but he was confident that it was real parallax. It suggested that Vega was a huge 1 600 000 AU from the Earth. Soon other ‘close’ stars were found. In 1838 Thomas Henderson, who had been working from the Cape of Good Hope, found that Alpha Centauri, a bright binary with a high proper motion, had a parallax of about 1 arcsec, putting it at about 200 000 AU. Alpha Centauri is the ‘Pointer’ furthest from the Southern Cross. In fact, it is a triple system with a third faint star called Proxima Centauri, for obvious reasons, being slightly closer. The modern value for the parallax of Proxima Centauri is 0.77 arcsec—making it the closest star to our Sun at a distance of 270 000 AU. This method of measuring the distance to stars is known as stellar parallax. It leads to a natural unit for the distance of stars, the parsec. The parsec (abbreviation pc) is the distance to a star that shows a parallax angle of 1 arc second. Clearly a star which has a parallax of only 0.5 arcsec will be twice as far away as one with 1 arcsec of parallax, and so on. Thus the distance in parsec is simply the reciprocal of the parallax angle.

1 Distance in parsec = parallax angle in arcsec 1 d= p As there is a simple geometrical relationship between the parallax angle and the radius of the Earth’s orbit (which is 1 AU by definition), the parsec can be easily related to 1 AU. As shown in the adjacent Physics file, it is equal to 206 265 AU. Apart from the metre, the other unit of distance you will come across in astronomy is the light-year (l.y.)—1 light-year is the

distance that light travels in a year. It is only about a third of a parsec. The actual relationships between these various units are shown in Table 11.1.

Table 11.1 Relationships between distance units Distance in:

m

AU

Light-year

Parsec

11

15

=

1.496 × 10

1

0.000016

0.000005

1 light-year =

9.461 × 10

63 240

1

0.3066

=

3.086 × 10

206 265

3.2616

1

1 AU 1 parsec

16

Note: Numbers less than 1 are given for illustration only. Use the corresponding factors >1 in calculations. In the 20th century, although the parallax of many thousands of stars was measured, most stars were too far away to show any measurable parallax. As a result, astronomers looked for indirect means of determining the distance to a star. The most obvious was to find a way of determining the actual intrinsic brightness of the star and compare it to the apparent brightness as we see it. If, for example, we knew that a certain star had the same actual brightness as our Sun it would be possible to calculate how much further away it was by comparing its apparent brightness to the Sun’s. The problem of course is how to determine intrinsic brightness. This is the subject of the next part of our story. However, these indirect methods have to be ‘calibrated’ by testing them on stars whose distances are known through direct methods, and for this reason it is most important to use stellar parallax methods on as many stars as possible. One limitation to parallax measurements is the distortion, or ‘shimmer’, of the star’s image caused by the Earth’s atmosphere. In 1989 the European Space Agency launched a satellite called Hipparcos to measure the stellar parallax of as many stars as possible from above the atmosphere. Hipparcos is an acronym for ‘high-precision parallax-collecting satellite’, but the name also honours the ancient Greek astronomer Hipparchus, who was one of the first to map the stars systematically. Hipparcos (the satellite) measured the parallax of more than 100 000 stars to an accuracy of 0.001 arcsec, in other words out to nearly 1000 pc. This may seem a vast distance, but given that our galaxy is about 50 000 pc wide and that we are about 8500 pc from its centre, you can see that a lot of stars are still ‘out of range’.

Figure 11.3 The telescope on the Hipparcos satellite was able to measure the parallax of stars to one-thousandth of an arcsecond.

Starlight—how bright? Astronomers measure the apparent brightness of stars on a scale that actually originated with Hipparchus in the second century bc! He called the brightest stars he could see ‘first-magnitude’ stars (+1), those about half as bright ‘second-magnitude’ (+2) and so on to those barely visible, which were ‘sixth-magnitude’. When astronomers sailed into the southern hemisphere they discovered brighter stars and so the scale had to be extended ‘upwards’ to 0 magnitude and then –1 magnitude and so on. When the telescope was invented the scale went ‘downwards’ to +7 and beyond. The scale is referred to as the apparent magnitude scale. Don’t be confused by the fact that the scale seems ‘backwards’—dimmer stars have a numerically higher magnitude. In the 19th century when astronomers were better able to measure the brightness of stars, they defined apparent magnitude more precisely by


393

Sun (–26.7)

–20 –15

brighter

–25

full moon (–12.6) –10 –5

Venus (brightest) (–4.4) Sirius A (–1.4) Betelgeuse (+0.5) E. Crucis (+3.6) naked eye limit (+6) Sirius B (+8.4) binocular limit (+10)

0 +5 +10

+20

dimmer

+15 visual—large telescope (+21)

+25 +30

photographic —large telescope (+30)

Figure 11.4 The apparent magnitude scale is a concoction of an ancient Greek scale with a 19th century mathematical redefinition! Visible stars range from –1.44 (Sirius A) ‘down’ to about +6 for stars barely visible under the best conditions.

Physics file

saying that a difference in magnitude of 5 corresponds exactly to a factor of 100 times in apparent brightness—which agreed roughly with the old values. In other words, it would take 100 stars of magnitude +6 to equal the brightness of one star of magnitude +1. Mathematically (see the Physics file on this page) this means that each level of magnitude represents a change in brightness of about 2.5 times (instead of Hipparchus’s ‘double’). In other words, it would take 2.5 stars of magnitude +6 to equal the brightness of a single +5 star, for example. Or it would take 2.5 × 2.5 = 6.3 stars of magnitude +6 to equal the brightness of a +4 star—and so on. Figure 11.4 shows the full extent of the scale with some examples. You might like to check with your calculator that it would take about 13 billion Sirius stars to equal the apparent brightness of the Sun! Clearly there must be a relationship between the apparent brightness of a star, its distance from us and its intrinsic brightness, the actual brightness of the star. Once the actual distances of a number of stars were known it was possible to determine their intrinsic brightness. It was soon found that stars vary enormously in intrinsic brightness. For example, Sirius and Canopus are the two brightest stars in the sky; however, although Canopus appears almost as bright, it is actually about 36 times further away than Sirius. To make up for this big difference in distance, it can be calculated that it must be about 3000 times brighter than Sirius. On the other hand, the very closest star, Proxima Centauri, is not even visible to the naked eye and so must be a very dim star. In fact it is intrinsically only about 1/30 000 as bright as Sirius. The intrinsic brightness of stars varies over a huge range. We will see later that the range is even greater than these examples suggest! The intrinsic brightness is often given on a scale of absolute magnitude, where this is defined as being equal to the apparent magnitude that the star would have if it was at a distance of 10 pc. So the apparent and absolute

Because of the way magnitudes are defined, they are effectively a logarithmic scale. The change of 5 in magnitude corresponding to a factor of 100 is equivalent to the statement b5 = 100, or 5 = logb100, where b is the base of the log scale. To make this true, b has to be close to 2.512 because 2.5125 = 100 (try it on your calculator). This means that an increase in the apparent brightness of 1 on the magnitude scale corresponds to about 2.5 times the brightness (instead of two times as in Hipparchus’s original scheme).

Sirius

Canopus

Figure 11.5 While Sirius (mid right) is the brightest star in the sky as we see it, Canopus (mid left) is intrinsically 3000 times brighter. However, it is 36 times further away. Note Orion (lower right) and the Large Magellanic Cloud (lower left).

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magnitudes of a star at 10 pc are the same. If the star is further away it will appear fainter and so the absolute magnitude would be brighter (i.e. more negative!) than its apparent magnitude. If, for example, Canopus were at 10 pc instead of almost 100 pc, it would have an apparent magnitude of a whopping –5.5, much brighter even than Venus at its best. On the other hand, if Sirius were at 10 pc instead of 2.6 pc, it would be just an ordinary +1.5 star.

d=1 d=

2

Brightness and luminosity While the apparent magnitude and absolute magnitude scales are convenient for observational purposes, the astrophysicist actually needs to know the apparent brightness and intrinsic brightness in SI units; that is, in watts per square metre of received radiation, and watts of total radiated power, respectively. When measured in this way the intrinsic brightness is called the luminosity (L) and is measured in watts. (There is no point in using multiple units such as megawatts because the power is so huge it is well beyond any normal multiple.) It may seem a tall order to measure these quantities, but in fact it is not so difficult. We know how far away the Sun is and we know how much energy we receive from it. For example, when thinking about solar panels, engineers assume that every square metre of roof facing the full Sun will receive about 1 kilowatt of power; that is, 1000 joules of energy every second (1000 W m-2). Now this is just what we mean by the apparent brightness (b) of the Sun. Actually the atmosphere absorbs some of the Sun’s radiation, but instruments aboard spacecraft have found that, at the Earth’s distance from the Sun, the apparent brightness is close to 1370 watts per square metre. Now if 1 square metre receives that much power, so must every other square metre around the gigantic sphere with radius 1 AU. Multiply the 1370 W m-2 by the total area of that sphere in m2 and we have the total power output of the Sun! We shall do this in Worked example 11.1A.

Worked example 11.1A What is the luminosity (L) of the Sun, given that the radiation received at the Earth, the apparent brightness, is 1370 W m-2?

Solution The area of a sphere is given by 4πR2 and so the total area over which the Sun’s radiation is spread at the distance of the Earth is a sphere of area 4πR2, where R is 1 AU, which we must express as 1.50 × 1011 m. As each square metre of this area receives 1370 W, the total power produced by the Sun must be given by this power times the area: L= 1370 × 4π × (1.50 × 1011)2 = 3.9 × 1026 W You can see from this example that we can write this relationship in a general way. !

L 4pR2

LUMINOSITY L = b × 4πR2 or b =

where L is the luminosity in watts b is the apparent brightness in W m-2 R is the distance to the star in m.

Figure 11.6 The Sun’s radiant power spreads out uniformly. These two spheres would receive the same total radiation, but because the larger one has four times the area the apparent brightness would only be one-quarter as much at that distance.

PRACTICAL ACTIVITY 52 The inverse square law

Physics file Inverse square laws occur in a number of situations in physics. The relationship describing luminosity is known as an inverse square law because the apparent brightness (b) decreases with the square of the distance (R2) from the Sun. If we were to go to half the distance from the Sun, as shown in Figure 11.6, our 1 m2 would receive not twice, but four times the amount of solar radiation, as the total radiation is spread over an area one-quarter the size. An inverse square law has the general form: 1 y∝ 2 x It can also be written in ratio form: y1 x2 2 = y2 x1 In our expression for apparent brightness, note that the brightness is also directly proportional to the luminosity.

( )


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Physics file Be careful with the term ‘brightness’. Sometimes it refers to the intrinsic brightness (the luminosity, L) and sometimes the apparent brightness (b). If the adjective is not given, the context will make clear which brightness is meant.

When determining the luminosity of other stars, astronomers compare the apparent brightness of the star with that of the Sun, and then make corrections for the big difference in distance. Quantities relating to the Sun are shown with the subscript , and so we can write that for the Sun L  = b  × 4πR 2. The same equation holds for the other star, but we don’t use any subscripts: L = b × 4πR2. Divide this equation by the first and we have: L b R 2 = × L b R

()

This expression holds for any two stars and so we can simply say: for two stars the ratio of the luminosity is equal to the product of the ratio of the apparent brightness and the square of the distance ratio. For most purposes astronomers don’t worry about the actual value of L , they simply give the other star’s luminosity in terms of L . (The actual luminosity of the Sun is taken to be L = 3.86 × 1026 W.) The brightness ratio can be worked out from the apparent magnitudes, but that is a little messy so we shall just quote the brightness ratio directly.

Worked example 11.1B What is the luminosity of Sirius if its apparent brightness is only 8.8 × 10-11 that of the Sun, and its distance is 8.61 l.y.?

Solution First we must convert the distance into AU. This will give us the distance ratio directly, as R = 1 AU. From Table 11.1, 1 l.y. = 63 240 AU and so the distance ratio is R/R = 63 240 × 8.61 = 5.44 × 105. We know the brightness ratio b/b and so the luminosity ratio is: L b R 2 = × = 8.8 × 10-11 × (5.44 × 105)2 = 26 L b R

()

so L = 26L So Sirius has an intrinsic brightness of 26 times the Sun, i.e. it is 26 times as powerful. It is helpful to think about the physical situation in this example: Sirius appears dimmer than the Sun by a huge amount (about 1010). This is mostly because it is about 5 × 105 times further away. But we compensate for the extra distance by multiplying by the square of the distance ratio (25 × 1010). In fact this more than compensates, indicating that Sirius is intrinsically about 25 times brighter than the Sun. In this way astronomers have been able to determine the luminosity of the thousands of stars for which stellar parallax has been measured. They find an enormous range: from about a ten-thousandth of the Sun’s luminosity up to a million times. This means that the brightest stars are about 10 billion times more powerful than the dimmest. Stars certainly come in a huge variety! Our Sun is somewhere a little below the middle in the range of luminosity, but the stars are not distributed evenly along this range. Most stars are dimmer than the Sun. Only about 10% of stars are brighter than the Sun. Clearly the relationship between luminosity, apparent brightness and distance suggests that if we could determine the luminosity of a star without knowing its distance we could use its apparent brightness to determine the distance to the star. Fortunately, astronomers have discovered that there is a way! The clue is in the colours of the stars. Perhaps surprisingly the colour also enables us to find the size of the star.

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Stars come in many colours... A casual look at the sky suggests that stars are all pretty much the same ‘white’ colour. A more careful examination reveals that there are differences. Consider the two bright stars that make up Orion’s shoulders—Betelgeuse and Bellatrix. They are the two bright stars a little below the saucepan. Bellatrix is bluish while Betelgeuse appears redder. Compare them with Sirius, which is quite white. Because the stars are so bright against the dark background our eyes find it hard to distinguish the colours of the stars, but a simple photograph of the stars reveals a different story.

Figure 11.7 This photograph was a half-hour exposure on 400-speed film at f/8. Can you find the Pointers and Southern Cross? The South Celestial Pole is a little below the left side of the picture. Notice the different colours of the stars.

The photograph in Figure 11.7 was taken with a normal SLR camera. It shows that the colours of the stars vary considerably, from quite blue, through greenish yellow and white to red. The explanation is quite simple. If you put a poker in a fire it becomes red hot. As the fire becomes hotter the poker becomes brighter and whiter. A similar change can be noticed as you turn up the dimmer control on electric lights. This change of colour as temperature increases is not confined to pokers and light-bulb filaments. Anything that becomes hot gives off light, which starts as dull red and gets whiter and then bluer as the temperature increases. People who work with kilns, for example, can tell the temperature of the kiln just from the colour of the light. Astronomers interpret the colours of the stars in the same way— the colours are an indication of the surface temperature. Astronomers can find the temperature of the star by analysing the starlight. Physicists have long studied the colour of light from hot objects and there is a well-known relationship between it and the temperature of the body emitting the light. We will not go into the details here, but suffice it to say that the temperature of a so-called ‘black body’ (and this includes stars!) determines the spectrum of the light emitted by it (see Physics in action ‘Black bodies are not always black!’ on page 399). Conversely, if astronomers can plot the spectrum, they can determine the temperature. Although stars are not perfect ‘black bodies’—there are dips and peaks in the radiation spectrum—the overall shape of the curve fits the black-body curve well. We will also see that the differences tell us other important things about the star. To analyse the spectrum of starlight, astronomers measure the brightness of stars as seen through special filters that allow only part of the spectrum through. One system uses three filters that separate the spectrum into three


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sections called U, B and V, as shown in Figure 11.8. The U filter allows only ultraviolet light through, the B filter allows violet and blue, while the V (for visible) allows a spectrum that more or less corresponds to the sensitivity of the eye, mostly green-yellow. The apparent brightness of the star is recorded for each filter. There are different ways of interpreting the results of this process. A simple one is to determine the ratio of the brightness in V (bV) to the brightness in B (bB). The hotter the star, the more its radiation will be in the blue part of the spectrum compared to the yellow-green, and hence the smaller the ratio bV/bB. The ratio in B and U, bB/bU, can be used as well. Table 11.2 shows the relationship between the colour and the surface temperature for a number of stars. 100

Figure 11.8 The wavelength range transmitted by the three UBV filters astronomers use to determine the colour, and hence temperature, of a star’s surface. U transmits only ultraviolet light, B transmits only the violet-blue end of the visible spectrum, and V transmits a range similar to the sensitivity of the eye.

Transparency (%)

80

U

60

B

V

40 20 0

300

400

500 Wavelength (nm)

600

700

Table 11.2 Colours and temperatures of stars Star

Apparent colour

bV/bB

Surface temperature

Bellatrix (Orion)

Blue

0.81

21 500 K

Regulus (Leo)

Blue-white

0.90

12 000 K

Sirius (Canis)

Blue-white

1.00

9 400 K

Altair (Aquila)

Yellow-white

1.23

7 800 K

Sun

Yellow-white

1.87

5 800 K

Betelgeuse (Orion)

Red

5.55

3 500 K

...and sizes Figure 11.9 Orion contains both a very blue star (Bellatrix) and a very red star (Betelgeuse). Can you identify them in this photograph of Orion?

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If we know a star’s surface temperature and its luminosity we can work out its size. This may seem surprising, but picture the hot poker we had in the fire earlier. When you pull it out of the fire you will feel it radiating a lot of heat. The hotter it is, and the bigger it is, the greater the amount of heat radiated. There is a relationship between the total amount of heat radiation, the area of the surface and the temperature (see the Physics in action, page 399). This enables astronomers to calculate how much surface area is required at the temperature of the star, to radiate the amount of heat energy the star is known to be producing (the star’s luminosity). Once the surface area of the star is known, the radius can easily be calculated (as A = 4πR2). This way of finding the size of a star depends on knowing its luminosity, which in turn depends on knowing its distance (and apparent brightness). While this can be done for ‘parallax stars’, we still have the problem of

determining the distance to most of the stars in the sky! Once we have either the distance or the luminosity we can determine the other (because we know the apparent brightness)—so is there some other way to find the luminosity of distant stars? There is also a problem with our method of determining the temperature from the colour. Just as a sunset is red because the light has travelled through a lot of atmosphere, the light from some stars has travelled through huge clouds of gas and dust floating around in interstellar space. These are not always easy to see but they can alter the colour of stars considerably, and so make the determination of temperature unreliable. You have probably guessed by now that astronomers have solved both these problems! It involves a careful analysis of the spectrum of stars. We return to this in section 11.3. To help in our understanding of what we can see in the spectrum of a star we shall first take a closer look at our nearest star.

Physics file Another system for categorising the colour of stars is the ‘colour index’, which consists of the difference in the absolute magnitudes of stars as measured through the blue and visible filters, B–V. In this system blue stars have a negative index (remember that brighter stars have a lower magnitude). Bellatrix has an index of –0.22, Sirius is +0.01, Betelgeuse is +1.50 and our Sun is +0.6.

Physics in action

Black bodies are not always black! As well as the total energy, the colour of the radiation changes with temperature. In 1893 Wilhelm Wien derived a law that related the wavelength peak of the radiation to the temperature—wavelength is inversely proportional to the absolute temperature. For example, if the temperature doubles, the wavelength halves. The hotter the object, the more the radiation emitted moves to the blue end of the spectrum. Figure 11.11 shows the spectrum of radiation for several temperatures. Notice that the shape of these curves depends only on the temperature and not on the body. The Stefan– Boltzmann law tells us that the total area under the graph increases with temperature, and the Wien law tells us that the peak moves to the left with temperature. Astronomers use both these laws to relate the temperature and luminosity of the star to its size. Visible light

Intensity

The term ‘black body’ means that we are concerned with an object’s radiation due only to its heat, not some other factor such as its colour or spectral lines. A poker in a fire is a good example. It may be black to start with, but can become red hot. The light-bulb filament is another. Even though they are totally different sizes, if the glowing poker and light-bulb filament are the same colour, it turns out that they must be at the same temperature. (This is actually the basis of ‘radiometric’ measurements of the temperature of very hot objects.) Two important relationships enable physicists to relate radiation emitted to temperature. The Stefan–Boltzmann law (from around 1884) relates the total amount of energy emitted per second from each square metre of a surface to the temperature of the surface. Clearly, the hotter a surface the more energy that must be being radiated from the surface. As you can imagine, the energy radiated rises steeply with temperature. In fact, the energy radiated is proportional to the fourth power of the absolute (kelvin) temperature.

12 000 K 6000 K 3000 K

0

500

1000

2000

3000

Wavelength (nm)

y is an accurate al in this foundr et m t ho e th e colour of Figure 11.10 Th temperature. its of n indicatio

Figure 11.11 As the temperature of a black body increases, the amount of energy radiated (shown by the area under these graphs) increases, and the peak wavelength decreases (it becomes bluer).


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11.1 summary The stars—how far, how bright? • The distance to stars can be measured by the parallax movement they show as a result of the Earth’s revolution around the Sun, but even the largest parallax found is less than 1 arcsec. • 1 pc is the distance to a star that would show 1 arcsec of parallax. 1 pc is about 3.26 l.y. or 206 265 AU. • The distances to more than 100 000 stars have been measured by stellar parallax, but most stars are too distant for this method. • The apparent brightness of stars can be measured on a scale of apparent magnitude. On this scale the brightest stars are around –1 and those barely visible are around +6.

11.1 questions The stars—how far, how bright? 1 In the late 1500s Tycho Brahe argued that the Copernican (Sun-centred) system could not be correct because if the Earth was moving around the Sun we should notice parallax in the stars. Despite his careful observations he detected none. a Explain what is meant by parallax. b Was his argument basically sound? c Was he correct in his conclusion? 2 An arc second is a very small fraction of a degree. a How many arc seconds are there in 1°? b The angular diameter of the Moon is about 0.5°. How many arc seconds is that? c How far away would a 5-cent coin need to be to subtend an angle of 1 arcsec? 3 As you watch a car travelling along a distant road you find that it moves an angle of 1° in 4 s. If the road is 6 km away from you, how fast was the car travelling? You then notice that a police car stops the car for speeding in a 100 km h-1 zone. What could be an explanation for this? 4 a Astronomers use a distance measurement called the parsec. What is it, how is it found, and why do they use it? b If a star shows a parallax of 0.44 arcsec, how far away is it in pc, l.y., AU and m? 5 Explain why the scale for the apparent magnitude of the brightness of a star has negative values for the brightest stars.

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Detailed studies

• The absolute magnitude is equal to the apparent magnitude that a star would appear to have if at a distance of 10 pc from us. • The luminosity of a star is the total power radiated in watts. The Sun has a luminosity of L = 3.86 × 1026 W. Other stars may have up to a million times greater or down to one ten-thousandth of L , but most stars are less bright than the Sun. • Stars vary considerably in colour. This is due to differing surface temperatures. Measurements of the brightness at different colours (UBV) enable the temperature to be deduced. • For stars whose luminosity and surface temperature are known it is possible to use basic laws of physics to deduce the radius of the star.

6 Imagine that we can put a completely absorbent black sphere around a 100 W light bulb. a If the sphere had a radius of 2 m what would be the total radiation intensity (in W m-2) on the inside? b How much heat energy would be absorbed by the sphere? c What type of radiation would most of this be? 7 If the intensity of sunlight on Earth is 1370 W m-2, what would it be on: a Venus, which is at 0.72 AU? b Jupiter, which is at 5.2 AU? 8 From Earth the stars Deneb and Mimosa appear to be the same brightness. However, Deneb is at a distance of 3230 l.y. and Mimosa at 353 l.y. Which star has the greater luminosity, and by how much? 9 Evan has two hot pokers side by side in the fire and glowing a dull red. One has twice the mass of the other. When he pulls them out of the fire and puts his hand near them he feels more heat from the larger one. The best explanation for this is that: A the radiation emitted from the larger one will be of a longer wavelength. B the radiation emitted from the larger one will be of a shorter wavelength. C more radiation will be emitted by the larger one because it is at a higher temperature. D more radiation is emitted from the larger one because it has more surface area.

11.2 Our favourite s ta

r

The brightest and most important star in our sky is the Sun. Without it the Earth would be a frozen wasteland of ice and rock. Our planet is at just the right distance to receive enough radiation to keep water liquid. This in turn enables the sea and atmosphere to distribute heat around the Earth in such a way that the planet is habitable. Any closer and we would have the nightmarish greenhouse of Venus; any further away and the oceans would freeze. For humans, the Sun has always been an object of fascination, even worship. We will look at it through the eyes of the astrophysicists, who specialise in trying to understand the processes that drive the universe.

Our best-known star It was only after Galileo and Newton that we began to realise that the stars were actually Sun-like objects a very long way away. The best way to learn about stars, then, was to look at the closest one. After Galileo’s discovery of sunspots it was realised that the Sun rotates on its axis. Curiously enough, however, the equator of the Sun was seen to have a period of about 25 days while regions of higher latitude took several more days for a full cycle, indicating that the Sun is not a solid body like the Earth. Because it is obviously so hot it was assumed to be gaseous. While the relative distances between the Sun and planets were known quite accurately as a result of the work of Tycho Brahe and Johannes Kepler, it was a far harder task to find the actual distances. The problem was to find the scale of the ‘map’ of the solar system. The only way to do that was to use a triangulation method somewhat akin to the parallax methods we have discussed for the stars. If a planet can be observed from opposite sides of the Earth at the same time, against the background of stars, the distance to the planet can be calculated in terms of the Earth’s known diameter. As this distance will also be known in astronomical units, the scale can be found. This was attempted with some success in the late 1800s, but it was not until the advent of astronomical photography and better transport that reliable determinations could be made. One of the first sightings used the asteroid Eros, which came within 0.27 AU of the Earth in 1901. The advantage of using Eros was that because it was small it was easier to pinpoint its location at a certain time. In the later 20th century it became possible to bounce radar pulses off the closer planets and hence find the time for light to travel the distance to the planet. This enabled very accurate determinations of the scale of the solar system map, and hence the value of 1 AU.

d

x

Figure 11.12 The brightest star in our sky—our Sun. It is a huge ball of very hot, very dense gas.

Eros

Figure 11.13 In order to determine the actual, as distinct from relative, distances in the solar system, some must be measured in terms of a known distance—such as the Earth’s diameter.


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Physics file

Table 11.4 Astrophysical constants Constant

Symbol

Value

Mass of Earth

ME

5.97 × 1024 kg

Radius of Earth

RE

6.38 × 106 m

Mass of Sun

M

1.99 × 1030 kg

Sun–Earth distance

1 AU

1.496 × 1011 m

Luminosity of the Sun

L

3.86 × 1026 W

Knowing the distance to the Sun, it is possible to determine its size very accurately. Furthermore, by using Newton’s law of gravitation it is possible to determine its mass (see the Physics in action, below). These two values can be used to determine the average density of the Sun. These values are summarised in Table 11.3. We can see that the Sun is gigantic compared with the Earth, except for its average density, which is only about 1.5 times that of water. On the other hand, as the Sun is basically composed of gas, we can see that it is a very dense gas!

Table 11.3 The physical properties of the Sun Average distance to the Sun (1 AU)

1.496 × 1011 mm (390 times as far as the Moon)

Angular diameter from Earth

0.5º (same as Moon)

Mass of Sun

1.99 × 1030 kg (300 000 times the Earth)

Diameter of the Sun

1.4 × 109 m (109 Earth diameters)

Average density of the Sun

1.4 × 103 kg m-3 (Earth 5.5 × 103 kg m-3)

Physics in action

Newton, gravity and the mass of the Sun As we have seen in our core studies, Newton realised that the gravity that held apples, and us, to the Earth also held the Moon to the Earth, and the planets to the Sun. By analysing the motion of the Moon around the Earth, and the planets around the Sun, he was able to find his ‘universal law of gravitation’, which stated that the force of gravity between any two objects, whether people or planets, depended on the product of the two masses and the inverse square of the distance between them. (This is another example of an inverse square law similar to the one for starlight discussed earlier.) Newton’s gravitation law can be written F = GMm/R2, where F is the force of gravity between the two masses M and m, a distance R apart, and G is the universal gravitation constant. G has the very small value of 6.7 × 10-11 N m2 kg-2. At least one of the masses involved needs to be very big in order to produce any significant force. If you use the Earth’s mass for M, your own mass for m, and the Earth’s radius for R, you can calculate the force of Earth’s gravity on you—try it. (The values for these numbers can be found in Table 11.4.) Once Newton had worked out the mechanics of the Earth’s motion around the Sun, he could find the force needed to hold the Earth in its orbit—which depended on the Sun’s mass. He could then express the mass of the Sun in terms of G. The problem was that he could only guess at the value of G because he could not know the Earth’s mass. However, Figure 11.1 4 G was found experimentally by Cavendish in 1798. Newton re It is said that while alised tha sitting in h t th is (This, in turn, also enabled the Earth’s mass to be held the M oon to the e gravity that held grandmother’s orc h apples to Earth. (It is found from the known strength of gravity at the a falling a the Earth ard, pple hit him also said also that he re surface.) on the he a

li ad—proba bly a myth sed this when !)

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Detailed studies

Where does the energy come from?

Figure 11.15 The Sun, photographed by a spacebased X-ray telescope. The bright patches are regions that are producing solar flares.

For a long time this was one of the great mysteries of physics. It was known that life had existed on the Earth for at least several hundred million years and so that meant that the Sun must have been radiating energy at much the same rate for at least that time. However, any known mechanism for producing heat could not possibly have generated so much energy. If the Sun was fuelled by even the most efficient chemical fires it would have lasted only around 10 000 years. The English physicist Lord Kelvin (of absolute temperature fame) and the German physicist Hermann von Helmholtz suggested that vast amounts of heat would be generated from the enormous weight of gas collapsing into the Sun. As the gas falls, the potential energy is converted into kinetic energy—just as compressing air in a bicycle pump gives it kinetic energy, which heats it. This was good physics, but again it could not possibly last for the billions of years that the Sun has been producing energy. However, this process, now known as Kelvin–Helmholtz contraction, does turn out to be important in the formation of new stars. Reasonably straightforward calculations based on the mass and energy output of the Sun showed astrophysicists that the amount of energy being produced for each atom in the Sun was hundreds of millions of times greater than the energy produced by each atom in chemical reactions. Something very different was going on in the Sun! The clue came with Einstein’s theory of special relativity in the years after 1905: we all know that he came up with the equation E = mc2. The real meaning of this equation is not simple (we deal with it in the ‘Einstein’s Special Relativity’ study next year); however, a reasonable interpretation for our present purposes is that there is a huge amount of energy locked up in mass. The factor c2 is the speed of light squared and so is a very large number (9 × 1016 m2 s-2). This means that if this energy can be released, a huge amount of energy will be produced at the expense of a small amount of mass. This suggested to the astrophysicists that the Sun may somehow be tapping this so-called ‘mass energy’. But what was the process involved? Because of the enormous amount of energy radiating from the Sun, and assuming the energy was being produced in the centre, it could be calculated that the temperature at the centre of the Sun must be millions of

Physics file Why is so much energy released when hydrogen atoms fuse to form helium? We can think of it this way: if two ordinary magnets are placed near each other so that they attract, and then let go, they will ‘fall’ together, gaining speed. When they collide, this kinetic energy is released as a little heat and sound energy. Something like this occurs when hydrogen nuclei ‘fall’ together to form a helium nucleus. The nuclear force that pulls the protons and neutrons together to form helium is enormously strong, but operates only over a short range (otherwise all hydrogen atoms would collapse together, forming helium). In order for the nuclear force to take effect, the huge repulsion between the positively charged protons has to be overcome. This is where temperature is important. Only with the sort of kinetic energy the particles have inside the core of the Sun can the particles come close enough, against the electrical repulsion, for the nuclear force to take over. When they do, however, they ‘fall’ together with incredible force, releasing huge amounts of energy.


403

Figure 11.16 Einstein discovered that mass and energy were interrelated by his famous equation … = mc2.

degrees. This would mean that the atoms at the centre would be stripped of their electrons and there would be just a frenzied mass of nuclei and electrons flying around at enormous speeds. Because the density of the Sun was relatively low, it was assumed that it must be composed mostly of hydrogen with perhaps some other light atoms. It was already known that the Sun contained helium, in fact it had been discovered in the Sun before it was known on Earth. In the 1920s British astrophysicist Robert Atkinson suggested that under the conditions at the centre of the Sun the hydrogen nuclei might ‘fuse’ together, creating helium nuclei. If this were the case huge amounts of energy would be released (see Physics file on page 403). That the energy released in this ‘hydrogen fusion’ is so much more than that released in chemical reactions can be seen from the fact that chemical reactions involve forces between the electrons around the outside of the atom, whereas the nuclear fusion involves the electrical (and nuclear) forces between the particles in the nucleus. As the electrical force increases with the inverse square of the distance between charges (that inverse square law again!) and the charges in the nucleus are about 10 000 times closer than those in the outer regions of the atom, the forces between nuclear particles are about 10 0002 or 100 million times larger—and so is the energy involved. This is the source of the energy that powers the Sun.

Physics file Is mass converted into energy? It is often stated that in nuclear reactions ‘mass is converted into energy’. This is really an oversimplification of Einstein’s ideas. There is no mass ‘shaved off’ the particles and somehow mysteriously turned into energy. What is true is that the total mass of the particles in a helium nucleus is a little less than the total mass of the equivalent particles in hydrogen nuclei. However, they are exactly the same particles as before, except that they now have a little less total potential energy than previously (that was what went off to power the Sun). Einstein showed that mass is actually a property, not of the individual particles, but of the system of particles including the energy bound up in the forces between them. As some of the energy has been lost, so also has the equivalent mass (as given by Einstein’s equation). In fact there is no fundamental difference between a nuclear reaction and a chemical reaction in this sense. A group of hydrogen and oxygen atoms will also have a slightly greater mass than the equivalent group of water molecules formed when the hydrogen ‘burns’ with the oxygen, releasing energy. It is just that in this case the mass equivalent of the energy released is so tiny it is completely unmeasurable. But E = mc2 applies just as surely in this case as in the nuclear case.

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Detailed studies

Figure 11.17 Nuclear reactions involve forces that are about 100 million times as great as those involved in chemical reactions. This is about the same ratio as the force required to keep a jumbo jet in the air to that required for a mosquito.

Worked example 11.2A The Sun is giving out about 4 × 1026 J of energy every second as visible and invisible radiation. At what rate is it losing mass due to this energy loss? How significant is this mass in proportion to the total mass of the Sun?

Solution This energy comes from the fusion of hydrogen into helium with a corresponding loss in the potential energy of the nuclei. This loss of energy will correspond to a mass loss of m = …/c2 = 4 × 1026/9 × 1016 = 4.4 × 109 kg every second, or about 4 million tonnes every second. As the mass of the Sun is 2 × 1030 kg this is only about 10-21 of the total mass! At this rate it would take 2 × 1030/(4.4 × 109 × 365 × 24 × 3600) = 1 × 1013 or about 10 000 billion years to consume itself. Actually only a little under 1% of the mass of hydrogen nuclei is lost in fusion so that reduces this value to under 100 billion years. However, as the Sun is ‘only’ about 5 billion years old we are in no danger of it running out of fuel for some time!

Modelling the Sun It was the combination of Einstein’s theory of relativity and the improved model of the atom (Bohr’s quantum atom was suggested in 1913) that enabled astrophysicists to understand the processes occurring in the Sun. Today, computer models have enabled us to gain considerable insights into the mechanisms that keep the Sun ‘burning’ in our sky. These models are based on the same laws of physics that govern the transfer of heat from a log fire to you, or the laws that tell us that the pressure will increase as we dive into the ocean, and so on. The basic principle used in modelling the Sun is that any part of it must be in what is called hydrostatic equilibrium. That means that on any ‘piece’ of Sun, the inward pressure from the weight of all the material above it must be balanced by the outward pressure of the radiation released by the nuclear reactions in the core below. The origin of the energy is rather different from the log fire, and the Sun’s interior is much hotter than the ocean, but the same basic laws of physics apply. The models take an imaginary piece of the Sun, put in assumptions about quantities such as its mass, temperature, heat content and ability to transfer heat, the pressure gradient across it and much more. Then the conditions to ensure that it is in hydrostatic equilibrium are determined. This is repeated over and over until the whole Sun is accounted for, each ‘piece’ interacting with its neighbouring pieces. A computer then ‘runs’ the Sun to see what would happen. If the computer Sun blows up or dies out we know some of our assumptions were wrong! Many such scenarios can be tested until they are found to be consistent with the known facts about the Sun, such as its mass, size, energy output and surface temperature— and the fact that the Sun has produced a very constant output over the last few billion years. These models suggest a number of interesting ‘facts’ about the Sun. As it is impossible to test them, they must be seen as theoretical predictions rather than facts, but astrophysicists have a lot of confidence in their model. The main features of the model are that the nuclear fusion occurs in a zone that extends from the centre out to about 0.25 of the radius (R ). The temperature in this region is above 10 million degrees, the density about 160 000 kg m-3 (that’s 14 times the density of lead!), and the pressure about 340 billion times the Earth’s atmospheric pressure. The energy flows outward from this zone by a combination of convection and radiative diffusion. The convection is just like that from a room ‘convection heater’— hot air rises, cooler air sinks. Radiative diffusion is a process in which light bounces around, transferring heat in the process. Radiative diffusion is the main mechanism for energy transfer out to about 0.7R where convection takes over. At this point the temperature is down to about 1 million degrees and the protons and electrons come together to form hydrogen atoms, which absorb the light more effectively, and so the radiative diffusion becomes less effective. The density is only 80 kg m-3 (much less than water, but about 60 times that of the air around us) and the pressure is only about 10 million ‘atmospheres’! Most (99%) of the Sun’s mass is below this level. By the time we reach the surface (in this very imaginative journey) the temperature is down to the 5800 K we ‘see’ as we look from our more comfortable position on Earth. There is of course no real ‘surface’, just a point where the churning hot gases start to sink again as they lose their energy by radiation out into space. The motley appearance of the Sun in detailed photographs (such as Figure 11.19) is due to the convection currents reaching the surface, cooling and then sinking again. This may

photosphere 6000 K

visible, IR and UV radiation

10 000 K radioactive zone

coronal loops

neutrinos core thermonuclear reactions 14 000 000 K

radio emissions

prominence

X and gamma radiation

Figure 11.18 The astrophysicists’ model of the Sun. The nuclear fusion occurs in the central core, from which energy slowly travels by radiative diffusion out to about 0.7R where the nuclei and electrons recombine. From here out the energy transfer is by convection.

Figure 11.19 The granular appearance of the Sun’s surface is due to the convection currents reaching the surface. Each granule is about 1000 km wide. They last a few minutes before disappearing and re-forming.


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all sound like a rapid process but in fact it has been calculated that energy produced by the nuclear fusion in the core takes about 170 000 years to travel out to the surface. Then it only takes 8 minutes to reach the Earth. The light energy we see by today was generated inside the Sun when humans were just beginning to diverge from the apes!

The Sun’s atmosphere... As the Sun is all gaseous, there is no real surface, but the thinner outer layers are often referred to as its atmosphere. The layer that we see in photographs such as Figure 11.12 is called the photosphere because it is the hot (5800 K) thin layer from which the Sun’s visible light is emitted. In fact it is only about 400 km thick, in the sense that we can see only about that far into it. Note that we are doing very well to see that far in the Earth’s atmosphere! Compared with the Sun’s diameter, however, the photosphere is very thin and so we see it as a sharp clear surface. There are two more layers above the photosphere. Normally we can’t see them, but during an eclipse when the Moon blocks the light from the photosphere, we can see a pinkish layer just above the photosphere. This is called (because it is coloured) the chromosphere. It is extremely thin, only about one hundred-millionth as dense as our atmosphere, and extends to about 2000 km above the photosphere. The chromosphere glows with coloured light for reasons we shall discuss shortly. Curiously enough, the temperature of the chromosphere increases with height, reaching about 25 000 K at the top. The top layer of the Sun’s atmosphere is the corona, which is the spectacular, but faint, whitish glow around the Sun that becomes visible only in an eclipse (or a specially designed telescope in space). It is so thin that we would regard it as a very high vacuum if we produced it in the laboratory. However, it extends for millions of kilometres out from the Sun’s surface. It consists of highly ionised ions (i.e. atoms that have lost many of their electrons) and free electrons at temperatures over 1 million degrees. Because particles at this temperature are moving so fast, some of them escape the Sun’s gravity and stream out into space. This is the origin of the ‘solar wind’ which, when it interacts with the Earth’s magnetic field and atmosphere, creates the auroras seen in polar regions. While the amount of material streaming from the Sun in the solar wind amounts to about a million tonnes every second, at this rate the Sun will only lose well under 1% of its mass in its entire lifetime. Most of the solar wind consists of protons (hydrogen nuclei) and electrons, but a small percentage contains ions of heavier elements. Figure 11.20 The corona we see around the Sun in an eclipse is there all the time but masked by the brilliance of the Sun. It is the origin of the ‘solar wind’ that, among other things, causes the beautiful auroral lights on Earth.

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...and its magnetic field

Average number of sunspots

The Sun’s atmosphere and magnetic field interact to produce a number of spectacular effects including sunspots, solar flares and coronal mass ejections. The Sun’s magnetic field appears to be much more dynamic than the Earth’s. While the Earth’s field may reverse a few times in a million years, the Sun’s does it on a regular 11-year cycle. This is the underlying reason for the well-known 11-year sunspot cycle. Models of the Sun’s magnetic field suggest that it becomes distorted because of the differential rotation of the Sun, in effect producing ‘knots’ that cause large disturbances in the photosphere, which can last several days. These disturbances are cooler, at 4300 K, and look dark against the average 5800 K surface. We see them as sunspots. Eventually the magnetic field gets itself in such a knot that it reverses and repeats the cycle all over again. 300 200 100 0 1880

1900

1920

1940

Year

1960

The ‘magnetic storms’ that give rise to sunspots are also the origin of the solar flares and other emissions from the Sun’s surface that affect the chromosphere and corona. A coronal mass ejection can blast a billion tonnes of coronal gas into space at speeds of hundreds of kilometres per second. These events can produce large increases in the solar wind which, when they reach the Earth a few days later, are hazardous to astronauts and even to communication and electric power systems on Earth.

1980

2000

2020

Figure 11.21 Sunspots come and go with an 11‑year cycle, with another maximum predicted for 2018.

Figure 11.22 Magnetic storms on the Sun’s surface create solar flares that result in ‘coronal mass ejections’. These storms result in increased solar wind activity that can be hazardous for astronauts and can sometimes cause interference to electrical systems on Earth.


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11.2 summary Our favourite star • The relative sizes of the orbits in the solar system have been well known since Kepler, but it required triangulation using the Earth’s diameter to determine the actual distances. • The distance to the Sun, its size, mass and density have been determined very accurately. • The source of the Sun’s energy was a mystery until the nature of nuclear forces was understood. • The Sun produces energy by the fusion of hydrogen nuclei to form helium nuclei.

• Computer modelling of the processes in the Sun enables us to deduce much about the mechanisms that drive it. • The light we see from the Sun is emitted from the photosphere, the temperature of which is 5800 K. • The solar wind originates in the Sun’s uppermost atmosphere, the corona, which we can see only during an eclipse. The solar wind influences the Earth’s upper atmosphere.

11.2 questions Our favourite star 1 The Sun and Moon appear to be almost the same angular size in our sky (about 0.5°). Use this fact and the data in Table 11.3 to work out the diameter of the Moon. 2 Even before the process of nuclear fusion was known, astrophysicists deduced that the temperature inside the Sun must be millions of degrees although the surface temperature was ‘only’ about 5800 K. Can you suggest reasons why they would have come to this conclusion? 3 Astrophysicists of 100 years ago were very mystified about the origin of the Sun’s energy. Why were they so mystified? 4 Why is it that the nuclear reactions involving hydrogen in the Sun can produce so much more energy than hydrogen burning in oxygen on the Earth? 5 If 1 g of hydrogen is burnt with oxygen, about 1.4 × 105 J of energy are released. If 1 g of hydrogen are fused to produce helium, about 6 × 1011 J of energy are released. In 1 g of hydrogen there are 6 × 1023 atoms. a How much energy is released per atom in each of the two types of reaction? b How much greater is the amount of energy released per atom in nuclear fusion than in hydrogen–oxygen burning? c Burning 1 kg of coal releases about 30 MJ of energy. How much coal would need to be burnt to produce the energy released by the fusion of 1 g of hydrogen?

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6 If 1 kg of hydrogen is burnt with oxygen, about 1.4 × 108 J of energy is released. If 1 kg of hydrogen is fused to produce helium, about 6 × 1014 J of energy is released. After the reaction, if all the products were collected, how much mass would have been lost in each case? 7 If we could collect all of the gases and ash produced by burning a tonne of coal would they have exactly the same mass as the tonne of coal? 8 Astrophysicists model the Sun by using powerful computers. Briefly outline the basic assumptions that are programmed into the computers. 9 Briefly describe the processes that bring the heat generated in the core of the Sun to the Earth. 10 a Which of these statements best describes the nature of sunspots? A Sunspots are due to disturbances in the nuclear reactions in the core of the Sun. B Sunspots result from convection currents being formed in the Sun’s outer layers. C Sunspots result from magnetic disturbances in the photosphere of the Sun. D Sunspots result from holes in the Sun’s upper atmosphere (the chromosphere and corona). b Which layer of the Sun’s atmosphere gives rise to the sunlight we see arriving on Earth? A Photosphere B Chromosphere C Corona D All layers

11.3 We know the st ars by

their light

Because the Sun is ‘only’ 150 million kilometres away we can study it in many ways, including spacecraft that have flown as near as they dare. However, the only way we can know most stars is by their ‘light’. The word light is in inverted commas because astrophysicists often include the invisible components of the electromagnetic radiation spectrum as well as the small visible ‘rainbow’ section that we can see. The invisible ‘light’ includes the radio and infrared waves that are longer than visible light, as well as the ultraviolet and X-rays that are shorter. In this section we shall look at what we can find out about stars, including the Sun, by analysing the light that reaches us.

Analysing starlight Astronomers use radio telescopes, such as the one at Parkes in New South Wales, to collect the longer wavelength (centimetres to metres) radio and ‘microwave’ radiation. Because much infrared (IR) radiation is absorbed by the Earth’s atmosphere, satellites equipped with special IR telescopes are used to photograph the sky at those wavelengths. Although the atmosphere is transparent to visible wavelengths it does cause distortion that ultimately limits the power of Earth-based telescopes. For this reason space-based telescopes such as the Hubble Space Telescope (HST) have proved to be of immense value to astronomers. At shorter ultraviolet and X-ray wavelengths we again need to leave the Earth’s atmosphere behind in order to see clearly.

Figure 11.23 The Hubble Space Telescope was placed in orbit 600 km above Earth by the space shuttle Discovery in 1990, but had an imperfect main mirror. In 1993, astronauts on a second space mission installed correction mirrors that exactly compensated for the faults. Since then it has been returning incredibly detailed pictures to astronomers on Earth.

Figure 11.24 ‘The Dish’, the 64 m diameter Parkes radio telescope, has been an invaluable tool to astronomers worldwide since it was built in 1961.

Even though photographs at all these wavelengths are valuable, one of the most important tools available to astronomers has been the spectrograph, a means of spreading out the spectrum of the light received by a star. A simple way to do this is with a prism, as we have seen in our light studies in chapters 7–9. Astronomers now tend to use diffraction gratings rather than prisms but the idea is the same. One way to get a spectrum of starlight is simply to put a prism or diffraction grating


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over the lens of a telescope. A picture such as that in Figure 11.26 is produced for each star in the photograph. Modern telescopes use more sophisticated arrangements, which focus the light from a star onto a device that analyses the wavelengths and produces a spectrum. This may well take the form of a set of digital data that can be examined directly by powerful computer software—the astronomer may never even see the colourful spectrum of the older methods. Nevertheless a spectrum such as that shown in Figure 11.26, in this case of the Sun’s light, is still wonderful to examine!

Figure 11.25 The stars as seen through a telescope with a prism over the lens. Each star produces a small spectrum.

PRACTICAL ACTIVITY 53 Spectra of different elements

Figure 11.26 This is a high-resolution spectrograph of the Sun’s visible light from a wavelength of 400 nm (violet) to 700 nm (red). The many black lines are the telltale spectral lines, not only of the many different elements that are present in the Sun, but also of the conditions on the Sun.

Before looking closely at this spectrum we need to understand how spectra such as this are produced. Figure 11.27 shows the basic principles. Light from a hot object produces a continuous ‘rainbow’ spectrum as shown in (a). If the light then passes through a cloud of gas, this gas will absorb certain wavelengths, leaving dark bands in the spectrum as seen at (b). The absorbed light temporarily puts the atoms in the gas into an ‘excited’ state, but they release it again shortly afterwards. The re-emitted light radiates out in all directions, however, and so if seen from the side, as shown in (c), appears as bright bands against a dark background. hot black body

(b)

cloud of gas absorption line spectrum

prism

Figure 11.27 (a) The hot object at the top emits

a continuous range of wavelengths depending upon its temperature. (b) If the light passes through a gas, some wavelengths are absorbed, giving rise to an absorption spectrum. (c) Re-emitted light from the gas produces an emission spectrum.

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(a)

continuous spectrum

(c)

emission line spectrum

We have already considered the continuous ‘black body’ spectrum as a way of determining the surface temperature of a star. In this section we concentrate on the significance of the emission and absorption spectra. The importance of these spectra is that the lines emitted or absorbed are characteristic of each of the 92 chemical elements. For example, sodium always emits or absorbs two very characteristic lines in the yellow region— which is why the barbecue flame turns yellow when you throw salt on the meat. If these two lines are seen in a spectrum we know that sodium is present. Whether it is present as part of a compound (salt is sodium chloride) or as a vaporised gas makes no difference, the lines always have the same position in the spectrum. Thus, by examining the spectrum of a star we can tell what elements are present—at least in the photosphere and surrounding gases. Indeed you can see the two yellow sodium lines near the centre of one of the yellow bands of the solar spectrum in Figure 11.26. The Sun’s spectrum has been examined in great detail and many elements identified. The dark lines in the solar spectrum are often referred to as Fraunhofer lines as it was Joseph Fraunhofer who discovered them in 1817. In 1868 some lines were discovered that did not seem to correspond to any element known on Earth. As it appeared to be exclusive to the Sun it was called ‘helium’. As we know, helium was eventually found on Earth in certain rocks and in very small quantities in the atmosphere. The absorption spectra from other stars are quite similar in most respects to the solar spectrum. This confirms that the basic chemistry and physics that we understand on Earth seems to apply right through the universe. No strange new lines have ever been discovered in any other stellar spectra. There are subtle differences, however, and it is these differences that tell astronomers an enormous amount about other stars. We have already seen in section 11.1 that the spread of colour through the continuous spectrum is an indication of the surface temperature of the star—the greater the proportion of blue and UV, the hotter the star. As was indicated in that section, this method of determining the temperature can be jeopardised by patches of interstellar dust or gas, which can absorb some of the bluer end of the spectrum. Fortunately there are other ways to determine the temperature that involve careful examination of the dark lines in the spectrum. In the 1890s a scheme for classifying the spectra of stars was developed. It arranged the spectra in an order determined by the presence or absence of certain lines associated with hydrogen, and labelled them with the letters A–O. It seemed to be a logical arrangement at the time, but later it became apparent that there was a better way to arrange them. The new method involved placing similar spectra adjacent to one another and finding a pattern that smoothly changed from one class of spectrum to the next over the whole range. As a result, some of the original classifications were dropped and the rest were re-ordered into a new sequence, which went OBAFGKM—known to all students of astronomy as Oh Be A Fine Girl (or Guy in these days of gender balance) Kiss Me. Later modifications divided each class into smaller subdivisions by adding the digits 0–9 after the letters. The real significance of this pattern was not known at the time, but later it was found that the changes corresponded to differing temperatures, with the O0 stars being hottest and the M9 stars cool (relatively!). It was


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Physics file It is worth thinking about the relationship between temperature, luminosity and size. Clearly the higher the temperature of the star the greater the amount of energy radiated every second from each square metre. (A hot poker radiates more heat than the same one when cooler.) However, the total radiation (the luminosity) depends on the size as well. For example, a large poker may radiate more total heat than a hotter, but smaller, poker. We can deduce from all this, for example, that if two stars have the same luminosity but one has a higher surface temperature then it must be smaller than the cooler one—as it is giving more energy from every square metre. The laws of physics actually enable the total surface area to be found if the temperature and total energy radiated are known. From this the radius can easily be found (from A = 4πR2).

parallax (p)

shown that the reason for the changes between the classes was associated with the fact that different atoms become ionised (lose electrons) at various temperatures and that at cooler temperatures the light may not have enough energy to excite the atoms sufficiently to create some of the lines. Therefore, lines may appear over a certain temperature range but not at higher or lower temperatures. This was an extremely important discovery, as it provided a new way to determine the temperature of a star, and was not subject to the problem of the overall colour change as light travelled through interstellar gas. Indeed the difference between the two methods gave astronomers useful information about interstellar gases. We do not have space to go into the details of stellar spectral analysis, but a careful analysis of a spectrum reveals much important information about the star. As well as the temperature, it reveals the elements that are present and the state of those elements (whether ionised or not, for example). The lines can also reveal information about the pressure in the gas emitting the light as well as any magnetic fields present. Even the size of a star can be determined with the help of the spectrum. (Remember that any star is only seen as a point of light in even the largest telescopes.) If an accurate surface temperature is known from the spectrum, a basic law of physics (the Stefan–Boltzmann law mentioned in section 11.1) can be used to determine the amount of energy given off (each second) by each unit area of the surface. Knowing this and the luminosity (the total energy given off by the star) enables the total surface area to be calculated— and hence the radius (as A = 4πR2). This depends on a knowledge of the luminosity, which is found by comparing the apparent brightness and the distance determined by stellar parallax. But is there a way to find the luminosity of the star if it is not within parallax range? apparent brightness (b)

spectrum

spectral type 1 d=— p

chemical composition

surface temperature (T) distance (d)

inverse square law

luminosity (L)

Stefan–Boltzmann law

radius (R)

Figure 11.28 A determination of stellar parallax, apparent brightness and the spectrum of a star can tell astronomers much about the nature of a star.

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Surface temperature (K)

SUPER GIANTS Betelgeuse

Rigel 104

–5 Antares

GIANTS Regulus 10 Vega Sirius A Altair

Arcturus Aldebaran

2

Luminosity (L . )

–10

0

MA

IN

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Sun

SE

QU

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Absolute magnitude

106

25 000 10 000 8000 6000 5000 4000 3000

EN

CE

10–2

+10 Sirius B

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+15 WHITE DWARFS O5 B0

A0

F0 G0 K0 Spectral type

M0 M8

Figure 11.29 The Hertzsprung–Russell (H-R) diagram is one of the astronomer’s basic tools. The luminosity scale takes the Sun’s value (L ) as 1, which corresponds to an absolute magnitude of about +4.8. The temperatures corresponding to the spectral types are plotted along the top. Notice that hotter stars are to the left, the background colour indicating the equivalent colour.

Types of stars—the Hertzsprung–Russell diagram Once data for many stars had been collected it was natural to look for patterns in the data. The usual procedure in science is to make a graph and look for relationships between the quantities involved. Sure enough, in 1911 and 1913 the astronomers Ejnar Hertzsprung in Denmark and Henry Russell in America independently discovered a pattern that brought order to the apparent chaos. The graph they produced is now known as the Hertzsprung–Russell diagram, or H-R diagram for short. In effect they plotted the luminosity of a star against its temperature and found a distinct pattern, or rather a set of patterns. Generally speaking, it was found that hot stars are brighter, but there were important exceptions to this rule. Figure 11.29 shows the common way of arranging the H-R diagram with the luminosity on the vertical axis (brighter at the top) and temperature on the horizontal axis (hotter to the left, not right as you might have expected—this seems to have been a historical accident). Each axis is labelled with two different scales, as temperature and luminosity are derived from the spectral type and the absolute magnitude, respectively. The first thing to notice from the H-R diagram is that the stars are not randomly distributed; there is a relationship between the brightness and temperature of a star. The band of stars from the lower right corner to the upper left corner is called the main sequence and includes 90% of the stars in the sky. Apparently there is some physical reason why a duller star will be cooler and a brighter star hotter. Although there are some stars that are hot but dull and others that are cool but bright, most stars are brighter if they

Physics file The H-R diagram can be a little misleading until it is realised that the scales are non-linear. (They are what you may recognise as log scales.) That is, the brightness increases very rapidly as we go up the graph. A star at the top is not twice as bright as one in the centre, it is about a million times brighter. From bottom to top the change is more than a billion times! The temperature scale does not have such extremes, but as you go from right to left the temperature goes up by the same factor (not amount) with each equal distance interval.


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are hotter. Our Sun is approximately in the middle of the main sequence. As we saw earlier, however, most stars are both cooler and dimmer than the Sun—as represented by the greater density of stars on the lower right of the main sequence. As well as indicating the brightness and temperature of the stars, the H-R diagram also indicates the relative sizes of stars. As pointed out above (and in the Physics file on page 412) a hot but dimmer star must be small compared with a cool but bright star. In other words, the stars at the top right of the diagram must be large, while those at the bottom left must be small. Roughly speaking, the top right stars have a radius about 1000 times that of the Sun (1000R ) while the bottom left stars are 1000 times smaller (0.001R ). For this reason the group of stars at the lower left is termed white dwarfs while those at the opposite extreme are red giants. In fact there are giants and supergiants, with giants being around 10–100R and supergiants being up to 1000R . The giants and supergiants together only make up about 1% of the stars in the sky. The other 9% of stars are the small white dwarfs. Sirius B was only discovered about 150 years ago when it was noticed that Sirius (A) had a slight wobble. Careful observation then revealed a very dim companion star (B) in orbit with it. Despite its lack of luminosity Sirius B turned out to have about the same mass as our Sun, and yet a diameter only one hundredth of the Sun’s. It is extremely dense! No white dwarfs are visible to the naked eye. Another category of dwarfs exists to the lower right (but off the scale)— the brown dwarfs. These are not really stars and have only been discovered relatively recently. They appear to be objects of about the size of Jupiter that were probably not quite large enough to become genuine stars.

Placing stars on the H-R diagram If we find a new star, how do we know where to put it on the diagram? Provided it is within stellar parallax range there is no problem; we can determine the luminosity and temperature directly. If the new star is beyond parallax range we can still determine its temperature from its spectrum, but what of its luminosity? Given the temperature, we can see that we can place it on a vertical line on the graph, but how far up? We need a way to determine whether the star is a white dwarf, a main sequence star or a giant. Once that is determined we can put it in the appropriate group—and hence determine its luminosity. You guessed it—yes, there is a way. As suggested earlier, the spectral lines in stellar spectra tell more than just the elements present. For example, magnetic fields tend to ‘split’ lines slightly. The pressure of the gas from which the light originated also has a subtle effect on some of the lines but not others. Certain lines in the hydrogen spectrum are sensitive to the number of collisions the hydrogen atoms are experiencing. The collisions tend to slightly increase or decrease the energy and hence wavelength of the emitted light. This in turn slightly broadens the spectral line. Because of the size of a giant star, the photosphere tends to be less dense, and hence the atoms will experience fewer collisions. This means that the spectral lines are less broadened than those of a main sequence or dwarf star. These types of techniques with the more subtle characteristics of the spectrum enable astronomers to classify the star and hence place it in the appropriate place on the H-R diagram.

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Once the new star is placed on the diagram its luminosity can be determined. Given the luminosity, as well as the apparent brightness, the distance to the star can be determined by reversing the procedure for obtaining the luminosity (intrinsic brightness) from the distance and apparent brightness. This is an enormously valuable tool as it enables astronomers to determine the distances to most of the stars in the sky, instead of just the small number within stellar parallax range. The technique of determining distance from the spectrum, rather than from parallax, is unfortunately given the rather misleading name spectroscopic parallax. It is misleading because it applies to stars for which parallax can not be used! The word parallax was used simply because it refers to the process of determining distance, but it is not to be confused with the process that uses genuine parallax called stellar parallax. (If one was to be a little unkind one could suggest that astronomers seem somewhat unable to let go of old but misleading names and techniques—you can possibly think of some others we have met along the way!) apparent brightness (b)

spectrum type of star: main sequence, giant, dwarf

spectral type

chemical composition

surface temperature (T) H-R diagram

luminosity (L)

inverse square law

Stefan–Boltzmann law

distance (d)

radius (R)

Figure 11.30 Spectroscopic ‘parallax’. Knowing the apparent brightness and correctly interpreting the stellar spectrum to decide whether a star is a main-sequence, giant or dwarf star enables astrophysicists to deduce most of the important characteristics of a star.

Worked example 11.3A If two stars have the same radius and are the same distance from us, but one looks red and the other blue, which one will be brighter? Explain. (See if you can answer this question before reading on!)

Solution To answer a question like this, think of what is implied by the difference between the stars; that is, their colour. The colour is an indication of the surface temperature of the star, hotter stars being bluer according to Wien’s law. Thus the blue star must be hotter. The Stefan– Boltzmann law tells us that, because it is hotter, the blue star will be radiating considerably more energy (as the radiation varies with T 4) and thus it will appear significantly brighter.


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One vital piece of information is still missing! As Figure 11.30 suggests, we can tell a lot about a star from the apparent brightness and the spectrum. But one important quantity does not appear on that chart: the star’s mass. We can see a definite pattern relating the luminosity and temperature, but what is it about most stars that means hotter stars have to be brighter? What is it that causes some stars to be white dwarfs or red giants? If we could find the mass of the various types of stars it may give us clues to the answers. The Sun’s mass was effectively determined by studying its effect on the planets. If we imagine what would happen if the Sun had a different mass we can get a feel for how this works. Let’s imagine the Sun suddenly lost half its actual mass—the Earth would not be held as strongly to the Sun and so would move off into a bigger and slower orbit, or even escape from the Sun altogether. So in order to tell the mass of a star we need to observe something in orbit around it—the faster and closer the orbit the heavier must be the star. Planets? While we believe that most stars do have planets, and in fact have detected some, they are far too distant to be seen and measured. However, yet again, nature does enable us to find the missing information! It turns out that many stars are binary stars—that is, one star revolving around another, or more particularly, two stars revolving around their common centre of mass. If the two stars can be seen as separate stars, and many of the closer ones can, their orbits can be found—after many years of patient watching and measuring. Clearly these orbits will depend on the masses of the two stars. Heavy, close stars would revolve around each other quickly while less massive stars, or those separated by greater

Figure 11.31 Binary stars revolve around each other. Measurements of the period and size of the orbits enable the masses to be found. These three photographs, made from data from the European Space Agency Hipparcos satellite, show one of the many binary stars discovered by Hipparcos, HIP 46 706 in Hydra. The precision of the data from Hipparcos is astonishing—these stars are 34 l.y. away and yet they are as close as the Sun and Saturn. Astronomers compute the masses of the two stars at 0.41 and 0.42 times the mass of the Sun. They orbit around each other every 18.3 years. (The ripple is a data-processing effect.)

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distances, would move more slowly. If we know the distance to the stars it is just a matter of geometry to calculate the size of the orbits. Finding the period and size of the orbits is just a matter of patient observation over many years. Newton could do the rest. His laws of motion, and of universal gravitation enable us to determine the mass of both of the stars involved. It really isn’t quite that easy—there are many problems to be solved before a reliable measurement can be obtained. In fact there were not a great number of suitable stars within parallax range, but astronomers found that small shifts in the spectral lines of some stars showed that they were revolving. Careful analysis of these spectroscopic binaries enabled many more determinations of mass. So what does knowing the mass of stars tell us? Perhaps not surprisingly there is a simple mass–luminosity relationship. It turns out that the luminosity of a star varies roughly with the mass cubed (L ∝ m3); that is, a star with twice the mass of the Sun would be about eight times brighter. This fits well with the computer modelling of stars that suggests that a more massive star will burn much more brightly and more rapidly because of the much greater pressure and temperatures in its central core. The giants and dwarfs, however, do not obey this relationship—we will see why shortly. The fact that all main sequence stars seem to conform to the same pattern confirms the notion that all stars are basically similar in composition and are powered by the same hydrogen-to-helium nuclear fusion reactions in their core.

Interpreting the H-R diagram—stellar evolution Why are the stars distributed on the H-R diagram as they are? This was the obvious question to ask once the nature of the diagram became clear. The story of how astrophysicists have come to interpret the diagram and build up a picture of the life and death of stars is one of the most fascinating in physics. We can only touch on it here, but hopefully you will have the opportunity to explore it further in your future studies. Before the 20th century it was generally thought that the stars were basically permanent features of an unchanging eternal universe. As our understanding of the processes that fuelled the nuclear fires developed, it became obvious that while stars could last a very long time, they could not last forever. For example, at the rate the Sun is consuming its hydrogen fuel it will run out of fuel in about 100 billion years—a long time, but not forever (see Worked example 11.2A). Models of the nuclear processes predict a much shorter life, however. As the heavier helium nuclei build up in the core of a star, the nuclear fusion reaction zone moves outwards. Once about 10% of the hydrogen is consumed the star becomes unstable and the outer layers expand until the star becomes about ten times the size, in other words a giant. The Sun is due to do this in about 6 billion years and when it does it will engulf the Earth—hopefully that’s enough time for us to find ways of travelling to another star system! For these reasons it was thought that the stars are ‘born’ on the main sequence and then eventually move into the giant phase. But was there a way of testing this idea? As we have seen, heavier stars burn their fuel more rapidly than lighter stars. Although they have more of it, models


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predict they will run out sooner; that is, they should have a shorter lifespan. Clearly we have no chance of watching what happens to a particular star through its lifetime. We have to infer from the different types of stars that we can see, at which stages of their lifetime they are. Fortunately there are ways to test our models. There are groups of stars, called star clusters, which seem to have all been born at about the same time out of the same cloud of ‘dust’ and gas. The photograph of the Omega nebulae at the beginning of this study shows this process in action. The even more famous ‘Starbirth’ photograph is another (Figure 10.1, page 335). If we look at the individual stars in such a cluster, because they are using their fuel faster, we should find that the heavier, brighter, bluer stars are moving off the main sequence (becoming giants) before the lighter stars further down the sequence. Many clusters were studied in this way in the 1950s and this is just what was found. Stars at the upper end of the main sequence were more likely to have moved towards the giant area than those further down. The difference between an older cluster and a younger cluster is illustrated in Figure 11.33. The number of stars seen in the various areas on the H-R diagram presumably is an indication of the amount of time stars spend in that area. Because there are many more stars in the main sequence than in the giant phase we can assume that stars spend most of their life as reasonably stable main sequence stars. There are few stars between the main sequence and the giant phase and so it can be deduced that after reaching the end of their main sequence lifetime they fairly rapidly expand into giants. With the aid of computer modelling, a fairly complete picture of the birth, life and death of stars has been built up which agrees well with the observational evidence. Stars are seen to begin their lives as ‘protostars’, large masses of gas and dust that have come together through gravitational attraction. The heat generated in this gravitational collapse (the so-called Kelvin–Helmholtz contraction discussed earlier) causes the protostar to become very hot. As the gravitational collapse continues, gravity becomes (a) Young cluster Surface temperature (K) 25 000 10 000 8000 6000 5000 4000 3000

(b) Old cluster Surface temperature (K) 25 000 10 000 8000 6000 5000 4000 3000

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Figure 11.33 (a) In a young cluster of stars most are still on the main sequence, or are moving towards it, but a few of the brighter stars have started to move into the giant area. (b) In the older cluster many of the brighter stars have become giants and others further down are moving toward the giants. Some of the giants have burnt out and are moving to the white dwarf area.

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Figure 11.32 Star clusters are thought to be groups of stars that were all ‘born’ about the same time. By looking for those that have moved off the main sequence we can confirm our models of the evolution of stars. This is the Pleiades cluster, which can be seen in Taurus, and is a lovely sight through binoculars.

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stronger, and accelerates the collapse even further. Eventually the interior of the new star becomes hot enough, about 10 million kelvin, to ignite the nuclear fusion reactions. The extra heat generated by these eventually stops the collapse as a result of the outward pressure of the radiation from the hot gas. At this point the protostar has become a new main sequence star and remains in a fairly stable condition for a time that depends on how large the star is. This time will be measured in millions of years for very massive stars, billions of years for stars like the Sun, and tens of billions of years for the majority of stars lower down the main sequence. Once a star becomes a red giant a new process starts in the core. The further gravitational collapse of the heavy core heats it to even higher temperatures. At around 100 million degrees new nuclear reactions start to occur: three helium nuclei combine to form carbon. This reaction more or less stabilises the giant star in its position on the H-R diagram. Eventually, however, the nuclear fuel begins to run out and the star begins to contract, a process that can keep it hot and actually raise its temperature so that it moves back towards the left of the H-R diagram. The overall brightness decreases, however, and so it gradually moves down the left side of the diagram to the region of white dwarfs. At this stage the star has collapsed into electron degenerate matter—the atoms have collapsed into a mass of protons, neutrons and electrons with a density about a million times that of water. Their remaining heat gradually radiates away, and they become cooler and redder, and slide down and out of the H-R diagram at the lowerright as black dwarfs. This picture of stellar evolution is appropriate for most stars, but a more spectacular fate awaits some! The clue is that observers through the ages have seen novae and supernovae, stars that have suddenly appeared in the sky (in a matter of days) and then faded away (in a matter of months). We now understand a supernova to be the explosive end of a massive star. There has been a number recorded in history. There was a notable supernova in the constellation Cassiopeia in 1572 which was observed by Tycho Brahe—who was so fascinated he went on to become one of the great astronomers of the period. A 1604 supernova had a similar effect on another great astronomer, Johannes Kepler. In 1987 your parents possibly witnessed the first supernova since 1604 visible to the naked eye! A new star suddenly appeared in the Large Magellanic Cloud, a fuzzy area near the Southern Cross, which is actually a satellite galaxy of the Milky Way. It was a great opportunity for astronomers to study a supernova, almost in their own backyard. Because of the huge pressures inside a star with a mass of about four times the Sun’s mass (4M ), after the initial hydrogen-burning phase the temperatures created by the contracting star can reach 600 million degrees, at which point new nuclear fusion reactions can take place. In these reactions the carbon produced by the helium fusion is ‘burnt’ to produce oxygen, neon, sodium and magnesium. In still more massive stars various heavier elements are produced and burnt in a series of reactions that occur at temperatures up to billions of degrees. The effect of these is to move the star around on the H-R diagram as it changes brightness and temperature with each new set of reactions. Stars over about 25M eventually become so large and bright that they form the supergiants at the top right of the H-R diagram. Betelgeuse and Rigel in Orion are supergiants.

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Figure 11.34 Once stars reach the end of their life on the main sequence they move up into the giants. Depending on their size they may fade to a white dwarf or become a supernovae.

Figure 11.35 Seven years after its explosion, Supernova 1987A had developed these rings. They are thought to be glowing gas that was ejected when the star was a red giant but which was ionised by the intense UV light from the explosion. This photo was taken by the Hubble Space Telescope in 1994.


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PRACTICAL ACTIVITY 54 LEA and CLEA

Figure 11.36 Large stars may shed some of their mass by ejecting layers that become planetary nebulae. This is the Helix Nebula in Aquarius. It is actually about the same angular size as the full moon—but not visible to the naked eye.

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There is a limit to this process, however. Once silicon is being burnt to produce iron the fusion reactions don’t generate energy any more and so the star begins to collapse. As they do, more heating occurs and as a result, stars up to about 8M tend to shed some of their mass by ejecting what are called planetary nebulae. These are basically layers formed as a result of the various stages of fusion reactions and may involve a considerable proportion of the star’s mass. They create a wonderful sight for a large telescope! For the true heavyweights over 8M their huge gravity doesn’t allow the ejection of planetary nebula and so they continue to collapse, generating yet more heat. And then in an amazing sequence of events not totally understood, the core, largely of iron now, rapidly heats to many billions of degrees in a fraction of a second. This results in the protons and electrons in the core being forced together with such ferocity that they combine to form neutrons and small massless particles called neutrinos. The core is now only about 20 km in diameter but with a mass several times that of the Sun! In this state the core can collapse no further, but the rest of the mass of the star is still collapsing in on it, building up enormous pressure, which, because the core can compress no further, creates a monumental ‘bounce back’. This sends the outer layers of the star flying off into space at enormous speeds, partly propelled by the neutrinos trying to escape from the core. The energy released in this explosion is an incredible 1046 J, which is far more than the Sun will produce in its entire life. One hopes that a nearby star is not thinking of becoming a supernova! (Be reassured that they are very rare events and no nearby stars seem to be candidates.) What is left of the star after this gigantic explosion is the extremely dense core made of ‘neutron matter’; that is, matter in which all the electrons and protons have collapsed to form neutrons. A neutron star (as this object is called) with a diameter of 30 km would have the same mass as the Sun. Put another way, a few cubic centimetres would have a weight of hundreds of millions of tonnes—if it could be supported at the Earth’s surface!

In the 1960s a number of pulsars were found. These were objects producing extremely regular pulses of radio emissions. Later, it was found that these flashes could also be seen as visible light. They are now thought to emanate from rotating neutron stars about 20 km in diameter but with about 1.4 solar masses. Any rotation that a star has will be greatly increased as it collapses, in just the same way that a skater or dancer can spin faster by pulling her arms closer to her body as she spins. The periods of pulsars are measured in seconds or even fractions of a second and so they are spinning at an incredible rate for their size. As they spin, leftover charged matter interacts with their powerful magnetic field and produces a beam of intense radiation that sweeps around the sky like the beam from a lighthouse. If we happen to be in line with the beam we see a flash each time it sweeps past us. The other ‘leftovers’ from the supernovae are the wonderful remnants such as those shown in Figure 11.37. These are the ejected gases that glow for possibly hundreds of years after the supernova, but gradually fade as they disperse into space to form more of the dust and gas clouds out of which new stars will be born. Many of these can only be seen by radio telescopes as their temperature drops and the wavelength of the radiation becomes longer. Apart from being a marvellous spectacle, supernovae are vital to our existence. As we saw, the processes in large stars can produce all the elements up to the atomic number of iron (26) because the fusion of nuclei larger than iron results in a net loss of energy rather than a gain. The only situation in which sufficient energy is produced to create the larger nuclei is a supernova, and that is exactly what happens. Among the material spewed into space by the supernovae are all the elements of the periodic table. Billions of years later they may become part of a new planetary system, a so-called second-generation system, which will include all the elements needed for life to emerge—a system such as our Sun and its planets. Many of the atoms in your body were created in a supernova! The 1987 supernova mentioned earlier was not even in our galaxy (although it was in a close one) but was visible to the naked eye. If one occurred in our vicinity in our own galaxy it would be a very spectacular event indeed. Indeed we would not want one too close! It is thought, from studies of the number of supernovae in other galaxies, that there should be around five supernovae in our galaxy per century, but we have only recorded a handful in the last 1000 years. It is thought that the stars and dust towards the centre of the Milky Way obscure much of the rest of the galaxy, but much research awaits keen new astrophysicists.

Figure 11.37 After a supernova explosion the leftover gas and dust create an impressive sight for perhaps hundreds of years. This is the Crab Nebula, which is the remnant of the supernova of AD 1054. It has expanded to about 3 pc (10 l.y.) across and is about 2000 pc from us.


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11.3 summary We know stars by their light • The continuous spectrum of a star provides information about the temperature of the surface, and the absorption lines tell us what elements are present and under what conditions. • The spectra of stars are quite similar to the spectrum of the Sun, indicating that they have similar character istics and processes. • Stellar spectra are classified according to a system that grades them by surface temperature (hot– OBAFGKM–cool). • The Hertzsprung–Russell (H-R) diagram plots stars according to temperature and luminosity. All stars fall into three main groups: main sequence, giants and supergiants, or white dwarfs.

• As we move up the main sequence on the H-R diagram the stars become hotter, brighter, larger and shorter lived. • The mass of binary stars can be determined from their motion and separation. There is a simple mass– luminosity relationship (roughly L ∝ m3) for main sequence stars. • Most stars begin their lives on the main sequence, eventually become giants and then move downwards to become dwarfs. Some very large stars become supergiants and end their lives as supernovae, which create the heavier elements.

11.3 questions We know stars by their light 1 The ‘light’ from stars includes more than the visible light we can see. What other types of ‘light’ do astronomers use to look at stars with, and what are some of the problems associated with using it? 2 There are three basic types of spectra that astrophysicists analyse to find out about stars and other objects in the universe. What are they, what is their value, and what are the basic similarities and differences between them? 3 Why is it that a determination of a star’s surface temperature is crucial to finding the size of a star? 4 a Which of the following best describes the Hertzsprung–Russell diagram? It is a graph of the: A absolute magnitude against the luminosity of the stars. B apparent magnitude against the temperature of the stars. C temperature against the spectral type of the stars. D luminosity against the spectral type of the stars. b In general it is found that for most stars the: A luminosity increases with the surface temperature. B luminosity decreases with the surface temperature. C luminosity is not related to the surface temperature.

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D luminosity is directly proportional to the surface temperature. 5 How does the H-R diagram help astronomers to determine the distance to a star that is too far away for stellar parallax? 6 The mass of a star is one piece of information that cannot be found from its spectrum. However, astronomers have found ways to directly measure the masses of certain types of stars and then to indirectly infer the masses of most other stars. How do they do this? 7 Main sequence stars are found to obey a mass– luminosity relationship. If a star is found to have a luminosity eight times that of the Sun, its mass is likely to be about (M is the mass of the Sun): A 8M B 4M C 2M D √2M 8 What evidence do astronomers have that all stars are fuelled by the same hydrogen–helium fusion process and don’t use other processes that we don’t know about? 9 The time for which astronomers have been observing stars is absolutely insignificant in terms of the lifetime of a star. How can we possibly study the life and evolution of stars given that they are measured in millions and billions of years? 10 What is a supernova, and why is it that only very massive stars will become supernovae?

11.4 Whole new wor lds

Figure 11.38 This is NGC 2997, a spectacular gas-rich spiral galaxy. The Milky Way is thought to be somewhat similar to this galaxy.

105 Luminosity (L . )

Since the first observations with telescopes astronomers had seen small fuzzy patches in the night sky. Many appeared irregular in shape, some round, and some even had a spiral appearance. The nature of these nebulae, as they became called, remained mysterious. Most astronomers thought of them as clouds of some kind relatively close by. Several astronomers, however, had suggested that at least some of the nebulae might be ‘island universes’. The matter was still being hotly debated in the 1920s when a young American astronomer named Edwin Hubble found the solution to the dilemma. The problem was to determine just how far away the nebulae really were. If they were at similar distances to most stars that would show that they were objects within our Milky Way. But they were known to be beyond stellar parallax range, and spectroscopic parallax could not be used on them. Hubble was fortunate in that a few years earlier, Henrietta Leavitt had discovered an important relationship between the period and the brightness of a type of variable star known as Cepheid variables. They are named after the star Delta Cepheid, which had been discovered to vary in brightness over a period of about 5 days in 1784. In 1894, after a careful examination of the star’s spectral lines, Russian astronomer Aristarkh Belopol’skii deduced that this was due to the whole star expanding and contracting at this rate. Astrophysicists now have a good model to explain this behaviour but it need not concern us here. Cepheids vary in brightness at a very regular rate, with a period of anything from a day to some weeks. Leavitt found that their average brightness was directly related to the period of the fluctuation in brightness—the longer the period, the brighter the star. The significance of this was that they could be used as distance markers because a measurement of their period enabled their intrinsic brightness to be found. By comparing this with the apparent brightness the distance could be calculated. On careful examination of his photographs of the Andromeda ‘nebula’, Hubble discovered some Cepheid variables. Hubble immediately realised

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Figure 11.39 The period–luminosity relation for Cepheid variables. There are two types that can be distinguished by the metal content of their spectra. L is the luminosity of the Sun. Fortunately Cepheids are very much brighter than our Sun, enabling them to be seen even in distant galaxies.


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Physics file Galaxies are often referred to by numbers such as M31 or NGC 224 (both of these refer to the Andromeda galaxy). The M numbers are ‘nebulae’ catalogued by French comet hunter Charles Messier (1730–1817). He listed them because they were sometimes confused with comets. NGC stands for the ‘New General Catalogue’ published in 1888 by the Armagh Observatory in Ireland. It lists over 13 000 galaxies, star clusters and nebulae. Both catalogues include both nebulae and galaxies, as the distinction was not known at that time.

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that as the Cepheids in Andromeda were very much dimmer than any others that had been seen, they were very much further away. His calculations put them at about 750 kpc (750 000 pc), a much greater distance than had been measured for any other star. Andromeda was no nebula; it was a whole new galaxy of stars. Based on its angular size in the sky and this estimate of its distance, it must be about 70 kpc in diameter (bigger than our own galaxy, which is less than 50 kpc). With Hubble’s announcement of his discovery in 1924, whole new worlds opened up, both literally and figuratively. The universe suddenly became even larger—and by a huge factor, as it was realised that many more ‘nebulae’ would turn out to be new galaxies at even greater distances. Fortunately, as you can see from the period–luminosity graph in Figure 11.39, Cepheids are very bright stars and so can be seen from a great distance. This enabled the distances to many more galaxies to be calculated. Hubble found that there seemed to be four main types of galaxies: the spirals, the barred spirals, the ellipticals and the irregulars. The spiral and barred spiral galaxies constituted about 77% of all known galaxies, most of the others being the elliptical galaxies. The diameters of galaxies varied from a few kiloparsecs to over 200 kpc, although the diameters of the irregular galaxies were generally smaller than 10 kpc.

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Figure 11.40 (a) NGC 4321 is a spiral galaxy with particularly pronounced arms. Many have larger central bulges and smaller arms. (b) NGC 1365 is a barred spiral because the spirals seem to originate at the ends of the bar-shaped region. (c) NGC 5010 is a typical elliptical galaxy. (d) The Large Magellanic Cloud (LMC) is a typical irregular galaxy with no obvious shape to it, as is the Small Magellanic Cloud (SMC).

The Cepheid variables visible in galaxies make them an ideal tool for distance measurements, but unfortunately they are not visible past about 30 Mpc (megaparsec). However, at greater distances astronomers have found that they can use supernovae as measuring sticks. There are different types of supernovae, but certain classes always seem to have about the same brightness. If one of these is seen in a galaxy its apparent brightness can be used to estimate its distance. This technique depends on there being the right sort of supernovae explosion in a galaxy, but as time goes by more are seen.

Our own galaxy—the Milky Way It soon became clear that the Milky Way was a galaxy like the others and of similar proportions. Instead of being populated by a vast number of stars the universe was suddenly populated by a vast number of galaxies. Indeed we now know there are about as many galaxies in the universe as there are stars in our galaxy—about 100 billion.

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The existence of the Milky Way as a narrow band of stars across the sky suggested that our galaxy was a huge disc, rather than a sphere. As we look at the Milky Way in our night sky it appears to encircle our part of the galaxy and is fairly uniform in all directions. For this reason many people believed that the Sun was at the centre of the galaxy. However, Harvard astronomer Harlow Shapely found that a number of globular clusters of stars appeared to be above and below the plane of the galaxy in one region of the sky only. Was that the centre of the galaxy? He suggested that the reason we saw roughly equal numbers of stars in all directions was that our view in the plane of the galaxy was obscured by dust and gas that absorbed the light from more distant stars, including those in the centre of the galaxy. Shapely assumed we could see the clusters better because they were out of the plane of the galaxy. In 1920 he calculated the distance to the globular clusters and so put the centre of the galaxy at about 14 kpc (modern measurements put it at about 8 kpc).

Figure 11.41 We can only see the Milky Way from the inside. This is the whole sky view of it.

Physics file Where is the centre? Shapely’s method for locating the centre of the galaxy could be likened to a motorist lost near the centre of a city. He can’t see the city because of the buildings around him, but can see the light in the sky above it. Shapely assumed that there would be more stars, and in particular more clusters of stars, above and below the plane of the galaxy near its centre.

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Modern astronomers, particularly those working with radio telescopes that can ‘see’ through the interstellar dust, have confirmed Shapely’s view. We now believe the Milky Way to be disc-shaped with spiral arms, rather like Andromeda. Its diameter is about 50 kpc (although the vast majority of stars are within 15 kpc of the centre) and its thickness a little less than 1 kpc. The central bulge is about 2 kpc in diameter and located in the region of Sagittarius. However, although this is certainly a rich area of the Milky Way to observe, we cannot see the central bulge because of the obscuring effects of the increasing amount of ‘dust’ towards the more central area of the galaxy.

Figure 11.42 Shapely assumed that the globular clusters would be concentrated around the centre of the galaxy.

Remember that 1 kpc = 3262 light-years.

Hubble, Doppler and the redshift In the 1920s Hubble continued his observations of galaxies. He found that when he looked at the spectrum of a galaxy it was ‘redshifted’. This means that it had the same pattern of lines as normal spectra but they were all moved towards the red (long wavelength) end of the spectrum. This


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Figure 11.43 Hubble discovered that the galaxies were receding from us at velocities proportional to their distance.

was interpreted as the well-known ‘Doppler shift’ that occurs whenever a source of waves is moving towards or away from us. It is most obvious as a police car goes past us. The frequency of the siren sounds higher than normal as it approaches us and then lower than normal as it recedes from us. In a similar way, if a source of light waves is travelling away from us the frequency of the light appears to drop. As light always travels at the same speed (and c = fλ) this means that the wavelength must increase, that is, become redder. This is explained in more detail in the Physics in action on page 427. As shown there, the speed of the galaxy can be found simply from the amount of redshift: Dl vg = c l Hubble also measured the distance of the galaxies by using the Cepheid technique. He found a remarkable pattern. The more distant galaxies had greater redshifts, and indeed the redshift was directly proportional to the distance of the galaxy. In other words, the galaxies seemed to be speeding away from us, and those at the greater distances were receding at greater speeds. This became known as Hubble’s law and is expressed as a simple equation: v = H 0d where v is the recession velocity of the galaxy, H0 is Hubble’s constant, and d is the distance of the galaxy. As you can see in the graph in Figure 11.43 the value of Hubble’s constant is about 70 km s-1 Mpc-1. It has proved difficult to find an accurate value for the constant because of the huge distances of the galaxies involved, but recent results from a satellite called WMAP (Wilkinson Microwave Anisotropy Probe) suggest a value of 71 km s-1 Mpc-1 to within 5%. One of the problems is that not all galaxies are receding from us. Some of the closer ones are moving towards us (and so have blueshifts) owing to their gravitational interactions. These irregular motions can mask the Hubble recession for the closer galaxies—unfortunately those for which more accurate distance measurements can be made. However, the further galaxies are all moving away at greater and greater speeds, which mask the speed of the irregular motion. It turns out that galaxies themselves come in clusters, some small, some much larger. Within the cluster the galaxies may move around somewhat randomly, but the clusters themselves appear to be moving apart at a very regular rate—this is the Hubble constant. The Hubble constant can be used to determine the distance to galaxies too far away to be measured by other techniques. Provided a spectrum can be seen, and thus a redshift measured, the recession speed can be determined and hence the distance. For this reason an accurate value of the Hubble constant is one of the most important quests of modern astronomy. As we will see shortly it is tied up with one of the biggest questions—how old is the universe?

Worked example 11.4A The spectral line of singly ionised calcium has a wavelength of 393.3 nm when measured in the laboratory, but can be seen to have a wavelength of 401.8 nm in the light from the elliptical galaxy NGC 4889. Use Hubble’s law to find the distance to this galaxy. (H0 = 70 km s-1 Mpc-1 and c = 3 × 108 m s-1)

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Solution The shift in wavelength is 401.8 - 393.3 = 8.5 nm. This is a ‘redshift’ because the shift is to a longer wavelength, and so the galaxy is receding from us. The proportional redshift is 8.5/393.3 = 0.022. We can find the recession speed from: vg = ∆λc/λ = 0.022c = 0.022 × 3 × 108 = 6.6 × 106 m s-1 = 6600 km s-1 Using the Hubble relation v = H0d, we can find the distance d = 6600/70 = 94 Mpc. This distance is 94 × 3.26 = 300 million light-years and so we are seeing the galaxy as it was 300 million years ago.

Where is the matter?

PRACTICAL ACTIVITY 55 The Doppler effect

The discovery of galactic clusters deepened a mystery already posed by galaxies themselves. Presumably it is gravity that holds all the stars to a galaxy, in much the same way that the planets are held to the Sun, and so the laws of mechanics that apply to the solar system can be also applied to galaxies. When this is done, the total mass of the galaxy can be found. The problem is that if we add up the mass of all the stars, along with the dust and gas that we can see, we get nowhere near the calculated total mass! The Physics in action

The Doppler effect When a source of waves, whether sound, light or any other, is moving towards an observer, the apparent wavelength seen by the observer will be shorter (girl in Figure 11.44a). This is because each wave is emitted a little closer to the observer than the previous one and so is not so far behind the previous wave as it would be if the source was stationary. The reverse is the case if the source is moving away from the observer (boy in Figure 11.44a). In Figure 11.44b two consecutive waves from a source moving with speed u are shown. The second wave has just been emitted at the position shown. The natural wavelength is λ and the period T. As the velocity of the wave is v, the wavelength λ = vT. In the time between emitting the two waves the source has moved a distance uT. This means that the distance between the waves in the forward direction is vT – uT, while that in the backward direction is vT + uT. These distances are the wavelengths as perceived by the observers to the front and rear, respectively.

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In the case of light waves travelling from moving sources, we are interested in the apparent change in wavelength ∆λ. As can be seen in Figure 11.44b, this is simply uT, so ∆λ = uT. As T is also given by λ/v, we can rewrite this expression as: u ∆λ = l v That is, the wavelength change, or ‘shift’ in the terminology of astrophysics (redshift or blueshift depending on the direction), is equal to the wavelength times the ratio of the speed of the source to the speed of the wave. If we replace v by the speed of light waves, c, and u by the speed of the galaxy, vg, and then do a little algebraic reorganisation the expression can be written as: Dl vg = c l So the speed of the galaxy is equal to the fractional wavelength shift (∆λ/λ) times the speed of light.

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Figure 11.44 The Doppler effect.


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Figure 11.45 The Horsehead, a dark nebula near Alnitak, the easternmost star of Orion’s belt.

Physics file Astrophysicists often tend to use the term ‘light’ in the broad sense of any electromagnetic radiation. This includes radio, microwave, IR and UV as well as X-ray and gamma ray radiation. So when we speak of the apparent brightness of quasars it may not necessarily mean that they will appear to be bright objects in an optical telescope, but that the total energy we receive at all wavelengths of electromagnetic radiation is high.

same process can be applied to the galaxies in a cluster, and we find the same problem. So where is the missing matter? On the way to this problem many assumptions are being made: notably, that the same laws of mechanics that apply (very successfully) to our solar system apply on the much larger scale, and that most of the matter in a galaxy is visible. Astrophysicists are reluctant to give up the first assumption and so tend to concentrate on looking for what they call dark matter—matter that cannot be seen through a telescope. Nevertheless, there are some indications that perhaps the laws need modification, and so some physicists are looking for solutions to the ‘missing matter’ problem in that direction. It is not too surprising that all of the matter in a galaxy may not be visible. In order to see anything it must either emit its own light or reflect the light from another source. Some nebulae emit their own light, but many glow because of the light from nearby stars. There may well be many nebulae not illuminated in this way. Sometimes we can detect such nebulae because they block our view of the stars behind them (as in Figure 11.45). Unfortunately, however, even if estimates of this type of dark matter are included, the missing matter is still something like 90% of the total calculated mass of the galaxies. That is a lot of missing matter! At present there is no good solution to this puzzle; it is one of the ongoing mysteries of astrophysics. Various forms of dark matter have been postulated including WIMPs (weakly interacting massive particles—these are nuclear-sized particles) and MACHOs (massive compact halo objects—these are planet or larger-sized objects). Even if they have no idea what they are looking for, astrophysicists are never at a loss for creating imaginative mnemonics!

Quasars, black holes and early galaxies Trying to find invisible dark matter somewhere out there might make looking for the needle in the haystack seem like child’s play, but by now you will be used to the fact that this doesn’t put astrophysicists off. To find clues to this puzzle we really need to look back in time to the early stages of the universe. But that is just what astronomers are doing when they

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are looking at hugely distant galaxies. If a galaxy is at a distance of, say 2000 Mpc, which is about 6600 million light-years, in effect we are seeing it as it was 6.6 billion years ago. That is a long time ago, but astrophysicists are interested in galaxies and other objects even further away, which are even further back in time. The Hubble Space Telescope and other very large telescopes have indeed enabled them to see objects that astrophysicists believe are actually quite ‘young’ in the sense that they are seeing them as they were not so long after the beginning of the universe. For example, the Hubble telescope has photographed what are believed to be possible subgalactic gas clouds some 3400 Mpc, or 11 billion light-years, away. This means they are ‘only’ about 3 billion years old. One such class of ‘young’ objects are the quasars, starlike objects that have very large redshifts and hence, it is assumed (although not entirely without question), they are at huge distances from us. One quasar, known as PC1247+3406 has a recession velocity of 94% of the speed of light, putting it at 3800 Mpc (over 12 billion l.y.). The term quasar is short for quasistellar radio sources because they were first discovered by radio telescopes because of the large amount of radio energy they emitted. Later, optical quasars were discovered as well, but many emit both types of radiation and so they are now all grouped together. The fascinating thing is that they seem to be so far away, and yet appear brighter than many stars in our own galaxy. This means that their intrinsic brightness is millions of times greater than that of most galaxies! What puzzled astronomers about quasars was that although they seemed quite numerous at vast distances there were none at relatively closer distances—they would be very obvious if they were! The redshift puts the closest at about 250 Mpc. This is why many astronomers thought they may actually be objects in our own galaxy and that the redshift may be due to other unknown reasons. However, other evidence gradually convinced them that they were indeed very distant, and therefore extraordinarily bright. It appeared, then, that quasars were something that existed a long time ago (a very long time ago!) but did no longer. Eventually it was found that some quasars seemed to be in the centre of galaxies—but these were very difficult to see because of the brightness of the quasar. It had also been noticed (even before the discovery of quasars) that some distant galaxies had extremely bright centres. In the last couple of decades astrophysicists have concluded that quasars are somehow involved in the formation of galaxies, or more particularly were involved, as none seem to have survived into the last few billion years. Such powerful objects must have been extremely massive, so where did all the mass go? The answer seems to be that it ended up as black holes in the centre of most galaxies. If a huge star stops producing enough energy, its gravity will cause it to collapse. We have seen that large stars end up as supernovae, which leave a superdense neutron star. The enormously powerful quasar objects would have left an even larger superdense mass. This mass would have been so dense that even the neutron matter would collapse into what astrophysicists have described as a black hole. This term basically describes a mystery! Once the density of the mass becomes so great, not even light can escape from it, hence the word ‘black’. What happens inside a black hole is not known—as no radiation can escape we cannot see into it. Various theories have been put forward including such fanciful ideas that it may be a passageway to another universe, but such thinking is pure speculation at present. A number of black holes have been found, not by seeing them

Figure 11.46 Quasar PG 0052+251 appears to be at the centre of a young galaxy. Quasars may be extremely powerful objects in the formation of galaxies.


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directly of course, but by seeing the radiation that is given off as matter is sucked into the black hole. Because of the enormous forces involved the matter is totally ionised and accelerated at such a rate as to give off a very characteristic type of electromagnetic radiation. It is now thought that black holes probably exist in the centre of most, if not all, galaxies.

11.4 summary Whole new worlds • Hubble found (using Cepheids) that certain ‘nebulae’ were not dust clouds but distant galaxies. • Cepheid variables are bright stars with a predictable luminosity–period relationship and can be used to find the distance of galaxies. • Galaxies come in four major types (spiral, barred spiral, elliptical and irregular) and tend to group in clusters. The Milky Way is spiral with a diameter of about 50 kpc and thickness less than 1 kpc. The Sun is about 8 kpc from the centre.

• Their redshift shows that galactic clusters are moving away from us at speeds proportional to their distance from us: v = H0d, where H0 is the Hubble constant. • Galaxies have much more mass than the visible matter in them. Astrophysicists are searching for the missing dark matter. • Quasars, extremely bright objects at vast distances, may be the beginnings of galaxies. Black holes may be the remnants of quasars.

11.4 questions Whole new worlds 1 Many faint ‘nebulae’ were known 100 years ago but it was thought they were some sort of dust clouds at similar distances to most stars. How was it discovered that they were in fact other galaxies at vastly greater distances than most of the stars? 2 Andromeda is a relatively close galaxy. In comparison to the diameter of our own Milky Way galaxy, how far away is Andromeda? How large would it appear in the sky if we could see it? Why can’t we see it? (Use the information at the start of this section.) 3 The Milky Way appears to be a fairly uniform band of stars right around the celestial sphere. This was taken to suggest that it was disc-shaped and that we were close to the centre of it. Later it was found that we were not in the centre. How was this discovered and how can it be reconciled with the first observation? 4 Two similar Cepheids, which have the same period, are found in two different galaxies, but one Cepheid appears to be four times as bright as the other. What is the relative distance of the two galaxies? 5 a Given that our galaxy is about 50 kpc in diameter and 1 kpc thick, what is its approximate volume? b If there are about 100 billion stars in the galaxy, what is the approximate volume for each star? c From this figure estimate the average distance between stars.

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d Compare this figure with the distance of the nearest stars to the Sun and explain any apparent anomalies. 6 The Doppler effect describes the apparent change of frequency we hear when a source of sound is approaching or receding from us. Often people say that an aircraft seems to be slowing its engines as it flies over us. How does the Doppler effect help us to understand that this is not actually the case? 7 If a galaxy has a redshift of 10%, at what velocity is it moving relative to us? 8 Hubble found that galaxies were receding from us at a rate that depended on their distance from us. His equation is v = H0d. Although H0 is often quoted as about 70 km s-1 Mpc-1, in SI units its unit would be s-1. Show that this is the case. What is H0 in s-1? 9 The ‘Coma Cluster’ of galaxies is about 90 Mpc from the Earth. At what speed are they receding from us? Is this a significant fraction of the speed of light? (c = 3 × 108 m s-1) 10 Why do astronomers want to look at very distant objects in the universe? Are the distant objects different from those closer to us?

11.5 The expanding un

iverse

If all the galaxies are speeding away from each other they must have been much closer at some time in the past. We have all heard of the expanding universe, but it must be realised that this is a relatively recent discovery. Before that the universe was thought to be infinite and unchanging— except for the occasional ‘new star’ (nova). At least that was the case after the time of Galileo and Newton. You may remember poor Giordano Bruno was burnt at the stake in 1600 for saying that the universe was infinite. Newton postulated that the universe must be infinite because if it were not, in time, gravity would have pulled all the stars together into one great mass. Besides, if it was not infinite, what was beyond its edge? There was a difficulty with this view, however. Kepler was one of the first to realise that if the universe was infinite the night sky should not be dark. Whatever direction one looked one should eventually see a star. Even if there was some ‘dust’ in the way, the dust should be illuminated from all sides. The German astronomer Heinrich Olbers brought this situation to attention again in the early 1800s and so it became known as Olbers’s paradox. Even when Einstein produced his general theory of relativity in 1915 he was dismayed to find that it did not allow for an infinite universe and so amended it by adding what he called a ‘cosmological constant’ so that an infinite universe could exist. (Later he said this was his ‘greatest blunder’—because if he had believed his own equations it would have been he who discovered the expanding universe.) It was Hubble who eventually resolved Olbers’s paradox. He realised that because all the distant galaxies were receding from us, unless we just happened to be in the ‘real’ centre of the universe, the whole universe must be expanding and therefore the light from extremely distant stars will never reach us. It is important to realise that it is not the galaxies all moving away from one point in space, it is space itself that is expanding. This is just what Einstein’s equations had predicted, but even he had not wanted to believe such an outrageous idea. A helpful analogy is to imagine yourself as an ant on a balloon that is being blown up. This balloon happens to have had little paper stars stuck onto it (not drawn on it). You would see all the paper stars moving away from you. Furthermore, wherever you wandered on the balloon’s surface you would see exactly the same thing. This is a two-dimensional analogy for what in reality is a three-dimensional situation, and so we have to bend our minds a little to imagine what this three-dimensional space would be like. One thing to realise is that just as the ant can travel in a ‘straight’ line and end up back where it started, so would we, if we held rigidly to what we believed was a straight course through space (and given more time than the lifetime of the universe!) eventually end up back where we started. The stars on our balloon were stuck on and not drawn on for a good reason. As the balloon expands, drawn stars would expand also, but the stick-on stars stay the same size. This is how astrophysicists picture the expansion of space. We don’t expand, even galaxies don’t expand; it is the space between them that is expanding. It is important to make the distinction between this expansion of space and a conventional explosion, in which everything flies out from one point in space. To see uniform expansion in the explosion we would have to be

Figure 11.47 Why is the night sky dark?

Physics file Another analogy often used to help understand the expanding universe is that of a rising fruit loaf in which all the raisins will be seen to be moving apart from each other. Apart from the difficulty that our little ant will have in moving through the dough, the real problem is that it can come to the edge of the loaf. The whole point about expanding space is that it has no ‘edge’, it is not expanding into anything. The ant on the balloon can wander as far as it likes, it will never come to an ‘edge’ (given that the neck of the balloon is covered over!).

Figure 11.48 As the balloon is blown up, the stars all move away from the ant. The more distant stars will move away faster than the closer ones. In the real universe it is the galaxies that are moving away from each other.


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Did it all start with a big bang?

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Figure 11.49 Curved space. These three circles all intersect at O. If we were to look at point O with a microscope we would see what could appear to be a set of xyz-axes. We have to imagine our usual view of space, as defined by xyz coordinates, is actually a micro-view of real space, which we can imagine as ‘curved’.

Figure 11.50 Fred Hoyle suggested that the universe was in a steady state, with matter being continuously created to compensate for the expansion and fuel new stars. He derided the alternative, which would seem to be that the universe must have started with a ‘big bang’.

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right at the centre. This is not the case for the expansion of the universe (or the ant on the balloon). Astrophysicists talk of this as the ‘cosmological principle’. This states that the universe is ‘isotropic and homogeneous’. This just means that it will appear uniform in whatever direction we look and from wherever we look. Just as there is no ‘centre’ on the two-dimensional surface of the expanding balloon, there is no ‘centre’ of the three-dimensional space of the universe. Mind boggling? Isn’t astrophysics fun!

Detailed studies

If space is expanding, then at some time in the past it must have been just a dot. Well not necessarily, said astronomer Fred Hoyle back in the 1950s. He put forward what became known as the ‘steady state’ theory. His idea was that the universe really was both infinite and expanding. If it was infinite the expansion was not a problem, the ‘outer’ stars would never reach infinity and so could go on moving away from us forever. But wouldn’t that mean that the universe was becoming ‘thinner’? And if this process had been going on forever, by now it would be ultimately thin! ‘No problem’, said Hoyle, ‘matter is being created all the time at just the right rate to keep the density of the universe constant’. This was not really such an outrageous idea. Quantum mechanics had already suggested that matter was less ‘permanent’ than we had thought. And what was the alternative: everything created in one ‘big bang’? After all, which is harder to believe? It was Hoyle who first used the expression ‘big bang’ to describe a universe that started off from a tiny dot. The astrophysicists of the day really were caught in a knot. Either way they had to accept the unacceptable—that matter just came into being out of … what, nothing? It seemed like a really impossible task (even for astrophysicists) to try to resolve this dilemma. If Hoyle was right where do we look for the new matter being created? If it really was a big bang, how could we possibly know? The observational evidence, an expanding universe, was consistent with both theories. If matter was being created at a steady rate it was not too difficult to calculate that rate from the observed density of the universe and the expansion. It turned out that about two or three atoms of hydrogen would need to be created every day in a volume about the size of a large sports arena. There was not much point in trying to look for that! If the big bang idea was correct, what difference might it make to what we can see around us now? Not much. It seemed that only some means of looking back in time could resolve the problem. You may realise that we have already shown that this is possible, but remember that in the 1950s and 1960s there was no Hubble Space Telescope or telescope with adaptive optics. It is important to reiterate that the big bang was not to be seen as an explosion from a small point in space. It was more an explosion of space. In fact it was more correctly an explosion of spacetime. Einstein had already showed that time and space were not the separate entities we normally think of them as (more about this in the Einstein’s Special Relativity study next year). So the big bang would not have occurred at some point in time any more than it would have occurred at a point in space. Time, space and matter were all created together. It was well known by this time that stars had limited lifetimes, but that was not a problem for the steady state theory. New hydrogen was being created all the time, from which new stars could gradually form to replace

those that died. Given that space would effectively appear to be infinitely big in either scenario, and that there did not appear to be any processes that would ‘run down’ the universe with time, it seemed a daunting task to test either hypothesis. However, and not for the first time in science, the eventual answer came from an ‘accidental’ discovery. In the early 1960s, Arno Penzias and Robert Wilson at the Bell Telephone Laboratories in New Jersey were trying out a new directional radio antenna designed to communicate with the new satellites that had just been placed in orbit. However, they found that wherever they pointed their antenna they seemed to pick up some ‘radio noise’, microwave radiation with a wavelength of a few millimetres. After testing everything about their apparatus they could think of they decided that the radiation was coming from outer space. Fortunately, they had heard of a new prediction made by physicists Robert Dicke and P.J. Peebles at Princeton University, just a few miles away. These physicists had calculated that in the big bang, if it occurred, the temperature near the beginning would have had to be hot enough to create a lot of helium from fusion of hydrogen nuclei. As we know, any hot object, including a hot new universe, gives off ‘black body’ radiation. This would have been intense high-energy, very shortwavelength gamma radiation. This radiation would have filled all of early space and simply kept radiating around ever since. So was it still present? Now we know that as a gas expands, it cools. The comparison is not quite fair, but in an analogous way, as the universe expanded, the ‘temperature’ of the early radiation would also decrease. (Radiation from cooler objects has a longer, or redder, wavelength.) This is because the wavelength of the radiation would expand with the space it occupied. Because of the huge expansion of space since the radiation was created, the wavelength should have increased by a huge factor up to around a millimetre, corresponding to a temperature of just a few kelvin. Now this was just the sort of radiation Penzias and Wilson had discovered! Needless to say, there were drinks all round, not to mention a great new area for research, and the death knell for Fred Hoyle’s steady state theory. While some of this cosmic microwave background radiation, as it is called, reaches Earth’s surface, most is absorbed by the atmosphere— fortunately, or it would create a lot more radio interference. In 1989 a special satellite called the Cosmic Background Explorer (COBE) was launched. It was designed to detect and measure radiation in the range of 1 µm to 1 cm from all directions in space. If it really was the leftover radiation from the big bang it should come uniformly from all directions. COBE performed wonderfully. Its instruments showed that the spectrum of background radiation corresponded exactly to a black body with a temperature of just 2.725 K (i.e. just a little above absolute zero). More recently another satellite, WMAP (Wilkinson Microwave Aniso tropy Probe), has mapped the radiation in even more detail. While con firming the COBE results, it also measured very small variations in the temperature of the microwave background. These were only of the order of thousandths of a degree, but they were very significant to astrophysicists. If the early universe really was totally uniform, then galaxies would never have formed. In this case the gravitational pull on any piece of matter would completely balance out from all directions and so everything would just stay where it was. We would have a completely bland, and very boring universe—not good for life at all!

Physics file Time, space, matter and … Even the laws of physics may have been created in the big bang, say some physicists. This is almost more a matter of philosophy than physics, but that’s another story. It is hard not to think of the big bang as a point in time, so to avoid talking of the big bang as though it happened in time physicists talk of it as a ‘singularity’; a strange term, but it is a strange happening. Perhaps we could think of a singularity as an event that can only happen once and from which everything arises—including space and time. Asking the question, what happened before the big bang is rather like asking ‘what is south of the south pole?’ The answer is not that there is there nothing south of the south pole. There is, in fact, no answer to the question because the question itself is meaningless.

Figure 11.51 The WMAP satellite mapped the cosmic microwave background and discovered tiny fluctuations in it, which probably marked the beginning of the formation of galaxies.


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In order for the galaxies to form, there had to be some anisotropy (which just means variation) in the structure of the early universe. This would enable local clumps of matter to start to coalesce and form what would become galaxies. Once a clump of matter starts to form, its gravity will accelerate the process. As the matter falls inward temperatures rise and we have the conditions for the formation of stars. However, the early anisotropy would have resulted in slight variations in the radiation produced, and hence also in its cooled-down form we see as the microwave background. This is what WMAP was looking for, and very successfully found.

When was the bang? Given that the universe started with a big bang, we should be able to say when it started; in other words, how old it is now. If Hubble’s law is right, we simply need to run the rate of expansion backwards to see when all the galaxies were in the same place. This is not so difficult. Hubble’s constant H0 is the proportionality constant between the distance from us to a galaxy and its recessional velocity (v = H0d). If the galaxy has been receding from us at a speed v, and in that time has travelled a distance d, the time it has taken is T = d/v. Now the value of d/v from Hubble’s equation is just 1/H0, so the age of the universe is simply the reciprocal of Hubble’s constant. Notice that the distance cancelled out in this expression—all galaxies, no matter how far away they are now, started off from the same point. Of course we need to convert the units into ones we recognise. We do this in Worked example 11.5A, and the answer is 14 billion years.

The age of the universe = 1/H0

Worked example 11.5A What is the age of the universe if the Hubble constant is taken as 70 km s-1 Mpc-1?

Solution The age of the universe is the reciprocal of the constant: 1 1 T = = Mpc km-1 s H0 70 The expression for the constant includes two different units for distance, km and Mpc, so we need to convert Mpc to km. Table 11.1 tells us that 1 pc = 3.086 × 1016 m. This means that 1 Mpc = 3.1 × 1019 km (to two significant figures) and so 3.1 × 1019 4.4 × 1017 T= = 4.4 × 1017 s, which is equal to = 1.4 × 1010 years 70 365 × 24 × 3600 That is 14 billion years. This method of determining the age has assumed a constant rate of expansion and ignored the effects of gravity. Neither of these is a totally valid assumption, but astrophysicists now have more complex models that do take these effects into account. Curiously enough the other effects tend to cancel each other out and the models still produce an age close to 14 billion years. The value of 14 billion years agrees well with other estimates of the age of the universe. As we saw earlier it is possible to estimate the age of

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stars from their position on the H-R diagram and other information. The oldest stars all seem to be well under 14 billion years. Stars like our Sun are much younger. Because of the presence of heavy elements in both the Sun and Earth, we know that our solar system must have formed from dust produced by a previous supernova, so it is at least a second-generation star if not a third. Estimates of the age of the Sun put it at about 5 billion years, with the Earth and planets forming around half a billion years later. It has taken about 4 billion years for intelligent life to emerge on the Earth. Given that there are about 100 billion stars in our galaxy alone, and that many of them do seem to have planets, perhaps there are many extraterrestrial beings looking up in their sky and wondering if there are other intelligent beings out there. There are various SETI (Search for Extraterrestrial Intelligence) projects looking for evidence of life ‘out there’. Mostly they involve listening to likely radio frequencies that could be used to broadcast information across the galaxy. It is fascinating to wonder about the likely content of such broadcasts. Whatever we may think of such searches, as Ellie (Jodie Foster) said in the 1997 movie Contact, ‘If there are no other intelligent beings out there, it’s an awful waste of space!’

Figure 11.52 Ellie Arroway (Jodie Foster) searches for aliens in Contact, the film based on the book by Carl Sagan.

Physics in action

The future of the universe If the universe is expanding, will it go on expanding forever, or will it eventually slow and perhaps start collapsing again? Naturally this question has fascinated astrophysicists ever since the idea of the big bang was suggested. If a rocket is fired upwards it may reach a maximum height and fall back, or it may have enough energy to escape from the Earth and sail on into space forever. Or it may have just enough energy to escape but none left over for space adventures. The question the astrophysicists asked was ‘Did the universe have enough energy to overcome the mutual gravitational attraction of all its mass and expand forever?’. Or would gravity eventually win out and pull it all back into a ‘big crunch’—and perhaps start the cycle all over again with another big bang? Actually, because of Einstein’s theory of general relativity, which links space, matter and gravity, the question was put a little differently. Mass, Einstein said, distorts space. We see the effects of the ‘curved space’ as the effects of gravity. In the language of relativity the question becomes whether space has positive, negative or zero curvature. Positive curvature implies that there is enough matter in the universe to fold it in on itself—producing a closed universe that will eventually collapse to the big crunch. Negative curvature is the opposite—space curves outwards in an open-ended way and the expansion can continue forever. In between positive and negative curvature there is the ‘flat’ universe—equivalent to the expansion just being sufficient to go on forever, but only just. These three possibilities are sometimes represented by the two-dimensional analogies shown in Figure 11.53. There are different approaches to answering this question. One is to try to measure all the mass in the universe and see if it is sufficient to curve space ‘inwards’. The difficulty here is to detect the mass. We have already seen that most of the

mass of the universe appears to be so called ‘dark matter’, which by its nature is very hard to detect. Another approach is to measure the rate of expansion at different times in the history of the universe to see if it is slowing down. This can be done by looking at galaxies that are vast distances from us—the further away, the further back in time we are looking. It is the cosmic microwave background radiation (CMB) that has given us our best clue to the answer. The CMB has been travelling through space since the time of the hot early universe and has therefore been affected by the space it has travelled through. It is possible to simulate the effects of positive, zero or negative space curvature on the radiation and then compare these with the actual radiation detected by satellites such as WMAP. The differences are found in the very small irregularities we discussed earlier. The results of this research seem fairly clear—the CMB pattern is consistent with a flat universe; that is, zero curvature. Given that the universe is flat, the total amount of mass and energy in it can be calculated (the reverse of the process of trying to determine if it is flat from the amount of mass). Remember at this point that Einstein said mass and energy are inextricably linked (E = mc2). It turns out that even with the estimated amount of dark matter, there is still a huge amount of missing ‘mass-energy’. Visible matter accounts for about 4% and dark matter for about 26%. This would appear to leave 70% of the total mass-energy content missing. However, you will not be surprised to find that astrophysicists now think they know where it is! Perhaps the most surprising results have come from the studies of very distant galaxies. They have suggested that rather than slowing down, the rate of expansion of the universe has been accelerating. At first this seemed to be quite contrary to all expectations! Even our spacecraft with plenty of energy to escape the Earth will still slow down


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(a)

(b)

(c)

as gravity tugs on them. Similarly, it was thought that the mass of the universe would eventually slow its expansion rate. However, when this acceleration was put together with the picture of a flat universe, as well as the implications of Einstein’s ‘cosmological constant’, a new idea emerged. It seemed that the missing 70% of the mass-energy content of the universe might be in the form of what has been called ‘dark energy’. This energy, it is thought, is what is actually accelerating the expansion—it is a sort of ‘anti-gravity’. As the effect of gravity (space curvature) becomes less with the increasing size of the universe, the effect of the dark energy becomes more significant and the acceleration increases. So how will the universe end? Perhaps unfortunately, it does not seem that it will end with a big crunch after all. At least that may have given rise to another big bang. Rather, as everything accelerates apart it will get thinner and thinner until everything is so far away that it will become a very boring place with a very dim and uninteresting night sky—not much good for astrophysicists at all! Oh well, perhaps by then we’ll have found a way to jump to one of those ‘parallel universes’ we hear about—one with a different, but no less interesting, set of problems for astrophysicists to solve.

Size of universe

Figure 11.53 These diagrams are two-dimensional analogies for what is really three-dimensional space—or four, if time is included! In (a), positive curvature, the surface eventually folds in on itself, while the negative curvature of (c) can go on expanding forever. Surface (b) is the in-between ‘flat’ case.

(c)

(b) (a) Past 10

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Figure 11.54 (a) This curve is for a flat universe without dark energy, and until recently seemed the most likely model. (b) This curve shows the expansion if we simply assume the Hubble constant has been constant from the beginning. (c) This curve is the latest model, which takes account of dark energy.

11.5 summary The expanding universe • The implication of a stable, infinite universe was that the sky should not be dark at night. This was a problem not resolved until Hubble discovered that the universe was not infinite but expanding. • It is space itself which is expanding, not the objects within it. Hoyle’s steady state theory explained this by postulating the continuous creation of matter. The alternative theory suggested the universe all came into being in one ‘big bang’.

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• The big bang theory was confirmed by the discovery of the cosmic microwave background, which is radi ation left over, and cooled, from the early universe. • The age of the universe is given by the reciprocal of the Hubble constant and is found to be about 14 billion years.

11.5 questions The expanding universe 1 Why was it that most physicists from Galileo and Newton to Einstein thought that the universe must be infinite and unchanging? Was there any difficulty with this idea? 2 Describe Olbers’s paradox and its significance to discussions about the nature of the universe. How was the paradox eventually resolved? 3 To account for the expansion of the universe as observed by Hubble, two competing theories were put forward. What were they, what was the essential difference between them, and why was it seen as extremely difficult to test them? 4 Given that the density of air is about 1.2 kg m-3, roughly estimate the number of molecules of air in a large sports stadium. The mass of a nitrogen molecule is 4.7 × 10-26 kg. How does this number compare with the number of molecules created each day according to the steady state theory? 5 Three ants, A, B and C, are standing still on a balloon being blown up. They are in a straight line and initially 5 cm apart (so A is 10 cm from C). After 10 s the radius of the balloon has doubled. At what speed does ant A see ants B and C receding from it?

6 What was happening in the universe before the big bang occurred? 7 In the 1960s two radio engineers discovered radio waves coming from space. In what way did this discovery contribute to resolving the question of the competing theories of the origin of the universe? 8 The radiation produced by the extreme conditions at the beginning of the universe was extremely shortwavelength gamma radiation. What is the relation ship between this and the much longer-wavelength cosmic microwave background radiation, that con firmed the big bang theory? 9 Satellites designed to measure the cosmic microwave radiation have found that although it is extremely uniform, there are minute variations on a small scale. What is the significance of these small variations to astrophysicists? 10 Hubble’s original value for the constant in his law was 500 km s-1 Mpc-1. What would that have sug gested about the age of the universe? Do you see a problem with this age?

chapter review Where necessary, take the intensity of radiation from the Sun at the Earth as 1370 W m-2 and the luminosity of the Sun as L  = 3.9 × 1026 W m-2. 1 Many people objected to Galileo’s heliocentric system on the grounds that if it were true then we should notice parallax among the stars over the course of the year. What did they mean by this, and what do you think Galileo’s reply would have been? 2 The star 61 Cygni was said to have a proper motion of about 5 arcsec per year. Given that it is about 3.5 pc away from us, what is its actual speed through the galaxy? How does this compare to the Earth’s speed around its orbit? 3 You are driving along a road at 100 km h-1 when you notice that an aeroplane travelling in the opposite direction appears to be stationary over a hill that is 2 km from you. In fact you know that the aeroplane is actually travelling at 800 km h-1 in the opposite direction. How far away from you is it?

4 While Hipparchus said it would take two +2 magnitude stars to equal the brightness of one +1 star, the modern scale was redefined so that it now takes 2.5 magnitude +2 stars to make up one +1 star. a How many +6 stars did Hipparchus assume would be needed to equal the brightness of one +1 star? b How many +6 stars are needed on the modern scale, and why the difference? 5 a Captain Blake and crew have just landed on a planet around the star Alpha Centauri B, which they knew from their astrophysics course has a luminosity of 0.55L . They find that the intensity (above the atmosphere) at the planet is 1000 W m-2. How far is the planet from the star? b They then travel to a planet at twice the distance from Alpha Centauri B. What will be the intensity of the radiation there? c Next our intrepid travellers go into orbit around Alpha Centauri A. They find that in their orbit at 4 AU the radiation


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intensity is only 150 W m-2. What is the intrinsic luminosity of Alpha Centauri A? 6 Two identical pokers are heated in a fire until one is at 500 K and the other is at 1000 K. Although one is only twice the temperature of the other it is seen to be glowing red while the other is not glowing at all, and the heat radiated from the hotter one seems very much stronger than the other. Explain these facts in terms of the laws of physics involved. 7 Although even the closest stars are no more than points of light in a telescope, astronomers can tell the size of a star from the nature of the light it emits. How is this possible? 8 The bV/bB colour ratio for Aldebaran is found to be 4.1. a If its relative brightness through a B filter is found to be 0.2, what would it be through a V filter? b Estimate the surface temperature of Aldebaran (see Table 11.2). What colour does it appear to be in our sky? 9 The diameter of the Sun is about 109 times that of the Earth. How many Earths would fit inside the Sun? 10 The energy output of the Sun is about 4 × 10  W and its mass is about 2 × 1030 kg. a If it were made of coal, which gives out about 30 MJ per kilogram burnt, how long would it last? b In fact it is powered by hydrogen fusion, which releases 6 × 1014 J for every kilogram fused to form helium. How long could it last while powered this way? 26

11 The picture that astrophysicists have built up of the Sun is most like which of the following? A An extremely dense solid core of helium gradually burning in the hydrogen around it. B A uniform gas about the density of water, which is gradually undergoing fusion of hydrogen throughout. C A gaseous sphere extremely dense in the centre where high temperatures initiate hydrogen fusion reactions. D A liquid sphere with enormous pressure in the centre and hydrogen fusion occurring on the outer layers. 12 Detailed photographs of the Sun’s normal surface reveal a motley appearance. What is the cause of this motley surface, and how permanent is it? 13 As the stars are so far away, how can we be sure that they are not made of totally new elements that we have never seen on Earth?

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14 Astronomers depend heavily on an analysis of the spectrum of a star to determine its various properties. What are some of the features of a star that can be determined from an examination of its spectrum? 15 ‘The Sun is an average star’. In what sense is this statement true in regard to the H-R diagram, and in what sense is it not true? 16 If a main sequence star has a mass of 10 times the Sun, how much brighter will it be? 17 In terms of the H-R diagram, where are stars ‘born’? 18 When a normal star reaches the end of its main sequence lifetime it becomes a red giant. Briefly describe the processes that cause this change. Why are giants red? 19 What evidence do astronomers have of the remains of supernovae explosions? 20 What is it about Cepheid variables that makes them a useful tool with which to find astronomical distances? 21 Most of the 100 billion stars in the Milky Way are within about 15 kpc of the centre. Mostly the galaxy is about 1 kpc thick although there is a bulge of about 2 kpc diameter in the centre. a What is the average distance between stars on the basis of these dimensions? Would all stars be about this far from their neighbours? b How does this distance compare with the distance to the stars nearest the Sun? 22 The speed of sound in air is 340 m s-1. If we find that the frequency of the sound of a police car is 10% higher than normal, at what speed is the car moving relative to us? 23 The closest galaxies to the Milky Way are the Magellanic Clouds. They are at distances of 50 kpc (LMC) and 63 kpc (SMC). Because they are in orbit around the Milky Way they are not receding from us, but if they were obeying Hubble’s law, at what speed would they be receding? Comment on the answer you get. 24 The universe is said to be expanding, and yet some galaxies have a ‘blueshift’ when their spectrum is examined. What is the implication of a blueshift and how does this fit into the picture of an expanding universe? 25 Using the value of the Hubble constant is one way to find the age of the universe, but there are other methods as well. What are some of these and are they consistent with the age as found from the constant?

Energ

y from

the n uc

I

leus

n 1905 Albert Einstein theorised that mass and energy are equivalent through the equation … = mc2. This led to the realisation that vast amounts of energy lay unharnessed within the nuclei of atoms. However, it was not until 1932 that Einstein’s theory could be verified. Chadwick’s discovery of the neutron in that year enabled scientists to fire these particles at stationary atomic nuclei and, in some circumstances, to ‘split the atom’. The ramifications of Einstein’s work and this discovery of nuclear fission were not seen until 1945, when the first atomic bomb was exploded in the desert near Alamogordo in New Mexico, USA. Today a more peaceful application of nuclear fission exists, with the use of nuclear power in many countries around the world. The use of nuclear energy is an issue that has been debated for many years. Although it yields a great deal of energy from a small amount of fuel, there are obvious problems associated with the use of nuclear energy for power production. The waste from nuclear reactors is highly radioactive, and serious nuclear accidents have occurred at Chernobyl and Three Mile Island. Against this there are positive uses for nuclear energy. Nuclear fission reactors produce radioisotopes that have many applications in industry, medicine, agriculture and scientific research. They also generate electricity without releasing the vast amounts of greenhouse gases that coal-fired power stations emit. For this reason, many people who are concerned about climate change would like to see more nuclear reactors used for generating electricity. Most green groups, however, see the problem of storing radioactive waste as being insurmountable and would like to see clean, renewable energy sources such as solar, wind, geothermal and tidal power used instead. Inside our Sun another type of nuclear reaction is occurring. Hydrogen nuclei are being squeezed together in a process known as nuclear fusion. Enormous quantities of energy, but no radioactive waste products, are released during fusion. It is possible that, within a few decades, fusion reactors will be generating electricity on Earth. We live in a nuclear age. This presents both possibilities and difficulties. An understanding of nuclear fission and nuclear fusion is important when developing an informed opinion on these issues.

by the end of this chapter you will have covered material from the study of energy from the nucleus including: • the nuclear model of the atom • nuclear fission in uranium-235 and plutonium-239 • nuclear fusion reactions and stars • conditions for fission chain reactions • neutron absorption in uranium-238 and plutonium-239 • the transformation of nuclear energy into electrical energy.

outcome On completion of this chapter, you should be able to describe and explain typical fission and fusion reactions, energy transfer and transformation phenomena of importance in stars and in the production of nuclear energy, and the benefits and risks of the use of nuclear energy as a power source for society.

ion

ear fiss l c u n — m o t a 12.1 Splitting the Physics file

(a)

You will be familiar with the basic model of the atom as discussed in Chapter 1. Nucleons (protons and neutrons) are in the nucleus of the atom. Electrons are in orbit around the nucleus and the electron cloud makes up nearly all of the volume of the atom. The nucleus contributes most of the mass of the atom but accounts for almost none of the atom’s size. The diameter of a nucleus is typically about one ten-thousandth of the diameter of the electron cloud. In this section, we will be examining the structure of the nucleus and the forces that exist within it.

(b)

Nucleus

electron cloud proton

neutron

Figure 12.1 This model of the atom was first developed in the early decades of the last century. Protons and neutrons are found in the nucleus. Electrons orbit the nucleus and this electron cloud comprises most of the atom’s size. If a tennis ball were used to represent the nucleus of an atom, the electron cloud would have a diameter of about 600 m!

(c)

Inside the nucleus

Figure 12.2 (a) In 1896 Henri Becquerel discovered radioactivity in uranium. This marked the birth of nuclear physics. (b) In 1911 Ernest Rutherford and his students Hans Geiger and Ernst Marsden performed an experiment which suggested that atoms had a small, heavy nucleus. In 1914 Rutherford also discovered the proton. (c) In 1932 James Chadwick discovered the neutron.

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Uranium is one of the largest atoms. It has 92 protons and over 140 neutrons in its nucleus. Even so, about 100 000 000 000 such uranium nuclei could be lined up across 1 mm! This gives you an idea of how small atomic nuclei are and why it has been so difficult to examine them. Our current understanding of the basic properties and structure of the nucleus is the result of intense scientific investigation in the early part of the 20th century. Physicists such as Geiger, Marsden, Rutherford, Harkins and Chadwick were instrumental in the development of the model of the nucleus that we are familiar with today. A model known as the ‘liquid drop model’, proposed by Niels Bohr in 1936, is a useful way of visualising a nucleus. Nuclei are generally spherical in shape, but are quite fluid and can wobble like jelly. Within the nucleus, there are protons in close proximity with other protons. This should seem odd since protons exert strong electrostatic forces of repulsion on each other. Shouldn’t the nucleus simply blow apart? It obviously doesn’t—so there must be something else going on. Another

force, known as the strong nuclear force, is also acting and this is a force of attraction that acts to hold the nucleus together. • Electrostatic forces act between charged particles and can act over relatively large distances. Within a nucleus, this means that each proton is strongly repelling every other proton, so this force is trying to make the nucleus disintegrate. Neutrons are unaffected by electrostatic forces. • The strong nuclear force is a force of attraction that acts between every nucleon. This force acts like a nuclear cement. However, this force only acts over relatively short distances. Neutrons are attracted to nearby neutrons and protons. Protons are similarly attracted to nearby neutrons and protons. However, for nucleons on opposite sides of a large nucleus, this force is not significant. When two protons are side by side in a nucleus, forces of attraction and repulsion are acting at the same time. The strong nuclear force of attraction is much greater than the electrostatic force of repulsion that they exert on each other. However, when the protons are on opposite sides of a large nucleus, the short-range strong nuclear force will be weaker than the electrostatic force. In small nuclei, where the protons and neutrons are all relatively close together, the strong nuclear forces are dominant. The nucleus is held together very strongly and so is very stable. In large nuclei, the size of the nucleus reduces the influence of the strong nuclear force. For a proton located on the outside of the nucleus, the forces of attraction and repulsion would be smaller. This nucleus would be less stable as a result. Now let us imagine that you were trying to construct a uranium nucleus, so you attempted to combine 92 protons and 92 neutrons. As shown in Figure 12.4, your attempt at a nucleus would spontaneously blow apart. It is inherently unstable because there are 92 protons all repelling each other and the size of the nucleus means that the effect of the strong nuclear force of attraction is diminished. To make your nucleus reasonably stable, more neutrons are required. In nature, uranium nuclei are only reasonably stable when they have over 140 neutrons. These isotopes of uranium are still unstable and are radioactive. In fact, every nucleus with more than 83 protons, i.e. above bismuth (Bi) in the periodic table, is similarly radioactive. (a)

(b)

(c)

neutron proton

C A B

Figure 12.3 Proton A both attracts and repels proton B, but the attraction is greater because the strong nuclear force is very strong over short distances. Proton A also both attracts and repels proton C, but because of the greater distance between them, the force of repulsion is larger.

Physics file The existence of the strong nuclear force was first proposed by Japanese theoretical physicist Hideki Yukawa in 1935. The properties of this force are so complex, however, that it took until 1975 for physicists to develop a mathematical model that could successfully describe it.

(d)

6p, 6n 92p, 92n

92p, 146n

Figure 12.4 (a) A small nucleus such as carbon-12 is stable. This is because the electrostatic force of repulsion that acts between the protons is overwhelmed by the strong nuclear force of attraction. (b) and (c) A large nucleus with equal numbers of protons and neutrons cannot exist. The electrostatic forces of repulsion between the protons would overwhelm the strong nuclear forces. (d) Additional neutrons increase the stability of large nuclei. The extra neutrons increase the influence of the strong nuclear force and act like a ‘nuclear cement’, holding the nucleus together.

Chapter 12 Energy from the nucleus

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Nuclear fission

Physics file

Neutron number (N)

120 N=Z

80

80 Proton number (Z)

Figure 12.5 There are about 400 stable nuclei. The dots on this graph show the number of protons and neutrons that they have. For small stable nuclei, it can be seen that the numbers of protons and neutrons are roughly equal. In larger stable nuclei, there are many more neutrons than protons. The largest stable nucleus is bismuth-209 with 83 protons and 126 neutrons.

Until 1932, scientists used alpha particles from radioactive sources as probes to explore the nature of atomic nuclei. The alpha particles, acting as highspeed ‘bullets’, bombarded target nuclei, and the resulting interactions were analysed. The problem with this method was that both the alpha particles and target nuclei were positively charged and so repelled each other. It was possible for very energetic alpha particles to actually smash into the nuclei of small atoms such as nitrogen and aluminium. However, target nuclei with atomic numbers above 19, i.e. larger than potassium, repelled the alpha particles so strongly that collisions were not possible. The discovery of the neutron by James Chadwick in 1932 enabled scientists to explore the behaviour of larger atomic nuclei. Being neutral, a neutron is not repelled by the target nucleus, and can be absorbed into the nucleus of the target atom. This makes it very useful as a form of bombarding radiation, and it is used in many experiments to artificially transmute different isotopes (e.g. 10n + baX → b + 1aX). Enrico Fermi, an Italian-born scientist working in the United States, was conducting one such experiment in 1934. He bombarded uranium nuclei with neutrons and obtained some unexpected results. Some of the uranium nuclei absorbed the neutrons and split in two! Fermi was the first to observe nuclear fission, although the mechanics of this process was not determined until 1938.

NUCL…AR FISSION occurs when an atomic nucleus splits into two or more pieces. This is often triggered by the absorption of a neutron.

(a)

α 2+

(b)

1n 0

Figure 12.6 (a) Alpha particles are repelled so strongly by large atomic nuclei that collisions are not possible. (b) Neutrons, having no charge, are capable of colliding directly with the nucleus of an atom.

Nuclides that are capable of undergoing nuclear fission after absorbing a neutron are said to be fissile. Fissile nuclides are very uncommon. Uranium-235 and plutonium-239 are readily fissile. Uranium-238 and thorium-232 are only slightly fissile, requiring a very high-energy neutron to induce fission.

The release of neutrons during fission Uranium-235 and plutonium-239 are the fissile nuclides most commonly used in nuclear reactors and nuclear weapons. This is because they release

incident neutron fissile nucleus

fission fragments released neutrons

Figure 12.7 When a fissile nucleus absorbs a neutron, the nucleus splits in two, releasing a number of neutrons. This is an example of nuclear fission.

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energy upon fission. When a uranium-235 or plutonium-239 nucleus absorbs either a slow- or a fast-moving neutron, it becomes unstable and spontaneously undergoes fission. However, fission is more likely to be induced by a slow-moving neutron. A typical fission reaction for uranium-235 is: 1 n + 235 U → 236 U → 91 Kr + 142 Ba + 310n + energy 0 92 92 36 56 Krypton-91 and barium-142 are known as the fission products or fission frag ments. Three neutrons are freed from this uranium nucleus when it splits. 91

Physics file (a)

1 0n

fissile nucleus

(b)

Kr (c)

fission fragments 1 0n

(d) 235

U

three neutrons 142

Ba

Figure 12.9 When a uranium-235 nucleus absorbs a neutron, nuclear fission occurs.

The uranium nucleus splits in two, forming in this example krypton-91 and barium-142. Three neutrons are released.

A uraniun-235 nucleus may split in many different ways, so when a sample of uranium-235 undergoes fission, a wide variety of fission products are produced. Usually either two or three neutrons are released. For uranium-235, an average of 2.47 neutrons per fission has been determined. Over 40 different pairs of fission fragments of uranium-235 have been found, and most of these are radioactive beta emitters. Other fission fragment pairs produced in the fission of uranium-235 nuclei include xenon-54 and strontium-94, and tin-132 and molybdenum-101. Some of these fission fragment isotopes can themselves be struck by and absorb a neutron, forming a different radioisotope again. In the end, the decay products of the nuclear fission process form a lethal cocktail of radioactive isotopes. It is these radioactive fission fragments that comprise the bulk of the high-level waste produced by nuclear reactors. Radioactive waste is discussed in detail in section 12.3. Plutonium-239 will also undergo fission in a variety of ways. It releases an average of 2.89 neutrons per fission, slightly more than uranium-235.

neutrons released

Figure 12.8 When a fissile nucleus such as uranium-238 is (a) struck by a neutron, the nucleus actually (b) wobbles about like a water droplet and (c) elongates. (d) If the nucleus stretches beyond a critical point, the electrostatic forces will overcome the strong nuclear forces and it will break into two. This is nuclear fission. Several neutrons are also released during this process. The fission fragments and neutrons have a large amount of kinetic energy.

The energy released during nuclear fission The chemical reactions that you have probably performed at school typically release only a few electronvolts of energy. Compared with this, an enormous amount of energy, about 200 MeV, is released during each fission reaction. This energy is mainly in the form of the kinetic energy of the fission fragments, with the neutrons and emitted gamma radiation also having some energy. It was Albert Einstein who provided the explanation of the origins of this energy. He showed that mass and energy, instead of being completely independent quantities, were in fact completely equivalent.


443

This does not mean that mass can be transformed into energy, rather that mass has energy and energy has mass. In any fission reaction, the combined mass of the incident neutron and the target nucleus is always greater than the combined mass of the fission fragments and the released neutrons. For example, in Figure 12.9 the mass of the incident neutron and the uranium-235 nucleus is greater than the combined masses of the fission products—barium-142, krypton-91 and three neutrons. The energy released as a result of this mass decrease is given by Einstein’s famous equation: E = mc2. It is important to note that only a very small proportion of the original mass of nuclei is available as usable energy—typically around 0.1%. The energy released during the fission process is usually expressed in either joules (J) or electronvolts (eV).

Worked example 12.1A Figure 12.10 Albert Einstein displayed little of his intellectual capability as a schoolboy, and was once told by a teacher that he would never amount to anything. However, he was extremely curious about physics and mathematics and used thought experiments to help him understand difficult concepts and situations.

Plutonium-239 is a fissile material. When a plutonium-239 nucleus is struck by and absorbs a neutron, it can split in many different ways. Consider the example of a nucleus that splits into barium-145 and strontium-93 and releases some neutrons. The nuclear equation for this is: 1 n + 239 Pu → 145 Ba + 93 Sr + x10n + energy 0 94 56 38 a How many neutrons are released during this fission process, i.e. what is the value of x? b During this single fission reaction, there was a loss of mass (a mass defect) of 3.07 × 10-28 kg. Calculate the amount of energy that was released during fission of a single plutonium-239 nucleus. Answer in both electronvolts and joules. c The combined mass of the plutonium nucleus and bombarding neutron was 3.99 × 10-25 kg. What percentage of this initial mass was related to the energy produced during the fission process?

Solution a Neutrons have no effect on atomic numbers, so the mass numbers must be analysed: 1 + 239 = 145 + 93 + x x=2 i.e. two neutrons are needed to make the mass numbers balance. b The energy released during the fission of this plutonium nucleus can be found by using … = mc2. … = 3.07 × 10-28 × (3.0 × 108)2 = 2.76 × 10-11 J To determine the energy released in electronvolts, it is necessary to convert joules into electronvolts. (2.76 × 10-11) The energy released is = = 1.73 × 108 eV or 173 MeV (1.6 × 10-19) c Percentage of mass decrease = (mass defect) × 100 (initial mass) 1 (3.07 × 10-28) 100 = × 1 (3.99 × 10-25) = 0.0769%

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Physics in action

Enrico Fermi and nuclear fission Enrico Fermi was born in Italy in 1901. He completed his doctorate and postdoctorate work in physics at the University of Pisa and in Germany. Fermi had emigrated to the United States by the time the nuclear age dawned in the 1930s. The neutron had just been discovered in 1932, which enabled scientists for the first time to fire neutral particles at atomic nuclei. Fermi was at the forefront of this research. He bombarded uranium atoms with neutrons and found that uranium nuclei absorbed the neutrons and formed a radioactive isotope of uranium. This isotope then decayed to neptunium and then plutonium, two completely new elements. Fermi had successfully produced the world’s first artificial and transuranic (i.e. after uranium) elements. The nuclear reactions for this process are:

barium (Z = 56). Hahn wrote to his colleague Lise Meitner about this unexpected result. She then discussed this with her nephew Otto Frisch, a nuclear physicist, and realised that the bombarding neutrons were causing the uranium nuclei to split. If barium (Z = 56) was one of the products, then krypton (Z = 36) must be another. This was found to be the case. It was Frisch who coined the term ‘fission’ and Meitner who proposed that energy would be released during this process.

n + 23982U → 23992U (neutron absorbed) U → 23993Np + -10β (β decay) Np → 23994Pu + -10β (β decay)

1 0 23 9 92 23 9 93

process ining the la p x d e in l as awarde umenta was instr eague Otto Hahn w y r e n it e M iall 2 Lise her coll ontrovers Figure 12.1 ion. Even though tion was c eitnerium, Mt. u s ib s tr fi n r o a c le of nuc ble is m eitner’s ze, Lise M periodic ta a Nobel Pri lement 109 in the d. E overlooke Figure 12 .1

producin g

1 Enrico Fe the first rmi was awarded transura nic elem the Nobel Prize in ents. physics for

In 1938, following on from Fermi’s work, two German scientists Otto Hahn and Fritz Strassmann were also bombarding uranium (Z = 92) in an attempt to produce some transuranic elements (Z > 92). They found that, rather than producing larger elements, they were getting isotopes of

After the commencement of World War II, Enrico Fermi was commissioned by President Roosevelt to design and build a device that would sustain a nuclear chain reaction. In 1942, Fermi succeeded in this task. A squash court at the University of Chicago was used as the site for the world’s first nuclear reactor. It produced less than 1 watt of power— not even enough to power a small light globe! The reactor was later modified to produce about 200 W. Fermi died of cancer in 1954. One year after his death, the element with atomic number 100 was artificially produced and named fermium, Fm.


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12.1 summary Splitting the atom—nuclear fission • Within a nucleus, forces of repulsion and attraction are acting. An electrostatic force of repulsion acts between the protons over relatively long distances. A strong nuclear force of attraction acts between every nucleon over relatively short distances. • Nuclear fission occurs when a nucleus is caused to split and release a number of neutrons. This can happen after a fissile nucleus has been struck by a neutron. A relatively large amount of energy is released during this fission process. • Elements that are capable of undergoing nuclear fission are known as fissile materials. Only a handful

of isotopes have this property, including uranium-235 and plutonium-239. These are the most common nuclear fuels. • When a nucleus splits during nuclear fission, a number of neutrons are released. There are many different fission fragments that are produced, including isotopes such as 91Kr, 94Sr, 142Ba and 85Br. These fission fragments are usually radioactive. • When a nucleus undergoes fission, the mass of the fission fragments is always less than the mass of the original particles. This decrease in mass is equivalent to the energy that is released during each fission and can be determined by using E = mc2.

12.1 questions Splitting the atom—nuclear fission 1 In stable nuclei, the forces acting between the nucleons are relatively large. a Comment on the relative strength of these forces in fissile atoms. b Explain why a collision with a single neutron is enough to split a fissile nucleus. 2 Tin has an atomic number of 50. Is an isotope of tin 100 Sn likely to exist? (You may need to refer to Figure 12.5.) Explain. 3 Determine the number of neutrons, x, released during each of these fission reactions: a 01n + 235 U → 148 La + 85 Br + x01n 92 57 35 b 01n + 235 U → 142 Xe + 90 Sr + x01n 92 54 38 c 01n + 235 U → 127 Sn + 104 Mo + x01n 92 50 42 4 Determine the value of the unknown mass number x and atomic number y in this fission reaction: 1 n + 94xPu → 130 Xe + 106yZr + 401n 0 54 5 Which one of the following particles is best able to split a fissile atomic nucleus? A A proton B A neutron C An electron D An alpha particle 6 Which one of the following statements is correct? A Nuclear fission is more likely to occur in small nuclei. B It is possible to produce fission in any nucleus provided the bombarding particles have enough energy. C All radioactive nuclei are fissile. D Only a few nuclei are fissile.

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Detailed studies

7 Einstein said that mass and energy are equivalent. In one particular nuclear fission reaction, a decrease in mass of 3.48 × 10-28 kg occurs. a Express the energy equivalent of this in terms of joules and MeV. b The highest amount of energy that is released during a single radioactive decay is about 10 MeV. Comment on the energy released during nuclear fission in comparison to this. 8 During the radioactive decay of cobalt-60, a gamma ray with 1.33 MeV of energy is released. Calculate the decrease in mass (in kg) of the cobalt-60 nucleus as a result of this emission. 9 When James Chadwick discovered the neutron in 1932, he was irradiating a sample of beryllium-9 with a radioactive emission. The equation for this is: 9 Be + X → 126C + 01n 4 By balancing the equation, determine the nature of the bombarding particle. 10 When a uranium-235 nucleus absorbs a slow-moving neutron, one of its possible fission reactions is: 1 n + 235 U → 148 La + 85 Br + 301n + energy 0 92 57 35 The mass defect during this fission reaction is 2.12 × 10-28 kg. a Calculate the energy (in joules) that has been released during this fission process. b Express this energy in electronvolts (eV). c Enrico Fermi used uranium-235 to establish the world’s first nuclear reactor in 1942, he was able to generate just 0.60 W of power. How many fission reactions were occurring each second?

12.2 Aspects of fiss io

n

In the years after the discovery of nuclear fission, an intensive scientific and military endeavour was undertaken to investigate all of the ramifications. The two main aspects of this investigation related to the use of nuclear fission for nuclear weapons and in the generation of electricity. Nuclear fission weapons release an enormous amount of energy in a split second; they are capable of causing death and devastation on a massive scale. This was tragically evident in the bombing of Hiroshima and Nagasaki in 1945. The energy released in the one bomb dropped over Hiroshima was equivalent to 20 000 tonnes of TNT but involved a fission reaction in only 40 kg of uranium. The explosion is estimated to have killed about 150 000 people.

Figure 12.13 Atomic bombs use the process of nuclear fission to release vast amounts of energy with devastating effects.

The properties of uranium-235, uranium-238 and plutonium-239 Uranium-235 is most likely to undergo fission when struck by a slowmoving or thermal neutron with energy as low as 0.01 eV. A slowmoving neutron can be absorbed into a uranium-235 nucleus, forming the highly unstable uranium-236 isotope. This then undergoes fission and releases energy. A fast, high-energy neutron is not easily captured by a uranium-235 nucleus and so is less likely to induce fission. It is difficult for the nucleus to capture a fast neutron because the neutron does not stay close to it long enough for the strong, short-range nuclear force to drag it in.

slow neutron 1 0n

235U

+

235 92U

neutron absorbed 236 92U

fission!

Figure 12.14 A slow neutron is absorbed by a uranium-235 nucleus, converting it into uranium-236, which is highly unstable. This nucleus then undergoes nuclear fission.


447

Uranium-238 is only slightly fissile. It requires a neutron with a large amount of energy (about 1 MeV) to cause fission in a uranium-238 nucleus. Because of its effectively non-fissile nature, uranium-238 is not suitable for use as a nuclear fuel, but it does have a role to play in nuclear energy production. A uranium-238 nucleus is far more likely to simply capture a neutron and become uranium-239. This then goes through a series of radioactive decays to become plutonium-239, itself a fissile substance. Uranium-238 is known as a fertile material because of this ability to capture a neutron and transform into a fissile substance. Plutonium-239 is fissile in the same manner as uranium-235. However, plutonium nuclei require fast neutrons to bring about nuclear fission. Slowmoving, or thermal, neutrons do not cause fission in plutonium-239 nuclei. (a)

slow neutron 1 0

n

Pu

239

+

239 94

Pu

Pu

240

240 94

Pu

(b)

Figure 12.15 (a) When a plutonium-239 nucleus absorbs a slow neutron, the isotope plutonium240 is formed. (b) The additional energy possessed by a fast-moving neutron causes the plutonium-240 nucleus to distort and split into fission fragments.

fast neutron 1 0

n

+

239 94

Pu

Pu

240

240 94

Pu

fission!

Chain reaction

uranium-235 nuclei

etc.

1

2

4

8

1

2

3

4

Number of neutrons 16 ... 5 Generation

Figure 12.16 A single slow-moving neutron causes fission, and two neutrons which are both capable of splitting another nucleus are released. After five nuclear generations, 16 neutrons are capable of triggering fission.

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Pu

239

Detailed studies

The scientists working on the Manhattan Project during World War II knew that nuclear energy could be released from a single fissile nucleus. The problem that they faced was how to obtain energy from a vast number of fissile nuclei. The nuclear fission bomb that was dropped with such devastating effect over Hiroshima in 1945 exploded as a result of an uncontrolled chain reaction in its uranium-235 fuel. When uranium-235 undergoes fission, it releases two or three neutrons each time. Each of these neutrons is then able to cause fission in another uranium-235 nucleus, which in turn will also release two or three neutrons. Within a very short time, the number of released neutrons and fission reactions has escalated in a process known as a chain reaction. In Figure 12.16, two neutrons are released during the fission reaction. The number of nuclei undergoing fission doubles each generation and within a small fraction of a second an enormous number of nuclei have undergone fission. Only a miniscule amount of energy (of the order of 10-13 J) is released by each fission reaction, but in this uncontrolled chain reaction there are so many reactions occurring in such a short time that an explosion results. In 1 kg of uranium-235, so many reactions occur that about 8 × 1013 J of energy is released in just over one-millionth of a second!

Nuclear fuel In the 4.5 billion years since the Earth was formed, the radioactive isotopes that are naturally present in the Earth itself, its oceans and the atmosphere have been decaying to form more stable isotopes. By far the two most common isotopes of uranium present in the Earth’s crust are uranium-238 and uranium-235. These have half-lives of 4.5 billion years and 710 million years respectively, and so uranium-235 has been decaying at a faster rate than uranium-238. This means that far less uranium-235 remains in the Earth’s crust so that today the uranium that is mined from the ground consists of: • 99.3% uranium-238—the non-fissile isotope • 0.7% uranium-235—the readily fissile isotope. This means that a chain reaction cannot occur in a sample of uranium taken from the ground because the proportion of fissile uranium-235 is too low. To be useful as a nuclear fuel, uranium ore has to be enriched. This involves increasing the proportion of uranium-235 relative to uranium-238, a very difficult and expensive process. The slightly different masses of the isotopes enables separation to be achieved. The three common enrichment methods are the ultracentrifuge, electromagnetic and gaseous diffusion separation techniques. Nuclear weapons require fissile material that has been enriched to over 90% purity. The bomb that was dropped over Hiroshima contained 40 kg of 95% purity uranium-235. Nuclear reactors require fissile material enriched to about 3% uranium-235.

Critical mass In their efforts to produce an explosion, the scientists working on the Manhattan Project during World War II had to establish a nuclear chain reaction in a sample of nuclear fuel. They found that the explosive ability of a sample of fissile material depended on its purity and size. In a sample of nuclear material in which the concentration of uranium-235 or plutonium-239 is too low, a chain reaction cannot be established. This is because the neutrons have only a small chance of being absorbed by a fissile nuclei and causing a further fission reaction. The chain reaction will die out. The fuel used in nuclear fission weapons is enriched to a high degree of purity so that a chain reaction can be sustained. The shape of the nuclear fuel is an important factor in its explosive ability. A 20 kg sample of enriched uranium-235 in the shape of a sphere will spontaneously explode, whereas 20 kg of enriched uranium-235 flattened into a sheet will not. The flat piece has a very large surface area and so an enormous number of neutrons are able to escape from the uranium into the air. These neutrons do not cause further fission reactions and so the chain reaction will die out. In the spherical piece of uranium, the surface area is much smaller and a greater proportion of neutrons remain in the uranium to sustain the chain reaction. (a)

(b)

20 kg 235U: subcritical

20 kg 235U: supercritical!

Figure 12.17 (a) The large surface area of the flat piece of uranium-235 enables a large proportion of neutrons to escape into the air, causing the chain reaction to die out. (b) In the spherical piece of uranium-235, a sufficient proportion of neutrons remains inside the material to maintain the chain reaction, leading to an explosion.


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The explosive ability of a fissile material also depends on its physical size. For example, a piece of uranium-235 the size of a marble will not explode but a piece the size of a grapefruit most definitely will. The small piece has more surface area compared with its volume than the large piece. In the marble-sized lump, a greater proportion of neutrons escape into the air and so the chain reaction dies out. This is a subcritical mass. In the sample the size of a grapefruit, a higher proportion of neutrons is available to continue the chain reaction within the material. This is a supercritical piece, capable of causing a nuclear explosion.

The minimum amount of enriched fissile material in the shape of a sphere that leads to a sustained chain reaction is known as the CRITICAL MASS.

(a) Subcritical

neutron escapes

incoming neutron

Figure 12.18 In each of these fission reactions, two neutrons are released. (a) When a high proportion of these neutrons escape or are absorbed by atoms other than 235U, the chain reaction is not sustained. This is a subcritical situation. (b) When there are just enough neutrons available to keep the chain reaction going at a steady rate, the situation is described as critical. This is what should happen in a nuclear reactor. (c) When a high proportion of released neutrons are available to continue the fission process so that the chain reaction grows, it is described as supercritical. Nuclear weapons make use of supercritical reactions.

Uranium-235 atom neutron escapes

Neutron

neutron absorbed

(b) Critical

neutron escapes

incoming neutron

neutron absorbed neutron absorbed

neutron absorbed

neutron escapes neutron escapes

(c) Supercritical incoming neutron

Physics in action

Nuclear weapons The design of a fission bomb is really quite simple. It has to contain separate subcritical masses of a fissile material, which are then combined at the desired time to make one supercritical mass when the explosion is required. The nuclear bomb that was dropped on Hiroshima contained two hemispherical subcritical pieces of 95% pure uranium-235. This bomb, known as Little Boy, was dropped by a B29 bomber called the Enola Gay on 6 August 1945. When it reached an altitude of 580 m, an explosive charge fired one piece into the other creating one supercritical mass of uranium-235, which exploded within one-millionth of a second after this.

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Three days after the bombing of Hiroshima, a second bomb was dropped over Japan. The target city was originally Kokura, but low cloud and fog resulted in the attack being changed to Nagasaki. This bomb, nicknamed Fat Man, contained many small subcritical pieces of plutonium-239. It used a spherical implosion to force these pieces together and make one supercritical mass. As with Little Boy, this led to an uncontrolled fission reaction and consequent nuclear explosion that killed or maimed almost 100 000 people. Twelve hours after the explosion, pilots could see Nagasaki burning from over 300 km away.

(a)

(b)

(a)

explosive propellant

subcritical masses of uranium-235

chemical explosive charges

(b)

beryllium shell

before firing separator

propellant detonates produces a supercritical mass

assembly of subcritical masses of plutonium-239 and separator material

before firing

compressed assembly becomes supercritical

implosion

an atomic explosion

Figure 12.19 (a) A replica of Little Boy, the uranium-235 bomb that was dropped on Hiroshima on 6 August 1945. (b) A simple fission bomb contains two subcritical pieces of uranium-235. The two pieces are kept separate until they are forced together, creating a supercritical mass and leading to an explosion within one-millionth of a second.

an atomic explosion

Figure 12.20 (a) A replica of Fat Man, the plutonium bomb that was dropped on Nagasaki on 9 August 1945. (b) Fat Man was a spherical fission bomb. In this type of bomb, an implosion forces a large number of subcritical pieces of plutonium-239 into one supercritical mass, producing a nuclear explosion. The beryllium shell reduces the loss of neutrons from the core.


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12.2 summary Aspects of fission • Uranium-235 is readily fissile when it is struck by a slow neutron. Plutonium-239 is fissile when struck by a fast-moving neutron. • Uranium-238 is effectively non-fissile. It is more likely to absorb stray neutrons and become plutonium-239, itself a fissile material, by radioactive decay. For this reason, uranium-238 is known as a fertile material.

• If a fission reaction that releases at least one neutron per fission is taking place, a chain reaction may be established. • The critical mass is the least amount of material that will sustain a chain reaction. The critical mass for a material depends on its concentration, shape and size.

12.2 questions Aspects of fission 1 Which one of the following isotopes is most suitable as a fuel for a nuclear reactor? A 238U B 235U C 234U D 236U 2 The uranium ore that is dug from the ground contains two different isotopes: 235U and 238U. Explain why uranium ore is not immediately suitable as a fuel for a nuclear reactor. 3 The uranium that is used as the fuel for a nuclear reactor has been enriched so that its uranium-235 content is around: A 0.7%. B 3%. C 10%. D 95%. 4 Uranium-238 is not really fissile, yet is known as a fertile material. What does this mean? 5 Plutonium-239 is fissile when struck by a fastmoving neutron, but non-fissile when struck by a slow-moving neutron. Use the liquid drop model to explain why this is so.

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6 Uranium-235 releases an average of about 2.5 neutrons during each fission, whereas plutonium-239 releases an average of about 2.7 neutrons. Which isotope will have the lower critical mass? 7 Discuss how the chain reaction that releases the energy in a nuclear fission explosion is established. Explain why the chain reaction grows rather than dies out. 8 The critical mass of uranium-235 is about 1 kg. Explain then why it is that a 5 kg piece of 235U that is flattened like a sheet is not capable of exploding. 9 Thorium-232 does not undergo nuclear fission, but absorbs a bombarding neutron into its nucleus and ends up transforming into uranium-233. U-233 is capable of capturing a neutron and undergoing nuclear fission. Circle the correct answers: a Th-232 is fissile/fertile/infertile. b U-233 is fissile/fertile/infantile. 10 Used nuclear fuel rods from Australia’s nuclear reactor at Lucas Heights were shipped to the UK for reprocessing. These uranium fuel rods were considered to be at no risk of exploding like a nuclear weapon. What does this suggest about the proportion of fissile uranium in the fuel rods?

12.3 Nuclear fission re

actors

Since the 1950s, it has been possible to control nuclear fission within a nuclear reactor for the purpose of producing electrical power for domestic use. Australia does not produce any electricity in this way, but in over 30 countries around the world there are about 400 nuclear power plants in operation. About 230 of these are pressurised water reactors (PWR). Many more reactors have been constructed for medical, military and research purposes. A nuclear power plant will produce electricity in much the same way as a coal-burning power plant. The primary difference is how the heat is produced. The power stations in the La Trobe Valley generate electricity by burning coal to produce heat that creates the steam which is used to turn the generator turbines. A nuclear power station simply has a different way of producing heat—nuclear fission.

Table 12.1 Percentage of electricity produced by nuclear power plants in 2007 Country

Percentage of electricity produced by nuclear power plants

France

78

South Korea

39

Japan

30

USA

19

UK

18

Canada

16

China

2

Australia

0

Physics file Australia’s only nuclear reactor is located at Lucas Heights, a suburb of Sydney. The HIFAR (High Flux Australian Reactor) operated here from 1958 to 2007, but has now been replaced by a new research reactor known as OPAL (Open Pool Australian Lightwater reactor). About 7 kg of uranium enriched to 20% U‑235 is immersed in a pool of water almost 13 metres deep. Radioisotopes that are used in industry and in the nuclear medicine departments of Australia’s hospitals are created here. Another important application is the irradiation of silicon chips, creating high conductivity silicon for the computer industry.

Thermal nuclear reactors In their endeavours to harness the energy from the nuclear fission reactions, designers of nuclear reactors had to overcome three major difficulties. First, the neutrons released from uranium-235 when it undergoes fission are travelling at very high speeds—about 20 000 km s-1. Uranium-235 is most fissile when irradiated by slow-moving neutrons. Thus, these emitted neutrons needed to be slowed down. Second, the fission of each uranium-235 nucleus releases an average of 2.47 neutrons. This can lead to a chain reaction that results in an explosion. A way had to be found to absorb some of these emitted neutrons and maintain a steady chain reaction. Third, the heat generated in the reactor by the fission process had to be somehow collected and used to create steam to drive the turbine and generate electricity. A thermal nuclear reactor generates energy through the fission of uranium-235, an isotope that is most likely to undergo fission when it is hit by slow-moving, or thermal, neutrons. There are many different varieties of thermal nuclear reactors, but they all include the following design elements: • fuel rods—long, thin rods containing pellets of enriched uranium • a moderator—a material that slows the neutrons


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• control rods—a material that absorbs neutrons • a coolant—a liquid or gas to absorb heat energy that has been produced by nuclear fission • a radiation shield—a thick concrete wall that prevents neutrons escaping from the reactor.

Nuclear fuel rods

Figure 12.21 Loading a nuclear fuel rod with uranium pellets. The pellets are in the foreground.

Uranium-235 is used as nuclear fuel since it is readily fissile with slowmoving neutrons. However, this isotope comprises only 0.7% of naturally occurring uranium, not enough to sustain a chain reaction. The predominant (99.3%) isotope, uranium-238, is effectively non-fissile and has the property of capturing slow-moving neutrons. Therefore, the proportion of uranium-235 in the ore has to be increased. That is, the uranium ore has to be enriched before it can be used as reactor-grade fuel. Uranium enriched for use in a thermal nuclear reactor contains about 97% uranium-238 and 3% uranium-235. The proportion of uranium-235 is increased from 0.7% to about 3%, i.e. the uranium is enriched by a factor of about four. By comparison, the fuel for a nuclear weapon is enriched to at least 90% uranium-235—a far more expensive and technically difficult procedure. The fuel used in a nuclear reactor is thus not suitable for use in nuclear weapons. The enriched uranium, in pellet form, can then be packed into a thin aluminium tube, known as a fuel rod. This is usually 3–5 m long. A large nuclear reactor has thousands of fuel rods in its core. A fuel rod will eventually become depleted in uranium-235. This means that over time the concentration of 235U falls to a level where it is not able to sustain the fission chain reaction. Each fuel rod needs to be replaced every 4 years or so, and a typical 1000 MW reactor produces around 30 tonnes of spent fuel each year. The disposal and reprocessing of this waste is discussed more fully at the end of this section.

The moderator In a nuclear reactor, the problem of fast-moving neutrons is overcome by including a material that slows down, or moderates, the speed of the free neutrons. It has been found that substances whose nuclei are small will slow the neutrons down to speeds at which neutron capture will occur. When the emitted neutrons collide with these small nuclei, the neutrons lose most of their kinetic energy and so slow down. After many collisions, the neutrons have been slowed down to about 2 km s-1 and have less than 1 eV of energy. Some materials that are commonly used as moderators are: • normal water—H2O • heavy water—containing deuterium (21H), an isotope of hydrogen • carbon dioxide—CO2 • graphite—consisting of carbon atoms. Each of these materials works well as a moderator because not only does it slow the neutrons, but it does so without absorbing a significant number of them. Heavy water is the most effective moderator, but is also the most expensive. Water is obviously the cheapest material, but absorbs more

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neutrons than the others and so reduces the extent of the chain reaction. Nevertheless, water is the moderator that is used in the most common type of reactor in use today—the pressurised water reactor (PWR). Graphite is less effective than water because carbon nuclei are heavier than hydrogen nuclei. In losing their energy, the emitted neutrons have to collide with about 120 carbon nuclei, but only about 25 water molecules.

Control rods A nuclear fission bomb can produce vast quantities of energy. The number of neutrons released during the uncontrolled fission chain reaction grows exponentially, releasing enormous amounts of energy in a split second. A nuclear reactor can also produce great amounts of energy. However, a steady controlled energy release is required. This is achieved by controlling the number of neutrons that are involved in the fission chain reaction of uranium-235. This task is performed by the control rods. A control rod contains material that has the ability to absorb neutrons. Cadmium and boron steel are commonly used in control rods. When a neutron strikes the nucleus of either of these atoms, it is absorbed into the nucleus and so takes no further part in the chain reaction.

Physics file Most modern pressurised water reactors do not have individual fuel rods and control rods in their core. They use fuel assemblies. Each fuel assembly consists of a tall, slender array (often 14 × 14) of fuel rods and about 20 control rods. A typical PWR may have around 20 of these fuel assemblies immersed in the water in its core. The outdated reactor that blew up at Chernobyl in the then Soviet Union had a graphite core with about 1600 fuel rods and more than 200 control rods.

Putting it all together In pressurised water reactors, the core of a thermal nuclear reactor consists of the moderating material with fuel rods and control rods placed in it. These rods are immersed into a volume of water. When a neutron is released during the chain reaction, it is slowed by the moderating material. This enables it to be absorbed by a further uranium-235 nucleus, induce fission, and so continue the chain reaction. The rate of the chain reaction is controlled by raising or lowering the control rods. If the operators wish to reduce the energy output of the reactor, or even shut it down completely, they will lower the control rods further into the core. This has the effect of absorbing more neutrons and reducing or stopping the chain reaction. The fission reaction in the reactor core produces an enormous amount of heat energy. The core of a reactor is typically at temperatures of 500–1500°C. Heat energy is removed from the core by pipes that contain a coolant. Water is usually used as a coolant, although advanced gas-cooled reactors use carbon dioxide gas and some fast breeder reactors use liquid sodium. Reactors are designed with a back-up coolant system should the primary coolant system fail. This greatly improves the safety of the reactor by stopping the coolant from overheating and therefore preventing meltdown of the core from occurring. Another safety aspect of modern pressurised water reactors is that the moderator (water) also acts as the coolant, and so if there is a loss of coolant from the core, the nuclear reaction will also stop. The heat energy that the coolant extracts from the core is used to generate electricity. A heat exchanger first transfers this energy into pipes containing water. The water is converted into steam which is then used to rotate the turbines that drive the generator.


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(a) containment building

(b)

pressuriser (prevents water boiling)

primary coolant circuit (high pressure water)

Nuclear energy of fuel

boiler

pump

water

condenser

turbines and electricity generator to grid

Heat energy of coolant

Kinetic energy of steam and turbine

control rod

sea water to condense steam

fuel rod

water acts as coolant and moderator in reactor core

Electric energy

Light, heat, sound, etc.

Figure 12.22 (a) A schematic diagram of a pressurised water reactor. The heat is removed from the core of the reactor by the coolant, water. The coolant then heats the water in the boiler (heat exchanger) turning it into steam that turns the turbine that drives the generator. In this way, electricity is generated. The primary difference between this and a coal-fired generator is in the way the heat is produced—a nuclear reactor uses the fission process and a coal generator burns coal. A typical 1000 MW power plant consumes about 6 000 000 tonnes of black coal each year, or about 25 tonnes of enriched uranium that has been obtained from around 75 000 tonnes of ore. (b) The energy stored in the nuclear fuel undergoes a number of transformations.

The core of the reactor is encased in a protective radiation shield about 2 m thick. This consists of layers of concrete, steel, graphite and lead. The function of this shield is to prevent neutrons and gamma rays from escaping from the reactor core, and so protect the workers at the plant from damaging radiation. The layers of graphite in the shielding act to reflect escaping neutrons back into the core to take part in the chain reaction. The workers at a nuclear power plant are continually monitored to ensure that they are not exposed to unacceptably high levels of radiation. However, their allowed dose would still be much higher than that of the general population.

Fast breeder reactors

control rods

coolant

neutron shield plutonium fuel uranium blanket

coolant

Figure 12.23 The core of a fast breeder reactor does not include a moderator. The plutonium in the fuel rods is surrounded by uranium. It is here that new plutonium is bred.

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There are over 400 nuclear reactors operating in the world at present. A handful of these are fast breeder reactors. When they were first designed, it was thought that fast breeder reactors would be the solution to the Earth’s energy problems. They make use of the most abundant isotope of uranium, uranium-238, to breed another fissile element, plutonium-239, for use in further nuclear reactions. However, they have been beset by safety and technical problems and have proved very expensive to operate. The political climate and concerns about terrorist groups obtaining the plutonium and producing nuclear weapons have also been major factors in governments deciding not to construct this type of reactor. Uranium-238 is only slightly fissile and comprises over 99% of naturally occurring uranium. It is not suitable for use as a fuel in a thermal nuclear reactor because it is only likely to split when it captures a high-energy neutron. There is only a low probability of this happening. However, despite its non-fissile properties, uranium-238 can play an important role in the production of nuclear energy. This is due to its ability to absorb neutrons. In a thermal nuclear reactor, there are many slow-moving neutrons

flying around. About 40% of these are captured by uranium-235 nuclei and induce fission, but 35% are captured by the non-fissile uranium-238. The remainder of the thermal neutrons are absorbed by other materials or escape from the core. The neutrons that are captured by uranium-238 nuclei may cause the uranium to change into a highly unstable isotope, uranium-239. This decays almost immediately by beta emission to form neptunium-239, which has a half-life of 2.35 days and decays by beta emission to form the artificial element plutonium-239. The decay series for this process is: 1 n + 238 U → 239 U t1/2 (239U) = 23.5 minutes 0 92 92 239 U → 239 Np + –10e t1/2 (239Np) = 2.35 days 92 93 239 239 Np → 94Pu + –10e t1/2 (239Pu) = 24 400 years 93 In other words, plutonium is a by-product of nuclear reactors; but it is a by-product that has great significance. Plutonium is a fissile material but only when it captures fast-moving neutrons. The slow-moving neutrons in most nuclear reactors do not induce fission of the plutonium nuclei that have formed there. Plutonium is extracted from the spent fuel rods of these nuclear reactors and used as the fissile material in a different type of reactor—a fast breeder reactor—which uses fast neutrons and actually breeds more plutonium than it uses! A fast breeder reactor has many similarities to a standard nuclear reactor. It has control rods to control the rate of the fission reaction. A coolant flows through the core of the reactor, removing the heat energy. This energy is then used to produce the steam that drives the turbines to generate electricity. However, a fast breeder reactor does not require a moderator since fast neutrons are needed to trigger fission in the plutonium. The fuel rods in a fast breeder reactor contain a core of plutonium and it is in this core that the fission process takes place. Fast neutrons released during fission are captured by plutonium nuclei to produce further fission. A great deal of heat energy is generated by this chain reaction. The breeding of new plutonium occurs in the following way. The plutonium core of the fuel rods is surrounded by a ‘blanket’ of uranium-238. Many neutrons from the chain reaction escape into this uranium-238 and are absorbed by these nuclei, leading to the formation of plutonium-239. Thus, as the plutonium in the core of the fuel rod is used up as it undergoes fission, more plutonium is bred in the surrounding uranium-238. This plutonium is eventually extracted for use in the next generation of fuel rods.

Disposing of nuclear waste—the nuclear fuel cycle A major problem facing the nuclear power industry is the disposal of the unstable radioactive waste. There are about 400 nuclear reactors generating electricity around the world, producing large quantities of radioactive waste. A typical 1000 MW reactor will produce about 25 tonnes of spent fuel rods annually. The safe disposal of this waste is a serious concern to many around the world. However, it needs to be remembered that the coal-burning power stations that we use in Australia also have a significant detrimental effect on the environment. A 1000 MW coal-fired power station

Physics file Plutonium is not a naturally occurring element in the Earth’s crust. Any plutonium that exists has been created in a nuclear reactor. Since there are only a few fast breeder reactors in operation, most of the plutonium that is produced by thermal nuclear reactors is not re-used as nuclear fuel, but is stored as nuclear waste. Plutonium is an alpha emitter and so it is not difficult to protect against its emissions. However, it has a half-life of more than 24 000 years. Furthermore, plutonium is an extremely toxic substance. This means that it must be safely stored for many thousands of years. At present, the plutonium is kept underground in bunkers and salt mines. However, many people are concerned that this is not an entirely satisfactory solution. Technically, it is relatively easy to use plutonium as fuel in a nuclear weapon. There is great concern that a terrorist group could steal a quantity of plutonium from a reactor, or while it is being transported for disposal, and use it to manufacture an atomic bomb. Countries that use plutonium are required by the Atomic Energy Commission to account for every gram of the material, yet despite every safeguard, quantities of plutonium have gone missing.


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Figure 12.24 Containers of low-level radioactive waste, being permanently housed in an underground storage facility in the United States. Many people are concerned about the long-term use of this practice.

Figure 12.25 Forty per cent of the world’s spent fuel rods are permanently stored in cooling ponds.

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produces millions of tonnes of carbon dioxide and sulfur dioxide annually, gases that are major contributors to the greenhouse effect. Radioactive waste products are classified into the following three categories: low level, intermediate level and high level. Low-level waste is generated primarily from hospitals, industry and laboratories and consists mostly of tools, clothing, used wrapping material and other items that have been contaminated with short-lived radionuclides. These materials make up 90% of the volume but only 1% of the radioactivity of all radioactive waste. Low-level waste solids are usually compacted or incinerated, then buried in shallow pits on land or at sea. Low-level liquid and gas waste is usually released into the environment. Intermediate-level waste typically consists of reactor components, chemical sludges, and contaminated materials from reactors that have been decommissioned. Intermediate-level wastes are solidified in bitumen or concrete, then buried or stored in deep trenches. High-level waste is waste from contaminated reactor parts, as well as liquid waste from the reprocessing of spent fuel rods. This waste contains highly radioactive fission fragments and transuranic elements, and so requires special shielding during handling and transport. High-level waste holds 95% of the radioactivity but only 3% of the volume of all radioactive waste. It is the disposal of high-level waste that is of major concern to governments around the world. Initially, spent rods from a nuclear reactor must be stored in water for several years while the level of radioactivity decreases. Their high level of radioactivity generates heat that must be removed in a cooling pond. In the United States and Canada, spent fuel rods are regarded as waste and they are stored permanently in cooling ponds. This is known as an open fuel cycle. The problem with this system is that large amounts of waste are produced. Only about 1% of the energy content of the uranium is actually converted to electricity in this open fuel system.

France, Japan and the UK have adopted a closed fuel cycle in which the spent fuel is reprocessed. This system doubles the amount of energy that is recovered from the fuel and removes most of the long-term radioactive isotopes from the waste. During reprocessing, the spent fuel is first dissolved, then chemically separated into uranium, plutonium and radioactive fission fragments. The uranium has been depleted to 0.8% uranium-235 and so must be sent to an enrichment plant before it can be re-used as fuel. The plutonium is also removed and used in new fuel rods, stored as waste, or used as fuel in a fast breeder reactor. This leaves about 700 kg of liquid high-level waste from the 25 tonnes generated annually. These liquid wastes are currently stored in multiple-walled stainless steel tanks inside reinforced concrete chambers. This is recognised as being a potentially hazardous arrangement and so much research has been devoted to developing a solidification process. This involves the liquid being evaporated, mixed with glass-forming materials, melted and then poured into stainless steel drums. This solidification method has been used in Europe for over 25 years. reactor

spent fuel

fuel

temporary storage

fuel fabrication recycling enrichment

conversion to uranium hexafluoride

recovered uranium spent fuel

mixed oxide fuel fabrication plutonium dioxide

high-level waste and transuranic waste

uranium ore mines and mills uranium dioxide and plutonium dioxide ‘once-through’, or open, fuel cycle

closed fuel cycle

geological waste repository

Figure 12.26 The open fuel cycle in which the uranium is used once in a reactor, then disposed of as waste is shown in red. This system is used in the USA and Canada. The closed fuel cycle is shown in blue. In this system, spent fuel rods are reprocessed and the retrieved uranium and plutonium are re-used as nuclear fuel. This approach of recycling uranium is used in France, UK, Russia, Japan and Germany.


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Physics in action

Chernobyl In the early morning of 25 April 1986, electrical engineers at Reactor 4, Chernobyl, in the then Soviet Union, began conducting a series of tests. They were trying to determine whether the generator would, as it was running down, continue to power the emergency systems in the plant. Over the next 24 hours, in performing the tests, they violated six different safety procedures. At one stage during the tests, the control rods were almost completely withdrawn from the core of the reactor. At 1:23 a.m. on 26 April the power output began to rise rapidly, so the engineers decided to fully insert the control rods and shut down the reactor. However, each control rod was designed with 5 m of graphite on its end. This acted as an additional moderator and sped up the reaction even more. Within 4 s, the power had surged to 100 times the reactor’s capacity, and an enormous steam explosion occurred. This lifted the concrete and steel dome many metres into the air and started a fire that burned out of control for 5 days. More than 50 tonnes of radioactive fallout was carried by the northerly winds across Ukraine, Belarus, Scandinavia and Europe. This was almost 100 times that of the waste from the bombs dropped at Hiroshima and Nagasaki in 1945. The Chernobyl reactor stands in what is now Ukraine, close to the border of Belarus. Vast regions within these countries now stand empty. Approximately one-quarter of Belarus is considered to be uninhabitable. Thousands of villages lie abandoned, livestock has been slaughtered and schools have been closed. The immediate toll from the explosion was 31 deaths. This included firefighters and workers at the power plant. Far more serious has been the toll from the radioactive fallout. The incidence of cancer in the region has risen dramatically. For example, in the 1980s there were just seven reported cases of thyroid cancer among children in Belarus. From 1990 to 1996, this increased to more than 300. Not many of these have been fatal, but it does indicate that the problems will be long term, and projections vary from 4000 to 75 000 deaths. There is currently a 30 km exclusion zone around the plant; however, this did not apply to workers who were still operating the other reactors at the power plant. Reactor 2 was shut down after a fire in 1991, Reactor 1 was closed in 1997 and Reactor 3 was closed in 2000 due to international pressure related to its safety.

Norway Scotland

Oslo

Helsinki

Ireland England Atlantic Ocean

London

Moscow

Sweden

North Sea

Latvia

Russia

Minsk Belarus

Germany Berlin

Poland

Czech. Paris Slovakia Austria France Switzerland Hungary Italy Spain

Ukraine Chernobyl

Romania Bulgaria

Rome Greece

Figure 12.27 This map shows how the radioactive fallout from the nuclear accident at Chernobyl spread across Europe. At present, the remains of Reactor 4 are buried under 5000 tonnes of material designed to absorb neutrons, act as a radiation shield and prevent the graphite core from burning. This is further encased in a huge concrete shell, or sarcophagus. However, the shell is already leaking radiation, so the original sarcophagus is to be entombed within a new shelter. This involves building a 12 m thick concrete arch more than 100 m high over reactors 3 and 4 and is known as the Chernobyl SIP (Shelter Implementation Plan). The contract for the construction of this massive engineering project was signed in August 2007 and construction is scheduled for completion in 2012. When construction is completed, the deconstruction of the Reactor 4 site will continue. Four large cranes will be used to dismantle the old structure and remove the radioactive waste materials, including 180 tonnes of nuclear fuel, that would then be treated and stored on site. It is hoped that the new shelter will last around 100 years.

New shelter 12 m 105 m

Existing sarcophagus shelter 259 m 1

sarcophagus

new shelter

2

foundation The new shelter will be built alongside the old sarcophagus

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Detailed studies

Once completed, it will be moved on teflon bearings to enclose the reactor site

Figure 12.28 In what could be the world’s most expensive environmental clean-up operation, an international consortium plans to entomb the stricken Chernobyl reactor in a giant containment shelter.

12.3 summary Nuclear fission reactors • A nuclear reactor uses enriched uranium as its fuel. The fuel rods in a nuclear reactor contain uranium that has been enriched to about 3% uranium-235. • The core of the reactor consists of a material (e.g. graphite, water) that acts to slow the neutrons that are emitted during fission. This material is called a moderator. These slowed neutrons are then able to induce fission in the uranium-235 nuclei. • The rate of the nuclear reaction in the reactor core is determined by the control rods. These consist of a material (e.g. cadmium, boron steel) that absorbs neutrons. The control rods are raised and lowered to control the chain reaction and so produce a steady release of energy. • The coolant is a fluid that flows through the reactor core. It extracts heat energy from the core. This energy is then used to produce steam that drives turbines to produce electricity.

• A fast breeder reactor uses plutonium as its fuel. Plutonium-239 is fissile when struck by fast-moving neutrons. A fast breeder reactor does not need a moderator. • Uranium-238 is also placed in the core of a fast breeder reactor. It absorbs neutrons and transmutes into plutonium-239 which can then be used as fuel for a fast breeder reactor. • Large quantities of nuclear waste are produced by the nuclear industry each year. This is to be weighed up against the millions of tonnes of CO2 and other greenhouse gases that are released into the atmosphere each year by coal-fired power stations. • Low-level radioactive waste is generally burnt or buried in pits, intermediate-level waste is buried in deep trenches, and high-level waste is either permanently stored in ponds of water or reprocessed before being stored in reinforced stainless steel drums.

12.3 questions Nuclear fission reactors 1 The moderator in a nuclear reactor has the ability to: A absorb neutrons. B slow neutrons. C release energy. D remove heat energy. 2 The function of the control rods in a nuclear reactor is to: A reduce the energy of the neutrons. B make the neutrons more likely to cause fission. C prevent the reactor core from overheating. D absorb neutrons and maintain a controlled chain reaction. 3 a Outline the process by which a nuclear power plant produces electricity. b Discuss the primary difference between how electricity is produced in a nuclear power station and in a coal-burning power station. c What aspects of the production of electricity do coal-fired and nuclear power stations have in common? 4 Explain why lead (Z = 82) would be unsuitable for use as a moderator.

5 Describe the effect on the operation of a nuclear reactor if the number of neutrons per fission that can continue the chain reaction is: a equal to one b less than one c greater than one. 6 The fissile material that is used in a nuclear reactor is uranium-235. What is the effect on the nuclei of this isotope when bombarded with: a fast neutrons? b slow-moving neutrons? 7 Approximately 97% of the uranium in the fuel rods of nuclear reactors is uranium-238. When struck by a neutron, a uranium-238 nucleus is most likely to absorb the neutron. a In what way does this change the uranium nucleus? b Why does this lead to problems in disposing of the nuclear waste from the reactor? 8 A fast breeder reactor is a simpler design than a thermal nuclear reactor. a What fuel is used in fast breeder reactors? b Why is it called a ‘fast’ reactor? c Why is it called a ‘breeder’ reactor?


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d In what way is the design of a fast breeder reactor different from that of a thermal nuclear reactor? 9 During the fission of plutonium-239, the average number of neutrons released is 2.91 per fission. This is higher than the average of 2.47 released during the fission of uranium-235. How is this of benefit in a fast breeder reactor? 10 During the lifetime of a reactor, the control rods need to be gradually removed over a period of months in order to maintain the energy production at a constant rate. Explain why this procedure is necessary. 11 The activity of 1 tonne of high-level nuclear waste is shown in the graph. The activity of the waste is the sum of the fission products and transuranic elements that it contains. By way of comparison, the activity of 1 tonne of uranium ore is 8 × 1011 Bq. 1016

total activity

1015 Activity (Bq)

fission products

1014

transuranic elements

1013 1012 1011

10

462

100 1000 10 000 100 000 Years after reprocessing

Detailed studies

106

107

a Approximately how long does it take for the high-level nuclear waste to return to the level of radioactivity of an equal quantity of uranium ore? b The activity decreases rapidly during the first thousand years. What is the reason for this? 12 Two different approaches have been adopted in the treatment of used fuel rods—the open and closed fuel cycles. a What happens to uranium from spent fuel rods in the open fuel cycle? b What happens to uranium from spent fuel rods in the closed fuel cycle? c What are the benefits of the closed fuel cycle?

12.4 Nuclear fusion (a)

Kr

91

three neutrons

1 0

n U

235

Ba

142

(b)

235U

fission fragments : energy released

Figure 12.29 (a) Each fission of a uranium-235 nucleus releases about 200 MeV of energy. (b) The fission fragments have a lower combined mass than the uranium nucleus. The missing mass is related to the 200 MeV of energy by the equation … = mc2.

In our study of nuclear fission, we saw how Einstein’s idea of mass–energy equivalence was used to explain why energy was released during the splitting of a nucleus. When fission occurs, the mass per nucleon of the particles in the fission fragments is lower than that of the parent nucleus. Einstein’s famous equation, E = mc2, can be used to calculate the amount of energy that is related to this missing mass. Recall that m in this equation is called the mass defect and represents the difference in mass of the nucleons in the parent and daughter nuclei.

Binding energy In order to change water into vapour, it is necessary to add lots of energy to the water. This energy acts to break the bonds between the water molecules so that they are able to move around freely. This added energy is a kind of ‘binding energy’. The free molecules of water vapour have more energy than the liquid water molecules and so, as described by Einstein, have slightly more mass. In order to change a helium nucleus into individual protons and neutrons, you would also need to add an amount of energy. This energy would act to separate the particles, enabling them to move around freely, and is called the binding energy of the nucleus. The free particles have

(a) He

+

energy

(b)

2p, 2n

He: energy released

Figure 12.30 (a) When two isotopes of hydrogen (deuterium) are fused together to form a helium nucleus, energy is released. (b) The binding energy of the nucleus appears as a loss in mass, which can be calculated using … = mc2.


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more energy and so, according to Einstein, have greater mass. Conversely, it also means that if you could force the individual nucleons together to form a helium nucleus, you would lose some mass. This lost mass appears as energy and can be calculated from E = mc2. This is the principle behind nuclear fusion. Fusion is taking place at an awesome rate in the stars around the universe.

(a)

(b)

fusion!

Figure 12.31 (a) Slow-moving nuclei do not have enough energy to fuse together. The electrostatic forces cause them to be repelled from each other. (b) If the nuclei have plenty of kinetic energy, they will overcome the repulsive forces and move close enough together for the strong nuclear force to come into effect. At this point, fusion will occur and energy will be released.

Potential energy

energy barrier electrostatic force dominant

strong nuclear force dominant

Distance

Figure 12.32 If hydrogen (deuterium) nuclei are to get close enough for the strong nuclear force to act, they must overcome the energy barrier presented by the electrostatic force. Temperatures of around 400 000 000 K are required to provide this amount of energy!

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Detailed studies

NUCL…AR FUSION occurs when two light nuclei are combined to form a heavier nucleus. This is the process by which energy is produced in stars. Even though the number of protons and neutrons is the same both before and after fusion, the mass per nucleon is lower after fusion has occurred. The binding energy of the nucleus corresponds to this missing mass and can be calculated by using E = mc2. It would seem as though obtaining energy from nuclear fusion should be quite straightforward. Unfortunately this is not the case. The main problem with nuclear fusion is that nuclei are positively charged. They exert an electrostatic force of repulsion on each other and so it is not easy to force the nuclei together. Remember that the electrostatic force is a long-range force and the strong nuclear force of attraction acts at much shorter distances. As two nuclei approach each other, the electrostatic force will cause them to be repelled. For slow-moving nuclei with relatively small amounts of kinetic energy, this means that they will not be able to get close enough for the strong force to come into effect and so fusion will not happen. If the nuclei are travelling towards each other fast enough, however, they may have enough kinetic energy to overcome the repulsive force and get close enough for the strong nuclear force to start acting. If this should happen, fusion will occur. The graph in Figure 12.32 shows the effect of the electrostatic force and the strong nuclear force on the potential energy of a pair of deuterium (2H) nuclei. At large separations, the electrostatic force dominates and the nuclei repel each other. At small separations, the strong nuclear force dominates and the nuclei fuse together. To get the nuclei to this point, however, they need an enormous amount of energy. Temperatures of the order of hundreds of millions of degrees are required! This enormous amount of energy enables the nuclei to overcome the energy barrier or energy hill shown in Figure 12.32 and fuse together.

Fusion in the Sun and similar stars On the Sun, there are many different fusion reactions taking place. The main reaction is the fusion of hydrogen nuclei to form helium. Each second, about 657 million tonnes of hydrogen and hydrogen isotopes are fused together to form about 653 million tonnes of helium. The missing 4 million tonnes is related to the energy released by the equation E = mc2. A tiny proportion of this energy reaches Earth and sustains life as we know it. The sequence of fusion reactions shown in Figure 12.33 has been occurring on the Sun for the past 5 billion years and is expected to last for another 5 billion years or so. Hydrogen nuclei are fused together and after several steps a helium nucleus is formed. This process releases about 25 MeV of energy.

Our Sun is a second- or third-generation star and was formed from the remnants of other stars that exploded much earlier in the history of our galaxy. As this giant gas cloud contracted under the effect of its own gravity, the pressure and temperature at the core reached extreme values. This is what enabled nuclear fusion to begin about 5 billion years ago. When much of the hydrogen in the Sun has been converted into helium, the Sun will collapse in on itself, raising the temperature, and helium will then undergo fusion to form larger elements such as carbon and oxygen. Further collapses will follow and elements as large as iron will be formed. Elements larger than iron are only formed when stars very much larger than the Sun end their days in a spectacular explosion called a supernova!

(a)

neutrino positron 1 1H

2 1H

+ +10 e + 00 ν

(b)

gamma ray

1 1H

+ 21H

3 2 He

+ 32He

4 2 He

+ 00 γ

(c)

3 2 He

Figure 12.34 The gold atoms in your jewellery and the nickel atoms in the coins in your pocket were formed during stellar explosions. One such exploding star or supernova occurred in a neighbouring galaxy in 1987.

+ 11H

+ 11H + 11H

Figure 12.33 These are three of the main fusion reactions that are taking place on the Sun right now. (a) Two protons (hydrogen nuclei) fuse together forming a hydrogen isotope, deuterium. Note that one of the protons has decayed, forming a neutron, a neutrino, a positron and releasing energy. (b) During the fusion of a proton and deuterium (2H) into helium-3, the nucleons lose mass and energy is released. (c) The fusion of helium-3 nuclei results in the formation of a helium-4 nucleus and releases two protons and energy.

Nuclear fusion reactors Since the 1950s, a great deal of research has been devoted to recreating nuclear fusion in a laboratory. The attractiveness of nuclear fusion as a source of energy is that few radioactive by-products are created. However, major technical problems have been encountered in trying to initiate ongoing fusion reactions on Earth. Replicating the reactions that take place on the Sun is extraordinarily difficult. This is because extremely high densities, temperatures and pressures are required. Fusion researchers are instead using two isotopes of hydrogen, deuterium 12H and tritium 3 H as fuel. Deuterium can be extracted in vast quantities from lakes and 1 oceans, but tritium is radioactive with a half-life of 12.3 years and must be artificially produced. The nuclear reactions used in current fusion reactors are as follows: 2 H + 21H → 32He + 10n 1 2 H + 12H → 31H + 11H 1 2 H + 31H → 42He + 10n 1 The main obstacle to the development of a successful fusion reactor is related to the extremely high temperatures that must be achieved before fusion can commence. This has been the major difficulty facing researchers. Temperatures of over 100 million degrees are needed to trigger a selfsustaining fusion reaction and this must then be contained inside the

Physics file In both fission and fusion reactions, less than 0.1% of matter is related to energy production. The fission of each uranium235 nucleus releases about 200 MeV of energy. When hydrogen fuses to form a helium nucleus, about 25 MeV of energy is released. In 1952, a fusion reaction was used to power the world’s first hydrogen bomb. It had five times the destructive power of all the conventional bombs that were dropped during the whole of World War II.


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Physics file The temperature at the surface of the Sun is about 5500°C, hot enough to vaporise any known material. The Sun is much, much hotter at the core where temperatures reach around 15 million degrees. As hot as this is, however, it is still well below the hundreds of millions of degrees needed to sustain nuclear fusion! The fact that fusion occurs at these low temperatures cannot be explained by classical physics. Quantum mechanics must be used—treating the protons as waves rather than particles. The process by which relatively slowmoving protons can undergo fusion is called the tunnel effect and is beyond the scope of this course.

reactor. Current fusion reactors have achieved temperatures of about 100 million degrees, but only for very short periods. A commercial nuclear fusion reactor is perhaps four or five decades into the future. At present, research is being carried out into tokamaks— doughnut-shaped reactors that use magnetic fields to contain the plasma of fusion reactants away from the reactor walls. The Joint European Torus, or JET, is in England. Another, called the Mega Amp Spherical Tokamak, or MAST, is also being built in the UK. steam boiler

turbine and generator superconducting coils

power

deuterium tritium

tritium extraction

Physics file Nuclear fusion reactions have been achieved on Earth during the explosion of nuclear weapons. In hydrogen bombs, the high temperatures achieved by the explosion of fissile fuel were used to initiate the fusion reaction. In other words, an atomic bomb was used as the fuse for a hydrogen bomb.

Physics file In 1989 chemists at the University of Utah reported that they had performed an experiment in which nuclear fusion had taken place at room temperature. When other scientists attempted to repeat this experiment, they were mostly unsuccessful and so the notion of ‘cold fusion’ was widely discredited. There are, however, groups of scientists who have made some quite puzzling observations that no existing theory can explain. These scientists are still investigating cold fusion and trying to make sense of these observations.

Figure 12.36 The doughnut shape or torus of the JET fusion reactor can be clearly seen in this photograph. Temperatures of hundreds of millions of degrees have been achieved in the torus.

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Detailed studies

heat reactor vacumm lithium bearing exchanger core vessel blanket

helium exhaust

Figure 12.35 This design of a tokamak reactor shows the doughnut shape of the core where the plasma of fusion fuel and products should be situated. The fusion reaction is held in place by a powerful magnetic field and lithium and water are used to extract the heat energy from the core. It is intended that this energy could then be used to generate enormous amounts of electricity.

Another technique called inertial confinement is also being investigated. Here, a pellet of fusion fuel is zapped by powerful lasers, causing the fuel to implode and initiating the fusion reaction. Nuclear fusion may perhaps be the ultimate energy solution. The fuel for nuclear fusion, deuterium, can be readily obtained from seawater. Vast quantities of energy are released during the fusion process, yet a relatively small amount of radioactive waste—consisting of the reactor parts that have suffered neutron irradiation—is created. Of course, the easiest way to use fusion power is to collect solar energy!

12.4 summary Nuclear fusion • Nuclear fusion is the combining of light nuclei to form heavier nuclei. Extremely high temperatures are required for fusion to occur. • A small amount of mass is lost during a fusion reaction. This mass is related to the energy produced according to E = mc2. • Nuclei are positively charged and so repel each other due to electrostatic forces. Approaching nuclei must have enough speed to overcome the electrostatic forces and get close enough for the strong nuclear

force to take effect. The energy that is required is called the energy barrier. • Fusion occurs in the Sun and the other stars in the universe. Hydrogen nuclei are being fused together in several steps to form helium nuclei. This process is releasing energy on a massive scale. • Nuclear fusion reactors are being tested, but face enormous technical difficulties before the successful production of electricity can be established. This may take decades.

12.4 questions Nuclear fusion 1 Which one of the following is an example of nuclear fusion? A A plutonium nucleus splits into two smaller nuclei. B A boron nucleus absorbs a neutron. C A uranium nucleus emits a helium nucleus. D Two deuterium (21H) nuclei form a helium nucleus. 2 The fusion reaction that is the most promising for use in nuclear fusion reactors is: 12H + 31H → 42He + 10n Which one of the following best explains why energy is released during this fusion reaction? A Nucleons are created during this reaction. B Nucleons are lost during this reaction. C The nucleons lose mass during this reaction. D The nucleons gain mass during this reaction. 3 Where is deuterium obtained from? 4 Discuss the major difficulties that are facing the designers of nuclear fusion reactors at present. 5 Describe the two techniques that are currently being used in nuclear fusion reactors. 6 In its early stage of development before it began to shine, the Sun was once a protostar. Its matter was spread out over a vast area many times larger than the solar system. a If two hydrogen nuclei approached each other in such a protostar, would nuclear fusion occur? Explain.

b Discuss the effect of the electrostatic force and strong nuclear force on these hydrogen nuclei as they approach. c Discuss the effect of the electrostatic force and strong nuclear force on two hydrogen nuclei as they approach each other inside the Sun as it is now. 7 Stars that are much more massive than the Sun glow much brighter, but do not last for nearly as long as the Sun. Explain why this happens. 8 Consider this nuclear fusion reaction: 1H + 3He → 4He + +10 e + ν Hydrogen and helium-3 nuclei are being fused together and a helium-4 nucleus is being created, along with a positron and a neutrino. 21 MeV of energy is also released. a How does the combined mass of the hydrogen and helium-3 nuclei compare with the combined mass of the helium-4 nucleus, positron and neutrino? b Where has the energy come from? c Convert the energy into joules. d What is the mass decrease during the fusion of this one helium nucleus? e One of the original protons is no longer present after fusion. What has happened to it? 9 Discuss two major advantages of using nuclear fusion rather than nuclear fission to generate electricity. 10 How do scientists stop the enormously high tempera tures that are achieved in tokamak fusion reactors from melting the reactor itself?


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chapter review 1 Which one of the following explains why it is easier to trigger nuclear fission with a neutron rather than an alpha particle? A Neutrons are more massive than alpha particles. B Neutrons have more energy than alpha particles. C Neutrons are uncharged and are not repelled by the nucleus. D Neutrons are not impeded by electrons but alpha particles are. 2 Uranium ore that is mined from the ground contains uranium-235 and uranium-238. a Which of these isotopes is the more prevalent? b Which of these isotopes is the more fissile? c Explain why a chain reaction does not occur in uranium ore.

a What is the source of this heat energy? b How is this heat energy converted into electrical energy? 10 If a slow-moving neutron strikes a uranium-238 nucleus, the neutron is likely to be absorbed. After two beta decays, plutonium is formed. a Write the nuclear equations for these nuclear transformations. b The daughter nucleus, plutonium-239, does not exist naturally on Earth. Explain why not. 11 The diagrams below show the nuclei of carbon-12 and uranium-238. One of the protons in each nucleus is labelled. proton X

proton Y

3 Determine the number of neutrons released as a result of this fission reaction: n + 235 U → 140 Xe + 94 Sr + x10n 92 54 38

1 0

4 One kilogram of uranium-235 is capable of releasing 6.8 × 10  J of energy during nuclear fission. In comparison, the burning of 1 kg of coal releases around 2.5 × 107 J. Calculate the number of tonnes of coal that is burnt to provide an energy equivalent to that released by 1 kg of uranium-235. 13

5 The fuel rods that are used in a thermal nuclear reactor: A consist mostly of uranium-235. B consist mostly of uranium-238. C contain about 50% uranium-235 and 50% uranium-238. D consist mostly of plutonium-239. 6 Only about 40% of the neutrons that are released during the fission process in the core of a thermal nuclear reactor continue to produce further fission. What becomes of the remaining 60%? 7 Imagine that uranium-238 was the highly fissile isotope, instead of uranium-235. What would be the ramifications of this in a uranium mine? 8 a Which has the greater surface area: a whole orange or an orange sliced in two? b Which will lose more neutrons to the surrounding environment: a spherical sample of uranium or the same sample cut into two hemispherical pieces? c In which is a chain reaction more likely to continue: a spherical 2 kg lump of uranium or two spherical 1 kg lumps of uranium? d Discuss the significance of your answer to part c in the design of a weapon relying on nuclear fission. 9 When nuclear fission takes place in the core of a nuclear reactor, a great quantity of heat energy is produced.

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Detailed studies

uranium-238 carbon-12 a Discuss the proportion of protons and neutrons in a small stable nuclide such as carbon-12. b Discuss the proportion of protons and neutrons in a large unstable nuclide such as uranium-238. c Discuss the sizes of the electrostatic and strong nuclear forces that are acting on proton X. d Discuss the sizes of the electrostatic and strong nuclear forces that are acting on proton Y. How do these compare with the forces acting on proton X?

12 Some low-level radioactive waste is currently sealed in drums and dumped in the ocean. a Why do the proponents of this method consider it to be a safe means of disposal? b Discuss some of the arguments against this method of disposal. 13 About 16% of the world’s electricity is generated in nuclear reactors that use uranium for fuel. Many of these power plants use water under high pressure as the coolant. a What is the function of the control rods in a reactor? b Explain how a reactor produces electricity. c What is the advantage of keeping the coolant under high pressure? 14 The liquid drop model of the nucleus was proposed by Niels Bohr in 1935. a Describe the main features of the liquid drop model of the nucleus. b Use the liquid drop model to explain why a uranium-235 atom undergoes fission when struck by a neutron.

15 Explain the meaning of: a fissile b fertile. 16 A typical fission of uranium is: 92 10n + 235 U → 141 Ba + 36 Kr + 310n 92 56 a For the fission to occur, should the incident neutron be slow or fast? b Why is uranium used in this process rather than a less dangerous element such as iron? c A large amount of energy is released by this reaction. What is the source of this energy? d Explain the importance of the three neutrons that are released.

17 During one of the fusion reactions that is taking place in the Sun, two helium-3 nuclei fuse together to form helium-4 and two protons. a How does the mass of the reactants compare with that of the fusion products? b Energy is released in this reaction. What form does it take? c Discuss the forces that are acting as the two helium-3 nuclei approach each other and fuse together. 18 Consider this fission reaction of uranium-235: 93 10n + 235 U → 141 Cs + 37 Rb + 210n 92 55

a How much energy (in joules) is released by this fission? b What is this energy release in MeV? c Caesium-141 and rubidium-93 are the fission fragments or decay products of this reaction. Why do fission fragments present a problem in the disposal and treatment of reactor fuel rods? 19 Which one of the following best explains why fast breeder reactors do not need moderators? A Fast breeder reactors use uranium as fuel, and so rely on slow neutrons. B The control rods in fast breeder reactors slow the neutrons down. C The neutrons that are released during the fission of plutonium-239 are slow moving. D Plutonium requires a collision with a fast-moving neutron in order to induce nuclear fission. 20 a Two slow-moving protons are travelling directly towards each other. Will the protons collide and fuse together? Answer this question by making reference to the forces acting on the protons and the energy barrier. b Two fast-moving protons are travelling directly towards each other. The protons collide and fuse together. Discuss the forces that act on the protons and make reference to the energy barrier in your answer.

During this fission reaction there is a mass defect of 4.99 × 10-28 kg.


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chapter 13

t h g i ns: fl

o i t a g i t s e Inv outcome

On completion of this chapter, you should be able to design, perform and report on an experimental investigation related to an aspect of flight, and to explain results and conclusions by including reference to Newton’s laws of motion and Bernoulli’s principle.

I

f you’ve ever flown on a large, modern jet then perhaps you too have been inspired to wonder at just how such an extraordinary machine stays in the air. Carrying more than 500 passengers, and with a mass of almost 400 000 kg at take-off, it flies with an ease that would have been considered nothing short of miraculous a bare 100 years ago. And since, as you’ll soon find, the main emphasis of this study is investigation, let’s start with a question. Who actually invented powered flight just over one lifetime ago? If you said Wilbur and Orville Wright with their ‘Wright Flyer’ then there are many who would agree with you. Others, with a more Eurocentric view of history, may refer to Alberto Santos Dumont’s ‘14bis’, recognised as the first to achieve powered, controlled flight in Europe. From a scientific standpoint perhaps we need to look at the question in a broader sense. Both the Wright Brothers and Dumont paid tribute to the many others working in aviation who had pioneered aspects of their basic designs; people such as Gabriel Voisin, Otto Lilienthal and, notably, Australia’s Lawrence Hargrave. More important to their whole work was an understanding of the basic principles upon which flight was based—principles developed by scientists including Galileo, Newton and Bernoulli. For the past 500 years, science has been developed by a community of scholars in which individual contributions became essential to the process. Powered flight is one very visible and practical achievement of that process. This study is an investigation of forces in flight. It is an opportunity to develop an understanding not only of the basis of the physics behind flight but also how seemingly disconnected ideas can be brought together to achieve new and exciting ideas through the continuous interaction between theoretical development and empirical results.

by the end of this chapter you will have covered material from the study of movement including: • applying earlier ideas on forces to the principles of flight • applying concepts of forces, torques, centre of mass and equilibrium to balancing an aircraft • an explanation of the forces inherent in flight • an explanation of lift • modelling, experimentally, aircraft performance • an analysis of the performance of aircraft.

13.1 The four forces o

f flight

Whatever the form of an aircraft, its motive power or purpose, there are four basic aerodynamic forces: weight, lift, thrust and drag. These can be considered as two pairs of opposing forces. Lift raises the aircraft upwards and weight pulls it down; thrust propels it forwards and drag slows it down. In order for an aeroplane to fly level and straight at a steady speed these forces are balanced and Newton’s first law of motion applies.

The remainder of this investigation assumes that you have studied Newton’s laws of motion including the material on forces and vectors covered in Chapter 5 of the core material. Before going any further make sure you review this area of study.

An object remains at rest or will travel at constant speed in a straight line unless an unbalanced force acts upon it. For an aircraft travelling at constant velocity that means that: • the size of the thrust force = the size of the drag force • the size of the lift force = the size of the weight force. In a vertical direction, if the amount of lift falls below the weight, the aircraft will begin to fall or descend. Increasing the lift allows the aircraft to climb or increase its altitude. In practice, the forces are difficult to consider in isolation, since speed is related to lift and drag. However, considering the effective size of each of these forces at any point in time is the basis of understanding flight.

Lift While the development of flying machines can be traced back to Leonardo Da Vinci’s drawings and musings, there is no doubt that many before him gazed at birds in awe and wonder, and wished that they too could fly. The apparent freedom of birds to lift on air currents has entranced people through the ages. Lighter than air craft—principally hot air balloons—realised part of the dream by the 19th century. However, it wasn’t until comparatively recently that two parallel lines of development produced the discoveries that were needed to achieve the dream of true aerodynamic, powered flight. One group developed sufficient understanding of the underlying forces involved in flight and another invented the comparatively light and efficient internal combustion engine. Converging with the development by Karl Benz, and others, of the internal combustion engine was an apparently unrelated investigation into fluid flow; that is, the flow of water around a propeller. It was the early 1800s and paddle steamers were taking the first tentative steps toward replacing sailing boats. In 1842, Isambard Kingdom Brunel was working on the plans for the largest ship yet to be built—the steam-powered Great Britain. He conceived the idea of replacing cumbersome stern- or side-mounted paddles with a single huge propeller. The fluid dynamic principles on which he based the successful development of the propeller for his ship are the same as apply to the lift of an aircraft wing or thrust of an aircraft propeller. A Swiss mathematician who knew nothing about flight developed those principles more than 150 years before the Great Britain was to sail the seas. His name was Daniel Bernoulli and he died in 1782, shortly before even the first hot air balloon took off.

lift

thrust

weight

drag

Figure 13.1 The four basic aerodynamic forces are thrust, drag, lift and weight. In this illustration, the size of each pair of forces is equal and opposite: thrust = drag and lift = weight. Hence, the aeroplane will fly straight and level at a constant speed.

Chapter 13 Investigations: Flight

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Figure 13.2 The Great Britain set many firsts for steam-powered ships, including records for size, speed and luxury. One of the most lasting initiatives was the use of a single bronze propeller at the rear of the ship. It was an early practical application of Bernoulli’s principle.

Bernoulli was investigating the flow of air through chimneys and tunnels—another example of fluid flow. In essence, Bernoulli’s principle states that the faster any fluid—whether liquid or gas—moves, the less pressure it exerts on a surface over which it passes.

B…RNOULLI’S PRINCIPL…: Where the velocity of a fluid is high, the pressure on a surface over which it flows is low, and where the velocity is low, this pressure is high. That means that for an object that has fluid flowing either side of it, and the flow on one side has a higher velocity than the flow on the other, the slower-moving fluid will exert a larger force on the object. That might seem a little strange but the higher pressure, and hence force, must be associated with the slower-moving fluid. If it were associated with the higher velocity it would slow the fluid down. Bernoulli also developed an equation that expresses the principle quanti tatively. Bernoulli’s equation is an expression of the law of conservation of energy described in Chapter 5 and can be derived by considering that the net work done on a system is equal to its change in kinetic energy.

B…RNOULLI’S …QUATION: Assuming a fluid with constant density, ρ, then, within a system: P1 + 12 ρv12 = P2 + 12 ρv22 where P1 and P2 are the fluid pressures in N m-2 on either side of a relatively thin, horizontal surface where there is no appreciable change in height, ρ is the fluid density in kg m-3, and v1 and v2 are the respective fluid velocities in m s-1. The equation ignores the effects of fluid friction (viscosity) and the compressibility of the fluid. Bernoulli’s equation tells us quantitatively that where the speed of gas or fluid moving over an airfoil is high the pressure is low, and vice versa. The principle is important for aircraft wings and explains many other common phenomena. It also explains why so many early attempts to fly met with failure. Too often these early attempts revolved around people trying to imitate birds by flapping wings. No one had separated thrust from lift. Birds flap to move forward, not simply to lift themselves into the air. The

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Detailed studies

aerofoil wing shape of a bird’s wing provides the lift once they are moving. The first person to demonstrate that important factor was English scientist Sir George Cayley. In 1804 he built the world’s first successful glider using wings with an aerofoil shape. The aerofoil shape is successful because of the comparative distances that air must flow as it moves through the air. Air flowing over the top of the aerofoil shown in Figure 13.3 must travel further and faster than the air flowing under it. The slower moving air underneath the wing has a greater pressure and pushes the wing up. Similar aerofoils can be seen in the propellers of aircraft and boats, the sails of yachts and on boomerangs— surely the oldest human invention to successfully apply Bernoulli’s principle to simple flight. Upside down aerofoils can be found on the back of racing cars and sports cars. The air pressure in this application is pushing down and assists in keeping the car on the ground while also improving traction between rear wheels and the ground.

v1, P1 aerofoil v2, P2

direction of air flowing over aerofoil

Figure 13.3 An aerofoil section generates lift for any wing, be it bird or aircraft. The pressure, P1, of the fluid with a higher velocity, v1, will be lower than pressure P2 where the fluid has a lower velocity, v2. The result is an upwards force or lift that allows an aircraft to take off and fly.

In practice, calculating the pressure distribution around a wing and the exact amount of lift it produces is more difficult than simply applying Bernoulli’s equation. Wings are usually mounted with a slight upward tilt referred to as the angle of attack. Air striking the bottom surface is deflected downwards, the resulting change in momentum of the rebounding mole cules providing an additional upward force on the wing. The lift coefficient of an aerofoil is a number that relates its lift-producing capability to air speed, air density, wing area and the angle of attack. Turbulence also plays an important role. The lift coefficient can be found from a derivation of Bernoulli’s equation: L = Cl × 12 × ρv2A where L is the lift in newtons, Cl is the coefficient of lift, ρ is the fluid density in kg m-3, v is the velocity in m s-1 of the aircraft, and A is the wing area in m2 at right angles to the direction of travel. The lift coefficient is dependent upon the angle of attack and must be recalculated for any change in the angle. Scientists conduct wind tunnel tests to measure the lift generated for a particular shape of aerofoil and use the results to publish tables of coefficients versus angle of attack. The lift of any wing can then be calculated. One thing immediately apparent from Bernoulli’s equation and the equation for lift is the high dependence on velocity. A wing going twice as fast as another can generate four times as much lift.

Chapter 13 Investigations: flight

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Blow across the top of the glass tube (point straw slightly upwards). ping pong ball

Blow hard down the spout of the funnel.

water

Figure 13.4 While more often associated with flight in modern times, Bernoulli’s principle has much broader applications. Try these simple experiments to see Bernoulli’s principle in action. Can you identify the positions of higher and lower pressure in each case?

Blow gently between strips of paper. bobbin thin card Hold the card in place and blow down the hole. Let go of the card as you are blowing. drawing pin

Physics in action

Developing the A380 During the development phase of the A380, Airbus filed more than 380 patent applications for technologies developed for what is currently the world’s largest passenger aeroplane. Since the programme was launched in December 2000, significant innovations have been achieved in aerodynamics, cabin design, engine integration, flight controls, aircraft systems and manufacturing techniques, and in the extensive use of advanced lightweight composite materials, making the A380 not only the world’s biggest but also its most efficient airliner per passenger carried. To lift such a heavy aircraft and achieve improved efficiencies of up to 17% over existing designs, new approaches needed to be taken to the design of the wing in particular. The wing was designed with a flexible structure, incorporating a range of elements that allow it to be virtually reconfigured for various stages of flight. The wing needed to

have sufficiently high lift and low drag for take-off, provide good control during flight and yet have low total weight and complexity. Compared with other wing designs, the outer wing profiles are thinner with a smaller leading edge radius. A thin rear profile allows high rear loading of the wing during cruising. Other design elements, such as a ‘taberon’ in the in-board section of the wing and a hydraulically controlled ‘droop nose’ on the leading edge, allow the high lift required during take off and the control needed during landing. The final design emphasised the need for both modern computer flight design (CFD) approaches and the reassurance of traditional investigation methods such as wind tunnels. The designers found that while CFD was able to deliver fast design, it was often with partly questionable results. Although wind tunnel testing is slow and extremely expensive (multiple scale models were required during testing), it was the practical results from the wind tunnel that proved vital in validating the data from the CFD. clean

take off flap 16n

Figure 13.5 The size of the A3

80 posed new ch allenges for aero

dynamic design.

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Detailed studies

landing flap 34n

Testing aerofoils The wing shapes of early aircraft were tested, like many other things, through a process of trial and error. The aircraft was built and flown and the results, sometimes humorous but too often tragic, were observed. Incremental improvements were made and the process repeated. It was a long drawn out process and often costly in terms of materials and test pilots. In the case of large aircraft, it was usual to build a smaller-scale proof-of-concept prototype; this was less costly in terms of materials but still very dangerous for the test pilots. One advantage of the process was that it was a real aircraft that was being tested and it was being done in real conditions. If it worked as a prototype designers could be pretty confident that the production model would also work. A big step forward in the design process, and a huge relief to pilots, was the invention of the wind tunnel. Scale models of aircraft, and even some full-scale components, could be tested on the ground before the complete aircraft was built. Today wind tunnels are used for pre-testing just about all aerodynamic shapes, with considerable savings in time, cost and human life. On a simple level, a wind tunnel is just a large chamber with a massive fan at one end. The model is suspended in the chamber in as near to true flying position as possible. Smoke released into the airstream can give a visual indication of airflow, while sensors attached to the model give realtime readings of pressure, temperature and strain. Vast amounts of data can be collected on just how the model performs in these simulated conditions. Similar ‘tunnels’ filled with water are used for testing fluid flow past the hull shapes of boats and their keels. Testing in a wind tunnel is a vast improvement over testing the real thing, but it is still expensive. The facilities are large and complex, the models still relatively expensive to make, and testing still needs to happen in real time for meaningful results to be achieved. At the higher end of performance it is also difficult to replicate the high air speeds and differing air densities encountered by fighter aircraft and rockets as they soar through the upper reaches of the atmosphere at supersonic speeds. So it is becoming

For details on how to build your own simple wind tunnel see the instructions given at the end of this study. Manufactured versions are also available.

KEY KEY

motor room

heat bench pipes fan anechoic turning valves diffuser

flow contraction test section turntable

car entrance retractable turning valves

heat bench system

anechoic turning valves turning valves

control panel

Figure 13.6 RMIT’s low-speed wind tunnel allows complex aerodynamic testing to be performed on scale and full-sized models before complex and expensive tooling are committed to producing working prototypes. More recent advances now see much of the initial testing done by computer modelling.


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increasingly common for simulations to be carried out by computer. Tests can be done by measuring the virtual flow of air over a virtual wing. The design can be rapidly changed on-screen to produce the best shape. The time for the design process is reduced from years with a real aircraft, to months with a wind tunnel, through to days with computer modelling. Of course, computer modelling is only as reliable as the original data that the models of air movement are built upon, so further testing in wind tunnels is performed both to verify and improve the computer model and to confirm the virtual results.

Weight According to Newton’s first law of motion, for an aircraft to be flying level when the aerofoil is applying lift there must be a balancing force. Weight, the force of gravity, provides that balancing force. The weight is equivalent to the value predicted by Newton’s second law.

You can read more on this important concept in Chapter 5 of this text.

N…WTON’S S…COND LAW: F = mg where F is the force in newtons (N) acting on the aircraft as a result of a gravitational field of strength g and m is the mass in kilograms (kg). In this case g has a value equivalent to the acceleration an object will experience in free fall, a value of approximately 9.8 m s-2 at the Earth’s surface.

Physics in action

The vomit comet Astronauts preparing for space travel are prepared for the feeling of apparent weightlessness in orbit by travelling in a specially prepared Boeing aircraft. The aptly named ‘vomit comet’ performs a series of loops. At the top of each loop both the aircraft and the people in it are in free fall along a parabolic path. Because everything is falling at the same rate, the astronauts experience a few seconds of apparent weightlessness. The effect of this on the stomach has led to the aircraft’s nickname.

Figure 13.7 The aptly named ‘vomit comet’ allows astronauts to experience apparent weightlessness, as both aircraft and occupants free fall at the same rate through a parabolic path from the top of a loop.

You too can demonstrate apparent weightlessness: • Attach a spring balance or force sensor to the top of a box by a hook or other secure fastener. • Attach a 1 kg mass to the spring balance. The balance should be showing a value equivalent to a 1 kg weight force; that is, approximately 9.8 N. • When you’re ready, drop the box, mass and spring balance.

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Detailed studies

During the fall, the spring balance and the mass are falling at the same rate. The recorded mass on the spring balance will be zero (a force sensor attached to a datalogger will show this clearly), as both mass and spring balance will be accelerating at the same rate. The resultant force between mass and force measurer is zero. Since the mass has obviously not lost any mass, this is a good example of apparent weightlessness.

Thrust Thrust is the force that propels a bird, yacht or aircraft forward. The aerofoils of a wing will not generate lift when an aircraft is stationary. An aircraft needs to travel very quickly along a runway so that air rushes under and over the aerofoil in order to produce the lift predicted by Bernoulli. Once in the air thrust is required both to keep the aircraft in the air and to carry it forward (you’ll recall that we noted that no single force in aeronautics can be considered in isolation). In modern times, two basic types of engine are used to provide the required thrust: jet engines and propeller engines. Turbojets are hybrid, largely propeller-driven engines that include some of the economy and thrust benefits of jets. While seemingly quite different in concept, both jets and propellers create thrust according to Newton’s third law.

propeller blade

Figure 13.8 Take a series of slices through one blade of a propeller and its relation to an aerofoil is clearly visible.

N…WTON’S THIRD LAW: For every action there is an equal and opposite reaction. Propellers are really just like a twisted wing. Brunel’s massive brass propeller for the Great Britain was simply three aerofoils twisted at an angle to each other. Take a series of slices through the blade of a propeller and this is clearly visible. The propeller of most aircraft drives a stream of air backwards and, by doing so, pushes itself forwards pulling (or pushing in some instances) the aircraft with it. The angle of attack along the propeller section varies along the length of the propeller. The angle is greatest at the centre because the speed of the propeller is slowest close to the hub. More elaborate three- and four-blade propellers have adjustable pitch mechanisms, allowing the pilot to adjust the propeller’s angle of attack, depending on air speed and altitude. The thrust of a propeller can be observed and measured through two simple models. One option is to attach an electric motor to a propeller and mount this unit on a low-friction trolley. Clearly the propeller moves the trolley forward. A simpler arrangement is the rubber-band-propelled model aircraft. Simple measurements should permit the thrust of either arrangement to be investigated. The thrust of propellers with two, three, four or more blades can be compared. The effect of the pitch or angle of the blades can also be readily investigated with this simple equipment. Jet engines work on the same basic principle as propellers—air rushes out the back and the aircraft is pushed forward. A turbojet engine has three main sections as shown in Figure 13.9: compressor, combustion chamber and turbine. Air is drawn into the engine and compressed, making the air hot (you may have experienced the same basic effect when pumping air through a bicycle pump). In the combustion chamber the air is mixed with kerosene fuel. The air and kerosene mix burn explosively, and a highpressure stream of hot gas rushes out the back. The gas turns the blades of the turbine, which pushes the aircraft forward. Aircraft engines exert thrust to overcome the drag from air resistance, to climb against the weight force of gravity, and to accelerate. An aircraft’s performance is limited by the rate at which it can do work. Even when an aircraft is travelling along in level flight at constant speed, it needs power just to provide the lift needed to balance the weight force and the retarding force of drag. These forces depend on the size, mass and shape of the aircraft and on the flying conditions, and they can be enormous when compared


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(a)

(b) stage wind flow compressor fuel

turbine

exhaust

compressor blades

combustion chamber

Figure 13.9 (a) The Messerschmitt Me 262 first flew with jet engines in November 1941. While ultimately highly successful, it was fortunate that this test flight still had a nose-mounted piston engine installed, as both jets failed on the point of lift off! (b) The basic principles of the turbojet engine haven’t changed since the first production jet reached the skies during World War II.

For a full explanation see Chapter 6 of this text. (a)

bottle

(b)

with the forces acting on early aircraft. A Boeing 747 with a mass of more than 390 000 kg is powered by four turbo-fan engines, each exerting over 250 kN of thrust. The power generated by the engine can be expressed in terms of the net force, F, applied to an object and its speed, v. As: x = vt where x is the effective distance moved, then: W Fx P = = = Fv t t where P is power in watts (W), F is the net force in newtons (N), and v is the average speed of the object in m s-1.

compressed air

Worked example 13.1A

water

(c) nylon fishing line

If the average air resistance encountered by a particular aircraft is 5 kN when it is travelling at an average velocity of 720 km h-1 at an altitude of 5000 m, what power must the engines develop solely to overcome air resistance?

short piece of drinking straw

Solution

720 × 1000 m = 200 m s-1 3600 s and P = Fv = 5000 × 200 = 1 000 000 W or 1 MW (Of course, the engine must develop considerably more power than this in order to also keep the aircraft aloft!) v = 720 km h-1 =

sticky tape

balloon bulldog clip to hold air in until ready to launch

Figure 13.10 Realistic testing of jet engines is

impractical outside a jet propulsion laboratory. However, the basic principle can easily be investigated using various sources of compressed air. Both commercial models and home-made alternatives allow measurement of force versus thrust.

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Detailed studies

Drag Drag—also known as air or fluid resistance—simply slows down any object moving in a fluid, whether it be a gas or a liquid. Riding your bike, you can feel the drag that your body creates. The amount of drag will depend on the shape, surface area, surface material (so-called skin friction), its speed and the fluid the object is moving through. The faster the object moves, the greater the drag experienced. Drag is produced by the work required to move the fluid out of the path of an object. If drag is to be reduced, then an aircraft needs to have a shape that best allows it to move through the air smoothly. A great example of drag reduction is the position adopted by speed skaters at the winter Olympics. Whenever they get the chance, they crouch down and reduce their frontal profile as much as possible. The drag

the skaters create is decreased and that allows them to move faster and more efficiently around the track. Modern aircraft retract the undercarriage as soon as possible after take-off. Just like the skater, the pilot is trying to minimise drag by making the profile of the aircraft as small as possible. The amount of drag produced by the undercarriage of a large jet is huge. The drag coefficient is a derivation of Bernoulli’s equation: D = CD × 12 × ρv2A where D is the drag in newtons, CD is the coefficient of drag, ρ is the fluid density in kg m-3, v is the velocity in m s-1 of the aircraft, and A is the wing area in m2 at right angles to the direction of travel. Like lift, the equation for drag is highly dependent on velocity. A wing going twice as fast as another can generate four times as much drag. Drag is caused by the turbulence created as an aircraft moves through the air, breaking up of the smooth flow of the fluid into swirls around and behind the aircraft. The faster an aircraft travels the greater the drag produced. Ultimately drag will equal thrust and the aircraft will no longer accelerate. You can see the effect by introducing a little smoke or the vapour of dry ice into a wind tunnel, or through a simple activity with a flame and a small beaker. A special example of the effect of drag is terminal velocity. Drop any object and as its velocity increases due to the acceleration, but so too does drag, just as in level flight. Eventually the force due to drag will equal that of gravity and the object will no longer accelerate. It will continue to fall at a constant velocity referred to as terminal velocity. Skydivers stretch their arms and legs to increase drag during free fall, to control their rate of descent, and to perform aerobatics. (a)

Physics file Try dropping different shaped objects through a thick liquid, such as glycerol, in a tall measuring cylinder. More streamlined shapes will continue to accelerate quite uniformly while those that are less streamlined will not reach the same speed by the time they reach the bottom of the cylinder. Video analysis of the acceleration can permit the investigation and calculation of terminal velocities for different shaped objects in the fluid. Place a candle behind the beaker as shown in Figure 13.11. Blow on the front, or opposite side, of the beaker and the flame should move in toward the beaker as the air curls into turbulence behind the smooth sides of the beaker. Try streamlining the back of the beaker and observe the changes in the movement of the flame.

(b)

Figure 13.11 (a) Blowing on one side of the beaker causes the flame of the candle to move back toward the beaker. (b) The blown air curls into turbulence behind the smooth sides of the beaker.

13.1 summary The four forces of flight • Four basic aerodynamic forces affect aircraft in flight: drag, thrust, weight and lift. • In order for an aircraft to fly straight and level, the size of the thrust must equal the size of the drag force, and the size of the lift force must equal the size of the weight force. • Drag is created by the work required to clear the air from the aircraft’s path. • Thrust is the force that overcomes drag and keeps the aircraft moving through the air. Aeroplanes generate thrust with propellers and jets.

• Weight is the force of gravity and is the product of the aircraft’s mass and its acceleration due to gravity: F = mg. • Lift is produced by a combination of factors, includ ing the angle of attack and the lift generated by the aerofoil shape of the wing. • Bernoulli’s principle explains the lift generated by an aerofoil. Air moving at low speed under the wing generates a higher pressure than air moving at higher speed over the top of the wing.


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• Bernoulli’s principle:

equation

quantifies

the

Bernoulli

P1 + 12 ρv12 = P2 + 12 ρv22 where P1 and P2 are the fluid pressures in N m-2 on either side of a relatively thin surface where there is no appreciable change in height, ρ is the fluid density in kg m-3, and v1 and v2 are the respective fluid velocities in m s-1.

• Wind tunnels provide a simple and effective means of testing aerofoils and aerodynamic models for improving performance without the risks and expense of flight testing. • The power an aircraft engine must develop to maintain a velocity, v, is: P = Fv where P is the power in watts (W), F is the net force (N), and v is the aircraft’s velocity in m s-1.

13.1 questions The four forces of flight 1 Explain, using a suitable vector diagram, the prin ciples by which a jet engine provides an aircraft with forward thrust. 2 A plane has a designed stall speed of 200 km h-1. Flying into a sudden wind shear (a sudden change in wind direction and speed relative to the aircraft) of 70 km h-1 relative to the ground at an indicated air speed of 300 km h-1 the aircraft may still stall. Explain how this could happen. 3 Stunt planes often fly upside down. Since the aerofoil shape of the wing is what generates lift when the plane is flying upright, explain how the plane can continue to generate lift and fly when upside down. 4 A plane doubles its indicated air speed. What is the proportional increase in the lift generated by the wing due to the change in speed? 5 The performance of a glider can be best measured by its glide ratio. This is the ratio of horizontal distance travelled compared with fall in altitude. Modern gliders have glide ratios of more than 60:1 compared with a jet liner’s 10:1. Drag reduction is the most important means of increasing glide ratio. Suggest three ways this can be achieved in glider design. 6 A number of modern fighter jets have wings whose geometry can be altered during flight—swept back for high speed flight, brought forward to a straight position for landing. What advantages are there to this form of design? 7 A frequent problem in older aircraft was the surface of the upper wing ‘lifting off’ at higher speeds. (Modern testing techniques usually highlight the problem before flight.) Explain this phenomenon using Bernoulli’s principle. Is it ‘lifted’ off or ‘pushed’ off? 8 Dangle two pieces of paper vertically, a few centi metres apart and blow between them. Try it and explain what you see.

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Detailed studies

9 Airplanes normally take off into the wind. Why? 10 A hummingbird drinks pollen from a flower while hovering in front of the flower—it doesn’t land while doing so. It expends about 15–20 times as much energy in hovering than it does in normal flight. Explain why. 11 Try this: Sit a 5-cent coin on a tabletop. Blow (very hard!) over the top of the coin. Explain what you see happen to the coin. Why does this occur? 12 What is the lift, in newtons, due solely to Bernoulli’s principle, of a wing of area 100 m2 if the air passes over the top surface and bottom surface at speeds of 960 km h-1 and 800 km h-1 respectively? (Use density of air = 1.29.) Note, you will need to work out the relationship between pressure, and lift and area, from their units. The following information applies to questions 13–16. A jet plane is travelling at a speed of 200 m s-1 relative to the air around it. Each second the engines of the plane take in a total of 150 kg of air. The air is used to burn 2.0 kg of fuel each second. Finally, the air and burnt fuel mix is expelled from the rear of the engines at a speed of 450 m s-1 relative to the plane. Use the following relationships: change in momentum (impulse), Δp = mΔv = FΔt (N s) power, P = Fv (W) 13 What is the magnitude of the total force exerted on the plane by the intake of the air? 14 What is the magnitude of the force exerted on the plane by the air and burnt fuel being expelled from the engines? 15 What is the net thrust produced by the engines? 16 What is the power developed as a result of the net thrust?

13.2 Modelling force s

in flight

So far we have discussed and developed the four basic aerodynamic forces as if they were acting at a single point. The initial advantage of considering the basic forces as acting on a point source is that it allows us to isolate the effect of these forces from other complexities. This was one of the strengths of Newton’s approach to understanding forces. He was able to isolate cause from effect. The analysis of many complicated systems can be simplified in a similar manner. FL FT

FD FW

FL

FL FT

FD

FT

FW

FD FW FL

FL FT

FD

FW

FW landing

FD

FT

descent

level flight

climb

take-off

Figure 13.12 The relative size of the four basic aerodynamic forces during take-off, climb, cruise, descent and landing acting about the aircraft’s centre of gravity.

Centre of mass and gravity The mass of the aircraft, no matter how complex the shape, can be considered to be concentrated into a single point whose path will follow exactly the same path as would a single point under the same net force. This single point is called the centre of mass.

The motion of any extended rigid body can be simplified to the motion of a particle of the same mass located at the centre of mass when it is subjected to the same forces. If a body is no more than a uniform length of wire, its centre of mass will lie exactly at the centre. In two or three dimensions, the centre of mass will be central for each dimension. It is even possible for the centre of mass to lie outside the body, as with a doughnut (the centre of mass is in the hole!). For a simple non-uniform system, the position of the centre of mass can be calculated easily. For the more complex shape of an aircraft more calculations are involved.


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pivot 1

plumb line 1

centre of gravity pivot 2 plumb line 1

plumb line 2

Figure 13.13 In order to determine its centre of gravity, a model aerofoil is suspended from a pivot, and a vertical plumb line is drawn through the pivot point.

A concept closely related to centre of mass is the centre of gravity. Instead of being a point particle whose motion equates to the whole extended body or system, the centre of gravity is the position from which the entire weight of the body or system is considered to act. As a consequence, the centre of gravity is the position at which the body will balance. For almost all practical purposes, the centre of gravity is exactly the same as the centre of mass. It is only when a body is so massive that its own gravitational field becomes significant, that the centre of gravity no longer coincides with the centre of mass—hardly a concern for an aircraft. The centre of gravity of an aircraft can be found more easily by experiment than by calculation with the mathematical tools available to us at this level. Figure 13.13 shows a simple model of a thin aerofoil suspended vertically from a pivot point and free to move. When released, the model will swing until its centre of gravity lies directly beneath the pivot point. A plumb line is then drawn vertically down from the pivot. If the aerofoil is hung again from a second, different pivot point, the centre of gravity will again lie on a line beneath the pivot, so a second plumb line can be drawn. The aerofoil’s centre of gravity will be at a position where the two plumb lines intersect. If the process is repeated, all such plumb lines will intersect at the same point. This method will work for any plane object. For a three-dimensional model of an aircraft, the centre of gravity will be found using a third line (although it may be hard to draw the plumb lines through the model). Suspending an aerofoil from its centre of gravity in a wind tunnel will keep it in a stable, flying attitude while the forces created by airflow are investigated.

Equilibrium We’ve already seen that when the two pairs of basic forces of flight are balanced the aircraft will remain at rest or at constant speed. This was the situation described by Isaac Newton in his first law of motion. When all the forces acting on the aircraft add up to a zero net force, the aircraft is said to be in translational equilibrium.

A body is said to be in TRANSLATIONAL …QUILIBRIUM when the sum of the forces acting on the body is zero; i.e. ΣF = 0. So the motion of the centre of gravity of the aircraft can be described and predicted. Increase, or decrease, any one of the four basic forces and the aircraft as a whole will accelerate. But an aircraft may be in translational equilibrium and still not be stable. The turning propeller, ailerons, wings, flaps, stabilisers and rudder, even the body of the aircraft itself, all exert additional turning, twisting forces called torques that will rotate the aircraft. They are an essential part of the flying and control of an aircraft and their individual effects on the rotation of the aircraft must be considered in any comprehensive investigation of forces in flight. For an aircraft to be completely in equilibrium it must also be in rotational equilibrium. You can imagine an aircraft quickly falling into a spin if these forces were not controlled. In other instances, the application of a rotational force is what steers an aircraft and is highly desirable.

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Detailed studies

vertical stabiliser

aileron up — lift decreases

rudder

flaps elevator wing drops

wing lifts

horizontal stabiliser

aileron down — lift increases

Figure 13.14 An aircraft in flight is affected not only by the translational forces of lift, weight, drag and thrust, but also by the rotational forces introduced by control surfaces and other key components of the aircraft.

For an aircraft to be in rotational equilibrium all of the rotational forces, introduced by control and other surfaces, acting about a point must be zero. The net clockwise rotational force or torque must equal the net anticlockwise torque.

For an aircraft to be in ROTATIONAL …QUILIBRIUM, the sum of all torques acting about a point must be zero; i.e. Στ = 0 where τ is torque in newton metre (N m). If the aircraft is not experiencing any translational acceleration or rotation then it is said to be in static equilibrium. That’s not to say it is not moving. It could be stationary or it could be moving at high, but constant, speed in a straight line without any rotation.

For an aircraft to be in STATIC …QUILIBRIUM, it must be in translational and rotational equilibrium; i.e. ΣF = 0 and Στ = 0.

Torque The size of a torque exerted by a control or other surface will depend on three factors. The first is the magnitude of the force. If all other things are equal, a larger force will result in a larger torque. The second is the point at which the torque acts. When analysing the rotation of a system, the position of the axis of rotation is an important reference point. A door, for example, moves in a circular path around its hinges. The line of the hinges is the axis of rotation. A force applied at the hinge itself will not create a turning effect; the maximum effect will be achieved by applying a force to the door as far from the hinge as possible. That is why an aircraft’s vertical stabiliser and rudder are positioned firmly at the rear of the aircraft; this is where the stabiliser will have maximum effect in overcoming any horizontal rotation of the aircraft along the line of the fuselage. An aircraft will experience turning effects in all three dimensions about its centre of gravity.

Physics file A propeller of an aircraft introduces considerable torque forces that must be controlled in some way. In single-engine aircraft the pilot must use the control surfaces to balance the tendency of the aircraft to roll in the direction in which the propeller is spinning. In multiengine aircraft the propellers are often ‘handed’. That is, the propellers on one side will be spinning clockwise while the propellers on the other will be spinning anticlockwise. The torque effects of the propellers balance themselves.


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Finally, the turning effect of a force depends on the direction in which it is applied. In closing a door, for example, the maximum effect is achieved if the force you apply is at 90º to the door surface. If the angle is reduced, a smaller component of the force is perpendicular to the door and so a smaller torque is produced. If the force is applied along the line of the door (i.e. 0°, directly towards the hinges), the door will not rotate no matter how great a force you apply. These three factors combine to form the definition of torque.

F⊥ is used to denote the perpendicular force. The small notation ‘⊥’ is used to represent a right angle.

The TORQU…, τ, acting on a body is given by the product of the component of the applied force acting perpendicular to the lever arm, F⊥, and the distance from the axis of rotation to the applied force, r: τ = r F⊥ sin θ Where the force is perpendicular to r, as in the vertical stabiliser, sin θ = 1, so: τ = r F⊥ The symbol for torque is the Greek letter ‘tau’, and its unit is the newton metre (N m). Torque is a vector quantity. A clockwise rotation is considered to be negative and an anticlockwise rotation to be positive. This convention can be useful when a number of torque forces are acting and the net torque has to be calculated.

Worked example 13.2A A small plane of length 10.0 metres has a mass of 8000 kg. A large container of mass 200 kg is placed in the rear cargo bay, 4.2 metres behind the centre of gravity of the unladen aircraft. A reserve fuel tank is located 1.8 metres forward of the centre of gravity. (The mass of the fuel tank is included in the mass of the aircraft.) What mass of fuel must be pumped into this tank to offset the torque exerted by the cargo?

Solution The torques from each object will act in different directions around the centre of gravity since one is forward and the other aft of the centre of gravity. (Let t stand for fuel tank and c the container.) So: τt = τc As both forces will be acting at right angles to the centre line of the aircraft: mt × g × dt = mc × g × dc Simplifying: mt × dt = mc × dc mt × 1.8 = 200 × 4.2 and mt = 467 kg So, 467 kg of fuel must be pumped into the tank to offset the torque exerted by the cargo.

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Detailed studies

13.2 summary Modelling forces in flight • The centre of gravity of an aircraft is the position from which the entire weight of the body or system is considered to act. For all practical purposes this is the same as the centre of mass. • An aircraft is said to be in translational equilibrium when the sum of the forces acting on it is zero. • An aircraft is said to be in rotational equilibrium when the sum of all rotational forces or torques is zero; i.e. Στ = 0. • An aircraft, whether moving at stable and constant speed or stationary, is said to be in static equilibrium

when it is in both rotational and translational equilib rium; i.e. ΣF = 0, Στ = 0. • The product of the applied force, F⊥, and the perpen dicular distance to the axis of rotation, r sinθ (the lever arm), gives the torque acting on an aircraft: τ = rF⊥ sinθ. • In analysing torque in flight, torque acting in a clock wise direction is considered to be negative; torque acting in an anticlockwise direction is considered to be positive.

13.2 questions Modelling forces in flight 1 Although no longer flying, the Concorde supersonic airliner set speed records for commercial airliners between Europe and North America. Its prodigious appetite for fuel and small payload led to its demise. The Concorde used fuel not only for keeping the engines running but also to restore aircraft stability. The pilots adjusted the fuel backward and forward in ‘trim’ tanks. Explain the need to do so. 2 Determine, by geometry or otherwise, the centre of gravity of the two delta wing forms shown. Assume, for the purposes of this exercise, that the wings are of uniform density and composition. (a)

(b)

3 What implications do your solutions for question 2 have for aircraft incorporating the wings shown in the figure? 4 An aircraft starts a flight with 3000 kg of fuel evenly distributed between two fuel tanks to balance the weight about the centre of gravity of the plane. One of the fuel tanks is 2.0 m forward of the aircraft’s centre of gravity, the other is 5.0 m to the rear of it.

The rear tank is used and emptied first. How much fuel must the pilot pump to the rear tank in order to restore the plane’s trim? 5 Explain, with a suitable force diagram or otherwise, the role of ailerons in turning an aircraft (see Figure 13.14). 6 An aircraft of length 70 m has a mass of 200 000 kg when unladen, which can be considered to be acting at a centre of gravity 30 m from the front of the aircraft. Two large cargo containers are to be loaded onto the aircraft. One of mass 15 000 kg is to be loaded into a cargo bay 20 m behind the aircraft’s centre of gravity. Where must the second container, of mass 25 000 kg, be located in order for the combined torques to offset each other? 7 During flight the aircraft in question 6 encounters severe turbulence, causing the rear load to shift a further 2.5 m towards the rear of the aircraft and the rear of the aircraft to fall slightly. The pilot can quickly restore the aircraft’s flying attitude by altering the position of the elevators on the rear wing while the cargo is re-secured. What lift must the rear wings and elevators exert in order to restore the aircraft’s original flying attitude? (Assume that the elevators are at the rear of the aircraft.) 8 In practice a pilot may deliberately alter the aircraft’s flying attitude or angle of attack by altering the position of the elevators. Why is this done?


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13.3 Investigating flight Principles of investigation The technology involved in many areas of aerospace is well beyond the understanding and equipment available in a secondary physics course. However, the basic principles of flight—lift, thrust, drag and weight—invite investigation. A simple wind tunnel can be made in the course of a single lesson. This is a process we can model on a small scale in a meaningful investigation of our own design. The process of investigation is quite different from the traditional teacher-structured practical activity for which often the outcome is already known. It is quite probable that your results will not have been duplicated nor be available in any published resource. A few simple guidelines will assist you in making those results meaningful. The following guide will assist you in planning your investigation.

Getting started The choice of topic is often the biggest hurdle to overcome. Be imaginative and investigate, within the constraints of the study of aerospace and aircraft design, a topic of genuine interest to you. Your investigation should concentrate on the physics involved, and make use of the established principles of scientific investigation. The topic should be simple enough for you to begin immediately. Here are some simple guidelines that can help you make a good choice. • Straightforward topics are almost always better than more complicated ones when just getting started. Don’t choose a topic for which the background research occupies most of the available time. Some ideas for investigation were suggested throughout the text in section 13.1. You’ll find some more ideas for starting points in section 13.4. • Develop some preliminary aims as part of your initial topic choice. They may change as your investigation develops, but they will be useful for checking that your topic is not too complicated. Focus your aims on particular, measurable tasks. Broad aims are hard to achieve. • Think about what you intend to do before you start investigating. Identify potential problems and plan ways of overcoming them. Thinking ahead will make it easier to identify and isolate other problems along the way. • Check that you have sufficient resources available for both research and practical tasks. • Reference material can be helpful, but don’t expect to find all the answers. You are carrying out a practical investigation. Try and generate the answers from your own activities. • Use simple or readily available equipment that can be easily modified or adapted as the investigation evolves. Don’t spend the whole time making a piece of equipment that leaves no time or opportunity for investigation. • Establish a timeline for each task and stick to it. Your time is limited and can’t be extended. Ask for advice quickly if you feel the investigation is getting bogged down with problems you hadn’t anticipated.

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• Make good use of a written record. Write everything down. Record ideas, revise your aims, develop a new hypothesis—think as you go. Record both success and failure. You don’t know what will become valuable when it comes time to complete the report. A thorough and easily followed logbook of your investigation will make it easy to write the report.

Your investigation So let’s begin. You may like to refer to the suggested starting points in section 13.4, those on the pearsoned.com.au website, or you may have an alternative source of ideas. Some preliminary reading prior to beginning is essential. And make sure that throughout the investigation and report you use correct scientific standards for units, significant figures and other conventions.

1 What is your topic area of choice? Is it aerofoil design, power versus thrust of a propeller, the angle of attack of a wing?

2 Define your topic in terms of a measurable task. For example, ‘Investigating the lift of an aerofoil under varying wind speeds’ or ‘A determination of the optimum number of blades and the optimum blade angle for a propeller’ or, more simply, ‘Comparing the performance of three different aerofoils with the same cross-section but different chord length’. Make sure that your intent is clear.

3 Write a brief introduction to clarify your intentions both to the reader and to yourself. Consider the propeller example:

‘Early propellers were simple fixed-pitch two-bladed affairs. Later aircraft have used three- and four-blade propellers with variable pitch, and some aircraft have used propellers with more blades. This investigation will compare three propellers using blades of the same length and section, but of varying blade-number, from two through to four. The angle of attack of the two-blade propeller will be varied to investigate the effect of this variable.’

4 Break your topic down into two or three distinct, achievable aims. For our example they may be: a To investigate the comparable effectiveness of two, three and four blades of the same length and section for a constant wind speed. b To investigate the effect on propeller effectiveness of the angle of the propeller blades to the wind direction. c To consider the implications of the practical findings in the design and construction of an aircraft propeller.

5 What variables are being considered in your investigation? Group the quantities you are going to measure: a What you are going to try to measure or change (e.g. blade number, blade angle, wind speed)


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b Investigated variables. The constants of your investigation (blade length, blade profile, wind direction relative to the propeller are some that could be included in our example) c Controlled variables. Outside influences that you will attempt to minimise but which are beyond your immediate control. You may also like to note how you are going to try to minimise or control the effect of these influences. In our example, air temperature may alter air density and affect the results a little with large temperature changes. Height of the blades above a surface and other factors affecting turbulence around the blades would have a marked effect. We’ll attempt to test under similar conditions.

6 Research relevant background material. In our propeller example the following points would be worth discussing in brief: • the history of the development of propellers • how a propeller works • the application of Bernoulli’s equation to propeller design

7 Write up your experimental method and results. Get away from the

An explanation of the use of significant figures, graphs and errors in analysing physics results can be found in Appendix C ‘Understanding Measurement’.

standard recipe style here. You are not instructing someone on what to do but rather telling them what you did. A narrative style interspersed with your raw data, diagrams of equipment arrangement, perhaps photos of special equipment would all help to tell the story. This is the stage where a well-kept logbook will help you.

8 Analyse your results. Make sure tables are constructed to show clearly the controlled and measured variables. Graphs are a very useful means of analysing results. Electronic measuring can have significant advantages. By producing graphs as you work, trends and errors are highlighted, and you can make changes to your experimental design as you go.

9 State your conclusion. Your conclusion should simply state your results in a manner that answers your original aims. A valid comparison could be included.

10 Evaluate your results. Were your results what you expected? Are four blades really best for aircraft or are other variables influencing that choice? How does pitch need to be varied with wind speed? Your evaluation should discuss to what degree and with what degree of certainty your investigation achieved your aims. Avoid the ‘It was really good and I learnt a lot’ style. Explain the science and the implications your results have on the larger scale.

11 List your references and/or bibliography. Credit where credit is due. Your references should be correctly and completely annotated together with appropriate recognition of partners. References are usually listed alphabetically based on the (first) author’s last name, with the title of the published work in italics: Adams, Phillip 1987, ‘Black and white and read no more?’, Weekend Australian Magazine, 7–8 Feb., p.2. Angelucci, Enzo 1984, World Encyclopedia of Civil Aircraft, Willow Books, London.

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13.4 Investigation s tar ting

points

Throughout this study a number of suitable starting points or ideas for investigation have been suggested. While by no means comprehensive, the following are some suggestions for investigation requiring little equipment beyond what could reasonably be expected in a senior secondary physics room. Further ideas, background and information are available from a variety of sources. The website pearsoned.com.au/physics will publish regular updates and links to selected worthwhile resources. Video analysis, or force sensors and loggers, provide a more reliable method of recording real-time maximum and minimum forces than will simple spring balances or beam balances. A simple and cheap video analysis can be achieved using a video played back through a television and traced onto graph paper. Check on the availability of other electronic measuring alternatives or simulation software with your teacher. • Investigating airflow in a wind tunnel—Investigate the effect of shape, speed and basic aerofoil support structures on the flow of air through a simple wind tunnel. Improve the design of the wind tunnel and observe the effect on airflow. • Power versus thrust of a propeller (I)—Vary the voltage supplied to an electric motor driving a small propeller. Measure the resulting thrust produced by measuring the displacement of an object suspended in the airflow from the propeller. Graph and compare power versus thrust for two different propellers. • Power versus thrust of a propeller (II)—Use a rubber-band-powered aircraft to investigate power versus thrust. Graph the number of twists of the rubber band versus flight range and comment on the results. • Efficiency of a propeller—Compare length, pitch, number of blades, or any other variable associated with propellers. Graph speed of rotation against airspeed for each propeller and comment on comparative efficiencies. • Modelling and testing aerofoils—Construct a simple wind tunnel using the plans provided. Test varying size, shape and thickness of aerofoils. Graph lift against airspeed for each aerofoil. Use your graphs to compare the flight characteristics of each aerofoil. • Angle of attack—Construct a simple wind tunnel and a standard aerofoil shape. Vary the angle of attack and measure the lift generated for a constant airspeed. Graph angle of attack versus lift and comment on the results. • Control surfaces—Suspend a simple model aircraft in a wind tunnel. Attach measuring lines (i.e. lines attached to force measurers) to the sides of the model. Vary the position of the vertical stabiliser and/or rudder. Graph rotational force measured versus wind speed and comment on the results. Vary the profile of the model or the size of the stabiliser. Graph size versus force and comment on the results. • Simulating force—Use a force simulation program to simulate the forces acting on an aircraft. Vary the size of the forces and investigate the effect on take-off, climb, cruise and descent. Comment on your results.


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Making a model wind tunnel Introduction Wind tunnels need not be expensive or large in order to achieve meaningful results. A simple wind tunnel that will allow testing of aerofoils and other simple aerodynamic models can be constructed in just one lesson. The shape of an aerofoil is the most important factor in modern aircraft design. A well-designed wing should be able to do three things: • lift the aircraft into the air • keep the aircraft stable in flight • ensure maximum fuel efficiency during flight. Consider working as a team; while you are making the wind tunnel, other team members can be making aerofoils for testing. Plan carefully before starting. Allocate tasks, set a time limit and stick to it. Your investigation is judged on results not on the model!

Requirements Substitute similar items where necessary. (Refer to Figure 13.15 to ascertain the purpose of each item.) • large, strong cardboard box • small electric fan • two or more pulleys as guides for the measuring line • triple beam balance or spring balance and retort stand (Substitute a force sensor and logger for real-time measurement and recording of instantaneous values as carried out in industry. Video cameras and video analysis through software tools such as Video Point can also be useful.) • light fishing line • drinking straws or other rigid tubes • balsa wood, perspex, card or other material for making aerofoils

Procedure 1 Open the box at both ends to create an open ‘tube’ (the squareness doesn’t matter).

2 Locate and set up a small electric fan to provide the airflow. A fan with variable speed settings is ideal, as it increases the variables available for investigation. You may find it useful to cut some windows in the sides of your ‘tube’ and cover them with clear plastic to allow viewing or videotaping of the test in real time. 3 Find the centre of gravity of your aerofoil. Attach two short straws or other rigid tubes to the sides as shown in Figure 13.15b. 4 Fix the aerofoil inside your wind tunnel by threading two pieces of fishing line through the drinking straws and fixing the line securely to the top and bottom of the cardboard box. It is essential that these lines are taut and as straight as possible. 5 Fix another piece of line to the bottom of the aerofoil, pass it through the pulleys and attach it to the force measurer you’ve chosen. The pulleys are used to ensure that the fishing line is free to move with the application of small forces. Check that the line is free to move.

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6 Record the following measurements for each aerofoil: • the length of upper and lower surfaces of the aerofoil (a string laid along the surface may assist you with measurement along a curved surface) • the mass of each aerofoil • the lifting force recorded by the force measurer. 7 Introduce smoke from a burning candle or mist from the vaporising of dry ice (solid carbon dioxide) to make the airflow over the aerofoil visible. (a)

triple beam balance aerofoil

pulley

electric fan

(b)

aerofoil

Figure 13.15 A model wind tunnel showing (a) the final model and (b) the standard mounting of an aerofoil in the tunnel.

Analysis Graph the relationship between the upper length of the aerofoil and the measured lift force.

…xtension Consider how you can modify your basic arrangement to test the effect of the angle of attack of your aerofoil. chor d line

angle of attack

aerofoil

airflow

Figure 13.16 Angle of attack or pitch angle of an aerofoil.


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chapter 14

e l b a n i a t s u s s e : c s r n u o o i s t a y g g i r t e s e n v e n I outcome by the end of this chapter you will have covered material from the study of electricity including: • the concept of energy efficiency in energy transfers • differentiating between the terms sustainable and renewable as applied to energy resources • a comparison of the potential (commercial, environmental and technological) of renewable energy sources • a practical investigation of a renewable energy source • a possible implementation of a chosen renewable energy source • a determination of the potential energy efficiency of a chosen alternative energy system • an analysis of the potential of a renewable energy source for commercial implementation including costs, benefits, limitations, and social and environmental impacts.

On completion of this chapter, you should be able to use concepts of energy transfer and transformations to design, conduct and report on an experimental investigation into an aspect of a renewable energy supply system.

‘World at Crisis Level by 2030’ ‘Barrier Reef to dissolve beyond rate of recovery by 2030’

H

eadlines of doom—barely a month goes by these days in which some newly released report doesn’t foretell a very different world in the very near future. Although some sceptics remain, human endeavour is placing an increasing demand on fossil fuels and the environment, and our society is fast approaching crisis point. Whether it is climate change or the legitimate demands of developing nations to enjoy the life styles we have come to expect, our use of fossil fuels must be significantly reduced, or stopped, sooner rather than later. The ‘Oil Age’ needs to be nearing its end. In searching for alternative ‘environmentally friendly’ energy sources, the many constraints that need to be considered are often not fully appreciated. …very energy source has an environmental and/or social impact. Many are limited in terms of location or commercial potential. What works in a laboratory model is not always scalable to production levels. Any investigation of viable alternatives needs not only to apply science and technology but also to assess the impact on society as well as the environment. This study is your chance to investigate in a practical fashion what may be the defining issue for your generation—sustainable energy.

14.1 Energy transfo rm

ations

We all take steps to conserve energy whether by choice or ‘persuasion’. Turning off that extra light saves both money and electrical energy. In physics terms conserving energy takes on a different meaning. The principle of conservation of energy is one of the fundamental laws of physics on which many other ideas are based.

Energy cannot be created or destroyed; it can only be transformed from one form to another. The total energy in a system is always conserved. In other words, while we may look for alternative energy sources we can’t just make new energy on demand. We can only transform the stores of existing energy in new and/or more efficient ways. While the total energy in a closed system may remain the same, in any physical activity energy is continually being transformed. Even simple activities can result in many different transformations. Figure 14.1 illustrates some of the means by which energy can be transformed from one type to another. electric lamps

thermocouples

photoelectric cells

electric fires electrical energy

high grade energy

electric motors electrolysis dynamos

batteries

light energy

photosynthesis

explosions

chemical energy

metabolism chemical synthesis

burning fuels

mechanical energy

movement

muscular energy

friction

steam engines heat energy

low grade energy

Figure 14.1 Energy is neither created nor destroyed; it can be transferred or transformed. The changes in the form of energy that take place during various activities can be followed by tracing the direction of the arrows in the diagram.

flames

Furthermore, not all energy is regarded as having the same value or usefulness. Some forms, such as electrical and chemical energy, are more easily transformed to other useful forms. These are referred to as high-grade energy. Heat energy, on the other hand, can be quite difficult to transform to alternative useful forms. Heat is a form of low-grade energy. The total energy is still the same and the available energy can still do some useful work, but it is harder to harness for a particular task.

Chapter 14 Investigations: sustainable energy sources

493

A simple example is the dropping of a ball. As the ball falls towards the floor there will be a number of energy transformations taking place. Figure 14.2 illustrates those changes. While the ball falls, gravitational potential energy becomes kinetic energy, the energy of movement. When the ball collides with the ground, kinetic energy is converted into elastic potential energy by compression then back to kinetic energy and, subsequently, to gravitational potential energy as the ball rises. gravitational potential energy

kinetic energy

gravitational potential energy kinetic energy

Figure 14.2 In a simple energy transformation, a ball bounces and high-grade kinetic and potential energy are gradually transferred to surrounding objects as less useful low-grade energy forms, particularly heat.

Table 14.1 High grade forms of energy Energy form

Description

Mechanical

Includes kinetic energy, gravitational potential energy, elastic or spring potential energy

Nuclear

Fission and fusions reactions, radioactive decay

Chemical

Stored potential energy in fossil fuels

Solar

Radiation from the Sun

Electrical

Associated with movement of electric charges

For more on the study of mechanical energy refer to Chapter 6.

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heat and sound elastic potential energy transfer to surface

These are the obvious changes but they are not the only ones taking place. As the ball hits the ground, its temperature rises and both heat and sound are produced by friction. It can seem, by the gradually reducing rebound, that energy is being ‘lost’ from the system; but, in reality, on each successive bounce a percentage of the initial potential energy is being transformed, producing heat and sound. The ball eventually stops not because it has used all of its energy, but because the energy has been transformed to other forms and transferred to the surrounding environment.

Energy in review Energy is a measure of an object’s ability to do work. Increasing an object’s temperature, or lifting an object, is referred to as ‘doing work’. Energy, and work, are measured in joules (symbol J, named after James Prescott Joule in honour of his pioneering studies into heat as energy). Kinetic energy is the energy of movement. It is equal to the work required to bring an object to rest from its present speed. Potential energy is stored energy. It is the amount of energy an object possesses, which would, if used, allow the object to move from one speed to another. There are many forms of potential energy: gravitational, chemical and nuclear are three of importance in the investigation of energy sources. Fossil fuels are stored chemical energy. Work is done whenever energy is transformed from one form to another. Work is usually defined as the product of force and displacement: W = Fx Work done is a measure of the energy transformed and so is measured in joules. The total amount of energy remains the same. The rate at which energy is transformed from one form of energy to another is the power, P, developed, measured in watts (W): E P= t

A 100 W light globe will use 100 J of energy every second. The meter of an electrical switchboard actually reads the total consumption of electrical energy in kilowatt hours (kW h) rather than joules; this is the unit used by an electricity supplier in calculating the cost of the electricity consumed. One kilowatt hour is the energy used in 1 hour by an electrical appliance using energy at the rate of 1000 J per second (1 kW h). In joule: 1 kW h = 1000 J × 60 min × 60 s = 3.6 × 106 J or 3.6 MJ In terms of electrical energy, power is also equal to the product of potential difference and current: P = VI Other forms of energy supplied to households, such as gas, are charged in terms of equivalent heating values. Gas is charged per megajoule of heating. The heating values of some conventional fossil fuels are given in Table 14.2. These values represent the quantity of chemical potential energy released when the fuel is burnt. Australia is one of the largest users per capita of energy in the world. A typical household may use about 8000 MJ per month. That is well over 30 times the amount used in a household in a developing country.

Worked example 14.1A The energy transformed from chemical potential energy to heat and light by completely burning one match is approximately 2000 J. a If this takes 5 seconds, how much power does the match develop? b If the match were held upright so that it took twice as long to burn, what effect would this have on the power produced?

Solution a … = 2000 J, t = 5 s … 2000 P = , so P = = 400 W t 5 b With t = 10 s: 2000 P = = 200 W 10 That is, with the time doubled the power is halved.

For more information on electrical energy and electrical circuits that may be useful in your investigation see chapters 2 and 3.

Table 14.2 Approximate heating values of some conventional fossil fuels Fuel

Heating value

Gas

39 MJ m-3

Wood

17 MJ kg-1 depending on type and moisture content

Briquettes

22 MJ kg-1 depending on moisture content

Coal

25 MJ kg-1 depending on moisture content

Heating oil

45 MJ kg-1

The efficiency of energy transformations The percentage of energy that is transformed usefully by a device is the efficiency of the device. If we are to make existing fuels go further, and make alternative energy sources more viable for commercial development, then our machines need to be efficient in transforming energy from one form to another. Compact fluorescent lights and LEDs for lighting, and hybrid-powered cars are some current practical examples of what can be achieved. In each case less energy is transformed into wasted heat energy and more to the desired application (such as lighting).


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Joule 1025 per year energy received by Earth from the Sun per day

world energy consumption per year

1023

severe earthquake

hurricane

Table 14.3 Comparative efficiencies of common devices 1017

Device

Desired energy transformation

1015

Electric motor

electric—kinetic

90

Gas heater

chemical—heat

75

Compact low-energy fluorescent light

electric—light

50

Steam turbine

heat—kinetic

45

Solar cells

radiation—electric

20–35

Coal-fired electric generator

chemical—electric

30

Human body

chemical—kinetic

25

Car engine

chemical—kinetic

25

Open fire

chemical—heat

15

Incandescent globe

electric—light

10

first atomic bomb Apollo moon rocket

1013

energy from 1 tonne of coal

1011

109 energy from 5 litres of petrol

107 1 kW electric heater for 1 hour (1 kW h)

useful energy transferred Efficiency (%) = × 100 total energy supplied useful output = × 100 energy input

1021 1019

hydrogen bomb

Still, no everyday transformation can be considered to be 100% efficient. That can only occur at the atomic level. All practical transformations ‘lose’ some energy as heat and as other forms of energy that are of no value to the required application. The efficiency of an energy transfer from one energy form to another is expressed as:

106

1 MJ

Efficiency (%)

105 energy from a completely burnt match kinetic energy of a bullet

Worked example 14.1B 103

1 kJ

The energy input of a gas-fired power station is 1100 MJ. The electrical energy output is 300 MJ. What is the efficiency of the power station in achieving this energy transfer?

10 1

1J

10–1 10–3 moonlight on your face for 1 s

10–5 energy needed to lift a hair through 1 mm

10–7 10–9

energy from fission of one uranium atom

10–11 kinetic energy of an alpha particle

10–13

Figure 14.3 One joule is a very small amount of energy. It takes 4200 joule just to raise the temperature of 1 kg of water by 1°C. This diagram illustrates the comparative amounts of energy available from several energy sources. Particularly notable is the energy received from the Sun. The potential for the harnessing of just this one energy source is obvious.

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Solution

energy output Efficiency (%) = × 100 energy input 300 = MJ × 100 = 27% 1100

Entropy While some systems may be more efficient in their conversion of energy from one form to another, the ultimate outcome in any energy transfer is the loss of some high-grade energy. No energy is ever lost, but in any energy transfer some of the original energy becomes unavailable for useful work. As time goes on all forms of energy degrade from orderly, highgrade energy to disordered heat or thermal energy. In the 1860s, R.J.E. Clausius introduced the term entropy to describe or measure this process. Entropy can be interpreted as a measure of the order or disorder of the energy in a system. A natural outcome of this idea is a theory that describes the end of the universe. The energy available in the universe will become increasingly low grade. Heat will have flowed from high temperature regions to areas of low temperature until the whole universe is at one constant temperature. All the available energy will have degraded to thermal energy. No work will be possible and all change will cease. The universe will die not with a bang but a whimper, the so-called heat death.

It sounds like there really may be no future, but there remains an area for optimism. Conservation of energy is based on a closed system. Nothing is transferred in or out. That means that for the above scenario to be true the universe must be finite, i.e. have definite boundaries. Cosmologists aren’t so sure that assumption is true.

14.1 summary Energy transformations • The law of conservation of energy is a fundamental principle in physics and states that energy is neither created nor destroyed but can be transformed from one form to another and is always conserved. • Any energy transfer will result in a higher-grade energy form being at least partially transformed into

a low-grade, less useful form such as heat. This is referred to as entropy. • The efficiency of an energy transfer from one form to the required form is: energy output Efficiency (%) = × 100 energy input

14.1 questions Energy transformations 1 Identify the main transformation that occurs when: a an electric bell rings b a ball is thrown upwards c water falls from the top of a dam d sunlight strikes a solar panel.

The following information relates to questions 6 and 7. A diesel truck engine uses 1400 MJ of energy to do 550 MJ of useful work.

2 Energy forms can be loosely grouped as high-grade useful energy or low-grade non-useful energy. Label each of the energy types identified inquestion 1 as high grade or low grade.

7 How can you account for the ‘lost’ energy? The following information relates to questions 8–10. A filament light globe has an efficiency of approximately 10%.

3 Name the forms of energy represented at each change by the letters A, B, C and D in the following statement: ‘In generators fired by brown coal, the energy from the coal (A) is used to change water into steam (B). The steam drives a turbine (C), which drives a generator (D).’ The following information relates to questions 4 and 5. A particular model of reverse-cycle air-conditioner produces 12 kW of useful heat from 4.8 kW of input electrical power.

8 If 3600 J of electrical energy are supplied to the globe, how much is converted into light energy?

4 What is the efficiency of the device in producing useful heating? 5 Your answer to question 4 may appear unreasonable as it is based on higher heat output than input. How can this be?

6 What is the efficiency of the engine in performing this energy transfer?

9 What happens to the other 90%? 10 Why is this lost energy referred to as a lower quality energy than the original electrical energy? 11 When a moving car is brought to rest, what happens to the original kinetic energy of the car? Could this energy be used for a useful purpose? Explain. The following information relates to questions 12 and 13. An electric kettle uses 3.60 kJ of energy to heat a particular quantity of water. The efficiency of the kettle itself is 70%. 12 How much electrical energy is expended in actually boiling the water? 13 What has happened to the remainder of the energy?


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able— n i a t s u s r o e l Renewab 14.2 the key to our future In the debate over what alternative energy sources are best developed, one key factor is often forgotten: the development of any resource will always have an environmental and social cost. Choosing any source needs careful consideration of the net gain over societal and environmental costs. A simplistic investigation of alternative energy resources will be based on renewable sources. A renewable energy source, by the very name, can be used and replaced. A good example is wood—for fires to heat homes, to drive steam turbines, or to be pulped and utilised as biofuels in cars. It is readily renewable. Any large expanse of land with reasonable climate can be planted with fast-growing trees and harvested for fuel in about 10 years. Once harvested the land can be tilled and replanted, and so the cycle continues. Carbon that would otherwise be permanently effecting climate change is reabsorbed by the growth of each cycle of trees. One of humankind’s oldest fuel sources, wood is also the easiest to renew. But is wood a sustainable fuel? Growing trees requires large amounts of water and land, and leads to the long-term degradation of soils unless considerable amounts of fertiliser are added over time. Land needed for housing and food crops is otherwise set aside. There is now considerable concern that in developing countries people may starve while land, once used for food crops, is used for growing fuel crops for developed nations. So, it seems that no decision regarding energy sources is ever straightforward. There are always impacts, both societal and environmental, to be considered. Many countries have gone as far as to regulate what fuel sources can be labelled fully sustainable or just renewable. To fully qualify as a carbon energy offset source, a company must establish that an energy source is sustainable.

A renewable energy source is a naturally occurring raw material or form of energy that will be replenished through natural ecological cycles or sound management practices. A sustainable energy source must meet the needs of the present without compromising the ability of future generations to meet their needs. It is this focus on the future that is the key difference. An energy source does not need to be renewable in order to be sustainable. The debate over what constitutes sustainable fuels and how they can be managed will no doubt continue to be a key focus of government and world leaders for many years to come.

Questions The following questions are designed to provoke debate prior to starting a physics-based investigation, and as such they don’t have definitive answers. How you approach the answers will depend on how you weigh concerns regarding the range of potential impacts on society and the environment. Before choosing the topic of your investigation consider your answers carefully. The answers are up to you but be prepared to justify them. 1 List five current energy sources and define each as renewable and/or sustainable. Justify your decisions. 2 List five energy sources currently under development, consideration or small-scale implementation. Grade your list on the basis of the renewable or sustainable nature of each energy source. Justify your decisions.

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Investigating alternative 14.3 energy sources Principles of investigation In an age when the Earth’s resources are being seriously depleted, it is disturbing that Australia continues to rely so heavily on non-renewable fossil fuels. Coal continues to provide more than 40% of Australia’s total energy needs, including 80% of our electricity supply. It is currently, and perhaps unexpectedly, Australia’s main export. Australia has vast reserves of coal and natural gas but is becoming increasingly dependent on imports of oil. Within a few decades Australia will need to import all of its oil requirements, offsetting, to a large degree, our exports of coal. Not only are our reserves of fossil fuels unsustainable in terms of supply, but the burning of fossil fuels produces solid waste and gases that contribute to air pollution and the greenhouse effect, believed to be contributing to the current increases in global temperatures. While the coal industry may yield to international pressure to meet greenhouse gas emission targets, the Kyoto protocol may also cause a major rethink of energy production technologies. If we are to continue to enjoy the lifestyle we have come to expect in a habitable world, then alternative energies must play an increasing role in our energy sources. Renewable energy from water, Sun, wind, ocean, geothermal and biomass are the way of the future; but presently few of these energy sources can compete with fossil fuels on a large scale. Research into the commercial application of these and other energy forms is becoming an increasing priority.

Figure 14.4 This is a cooling tower at the coal-based power station in the Latrobe Valley. Australia’s reliance on fossil fuels, and particularly on coal, is likely to become unsustainable in the near future.

Thermal efficiency (%)

33 32 31 30 29 28 27 26

countries 1974

1979

1984

1989

Figure 14.5 The efficiency of Australia’s coal-fired power stations has increased over time. Australia now has some of the world’s most efficient coal-fired power stations; however, they still only extract about 33% of the energy from the coal they burn. Each 1% rise in efficiency has the potential to reduce carbon dioxide emissions by 2.5%. Research and investigation into improved burning methods continue. (Figures supplied by the Australian Department of Primary Industries and Energy.)

Simply adopting new technologies or energy sources is not enough. Alternative energies must be sustainable in terms of supply, development, economics and environmental outcomes. The process of investigation involved in determining the suitability of the various alternatives is one that must consider not only the science but also the viability of the technologies required and the effects on society. Ecologically sustainable development principles must underpin any sustainable energy project.

Table 14.4 National average thermal efficiencies for power generation for selected countries Country

Coal

Gas

Australia

33.3%

31.1%

Canada

36.8%

43.1%

West Germany

35%

32.4%

Italy

35.5%

14.31%

Japan

38.4%

43.6%

Sweden

17.3%

–

UK

36.8%

45.6%

USA

36.9%

31.6%

Source: Energy Balances of OECD Countries 1987–1988, IEA/OECD, 1990.


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This is a process we can model on a small scale in a meaningful investigation of our own design. The process of investigation is a quite different form of activity to the traditional teacher-structured practical activity for which often the outcome is already known. It is quite probable that your results will not have been duplicated nor be available in any published resource. A few simple guidelines will assist you in making these results meaningful. The following guide will assist you in planning your investigation.

Getting started The choice of topic is often the biggest hurdle to overcome. Be imaginative and investigate, within the constraints of the study of sustainable energy, a topic of genuine interest to you. Your investigation doesn’t need to be a purely physics-orientated one. It should consider commercial, environ mental and other more general measures of sustainability, but it should make use of the established principles of scientific investigation. The topic should be simple enough for you to begin immediately. Here are some simple guidelines that can help you make a good choice. • Straightforward topics are almost always better than more complicated ones when just getting started. Don’t choose a topic for which the background research occupies most of the available time. Some starting points for investigation are included in section 14.4. • Develop some preliminary aims as part of your initial topic choice. They may change as your investigation develops, but they will be useful for checking that your topic is not too complicated. Focus your aims on particular, measurable tasks. Broad aims are hard to achieve. • Think about what you intend to do before you start investigating. Identify potential problems and plan ways of overcoming them. Thinking ahead will make it easier to identify and isolate other problems along the way. • Check that you have sufficient resources available for both research and practical tasks. Fortunately, renewable energy is an area for which good material is easy to find; just don’t lose sight of your aim when researching your topic. • Reference material can be helpful, but don’t expect to find all the answers. You are carrying out a practical investigation. Try and generate the answers from your own activities. • Use simple or readily available equipment that can be easily modified or adapted as the investigation evolves. Don’t spend the whole time making a piece of equipment that leaves no time or opportunity for investigation. • Establish a timeline for each task and stick to it. Your time is limited and can’t be extended. Ask for advice quickly if you feel the investigation is getting bogged down with problems you hadn’t anticipated. • Make good use of a written record. Write everything down. Record ideas, revise your aims, develop a new hypothesis—think as you go. Record both success and failure. You don’t know what will become valuable when it comes time to complete the report. A thorough and easily followed logbook of your investigation will make it easy to write the report.

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Your Investigation 1 What is your topic area of choice? Select one form of sustainable energy. 2 Define your topic in terms of a measurable task. For example, ‘Investigating the efficiency of various fuels in an alcohol-based fuel cell’, or ‘A determination of the optimum number of blades and blade angle for a wind turbine’, or, more simply, ‘Comparing the performance of three different solar water-heaters’. Make sure that your intent is clear.

Many investigations into alternative energies have the potential for producing high temperatures, high currents or other potentially dangerous practical conditions. Make sure you read and follow appropriate safety procedures throughout the investigation.

3 Write a brief introduction to clarify your intentions, both to the reader and to yourself. Consider the wind turbine example: ‘Commercial wind farms generate electricity through the action of the wind on aerofoil blades. Modern designs have seen efficiencies rise from about 5% through to 35% or better. This investigation will use aerofoils attached to a small motor to investigate the efficiencies of different numbers of blades at different angles to determine the most efficient arrangement.’ 4 Break your topic down into two or three distinct, achievable aims. For our example they may be: a To investigate the efficiency of two, three and four blades of the same length and section for a constant wind speed. b To investigate the effect on turbine efficiency of the angle of the turbine blades to the wind direction. c To consider the implications of the practical findings in the design and construction of a commercial wind turbine. 5 What variables are being considered in your investigation? Group the quantities you are going to measure: a What you are going to try to measure or change (e.g. blade number, blade angle, wind speed) b Investigated variables. The constants of your investigation (blade length, blade profile and wind direction relative to the turbine are some that could be included in our example) c Controlled variables. Outside influences that you will attempt to minimise but which are beyond your immediate control. You may also like to note how you are going to try to minimise or control the effect of these influences. In our example, air temperature may alter air density and affect the results a little. Height of the blades above a surface and other factors affecting turbulence around the blades would have a marked effect. We will attempt to test under similar conditions. 6 Research relevant background material. In our wind turbine example the following points would be worth discussing in brief: • the causes of wind • the environmental credentials of wind power • basic physics ideas about energy, power and efficiency • the power available from wind energy • the fundamentals of wind turbine design.


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7 Draw a flow chart showing the energy changes taking place in your complete system. We’ve kept this one simple in case you’ve chosen wind power. See if you can consider all the energy changes that are taking place in your system. For example, what created the wind in the first place? Try and include the percentage of energy transformed in each case.

wind

wind

Figure 14.6 Some of the energy changes in a

wind turbine.

kinetic energy

rotation of turbine kinetic energy + sound lost from + heat system

}

generator electrical energy + heat (lost from system)

transmission lines electrical energy + heat (lost from system)

8 Write up your experimental method and results. Get away from the standard recipe style here. You are not instructing someone on what to do but rather telling them what you did. A narrative style interspersed with your raw data, diagrams of equipment arrangement, perhaps photos of special equipment would all help to tell the story. This is the stage at which a well-kept logbook will help you. 9 Analyse your results. Make sure tables are constructed to show clearly the controlled and measured variables. Graphs are a useful means of analysing results. Electronic measuring can have significant advantages. By producing graphs as you work, trends and errors are highlighted, and you can make changes to your experimental design as you go. An explanation of the use of significant figures, graphs and errors in analysing physics results can be found in Appendix C.

PRACTICAL ACTIVITY 56 Electrical power PRACTICAL ACTIVITY 57 Hydroelectric power PRACTICAL ACTIVITY 58 Sunlight intensity and reflectivity of Earth’s surface PRACTICAL ACTIVITY 59 Solar energy: generating electricity

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10 State your conclusion. Your conclusion should simply state your results in a manner that answers your original aims. A valid comparison could be included. 11 Evaluate your results. Were your results what you expected? Are two blades really best for wind turbines or are other variables influencing that choice? Is wind energy a viable alternative energy? Your evaluation should discuss to what degree and with what degree of certainty the investigation achieved your aims. Avoid the ‘It was really good and I learnt a lot’ style. Explain the science and the implications your results have on the larger scale. 12 List your references and/or bibliography. Credit where credit is due. Include any books or articles you have read in your bibliography. A list of references will only include those references that you have actually cited in your investigation. Your references should be correctly and completely annotated together with appropriate recognition of partners. References are usually listed alphabetically based on the [first] author’s last name, with the title of the published work in italics: Adams, Phillip 1987, ‘Black and white and read no more?’, Weekend Australian Magazine, 7–8 Feb., p.2. Angelucci, Enzo 1984, World Encyclopedia of Civil Aircraft, Willow Books, London.

14.4 Starting points Before starting your investigation try working through the five short practical investigations of energy efficiency. Key ideas about renewable energy, energy transfer and energy efficiency are established. Use the same approach or refer to these results to complete your own investigation into a renewable energy system.

PRACTICAL ACTIVITY 60 Solar energy: generating electricity II PRACTICAL ACTIVITY 61 Wind power

Practical activities

PRACTICAL ACTIVITY 62 Solar constant

• Generator: Transfer of gravitational potential energy to electrical energy • Hydroelectricity: Water spinning a small generator • Efficiency of energy storage: Charge and discharge of capacitors • Energy transfer—solar (direct): Incident radiant energy to thermal energy • Energy transfer—solar (photovoltaic): Incident energy to electrical energy through a solar cell

PRACTICAL ACTIVITY 63 Energy efficiency of a fuel cell

Table 14.5 Comparative energy-conversion efficiencies INPUT

OUTPUT Mechanical

Gravitational

Mechanical Gravitational

Electrical

Radiant

Chemical

99% (generator) 86% (water turbine)

Electrical

100% (brake drum)

85% (hydroelectric) 80% (pumped storage)

Radiant

Thermal

40% (gas laser)

72% (storage battery)

25% (solar cell)

100% (heating coil)

0.6% 100% (photosynthesis) (solar furnace)

Chemical

30% (muscle)

91% (dry cell battery)

15% (chemical laser)

Thermal

47% (steam turbine)

7% (thermocouple)

3% (light bulb)

Starting point: agricultural waste and bioenergy Organic wastes are produced as a result of a wide range of agricultural activities and include crop residues, animal manure and poultry litter. These wastes can be used to produce energy using a range of technologies, including direct combustion, gasification, anaerobic digestion (to produce biogas), etc. Crops, such as sugar cane, provide large quantities of material that can be converted to fuels and fuel additives. While the addition of ethanol to petrol gained some notoriety in the early 2000s here in Australia, in other countries such as Brazil cars have been specially adapted and ethanol is becoming an important component of the fuel that is available. Concerns remain about the environmental impact in Australia of largescale water-intensive farming.

88% (furnace of steam boiler)

Many of these starting points will require theory from the study of electricity. Refer to Chapters 2 and 3 to review the key ideas you will be applying in your investigation.


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Figure 14.7 Agricultural waste is now a serious alternative energy source.

Table 14.6 Typical energy value of fossil and biomass fuels Solid fuels

Net heating values (MJ kg-1)

Biomass fuels Wood (wet, freshly cut)

10.9

Wood (air dry, humid zone)

15.5

Wood (air dry, dry zone)

16.6

Wood (oven dry)

20.0

Charcoal

29.0

Bagasse (wet)

8.2

Bagasse (air dry)

16.2

Coffee husks

16.0

Rice hulls (air dry)

14.4

Wheat straw

15.2

Corn (stalks)

14.7

Corn (cobs)

15.4

Cotton stalks

16.4

Coconut husks

9.8

Coconut shells

17.9

Fossil fuels Anthracite coal

31.4

Bituminous coal

29.3

Lignite

11.3

Coke

28.5

Source: Bioenergy Australia, no. 17 (Sept 2002), Bioenergy Australia, Killara NSW.

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Starting point: wind (a)

(b) rotor blade

generator rotor

generator stator

wind measuring device (anenometer and vane)

pitch box

battery boxes

rectifier units

relay box

blade adapter

subdistribution control

pitch motor

winch

slip ring casing axle pin

fan main bearings

excitation controller

tower

main carrier sub

rotor subdistribution

stator jib carrier

yaw motors

Figure 14.8 (a) For windmills built before 1900, efficiency was less than 5%. Considerable improvements in the design of modern wind turbines have achieved efficiencies of about 30%. The theoretical maximum in most situations is 59%. (b) The main features of a modern wind turbine.

In August 1998, Australia’s first large-scale wind farm commenced opera tion near Crookwell in New South Wales. Since then, wind farms have been established at Toora in Victoria’s South Gippsland region, at Challicum Hills near Ararat, and at Codrington near Portland. Others are under consideration along much of Victoria’s southern coast. The world has never been short of wind. For thousands of years it has turned windmills, flown kites, cooled houses and filled sails. Now technological advances are allowing us to use wind as a clean, renewable, cost-effective means of generating electricity. A drawback to wind power is that wind can be highly erratic in both direction and speed. But many of the problems of wind power are now being solved; for example, locating wind turbines in areas where the wind blows regularly and at optimum speeds. In Australia, important advances have been made in this regard. Researchers at CSIRO have used computer models of wind flow over complex terrain, together with extensive wind measurements, to calculate potential wind energy at different locations. Other methods are being employed to catch the best of the available wind. For example, wind is slowed by friction with the land surface. Modern wind turbines are therefore mounted on towers 40–60 m high to expose the blades to a higher wind speed.

Physics file It is easy to fall into the trap of considering all fossil fuels as badly polluting and alternative energy as pollution-free. That is far from the truth. The demonstrations of the early 1980s over the hydroelectric dams in Tasmania, and the more recent concerns over wind farms, are two cases in point. While there is no suggestion by the demonstrators that they will pollute the air, concerns have been expressed regarding the potential effects on existing ecosystems, noise and tourism. A consortium was planning to build 78 wind turbines, each the height of a 30-storey building, near Port Campbell National Park. The proposal fell foul of tourism authorities and the National Trust, as the towers would have been clearly visible from the Twelve Apostles and the Bay of Islands, both popular tourist attractions on the Great Ocean Road. The Victorian Government was committed to generating 1000 MW of Victoria’s electricity from wind power by 2006. This figure is yet to be met. This compares with 3000 MW being generated by just one coal-fired power station.


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Physics in action

Nuclear energy Nuclear energy from the fission reaction has proven to be a viable and cost-effective alternative energy to fossil fuels. Countries poor in fossil and other alternative energy sources are large-scale implementers of fission reactors. France currently produces about 80% of its electrical energy from nuclear fuel. The basic principle of fission reactors is the same as for fossil-fuel plants. Water is heated to form steam. The steam is used to turn a generator and produce electricity.

Figure 14 .9

506

Nuclear p ower

accounts for

Detailed studies

approxim ately

80% of Fra

The emissions of greenhouse gases and solids from fission reaction are much less than the emissions produced by the use of fossil fuels. However, potential radioactive leaks and the long-term problems associated with the disposal of spent fuel rods offset the benefits in the minds of many. The future of clean, efficient power from nuclear energy lies in fusion, a process that is yet to be implemented on a commercial scale. Nuclear energy is not suitable for a practical investigation at secondary level. Students interested in finding out more background for research investigations can refer to Chapter 12 ‘Energy from the Nucleus’.

nce’s ele ctrical en ergy.

Starting point: hydroelectricity Hydroelectricity (or hydropower) is one of the oldest and most widely used forms of renewable energy. In reality, it is really another form of indirect solar power, relying heavily on the Sun to evaporate sufficient water to fill the dams to provide the falling water. It exploits the mechanical energy of falling water to drive a turbine, of which there is a wide variety of designs. Systems range from a low fall of less than 3 m to systems such as that of the Snowy River where water falls through more than 100 m.

Figure 14.10 In the 1950s the Snowy River Hydroelectric Scheme was the largest engineering project of its kind in the world.

Physics in action

Micro hydro installations Micro hydro is often the most cost-effective method of producing electricity in remote areas of Australia with a running water supply, and is generally the cheapest smallscale renewable energy option. Lemonthyme Lodge is located in the Tasmanian World Heritage Wilderness Area. Lemonthyme has eight lodge rooms, 18 luxury cabins with spas or hot tubs, and five self-contained family cabins. No mains electricity was available at the site, and the closest power lines were more than 2 km away. Clearing

surrounding forest for a transmission line would have been both destructive and expensive. A 25 kW diesel generator was initially installed, but the system proved expensive to run, potentially expensive to maintain, noisy and polluting. Environmental concerns about the impact of the micro hydro system were identified and addressed in the development. No canal or creek diversion was required, and enough water remains in Bull Creek to ensure it remains healthy.

gross head to turbine

gross head to tail water level

intake system

governer controls frequency

pipeline 2260 m

governer

turbine

208 m

generator 90 kVA

Tourist lodge load UÊÊÀivÀ}iÀ>Ì UÊÊ}Ì} UÊÊ>Õ`ÀÞÊv>VÌiÃ UÊÊÃ«>ÊÌÀÃ UÊÊÃiÊÃÌ>ÌÊi>Ì} UÊÊ}iiÀ>ÊiiVÌÀV>ÊiµÕ«iÌ UÊÊxä¯ÊvÊÌÊÜ>ÌiÀ

large capacity water heating system

tailwater level shunt loads

Figure 14.11 The micro hydro system at Lemonthyme Lodge.


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Starting point: ocean energy Energy in the oceans and seas can be harnessed in a number of ways. Tidal barrages capture water at high tide, and tidal-stream devices use tide-induced currents to generate power. Devices to capture the energy in waves can be located on the shoreline or some distance offshore. It is also possible to generate power from the thermal gradients in seawater at suitable locations. The Portland Wave Energy Converter (WEC) is being installed off the coast of Portland, Oregon, USA, and will be, once connected, the first commercial offshore wave power generator in the world to provide power to the general power grid. The PowerBuoy system developed by Ocean Power Technologies (OPT) is able to harness wave energy using a piston-like device in the top of a buoy. A computerised system housed in a watertight canister at the top of the buoy allows uniform power to be derived from the random motion of ocean waves. The power will then be carried via underwater cables to shore. The prototype wave energy converter was deployed in September 2007. Each buoy generates, on average, 20 kW of electricity, enough to power about 25 households. Depending on the needs of the area, the system can be ‘scaled up’ by connecting more buoys. The OPT system will usually be placed more than a mile offshore, and is mostly below the water, so it can’t be seen from the shore. The system’s minimal visual and noise impacts constitute one of its major advantages.

Figure 14.12 The Portland Wave Energy Converter is intended to be the first commercial offshore wave power generator in the world to be connected to the main grid should development be completed, as planned, in 2010.

Starting point: solar energy (heat)

Figure 14.13 Solar panels can be used to provide electricity and hot water for homes.

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Detailed studies

In an age when the Earth’s resources are being seriously depleted it seems incredible that the Earth receives 1.8 × 1017 J of radiant energy from the Sun every second. This means that the total energy output from the Sun in all directions is 3.9 × 1026 J s-1. Radiant energy from the Sun is distributed across all types of electromagnetic radiation, with the predominant amounts lying within or close to the visible spectrum. When the Sun is directly overhead, each square metre of the Earth’s upper atmosphere receives an average of 1365 J of energy per second. At the autumn and spring equinoxes, when the Sun is directly above the equator, on a relatively clear day every square metre of upper atmosphere at Melbourne’s latitude will receive about 1000 J of energy each second at midday. Further south, at about 60º latitude, this falls to 600 J. As spring

progresses into summer, the orientation of the Earth’s tilt brings the Sun more directly overhead and increases the amount of radiant energy received over each square metre. Energy from incident sunlight, in the form of heat, can be captured by devices that circulate a fluid (water, air or other fluid) and deliver the energy elsewhere. Solar water heaters range from systems supplying a single house or swimming pool to those supplying multiple dwellings or commercial applications. Solar energy can also be used to provide heat for industrial and agricultural processes. Solar cooling technologies have also been developed. Through the use of climate-sensitive building design, solar energy can be used for space heating, natural light and even ventilation. Buildings that exploit these designs range from individual dwellings to large commercial/industrial offices, factories, and even shopping centres such as Melbourne Central. The transfer of heat from solar energy (by convection, conduction, evaporation and radiation) is the most widespread application of sustainable energy.

Physics file Any object that is warmer than its surroundings will transfer heat to the environment through convection, conduction and radiation. Despite the great differences in mechanism and the share each method has in any exchange, a simple approximation can be made for the rate of cooling. This approximation is one of the many natural laws established by Sir Isaac Newton and is referred to as ‘Newton’s law of cooling’.

Q N…WTON’S LAW OF COOLING: = k (To – Ts) t where: To is the temperature of the object Ts is the temperature of the surroundings k is a measure of the rate of cooling (if negative) or heating (if positive) per degree temperature difference Q is the amount of heat energy transferred in joule, and t is the time. Time can be measured in any convenient unit: seconds in a rapid transfer; minutes, hours or days in other situations.

Physics in action

The Solar Tower The Solar Tower sucks warm air through the chimney-like tower to drive a turbine. To be economically viable, the Solar Tower power plant has to be large enough to generate 200 MW of electricity—5 km across and 1 km high. This capacity can power 200 000 homes. The plant is to be built near Buronga, New South Wales, and will cover an area of at least 20 km2, and it will be Australia’s—and the world’s— tallest and largest engineered structure. The 5 km diameter collector system will have a slightly rising roof of semi-transparent glazed material, standing between 2 and 20 m above the ground with steel supports at 6 m intervals. The 4 mm glass (or plastic) will be held by steel horizontal prop girders and transverse steel plates. Under the collector roof, air temperature is about 30°C— quite habitable—and wind speed will be about 32 km h-1 —equally benign. Air speed through the tower itself is constant. Thirty-two small (6.5 MW) turbines will be set vertically in the transition area of the tower, where the collector system meets the tower. The Solar Tower’s supply curve is approximately bell shaped, peaking with the Sun’s intensity. During summer, it is similar to the demand for air-conditioning, and the hotter the day, the more efficient the power plant. If a more even supply of electricity is required, energy can be stored under the collector during the day using black plastic tubing filled

Figure 1

4.14 A pro totype So generate lar d5 10 years 0 kW of electricit Tower successfu lly . It had a y for 15 0 co 0 tower 19 7 m high llector roof 240 m 0 hours over . in diame ter and a

with water. Annual production from the proposed power plant is just over 700 GW h. At time of writing, development was waiting for final investors to sign on.


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Starting point: solar power When certain materials are subjected to sunlight they develop an electric potential that can be used to generate electricity. These photovoltaic (PV) systems can be used for large scale centralised power generation, for decentralised grid-connected power supply, and as a remote power source when there is no grid electricity available. The energy in sunlight can also be concentrated sufficiently to achieve temperatures suitable for thermal power generation. A variety of designs have been used in areas that receive regular high-incident solar energy.

Figure 14.15 The Solar Collector Dish is an example of Australian world-leading design. The dish can be transported to wherever energy is required.

Australia leads the world in the development of photovoltaic cells with ever-increasing efficiencies. While still expensive when compared with fossil fuels, solar cells are now cost-effective enough for use in remote loca tions and even, with the benefit of government subsidies, in some urban installations. Surplus energy can be sold back into the main power grid, increasing the cost-effectiveness. CSIRO is also involved in the research and development of technology for capturing radiant solar energy more efficiently. The Solar Concentrator Dish is constructed in a manner that allows highly efficient collection of radiant solar energy. The dish can be transported to where energy is needed. Physics in action

SunRace sizzles across Australia A scorching 2300 km over 10 gruelling days, the SunRace includes solar, electric and hybrid cars, demonstrating the advances in real applications of photovoltaic cells. SunRace 2003 saw a battle between the reigning champion, Melbourne-based Aurora RMIT 101, and the University of NSW Sunswift II from Sydney. Sunswift II, winner of the inaugural 2002 SunRace Enterprise Award, was arguably the finest of the contenders in the Australian solar car stable. The team to beat, Aurora RMIT 101, winner of SunRace 2002, set the world record average speed for a solar car event at a sizzling speed of 94.5 km h-1 for the SunRace from Adelaide to Sydney. Since then commercial entries have further increased the speeds attained.

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Detailed studies

With cars travelling at speeds of about 100 km h-1 SunRace challenges the ideas of what we will be driving in the future. SunRace has three major objectives: • to create community, corporate and government awareness of sustainable technologies and greenhouse issues • to promote research, development and implementation of sustainable technologies • to engage and encourage young people in the pursuit of career paths in sustainable technology disciplines.

SunRace, winner of the National 2001 Banksia Environmental Award for Communications—Promoting Change Through Informed Participation—is a powerful vehicle for creating public awareness of sustainable energy and transport technologies, not as alternative technologies, but as imperative technologies at a time of global energy concerns. On a less costly and smaller scale, the model solar car competitions, held throughout Australia each October, allow secondary students to build cars using similar technologies for racing against other schools.

Figure

14.16 Su nRac technolo gy and e e demonstrates fficiency th achieved e huge advance s in in photov oltaic-ce ll design .

Starting point: hydrogen Hydrogen is emerging as a major component of clean, sustainable energy systems, relevant to all energy sectors (transportation, buildings, utilities and industry). Ultimately it is hoped to produce hydrogen from renewable energy sources. In the meantime, advances in fuel cell technology have raised the possibility of a viable passenger car before the end of the decade, and Mercedes continues to experiment with a purely hydrogen-powered car. The distribution of such a different (requires high pressure storage) and volatile fuel, together with the need for clean, efficient electricity for the production of the hydrogen from water, provides dual stumbling blocks for hydrogen as a stand-alone fuel. Fuel cells are electrochemical devices that operate at a high level of efficiency with little noise or air pollution. There are many potential applications for fuel cells, including electricity generation in stationary applications and power for a new generation of vehicles. All fuel cells operate on the same principle—they convert chemical energy directly into electricity and heat, rather than oxidise (burn) a fuel. In most, but not all, fuel cells the source of the fuel’s chemical energy is the chemical reaction of hydrogen and oxygen to form water. In some cases, the fuel may need to be processed, or ‘reformed’, before it can be used in the fuel cell. membrane electrode assembly

O2

gas flow channels H2

H2

endplate bipolar plate repeat unit

O2

Figure 14.17 Structure of a hydrogen fuel cell.


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An input fuel is catalytically reacted (electrons removed from the fuel elements) in the fuel cell to create an electric current. As shown in Figure 14.17, fuel cells consist of an electrolyte material that is sandwiched between two thin electrodes. The electrons go through an external circuit to serve an electric load while the ions move through the electrolyte toward the oppositely charged electrode. At the electrode, ions combine to create by-products, primarily water and CO2. Depending on the input fuel and electrolyte, different chemical reactions will occur, producing lesser or greater amounts of electrical power. Proton exchange membrane (PEM) polymer electrolyte fuel cells are the most promising technology for transport applications. Built from components of carbon and polymer and operating at 60–90°C, PEM fuel cells are lightweight and allow fast start up and shutdown. They are also very compact. Australia is competing with multinational companies for a slice of a market said to be worth billions of dollars worldwide. Ceramic Fuel Cells Ltd has developed a solid-oxide fuel cell the size of a 2 L milk carton that produces 1.5 kW; enough power to meet the needs of a typical household. The company has also demonstrated a larger, 5 kW unit, operating it continuously for 200 h. These solid-oxide fuel cells promise to achieve a very efficient conversion of fossil fuels to electricity, while producing very low levels of pollutants.

Figure 14.18 While fuel-cell-powered cars capable of replacing cars with combustion engines are still in their infancy, simple working models like this one demonstrate the principles. The proton exchange membrane is clearly visible, sandwiched between two pieces of perspex, in this working model from Hytec.

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Making a simple electrochemical cell Introduction Electrochemical cells convert chemical energy into electrical energy. Fuel cells are a type of electrochemical cell.

…quipment (per small group) piece of zinc (about 8 cm × 1 cm) piece of copper (about 8 cm × 1 cm) emery paper distilled water (about 40 mL for rinsing) 25 mL of 0.1 M zinc(II) nitrate in a 50 mL beaker 25 mL of 0.1 M copper(II) nitrate in a 50 mL beaker ammeter (centre zero, 0–1 mA with two leads, each with an alligator clip at the free end) or current sensor • filter paper strip (about 8 cm × 1 cm) • saturated solution of sodium nitrate (to make salt bridges) • • • • • • •

Preparation To prepare 1 L of 0.1 M zinc nitrate, weigh 24.3 g of Zn(NO3).3H2O and add distilled water up to 1 L. (For Zn(NO3).6H2O, weigh 29.7 g and add distilled water up to 1 L.)

Procedure +

ammeter

–

20

strip of filter paper soaked in sodium nitrate solution 30

40

50

0

10

copper zinc

solution of zinc(II) nitrate

solution of copper(II) nitrate

Figure 14.19 Experimental set-up for an electrochemical cell.

Set up the equipment as shown in Figure 14.19, using the following steps: 1 Place the piece of zinc into the zinc(II) nitrate solution. 2 Place the piece of copper into the 25 mL of copper(II) nitrate solution. 3 Place the beakers side by side, making sure the zinc and copper are not touching. 4 Connect the alligator clip of one lead to the zinc, and the alligator clip of the other lead to the copper. 5 Observe the ammeter needle or current sensor read out.


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6 Carefully immerse the filter paper strip in the saturated solution of sodium nitrate.

7 Suspend the wet filter paper so that one end is immersed in the zinc(II) nitrate solution and the other is immersed in the copper(II) nitrate solution. This will form a salt bridge. (Make sure that both ends of the strip are immersed.) 8 Observe the value shown on the ammeter. Was there a difference in the ammeter reading before and after you added the salt bridge? Explain your answer.

Background information Before the salt bridge is in place, there should be no ammeter deflection—no current flow. After the salt bridge is added there should be current flow. Electron flow is from zinc to copper. (Electrons are taken from the more reactive metal, zinc.)

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Medic a

l phys

ics

outcome On completion of this chapter, you should be able to describe and explain applications of radioisotopes, optical fibres, waves and lasers to medical diagnosis and treatment, and describe the production and/or simple interpretation of images of the human body produced by the processes of CT, ultrasound or X-rays.

A by the end of this chapter you will have covered material from the study of medical physics including: • the processes of medical imaging using ultrasound, X‑rays and CT • the interpretation of images produced by using ultrasound, X‑rays, CT, MRI and PET • applications of radioisotopes in medical diagnosis and treatment • the operation of optical fibres as endoscopes.

ustralia likes to view itself as a sporting nation. Certainly our climate and standard of living provide opportunities for many Australians to participate in a wide range of indoor and outdoor sporting activities. Our elite athletes can achieve the status of role models and media stars. However, all of this physical activity can take its toll. The medical response to sports injuries contributes a significant proportion of the diagnostic and therapeutic treatments taking place in Australia every day. Many of these procedures could only have been developed through our understanding of the physics on which they are based. Many healthy (just injured) people rely on medical physics. In this study we will take a particular look at the different ways in which images of the body are produced in the effort to diagnose disease and injury. Medical physics also offers a range of therapeutic treatments that are available for numerous diseases. Although ultrasound, X‑rays and gamma rays are often thought of in medicine in relation to their use in forming images of the body, they can also play a therapeutic role in treating an assortment of ailments.

made s i t i w o h d n ltrasound a

15.1 U

Your earlier study in this course has included an examination of waves. Sound is an example of energy travelling in the form of a wave, and the properties of sound waves are discussed below. Humans can hear sound waves in the frequency range 20–20 000 Hz. Infrasound is the term given to the very low frequencies below our hearing range; that is, below 20 Hz. The frequencies above 20 000 Hz are called ultrasound. Ultrasonography is the use of ultrasound in medical and industrial applications. In medical applications ultrasound is used in two different modes. The most common use of ultrasound is in the production of images of internal parts of the body. Here ultrasound is being used as a diagnostic aid. However, the uses of ultrasound as therapy are less well known. In therapeutic applications ultrasound may be used to treat an ailment such as kidney stones or to carry out delicate brain surgery.

Diagnostic medical ultrasound

Physics file Industrial items need some diagnoses too! The principles used in producing an ultrasound picture of the insides of the body can also be applied to many other items. In industry, ultrasound is used to check for faults, particularly in structural materials such as cast metals. SONAR (sound navigation and ranging) developed during World War I is still in use today, particularly in geological surveying.

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Diagnostic ultrasound relies on the reflection of sound waves. In your earlier studies of waves you will have seen that when waves travelling in a medium are incident upon the boundary of another medium, some of the wave energy will be reflected and some will be transmitted. Diagnostic ultrasound technologies rely upon the detection of sound waves that have been reflected from the boundaries between different types of tissue within the human body. For example, sound waves can be directed through the amniotic fluid within the womb and reflected from the unborn foetus. From various positions, the time taken for the reflected wave to return to the receiver indicates the location of the boundary between different types of body tissue. In order to make sense of the reflected waves the signal is analysed by computer and the corresponding visual image is constructed.

Sound is a longitudinal mechanical wave In your study of waves you will have seen that all mechanical waves involve the transfer of energy without the net transfer of matter. This means that as waves are carried though a medium, the particles of that medium simply vibrate around a mean (average) position. Only the energy travels. There is no overall displacement of the particles of the medium. Mechanical waves can be categorised into two groups: transverse waves and longitudinal waves. Both categories involve vibrating particles. Transverse waves involve vibrations that are perpendicular (at 90°) to the direction of travel of the energy. Longitudinal waves involve vibrations of particles parallel to the direction of travel of the energy. All sound waves must have a vibrating item at their source. For example, the vocal chords within the larynx vibrate as somebody speaks, stereo speakers include a vibrating cone, and an ultrasound technician holds a transducer containing vibrating crystals. A sound wave travelling in air is essentially the creation of alternate regions of high air pressure and low air pressure. To picture the creation of a sound wave, assume that a vibrating sound source initially moves forwards

(a) Transverse wave

crests

wave movement

particle movement

movement of hand from side–to–side troughs (b) Longitudinal wave

rarefactions

movement of hand backwards and forwards

compressions wave movement

particle movement

Figure 15.1 (a) Transverse waves; (b) longitudinal waves.

and the air molecules nearby are scooped up closer to one another; we say that an area of higher air pressure or a compression has been created. As the sound source moves backwards it leaves a region of space that is emptier than usual. That is, a rarefaction or area of lower air pressure is formed. This process is continued over and over so that a sequence of compressions and rarefactions is formed, as shown in Figure 15.2. These compressions and rarefactions travel away from the source. As the sound wave is carried though the medium, each individual air molecule will just vibrate to and fro around a central mean position. The air molecules do not undergo any overall change in their mean position. loudspeaker diaphragm vibrations

C = compression C

loudspeaker

W

R

R = rarefaction C

R

C

Y X Z direction of travel of sound waves

R

longitudinal displacement of air molecules

Figure 15.2 Compressions and rarefactions move through the medium and its particles vibrate.

Since sound waves are longitudinal they are capable of travelling through solids, liquids and gases. In your study of waves you saw that transverse waves could travel across the surface of liquids, but they do not travel readily within the liquid itself. Unless a solid material is particularly elastic it can be difficult to make a solid medium carry a transverse wave. The ability of longitudinal sound waves to travel through solids, liquids and gases makes them particularly suitable for imaging the human body.

Properties of sound waves In order to appreciate the details of the use of ultrasound you will need to be familiar with the following properties of sound waves. The frequency of a sound wave, f, is determined by the frequency with which the source of the sound wave vibrates. Frequency (measured in hertz)

Chapter 15 Medical physics

517

is the number of vibrations or cycles that are completed per second or the number of complete sound waves that pass a given point per second. Since the period, T, is the time interval for one vibration or cycle to be completed, it follows that: 1 f= T where f is the frequency of the sound wave in hertz (Hz) T is the period of the sound wave in seconds (s) The wavelength, λ, of a sound wave is the distance (within the medium that is carrying the sound wave) between successive points that have the same displacement and are moving in the same direction. This is also referred to as successive points being in phase. The amplitude, A, of a wave is the value of the maximum displacement of a particle from its mean position. If the sound wave is of an audible frequency its amplitude would determine the loudness of the sound. Even though ultrasound is inaudible, the amplitude of the sound wave affects the amount of energy it carries.

Displacement

wavelength, L amplitude

amplitude

Direction of travel of wave

Figure 15.3 The wavelength and amplitude of

a wave.

L

In our previous study we saw that the frequency of the source of a wave and the velocity of that wave in the medium together determine the resulting wavelength. In a given medium, sound waves of a higher frequency would result in waves that were closer together; that is, waves of a shorter wavelength. A low-frequency source would produce longer 1 wavelength sound waves. We say that for a given wave speed: λ ∝ . f Also note that a source that has a specific frequency of vibration is able to produce waves of different wavelengths, depending upon the medium that carries the wave. We say that for a source of a given frequency: λ ∝ v. The wave equation links the speed, frequency and wavelength of a wave:

v = f λ where v is the speed of the wave in metres per second (m s‑1) f is the frequency of the wave in hertz (Hz) λ is the wavelength of the wave in metres (m)

Worked example 15.1A Since the human body is largely made of water, the average speed of sound in the body is 1540 m s‑1, close to the speed of sound in water. When using ultrasound, the smallest item that can be detected is the same size as the wavelength of the sound wave used. If an ultrasound frequency of 1 MHz is used, what is the smallest foreign object that would be noticeable by ultrasound?

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Solution A frequency of 1 MHz = 1 × 106 Hz. The wavelength of a sound wave is given by: v λ = f 1540 = 1 × 106 = 1.54 × 10‑3 m ≈ 1.5 mm Therefore, the smallest sized object that can be detected is around 1.5 mm. Physics in action

The differences between ultrasound and everyday sound waves Sound waves within the human hearing range (20 Hz to 20 kHz) travel at around 340 m s‑1 through air and subsequently have wavelengths in the range 0.017–17 m. Most everyday sounds, however, are between 50 and 3000 Hz. Medical ultrasound typically uses frequencies ranging from around 1 to 10 MHz (that is, 1 000 000 to 10 000 000 Hz). These sound waves travel through body tissue much faster than they would through air and will have wavelengths of less than a few millimetres.

Table 15.1 Speed of ultrasound in various media Medium

Typical ultrasound velocity (m s‑1)

Air Water

340 1430

Fat

1470

Soft tissue

1540

Brain

1560

Blood

1570

Muscle

1590

Bone

Sound waves, like other waves, display a property called diffraction. Diffraction is the bending of the direction of travel of a wave as it passes through an aperture or around an obstacle. Diffraction is significant when the size of the obstacles or apertures in the path of the sound wave is similar to the sound’s wavelength. Since ultrasound utilises waves with very short wavelengths, the internal parts of the human body do not cause much diffraction. The ultrasound waves can travel in reasonably direct paths through the human body. Longer wavelengths, such as those of audible sounds, would not be as directional in their paths through the body, and so these wavelengths cannot be used to produce diagnostic images.

3360–4080

Producing ultrasound In diagnostic ultrasound the sound waves are both generated and detected by a small handheld probe (or transducer), which is moved around the surface of the body. The term transducer means any device that converts energy from one form into another. An ultrasound transducer converts electrical energy into a sound wave and vice versa. Between bursts of transmitted sound waves the transducer acts as a receiver, detecting the reflected sound waves. Initially the sound waves must be able to travel from the transducer into the body without much reflection at the skin’s surface, so a coupling (or joining) gel is used between the transducer and the patient’s skin. In order to produce ultrasound, just as in the production of any sound wave, a physical item must be made to vibrate. Within the transducer varying electrical signals are applied to a disc made of special crystals that produce the piezoelectric effect. This means that in response to a varying

PRACTICAL ACTIVITY 64 Ultrasound interactions: attenuation of sound


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electrical signal of a particular frequency, the disc undergoes mechanical vibrations of the same frequency; that is, they produce sound. The trans ducer can also act as a receiver, since any returning sound waves will cause the same disc to vibrate and produce a small varying electric signal voltage that can be analysed. The most common piezoelectric material in use in ultrasound transducers is a synthetic ceramic called lead zirconate titanate (PZT). This ceramic material can be made into discs of various shapes and thicknesses and so transducers can be designed for specific diagnostic procedures.

The piezoelectric effect Figure 15.4 The transducer acts as both a source

and receiver of ultrasound waves.

Physics file The common whistle such as that used by a sports umpire can be thought of as a transducer. It converts the kinetic energy of the blown air into the mechanical vibration that is sound energy. Therefore, dog whistles that produce sounds of frequencies above the audible limit of 20 000 Hz can be called ultrasound transducers.

Piezoelectric materials are materials whose electrical properties alter when they are stressed; that is, when they are compressed or stretched. This is due to the slight rearrangement of charges within the atomic structure of the material. Piezoelectric materials have a crystal structure. Their atoms are evenly arranged in a geometric lattice pattern with regular spacing between them. The positive and negative ions are evenly distributed throughout the lattice, making the material electrically neutral overall. A disc of piezoelectric material can have a thin metal plate placed on each side of it. These plates are called electrodes. If the material isn’t being compressed or stretched, the surfaces of the piezoelectric disc will be electrically neutral and there will be no potential difference (electrical voltage) between the plates.

How sound waves are detected by piezoelectric transducers A sound wave is made up of travelling regions of high and low pressure in the medium that is carrying the sound wave. If a sound wave strikes the surface of a metal plate (electrode) in the transducer, the piezoelectric crystal layer (the disc) will be alternately compressed and stretched slightly. The arriving high-pressure parts of the sound wave will compress the piezoelectric material and the lower pressure regions of the sound wave will cause the piezoelectric material to be stretched. (A low-pressure region can be thought of as creating a slight suction effect just like a vacuum cleaner creates.) The compression of the piezoelectric crystal disc will result in a net negative charge occurring close to one electrode, leaving the other plate with a slight positive charge. Hence a potential difference exists across the two electrodes. The stretching of the crystal layer (due to the arrival of a low-pressure region of the sound wave) will result in a potential difference across the plates of the opposite polarity. As the alternate high- and lowpressure regions of the sound wave arrive at the transducer, an alternating electrical signal is produced. This electrical signal is then amplified and used to create a meaningful image on the sonographer’s computer screen. Later in this section we will look at the aspects of the sound wave that will produce different features in this image.

The PI…ZO…L…CTRIC …FF…CT occurs in certain crystals which, when placed under stress, acquire equal and opposite charges on opposite faces. Hence an electrical potential difference is established between the surfaces.

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How sound waves are produced by piezoelectric transducers An electrical signal can be supplied to the transducer. If a small voltage is applied across the electrodes of a piezoelectric transducer, the charges in the atoms of the piezoelectric material will adjust their positions slightly. This affects the overall volume that these atoms occupy. Therefore, depending on the polarity of the applied voltage, the crystal disc will either expand or contract. The electrical supply needs to provide a small alternating voltage; that is, an electrical voltage that periodically reverses in polarity. The application of an alternating voltage to a piezoelectric transducer will cause the crystal disc to alternately expand and contract. That is, it behaves as a vibrating item and as such produces sound waves. Ultrasound transducers can also be made from magnetostrictive mat erials. Magnetostrictive materials exhibit similar properties to piezoelectric materials. When sound waves strike, the material produces a varying mag netic field. When a varying magnetic field is supplied, the transducer will oscillate, producing ultrasound waves.

Physics file Sound waves of maximum energy occur if the frequency of the alternating voltage matches the natural or resonating frequency of the crystal disc. Like guitar strings and pendulums (and all physical objects) the crystal disc within the transducer will have specific natural frequencies at which it can most readily vibrate. The values of these natural frequencies depend on the physical dimensions of the disc. The fundamental or lowest resonant frequency of a piezoelectric transducer will be designed to lie between 1 MHz (1 megahertz) and 15 MHz.

electrode plate crystal layer expands

+ –

electrode plate

electrode plate crystal layer contracts

– +

electrode plate

Figure 15.5 The pulsating crystal material will produce alternate high- and low-pressure regions in the adjacent medium. Hence a sound wave is produced.


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15.1 summary Ultrasound and how it is made • Sound waves with frequencies above 20 000 Hz are called ultrasound. • Sound waves are longitudinal waves that involve vibrations of particles parallel to the direction of travel of the energy. • The wavelength, frequency, period and amplitude are features of a wave. • v = f λ where v is the speed of the wave in metres per second (m s‑1), f is the frequency of the wave in

hertz (Hz), and λ is the wavelength of the wave in metres (m). • In diagnostic ultrasound the sound waves are both generated and detected by a small handheld probe (or transducer). • In response to sound waves, the transducer produces an alternating electrical signal due to the piezoelectric effect.

15.1 questions Ultrasound and how it is made 1 Define the following terms: period, frequency, wave length and amplitude of a wave. 2 a Identify the source of the sound wave in each of the following situations: i an opera singer holding a note ii a smoke detector alarm ringing iii the mating call of a male cicada. b What characteristic do all sources of sound have in common?

5 When a sound wave strikes the boundary between media of different density, what might happen to the sound energy? 6 Which pair of substances found in the human body (as listed in Table 15.1) would produce the greatest amount of reflection of ultrasound waves at their boundary? 7 What is the longest wavelength of a sound (in air) that could be classified as ultrasound?

3 State and briefly describe the two main categories of use of ultrasound in medicine.

8 Explain how the piezoelectric effect enables a transducer to pick up sound waves.

4 Use the speeds of sound shown in Table 15.1 to calculate the wavelength of an ultrasound wave of frequency 2.5 MHz that is travelling in: a soft body tissue b the brain c fat d blood.

9 How is the motion of a piezoelectric source different from an ordinary source of sound waves?

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10 If only items that are longer than around five wave lengths can be imaged by ultrasound, how large is the smallest object that could be detected in soft body tissue, using ultrasound of frequency 4.5 MHz? The velocity of sound in soft tissue is 1540 m s‑1.

15.2 Ultrasound int era

ctions

Measuring sound energy Sound waves deliver energy in a continuous stream, rather than discrete bundles, so it is convenient to think of sound waves in terms of the number of joules of energy delivered per second. Since the concept of ‘energy per second’ really is a measurement of power, the unit of the watt is employed (1 watt = 1 joule per second). Sound waves carry energy through a medium and so the amount of energy passing through a square metre area each second is called the intensity of the sound wave. Intensity is therefore the power (in watts) passing through a square metre of area. The unit of intensity is watts per square metre (W m‑2). In medical applications of ultrasound the potentially destructive effects of very intense ultrasound waves (discussed later) limit the allowable intensities of sound waves.

PRACTICAL ACTIVITY 65 Scanning techniques: the speed of sound

Physics in action

Amplitude, intensity and the decibel scale For a given frequency, the amplitude of the vibrating sound source (the transducer) determines the intensity of the sound wave produced. Doubling the amplitude results in a sound wave of four times the intensity. Tripling the amplitude results in an intensity that is nine times greater. That is, for a given frequency, intensity is proportional to the square of the amplitude, or I ∝ A2. The decibel scale does not give a measurement of the actual intensity of a sound; rather it allows a comparison of relative sound intensities to be communicated. The decibel is a subdivision of the unit the bel and it is not an SI unit. It assigns sound level values for a sound that is compared in intensity to the threshold intensity value, Io. Io is the intensity of the quietest sound that the average person can hear when a reference frequency of 1000 Hz is used. The formula for determining a given sound level, L, in decibels is: I L = 10log10 Io Hence the range of human hearing that lies between intensity values of 10‑12 and 102 W m‑2 (at 1000 Hz) corresponds to a decibel range of 0–140 dB. The log scale provides some useful features. Should the intensity of a reflected wave be reduced by a factor of 10 then the sound level decreases by 10 dB. If the intensity decreases by a factor of 100 then the sound Figure 15.6 Fo level decreases by 20 dB etc. vibrating so r a given frequency th ea u rc A typical absorption rate of ultrasound in soft decibel leve e determines the inte mplitude of the l. nsity and th body tissue is around 1.0 dB cm‑1. Therefore, erefore the regardless of its initial intensity value, a sound wave will have dropped by 20 dB by the time it has travelled through 20 cm of soft body tissue. This corresponds to a reduction in intensity by a factor of 100.

()


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Sound waves interacting with materials The attenuation of ultrasound waves As a sound wave is transmitted through a medium, it is inevitable that some of its energy will be absorbed by the particles of the medium and converted to internal energy. Much of this absorbed energy contributes to an increase in the average kinetic energy of the particles, which we call an increase in temperature. The gradual reduction in the intensity of a sound wave as it travels away from its source is called the attenuation of the sound wave. The amount of attenuation as a sound wave travels through a given medium depends largely on the physical properties of the medium. These properties determine how widely the sound wave will spread out, how much the energy is scattered and how much energy is absorbed. Choosing the most effective frequency of sound for use in medical diagnostic ultrasound is not straightforward. We have discussed that the inaudible ‘ultrasound’ range (above 20 000 Hz) is used, since these frequencies are less likely to diffract or spread inside the body. However, frequencies that are extremely high are also unsuitable because they are far more prone to be absorbed by the body. We say that very high frequencies are poorly attenuated. For example, if the frequency of a sound wave is doubled, it will be up to four times more likely to be absorbed. Hence if the selected ultrasound frequency is too high the body will readily absorb it and the waves may not penetrate all the way to the body part of interest. For example, in human soft body tissue a sound wave of frequency 1 MHz will drop off in intensity by one-half when it has travelled less than 7 cm. The frequencies commonly used in medical ultrasound range between 3 and 10 MHz, and these frequencies have a penetration distance of about 200 wavelengths in soft tissue.

Table 15.2 Commonly used medical ultrasound frequencies and their uses Frequency (MHz)

Estimated penetration (cm)

Medical diagnostic use

3–5

10–20

liver, heart, uterus

5–10

5

breast, thyroid

10–15

1

eye

incident sound

The reflection and transmission of ultrasound waves

transmitted reflected absorbed

Figure 15.7 At the boundary between body tissues of different density or that carry sound at different speeds, some of the incident sound wave energy will be reflected and some of the incident sound energy will be transmitted into the second medium.

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Anyone who has experienced a clearly defined echo would realise that sound waves that are sent through the air can reflect from surfaces. The harder and smoother the surface is, the more likely that the sound wave will be reflected. The large walls of brick buildings, cliff faces and caves can create some impressive echo effects. In fact when sound is incident upon the boundary between any two substances of different density, some of the energy will be reflected at the boundary and some of the energy will be transmitted into the second medium. These processes also occur if sound strikes the boundary between two surfaces that carry sound at different speeds. The reflection and absorption of sound waves as they are transmitted through the human body are unavoidable, but these properties of sound waves are put to good use in diagnostic and therapeutic ultrasound.

Acoustic impedance Since the passage of a sound wave through a medium involves the passing on of vibrations, it can be said that all media will hinder the progress of the wave to some extent. The level of efficiency with which sound waves can pass through a given medium is called the acoustic impedance, Z, of the medium. The acoustic impedance is strongly related to the speed at which sound can travel in the medium. The density of the medium is also a key factor.

The ACOUSTIC IMP…DANC… of a medium is given by: Z = ρv where Z is the acoustic impedance (kg m‑2 s‑1) ρ is the density of the medium (kg m‑3) v is the the velocity of sound in the medium

Worked example 15.2A A blood donor donates 600 mL of blood, which is found to have a mass of 636 g. Assuming that the velocity of sound in blood is 1570 m s‑1, calculate the acoustic impedance of blood.

Solution Working in SI units, the 600 mL of blood is equal to a volume of 600 cm3 or 0.0006 m3. The mass of the blood is 0.636 kg. Therefore, the density of the blood is given by: mass ρ= volume 0.636 = = 1060 kg m‑3 0.000 600 Acoustic impedance: Z = ρv = 1060 × 1570 = 1.66 × 106 kg m‑2 s‑1 Whereas the attenuation of a sound wave refers to the loss of energy of a sound wavefront as it passes through a medium, the acoustic impedance of the medium indicates how readily a sound wave is able to travel through a medium. A conductive medium may have a low impedance of sound, but if it allows a sound wave to spread out it will also have low level of attenuation.

Table 15.3 Typical acoustic impedance values in various media Medium

Typical acoustic impedance, Z (kg m‑2 s‑1)

Air

430

Water

1.43 × 106

Fat

1.38 × 106

Brain

1.58 × 106

Blood

1.59 × 106

Soft tissue

1.63 × 106

Muscle

1.70 × 106

Bone

5.6–7.8 × 106


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Table 15.3 shows some relevant acoustic impedance values. Acoustic impedance values for gases and liquids are very different. Note also that most of the substances within the human body differ only slightly in acoustic impedance. We have mentioned that when sound is incident upon the boundary between any two substances of different acoustic impedance, some of the energy will be reflected. If the difference in acoustic impedance, Z, across this boundary is large, the wave cannot readily cross into the new medium and a large proportion of the sound wave’s energy is reflected. In other words, a strong echo occurs if two substances have different Z values. If the two media have similar acoustic impedances then most of the incident wave energy will be transmitted into the new medium. This would be described as good penetration of the wave, but a poor echo. The proportion of the incident sound wave energy that is reflected at a boundary is called the reflection coefficient, α. We have seen that for sound waves carrying energy through a medium, the amount of energy passing through a square metre area each second is called the intensity (in W m‑2) of the sound wave. The reflection coefficient, α, is defined as the ratio of the intensity of the reflected sound wave (Ir) to the intensity of the sound wave that was initially incident on the boundary (Ii).

The R…FL…CTION CO…FFICI…NT, α, is the proportion of the incident sound wave energy that is reflected at a boundary. It is determined by the difference in the acoustic impedance values for the two media, Z1 and Z2. I (Z - Z )2 α = I r = (Z2 + Z1)2 1 2 1 The above equation shows that when the acoustic impedances of the two media are similar the proportion of reflected sound energy is small. However, if Z1 and Z2 are very different then the amount of reflection that occurs is increased. As Table 15.3 shows, there are only subtle differences between the acoustic impedances of the many different components of the human body. Although this produces very small quantities of reflected energy, it is sufficient for effective images to be obtained. As long as these reflections carry enough energy to be detected, the resultant electrical signals at the transducer can be amplified. You should not assume that a very high percentage of reflection of energy is desirable. This would mean that the sound wave couldn’t penetrate into the second medium and nothing behind this boundary could be imaged. This is why ultrasound imaging of items that lie inside the lung or behind or within a bone is very difficult. The differences in acoustic impedances between these items and the surrounding tissue result in too much reflection at boundaries, leaving little of the sound wave to continue onwards.

Table 15.4 Some typical reflection coefficients Boundary

Reflection coefficient, α

Air–soft tissue

0.99 (or 99% reflection)

Bone–soft tissue

0.40 (or 40% reflection)

Fat–muscle

0.011 (or 1.1% reflection)

Worked example 15.2B

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Use the values in Table 15.3 to calculate the reflection coefficient, α, that occurs when the inner lining of the bladder is imaged by ultrasound. Consider the lining to be a water–soft tissue boundary.

Detailed studies

Solution Find the relevant acoustic impedance values from Table 15.3. Z1 = Zwater = 1.43 × 106 kg m‑2 s‑1 Z2 = Zsoft tissue = 1.63 × 106 kg m‑2 s‑1 (Z - Z )2 α = 2 1 2 (Z2 + Z1) (1.63 × 106 - 1.43 × 106)2 = (1.63 × 106 + 1.43 × 106)2 = 0.0043 Therefore, 0.43% of the incident energy is reflected. This seems a small value, but it is large enough to be detected by modern transducers. The electrical signal produced would be amplified and an image formed from it.

15.2 summary Ultrasound interactions • The amount of sound energy passing through a square metre area each second is called the intensity of the sound wave. • The gradual reduction in the intensity of a sound wave as it travels away from its source is called the attenuation of the sound wave. • The efficiency with which sound waves can pass through a given medium is called the acoustic impedance, Z, of the medium.

• The proportion of the incident sound wave energy that is reflected at a boundary is called the reflection coefficient, a. It is determined by the difference in the acoustic impedance values for the two media, Z1 and Z2. That is, (Z - Z1)2 α= 2 (Z2 + Z1)2

15.2 questions Ultrasound interactions 1 Assuming that the velocity of sound in bone is 4000 m s‑1 and the density of bone is 1600 kg m‑3, calculate the acoustic impedance of bone. 2 A transducer produces a sound wave of intensity of 10 W m‑2, and its interface has a cross-sectional area of 5.0 cm2. How many joules of sound energy are produced per second? 3 Why are frequencies of sound higher than 15 MHz unsuitable for use in diagnostic ultrasound? 4 A 2.0 MHz ultrasound beam has a penetration of 200 wavelengths in soft tissue. Will it be suitable for use in imaging a small tumour that lies 5.0 cm beneath the skin’s surface? 5 Why isn’t ultrasound useful for obtaining a diagnostic image of an illness of the inner lung?

6 Using the velocity of sound and acoustic impedance values in Tables 15.1 and 15.3, estimate the density of your: a brain b fat c muscle 7 Why do bubbles of gas in the bowels show up very clearly on an ultrasound scan? 8 If a sonographer (ultrasound technician) forgot to use the coupling gel when imaging a foetus, what proportion of the incident sound energy would penetrate the air–skin boundary on the mother’s abdomen? 9 Use the data in Table 15.3 to calculate α, the ratio of the reflected intensity to the incident energy for a sound wave incident on the boundary between: a air and water b water and muscle c water and fat.


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s

15.3 Scanning technique

All diagnostic ultrasound techniques rely on the principles related to the reflection of sound. If a burst of sound waves is sent through a medium, when it meets the boundary with a medium of different acoustic impedance some of the sound wave’s energy will be reflected back towards the source. The transducer, which produced the sound wave, can then act as a receiver and detect the reflected sound wave. The properties of the reflected sound wave such as the time that it takes to return, the intensity and the phase (discussed later) can all provide information about the item from which the sound wave rebounded.

The A‑scan

Figure 15.8 After sending a burst of sound waves the transducer will cease sending and instead detect the reflected wave.

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The early ultrasound devices used a single transducer and only gave information about the time for the return of the pulse from the item being investigated. The A‑scan apparatus is essentially a range-finding device similar to SONAR (sound navigation and ranging) devices used for locating underwater objects such as submarines. As shown in Figure 15.9, a brief pulse is sent in the direction of the organ of interest, or into the region of the body that is being searched for a tumour, for example. The emission of the sound wave triggers the commencement of a timing mechanism, so that a measurement is obtained for the time, t, that it takes for the pulse to travel to the relevant boundary and back. The speed of ultrasound in the body is known and hence the distance to this boundary can be calculated. Since speed = distance travelled/time taken, the total distance that the sound wave travels is given by: total distance travelled = speed × time taken But because the sound wave has travelled to the boundary and back, the distance to the boundary is simply half of this calculated distance. The synchronisation of the sending of the pulse and the time-measuring instrument is critical. In older machines this was done by an oscilloscope and the reflected pulses were shown on an oscilloscope screen, similar to the CRO screens that you may have in your physics laboratory. Each time a reflection was detected a pulse would be shown on the screen. The ‘time reading’ for each of these pulses indicates the distance through which it has travelled (see Figure 15.9). The pulses returning from greater distances have greater attenuation and the electrical signals produced at the transducer subsequently require greater amplification. New ultrasound devices use smaller electronic circuits and computers to measure and analyse the various time values. If two surfaces lie too close to one another within the body—i.e. a millimetre or so apart—a reflected pulse may not pick them up as two separate layers. If the duration of the burst of sound waves is too long then the wave reflected from the second layer will overlap the wave reflected by the first layer. The use of short bursts of sound from the transducer—of around a millionth of a second duration—means that layers less than a millimetre apart can be resolved.

transducer Oscilloscope trace

skin A B

Boundaries of body parts producing reflections

C D E

reflection from boundaries

reflection from skin surface

Figure 15.9 The display produced by an A‑scan. The time between peaks is an indication of the distance to the respective tissue boundaries.

The B‑scan The B‑scan uses the same basic principles of the A‑scan, but the reflected pulse produces dots instead of peaks on the screen (Figure 15.10). The larger the amplitude of the reflected pulse, the brighter the dot. The dots are not all white—a grey scale is employed so that important weak reflections are not lost and subtle differences in tissue can be displayed. This means that a more meaningful image is produced. The B‑scan employs a probe that contains a line of transducers so that a sector scan can be obtained. In older devices the transducer is tilted or rocked so that the ultrasound waves sweep across a wide area inside the body. This effect is now simulated electronically. The resulting dots on the screen will then build a two-dimensional image of the item of interest. As Figure 15.11 shows, the sweep causes the frame of the final image to be wedge shaped. The image is a snapshot of all of the points of reflection in the particular plane that was aligned with the set of transducers in the probe. A-scan

A

B

C

B-scan

D

E

A

B

C

D

E

Figure 15.10 The B‑scan shows the basis from which ultrasound images are created. Each reflection results in a dot in the image.


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Obstetric imaging is a major application of ultrasound imaging today. The low-intensity sound waves do no apparent harm to the developing foetus. Real-time B‑scanning is commonly used (see below). In Australia a pregnant woman is likely to have a routine ultrasound at 12 or 14 weeks into her pregnancy, even if there are no particular health concerns. At this time the baby’s skull is measured, as the dimensions can be used to estimate the length of gestation and therefore the delivery date. Abnormalities in organ or bone development may also be diagnosed at this time. The size and position of the placenta is readily determined by ultrasound, as it is made up of soft tissue. It is during this procedure that parents can be surprised to discover the presence of more than one baby!

Figure 15.11 The sweep of the B‑scan produces the familiar wedge-shaped ultrasound ‘photograph’. Ask your mother if one of these images was taken of you.

Simple B‑scans are now not commonly used, as they are gradually being replaced by equipment that can produce clearer images (see discussion below). However, because they only require a small entry window to the body they are particularly suitable for scanning the brains of newborn babies through the gap in their immature skull at the top of their head (see Figure 15.12).

Interpreting B‑scans Figure 15.12 The fontanel is the ‘gap’ in the immature skull through which a premature baby’s brain can be scanned by ultrasound to check for normal development.

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Experienced sonographers, who know what they are looking for, provide clearer images for the doctor to use for his or her diagnosis. Have a look at the ultrasound image in Figure 15.13—a cross-section through the liver, which is located near the diaphragm. Any boundary that produces a strong reflection will result in bright spots in the image. Regions that are reflecting no sound are black. The diaphragm has produced strong reflections and the liver is seen lying against it. Throughout the liver are tissues of slightly varying acoustic impedance and so light reflections are displayed. In the region of the liver tumour the reflections are stronger. The large blood vessels that deliver blood to the liver are black inside, since the fluid inside them produces no reflections. If you look carefully though you can see that some reflection occurred on the outside wall of these vessels.

Figure 15.13 B‑scan ultrasound images require considerable skill to interpret.

Three-dimensional images, real-time scanning and phase scanning The equipment associated with B‑scans has been greatly improved in modern ultrasound apparatus, but the principles of image formation are the same. At one time only static images could be produced; that is, still photographs. The ongoing development of more sophisticated transducers and the application of computer technology to analysing the reflected sound waves have led to some amazing diagnostic capabilities. Probes, essentially operating in the B‑scan mode, may include hundreds of tiny transducers. Real-time imaging uses large stationary transducers that may sit on the patient’s skin and produce images of inside the body moment by moment. This is achieved by having multiple sets of transducers that are synchronised (using computer circuits) to trigger one set after another. The returning reflections can then produce apparently continual images of the relevant organs, even if the organs are moving. This process can be done so quickly that an expectant mother can sit and watch her unborn child’s heart beat. Image detail has recently been improved by the use of phase scanners. B‑scans produce an image that is made up of a set of separate dots—the more transducers present, the more dots making up the image. With phasescanning transducers the phase of the sound wave that is picked up by adjacent transducers will be analysed. That is, the transducer/computer apparatus can determine at which part of the high- and low-pressure cycle the reflected sound wave is. The computer then fills in the part of the image

Physics file Using ultrasound to obtain an image of an adult heart is difficult, since the lungs cover a large part of the heart and the lungs contain air, which is highly reflective.


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that would lie between these two imaged points. This results in much smoother and complete images. The expected progression to the production of three-dimensional images by ultrasound has occurred, but the equipment is very expensive. Since two-dimensional images often completely fulfil diagnostic requirements, the majority of diagnostic ultrasound images are, even now, only two dimensional. In order to produce a three-dimensional image a set of multiple-element B‑scans is taken. Each scan would be taken from a slightly different perspective and, in effect, a set of numerous twodimensional images is analysed by a computer and converted into one three-dimensional image.

Doppler scanning Figure 15.14 A three-dimensional image of a foetus

can be produced from the digitised output of a scan.

PRACTICAL ACTIVITY 66 The Doppler effect

All waves can create an interesting effect called the Doppler effect when there is relative motion between the source of the waves and the observer (or receiver). Consider the moving source of waves shown in Figure 15.15. As the source moves to the right the successive emitted sound waves are bunched closer together on the right of the source; and they are more separated on the left of the source. This bunching together results in a shortening of the wavelength in the direction of motion. If the source were moving faster, the sound waves would be even more bunched together. Therefore, the extent to which the wavelength is adjusted can be an indication of how fast the source is moving. If a moving item in the human body is considered to be a ‘source’ of sound waves, then its speed can be determined by an analysis of the Doppler effect.

V

Figure 15.15 In the Doppler effect the extent to which the wavelength (or frequency) alters indicates the speed of the sound source.

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Recently developed ultrasound apparatus gathers information about any change in the wavelength of the sound that is reflected back to the transducer from the moving item being investigated. This technique is commonly used to measure blood flow rates. If the blood cells in a given artery are moving towards the transducer then the wavelength will be slightly shortened. If the blood is moving away from the transducer then the wavelength of the sound will be slightly lengthened. The extent of shortening or lengthening is indicative of the velocity of the blood relative to the transducer. Of course the rate of blood flow in any vessel will vary during a period of one heartbeat. Figure 15.16 shows the variation in blood flow rates in the large (carotid) artery that travels up the neck to the brain during four heartbeats. Since any hardening of artery walls will produce irregularities in blood flow rates, Doppler imaging is invaluable in monitoring the cardiovascular system. Blood clots show up easily because it can be seen that there is no movement of blood in nearby vessels. Doppler imaging is used to monitor the rate of blood flow through the umbilical cord to an unborn child. Doctors can then determine whether an underdeveloped foetus is receiving enough nutrition and oxygen. Blood flow in organs such as the heart, liver, kidneys and spleen can be imaged. Sophisticated display screens allocate various colours to speeds of blood flow towards and away from the transducer. The sonographer can then tell from the display which regions in the organ have abnormal blood flow rates.

Ultrasound therapies Heat treatments

Figure 15.16 Doppler imaging of the carotid artery (in the neck) is easy, since this artery is near the surface. During each heartbeat the blood flow varies periodically.

Physics file It is very likely that you have experienced the Doppler effect with sound waves. Have you ever noticed that the pitch of the sound emanating from an approaching ambulance (with its siren screaming) is higher as it approaches you but then deepens as it goes past and away from you?

Diagnostic applications of ultrasound use low-intensity sound waves, so any effects on the cells of the body are minor. However, besides being used to diagnose a medical condition, ultrasound can also be used to treat some medical conditions. To understand the therapeutic use of ultrasound you will need to understand the effect of high-intensity sound waves on the cells of the body. Keeping in mind that sound waves are just mechanical vibrations in the particles of the carrying medium, during ultrasound procedures vibrations are created in the cells of the body. At low intensities these vibrations are hardly more energetic than they would normally be. However, if the sound waves are intense enough they can heat regions deep inside the body and damage or destroy cells. An example of the use of the heating effects of ultrasound is the treatment of sports injuries in muscles and joints. Intensity values of less than 30 000 W m‑2 must be used in order to avoid tissue damage. Ultrasound exposures of 5 or 10 minutes are often part of a physiotherapy program. It is believed that the heating effect increases metabolism in the treated site and accelerates healing. This is most effective in bone and denser muscles, which are highly absorbing of sound waves. Mild heating by ultrasound can also be used to treat blockages of the middle ear region. Ultrasound is also being investigated as a treatment for arthritis.


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Figure 15.17 Ultrasound is suitable for the treatment of particular injuries.

Destructive effects of ultrasound If the intensity of the sound waves that are sent into the body is increased markedly then they can be used to destroy certain cells. The intense vibrations cause overheating and large stresses, which can rupture cell membranes. This property can be used beneficially, for example in the treatment of gallstones (in the gall bladder) or kidney stones. High-intensity (~105 W m‑2) ultrasound waves physically break up the stones, and their component particles are then washed away by the normal removal processes of the body. Previously the removal of these items would have involved surgery. Some types of tumorous cells can also be treated with ultrasound. Again, an intense beam of sound waves is incident, the intention being to break the cells apart. When this process is applied to a very concentrated area, surgeons have a very effective cauterising tool. Neurosurgeons use extremely narrow beams of sound waves with an intensity of around 2.5 × 105 W m‑2 to ‘cut out’ brain tumours. Extremely intense ultrasound waves can produce the same cell damage as that caused by exposure to ionising radiation as discussed in Chapter 1. Although not fully understood, this is thought to be due to a process called cavitation. Bubbles (cavities) of vapour are produced in fluids and soft tissue in the body, and as these bubbles collapse the implosion breaks cell membranes and destroys cell components such as chromosomes and DNA.

Therapeutic uses of light …ndoscopes and laser light The presence of a tumour or injured tissue may also be discovered by a doctor using another non-invasive technique to look inside the body. In your study of light you took a detailed look at the use of optical fibres in communication. Optical fibres called an endoscope (also called a fibrescope) can also be used to construct an image. Like the optical communication cables you studied, fibrescopes are constructed from a number of thin optical fibres placed adjacent to one another. In a fibrescope

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some of these fibres will be used to deliver visible light to the region of the body that the doctor would like to see, and the other fibres are used to produce an image of this region. Parts of the digestive tract (such as the oesophagus, stomach and the small and large intestines), the bladder, the female reproductive system, knee joints, the trachea and the lungs are often investigated by using a fibrescope. Fibrescopes have recently enabled ‘keyhole’ surgery to be conducted. Surgeons no longer have to make large incisions in the skin to conduct a standard appendectomy, for example. Instead the insertion of a fibrescope (with special attachments) through a 2–3 cm incision allows the surgeon to see and remove the appendix. This is often done through use of a laser (as discussed below). Smaller surgical sites mean less risk and much faster recovery times for patients. Special fibrescopes are also used to obtain biopsy samples. For example, during a gastroscopy procedure a surgeon will take a look at a patient’s oesophagus, stomach lining and upper duodenum to check for ulcers or cancerous growths. A scraping of tissue (called a biopsy) is taken so that pathologists can check the cells for abnormalities. Fibrescopes can also include a water or gas supply pipe for washing the site to be viewed, or they can deliver a tiny laser beam to cauterise (cut and seal) tissue. Some heart surgery is carried out by this keyhole approach. Fibrescopes can have between 5000 and 50 000 fibres bundled together. Each fibre is between 5 and 25 µm wide. The more fibres there are, the more detail there is in the picture. Although the fibres used in fibrescopes don’t need to be as pure as the fibres used in communication, the bundle of image-producing fibres does need to be coherent. Each tiny fibre delivers a tiny part of the image, so the fibres must be arranged correctly relative to one another. If their positions were jumbled along the way the image would not make sense. As studied in Chapter 8, light travels along the fibre because of total internal reflections at the core–cladding boundary. A magnifying eyepiece or even a TV system can be used at the end of the fibrescope to view the image. Figure 15.18 shows a photograph of a stomach lining taken through a fibrescope. The laser light, delivered along optical fibres, is used in a number of different types of surgery. For example, when intense laser light is shone onto skin cells, the absorption of energy can result in the vaporisation of cells. This means that laser surgery can be used to remove birthmarks and other skin defects. Different lasers penetrate further and cause cells beneath the skin to coagulate (gel together). This can have the effect of puffing out sub-skin cells that lie beneath a wrinkled area, thus smoothing wrinkled skin. These are common types of cosmetic surgery. Less ‘cosmetic’ surgical procedures include the use of lasers to burn small passages in the heart muscles to allow an increase in blood supply to these muscles. Eye surgeons use lasers to alter the shape of the cornea. Their techniques can involve burning away cells from the surface of the cornea with a laser to change its overall curvature, thus helping patients who suffer from myopia or hypermetropia. Alternatively, computer-guided lasers can be used to cut and peel back the outer layer of the cornea, and then burn away some corneal tissue and replace the outer flap, to an extremely high level of precision.

(a)

(b)

Figure 15.18 (a) Endoscope; (b) polyps in the

stomach.


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15.3 summary Scanning techniques • In an A‑scan the time between peaks is an indication of the distance to the respective tissue boundaries. • The B‑scan employs a probe that contains a line of transducers so that a sector scan can be obtained. The larger the amplitude of the reflected pulse the brighter the dot that is produced. • Real-time imaging uses multiple sets of transducers that are synchronised (using computer circuits) to trigger one set after another. Organs can be seen in motion. • Doppler imaging uses an analysis of the speed of a moving item within the body (via an analysis of the

Doppler effect) to measure things such as blood flow rates. • Intense ultrasound waves are used therapeutically to administer heat or destroy cells. • Fibrescopes allow surgeons to see inside the body. Some optical fibres will deliver light to a site, and an image of the site is carried back to the surgeon along other fibres. Attachments to the fibre can allow surgery to be carried out without the need for a large entry incision.

15.3 questions Scanning techniques 1 This image shows a cross-section of the region around the gall bladder. Why is there a shadow region?

4 A highly specialised application of diagnostic ultrasound uses frequencies of 50 MHz to image the walls of arteries. Why is such a high frequency rarely used in ultrasound diagnostic procedures? 5 Explain the Doppler effect and the role it would play in diagnosing a heart disease such as a build-up of plaque in the main artery servicing the heart. 6 Which category of ultrasound scans has most in common with ocean depth-sounding techniques? Explain. 7 Is an ultrasound frequency of 10 MHz suitable for imaging the heart? Explain why or why not.

8 State some limitations to ultrasound scanning. Under what conditions are ultrasounds not useful?

2 A single pulse is sent into a patient being given an A‑scan. Give one reason why the three reflected pulses could each have different amplitudes.

9 What technological improvements in transducers allowed real-time scanning to be achieved?

3 An echo-sounding device in a ship sends a brief ultrasonic pulse in the direction of the seabed and 0.500 s later an echo is detected. The speed of sound in the seawater is 1500 m s‑1. a Calculate the depth of the sea at this location. b If the frequency of the pulse was 10 MHz, what was the wavelength of the sound wave in the water?

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10 Two-dimensional ultrasound images are far more common than three-dimensional images. If the three-dimensional technology exists, why are threedimensional images not employed more often?

15.4 Diagnostic X-ra ys The discovery of X‑rays In 1895 Wilhelm Konrad Roentgen was studying the newly discovered ‘cathode rays’ that could be produced inside a glass tube. These cathode rays were later found to be streams of electrons emitted from a metal plate when an electrical voltage was applied to it. He noticed that, even though the tube encasing the cathode rays was inside a box, whenever the cathode rays were being produced a screen in the room that was coated with a special fluorescing substance would start to glow. He presumed some sort of invisible rays must have been travelling from the tube to the screen. Not knowing what these rays were, he called them X‑rays. Investigations soon showed that these X‑rays could pass straight through a range of materials. They could penetrate some materials more easily than others and within a few weeks of their discovery the first X‑ray images of a human body part were produced. Roentgen produced an X‑ray photograph of his wife’s hand. The effects of overexposure to X‑rays were not revealed for some time. Early scientists working with X‑rays suffered hair loss, burns, skin ulcers, skin cancers and death from unprotected overuse. In the Balkan War of 1897, X‑rays were used to locate bullets in wounded soldiers, and Roentgen received the first Nobel Prize for Physics in 1901 for his work in this field. Today X‑rays are recognised as being part of the electromagnetic spectrum, with wavelengths of approximately 10‑11–10‑8 m. Hence, X‑rays have much shorter wavelengths than visible electromagnetic radiation (visible light). X‑rays are far more energetic than visible light too. They carry no charge. They are not affected by electric or magnetic fields. These properties mean that they are able to pass through many materials that are opaque to visible light and other less energetic forms of electromagnetic radiation. (The electromagnetic spectrum is discussed in Chapter 8 ‘Models for light’.)

X‑rays as electromagnetic waves X‑rays form only a very small portion of the entire electromagnetic spectrum. The electromagnetic spectrum is roughly divided into seven different categories according to how the radiation is produced and its wavelength. A detailed discussion of the true nature of electromagnetic radiation is well beyond the scope of this course. However, all electromagnetic waves are created by accelerating charges, which result in rapidly changing magnetic and electric fields travelling out from the source at the ‘speed of light’. The electric field component and the magnetic field component are at right angles to each other and to their direction of travel. In this respect electromagnetic radiation meets the description of a transverse wave as discussed in Chapter 7. Recall that for any wave the relationship between its frequency and wavelength is given by: v = f λ where, in the case of X‑rays, v is the velocity of the X‑ray (in a vacuum or air this is c = 3.00 × 108 m s‑1) f is the frequency in hertz (Hz) λ is the wavelength in metres (m)


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Worked example 15.4A Calculate the wavelength of X‑rays that have a frequency of 2.75 × 1019 Hz, if they are travelling in air.

Solution v = f λ v c Therefore λ = or λ = f f 3.00 × 108 = 2.75 × 1019 = 1.09 × 10‑11 m

Physics in action

The electromagnetic spectrum All forms of electromagnetic radiation (EMR) are essentially the same except for their different wavelengths (and frequencies). The highest energy, smallest wavelength radiation is a gamma ray, which is produced within the nucleus of an atom. X‑rays can pass through body tissue and be detected by photographic film, and so are used in medical diagnosis. Ultraviolet light is less energetic than gamma or X‑rays but it is known to cause skin cancer. Visible light enables us to see our world. Infrared radiation includes the wavelengths that our skin responds to. When you feel the warmth from the Sun or an electric bar heater you are actually detecting infrared radiation. All objects that are not at a temperature of absolute zero radiate electromagnetic radiation. The hotter the object, the more radiation is emitted, and the further along the spectrum the radiation is. Nightscopes and infrared spy satellites create an image by sensing infrared radiation and converting it into a visible picture. Accelerating a positive or negative charge can produce electromagnetic radiation. Electrons oscillating in a conducting wire, an antenna, produce the radio waves that bring music to your home. The microwaves that cook your dinner are produced by the spin of an electron or nucleus. All electromagnetic radiation travels at the extremely high speed of 3.0 × 108 m s‑1 in a vacuum. The speed of light in a vacuum has been given the symbol c, which comes from the Latin word celer, meaning fast.

Figure 15.19 The electromagnetic spectrum.

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Frequency

Wavelength

1022 Hz

10–14 m

Ultraviolet

1016 Hz

10–8 m

Visible spectrum

800–400 THz

400–800 nm

Infrared

1012 Hz

10–4 m

Microwaves

1010 Hz

10–2 m

TV

108 Hz

10 m

Radio

106 Hz

102 m

Gamma rays

X-rays

Producing X‑rays The production of X‑rays requires very specific apparatus, as shown in Figure 15.20. The process involves the use of an evacuated glass tube and within it a filament that is heated so that it gives off electrons. Placing a positive metal plate at the other end of the glass tube (to which the electrons will be attracted) then accelerates these electrons. The electrons will strike the metal plate and decelerate suddenly. As they decelerate energy is given off in the form of X‑rays. Modern X‑ray machines still have the three main components of the original X‑ray tubes, but almost the entire apparatus will be encased in lead shielding. The cathode filament is shown in Figure 15.20. This is the source of the electrons but it is now specially designed to send the electrons in a particular direction. The electrons are again accelerated by the use of a positive accelerating potential. Today in medical X‑ray machines the voltage (potential difference) used typically lies between 25 000 V and 250 000 V. The metal targets found in today’s X‑ray machines are often made of tungsten because this metal is stable in the high-temperature environment that occurs. The target also needs to have a high atomic number (lots of protons in the nucleus) since this increases the probability of the bombarding electrons losing a lot of energy and producing an X‑ray. Tungsten has an atomic number of 74 and a melting point of about 3900°C, making it the most common choice for a target material. The production of X‑rays from accelerated electrons is a very inefficient process. Even with the best-chosen target, the efficiency of the conversion of the electron’s energy into X‑ray energy is very low. Most of the energy of the moving electron is converted into heat energy rather than X‑rays. Efficiencies can be as low as 1%. In modern X‑ray machines the surface of the metal target is placed at a 45° angle to the incident electron beam so that the X‑rays are directed out of the tube. A gap in the lead shielding allows the X‑rays to be utilised (see Figure

PRACTICAL ACTIVITY 67 Diagnostic X-rays

evacuated glass tube cathode filament (heated)

X-rays

electron beam

large potential difference

metal target (positive)

– +

Figure 15.20 Modern X‑ray machines still include the three main components of the original X‑ray machines: the filament, an accelerating potential and a target metal.


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15.20). The tungsten target requires some sort of cooling system to remove the substantial quantity of heat that is generated with X‑ray production. Large copper heat sinks, or air or water cooling may be employed. Commonly the target is spun constantly by an electric motor so that no single section of the target is used for long enough for it to overheat.

Continuous and line spectra from X‑ray machines X‑rays can have a range of frequencies. Two different processes within the atom contribute to the production of X‑rays. A continuous spectrum of X‑rays is produced by the sudden deceleration of the incident electrons as they come near to the nuclei of the target metal (tungsten). Since the electrons pass by the nuclei at different distances, and lose energy according to this distance, a range of X‑ray energies results. This continuous spectrum can have X‑ray energy values up to the energy value of the original incident electrons. Hence, the accelerating potential (or voltage) of the X‑ray machine determines this maximum X‑ray energy value (see Figure 15.21). The sharp peaks in this spectrum that are shown in Figure 15.21 are called line spectra. Line spectra are produced because the incident electrons knock some of the electrons out of the target atoms (the tungsten atoms). As the electron arrangement of the tungsten atom re‑adjusts, X‑rays of very specific energy values are given off. Each type of target material produces a specific set of X‑ray line spectra energy values. continuous emission characteristic X-rays Intensity

Figure 15.21 The continuous and line spectra of X‑rays emitted from a tungsten target. The material in the target determines the line spectra. The accelerating voltage determines the maximum energy of the X‑rays in the continuous spectrum. A high voltage produces more X‑rays since more electrons will have enough energy for X‑ray production.

high voltage

low voltage

20

40

60

80

100

Photon energy (keV)

The energy of X‑rays Physics file If your physics teacher is old enough he or she may remember when shoe shops sometimes had little X‑ray machines in them. A small device in the corner of the shop would ‘show’ you the bones in your foot when you placed your foot inside a small box! Parents could check if there was room for a child’s foot to grow inside a new pair of shoes.

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When a beam of X‑rays is sent into a material, some of the X‑rays will be absorbed. This absorption occurs because of interactions between the X‑rays and the atoms of the material. When the operators of X‑ray machines, called radiographers, refer to the quality of an X‑ray beam they are referring to its penetrating ability. A beam containing high-energy X‑rays has greater penetrating ability than X‑rays of lower energy. Penetrating ability is discussed in more detail later. The most penetrating of X‑rays are called hard X‑rays and these have wavelengths less than 10‑10 m and high energy values. They undergo very little absorption in materials, even materials as dense as lead. Longwavelength, low-energy X‑rays are called soft X‑rays. They can be absorbed by lead and many materials and have wavelengths of more than 10‑10 m.

X‑ray photon energy Later in your studies of the nature of light you will come across the photon theory of light and study the work of Albert Einstein and Max Planck, carried out early last century in this field. The photon theory of light applies equally to other forms of electromagnetic radiation such as X‑rays. According to this theory electromagnetic radiation is considered to carry and deliver energy in little packets, which are called ‘photons’. The energy of a particular X‑ray photon is directly proportional to the value of its frequency. Hence, hard X‑rays are high energy and high frequency, and therefore have shorter wavelengths. Soft X‑rays are low energy and low frequency, and have longer wavelengths.

The energy of an X‑ray photon is given by: … = hf where … is the energy of the X‑ray photon in joules (J) h is Planck’s constant = 6.62 × 10‑34 J s f is the frequency in hertz (Hz)

Worked example 15.4B The optimal photon energy in many diagnostic applications of X‑rays is about 5.0 × 10‑15 J. What is the frequency of these X‑rays?

Solution … = hf Rearranging gives: … f = h 5.0 × 10‑15 = 6.62 × 10‑34 = 7.6 × 1018 Hz

The electronvolt as a measure of energy When dealing with the energy values of X‑rays, the joule is not often used as the unit of measurement. A smaller, non‑SI unit is used. This unit is the electronvolt, eV. It is defined as the amount of energy that one electron attains on moving through a potential of 1 V. It is a tiny fraction of a joule. Its name is a little misleading so keep in mind that it is a unit of energy and not voltage. The conversion factor from joules to electronvolts is 1.6 × 10‑19 and X‑ray energy values are also commonly given in keV and MeV.

The …L…CTRONVOLT is a (alternative) unit of energy: 1 eV = 1.6 × 10‑19 J For example, if an X‑ray machine had an accelerating voltage of 100 kV then the highest possible energy that an emerging X‑ray photon could have is 100 keV. However, recalling the shape of the curve of the continuous spectrum shown in Figure 15.21, we can see that the X‑rays will typically have energy at around a third to a half of this peak value.


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Filtering, heterogenous and homogenous X‑ray beams

Figure 15.22 A patient about to have an X‑ray. Note the lead apron worn by the radiographer.

The output of an X‑ray machine usually includes a full range of X‑ray frequencies (or energies) with some specific frequencies being present in larger quantities (the line spectra). This is called a heterogenous X‑ray beam. In the majority of medical applications it is important for the radiographer to be using X‑rays of a specific energy value (a homogenous beam), or X‑rays that have only a small range of energy values. Therefore, it is usual for the X‑rays produced at the metal target to be filtered. The lowest energy X‑rays are not useful in medical applications, as most do not possess enough energy to penetrate much further than the patient’s skin, where they will be absorbed without contributing at all to the formation of an image. The low-energy X‑rays that do penetrate the skin are likely to be scattered widely and produce blurring in the image. Since these X‑rays are counterproductive, they are removed by filtering in order to minimise the patient’s total exposure to X‑rays. The X‑rays that are produced at the metal target pass first through the glass wall of the evacuated tube within which they were made. They will then be passed through a layer of oil and some shielding. These materials act as filters and absorb a proportion of the lower energy X‑rays, regardless of the application that the radiographer is catering for. For most diagnostic procedures further filtering will probably be required. This further filtering is called ‘hardening’ the X‑ray beam. Recall that high-energy X‑rays are called hard X‑rays. Removing the low-energy components of a heterogenous X‑ray beam ‘hardens’ it because the average energy of the

(a)

Figure 15.23 (a) A compound fracture of the left radius near the wrist. (b) The metal plate produces strong contrast with the bone, owing to its greater attenuation.

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(b)

X‑rays will be increased. Filters are often made of metal. The amount of radiation absorbed by the filter will depend on the type of metal used, the X‑ray energy and the thickness of the filter.

Attenuation of X‑rays and creating contrast As X‑rays are passed through parts of the body it is inevitable that the body will absorb some of the energy of the X‑rays. X‑rays can also be scattered; that is, bounce off the atoms they meet. Also, if a diverging beam is used the X‑rays spread out as they get further from their original source. All of these mechanisms mean that the X‑ray beam will be less and less intense as it passes through more and more body tissue. The gradual reduction in the intensity of an X‑ray beam as it travels away from its source is called its attenuation. X‑rays attenuate differently through different materials, and it is this property that enables them to produce such useful images of the body. The radiographer’s objective is to produce images with optimal contrast; that is, to make the most of the different tissue attenuation levels. The amount of attenuation that occurs as an X‑ray travels through a given material depends on both the physical properties of the material and the frequency (or energy) of the X‑ray. Since this is such a complex matter, in order for radiographers to be able to compare the attenuation (and therefore penetration) of different X‑ray beam–tissue combinations, a term called the half-value thickness is defined. (This is discussed below.) The absorption and scattering processes that cause the attenuation of X‑rays in materials are complex and occur at an atomic level. In essence, X‑rays are able to interact with the electrons that are attached to atoms in the material and these interactions can produce various electromagnetic radiation emissions, electron emissions and positive ions. The interaction processes that occur in bone, for example, are different from the processes that occur in soft body tissue. As a result X‑rays attenuate very differently through bone than through soft tissue. This is why bones ‘show up’ on X‑rays.

Half-value thickness The penetration of a homogenous X‑ray beam in a given material is described by the term half-value thickness (HVT). The HVT of a beam–material pair is the thickness of the material that is required to reduce the intensity of the beam to half of its original value. Therefore, as a beam travels through each further HVT distance, the beam intensity would be one-quarter of its original intensity, then one-eighth, and so on. Dense materials, which have higher attenuation levels, will have smaller HVT values. The HVT of a particular beam–material combination is determined experimentally. Since X‑ray beam energies are especially selected to suit particular diagnostic procedures, HVT values are small and often quoted in millimetres. A heterogenous beam will not have a constant HVT as it passes through a material since the lower energy X‑rays in the beam are attenuated more rapidly. If you think of a heterogenous beam as made up of a multitude of photons of different energies, the low-energy photons will disappear more quickly. Therefore, the average photon energy within the beam is increased. (Recall that this is called hardening the beam.) This means that the HVT


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Figure 15.24 As the beam penetrates tissue, beam-hardening leads to an increase in the HVT.

exponential curve 0.5

Thickness of material

1 Fraction of original intensity

Fraction of original intensity

1

Homogenous X-ray beam

0.5

Heterogenous X-ray beam

HVT of beam increases as penetration increases

Thickness of material

value will also be increased. The smaller the range of frequencies that are present in an X‑ray beam, the more consistent the HVT as it penetrates a material.

Making an image Beam cross-sectional area If you’ve ever had an X‑ray you may remember the radiographer shining a light on the relevant body part first and checking that you were in the correct position. The size of the illuminated area is then adjusted by the use of little sliding doors, similar to the aperture adjustment on a camera. The illuminated area shows the radiographer exactly where the X‑ray beam will fall. The radiographer will always use as narrow a beam as possible to minimise your X‑ray exposure and reduce the blurring that scattered rays cause. Figure 15.25 An X‑ray of the lungs.

Film cassettes You are probably aware that when an X‑ray is taken an image is produced on a piece of photographic film. The photographic film used with X‑rays is a simpler version of the film that you would place in a camera to take black and white photographs. The patient must simply be positioned between the X‑ray machine and the film cassette. The image produced on the film is the result of the exposure of the film cassette to X‑rays. The more X‑rays that reach the film cassette the darker the film becomes. X‑rays passing through regions of low attenuation such as the lungs result in dark regions on the film. X‑rays passing through regions of relatively high attenuation, such as bone, produce white regions on the film. It is a common misconception that the film is responding purely to the X‑rays themselves. In fact X‑rays have very little effect on photographic films. They mostly pass straight through it; certainly any image would be too faint to see clearly. Instead the film responds to much lower electromagnetic radiation energy, around the energy levels of visible light. The X‑ray film sits inside a lightproof cassette with two screens sitting against either side of it. The screens contain a substance that fluoresces (gives off light) when an X‑ray strikes it. It is this fluorescence that causes a reaction in the film and produces the image (see Figure 15.26).

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X-ray fluorescent screen

X-ray

X-ray film fluorescent screen

film blackened by fluorescence

film blackened by X-ray

Figure 15.26 X‑rays cause some exposure of the film, but the image is mainly produced when the adjacent screens fluoresce and visible light is emitted onto the film.

The reliance on the fluorescence process means that the image is a little less sharply defined than it would have been if produced directly by X‑rays, but the image is much more intense. This means that it is much easier to distinguish the different types of tissue from one another since the image has greatly improved contrast. The use of fluorescence also means that exposure times for patients can be minimised. If X‑rays alone were used to darken the film, much longer exposure times would be needed to produce intense images.

Contrast enhancement There are some parts of the body where there is just not enough difference in X‑ray attenuation in the body tissues under investigation. The X‑ray images will not have enough contrast between the adjacent body parts for them to show up clearly. For example, it may be difficult for an X‑ray to show up the lining of the bowel, or the pathways through the urinary tract, or the walls of the stomach or blood circulating in the kidneys because they are all surrounded by soft tissue with similar attenuation values. In these circumstances a material can be introduced that will be more opaque to X‑rays.

Figure 15.27 A liquid containing an iodine compound has accumulated in the bladder. The isotope will have been injected into the bloodstream and a sequence of images will have checked its passage through the kidneys and ureter.


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Figure 15.28 An X‑ray of the stomach and upper small intestine with the use of a barium contrast medium. This is one of a series of images that would be taken that would track the path of the barium though the digestive tract. Note that the barium has begun to move into the large intestine too.

Aortic arch angiogram

If the digestive system is being imaged then a barium-based substance is introduced to the appropriate part of the digestive tract. If the circulatory system is being investigated then an iodine-based substance is injected. These substances are relatively dense and by blocking X‑rays while they lie in the vein or bowel, for example, they help to produce a high-contrast image of the outline of the item being investigated. An X‑ray of a bladder containing an iodine solution is shown in Figure 15.27. This X‑ray is likely to be part of a series of X‑rays that have tracked the passage of the iodine through the kidneys. A set of X‑rays, taken over a period of time, can monitor how substances are processed by the body.

Diagnostic procedure: a series of upper gastrointestinal X‑rays If a patient is experiencing particular symptoms related to the early stages of the digestion of their food, their doctor may request a series of upper gastrointestinal X‑rays. This is an X‑ray examination of the oesophagus, stomach and the very first part of the small intestine. As discussed earlier, since the attenuation of the soft tissue in this region is not very different for the different organs, a contrast agent such as barium would be used to coat the stomach wall so that it may be examined under X‑ray. The patient would need an empty stomach for the barium to successfully coat the stomach walls, so he or she would not be able to eat or drink for eight or so hours before the procedure. The barium substance is delivered in a cup of chalky, white liquid (a ‘barium milkshake’) that the patient must drink just prior to the X‑ray procedure. The patient is then given a substance that will help to produce gas in the stomach, to make it bloat and hold its full shape. This gas is just the same as the bubbles in fizzy soft drinks. The challenge here for the patient is to not burp, as a bloated stomach produces the best images! Different types of X‑ray machines may be used to take images of parts of the oesophagus and stomach. The barium substance will pass through the digestive tract in the next day or so.

Diagnostic procedure: the angiogram An angiogram is an X‑ray procedure that looks at the function of a blood vessel or organ. A contrast medium, such as iodine, is injected into blood vessels and the resulting X‑ray images show the fine details of the arteries or veins that are carrying the contrast medium. They are particularly useful in investigating the delivery of blood to organs, such as the kidneys, heart and brain. The image shown in Figure 15.29 looks at blood flow out of the aorta, the main artery that delivers oxygenated blood from the heart to the body.

ExtCl = left external carotid artery IntCl = left internal carotid artery CCR = right common carotid artery CCL = left common carotid artery VAR = right vertebral artery VAL = left vertebral artery SbCIR = right subclavian artery SbCIL = left subclavian artery BrcT = brachiocephalic trunk AA = aortic arch DesA = descending aorta

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CT scans During the 1970s a new diagnostic imaging technique that utilised X‑rays was developed called computer tomography (CT) scanning. The word tomography comes from the Greek term tomos meaning slice or knife. Figure 15.29 This angiogram shows blood vessels that deliver blood to the parts of the body above the heart. The largest arteries are the ones that eventually lead up the neck and deliver blood to the brain.

CT scans are an image of a slice of the body. To understand the usefulness of CT scans, imagine a large sausage of manufactured meat. By taking a set of thin slices of the sausage you could see exactly what it was constituted of. If you saw anything unexpected in your slice (or cross-section) you could determine exactly where in the sausage it was originally located (see Figure 15.30). The main principles behind obtaining a CT scan image are the same as for X‑rays. The main difference is that a set of thin X‑ray beams and detectors are used. The width of the X‑ray beam determines the thickness of the ‘slice’ of the body that is imaged, typically about 3–5 mm. The X‑ray source and detectors are located in a circle around the body, as shown in Figure 15.31. Beams passing through a cross-section of the body are detected by the array of detectors, and the intensity of the X‑ray beam that reaches each tiny detector is converted to digital information and sent to a computer. The computer allocates a corresponding shade from a black– grey–white scale. Typically each detector represents a square of tissue of 1 mm × 1 mm. The X‑ray source and detectors are then rotated through a small angle and the detectors record another set of readings. The computer uses a set of these readings to construct an image of the cross-section of the body for this narrow slice that has been analysed. An example is shown in Figure 15.32. While this relatively new technique provides much more detailed images for doctors, the doses of radiation received by the patient are much higher. Recall that a typical dose from a diagnostic chest X‑ray is 0.017 millisieverts, but a CT scan for the chest may provide a dose of up to 8 mSv. This is a considerably higher dose and reason enough to settle for the less sophisticated tool if it will do the job!

Figure 15.30 A slice of something provides lots of information about the location of its contents.

Figure 15.31 The CT scan.


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(a)

(b)

Figure 15.32 (a) The locations at which slices of the body are to be taken; (b) a cross-sectional view obtained via CT scan. The kidneys are easily identified. Can you identify anything else?

Magnetic resonance imaging Magnetic resonance imaging (MRI) can be carried out using elements that have an odd number of nucleons, such as hydrogen, fluorine‑19, sodium‑23 and carbon‑13. Since hydrogen is abundant in the body, it is used for most medical MRI. MRI uses non-ionising radiation and therefore it is considered safer, though much more expensive, than X‑ray imaging. If a person is placed in a very strong magnetic field (of around 1 tesla), certain protons in their body tissue will align themselves with the magnetic field, a little like the needle of a compass. If the patient is then exposed to a pulse of low-energy electromagnetic radiation (radio frequency), the protons shift their alignment. After the pulse has passed the protons move back to their magnetic field alignment. As they do so they emit low-frequency electromagnetic radiation, which is detected by the MRI machine. This emission is used to create images of the locations of the body tissue. The different tissues in the body have different concentrations of hydrogen; therefore, different body tissues will create contrasting shades on a magnetic resonance image (see Figure 15.33). Like CT scans, magnetic resonance images are cross-sections or slices of parts of the body. MRI scanners can take unusual oblique cross-sections of the body. Since they can discriminate very well between the different types of soft tissue, they are used for imaging the brain, heart, organs of the abdomen, reproductive system, muscular regions and joints, arteries and veins throughout the body. Later we will see that they are like PET scans in that they can show the functions of different parts of the body. They are particularly useful in showing the function of the brain since the patient is awake during the imaging. The patient could be asked to do some physics questions and the brain activity then tracked!

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Figure 15.33 MRI is used to investigate sports injuries, as every type of body

tissue can be clearly imaged, not just hard or soft tissue. Previously patients may have had an X‑ray and an ultrasound in order to image both types of body tissue.

15.4 summary Diagnostic X‑rays • X‑rays are part of the electromagnetic spectrum, with wavelengths between 10‑11 and 10‑8 m. • An X‑ray machine produces a continuous spectrum of X‑rays by the sudden deceleration of the incident electrons as they come near to the nuclei of the target metal (tungsten). Each type of target material also produces specific line spectra. • The energy of an X‑ray photon is given by E = hf, where E is the energy of the X‑ray photon in joules (J),

MRI shoulder coronal view

Del = deltoid Acr = acromion Cl = clavicle Trap = trapezius muscle HHm = head of humerus Hm = humerus Glnf = glenoid front of scapula SubS = subscapularis muscle

h is Planck’s constant (6.62 × 10‑34 J s), and f is the frequency in hertz (Hz). • The electronvolt (eV) is a (alternative) unit of energy: 1 eV = 1.6 × 10‑19 J. • The HVT of the beam–material pair is the thickness of the material that is required to reduce the intensity of the beam to half of its original value. • Contrast media such as barium or iodine can be used when imaging soft tissue.

15.4 questions Diagnostic X‑rays 1 Describe the function of the following parts of an X‑ray machine: a target b filament c electric field d lead cover e aluminium filter

a What energy (in eV) does the electron gain? b Convert this energy into joules.

2 Calculate the frequency of X‑rays travelling in air with a wavelength of: a 1 × 10‑10 m b 1 × 10‑9 m

7 Would a contrast-enhancing medium be used to produce an X‑ray image of: a the blood flow to the muscles of the heart? b the ribs? c the bladder? d the bones in the foot?

3 Calculate the photon energy of X‑rays with a frequency of 5.0 × 1018 Hz.

8 What characteristics must the element chosen as the target material in an X‑ray machine possess?

4 Why does an X‑ray tube require an extremely high voltage?

9 What changes would occur in the output of an X‑ray machine if: a the accelerating voltage was reduced? b the target was replaced with a different element?

5 An electron has 1.6 × 10‑14 J of energy when it collides with a tungsten target in an X‑ray machine. Assuming all of the electron’s energy is converted to an X‑ray, what is the frequency of the X‑ray? 6 An electron is accelerated through a potential differ ence of 100 000 V.

10 If a cathode tube has a potential difference of 10 000 V applied across it, is a single decelerated electron capable of producing a photon in the X‑ray region of the spectrum?


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pes o t o s i o i d a r , y Radiotherap 15.5 in medicine and PET Ionising radiation Cancer is a general term that actually incorporates hundreds of different diseases. The term tumour describes the growth of abnormal cells. Tumours are not all cancerous. Benign tumours are abnormal, but not spreading, and often grow slowly. They can be single growths that remain in one place. Sometimes doctors don’t even remove them unless they are affecting a nearby part of the body. Cancer cells (malignant tumours) can grow in just about any part of the body. Malignant tumour cells can grow into and invade the surrounding healthy tissue. Some cells may break away and be carried by the bloodstream; they then settle in other parts of the body, spreading the cancer. Their growth can be aggressive and they can severely affect the ability of the invaded body parts to function. It is well accepted that exposure to radiation can cause cancer, but it can often also provide a cure. A common method of cancer treatment is the use of ionising radiation to destroy the cancerous cells. Particular types of electromagnetic radiation, namely X‑rays and gamma rays, can kill living cells by ionising molecules in them. Ionisation means the forming of ions—positively and negatively charged particles. When cells are ionised cell death can result. Therapeutic ionising radiation with X‑rays or gamma rays targets the DNA in the cancerous cell. Radiation either breaks the DNA strand, resulting in cell death, or at least damages the DNA so that the cell cannot reproduce. Ionising radiation can produce ionised water molecules in the cells and highly reactive hydroxide ions that can damage parts of the cell, including DNA. Cancerous cells are rapid-growth cells. This means that they reproduce at a faster rate than most other cells in the body; but this also means that they are more susceptible to being destroyed by radiation. Generally, normal healthy cells recover from radiation treatment more quickly than cancer cells. Regardless, radiation therapy programs must be designed to provide the optimal dose of radiation to the cancerous cells while minimising the exposure of healthy cells. There are some relatively rapidly reproducing cells in the body that don’t cope with radiation exposure, such as the reproductive organs, the kidneys, the liver, bone marrow and the eyes. Radiation exposure of healthy cells can cause their ionisation and the production of malformed cells. So while the ‘benefit outweighs the risk’ for the patient with a malignant tumour, doses should always be kept to a minimum for the healthy cells of the patient.

X‑ray radiotherapy treatments The X‑rays used in therapy treatments are of a much higher frequency and energy than those used to produce diagnostic images. Recall that we discussed soft and hard X‑rays. Hard, high-energy X‑rays have deeper penetrating ability and these are produced by accelerating voltages of between 4 and 25 million volts (4–25 MV). Although the principle of X‑ray production from rapidly decelerated electrons still applies, owing to the extremely high voltages used these X‑ray machines will have a slightly different set‑up. Linac machines (an abbreviation of linear accelerator) are

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used. Parts of these machines rotate so that the beam of X‑rays can be sent into the patient from different directions. First, a CAT scan or other diagnostic procedure is carried out to accurately locate the malignant tumour. The surrounding anatomy of the patient is also carefully plotted and its state of health recorded for monitoring. The information about the tumour location is fed into the computer that controls the X‑ray machine (linac). The patient is then positioned so that the tumour is located at the isocentre (or pivotal point) of the X‑ray machine. A light source such as a low-wattage laser is often used to project an image of a grid onto the patient to assist their correct positioning. Sometimes reference lines are drawn on a patient’s skin with semipermanent ink so that their alignment is consistent over consecutive treatments. Alternatively, a cast may be made if a patient needs to have their head and neck in a particular position, for example. X‑ray treatment beams are so precise that it is worth accurately positioning the patient and then strapping them into position. A quick diagnostic X‑ray is then taken to check that the tumour lies in the centre of the X‑ray beam. The patient’s specialist doctor will have provided a treatment plan. A beam of the appropriate energy (or frequency) and penetration is produced. This beam of X‑rays is sent into the tumour for the prescribed time, but of course it must pass through healthy body tissue on the way. To minimise exposure of healthy cells, the X‑ray beam is sent into the body from a series of different angles. Usually the patient lies still and the linac machine (X‑ray source) and/or the table that the patient is lying on move around so that the beam is sent into the body from different directions. Any healthy cells in the path of the beam will receive a dose of X‑rays during each exposure. However, the tumour receives a dose during every exposure, accumulating a much larger dose than any of the healthy cells. Treatments aim to have the maximum dose occurring at the centre of the tumour and the entire tumour should lie in a region of less than 5% dropoff in dose size. To minimise the side effects of radiation (as covered in Chapter 1) the radiation treatment is delivered in a series of small-dose sessions, which are spread over a number of days or weeks. The dose at the tumour can be as high as tens of grays if required. With luck, the healthy cells, having received a much smaller dose, will have time to recover between treatments. Ideally, the cancerous cells do not recover as well, and each treatment diminishes their presence in the body. Radiotherapy can also be carried out using gamma radiation, which has a higher energy than X‑rays. The gamma rays are produced from a radioactive source (discussed below) and therefore the source of the radiation cannot be turned on and off. Instead the gamma source is located in a machine that has thick lead walls. When the machine is not being used for therapy it will be sealed. Because of the increased exposure risks with the storage of strong gamma sources, radiotherapy using X‑rays is far more common in hospitals.

Radioisotopes in medicine Your studies of radioactivity (Chapter 1) will have introduced you to the three different types of radiation that can be emitted by unstable or radioactive atoms—alpha (α), beta (β) and gamma (γ) radiation. The branch


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of medicine that uses radioisotopes is called nuclear medicine. Just as was the case with both X‑rays and ultrasound, radioisotopes can be used in two different ways in medicine—either to aid the diagnosis of disease or as therapy in order to treat disease. The widespread use of nuclear medicine did not start until the early 1950s. Today, there are over 100 different nuclear medicine imaging and therapeutic procedures. They can provide information about every major organ system in the body. The Australian Nuclear Science and Technology Organisation (ANSTO) is responsible for the production of the majority of medical radioisotopes used in Australia. This organisation supplies a wide range of radioisotope products and services to nuclear medicine and industry throughout Australia, New Zealand and Asia. These isotopes are mainly produced in Australia’s only nuclear reactor at Lucas Heights near Sydney and in the cyclotron at Camperdown, New South Wales. Many large hospitals have their own nuclear medicine departments where products from ANSTO are processed within the hospital immediately before use. Radioactive isotopes are produced in nuclear reactors and cyclotrons. There are some differences in the products that each produces, and so both types of facilities are needed. In reactors, medical radioisotopes are usually produced by bombarding a stable isotope of a substance with neutrons, which are plentiful in a reactor. This creates a radioactive isotope of the original substance. The most commonly used radioisotope, technetium‑99m, is the daughter product of an atom produced in reactors and is used in the diagnosis of brain and heart disease and for studying lung, kidney and liver function. Iodine‑131, which is used to detect and treat thyroid cancer, is also a reactor-produced radioisotope. In a thermal nuclear reactor a rod of the element tellurium‑130 is bombarded with slow-moving neutrons, forming the unstable tellurium‑131 isotope according to the equation: 130 Te + 10n → 131 Te + γ 52 52 The tellurium‑131 undergoes spontaneous decay that results in the production of iodine‑131: 131 Te → 131 I + –10β + v 52 53 After a month or so in the reactor, the rod is removed and the tellurium and iodine are chemically separated from one another. In cyclotrons, radioactive isotopes are produced by bombarding atoms with charged particles. Cyclotron-produced radioisotopes have only been produced in Australia since 1992. According to ANSTO, three of the most widely used cyclotron-produced radioisotopes in medicine are gallium‑67, thallium‑201 and iodine‑123. Gallium‑67 is used mainly in the detection of soft tissue tumours and infections. Thallium‑201 is instrumental in detecting heart disease. Iodine‑123 is used in imaging the thyroid gland and can be labelled onto a range of drugs to detect stroke, epilepsy, kidney and heart diseases, as well as some cancers.

Diagnostic imaging with radioisotopes The radioisotopes used in diagnostic procedures must be able to provide useful information such as an image of an internal structure in the body, or an image of the various stages in the function of an organ. We have already looked at the use of barium and iodine as non-radioactive providers of contrast in X‑ray images.

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In diagnostic procedures, the gamma radiation output by radioisotopes that have been put into the body is detected by gamma cameras (rather than the films used with X‑ray radiation). The radioisotope is first attached to a compound (drug) that will be taken up by specific organs or disease sites within the body. This is called a radiopharmaceutical. The compound is therefore called the carrier and the gamma radiation from the radioisotope produces the image.

A RADIOPHARMAC…UTICAL is a radioactive material combined with a chemically or biologically active compound that is taken up in the body. One particular radioisotope may be attached to a number of different types of compounds, each one designed for a particular application. For example, technetium‑99 is attached to one carrier compound when used to investigate the thyroid and another when it is used to investigate the brain. The brain consumes large quantities of glucose, whereas the thyroid takes up iodine.

Figure 15.34 Production of a pharmaceutical.

Once the radiopharmaceutical has delivered the radioisotope to the designated part of the body, the gamma camera is simply placed in the vicinity of the patient for a set time so that the emitted radiation produces a digitised image. Gamma cameras are designed for use with gamma radiation of a small range of energy values—a typical gamma camera will detect radiation between 126 and 154 keV. Scattered gamma rays are prevented from entering the camera, as its front plate is a thick layer of lead with narrow tubes drilled through it. These tubes ensure that only gamma rays travelling perpendicularly out from the body are imaged. Although this is necessary for the clarity of the image it means that a large portion of the patient’s radiation exposure does not contribute to the image.


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Figure 15.35 For image clarity a gamma camera prevents the entry of scattered gamma rays.

Choice of radioisotope for diagnostic imaging

saline central alumina column

molybdenum coating technetium-99m coating formed

saline containing technetium retrieved

Figure 15.36 Hospitals must order new samples of the mother isotope molybdenum each week, but this is frequently enough to ensure a continual supply of the widely used radioisotope technetium‑99m, which decays via a gamma emission with a half-life of 6 hours.

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According to the supplier ANSTO, for a radioisotope to be used for diagnostic imaging it must: • have a short half-life (hours), which is appropriate for the time taken for the diagnostic procedure • not emit alpha or beta radiation, because these particles would be trapped in the patient’s tissues and they would not be detected externally • emit gamma radiation of an energy that can be detected by a gamma camera • be available in the highest possible activity, but not be toxic to the patient or react with drugs used at the same time. According to ANSTO the reactor-produced technetium‑99m is used in more than 80% of the estimated 100 000 patient studies that are performed worldwide each day. After technetium‑99m, a series of cyclotron-produced radioisotopes, such as thallium‑201, gallium‑67, indium‑111 and iodine‑123, are the next most frequently used.

The production of technetium‑99m Technetium‑99 is the semistable daughter nucleus of the atom molybdenum (99 Mo), which is produced in a nuclear reactor at Lucas Heights. Each week 42 some molybdenum is brought to hospitals that have a technetium generator. Technetium generators consist of a central column of alumina to which the molybdenum is chemically bonded. As the molybdenum undergoes beta decay, technetium is formed on the outside of the molybdenum layer (see Figure 15.36). A saline solution is then washed over the technetium and this solution is collected in a syringe. This solution contains concentrated technetium and can be diluted for use. Of course all of this is housed within a lead-lined evacuated chamber so that the radioactivity is contained. A recent development in the use of the technetium‑99m isotope is the development of a new radiopharmaceutical that, when injected into the body, will accumulate at the site of any ulcerous infections in the body. It can be used to investigate local areas in the body or for whole body investigations. The pharmaceutical takes 5 minutes to prepare and the patient will be ready for scanning 1–2 h after the injection.

Table 15.5 Medical isotopes manufactured by ANSTO Reactor-produced radioisotopes Chromium‑51

Used to label red blood cells and quantify gastrointestinal protein loss

Copper‑64

Used to study genetic disease affecting copper metabolism

Iodine‑131

Used to diagnose and treat various diseases associated with the human thyroid. Also used in diagnosis of adrenal medullary tumours and for imaging suspected neural crest and other endocrine tumours

Iridium‑192

Supplied in wire form for use as an internal radiotherapy source

Molybdenum‑99

Used as the ‘parent’ in a generator to produce technetium‑99m, the most widely used isotope in nuclear medicine

Phosphorus‑32

Used in the treatment of excess red blood cells

Samarium‑153

Used with ethylenediaminetetramethylene phosphonate (Quadramet) to reduce the pain associated with bony metastases of primary tumours

Technetium‑99m

Used to image the brain, thyroid, lungs, liver, spleen, kidney, gall bladder, skeleton, blood pool, bone marrow, salivary and lachrymal glands and heart blood pool and to detect infection

Yttrium‑90

Used for liver cancer therapy

Cyclotron-produced radioisotopes Gallium‑67

Used in imaging to detect tumours and infections

Iodine‑123

Used in imaging to monitor thyroid function and detect adrenal dysfunction

Thallium‑201

Used in imaging to detect the location of damaged heart muscle

Carbon‑11, nitrogen‑13, oxygen‑15, fluorine‑18

Used in positron emission tomography (PET) to study brain physiology and pathology, for detecting location of epileptic foci and in dementia, and psychiatry and neuropharmacology studies; also used to detect heart problems and diagnose certain types of cancer

Source: www.ansto.gov.au/ari/brochures_misc/rad1.html

Diagnostic imaging with PET and radioisotopes Positron emission tomography (a PET scan) is used to study the function of the brain, heart and tissues throughout the body. PET scanning uses similar principles to diagnostic imaging with radioisotopes. A radioisotope is introduced to the body attached to a compound that will cause it to accumulate in a particular part of the body. Since PET scans are often used to scan the brain, the radiopharmaceutical often involves glucose (sugar) as the carrier substance. The radiopharmaceuticals used in PET scans are designed so that electrons from the investigated tissue combine with positrons (small positively charged particles) from the radioactive tracer to produce gamma rays (similar to X‑rays). These rays are recorded by a gamma camera and reconstructed into three-dimensional pictures by the computer. The radioactive materials that are used in the PET scan only last a short time, so a nuclear particle accelerator cyclotron is needed to make them in a location close to the patient. A cyclotron accelerates protons to a very high energy and these are then caused to collide with a stable gas or liquid isotope. This isotope is then temporarily radioactive. It will become a radiopharmaceutical by attachment to an appropriate carrier substance, for example sugar.

Figure 15.37 A PET scan.


555

A PET scan can produce very high resolution ‘slices’ of the brain. It can show which tissues are living and using energy. Therefore, it would show up any dead tissue, in the muscles of the heart for example. PET scans can also show live or dead brain tissue after a stroke and where seizures may be occurring in the brain. It can even show the functioning of the brain during thought processes. PET can be used to find tumours and then monitor how well they respond to treatment. PET scans are very expensive and involve patient exposure to ionising radiation, but they can decrease the patient’s overall medical costs by avoiding exploratory surgery or showing where surgery can do the most good.

Therapeutic treatments with radioisotopes

Physics in action

The medical approach of treating cancerous growths with ionising radiation has been discussed earlier. When using radioisotopes the source of the ionising radiation is placed inside the body; it is usually localised in the affected organ. This can be achieved in a number of different ways. A radiopharmaceutical can be selected that when ingested or injected will accumulate in a particular part of the body. Iodine‑131 or technetium‑99m will accumulate in the thyroid and radiate a cancerous growth there. Phosphorus‑32 will accumulate in the bone marrow. Technetium‑99m can be used in a number of radiopharmaceuticals that will cause it to accumulate in the lungs or kidneys or brain, for example. Samarium‑153 can accumulate in breast or prostate tumours. Brachytherapy uses physical sources that are temporarily implanted in the body so that they radiate and kill the cells in the local area. For example, iridium‑192 is produced in a thin wire that is then inserted via a catheter into the tumour. This is often used for tumours in the head and breast. When selecting isotopes for use in therapy, different criteria apply. Unlike diagnostic isotopes, isotopes used in therapy should be sources of alpha and beta radiation. These are ionising radiations and need to be of sufficient energy to penetrate the diseased section of the body. It is convenient if the isotope is also a weak source of gamma radiation so that its presence in the body can be monitored by the gamma cameras discussed earlier. Radioisotopes commonly used in therapy include iodine‑131, phosphorus‑32 and yttrium‑90, and several others are being investigated for possible application.

A nuclear approach to liver cancer Liver cancer is a very common disease and very difficult to treat. However, thanks to a new invention developed by an Australian company, SIRTeX Medical, incorporating products manufactured by the Australian Nuclear Science and Technology Organisation, effective treatment is now available in Australia. SIR‑Spheres®, SIRTeX Medical’s registered tradename for the product, are tiny cancer treatment weapons incorporating radioactive particles which are selectively placed in liver cancers. Quite simply, they emit radiation to kill cancer cells that in turn shield the rest of the healthy liver from most of the irradiation. SIR‑Spheres® have been approved by the Therapeutic Goods Administration for use in Australia and by the United States Food and Drug Administration for use in the US.

556

Detailed studies

Primary liver cancer is common in Asia and Africa and is linked to liver damage of some kind, for example from hepatitis B and C. Secondary liver cancer is common in all countries, including Australia, and occurs through the spread of cancer from other organs to the liver. Secondary cancers may form at other sites such as in the lungs or bone; however, it is frequently the liver that fails first, causing death. Conventional treatment of liver cancer by chemotherapy and conventional radiation beam therapy is not effective against primary liver cancer cells. However, the new invention developed by SIRTeX makes it possible to treat liver cancer effectively with radiation and offers a new option to patients who might otherwise be left untreated.

The aim is to increase the amount of radiation specifically targeted on the tumour, while sparing the normal liver. The treatment centres on tiny spheres of the radioactive isotope yttrium‑90. This isotope emits ionising beta radiation that destroys cancer cells. To make the necessary medicine, the nuclear research reactor at Lucas Heights is used to irradiate a nonradioactive isotope of yttrium (pronounced itt‑ree‑umm). ‘The process of irradiation usually takes three and a half days,’ and each atom of the isotope yttrium‑89 captures a neutron, forming the isotope yttrium‑90, which then emits beta radiation. The radioactive yttrium is chemically combined to form resin microspheres, and these tiny anti-cancer weapons are then transported to clinics.

The spheres, which measure approximately a thousandth of a millimetre in diameter, are injected into the patient’s main liver artery (the hepatic artery), one of the two vessels that supply blood to the liver. Since liver cancer cells get blood almost exclusively from the hepatic artery, these tiny spheres zero in on the cancer. Only a small number of the microspheres migrate to healthy liver tissue. As beta radiation has a short penetration distance, more than 98% of the radiation is contained within the liver, and most of this is delivered to the cancer. The treatment, often used in combination with chemotherapy, is more precise than conventional radiotherapy, which is based on external beam sources.

15.5 summary Radiotherapy, radioisotopes in medicine and PET • Gamma rays and X‑rays can kill cancerous cells by ionising them. • Radiation can break the DNA strand of a living cell, resulting in cell death or at least damage to the DNA so that the cell cannot reproduce. • The X‑rays used in therapy treatments are of a much higher frequency and energy than those used to produce diagnostic images. To minimise exposure of healthy cells the X‑ray beam is sent into the body from a series of different angles. • Radioactive isotopes are produced in nuclear reactors or in cyclotrons. • In diagnostic procedures the gamma radiation output by radioisotopes that have been put into the body is detected by gamma cameras.

• A radiopharmaceutical is a radioactive material combined with a chemically or biologically active compound that is taken up in the body. • Technetium‑99 is the most commonly used radioisotope in medicine. • Positron emission tomography (a PET scan) is used to study the function of the brain, heart and tissues throughout the body. • In radiation therapy a radiopharmaceutical can be selected that will accumulate in a particular part of the body and radiate a cancerous growth there.


557

15.5 questions Radiotherapy, radioisotopes in medicine and PET 1 a What is the most common radioisotope used in nuclear medicine? b How is it supplied to hospitals?

6 Which areas of the body need special protection from ionising radiation during nuclear medicine procedures? Why?

2 Technetium‑99m is the semistable daughter nucleus of the atom molybdenum (99 Mo). Write the decay 42 equation for the production of technetium‑99m.

7 Why are radiation treatments usually given in courses of lots of low doses spread over days or weeks?

3 What are the required characteristics of an isotope that will be used as a radioactive tracer?

8 State two reasons why a cancer patient undergoing radiation treatment should be given ultrasounds, CAT scans or X‑rays prior to their treatment.

4 Briefly explain how ionising radiation kills cancer cells. 5 Why do diagnostic radioisotope tracers need to be gamma emitters, but therapeutic radioisotopes need to be alpha, beta and gamma emitters?

9 What is a radiopharmaceutical? 10 Explain the apparent contradiction that ionising radiation can both cause and cure cancer.

chapter review 1 Why is a coupling gel needed between the transducer and the skin of the patient undergoing an ultrasound procedure? 2 When selecting isotopes for use in therapy, what criteria apply? 3 If an ultrasound frequency of 1.00 MHz was used to send a sound wave through the following substances, what wavelength sound wave would be produced? a fat b blood c bone (v = 3700 m s‑1) 4 What is a transducer? 5 Why are cancer cells generally more vulnerable to ionising radiation than most other healthy cells in the body? 6 Explain the piezoelectric effect. 7 What features do the original X‑ray tubes and modern X‑ray machines have in common? 8 Why does the electron accelerate inside the X‑ray machine? 9 What are the differences between the diagnostic and therapeutic applications of nuclear medicine? 10 Calculate the photon energy of X‑rays with a frequency of 4.0 × 1018 Hz in: a joules b electronvolts.

558

Detailed studies

11 a Explain the term acoustic impedance. b Calculate the acoustic impedance of muscle tissue that has a density of 1025 kg m‑3 and in which sound waves travel at 1590 m s‑1. 12 Why should a radioactive tracer have a short-half life? 13 What causes the deceleration of the electron inside the X‑ray machine? 14 State the seven categories of the electromagnetic spectrum. Quote one known feature of each category of radiation. 15 If X‑rays whose frequencies range between 1.0 × 1016 and 1.0 × 1021 Hz are created by an X‑ray machine, what range of wavelengths have been produced? 16 What mechanism is responsible for the presence of line spectra in the spectrum of X‑rays produced by an X‑ray machine? 17 Are hard X‑rays or soft X‑rays more suitable for use in a mammogram? Explain. 18 a Explain the term half-value thickness (HVT). What is it used to describe? b Why is the HVT not a constant value when heterogenous X‑ray beams are passing through a material? 19 What is the difference between a benign tumour and malignant tumour?

20 Describe two different processes that contribute to X‑ray attenuation as X-rays pass into or through a substance.

25 Which X‑rays have the higher energy values, X‑rays used in imaging or X‑rays used in radiotherapy? Why?

21 A transducer produces a sound wave of intensity of 100 W m‑2, and its interface has a cross-sectional area of 7.0 cm2. How many joules of sound energy are produced per second?

26 Why do the cassettes used by radiographers contain a fluores cent material?

22 Is the following statement true or false? X‑rays pass easily through the lungs and produce a black region on the film; X‑rays are partially blocked by bone and result in the film being underexposed and white. 23 Who is responsible for the production of the majority of medical radioisotopes used in Australia? 24 Construct a table showing the applications, advantages and disadvantages of using ultrasound, X‑rays, MRI and PET for diagnostic imaging.

27 Why are alternating voltages applied to the piezoelectric crystals in an ultrasound transducer? 28 If a patient has a leaky heart valve, some blood will flow backwards through the valve. Describe which ultrasound techniques would be used to diagnose this illness. 29 Why do radiography departments in hospitals have large posters that say to patients: ‘Please let us know if there is a chance that you are pregnant’? 30 Why are the targets used in X‑ray machines often rotating targets?


559

SKILLS R O T C E V A X I APPEND Scalars Many physical quantities are fully described by a magnitude (or size) only. For example, the distance travelled by an athlete when completing three lengths of a 100 m track is 300 m. There is no direction associated with this quantity—it would not matter if the athlete had started from the other end of the track. Distance travelled has been fully described by just one numerical piece of information—300 m. A quantity that requires a magnitude only to describe it fully is known as a scalar. Other examples of scalar quantities include speed, mass, time, temperature and refractive index.

Vectors

x = 50 m

_

Figure A1 A displacement of 50 m can be

represented by the displacement vector x.

+

Some physical quantities, by definition, require two pieces of information to describe them completely—a magnitude and a direction. For example, when describing the displacement of an athlete who has completed three lengths of a 100 m track, it is necessary to indicate how far the athlete finished from his/her starting point, as well as the direction in which he or she finished relative to the starting point. So, for example, the displacement of the athlete could be 100 m north or positive 100 m. Quantities such as this are called vectors. A vector can be represented as a directed line segment where the length of the line represents the magnitude or size of the quantity, and the arrowhead indicates the direction. In the text, a vector is indicated by bold italic type. This is why the symbol for displacement is shown as x. Some other vector quantities are velocity, acceleration and force.

Worked example A1 As Benjamin rides to school, he passes a petrol station and a milk bar. home

petrol station

0

250

500

milk bar

school

750

1000 m

Position

Figure A2 Benjamin’s route to school.

When he reaches the milk bar, he stops for an ice-cream, then realises that he has spent his lunch money and so returns home before continuing on to school. a What is the distance travelled in his journey? b What is the displacement as he travels from home to school? c Calculate Benjamin’s displacement as he travels from the petrol station to the milk bar and then returns home.

Solution a

The distance that Benjamin travels must take account of his back-and-forth route. Distance travelled d = 750 + 750 + 1000 = 2500 m b Displacement x = final position – initial position = 1000 – 0 = 1000 m c Displacement x = 0 – 250 = –250 m

560

Appendix A

Adding vector quantities When analysing movement, it is often necessary to add vector quantities. Consider the example in Chapter 4 of a swimmer, Sophie, doing laps of a 50 m pool. We will analyse her displacement as she travels from the warm-up position to the starting block, and then from the starting block to the far end of the pool. Her resultant, or total, displacement can be determined by vector addition of the two separate displacements. The addition of two vectors is performed by placing the tail of the second vector at the head of the first. The sum of these vectors gives the total displacement x, which runs from where the first vector starts to where the second vector ends; i.e. x = x1 + x2… Figure A4 shows how the resultant displacement of +60 m is obtained. (a) –10

0

10

20

30

40

50

60 m position

–10

0

10

20

30

40

50

60 m position

–10

0

10

20

30

40

50

60 m position

(b)

(c)

Figure A3 Sophie moves from behind the starting position to the starting position and then dives into the pool and swims her first 50 m lap.

(a)

x1 = 10 m

x2 = 50 m

_

+ start 10 m

50 m

+

finish

(b) total displacement = 60 m

Figure A4 (a) Sophie travels from her warm-up position to the 50 m mark. (b) Her total displacement, x, can be found by adding vectors x1 and x2. (c) Her displacement, x, is a vector that runs from the start of x1 to the end of x2.

In situations where only the magnitude of a vector quantity needs to be considered, the quantity will be represented in light italics.

Multiplying a vector by a scalar When a vector is multiplied by a scalar, the magnitude of the vector changes accordingly. For example, if x is a displacement of 5 m north, then 4x is: x + x + x + x = 20 m north If the vector is multiplied by a negative term, then the direction of the vector will reverse. So using x as 5 m north once again, –x will indicate a displacement of 5 m south, and –3x will be 15 m south.

Appendix A

561

Worked example A2 If x1 is a displacement of 10 m east and x2 is 6 m west, determine: a x1 + +x2 b x1 + +3x2.

Solution (a)

x1 = 10 m

x1 + x2 =

x2 = 6 m

10 m

= x

W (b)

+

x1 + 3x2 =

6m

E x1 = 10 m

+

3x2 = 18 m

x

=

10 m 18 m

Figure A5 Illustration of the addition of vectors x1 and x2.

a

The total displacement vector is: x = x1 + x2 = 10 m east + 6 m west As shown in Figure A5a, the total displacement vector x is 4 m east. b The total displacement vector is: x = x1 + 3x2 = 10 m east + 18 m west As shown in Figure A5b, the total displacement vector x is 8 m west.

Adding vectors in two dimensions Vectors can be used to analyse the motion of a person or body moving in two dimensions. Consider Jessica, a bushwalker, who hikes 5 km north then 8 km east. The distance travelled, d, is 5 km + 8 km = 13 km; however, her total displacement is: x = 5 km north + 8 km east 8 km

finish

5 km 5 km

+

=

x

N start

Figure A6 The displacement, x, of the hiker is a vector than runs directly from the start to the finish of her journey.

The magnitude of Jessica’s total displacement can be determined using Pythagoras’s theorem: magnitude of displacement = √52 + 82 = √89 = 9.4 km The direction of the displacement can be determined using trigonometry: tan θ = 8/5 = 1.6 θ = tan–1 1.6 = 58° This direction can be stated as 58°T. The resultant displacement of the bushwalker is 9.4 km 58°T. The distance travelled is 13 km.

562

Appendix A

APPENDIX B S

I UNITS

The standard units of measurement The accurate and easy measurement of quantities is essential in both everyday life and for scientific investigation. Over the centuries, many different systems of measuring physical quantities have been developed. For example, length can be measured in chains, fathoms, furlongs, yards, feet, rods and microns. Some units were based on parts of the body. The cubit was defined as the distance from the elbow to the fingertip, and so the amount of cloth that you obtained from a tailor depended on the physical size of the person selling it to you. The metric system was established by the French Academy of Science at the time of the French Revolution (1789–1815) and is now used in most countries. This system includes units such as the metre, litre and kilogram. Countries of the British Empire adopted the British Imperial system of the mile, gallon and pound. These two systems developed independently and their dual existence created problems in areas such as trade and scientific research. In 1960, an international committee set standard units for fundamental physical quantities. This system was an adaptation of the metric system and is known as the Système Internationale d’Unités (SI) system of units.

Table B1 The SI units identify the seven fundamental quantities whose basic value is defined to a high degree of accuracy Fundamental quantity

SI unit

Mass

kilogram

SI unit symbol kg

Length

metre

m

Time

second

s

Electric current

ampere

A

Temperature

kelvin

K

Luminous intensity

candela

cd

Amount of substance

mole

mol

Physics file The metric system was originally developed in France and is known as the Système International (SI). It was adopted in France in 1840 as the official system of units, although it had been developing in that country since 1545. It has remained in use ever since and has gradually been adopted by most other countries. It has been modified a little over the years and now, in Australia, we use SI units that have been standardised by the International Standards Organisation (IS0) since the 1960s. Some countries such as France, Italy and Spain use an earlier form of the metric system that is slightly different. The USA still measures almost everything in the old imperial units such as pounds for mass and feet for distance but, even there, scientists use the SI system of units. There are two major advantages of using the metric system. It is easier to use than other systems in that derived units are straightforward and various sizes of units are created using multiples of ten. The other very big advantage is the international nature of the standards and units. All units are standardised, making comparisons straightforward.

Mass The kilogram was originally defined as the mass of 1 L of water at 4°C. This is still approximately correct, but a far more precise definition is now used. Since 1897 the measurement standard for the kilogram has been a cylindrical block of platinum–iridium alloy kept at the International Bureau of Weights and Measures in France. Australia has a copy of this standard mass at the CSIRO Division of Applied Physics in Sydney. At times it is returned to France to ensure that the mass remains accurate.

Length The metre was originally defined in 1792 as one ten-millionth of the distance from the equator to the North Pole, i.e. approximately 10 000 km. This definition has changed a number of times since. In 1983, to give a more accurate value, the metre was redefined as the distance that light in a

Appendix B

563

1 second. This standard can be reproduced all 299 792 458 over the world, as light travels at a constant speed in a vacuum. vacuum travels in

Time Up to 1967, time had always been based on the apparent motion of the heavens. The second was once defined in terms of the motion of the Sun. 1 1 1 Until 1960, one second was defined as of of of an average day 60 60 24 in 1900. This reflected the rate of the Earth’s rotation on its axis; however, its rotation is not quite uniform. In 1967, a more accurate definition was adopted—one not based on the motion of the Earth. One second is now defined as the time required for a caesium-133 atom to undergo 9 162 631 770 vibrations. These vibrations are stimulated by an electric current and are extremely stable, allowing this standard to be reproduced all over the world.

Derived units As well as the seven fundamental quantities, a wide variety of other physical quantities can be measured. You may have encountered some of these, such as frequency, velocity, energy and density, already. A derived quantity is defined in terms of the fundamental quantities. For example, the SI unit for area is square metres (m2).

Table B2 Some derived SI quantities and their units Quantity Velocity

564

Appendix B

SI unit metres per second

SI unit symbol

Equivalent unit

ms

−1

—

−2

—

Acceleration

metres per second per second

ms

Frequency

hertz

Hz

s−1

Force

newton

N

kg m s−2

Energy/work

joule

J

kg m2 s−2

APPENDIX C UNDERSTANDI NG MEASUREM E

NT

Measurement and units In every area of physics we have attempted to quantify the phenomena we study. In practical demonstrations and investigations we generally make measurements and process those measurements in order to come to some conclusions. Scientists have a number of conventional ways of interpreting and analysing data from their investigations. There are also conventional ways of writing numerical measurements and their units.

Correct use of unit symbols The correct use of unit symbols removes ambiguity, as symbols are recognised internationally. The symbols for units are not abbreviations and should not be followed by a full stop unless they are at the end of a sentence. Upper-case letters are not used for the names of any physical quantities of units. For example, we write newton for the unit of force, while we write Newton if referring to someone with that name. Upper-case letters are only used for the symbols of the units that are named after people. For example, the unit of energy is joule and the symbol is J. The joule was named after James Joule who was famous for studies into energy conversions. The exception to this rule is ‘L’ for litre. We do this because a lower-case ‘l’ looks like the numeral ‘1’. The unit of distance is metre and the symbol is m. The metre is not named after somebody. The product of a number of units is shown by separating the symbol for each unit with a dot or a space. Most teachers prefer a space but a dot is perfectly correct. The division or ratio of two or more units can be shown in fraction form, using a slash, or using negative indices. Most teachers prefer negative indices. Prefixes should not be separated by a space.

Table C1 Some examples of the use of symbols for derived units Preferred

Correct also

WRONG!

m s-2

m.s-2

ms-2

kW h

kW.h

kg m

-3

Nm

kWh

kg/m

3

mm

m/s2 -3

kg.m

k Wh

-3

kgm mm

N.m

Nm

Units named after people can take the plural form by adding an ‘s’ when used with numbers greater than one. Never do this with the unit symbols. It is acceptable to say ‘two newtons’ but wrong to write 2 Ns. It is also acceptable to say ‘two newton’. Numbers and symbols should not be mixed with words for units and numbers. For example, twenty metres and 20 m are correct while 20 metres and twenty m are incorrect.

Appendix C

565

Scientific notation

Figure C1 A scientific calculator.

To overcome confusion or ambiguity, measurements are often written in scientific notation. Quantities are written as a number between one and ten and then multiplied by an appropriate power of ten. Note that ‘scientific notation’, ‘standard notation’ and ‘standard form’ all have the same meaning. Examples of some measurements written in scientific notation are: 0.054 m = 5.4 × 10-2 m 245.7 J = 2.457 × 102 J 2080 N = 2.080 × 103 N or 2.08 × 103 N You should be routinely using scientific notation to express numbers. This also involves learning to use your calculator intelligently. Scientific and graphics calculators can be put into a mode whereby all numbers are displayed in scientific notation. It is useful when doing calculations to use this mode rather than frequently attempting to convert to scientific notation by counting digits on the calculator display. It is quite acceptable to write all numbers in scientific notation, although most people prefer not to use scientific notation when writing numbers between 0.1 and 1000. An important reason for using scientific notation is that it removes ambiguity about the precision of some measurements. For example, a measurement recorded as 240 m could be a measurement to the nearest metre; that is, somewhere between 239.5 m and 240.5 m. It could also be a measurement to the nearest ten metres, that is, somewhere between 235 m and 245 m. Writing the measurement as 240 m does not indicate either case. If the measurement was taken to the nearest metre, it would be written in scientific notation as 2.40 × 102 m. If it was taken to the nearest ten metres only, it would be written as 2.4 × 102 m.

Prefixes and conversion factors Conversion factors should be used carefully. You should be familiar with the prefixes and conversion factors in Table C2. The most common mistake made with conversion factors is multiplying rather than dividing. Some simple strategies can save you this problem. Note that the table gives all conversions as a multiplying factor.

Table C2 Prefixes and conversion factors Multiplying factor 1 000 000 000 000

10

Prefix tera

1 000 000 000

109

giga

G

1 000 000

10

mega

M

1 000

10

kilo

k

0.01

10-2

centi

c

12

6

3

-3

Symbol T

0.001

10

milli

m

0.000 001

10-6

micro

m

-9

nano

n

-12

pico

p

0.000 000 001 0.000 000 000 001

10 10

Do not put spaces between prefixes and unit symbols. It is important to give the symbol the correct case (upper or lower case). There is a big difference between 1 mm and 1 Mm.

566

Appendix C

There is no space between prefixes and unit symbols. For example, onethousandth of an ampere is given the symbol mA. Writing it as m A is incorrect. The space would mean that the symbol is for a derived unit—a metre ampere.

Worked example C1 The diameter of a cylindrical piece of copper rod was measured at 24.8 mm with a vernier caliper. Its length was measured at 35 cm with a tape measure. a Find the area of cross-section in m2. b Find the volume of the copper in m3.

Solution a The area of cross-section is pr 2. The radius is calculated by dividing the diameter by two. Hence the radius is 12.4 mm. To calculate the area in m2, first halve the diameter and convert it to metres. The radius is 24.8/2 = 12.4 mm = 12.4 × 10-3 m. The radius is not written in scientific notation. This is not necessary. All you need to do is multiply by the appropriate factor. The conversion factor for mm to m is 10-3. Just multiply by the conversion factor and don’t bother to rewrite the result in scientific notation. This is because it is only going to be used in a calculation and is not a final result. The area of cross-section is pr 2 = p(12.4 × 10-3)2 = 4.8 × 10-4 m2. b The volume is pr 2h, where h is the length of the cylinder. The length is 35 cm = 35 × 10-2 m. Hence the volume is p(12.4 × 10-3)2(35 × 10-2) = 1.7 × 10-4 m2.

Worked example C2 a A car is traveling at 110 km h-1. How fast is this in m s-1? b Convert 35 miles per hour to metres per second. A mile is approximately 1600 m. Solution a 110 km h-1 is 110 × 103 metres per 3600 s. 110 × 103 = 30.6 3600 Hence 110 km h-1 = 30.6 m s-1. b 35 miles per hour is 35 × 1600 metres per 3600 s. 35 × 1600 = 15.6 3600 Hence 35 mph = 16 m s-1.

Data Physicists and physics students collect, analyse and interpret experimental data. This requires a good understanding of the meaning and limitations of measurement. You can expect to find questions in your examinations specifically designed to test your understanding of data and skill in analysing experimental data. You may well be studying this section before the end of the course and data analysis skills should be practised and refined by carefully analysing and reporting on your practical work. Don’t forget that this material is about the analysis of data from real investigations and experiments rather than a purely mathematical exercise.

Appendix C

567

5

10

15

Accuracy and precision

20

good tape measure

5

10

15

20

stretched tape measure

Figure C2 The diagram shows that a correctly manufactured tape measure correctly measures the cylinder to be 16 cm long while the stretched tape measure gives a wrong measurement of 15 cm. The stretched tape measure is inaccurate.

5

10

15

20

good ruler

5

10

15

20

broken ruler

Figure C3 The diagram shows that an undamaged ruler correctly measures the cylinder to be 16 cm long while the broken ruler gives a wrong measurement of 19 cm. The broken ruler is inaccurate but equally as precise as the unbroken ruler.

5

5 16

measuring wheel ±0.5 m

20 16

17

17

tape measure ±0.5 cm

metre rule ±0.5 mm

Figure C4 The measuring wheel has low precision and only measures to the nearest metre. It has an uncertainty of 0.5 m. The tape measure has more precision and has an uncertainty of 0.5 cm or 0.005 m. The metre rule has an uncertainty of 0.5 mm or 0.0005 m.

568

Appendix C

Two very important aspects of any measurement are accuracy and precision. Accuracy and precision are not the same thing. The distinction between the two ideas is only hard to grasp because the two words are defined in a similar way in the dictionary. We often hear the words used together and in general conversation they tend to be used interchangeably. Instruments are said to be accurate if they truly reflect the quantity being measured. For example, if a tape measure is correctly manufactured it can be used to measure lengths accurately to the nearest centimetre. Imagine that the tape measure is accidentally stretched during the manufacturing process. It would still be used to measure length to the nearest centimetre but all measurements would be wrong. It would be inaccurate. Suppose an accurate ruler had 3 cm snapped off the end. It would now give readings all too large by 3 cm if no allowance were made for the missing piece. This ruler measure would be inaccurate. In these two examples, the tape measure or ruler is used to measure to the nearest centimetre but is inaccurate. Inaccurate means just plain wrong. Instruments are said to be precise if they can differentiate between slightly different quantities. Precision refers to the fineness of the scale being used. Consider the metre rule, the tape measure and the measuring wheel used to mark out sports fields. All three measure distance. All three can be accurate. The metre rule is more precise because it measures to the nearest millimetre, the tape measure has less precision due to measuring only to the nearest centimetre, while the wheel measures only to the nearest metre (figure C4). The tape measure is a more precise instrument than the measuring wheel. Suppose two distances of 2673 and 2691 mm are being measured with these two instruments. Each distance would be measured as 3 metres, to the nearest metre, by the wheel. They would be measured differently as 2.67 and 2.69 metres, to the nearest centimetre, by the tape measure. The tape measure is more precise because it has a finer scale. We might also say that it has greater resolution. The measuring wheel has such low precision that it can’t be used to measure which of the two distances is greater or smaller. Measuring instruments with less precision give measurements that are less certain. The uncertainty in the measurement is due to a coarser scale. The measuring wheel gives less certain measurements than the tape measure even though both instruments may be equally accurate. All measurements have some amount of uncertainty. The uncertainty is generally one half of the finest scale division on the measuring instrument. The measuring wheel has an uncertainty of 0.5 m. The metre rule has an uncertainty of 0.5 mm. The tape measure has an uncertainty of 0.5 cm. The electronic balance set to measure grams to two decimal places has an uncertainty of 0.005 g. Sometimes this uncertainty is referred to as error. It is not error, in that it is not a mistake or something wrong. All measuring instruments have limited precision and, in general, the uncertainty is half of the smallest scale division on the instrument.

The uncertainty is, indeed, the measure of the precision of an instrument. It is not related to accuracy. A micrometer screw gauge, which measures length to the nearest one-hundredth of a millimetre and hence is very precise, may not be accurate. Usually they are, but if one has been badly manufactured or bent by being over-tightened repeatedly it most likely will be inaccurate. But its precision will still be ±0.000 005 m, or half of onehundredth of a millimetre. The uncertainty gives us the range in which a measurement falls. If we measured the length of a stick with a metre rule then we would get a measurement ‘plus or minus’ half a millimetre. Any stick between 127.5 and 128.5 mm long would be measured as 128 mm to the nearest millimetre. We would record this as 128 ± 0.5 mm. When using an analogue scale, you might think that you can ‘judge by eye’ fractions of a scale division and hence get greater precision than half a scale division. You should be able to judge to the nearest half a scale division. You might think you can judge to the nearest tenth of a division. You can’t. Research shows that despite the fact that people try to judge the spaces between scale divisions to better than half a division, as soon as this is done, inconsistent measurements are obtained. That is, different people get different measurements of the same thing. The best judgement you can definitely claim is one half of a scale division. The uncertainty we will still assume, however, is a full half scale division. Hence, you might measure another stick, one that has a length somewhere between 154 and 155 mm, as 154.5 ± 0.5 mm. Of course, you don’t have the option of adding an extra decimal place containing a 0 or a 5 if you are using a digital instrument. The uncertainty can be recorded as the absolute uncertainty as we have done above. The absolute uncertainty is the actual uncertainty in the measurement. In this case it is 0.5 mm. Alternatively, it is often useful to write the uncertainty as a percentage: 0.5 mm is 0.32% of 154.5. Hence, the above length would be recorded as 154.5 mm ± 0.32%. Percentage uncertainty is also called relative uncertainty. It is the size of the uncertainty relative to the size of the measured quantity.

stick

12

13 ruler

Figure C5 A stick anywhere between 127.5 mm and 128.5 mm would be recorded as having a length of 128 mm if measured by a metre rule with a scale division of 1 mm. Conversely, a measurement recorded as 128 mm could be of an object of length anywhere between 127.5 mm and 128.5 mm.

Estimating the uncertainty in a result An experiment or a measurement exercise is not complete until the uncertainties have been analysed. The report should include an estimate of the total uncertainty. This gives the reader of the report some idea of your confidence in the result. The following three processes are used for estimating uncertainty. They are demonstrated in Worked example C3. • When adding or subtracting data, add the absolute uncertainties. • When multiplying or dividing data, add the percentage uncertainties. • When raising data to power n, multiply the percentage uncertainty by n. In Worked example C3, the analysis of uncertainty reveals the precision of an experimental result.

Appendix C

569

Physics file Many people use the term ‘error’ to refer to uncertainty and many other things. The problem with referring to uncertainty as error is that it is not actually error. Things that are a normal consequence of the limitations of measuring instruments must happen, and are not mistakes. If they are not mistakes or ‘something gone wrong’ then it makes no sense to call them errors. Errors are the factors that limit the accuracy of your results. For example, if you perform a calorimetry experiment and do not use a good enough insulator, you will get inaccurate results due to heat losses to the environment. This will contribute to the error in your measurement. Suppose you measured the refraction of light in glass but did not place the protractor in the correct place when measuring angles. This would also cause error. Many different things can contribute to experimental error. Some are unavoidable. Some are factors in the design of experiments. Good experimental design seeks to eliminate or at least minimise potential sources of error. Never quote ‘human error’ as a source of error. Your data should be examined carefully and mistakes eliminated or at least ignored. So-called human errors, or lack of care, have no place in your experimental work. If you make mistakes then you should repeat the measurements.

Physics file In some classes, students are instructed to quote all results to two decimal places or to three significant figures. You should be able to see from Worked example C3 that these rules are not absolutely correct when applied to real data. For ordinary calculations in assignments, tests and examinations, you might just give your answers to three figures. If a calculation is done in several stages then you should not round off any intermediate results. This will add rounding error to your calculations. Use the memories on your calculator so that there is no rounding until the end of your calculation.

570

Appendix C

Worked example C3 Last year, you might have measured the specific heat of a metal. You could have calculated your result using: cmetal =

cwatermwaterDTwater mmetalDTmetal

Suppose you had the following data included in your table. Quantity cwater

Absolute uncertainty 4180 J kg-1 K-1 -3

5 J kg-1 K-1 -3

% uncertainty 0.120

mwater

72.5 × 10 kg

0.05 × 10 kg

0.069

DTwater

5°C

1°C*

20

mmetal

87.3 × 10-3 kg

0.05 × 10-3 kg

0.057

DTmetal

72°C

1°C*

1.389

*Note that the DT values have an absolute uncertainty of 1°C because they are calculated by subtracting one temperature measurement from another. You would calculate as follows: cmetal = 241 J kg-1 K-1 Uncertainty (%) = 0.120 + 0.069 + 20 + 0.057 + 1.389 = 21.6% Hence, you would obtain the following result: cmetal = 241 J kg-1 K-1 ± 21.6% cmetal = 241 ± 52 J kg-1 K-1 Once you have done all of this you can consider the relative success of your measurement exercise. Your result is: 189 J kg-1 K-1 < cmetal < 293 J kg-1 K-1 If measurements by other people, such as the constants published in data books, fall within this range then you can conclude that your experiment is consistent with established values. That is, within the precision of your technique, there are probably no significant errors although the final measurement is rather imprecise in this case. We might say that it is accurate within the limitations of the equipment. You are also now in a position to refine the experiment by reducing the larger uncertainties. In this case, the largest uncertainty was in the temperature change for the water. Hence, it would not be very helpful to measure the masses to greater precision because the limit to precision in this activity would be the temperature differences. Getting greater precision in the temperature changes would be a useful refinement. You could consider ways of getting larger temperature changes in the water and hence obtain a smaller percentage uncertainty in the temperature change. Alternatively, you might consider ways of measuring the temperatures to greater precision. If your measurement range does not include the result you expect, you should think about the origin of the errors. In other words, if you are sure that cmetal is less than 189 J kg-1 K-1 or more than 293 J kg-1 K-1 then there must be some error in your experimental technique or more uncertainty than you realised. When reviewing an experiment or a measurement exercise, it is a good idea to consider both errors and uncertainties.

Significant figures The number of significant figures in a measurement is simply the number of digits used when the number is written in scientific notation. Once you have done a calculation, your calculator usually has eight or ten digits in the display but most of them are meaningless. You must round off your answer appropriately. Consider the result of the experiment described in Worked example C3. It would make no sense to quote the result to two decimal places (or five significant figures) when clearly the precision of the experiment gives less than three significant figures. Calculated results never have more significant figures than the original data and might have fewer than the original data. If you are not doing a full analysis of the uncertainties, it is customary to give your answers to the same number of significant figures as the least precise piece of data. For example, in Worked example C3, the least precise data is the change in temperature of the water with only a single digit. The value for the specific heat might then be quoted simply as 2 × 102 J kg-1 K-1, but doing the full calculation of the uncertainty in the result is much more informative.

Graphical analysis of data A major problem with doing a calculation from just one set of measure ments is that a single incorrect measurement can significantly affect the result. Scientists like to take a large amount of data and observe the trends in that data. This gives more precise measurements and allows scientists to recognise and eliminate problematic data. Physicists commonly use graphical techniques to analyse a set of data. In this section, the basic techniques that they use will be outlined and a general method for using a set of data that fits a known mathematical relationship will be developed.

Linear relationships Some relationships studied in physics are linear, while others are not. It is possible to manipulate non-linear data so that a linear graph reveals a measurement. Linear relationships and their graphs are fully specified with just two numbers: gradient, m, and vertical axis intercept, c. In general, linear relationships are written: y = mx + c The gradient, m, can be calculated from the coordinates of two points on the line:

m=

=

rise run y2 – y1 x2 – x1

where (x1, y1) and (x2, y2) are any two points on the line. Don’t forget that m and c have units. Omitting these is a common error.

Physics file When plotting graphs, put the correct variable on the horizontal axis. The independent variable goes on the horizontal axis. The independent variable is the quantity being changed while performing the experiment. The dependent variable is the one that may change as a consequence of the changes to the independent variable. For example, if you are manipulating the potential difference across an electrical circuit by changing the ‘voltage’ setting on the power pack, then you plot the measurements of potential difference on the horizontal axis. When analysing data from a linear relationship, it is first necessary to obtain a graph of the data and an equation for the line that best fits the data. This line is often called the regression line. The entire process can be done on paper but most people will use a computer spreadsheet, the capabilities of a scientific or a graphics calculator, or some other computer-based process. In what follows, it is not assumed that you are using any particular technology. Starting with a graph of your data has a number of uses including identifying any data that you may wish to ignore or else repeat the measurements. Often, some data is obviously suspect and it is common practice to acknowledge that the data

Appendix C

571

was collected but eliminate it from the mathematical analysis. A graph of the data will also indicate when more data is needed. If there are gaps in the range of data then more data can be collected. The graph can also indicate that the relationship being studied is not linear and may need more complex analysis. Graphs should be created as the experiment proceeds. If you are plotting your graph manually on paper then proceed as follows: 1 Plot each data point on clearly labelled, unbroken axes. 2 Identify and label but otherwise ignore any suspect data points. 3 Draw, by eye, the ‘line of best fit’ for the points. The points should be evenly scattered either side of the line. 4 Locate the vertical axis intercept and record its value as ‘c ’. 5 Choose two points on the line of best fit to calculate the gradient. Do not use two of the original data points as this will not give you the gradient of the line of best fit. 6 Write y = mx + c, replacing x and y with appropriate symbols, and use this equation for any further analysis. If you are using a computer or a graphics calculator then proceed as follows: 1 Plot each data point on clearly labelled, unbroken axes. 2 Identify suspect data points and create another data table without the suspect data. 3 Plot a new graph without the suspect data. Keep both graphs as you don’t actually discard the suspect data but do eliminate it from the analysis. 4 Plot the line of best fit—the regression line. The manner in which you do this depends on the model of calculator or the software being used. 5 Compute the equation of the line of best fit that will give you values for m and c. 6 Write y = mx + c, replacing x and y with appropriate symbols, and use this equation for any further analysis.

572

Appendix C

Worked example C4 Some students used a computer with an ultrasonic detector to obtain the velocity–time data for a falling tennis ball. They wished to measure the acceleration of the ball as it fell. They assumed that the acceleration was nearly constant and that the relevant relationship was v = u + at, where v is the speed of the ball at any given time, u was the speed when the measurements began, a is the acceleration of the ball and t is the time since the measurement began. Their computer returned the following data: Time (s)

Speed (m s-1)

0.0

1.25

0.1

2.30

0.2

3.15

0.3

4.10

0.4

5.25

0.5

6.10

0.6

6.95

Find their experimental value for acceleration.

Solution The data is assumed linear, with the relationship v = u + at, which can be thought of as being v = at + u, which makes it clear that putting v on the vertical axis and t on the horizontal axis gives a linear graph with gradient a and vertical intercept u. A graph of the data is shown in Figure C6. y = 9.5714x + 1.2857 8 Speed (m s –1 )

Physics file (cont.)

7 6 5 4 3 2 1 0 0.00

0.20

0.40

0.60

0.80

Time (s)

Figure C6 Velocity–time profile for a falling tennis ball.

This graph of the data was created on a computer spreadsheet. The line of best fit was created mathematically and plotted. The computer calculated the equation of the line. Graphics calculators can also do this. Scientific calculators can find the equation of the line but not draw the graph. A scientific calculator or graphics calculator or spreadsheet gives the regression line as y = 9.5714x + 1.2857. If this is rearranged and the constants are suitably rounded, the equation is v = 1.3 + 9.6t. This indicates that the ball was moving at 1.3 m s-1 at the commencement of data collection and the ball was accelerating at 9.6 m s-2.

Manipulating non-linear data Suppose we were examining the relationship between two quantities B and d and had good reason to believe that the relationship between them is

B=

k d

where k is some constant value. Clearly, this relationship is non-linear and a graph of B against d will not be a straight line. By thinking about the relationship it can be seen that in ‘linear form’: 1 d ↑ ↑ ↑ y = m x + c

B = k

1 (on the horizontal axis) will d be linear. The gradient of the line will be k and the vertical intercept, c, will be zero. The line of best fit would be expected to go through the origin because, in this case, there is no constant added and so c is zero. In the above example, a graph of the raw data would just show that B is larger as d is smaller. It would be impossible to determine the mathematical relationship just by looking at a graph of the raw data. A graph of raw data will not give the mathematical relationship between the variables but can give some clues. The shape of the graph of raw data may suggest a possible relationship. Several relationships may be tried and then the best is chosen. Once this is done, it is not proof of the relationship but, possibly, strong evidence. When an experiment involves a non-linear relationship, the following procedure is followed: A Plot a graph of the original raw data. B Choose a possible relationship based on the shape of the initial graph and your knowledge of various mathematical and graphical forms. C Work out how the data must be manipulated to give a linear graph. D Create a new data table. Then follow steps 1–6 given in the Physics file on pages 571–572. It may be necessary to try several mathematical forms to find one that seems to fit the data. A graph of B (on the vertical axis) against

Worked example C5 Some students were investigating the relationship between current and resistance for a new solid-state electronic device. They obtained the data shown in the table.

Current, I (A)

Resistance, R (W)

1.5

22

1.7

39

2.2

46

2.6

70

3.1

110

3.4

145

3.9

212

4.2

236

Appendix C

573

According to the theory they had researched on relevant Internet sites, the students believed that the relationship between I and R is R = dI 3 + g, where d and g are constants. By appropriate manipulation and graphical techniques, find their experimental values for d and g. The following steps would be used: a Plot a graph of the raw data. b Work out what you would have to graph to get a straight line. c Make a new table of the manipulated data. d Plot the graph of manipulated data. e What is the equation relating I and R?

Solution a Figure C7 shows the graph obtained using a spreadsheet. It might be argued that the second piece of data is suspect. The rest of this solution supposes the students chose to ignore this piece of data. b We can see what to graph if we think of the equation like this: R = d I 3 + g ↑ ↑ ↑ ↑ y = m x + c

A graph of R on the vertical axis and I 3 on the horizontal axis would have a gradient equal to d and a vertical axis intercept equal to g. 250

Resistance (7)

200 150 100 50 0

0

1

2 3 Current (A)

4

c The data is manipulated by finding the cube of each of the values for current.

574

Appendix C

5

Figure C7 Current, resistance characteristic.

Current cubed, I3 (A3)

Resistance, R (W)

3.38

22

10.65

46

17.58

70

29.79

110

39.30

145

59.32

212

74.09

236

d The graph in Figure C8 was obtained from the spreadsheet. y = 3.1412x + 15.092

300 Resistance (7)

250 200 150 100 50 0

0

20

40 60 Current cubed (A3)

80

Figure C8 Current resistance characteristic (manipulated data).

e The regression line has the equation y = 3.1x + 15.1, so the equation relating I and R is R = 3.1I 3 + 15.1. Hence, the value of d is 3.1 W A-3 and the value of g is 15.1 W.

Appendix C

575

SOLUTIONS

1.1 Atoms, isotopes and radioisotopes 1 a 20 protons, 25 neutrons, 45 nucleons b 79 protons, 118 neutrons, 197 nucleons c 92 protons, 143 neutrons, 235 nucleons d 90 protons, 140 neutrons, 230 nucleons 2 a 27 protons, 33 neutrons b 94 protons, 145 neutrons c 6 protons, 8 neutrons 3 A radioisotope is an unstable isotope. At some time, it will spontaneously eject radiation in the form of alpha particles, beta particles or gamma rays from the nucleus. Three isotopes that are not radioisotopes could be any three stable isotopes, e.g. carbon-12, lead-206 and bismuth-209. 4 Yes, a natural isotope can be radioactive. For example, every isotope of uranium is radioactive. 5 Polonium-210 and uranium-238. These have atomic numbers of 84 and 92 respectively; and every isotope beyond bismuth (Z = 83) in the periodic table is radioactive. 6 a i 5.13 × 10-45 m3 ii 3.25 × 1017 kg m-3 b 3.25 × 1011 kg (325 million tonnes) c 325 million cars d The density of normal matter is far lower than the density of an atomic nucleus. 7 1.1 × 10-13 8 210 m 9 a There are no differences. b 89Kr has 5 more neutrons in its nucleus. 10 28Al

1.2 Radioactivity and how it is detected 1 a the nucleus b the nucleus c the nucleus 2 α is a helium nucleus, β is an electron, γ is electromagnetic radiation. 3 a beta particle b proton c alpha particle d neutron 4 a Z = 82, A = 214, lead b Z = 90, A = 231, thorium c Z = 89, A = 228, actinium d Z = 80, A = 198, mercury 5 a α b β- c β- d α e γ 6 lithium-7 7 a proton b neutron c neutron d alpha particle 8 a 7 protons, 7 neutrons, 1 electron, 1 antineutrino b A neutron has decayed. c 01 n→11p + –10 e + n d Kinetic energy carried by the beta particle, antineutrino and nitrogen-14 nucleus. 9 a 40Ca, 42Ca, 43Ca, 44Ca, 46Ca, 48Ca b one c beta 48 48 d 19 K → 20 Ca + –10 β, 48 Ca is stable 20 e K: 1.53, Ca: 1.40 f alpha-emitter g 217 Fr→213 At + 42 α→209 Bi + 42 α; bismuth-209 is stable 87 85 83 10 a 197 Au + 10 n → 198 Au b 198 Au → 198 Hg + –10 β 79 79 79 80

576

Solutions

1.3 Properties of alpha, beta and gamma radiation 1 a α, β, γ b γ, β, α 2 B 3 Gamma radiation is most suitable since its penetrating ability will enable it to reach the tumour. 4 A beta emitter would be best suited because its penetrating ability would enable it to irradiate a small volume of tissue around the source. Alpha radiation would not penetrate the tumour at all, and gamma radiation would pass out of the body irradiating some healthy cells along the way. 5 Beta and gamma radiation. Alpha particles are soon absorbed by air, but beta and gamma radiation can travel through air for metres. 6 a 1.4 × 10-12 J b 6.7 × 10-14 J c 8.0 × 10-14 J 7 a 3.4 × 106 eV b 1.65 cm 8 D 9 A 10 D

1.4 Half-life and activity of radioisotopes 1 C 2 a 10 g b 5 g c 2.5 g d 0.31 g 3a 400 350 Count rate (Bq)


300 250 200 150 100 50 0

10

20 30 40 Time (min)

50

b ~235 Bq c 20 min d 50 Bq 4 15 min 5 0.5 6 192 mg 7 a 10 half-lives b ~240 000 years 8 a uranium-235 b It has a much shorter half-life than uranium-238 and so has decayed much more rapidly since the formation of the Earth. 9 a From the activity graph, the time that the activity takes to halve from 800 Bq to 400 Bq is 10 minutes. This is the half-life. b 50 Bq 10 a Over time, the radioisotopes transmute by a series of decays to form lead-206, which is stable. The percentage of lead in the sample will increase over time. b 214Po has such a short half-life (160 ms) that when 214 Bi nuclei decay to 214Po, they almost instantaneously transmute to 210Pb.

1.5 Radiation dose and its effect on humans 1 a A, B, C, E b C, D, E 2 a 10 mSv b 0.50 mSv c 0.50 mSv 3 a 2.0 × 10-4 Sv b 1.6 × 10-2 J 4aB bA 5 a 2 days b 50 days c 807 mSv 6 Minimise your exposure to medical procedures involving ionising radiation, keep air travel to a minimum, and live in a weatherboard house at low altitude near the equator. 7 They must have received massive doses of at least 8 Sv. 8 a 90 hours b Beta particles have enough penetrating ability to irradiate the tumour, and the long half-life means that the activity of the sample would not decrease noticeably during the course of treatment. 9D 10 Effective dose = ∑(dose equivalent × W) = (35 × 0.20) + (35 × 0.05) + (50 × 0.12) = 15 mSv

Chapter 1 review 1 a 17 protons, 18 neutrons, 35 nucleons b 88 protons, 138 neutrons, 226 nucleons 2 a X-rays, infrared radiation, microwaves, gamma rays b alpha particle c alpha particle d beta particles e α, β, γ, X-rays f gamma radiation 3 a x = 82, y = 208 b x = 78, y = 176 4 a β- b α c β+ 5 60

Mass (g)

50 40

10 Her blood cells have been affected; this is a somatic effect. 11 7 alpha, 4 beta 12 a 3.0 × 1010 nuclei b 1.5 × 1010 nuclei c 7.5 × 109 nuclei d 9.4 × 108 nuclei 13 D 14 lithium-6 15 a The half-lives of the samples are equal. b The activity of A is twice that of B. 16 a A proton has transformed into a neutron, a positron and a neutrino. b Kinetic energy carried by the neutrino, positron and carbon-12 nucleus. 17 3.1 × 1019 Bq 18 a ~3 mSv b about 4 times background 19 Effective dose for patient A = 1080 mSv. This is greater than the dose received by patient B, so patient A is at a slightly greater risk of developing cancer from this exposure.

Area of study review (Nuclear physics and radioactivity) 1 Beta particles are electrons that are emitted from an unstable nucleus. Beta decay occurs in nuclei when a neutron spontaneously decays into a proton. 2 Beam ii is gamma since it is undeflected by magnetic fields. Beam i is alpha and beam iii is beta. Beta is much lighter and is deflected more strongly. 3 B 4 a nucleus b nucleus c nucleus 5 Time No. of K No. of Ar Ratio (G years) nuclei nuclei K : Ar 0

1000

0

20

1.3

500

500

1 : 1

10

2.6

250

750

1 : 3

3.9

125

875

1 : 7

30

0

8

16 24 32 Time (hours)

40

48

6 a 3.75 g b 84 Po → 82 Pb + 2 α c lead-214, bismuth‑214, polonium-214 d Lead-214; it has the longest half-life and so is the slowest to decay. This means that the atoms of this isotope remain in the sample for the longest time. 7 a 146 C→147 N + –10 e + ν b 4 per minute c ~11 500 years (2 half-lives) 8 a 2.2 × 10-14 J b 6 hours c 2.0 × 106 Bq d 0.5 × 106 Bq 24 9 a 24 Na → 12 Mg + –10 e + γ, magnesium-24 is produced 11 b Gamma rays can penetrate a great thickness of air, but beta particles can penetrate only a few metres. c β d Kinetic energy carried by the gamma ray, beta particle and magnesium-24 nucleus. 218

214

4

–

6 Radioactive isotopes of the same element will have identical chemical properties to the non-radioactive isotope. 7 Each atom of gold-198 has one more neutron in its nucleus meaning that these atoms have a slightly higher density. 8 73 Li + 11 H → 2(42 He) 9 a 8.8 MeV b 0.42 MeV c 0.50 MeV

Solutions

577

Activity (Bq)

10 a 1000 800 600 400 200 0

40 This represents about 15 times background radiation.

Chapter 2 Concepts in electricity 2.1 …lectric charge 0

5

10

15 Time (min)

20

25

30

11 ~320 Bq 12 10 min 13 100 Bq 14 B 15 Gamma rays, having no electric charge, and moving at the speed of light, are the most highly penetrating form of radiation. Gamma rays interact with matter infrequently when they collide directly with a nucleus or electron. The low density of an atom makes this a relatively unlikely occurrence. Hence, gamma rays pass through matter very easily. 16 Alpha particles travel through air at a relatively slow speed and with a double positive charge. They interact with just about every atom that they encounter. The alpha particle ionises many thousands of these atoms. Each interaction causes the alpha particle to slow down a little, and eventually it will be able to pick up some loose electrons to become a helium atom. This takes place within a centimetre or two in air. As a consequence, the air becomes quite ionised, and the alpha particles are said to have a high ionising ability. But alpha particles don’t get very far in the air, so they have a poor penetrating ability. 17 1.0 × 1020 Bq 18 1 p, 2 n 19 helium 20 B 21 A neutron has transformed into a proton and an electron. 22 185 Au → 181 Ir + 42 α 23 79 p, 106 n 79 77 24 No, the alpha particles would not penetrate into the living cells of your hand. 25 Gamma rays have the ability to penetrate the skull and pass through the target region. The half-life of 5.3 years means that the source would retain its activity for a number of years and would not need replacing very often. 26 300 mSv 27 60 s 28 2.3 g 29 26 Na → 26 Mg + –10 β + ν 11 12 30 radiation sickness, hair loss 31 testes 32 C 33 1.5 × 103 mSv 34 These bombarding electrons may not be product of radioactive decay. 35 The electrons would be strongly repelled by the electron clouds of the atom. 36 B 37 After 8 minutes, 211Bi will have half the activity of 215Bi. 38 a 380 J b 5.0 Sv c nausea, hair loss, leukaemia, possibly death 39 1.2 × 10-5 Sv

578

Solutions

1 He was providing lightning with an easy path to him. 2 Almost none of our modern technology would exist! 3 About 20–30 4 Both show repulsion and attraction. Magnetism is relatively permanent, electrostatic effects are temporary. Magnetic poles cannot be isolated, opposite charges can. 5 Charges either repel or attract, there are no other alternatives. 6 They will attract. 7 An excess of the fluid creates one type and a deficit the other. 8 Charge leaks from sharp points. 9 Lightning will strike tall objects and may jump to nearby objects. Find a spot away from tall objects. 10 Positive and negative can cancel each other, as do the effects of opposite charges; colours do not.

2.2 …lectric forces and fields 1 a 2F b 4F c –F d 4F 2 900 kN 3 It is not! 4 a 0.35 N b Charges are further apart. 5 a 0.01 N down, 0.025 N up b – 0.2 mC

2.3 …lectric current, …MF and electrical potential 1 a 3 A b 0.5 A c 0.008 A 2 a 5 C b 300 C c 18 000 C 3 a 9 A right b 2 A left 4 a 3 C b 1000 C c 1440 C 5 No 6 3 mA down belt 7 a 2.7 × 107 A b No, I = 0 8 Spark plugs 15 000 V 9 a i 4 V ii 4 V iii 2 V b i 10 A ii 1 A iii 1 A 10 D 11 20 V 12 9 J 13 167 C 14 The gravitational potential energy of the water

2.4 Resistance, ohmic and non-ohmic conductors 1 a M2 or M3 b M1 or M4 2 a 3 A b 50 V c assumed constant resistance d 2.5 Ω at all voltages graphed 3 a 0.71 Ω b ohmic 4 Both right at different voltages. 5 72 mA 6 a 2 Ω b 5 A 7 a 1.6 Ω b 0.2 Ω 8 a non-ohmic b 0.5 A c 15 V d i 20 Ω ii 13.3 Ω 9 Cathryn, lower slope means less current at given voltage. 10 a 0.11 Ω b 1.1 V

2.5 …lectrical energy and power 1 a 4.5 J b 4.5 J c No, some lost as heat. 2 2.4 × 10-19 J 3 a 0.6 W b 2400 W c 720 W 4 a 250 mA b 5 A c 7.5 A 5 a 25 V b 8.7 V c 417 V

6 21 kA 7 0.84 kW h, 3.0 MJ, $6.50 8 a 1.5 A b 0.075 A 9 AC oscillates, DC steady 10 Less current is needed.

Chapter 2 review 1 Negative: electrons; orbit nucleus. Positive: protons; in nucleus. It can lose / gain electrons 2 Ensure no p.d. and hence no spark. 3 i A +, B – ii same iii same iv neutral 4 No. 5 No, all the likes would collect together well away from the others. Atoms depend on the mutual repulsion of the electrons and their attraction to the nucleus. 6 90 N 7 9000 N 8 4 × 105 N C-1 9 0.001 s, comes as pulses. 10 Conducts in electrostatic experiments, not at mains and below. 11 240 J 12 a 3 V b 0 V c 3 V 13 a in the filament b same 14 a 18 C b 54 J c the battery 15 a 48 Ω b R increases with temperature c 1200 W 16 a 0.125 Ω b 1.25 V c 12.5 W 17 a no b 40 Ω, 27 Ω c 1.6 W, 9.6 W 18 a 2 MW b 5 A 19 a 0.079 Ω b no (twice) 20 440 kW h, 1600 MJ

Chapter 3 Electric circuits 3.1 Simple electric circuits 1 Itot = 0 at a point. Etot = ∆Vtot in a circuit. 2 2 V drop in circuit 3 +0.7 A 4 a 0.25 A b 2.4 V 5 a yes b yes 6 17 Ω 7 0.22 A, 45 Ω 8 a 0.01 A b 4 V and 1 V 9 a through (15V, 2A) b 7.5 Ω 10 15 Ω

3.2 Circuit elements in parallel 1 a 0.30 A b 3 V 2 a 20 Ω b 5 Ω 3 a through (10 V, 6 A) b 1.7 Ω 4 68 Ω 5 0.9 A, 11 Ω 6 a 6.7 Ω b 20 V c 1 A (in 20 Ω) and 2 A 7 20 W (in 20 Ω), 40 W 8 a 4.4 kΩ b 0.44 kΩ 9 a A1 = 7 A, A2 = 5 A, A3 = 2 A b 50 W and 20 W respectively c 1.4 Ω d 70 W (same as previous total) 10 Betty is correct.

3.3 Cells, batteries and other sources of …MF 1 Need different electronegativities. 2 + to + and – to – 3 No resistance across terminals. 4 15 V 5 a 6 V b batteries last longer 6 The heavy current causes a p.d. within the battery. 7 10 V 8 1 Ω 9 12 A 10 Increased I leads to lower V.

11 a 0.4 W b 0.33 V c No, most power is available at 0.42 V. 12 a 1.6 V b 5 Ω c 30 mW, 120 mW, 110 mW d 2 mW, 72 mW, 242 mW

3.4 Household electricity 1 C 2 Neutral and earth are common. 3 Make the circuit safe. 4 Ensure neutral is 0 V. 5 Yes. Live when off. 6 Case could become live. 7 safer 8 Less contact, more resistance. 9 2.4 mA 10 Ensure better ground contact.

Chapter 3 review 1 Car body is the other conductor. 2 12 v 3 no 4 a 2 A b 60 Ω c 1920 W 5 5 Ω 6 a 100 ma b 15 Ω c no definite current 7 a 220 mA b No, much more. 8 a 400 mA b 800 mA c 1.6 A 9 a 15.4 kΩ b 63 kΩ 10 0.1 mho, 0.2 mho 11 4.5 A 12 a Voltmeter has very high resistance. b burn out ammeter c M1 = A, M2 = V 13 a 1.3 A b 1.3 A c 3.2 A 14 a 4.9 V b 1.1 V c 6.0 V 15 A1 = 3.2 A, A2 = 0, V2 = 0 16 A1 = 3.2 A, A2 = 0, V1 = 6.0 V, V2 = 6.0 V 17 a 10 Ω b 25 Ω 18 a 4 A b 8 Ω 19 0.25 Ω 20 a 320 W b 232 V

Area of study review (Electricity) 1B 2A 3 a 1.25 × 1019 b 1.60 × 10-17 J c The electrical energy is converted into heat energy in the wire. d The EMF of a battery is defined as the number of joules of electrical energy the battery adds to each unit of charge passing through it. Therefore 100 V = 100 J C-1. This means that the battery adds 100 J to each coulomb of charge passing through it. e 200 W f 200 W g 200 W h Answers are the same because power is the energy given to each unit of charge (volt) times the number of charges that flow in the wire each second (amperes). 4 Charges have Ep not Ek. 5 a Charges escape at the same rate. b 1.2 × 1013 off c 0.80 W d 6.4 × 10-14 J 6 a 50 Ω b 1.44 × 10-18 J c 1.62 W d 1.1 × 1019 7 B 8 a 100 Ω b 1.0 A c i 80 V ii 10 V iii 8.0 V d 98 V e 100 W 9 C 10 a 60 Ω b 2.0 A c I1 = 1.20 A, I2 = 0.60 A, I3 = 0.20 A d 240 W e 240 W 11 a 6.7 kΩ b 2.5 kΩ 12 a 12 V b 11.4 V c The 9 V battery will go flat before the others. 13 a 2.0 W b 0.67 A 14 2 A 15 50 Ω, 50 Ω, 55 Ω, 67 Ω 16 409 W 17 Element is hotter.

Solutions

579

4.1 Describing motion in a straight line 1 a +40 cm, 40 cm assuming movement to the right is positive b –10 cm, 10 cm c +20 cm, 20 cm d +20 cm, 80 cm 2 a 80 km b 20 km north 3 a 10 m down b 60 m up c 70 m d 50 m up 4 displacement 5 a D b D c C d A 6 a 39 steps b 1 step west of the clothes line. c 1 step west of the clothes line. 7 a ~2 m s-1 b ~1 mm s-1 c ~10 m s-1 d ~5 m s-1 8 a 15 km h-1 b 4.2 m s-1 c No, she would probably be travelling faster or slower than this average speed. It depends on the traffic and the terrain. 9 a 22.2 m s-1 b 6.7 km h-1 s-1 c 1.85 m s-2 d 20 m 10 a –10 m s-1 b 40 m s-1 west c 800 m s-2 11 a 12 km h-1 s-1 b 3.3 m s-2 south 12 a 1500 m b 1.7 m s-1 c 0 d 0

4.2 Graphing motion: position, velocity and acceleration 1 a +4 m b A, C c B d D e 0.4 ms–1 2 The car initially moves in a positive direction and travels 8 m in 2 s. It then stops for 2 s. The car then reverses direction for 5 s, passing back through its starting point after 8 s. It travels a further 2 m in a negative direction before stopping after 9 s.

580

Solutions

Position (m)


3 a +8 m b +8 m c +4 m d –2 m 4 8 s 5 a +4 m s-1 b 0 c –2 m s-1 d –2 m s-1 e –2 m s-1 6 a 18 m b –2 m 7 a The cyclist travels with a constant velocity in a positive direction for the first 30 s, travelling 150 m during this time. Then the cyclist speeds up for 10 s, travelling a further 150 m. Finally the cyclist maintains this increased speed for the final 10 s, covering another 200 m in this time. b +5 m s-1 c +20 m s-1 d approx. +13 m s-1 e +15 m s-1 8aB bA cC dD 9 a running north at 1 m s-1 b increasing speed from 1 m s-1 to 3 m s-1 while running north c running north but slowing to a stop d stationary e accelerating from rest to 1 m s-1 while running south f running south at 1 m s-1 10 a 2 m north b 10.5 m north c 9 m north 11 a 2 m s–1 north b 1.7 m s–1 north 12 11 10 9 8 7 6 5 4 3 2 1 0

1 2 3 4 5 6 7 8 9 10 11 12 Time (s)

13 a 80 s b ~1.3 m s-2 c ~0.5 m s-2 d ~4.9 km 14 a +2 m s-2 b 4 s c 10 s d 80 m e +7 m s-1 15 a 2

Acceleration (m s–2)

18 a 200 Ω b 2.0 A c V1 = 200 V, V2 = 160 V, V3 = 36 V, V4 = 396 V d 800 W e 800 W f 396 V g 400 V 19 a I = 0.27 A. We would not expect twice the current at 4 V. If the curve is extended to 4 V we could expect around 0.31 A, but it might well be that the globe would have burnt out before 4 V is reached, as the graph implies that it is probably a 3 V globe. b At 1 V, R = 5 Ω; at 3 V, R = 10 Ω c This is not an ohmic conductor because the I–V graph is non-linear. The graph of an ohmic conductor must be a straight line because the current is directly proportional to the voltage applied. d 0.9 W 20 a The device is non-ohmic. The purpose of this device is to limit the current through a particular section of the circuit to a constant value regardless of the voltage across that part of the circuit. b The resistance of the device increases with voltage. c V1 = 150 V, V2 = 100 V d 0.30 W e 0.20 W f 0.50 W 21 53% 22 2.8 s 23 42 A 24 0.25 V 25 219 V 26 60 V, 60 W 27 1 A, same 28 Mary 29 4.3 Ω 30 a 12 V b 8 V c 16 Ω 31 A 32 D 33 C 34 A 35 a 50 Ω b 15 Ω 36 a 20 mA b 100 mA 37 A 38 D 39 A 40 a 8.0 V b 7.3 V

1 0

2 4 6 8 10 12 14 16 18 20 Time (s)

b 12 m s

-1

4.3 …quations of motion 1 a +2.0 m s-2 b +8.0 m s-1 c 64 m 2 a +3.1 m s-2 b 50 m s-1 c 180 km h-1 3 a 40 m s-2 b 1120 m c 580 km h-1 d 80 m s-1 e 124 m s-1 4 a –99.4 m s-2 b 0.284 s c 19.9 m s-1 5 Cars have greater accelerations when they are travelling slowly (i.e. when they are in a low gear).


When they are travelling fast, they may have a high speed, but this speed does not increase rapidly when the throttle is pushed. 6 D 7 a +0.40 m s-2 b 28 m c 5.3 m s-1 8 a 21 m s-1 b 5.3 m c 36 m d 42 m 9 a 4.0 m s-1 b 5.7 m s-1 c 2.0 s d 0.85 s 10 a 8.0 s b 16 s c 192 m

5.1 Force as a vector

4.4 Vertical motion under gravity 1 a The brick would fall much faster than the paper clip. b They would fall together. c They fall together. 2 The bubbles rise because they are an air element and are returning to their natural place in the atmosphere. 3 B 4 A, D 5 a i 9.8 m s-1 ii 20 m s-1 iii 29 m s-1 b i 14 m s-1 ii 20 m s-1 iii 24 m s-1 c 12 m s-1 6 a

b d

c v

x

t d

t e a

v t

t

7 a 3.9 m s-1 b 0.78 m c 0.20 m d 0.58 m 8 a 2.0 s b 20 m c 20 m s-1 d 20 m s-1 e i 9.8 m s-2 down ii 9.8 m s-2 down iii 9.8 m s-2 down 9 a 1.7 s b 3.2 s 10 6.7 s

Chapter 4 review 1 D 2 15 m s-1 up 3 11 m 4 Aristotle’s theories did not explain why light objects fell equally as fast as heavy ones or why a solid (earth element) such as wood floated on water. 5 a from 10 s to 25 s b from 30 s to 45 s c from 0 s to 10 s, 25 s to 30 s, and 45 s to 60 s 6 after 42.5 s 7 a 20 m s-1 north b 40 m s-1 south 8 a 5 m s-1 b 15 m s-1 9 C 10 2.0 s 11 15 km h-1 12 a 10 km h-1 north b 2.8 m s-1 north 13 a 4.0 m s-2 b 4.0 m s-1 c 6.0 m 14 a 0.80 m s-1 b 0.50 m s-1 c 0.67 m s-1 15 a 0.75 m s-1 b 8.0 m s-2 16 a 3.5 s b 2.9 s 17 1.0 s 18 ~10 m s-1 north 19 A 20 a 0 b ~2.0 m s-2 north c 7.0 m s-2 south

t

1 a B, D b i 150 N up ii 40 N west iii 0 iv 14 N NW 2 a 10 N b 10 N c 50–100 N d 400–800 N 3 1000 N, 10 000 N 4 B, C, D 5 a 60°T b 320°T c 240°T d 135°T e 22.5°T 6 a 60 N right b 50 N up c 70 N down 7 a 1 N west b 100 N 143°T 8 610 N in a direction that bisects the two ropes 9 a 50 N south, 87 N east b 60 N north, 0 N east c 282 N south, 103 N east d 1.5 × 105 N up, 2.6 × 105 N horizontal 10 150 N horizontal , 260 N upwards

5.2 Newton’s first law of motion 1 Aristotle felt that the natural state for any object was at rest in its natural place. This meant that any moving body would come to rest of its own accord. Galileo introduced the idea that friction was a force that could be added to other forces that act on a moving body, but it was Newton who explained that the moving object should continue to travel with constant velocity unless a net force is acting. 2aA bD cD 3 No forwards force acts on the person. In accordance with Newton’s first law of motion, the bus slows and the standing passenger will continue to move with constant velocity unless acted on by a force—usually the passenger will lose their footing and fall forwards. 4 20 N in a forwards direction, so that the net force will be zero 5 lift = 50 kN up, drag = 12 kN west 6 a 25 N b 25 N horizontally c 29 N at 30° to the horizontal 7 When the car or aircraft slows suddenly, a passenger will continue to travel with the same velocity as before, until being acted on by an unbalanced force. The purpose of the seat belt is to supply that force, but across the body where the effects of the force are reduced. 8 a gravitation b electric force c friction between the tyres and the road d tension in the wire 9 a If the cloth is pulled quickly, the force on the glasses acts for a short time only. This force does not overcome the tendency of the glasses to stay where they are i.e. their inertia. b Using full glasses makes the trick easier because the force will have less effect on the glasses due to their greater mass. The inertia of the full glasses is greater than that of empty glasses.

Solutions

581

10 The fully laden semi-trailer will have more difficulty pulling up. Its large mass means that more force is required to bring it to a stop.

5.3 Newton’s second law of motion 1aU bB cB dU 2 a 4.3 × 103 m s-2 b 260 N 3 The 1.5 kg shot-put is the larger of the two masses, and so for the same applied force, its acceleration will be lower. This means a lower speed on leaving the athlete’s arm, and so it cannot be thrown as far as the lighter 1.0 kg ball. 4 a 160 N b 141°T c 210 m s-2, 141°T (but this is for a very short time) 5 2.6 × 103 N in the direction of travel 6 a 50 kg b 490 N c 20 N south d 0.31 m s-2 7 m = 1.5 kg, Fg = 5.4 N 8 1.1 × 103 N opposing the motion 9 Net force on parachutist 700 N

is thrown directly away from the ship, hopefully the reaction force will propel him back to the craft. b 20 N act on each body, but in opposite directions c 0.10 m s-1 d 100 s or 1 min 40 s 4 12.5 m s-2 south 5 a B b C 6 a Weight acts vertically down, normal force is perpendicular to the incline, drag and friction act up the incline b Fg = 640 N downwards, FN = 410 N up and perpendicular to the track c ∑F = 490 N along the incline d 7.5 m s-2 along the track 7 280 N 8 The normal force will be smallest at A and become progressively larger as he travels through points B and C. 9 Tania is correct. The forces are equal in size but opposite in direction, but they are both acting on the lunchbox and so cannot be an action—reaction pair in the sense of Newton’s third law. 10 a 196 N down on the side of the student b 2.45 m s-2 down to the side of the student c 370 N

Chapter 5 review

0

Time

As the parachutist leaves the aircraft, his or her weight will be the net force acting, accelerating at 9.8 m s-2 but as the speed increases, the drag force (air resistance) opposing the motion also increases until it equals the weight. At this point in time, the net force will be zero and the parachutist will travel with a constant speed. 10 a 1.6 m s-2 b 0.80 m s-1 c 0.20 m s-2

5.4 Newton’s third law of motion 1 a Force that the racquet exerts on the ball; force that the ball exerts on the racquet. b Force of attraction (gravity) that the Earth exerts on the pine cone; force of attraction (gravity) that the pine cone exerts on the Earth. c Downwards force that the pine cone exerts on the ground; upwards force that the ground exerts on the pine cone. d Force of attraction (gravity) that the Sun exerts on the Earth; force of attraction (gravity) that the Earth exerts on the Sun. 2 a 140 N in opposite direction to leaping fisherman b 3.5 m s-2 in opposite direction to fisherman c fisherman: 1.0 m s-1, boat: 1.8 m s-1 3 a If no other forces act, there will be an action— reaction force pair in which the action force is the force on the tool kit, and the reaction force will act on the astronaut in the opposite direction. If the tool kit

582

Solutions

1 C 2 48 N 3 6.7 m s-1 4 • Kicking the tyre of a car hurts because you apply a force to the tyre (action) and the tyre will apply a reaction force to your foot (reaction). • Hot gases are forced out of a jet engine (action) and the gases push the engine forward (reaction). • The weight of any object can be considered an action force—the Earth pulls on the body. The reaction force acts on the Earth—the object pulls on the Earth. 5 a 5.0 m s-1 b i 540 N down ii 540 N down 6 a 85 kg b 85 kg c 306 N down 7 A block is struck with a sharp blow so that it overcomes the grip of the blocks with which it is in contact, and it is quickly ejected from the pile. The other blocks experience a frictional force that acts for a very short time and they stay in the vertical stack. 8 a 110 N horizontally, 41 N down b 110 N to the south c FN = 240 N upwards d When the trolley is pulled, the vertical component of the applied force is upwards rather than downwards, and so a smaller (upwards) normal force is needed—helping the wheels to rotate more freely. 9 7.0 m s-1 10 B 11 a 1.5 m s-2 in the direction of the force b 75 N 12 a 1 b 0.25 c 0.25 13 To maintain a tension of less than 100 N in the rope, the bucket must accelerate at greater than 1.5 m s-2 downwards. If the acceleration falls below this, the rope will snap.

14 a No b Both forces are acting on the water tank. Action–reaction forces act on different objects. 15 71 km h-1 16 30° 17 3.4 m s-2 down the incline 18 a force due to racquet

Fg

b

6.2 Conservation of momentum 1 2.0 m s-1 in direction of white ball’s original motion 2 4.7 m s-1 in the same direction 3 0.44 m s-1 in the opposite direction 4 11(.33) tonnes 5 B 6 a Only if Superman is fixed to the ground so that the momentum of the truck is transferred directly to the ground. b Final velocity = ptruck/total mass of Superman + truck 7 vcar = 100 km h-1 so yes! 8 8.0 m s-1 9 3 m s-1 opposite the direction of the exhaust gases 10 a Rocket is losing 50 kg over this 2 s period so use average mass of rocket is 225 kg and v = 40 m s-1. b 4.5 × 103 N c 10 m s-2

direction of travel

Fg

drag force due to air resistance

19 3.3 m s-2 down, on the side of the 10-kg mass; 65 N 20 a Therese is correct. The marble and the billiard ball exert equal and opposite forces on each other as described by Newton’s third law. b Therese is correct. The floor and the basketball exert equal and opposite forces on each other as described by Newton’s third law.

6.3 Work 1 2.6(5) × 102 J 2 1.5 × 105 J 3 Also travelled horizontally against frictional forces 4 D since i mentions distance rather than displacement 5 a 1.5 × 102 J b >1.5 × 102 J c 1.0(29) × 103 J 6 B, since the motor won’t convert all supplied energy to useful work 7 No work, since there is no change in position (don’t need to read the scale) 8 a 2.4 × 103 J b Nil 9a FN

Chapter 6 Momentum, energy, work and power 6.1 The relationship between momentum and force 1 a 100 kg m s-1 b 1.0 kg m s-1 c 28 kg m s-1 2 a 24 kg m s-1 b 64 kg m s-1 c 8.0 kg m s-1 d 40 N s e 8.0 N 3 object 2 4 a 41 kg m s-1 b 0.24 kg m s-1 c 500 kg m s-1 d 160 N s 5 a 4.0 kg m s-1 b 124 N 6 a 9.0 N s b 180 N in the direction of the ball’s travel c 180 N opposite the direction of the ball’s travel 7 a 1200 N b 63 N s 8 a 1.25 kg m s-1 opposite direction of flight b 1.25 N s opposite direction of flight c 1.6 × 103 N opposite direction of flight 9 a Based on idea of impulse. Stopping time increased by collapsing shell reduces impact force. b Stopping time would be reduced and force increased therefore not as successful. 10 Need to include mass and initial velocity then ∆p = m∆v. Impulse = ∆p = F∆t (units: N s). Estimate stopping time then F = ∆p/∆t (units: N).

FT 35n

Ff Fg

b 2.0 × 10 J c 1.6 × 102 J d 40 J 10 ~1.3 J 2

6.4 Mechanical energy 1 a 3(.125) J b 4 × 102 J c 2.6 × 105 J 2 a 49 J b 412 J c 1.2 × 105 J 3 a 2.5 J b 1.8 J c –0.69 J 4 A 5 6.9 × 103 N 6 a 0.98 J b 0.98 J c 4.9(5) m s-1 7 A: k = 4000 N m-1 (stiffest), B: k = 1300 N m-1, C: k = 570 N m-1 8 A: 1.3 J, B: 3.7 J, C: 8.7 J 9 a 3.13 J b 12.5 J c 28.1 J 10 9.13 cm

6.5 …nergy transformation and power 1 a kinetic energy to heat b Ek to Us to Ek to Ug c Us to Ek to Ug d kinetic to heat/sound 2 elastic potential → kinetic → gravitational potential → kinetic (+ heat/sound) → heat (+ sound, kinetic of water rebounding) 3 a 5.4 × 103 J b 4.5 × 102 W 4 a 73(.5) J b 73 J c 0.61 W

Solutions

583

d Coach lifting own body, losses to surrounding environment 5 a 2.4(5) J b 2.4(5) J c Transfer to the ground and transformed into other energy forms such as heat and sound. 6 a 4.3(2) m b Length of rope insufficient to allow height change, air resistance and other factors leading to transfer/loss to the surrounding environment. 7 6.0 m s-1 8 a 9.9 m s-1 b 8.9 × 10-1 N 9 a 5.0 J b 8.2 m s-1 c 6.1 cm 10 a 75 J b 2.2 m s-1 c 0.26 m

Chapter 6 review

Energy x mass (J)

1 5.0 J 2 5.0 J 3 10 m s-1 4 No. Using the area under the graph the work done is 1.25 J or one-quarter of the total work required. 5 2.5(6) × 102 J 6 1.1 × 103 N 7 a 2.4 × 104 J b gravitational potential energy → kinetic energy → elastic potential energy (and heat) c 6.9 × 103 J d 1.7 × 104 J e The gravitational potential energy and Ek values will diminish with each bounce. The total mechanical energy is not conserved as this is not an isolated system. This is because some of the work done in stretching the rope is converted to heat. 8 D (147 J) 9 B (10 m s-1) (B) 10 A (5.1 m) 11 13 m s-1 12 8.3 m 13 8 × 104 N 14 6.5 × 105 J 15 1 × 106 W 16 Gravitational potential + kinetic → kinetic (+heat/ sound) → kinetic of ground + heat + sound + gravitational potential of rebounding water 17 B: 25 m s-1, C: 11 m s-1, D: 19 m s-1 18 Total mechanical energy is constant. Total = 302 m A (30, 294)

B (0, 302)

300 250

D (12, 184)

200

C (25, 245)

Ep

Ek

150 100

D (12, 117.6) B (0, 0)

50 0 0

5

10

15

20

C (25, 57) A (30, 8) 25

30

Vertical displacement (m)

19 57 × mass J 20 81% 21 2.2 m s-1 in direction of moving player 22 2.2 m s-1 north 23 2.2 m s-1 in same direction 24 negligible movement but there is a change (v = 2.7 × 1021 m s-1) 25 7.1 m s-1 26 1.1 × 104 N s 27 14 × 103 N

Area of study review (Motion) 1 10.2 s 2 510 m 3 9.8 m s-2 down 4 The weight will travel three times as far during the 2nd second as during the first.

584

Solutions

5 10 m 6 5.0 s 7 60 m 8 4.99 s 9 4.42 s 10 platinum sphere 24.4 m s-1 down; lead sphere 31.7 m s-1 down 11 Yes, air resistance would affect the spheres over the distance. The sphere with the larger diameter would be affected more. 12 a 0.10 m s-1 b 0.30 m s-1 c 0.50 m s-1 13 a 0.10 m s-1 b 0.30 m s-1 c 0.50 m s-1 14 The marble is moving with constant acceleration. 15 0.375 m s-2 west 16 375 N west 17 575 N west 18 575 N east 19 70.6 km h-1 20 600 N 21 800 N 22 4.0 m s-2 up the incline 23 1.6 × 103 N 24 a 3.3 m s-2 clockwise; i.e. 3.3 m s-2 up for the 10 kg mass and 3.3 m s-2 down for the 20 kg mass b 1.3 × 102 N 25 a 300 N b 4.0 kJ c 1.0 kJ d 3.0 kJ e 3.0 kJ f 17.3 m s-1 g The frictional force is retarding the motion of the force and producing heat. 26 a 1.6 MJ b 1.6 × 104 N c 16.4 kN d 1.64 MJ e 328 kW f 40 kJ 27 140 J 28 80 J 29 6.4 m s-1 30 140 J 31 a 40 J b 250 J c 5.0 cm d The elastic potential energy stored in the spring is transferred back to the trolley as kinetic energy when the spring starts to regain its original shape. e The spring is elastic. This means it can retain its original shape after the compression force has been removed. f Pinball machines, watches, computer keyboards etc. 32 a The section from 2.0 cm to 5.0 cm because the cube loses kinetic energy in this section. b 7.1 m s-1 c 3.0 J d The kinetic energy has been converted into heat and sound. e 100 N 33 100 m 34 a 9.1 m s-1 b 8.0 m s-1 35 a A, B, C b D c E, F, G d A, E e C, G 36 a 160 m b 2.0 m s-1 37 a 1.2 m s-1 north b 6.0 m s-1 north c 3.0 m s-1 south 38 2.4 m s-1 39 0.63 J 40 Energy is converted into sound and heat when the mass collides with the pencil. 41 0.69 m s-2 42 5.7 × 102 N 43 2.5 × 102 m 44 Kinetic energy is converted into heat. 45 On Earth W = 1.96 × 105 N; on Moon W = 3.2 × 104 N 46 a The chair will obtain an initial velocity from the push, and then slow down to a stop due to sliding friction. b Castors will involve rolling friction, which is less than sliding friction, and hence the chair will travel further before coming to a stop. c In both cases the force of the initial push causes the chair to accelerate while being pushed, which is an application of Newton’s first and second laws. While the person pushes on the chair (action) the chair will push back on the person (reaction) with an equal and opposite force being an example of Newton’s third law.

9 a 0.80 m, 20 cm b Q: up, S: up c


10 a A = 0.5 × 10-8 cm, T = 2.0 ms, f = 500 Hz b i 0 ii 0.5 × 10-8 cm iii –0.5 × 10-8 cm c

1 A particle carrying its kinetic energy with it, or a wave transferring energy but not matter. 2 A continuous wave has a vibrating item at its source; a pulse is caused by a single input of energy. 3 a continuous b pulse c continuous d pulse e pulse 4 a transverse b In a longitudinal wave the particle vibrates parallel to the direction of travel of the wave. In a transverse wave the vibration is perpendicular to the direction of travel. 5 a transverse b pulse c

20 10

P

–10

R

0.4

–20

0.8

1.0

1.2

1.4

1.6

S

Q

Distance along spring (m)

Displacement of Q (x 10–8 cm)

7.1 Introducing waves

Particle displacement (cm)

47 The steady force applied by the engine is equal and opposite to the combined resistance forces such as air resistance, friction between the wheels and track etc. The net resultant force on the carriages is zero, and according to Newton’s first and second laws, constant velocity is the result. 48 a 85 N m-1 b 43 J c Estimate area under force– extension graph 49 78.4 kJ 50 78.4 kJ 51 80 kW

0.5

1

2

4

3

– 0.5

Time (ms)

d The y-axis is actually showing displacement parallel to the direction of travel of the wave, but when showing this on a graph the y-axis is always perpendicular to the x-axis. Therefore this reminds us of a transverse wave.

7.3 Waves and wave interactions

7D 8 a 3 b i right ii left iii right 10 For a transverse wave, where a water wave travels parallel to the surface the water, particles must be able to move up and down. Only the particles near the surface can do this. Below the surface, the wave wouldn’t possess enough energy for particles to lift the weight of the water above.

7.2 Representing wave features 1 a f = 0.80 Hz, T = 1.2(5) s b f = 2.8 Hz, T = 0.36 s 2 0.28 m s-1 3 It diminishes. Amplitude indicates energy. Wave energy is spread over a larger area as the wave moves further from the source, hence the amplitude is smaller. 4 F = 8 Hz, T = 0.125 s 5 a 8.3 Hz b 0.12 s c i central ii maximum displacement upwards iii central (moving downwards) 6 D 7 0.04 m 8 See Teacher’s Resource CD

9 2

Amplitude (cm)

6 a longitudinal b continuous c

1 a halved b no change 2 0.60 m s-1 3 5.0 Hz 4 0.040 m 5 D 6 yes 7 They have identical frequency (or wavelength), and they are in phase. 8

1 1

2

3

4

5

Distance (cm)

6

7

8

wavelength = 4 cm amplitude = 2 cm

Solutions

585

17 Displacement (cm)

2 1 1

2

4

3

5

6

Distance (cm)

7

8

wavelength = 4 cm amplitude = 1.7 cm

Displacement (cm)

10

10

20

30

40

50

60

Q 70

80

1

4

5

6

7

8

1.5

1

2

90

3

4

5

6

7

8

6

7

8

Distance (cm)

Distance (cm)

19

P

10

10

20

Displacement (cm)

t = 5.5 s

30

40

50

60

70

80

90

Distance (cm)

wave A and wave B

1

2

3

4

5

Distance (cm)

20

10

0.25 0.50 0.75

1.0

1.25 1.50 1.75

2.0

Time (s)

4 1D: wave in string or spring, 2D: water surface waves, 3D: sound waves 5 string = transverse, air = longitudinal (sound wave) 6 2 cm 7 4 cm 8 0.15 m s-1 9 f = 3.7(5) Hz, T = 0.27 s 10 C, D 11 a 4.0 s b 1.95 × 10-3 s 12 a 2.9 Hz b 250 Hz c 100 Hz 13 a destructive interference b The two pulses must have opposite amplitudes that are of the same size, and they must have identical but opposite shapes. 14 A 15 f = 86 Hz, T = 1.2 × 10-2 s 16 While maintaining tension in the spring, the spring can be shaken in a direction that is perpendicular to the axis of the spring. The windings of the spring will pass on the disturbance.

586

Solutions

wave A

Displacement (cm)

Displacement (cm) Displacement of Q (cm)

3

3

wave B

2.6

2

2

Distance (cm)

t=0s P

1.5

18

Chapter 7 review 1

wave A

Displacement (cm)

Displacement (cm)

10

1.5

1

2

3

4

Time (s)

Chapter 8 Models for light 8.1 Modelling simple light properties 1 shadows, eclipses, can’t see around corners 2 a 3.6 m b 0.66 m c 18 m 3 D, then B 4 a A reflective material is deposited onto one side of a piece of flat glass. This is usually covered by a protective layer of paint. b Some reflection occurs from the front of the glass of a mirror as well as from the silvered layer. The images produced are slightly offset and seen as separate.

5 a upright mirror i = 50°, r = 50°, horizontal mirror i = 40°, r = 40° b upright mirror i = 30°, r = 30°, horizontal mirror i = 60°, r = 60° 6 Looking out of a window you can see your own image and the scene through the window. The two sets of images overlap. 7 a Diffuse: v, vi Regular: i, ii, iii, iv b The paper will produce diffuse reflection, and so light from the same point will be scattered in different directions. 8 B, D 9 The ocean can be considered to be made up of many layers of water from the surface down to the ocean floor. Each layer will reflect and absorb a little light, allowing the rest to be transmitted. Eventually, all the light will have been absorbed before it can reach the floor of the ocean. 10 eye

8.2 Refraction of light 1 a i D ii B iii A iv E v C b It slows down since it is refracted towards the normal. 2 a A, C b B, D c C d A 3 As the wavefront enters the faster medium the wavelength will increase and, as a result, the direction of travel will bend away from the normal. 4 a The light rays travelling from your feet are refracted at the air–water boundary and appear to have come from a higher position, therefore you are shortened. b The path of light from a section of the sky is curved because of layers of air at different temperatures. Light enters the eye, travelling slightly upwards. Therefore the ‘sky’ is seen on the road ahead. This is interpreted as a puddle of water reflecting the sky. 5 C 6 a 1.5 b 2.26 × 108 m s-1 7 b 1.5 c 1.5 d 2.0 × 108 m s-1 8 a 1.81 b 35° 9 a 58.0° b 39.6° c 58.0° d 18.4° 10 a = 25.4°, b = 25.4°, c = 28.9°

8.3 Critical angle, TIR and …MR 1 a No b Yes c Yes d No 2 1A, 2C, 3C, 4D 3 a 24.4° b 38.7° c 48.8° d 62.5° 4 1.46 5 36.9° 6 a 10-2 m b 300 nm

7 b travels at speed of light (3 × 108 ms-1) in a vacuum, does not require a medium in which to travel 8 C 9 a magenta b white c white d white 10 Blending of colours of the spectrum to form white.

8.4 Dispersion and polarisation of light waves 1 a red b blue c blue 2 If light was corpuscular, the orientation of the polarising material shouldn’t affect the amount of light that can pass through. 3 red white light

violet

4 If they are polarising holding the lenses parallel to one another, rotating one lens 90° relative to the other should block any light from passing through. 5 n(diamond) > n(glass) therefore more dispersion as light enters and leaves diamond. ic(glass) ≈ 42°, ic(diamond) ≈ 24°. Therefore light is more likely to strike diamond–air boundary at an angle greater than the critical angle and totally internally reflect, spreading the spectrum even further. 6 Only light that has its electric field aligned with the filter will be able to get through, therefore polarised light emerges from the filter. 7 a violet b violet c 25.4° d 24.5° e 0.9° 8 red 43.4° 9 The pathway is determined by the wave property, frequency. 10 The first filter reduces the intensity by half, the second filter will reduce the intensity further, but not completely block the light.

Chapter 8 review 1 a speeds up b away 2 Take measurements of corresponding angles of incidence and refraction. Graph sin i versus sin r to determine the refractive index. 3 a radio waves, microwaves, infrared, visible light, ultraviolet light, X-rays, gamma rays b All EMR travels at 3 × 108 m s-1 and all involve electric and magnetic fields varying at right angles to one another. 4 Speed of light is calculated as 2.25 × 108 m s-1, so medium is not air as this is too slow. 5 a 1 b 2 6 a 32.0° b 53.7° c 21.7° d 1.97 × 108 m s-1 7 a radio waves, microwaves, visible light, X-rays b X-rays, visible light, microwaves, radio waves c radio waves, microwaves, visible light, X-rays

Solutions

587

8 Blue light is refracted most by water and therefore scattered most, making the water appear blue. 9 C 10 Light reflected from horizontal surfaces is quite polarised, so sunglasses can filter out a large proportion of reflected light. 11 a There is refraction by layers of the atmosphere. b See Figure 8.21. 12 reflection, refraction, total internal reflection 13 Black: no colours of light are reflected; grey: all colours partially reflected. 14 A 15 a 19.5° b 19.1° c 0.4° d 1.96 × 108 m s-1 16 2.1 × 108 m s-1 17 a particle or ray b electromagnetic wave c ray d electromagnetic wave 18 1.17 19 a i 24.4° ii 41.8° iii 48.8° b Closer n values result in a larger critical angle. 20 It is refracted twice, first towards the normal on entering the glass and then away from the normal on exiting the glass. It emerges parallel to its original direction of travel.

Chapter 9 Mirrors, lenses and optical systems 9.1 Geometrical optics and plane mirrors 1 Any two of: linear propagation, reflection, paths in lenses and mirrors, and intensity reducing with the square of the distance from source. 2 When the dimensions of the physical systems are much greater than the wavelength of light. 3 If light is particle-like, sharp-edged bright areas of illumination would occur opposite the pinhole. 4 a upright b same size c virtual 5 nature (real or virtual), orientation, position, size 6 a They all produce images which are virtual, upright and same size. b A virtual image is produced behind the mirror (see Figure 9.3). 7 a 5.5 m toward the mirror b 3 m s-1 8 a 0.8 m—exactly half her height b 0.75 m above the ground 9 b An inverted image of the ‘S’ is seen. 10 b The image would become inverted as there is now only one mirror.

9.2 Applications of curved mirrors: concave mirrors 1 a Both mirrors are curved, usually spherical. Concave mirrors are like the inside of a spoon (e.g. make-up mirrors). Convex mirrors are like the back of a spoon (e.g. security mirrors). b The mirrors are a part of a sphere. It is the radius of this sphere. c f = 25 cm d See Figure 9.10

588

Solutions

2 a See Figure 9.8 b torch, headlights c Spherical aberration is the blurring of an image near the edge of a mirror because most spherical concave mirrors are not capable of bringing parallel light rays to an exact focus. 3 a virtual b 72 cm c 6 mm 4 real, inverted, diminished, v = 30 cm 5 f = 2 cm 6 a 40 cm b 10 cm 7 4 mm, virtual, upright, enlarged 8 80 cm; less than 40 cm from the pole of the mirror 9 u = 26.7 cm or 53 cm. A real or virtual, upright image can occur with M = 3. 10 f = 6.7 cm

9.3 Convex mirrors 1 35 cm 2 See Figure 9.15 3 If a wide field of view is required. Both mirrors produce upright and virtual images. The convex mirror will produce a diminished image. 4 a The image of the ant is virtual, upright and slightly diminished; v ~–1 mm. b Reflected rays diverge in a convex mirror. 5 B 6 shaving and make-up: concave, f ≈ 50 cm; shop security: convex, f ≈ 70 cm; rear-view mirror: 1 convex, M ≈ 40 7 –3.3 cm; –2.5 cm; –1.4 cm 8 upright, virtual, diminished, v = –2.7 m, M = 0.09 9 0.4 m 10 2 m

9.4 Refraction and lenses 1a

optical axis

F

F principal axis optical centre

F = principal foci

b See Figure 9.25. 2 Rays must actually intersect to form a real image. Ray paths through convex lens only allows ray from above principal axis to intersect below principal axis. Therefore image is inverted. 3aB bA 4 Yes, if different radius of curvature. Higher n requires longer radius of curvature. 5 a Focal length increases since light no longer bent as severely, since change of speed is less. b Since light speeds up on entering air from water, rays are refracted away from the normal and a beam of light would be diverged.

6 20 cm 7 B 8 D 9 a real, inverted, enlarged b real, inverted, diminished c virtual, upright, enlarged 10 a 0.6 m b 10 cm

9.5 Concave lenses 1 See Figure 9.25. 2 Rays are diverged, hence no real intersection of rays. 3 C 4 b virtual c 10 cm d 1 cm 5 b virtual c –7.5 cm c 5 cm 6 a –30 cm b 3 c 15 mm 8 a 22.5 cm b 202.5 cm 9 a Image is no longer real, it is virtual. b Image is now on the same side of the lens as the object, and at a distance of 30 cm is twice as close as previous. c Image is half the original size. 10 3.87 cm

9.6 Optical systems 1 spherical and chromatic aberration 2 equal to 3 a 4 cm b 4 mm 4 a 5.2 cm b away from film or sensor 5 i a slow b slow c high ii a fast b fast c low iii a slow b slow c high 6 It has no spherical or chromatic aberration. 7 a 50 cm, 10 cm b objective 50 cm, eyepiece 10 cm c 60 cm d inverted 8 Concave mirrors can produce real images. Photographic arrangement replaces eyepiece lens. 9 a 20 cm from objective lens, 5 cm from eyepiece lens b 30 cm from eyepiece lens

Chapter 9 review 1 D 2 B 3 a 6 m b See Figure 9.3 4 Parallel rays not converged to a perfect focus, resulting in an image that is blurred near the edges. 5 All images are virtual and upright but diminished in size. The mirror will give a very wide field of view. 6 a v = 3 m b Hi = 0.5 m c real, inverted, diminished 7 a virtual, upright, magnified b v = –120 cm c M = 4 d Hi = 8 mm 8 a v = –8.6 cm b virtual, upright and diminished c M = 0.43 9 rear-view mirror, shop-security mirrors 10 Object can be inside F to produce virtual image 1 (at 2 f), or between C and F to produce real image. 11 radius of curvature, refractive index 12 b virtual c –6.7 cm d 5 cm 13 hypermetropic 14 C 15 a v = 120 cm, real, inverted, 6 cm high b v = 72 cm, real, inverted, 2 cm high c v = –13.3 cm, virtual, upright, 3.3 cm diameter 16 A, C, D 17 a Microscope objective has shorter focal length. b Microscope forms a virtual, enlarged image. Refracting telescope forms a real, inverted image.

18 at infinity 19 a upright b 0.01 c 0.1 m d –0.10 m 20 a Either u = 10 cm, v = 40 cm, or u = 40 cm, v = 10 cm b 8 cm or 0.5 cm

Area of study review (Wave-like properties of light) 1 A wave is a periodic disturbance that travels with a velocity, v, and continuously transfers energy through a medium without any movement of the medium as a whole. Wave motion involves the transfer of energy without transfer of the medium. 2 C 3 B 4 B 5 a 0.02 s b 1.0 cm c 200 cm d 100 m s-1 6 a 0.19 m b 1.33 s c no d temperature of the water. 7A 8 When light passes from a one transparent medium into another transparent medium its speed is altered. It is this change of speed that leads to the bending of light rays as they pass through media of different optical densities. This phenomenon is referred to as refraction of light. 9 2.42 10 a 29° b 42° 11 a 1.49 b 2.01 × 108 m s-1 c 42.15° d Total internal reflection will occur. e Medium 1 is more optically dense than medium 2 because it has a higher absolute refractive index. 12 1.56 13 a 1.94 × 108 m s-1 b 3.09 × 10-11 s c 5.08 × 1014 Hz 14 D 15 D 16 C 17 D 18 Sound waves are longitudinal, all the others are transverse waves. 19 Total internal reflection can only occur when the absolute refractive index of the material containing the incident ray is greater than the absolute refractive index of the medium containing the refracted ray. 20 In a given medium the velocity of light, v, is set. From f = vλ we know that the frequency, f, varies inversely with the wavelength, λ. For example, a higher frequency note will have a shorter wavelength. 21 2500 m s-1 22 790 nm 23 B 24 C 25 C 26 This phenomenon, referred to as dispersion, occurs because each frequency of light (colour) is refracted a slightly different amount as it passes through the glass. 27 Concave mirrors form inverted real images of objects placed beyond the principal focus. If the object is between the principal focus and the mirror, the image is virtual, upright and magnified. Convex mirrors produce only upright virtual images of objects placed in front of them. The images are diminished. 28 a v = 20 cm, Hi = 1.0 cm, the image is real, inverted and the same size as the object. b v = 30 cm, Hi = 2.0 cm, the image is real, inverted and magnified. c The image

Solutions

589

cannot be located. d v = –10 cm, Hi = 2.0 cm, the image is virtual, upright and magnified. 29 E 30 object mirror (concave) screen

9.0 m

u

(image) v

Since image is to be projected onto a screen it must be real; u = 2.25 m and v = 11.25 m, f = 1.88 m. 31 object mirror (concave) screen

12 m

u


(image)

10.1 Motion in the heavens

v

Since image is to be projected onto a screen it must be real; u = 3 m, f = 2.5 m and, hence, the mirror is concave. 32 v = –0.20 m, Hi = 1.5 cm; hence the image will be upright, virtual, 1.5 cm high, and appear to be 20 cm behind the mirror. 33 The focal length of a convex lens is determined by its index of refraction and its radius of curvature. 34 C 35 a i 8.0 cm ii real, inverted and same size b virtual, upright, magnified 36 a

F

0.75 m 3.0 m

1.0 m

b The image is virtual, upright and diminished, apparently 0.75 m behind the mirror. 37 A and C 38 D 39 D 40 a u = 361 cm or 3.61 m b M = –0.024 c real, inverted, diminished d If u is very large, then using the lens equation v = 8.40 cm, so the lens will need to be moved a distance of 0.20 cm towards the film. 41 Chromatic aberration occurs because higher frequency colours are refracted slightly more by a medium than colours with lower frequency. 42 Polarised waves consist of transverse waves vibrating in one particular plane. 43 A

590

Solutions

44 Lenses refract different frequencies of light through slightly different angles. White light would therefore be spread out into its component colours. 45 Polarising filters only allow light with an electric field varying in a particular orientation to pass through. Other orientations are blocked. We say that the orientation of the transverse wave in only one direction is called polarisation. 46 A luminous body emits a stream of particles that travel in straight lines at high speed. These particles are small and do not interact with each other. These particles travel at different speed in different media. 47 B 48 A 49 A 50 D 51 C

1 C 2 D 3 A 4 SE, rising 5 27° above horizon in Brisbane, 38° in Melbourne 6 a C b D 7 a Pollux b Fomalhaut 8 a 6 h 23 min, –53° b 16 h 27 min, –26° 9 Stars rotate 361° per day. Orion will be 7° W of N. 10 Aquarius, 7 h before Orion

10.2 The Sun, the Moon and the planets 1 The Sun has reached its highest or lowest noon point. 2 Dates are the same but reversed. 3 No, the appropriate pole tilts to the Sun. 4 77½° summer, 30½° winter 5 Area is inversely proportional to the sine of the angle (75½° and 28½°), and the ratio of areas is 2.03. 6 Ecliptic is the actual path of the Sun; Zodiac is the band of stars through which the ecliptic passes. 7 If Moon–Earth distance is 30 cm, Earth’s diameter is 1.0 cm, Moon’s diameter is 1.7 mm. 8 Full moon rises at sunset, opposite the Sun. New moon rises with the Sun. 9 a, b, c midnight 10 Earth would not move in the Moon sky but would rotate and go through phases (29½ Earth days).

10.3 Understanding our world 1 2290 km 2 Venus is inside the Earth’s orbit. 3C 4 a It won’t; it is in line with the Sun. b opposition; no, only the superior planets 5 235 m. 6 Brahe found no evidence of parallax motion of the stars. Brahe had the planets circling the Sun.

7 a Mercury, 102.1%; Earth, 100.1%; Mars, 100.5%; b 2 mm, 0.1 mm and 0.5 mm 8 a 2.8 years b 4 AU 9 These implied that the Sun and the Moon were not perfect. 10 Relationship between the phases and the apparent size.

10.4 The telescope: from Galileo to Hubble 1 a B b C 2 12.5° 3 Yes. Light gathering power 2.25 times that of the other. 4 No hole because light unfocused; image a little duller. 5 Telescope has to counter Earth’s rotation. 6 a The eye is unable to resolve planets. b Pupil is larger, and diffraction pattern smaller. 7 θ = 0.003 arcsec, so diameter over 40 m—far larger than the biggest telescopes. 8 250 m 9 It can use UV and IR, which are absorbed by the atmosphere, and gather light for longer. 10 In light-gathering power it is like a 16 m telescope. Resolution is like that of a 200 m telescope.

10.5 New ways of seeing 1 Very faint, difficult to tell from a star or comet. Sky maps drawn by hand and needed long, careful observation. Outer planets move very slowly through the stars. 2 Neptune was predicted from its gravitational effect on Uranus. 3 Long exposures enabled very faint objects to be seen, recorded and compared over time. 4 electromagnetic radiation outside visible light; UV wavelengths <400 nm, IR wavelengths >700 nm 5 They did, but the atmosphere strongly absorbs these wavelengths. 6 Maybe, but any heat source would create shimmer in a telescope. 7 Radio waves are recorded electronically, day and night. 8 Advantages: not looking through the atmosphere, continuous operation. Disadvantages: cost, difficult repairs and adjustments. 9 X-rays are absorbed by the atmosphere. Stars emit the X-rays, people don’t. 10 Basically the same, both use a curved mirror to focus. Radio waves are >>1000 times longer. Optical image is formed in a 2D plane; radio telescope has to scan the object.

Chapter 10 review 1 Many answers! 2 Stars appear to rotate on the inside of a huge sphere. 3 Diurnal is nightly rotation due to Earth rotation. Annual is due to Earth’s motion around orbit (1° per night). 4 52°; higher in Brisbane, lower in Hobart 5 due east and west everywhere 6 No night. In eclipse all stars rotated around the SCP by 180°. 7 Opposite halves of the sky visible. 8 8 min; actual difference about 7 min. 9 a i A ii N iii A iv P v P vi P b Orion 12 h, star v 3 h, star vi 21 h 10 a Arcturus b Betelgeuse c Alpha Centauri d Aldebaran 11 a RA 6 h 43 min, dec. –17° b RA 1 h 36 min, dec. –58° c RA 18 h 35 min, dec. +39° d RA 5 h 12 min, dec. –8° 12 It varies with position in time zone and time of year (±1 h). 13 A 14 Altitude of Sun is 52° at equinoxes. The 23½° tilt of the Earth is added or subtracted. 15 Concentric, but tilted at 23½°. Tilt rotates around the sphere once every 26 000 years. 16 A 17 A full moon is on the opposite side of the ecliptic to the Sun and so tilted the other way. 18 a Io b i Lunar eclipse is much more likely. ii Io appears about 5 times larger than the Sun and so a total solar eclipse is more likely and will last longer. 19 23°, about 1½ hours 20 5.5° 21 30.3 km s–1 22 The one with the 20× magnification gives the larger image, but the one with the 26 mm objective would have a brighter image. 23 a 44 times b 975× 24 a 8 × 30 b 7 × 50 by 2.8 times. 25 They use either photographic film or electronic charge-coupled devices. 26 Both are electromagnetic waves, but radio waves are over 1000 times longer.

Chapter 11 Astrophysics Note that while many of these questions require detailed answers, there is room here for only very brief answers. More complete answers are provided on the Teacher’s Resource CD.

11.1 The stars— how far, how bright? 1 a apparent movement of close objects relative to distant ones b yes c No, stars were very far away.

Solutions

591

2 a 3600 arcseconds b 1800 arcseconds c 4 km 3 Calculated speed is 94 km h–1, so car is not travelling perpendicular to line of sight. 4 a It is the distance to a star with parallax angle of 1 arcsecond: d = 1/p. Used because it is simply determined from parallax measurement. b 2.3 pc, 7.4 l.y., 4.7 × 105 AU, 7.0 × 1016 m 5 Hipparchus called brightest stars ‘first magnitude’ (+1) and dimmest ‘sixth magnitude’ (+6). Brighter ones were then ‘below’ +1. 6 a 2 W m–² b all c Mostly IR with a little visible light. 7 a 2640 W m–² b 50.7 W m–² 8 Deneb by 84 times 9D

11.2 Our favourite star 1 3600 km 2 Physics of laws of heat transfer was well known. 3 Known processes could only explain a few thousand years of Sun’s life. 4 Nuclear forces are 104 times greater so 108 times more nuclear energy is produced. 5 a 2.3 × 10–19 J, 1 × 10–12 J b about 4 million times c 20 tonnes 6 1.6 × 10–9 kg, 7 g 7 There is a very small theoretical difference. 8 Hydrostatic equilibrium—see full answers. 9 Conduction then radiation—see full answers. 10 a C b A

11.3 We know stars by their light 1 Radio, IR, UV and X-rays are used. The atmosphere blocks much of it. 2 continuous (temperature), emission (composition), absorption (gases in space) 3 Temperature gives rate of radiation per m², which is compared to luminosity to give surface area. 4aD bA 5 Determines type of star and hence gives luminosity, which is compared to apparent magnitude. 6 Binary star mass found from Newton’s laws. Gives mass–luminosity relationship for all main sequence stars. 7C 8 The same mass–luminosity and H-R diagram relations hold for all stars and these depend on the fusion processes. 9 Comparing ‘clusters’ and H-R position enables estimates of lifetime. 10 It is the explosion of a very large star. Requires huge gravitational forces.

592

Solutions

11.4 Whole new worlds 1 Hubble found Cepheids in them, and so found absolute magnitude and hence distance. 2 approx 15 Milky Ways; 5º, about 10 full moons; very far away, so too faint 3 Centre deduced from globular clusters. Dust clouds obscured the centre. 4 Distance to dimmer Cepheid is twice distance to other. 5 a 2 × 1012 pc3 b 20 pc3 per star c 2.7 pc d About 1 pc, because stars in outer reaches are much more sparse. 6 On approach the frequency is higher, then gets lower—as if engines are slowing. 7 30 000 km s–1 away from us 8 2 × 10–18 s–1 9 6300 km s–1, 2% of c 10 Looking far back in time to early universe. Quasars are not seen at closer distances.

11.5 The expanding universe 1 Gravity should collapse a non-infinite universe, but is appeared unchanging. Olber’s paradox 2 Sky should be bright because there should be stars in all directions. Expanding universe implied light from stars >14 billion l-y would not reach us. 3 Steady state (universe infinite, eternal) and big bang (universe expanding). In steady state matter is continuously created. 4 4 × 1031 molecules; would not detect formation of a few new ones 5 0.5 cm s–1 and 1 cm s–1 6 No meaningful answer possible! 7 Radio waves detected were the left over radiation from the big bang. 8 Wavelength has increased due to expansion of space. 9 Irregularities necessary for the formation of galaxies. 10 Implies age is 2 billion years—less than the age of the Earth.

Chapter 11 review 1 It is the apparent movement of stars as the Earth moves around the Sun. Galileo said stars must be very far away. 2 85 km s–1, about 3 times Earth’s orbital speed 3 18 km 4 a 32 b 100 5 a 0.87 AU b 250 W m–2 c 1.75L• 6 Radiation emitted increases very strongly with T and increases in frequency (into visible). 7 colour → temperature → radiation m–²; L → total radiation and thus → surface area → radius 8 a ~0.8 b ~4500 K, reddish 9 1093 = 1.3 million 10 a 5000 years b 10 billion years 11 C

12 Convection currents bring the heat from the core. Surface changes over minutes. 13 Spectra show lines similar to spectra of Sun and known elements. 14 temperature, elements and their state, pressure, magnetic field 15 Sun is close to centre of H-R diagram, but most stars are actually cooler. 16 1000 times (approx.) 17 the middle of the main sequence (approx.) 18 See full answer. 19 nebulae remnants and pulsars 20 Period is related to luminosity, so distance can be found. Also very bright. 21 a about 1.9 pc average; central stars are closer, outer ones are further apart b Sun–stars approx 1 pc; a little closer than average 22 34 m s–1 (or 122 km h–1) toward us 23 3.5 km s–1 (4.4 km s–1 SMC); a negligible speed astronomically. 24 Blueshifts imply motion toward us—only relatively close galaxies. Expansion is on the very large scale. 25 Position on the H-R diagram and nuclear processes; the cosmic microwave background radiation. All estimates around 14 billion years.

Chapter 12 Energy from the nucleus 12.1 Splitting the atom—nuclear fission 1 a In fissile nuclei, the forces that act between the nucleons are not as large. b The energy of the additional neutron makes the nucleus wobble and the electrostatic forces cause the nucleus to break apart. 2 No, extra neutrons would be needed to make the nucleus stable. 3 a 3 b 4 c 5 4 x = 239, y = 40 5 B 6 D 7 a 3.1 × 10-11 J, 196 MeV b About 20 times more energy is released in the fission reaction. 8 2.36 × 10-30 kg 9 alpha particle 10 a 1.91 × 10-11 J b 1.19 × 108 eV c 3.1 × 1010

12.2 Aspects of fission 1 B 2 It doesn’t have a high enough concentration of the fissile isotope, uranium-235. 3 B 4 238U itself is not very fissile. However, when 238U absorbs a neutron it decays to form 239Pu, which is fissile. 5 When a slow neutron strikes the nucleus, the nucleus is made to wobble slightly, but not enough to disturb the balance between the repulsive electrostatic and attractive nuclear forces. However, when a fast neutron strikes the nucleus, it has enough energy to make the nucleus wobble and elongate. This upsets the delicate

balance between the strong and electrostatic forces and the electrostatic forces cause the nucleus to separate. 6 239Pu, a smaller amount of this isotope is capable of exploding. 7 First, a neutron causes fission to occur in a uranium-235 nucleus, thus releasing 2 or 3 more neutrons. These then go on and induce fission in more uranium-235 nuclei, each resulting in the release of 2 or 3 neutrons and so on. The chain reaction grows very rapidly and energy is released in each fission reaction. 8 As a result of its shape, a very high proportion of neutrons can escape from the material, and so the chain reaction dies out. 9 a fertile; b fissile 10 It must be a low percentage of U-235.

12.3 Nuclear fission reactors 1 B 2 D 3 a The fission process in the reactor core produces heat. This heat energy is conducted into the coolant which is flowing through the core. The energy is used to produce steam which drives a turbine to generate electricity. b The difference is that the heat energy that makes the steam is produced by burning coal instead of a nuclear fission reaction. c They both use steam to turn a turbine to generate electricity. 4 The nucleus is too heavy. When a neutron collides with a lead nucleus, the neutron will keep almost all of its energy, and so not slow down sufficiently to be captured by a fissile nucleus. 5 a The chain reaction will be self-sustaining, i.e. critical, and a steady release of energy will result. b The chain reaction will die out because it is subcritical. This will lead to a decrease in the amount of energy produced. c The chain reaction will grow, causing an increasing amount of energy to be produced. This may be dangerous and could result in an explosion. 6 a Fast neutrons are most unlikely to be captured by the nuclei. b Slow neutrons are likely to be absorbed by the nuclei and cause fission. 7 a It results in the uranium-238 transmuting into plutonium-239. b Plutonium is highly radioactive and has a half-life of about 24 000 years. 8 a plutonium-239 b They rely on fast, high-energy neutrons to induce fission in plutonium nuclei. c They produce, or breed, more of their own fuel, plutonium-239, when neutrons are absorbed by uranium-238 nuclei. d Fast breeder reactors do not have moderators. 9 Since only one neutron is required to sustain the chain reaction, the remaining neutrons are able to breed more plutonium.

Solutions

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10 Over a period of months, the fissile nuclei in the fuel rods become depleted, the number of fissions decreases, and so fewer neutrons are flying around in the core. In order to maintain the chain reaction, the control rods must be gradually withdrawn. 11 a about 100 000 years b The fission fragments, which have relatively short half-lives, have decayed. 12 a The uranium stays in the spent fuel rod, which is initially stored in water, then kept underground in storage tanks. b The fuel rod is initially stored in water, then the uranium is extracted from the fuel rod. It then undergoes enrichment and fabrication before being used as fuel again. c More energy is extracted from the fuel and a much smaller quantity of high-level waste is produced.

12.4 Nuclear fusion 1 D 2 C 3 water in lakes and oceans 4 achieving temperatures high enough for fusion to take place and containing the extremely hot fusion material within the reactor 5 The tokamak style of reactor has a torus-shaped core where the plasma of the fusion fuel is located. This fusion material is held away from the reactor wall by magnetic fields. The heat energy produced is carried from the reactor core by lithium and water. Inertial confinement is a technique in which a pellet of fusion fuel is struck by high-intensity laser beams, creating temperatures high enough to sustain nuclear fusion. 6 a No—the hydrogen nuclei would not be travelling fast enough. b The nuclei would only experience the electrostatic force of repulsion as they approached. c When the nuclei are widely separated, only the electrostatic force will be acting. Then when the nuclei move close together, the strong nuclear force of attraction will overwhelm the electrostatic force and bring the nucleons together. 7 Inside a massive star, the gravitational pressure and so the temperatures will be greater than inside a star the size of the Sun. As a result, fusion will take place at a greater rate, causing these stars to shine more brightly, but also causing them to run out of nuclear fuel more quickly. 8 a The hydrogen and helium-3 nuclei have more mass than the products of the reaction. b The total mass of the products is less than that of the original nuclei. The difference in mass is the energy released. c 3.4 × 10-12 J d 3.7 × 10-29 kg e It has decayed into a neutron, a positron and a neutrino. 9 The fuel that is used for fusion is readily available and easily extracted from water. The waste products of fusion are not highly radioactive.

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10 They use magnetic fields to confine the plasma in which the fusion process is taking place to the middle of the doughnut-shaped reactor core—away from the wall of the reactor.

Chapter 12 review 1C 2 a uranium-238 b uranium-235 c The concentration of the fissile uranium-235 is too low. 3 2 4 2.7 × 103 tonnes 5 B 6 They are mostly absorbed by uranium-238 nuclei, absorbed by the control rods, or escape from the reactor core. 7 The uranium deposits would have spontaneously exploded millions of years ago. 8 a sliced b two hemispherical pieces c a 2 kg lump d The material is carried as small subcritical pieces, then combined at the time of detonation. 9 a The kinetic energy of the fission fragments. b The heat energy is removed from the reactor core by a coolant and is used to create steam, which is then used to turn a turbine and generate electricity. 10 a 10 n + 238 U → 239 U, 238 U → 238 Np + –10 e, 92 92 92 93 239 Np → 239 Pu + –10 e 93 94 b It has a relatively short half-life (24 400 years) and so has completely decayed since the formation of the Earth. 11 a There are about the same number of protons and neutrons. b There are more neutrons than protons. c The forces are relatively large, so the nuclide is very stable. d The forces are not as large as those on proton X. This nuclide is less stable. 12 a The oceans are so vast that any radioactivity leaks would be greatly diluted. The radioisotopes also have relatively short half-lives. b The radioactivity could leak and cause damage to fish and other marine life. 13 a The control rods absorb neutrons and control the rate of the fission process. b The coolant, after being heated by the fuel rods, is used to produce steam. This is in turn used to drive a turbine around and generate electricity. c It can reach higher temperatures when under pressure. 14 a The nucleus is not rigid. Rather, it can move and wobble like a drop of water. b The neutron causes the nucleus to wobble around and neck in the middle. At this point, the repulsive forces overwhelm the attractive forces and the nucleus splits. 15 a Able to split in two when hit by a neutron. b Able to absorb a neutron and become a fissile isotope. 16 a slow b Iron is not fissile. c the decrease in the binding energy during fission d They also cause fission and so produce a chain reaction.

17 a The reactants have more mass than the fusion products. b The difference in potential energy represented by the missing mass has appeared as the kinetic energy gained by the fusion products. c As the nuclides approach, an electrostatic force of repulsion is acting on them, but their speed enables them to overcome this force and fuse together. At this point, electrostatic forces of repulsion and larger nuclear forces of attraction are acting on the protons. The neutrons experience only strong nuclear forces of attraction. 18 a 4.49 × 10 –11 J b 280 MeV c Fission fragments are usually radioactive. 19 D 20 a Electrostatic forces of repulsion act on the protons. They do not have enough energy to overcome this force and fuse together and so the strong nuclear force does not come into effect. These protons have not jumped the energy barrier. b Electrostatic forces of repulsion act on the protons initially, but they have enough energy to push past these forces and get close enough for the strong nuclear forces of attraction to take effect. This force enables the nucleons to fuse together. These protons have overcome the energy barrier.

Chapter 13 Investigation: flight 13.1 The four forces of flight

7 The surface was pushed off by the increased pressure under the surface of the wing or from a high angle of attack. 8 The sheets of paper come together due to the creation of a low pressure region between the sheets as a result of the faster air flow. 9 Taking off into the wind increases relative air speed past the wings and increases lift. 10 The wings are not generating any lift as a result of relative air speed. For the same reason, helicopters use considerably more fuel when hovering. 11 The air over the top of the coin creates an area of low pressure. A little air under the coin and it will lift through this area of low pressure. 12 Fnet = 1.4 × 106 N 13 30 kN 14 38 kN 15 8 kN 16 1.6 × 106 W

13.2 Modelling forces in flight 1 The change in relative position of the large mass of fuel affected the aircraft’s angle of attack, changing the amount of lift generated by the wing. Moving the fuel allowed the pilot to restore the angle of attack without the need for large rear elevators. 2 b

1 Ff

centre of side

Fd Fa

Where Ff is the forward push of air on the jet, Fa is the push of the jet on the air, and Fd is air resistance. 2 The sudden change in relative air speed causes a large and sudden change in the lift generated by air speed past the wing. This can be enough to cause a stall. 3 The aerofoil shape is only part of the lift generated by the wing. The transfer of momentum from air incident on the surface of the wing also contributes large amounts of lift. The amount of lift from this source is determined by the angle of attack. A plane flying upside down will maintain an angle of attack sufficient to generate the lift required. 4 Lift will be four times that at the original speed. 5 small and light, small frontal area, very smooth outer skin, long and narrow wings 6 Broad wings provide lift at low speeds for takeoff and, with flaps extended, extra drag for landing. At high speeds swept back wings reduce drag and lift, which is proportional to speed.

a centre of side

centre of gravity

centre of gravity

3 Other aspects of the aircraft must be designed around the position of the centre of gravity to maintain stability. Other considerations include frontal area and increased drag for the wider wing compared with decreased glide ratio for delta wings. 4 612 kg

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5


aileron down—lift increased

15.1 Ultrasound and how it is made upwards force

aileron up—lift reduced

weight sideways force

downwards force

6 12 metres forward of the aircraft’s centre of gravity 7 937 N 8 Altering the angle of attack increases lift from the main wing.

Chapter 14 Investigations: sustainable energy sources 14.1 …nergy transformations 1 a electrical to sound b kinetic to gravitational potential c gravitational potential to kinetic (and sound) d light to electrical 2 a high grade, high grade b high grade, high grade c high grade, high grade d low grade (depending on wavelength), high grade 3 A is chemical potential B is heat energy C is kinetic D is electrical 4 250% 5 Additional heat energy is gathered from the surrounding environment. (Note: While the heat pump may still be powered by fossil fuels, the efficiency of heat pumps makes them a valuable contributor to energy conservation.) 6 39% 7 Most of the remaining energy is transformed into heat energy and lost to the surrounding environment through the cooling system. 8 360 J 9 It is mainly transformed into heat. 10 It is harder to harness into effective work. 11 The energy is transferred to the ground and transformed into heat energy. New energy-efficient cars recycle this energy to drive generators that are charging batteries, or to power devices such as superchargers. 12 2.52 kJ 13 It has been mainly transformed into heat; the material the kettle is made of will get hotter.

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1 The period, T, is the time interval for one vibration or cycle to be completed. The frequency of a wave, f, is the number of vibrations or cycles that are completed per second (Hz). The wavelength is the distance between successive points having the same displacement and moving in the same direction; that is, points that are in phase. The amplitude, A, is the value of the maximum displacement of a particle from its mean position. 2 a i vocal cords and air in the larynx vibrating ii small speaker disc vibrating iii the sides of the cicada’s thorax (body) vibrating b All sounds must have a vibrating item at their source. 3 Diagnostic ultrasound involves images taken of the inside of the body, and therapeutic ultrasound involves treatments for specific illnesses such as kidney stones. 4 a 6.16 × 10-4 m b 6.24 × 10-4 m c 5.88 × 10-4 m d 6.28 × 10-4 m 5 reflection, absorption, transmission 6 bone and air 7 1.70 × 10-2 m 8 The piezoelectric crystal layer (the disc) is alternately compressed and stretched slightly by the arriving sound wave. The compression results in a net negative charge occurring close to one electrode, leaving the other plate with a slight positive charge. The stretching of the crystal layer results in a potential difference across the plates of the opposite polarity. An alternating electrical signal is produced. 9 Rather than a whole item vibrating while maintaining a constant thickness, the piezoelectric crystal expands and contracts to produce sound waves. 10 1.7 mm

15.2 Ultrasound interactions 1 6.400 × 106 kg m-2 s-1 2 0.005 J 3 Very high frequencies are more readily absorbed by the body and therefore cannot penetrate deeply enough and reflect back to the transducer. 4 Penetration is ~15 cm so it is suitable. 5 There is too much reflection from bone (ribs) and reflection at air pocket boundaries. 6 a 1.01 × 103 kg m-3 b 9.39 × 102 kg m-3 c 1.07 × 103 kg m-3 7 The gas–soft tissue boundary has a large difference in acoustic impedance values; therefore, strong reflection occurs. 8 From Table 15.4, only 1% would penetrate. 9 a 0.99 or 99% b 0.00744 or 0.744% c 3.17 × 10-4 or 0.0317%

15.3 Scanning techniques 1 The gallstone is so dense that ultrasound waves reflect and there is little penetration beyond it. 2 The pulses may be reflected from different depths. Therefore, more absorption will occur for the ones that have travelled through more tissue. 3 a 375 m b 1.5 × 10-4 m 4 Very high frequencies have very little penetrating ability due to absorption. A frequency this high would achieve only a few millimetres of penetration. 5 The Doppler effect is the change in the wavelength of a periodic wave that is reflected from a moving object. The presence of plaque in the artery would be indicated by irregular blood-flow speeds in the region. 6 The A-scan, since only the timing of the emitted and reflected pulse is utilised. 7 No. This is a very high frequency and would not penetrate the 10 cm or so that is required to image the heart. 8 Ultrasound cannot image items that lie behind or inside bone or air sacs within the body, due to the large reflections at these items. Layers of soft tissue that are very thin may not be detected as separate layers. 9 Many tiny transducers are placed in the one probe; they are triggered one after another to produce a sequence of images. 10 Equipment that produces 2D images is far more common and cheaper for hospitals to purchase. The images produced are more than adequate for diagnosis. Three-dimensional images don’t necessarily greatly improve diagnostic capabilities and the machines are very expensive.

15.4 Diagnostic X-rays 1 a Target: electron strike results in electron deceleration and hence X-ray production b Filament: source of released electrons c Electric field: accelerates electrons to high speeds towards the targetd Lead cover: absorbs stray X-rays to protect workers e Aluminium filter: filters/removes higher energy X‑rays from the emergent beam 2 a 3.0 × 1018 Hz b 3.0 × 1017 Hz 3 3.3 × 10-15 J 4 to accelerate electrons to a high enough speed so that their sudden deceleration is large enough to produce an X‑ray 5 2.4 × 1019 Hz 6 a 100 000 eV b 1.6 × 10-14 J 7 a yes b no c yes d no 8 Large atomic number (i.e. heavy atom) and a very high melting point. 9 a The average and the maximum X-ray energy would be reduced. b The line spectra will have different frequency (energy) values.

10 No, each electron is given 10 000 eV. X-ray photon energies lie between 30 and 50 keV.

15.5 Radiotherapy, radioisotopes in medicine and P…T 99 1 a technetium-99m b as the mother molybdenum 42 Mo 99 99 0 2 42Mo → 43Tc + –1 β 3 It must have a short half-life (within hours), it must not emit alpha or beta radiation, it must emit gamma radiation and be available in the highest possible activity. 4 Ionising radiation can produce ionised water molecules in the cells and highly reactive hydroxide ions that can damage parts of the cell including DNA. 5 Gamma radiation will leave the body and be detected by camera; alpha and beta cannot leave the body but will reach nearby cells and destroy them. 6 Cells in reproductive organs, the eye and the bowel need protecting, because they reproduce relatively quickly. 7 to give healthy cells a chance to recover between treatments 8 to locate the exact position of a tumour and monitor health of nearby organs 9 a radioactive isotope attached to a carrier substance that will aggregate in a particular part of the body 10 The effect that ionising radiation has on DNA is that it can cause the mutation of the cell as it reproduces; the mutated cell can be cancerous. Enough ionising radiation can kill such a cell or at least damage it so that it can’t reproduce.

Chapter 15 review 1 If the gel is not used 99% of the sound wave would be reflected at the air–skin boundary. 2 They should be sources of alpha and beta radiation of sufficient energy to penetrate the diseased section of the body. It is convenient if the isotope is also a weak source of gamma radiation so that its presence in the body can be monitored by gamma cameras. 3 a 1.47 × 10-3 m b 1.57 × 10-3 m c 3.70 × 10-3 m 4 Any device that converts energy from one form into another. An ultrasound transducer can convert electrical energy into a sound wave and vice versa. 5 These cells are reproducing at a faster rate than normal cells. 6 When certain crystals are placed under stress (stretched or compressed) they acquire equal and opposite charges on opposite faces. Hence an electrical potential difference develops between the surfaces. 7 They both have a cathode and an anode target. 8 It is attracted by the large positive potential.

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9 Diagnostic applications use radioactivity to produce an image of the diseased part of the body. Therapeutic applications use radiation to kill cancerous cells. 10 a 2.6 × 10-15 J b 1.7 × 104 eV 11 a the level of efficiency with which sound waves are able to pass through a given medium b 1.63 × 106 kg m-2 s-1 12 A short half-life means that a patient’s exposures are limited and they are not emitting radiation when they leave hospital. 13 the collision of the electron with the heavy positive target 14 gamma (food irradiation), X-rays (medical imaging), ultraviolet (tanning booths), visible (sight), infrared (radiator-style heaters), microwaves (communication), radio waves (remote control garage doors) 15 3.0 × 10-13 to 3.0 × 10-8 m 16 electrons moving between shells in the atoms of the target material 17 soft (low energy), since they do not need to be deeply penetrating 18 a the thickness of a material that is required to reduce the intensity of an X-ray beam to half of its original value b Heterogeneous beams include X-rays of a range of energy values. On their way through the body the lower energy X-rays are absorbed into the tissue, the beam is hardened and therefore it becomes more penetrating —hence the HVT increases with depth. 19 Benign tumours are self-contained and nonspreading. Malignant tumours are invasive and often cells can be carried through the body. 20 absorption and scattering 21 0.07 J 22 true 23 The Australian Nuclear Science and Technology Organisation, ANSTO 25 radiotherapy, since these must be energetic enough to damage and destroy cells 26 X-rays alone do not have much effect on the photographic film. The fluorescent material emits light (where the X-rays strike) and produces stronger effects on the film. 27 To produce a sound wave, the applied voltage must be reversed periodically. One polarity will expand the crystal and the opposite polarity will contract the crystal. 28 Doppler imaging is used as it can provide information regarding the speed and direction of blood flow through the valve. 29 Rapidly reproducing cells, such as those of a foetus, are more prone to damage from radiation. 30 In order to prevent over-heating the target is rotated so that the same section of the target is not used for very long.

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GLOSSARY aberration A distortion of an image due to a physical property of the image-forming apparatus.

absolute quantities Physical quantities that remain constant within any frame of reference.

absolute refractive index An index or number that is equal to the ratio of the speed of light in a vacuum compared with the speed of light in a medium.

absorbed dose The amount of radiation energy absorbed per kilogram of irradiated material, measured in grays.

absorption Energy taken up by a substance and converted to heat (kinetic) energy.

acceleration The rate of change of velocity. Acceleration is a vector quantity.

accommodation The adjustment of the focal length of the lens of the eye in order to focus upon near or distant objects.

acoustic impedance, Z, The level of efficiency with which sound waves can pass through a given medium.

action–reaction forces Forces that bodies exert on each other during collisions or interactions as described in Newton’s third law. These forces are equal in magnitude but opposite in direction.

activity The number of nuclei of a radioactive substance that decay each second, measured in becquerels.

aerodynamic Having a shape that reduces drag when moving through the air.

aerofoil An object that is shaped like the cross-section of a bird’s wing, and experiences an upward thrust when a stream of air moves across it.

air resistance The retarding force (drag) caused by collisions between air and moving objects.

alpha particle (α particle) Two protons and two neutrons ejected from the nucleus of a radioactive nuclide.

alternating current (AC) An electric current that continually reverses direction. The mains electricity supply is AC and, in Australia, alternates direction 50 times per second.

alternating voltage An electrical voltage that periodically reverses in polarity.

ammeter A meter that displays the amount of electric current that is flowing through it. It must be connected in series with the appropriate circuit elements. Compare voltmeter.

ampere The unit for electric current; a charge of one coulomb per second.

amplitude The amplitude of a wave is the value of the maximum displacement of a particle from its mean position.

angle of incidence The angle made between an incident ray and the normal to the surface that the ray strikes.

angle of reflection The angle between the reflected ray and the normal to the surface that it strikes.

antineutrino Neutral sub-atomic particle having very weak interacting ability with other matter. Antimatter particle of neutrino.

astronomical unit (AU) The average distance from the Sun to the Earth; 150 million km.

atomic number The number of protons in a nucleus. attenuation The energy losses that occur as a medium carries a signal or wave.

bandwidth The range of wave frequencies that is carried by a medium.

battery A set of cells arranged in series or parallel so as to provide an EMF to an electric circuit.

Bernoulli’s principle Where the velocity of a fluid is high, the pressure on a surface over which it flows is low, and where the velocity is low, this pressure is high.

beta particle (β particle) An electron ejected from the nucleus of a radioactive nuclide.

big bang The explosion of space and time that occurred about 14 billion years ago and brought our universe into existence.

celestial poles (North and South) The points in the sky around which all the stars appear to rotate. They are actually the extensions of the Earth’s axis.

Glossary

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celestial sphere The imaginary huge sphere of stars that appears to rotate around us (because of the Earth’s rotation). It rotates around the celestial poles, and the celestial equator separates the northern and southern skies.

cell A combination of different chemicals which convert chemical energy to electrical potential energy of electrons; that is, it provides an EMF. Usually part of a battery.

centre of mass The point at which the total mass of an extended object can be considered to be located for purposes of motion analysis.

chain reaction A series of nuclear fissions that may be controlled or uncontrolled.

charge concentration Any source of EMF produces electrical potential energy by doing work pushing charges close together (or pulling them apart); i.e. concentrating charges. The greater the concentration the greater the voltage. A Van de Graaff machine has a very high concentration of charges on its dome.

circuit elements Items, such as resistors and light bulbs, in an electric circuit which convert electrical energy into some other form of energy.

circumpolar stars The stars that are close to one of the celestial poles and which are therefore always in our sky.

compression An area of higher air pressure in the transmission of a sound wave.

concave A shape that is inwardly curved. conductor A material through which charges (often electrons) can travel relatively easily when subject to an electric field.

conservation of charge Electric charge cannot be created or destroyed, only moved from one place to another and so the total charge in a closed system remains constant. Remember that equal positive and negative charges balance one another.

constructive interference This occurs when two waves that have particle displacements in the same direction meet.

contact forces Forces that require one object or material touching another. Friction, drag and normal reaction forces are contact forces.

600

Glossary

continuous waves Waves created when there is a repetitive motion or oscillation at the wave source. These are also called periodic waves.

control rod A material, commonly boron steel or cadmium, that absorbs neutrons in a nuclear reactor.

conventional current Basically the same as electric current. Conventional current is in the opposite direction to electron flow.

convex A shape that is outwardly curved. coolant A substance, commonly water, carbon dioxide or liquid sodium, used to transfer heat energy from the core of a nuclear reactor.

core The part of a nuclear reactor where nuclear fission occurs and heat energy is produced.

corpuscular Particle-like in nature. cosmic microwave background The leftover radiation from the big bang, which has cooled to about 2.7 K and permeates the entire universe.

coulomb The SI unit of charge: 6.24 × 1018 times the charge of an electron. Defined as the charge carried by a current of one ampere in one second.

Coulomb’s law The law of force between two point charges. The force is proportional to the product of the two charges and the inverse square of the distance between them.

critical angle The angle of incidence that produces an angle of refraction of 90°.

critical mass The minimum amount of enriched fissile material in the shape of a sphere that leads to a sustained fission reaction.

decay series A sequence of radioactive decays that finally results in the formation of a stable isotope.

declination (dec) One of the two coordinates describing the position of a star. It is the angular separation from the celestial equator.

de-modulated A signal bearing information is removed from its carrier wave for processing.

destructive interference This occurs when two waves with particle displacements in opposite directions meet.

diffraction The bending of the direction of travel of a wave as it passes through an aperture or as it passes by an obstacle.

diffuse reflection When light is reflected from a rough or uneven surface and is scattered in all directions.

diode A semiconductor device that allows current to flow only in one direction. There is always a small potential drop of around 1 V across a diode when conducting in the forward direction.

dioptre The unit with which the power of a lens is measured. It is equal in value to the reciprocal of the focal length of the lens.

direct current (DC) An electric current that continually flows in the same direction. Batteries provide DC. Compare alternating current (AC).

dispersion of light On refraction, different colours of light will take slightly different paths. This results in the spreading out of white light into its component colours. This is known as dispersion.

displacement The change in position of an object. Displacement is a vector quantity.

distance How far an object travels during a particular motion. Distance is a scalar quantity.

dose equivalent A measure of radiation that takes account of the biological effect of the dose. It is measured in sieverts, Sv.

drag Resistance of the air (or some other fluid) to something moving through it.

dynamics The study of the forces that affect the motion of objects.

ecliptic The annual path of the Sun through the stars. The band of stars along the ecliptic is known as the zodiac.

effective dose A measure (in sieverts, Sv) used to compare the risk of a non-uniform exposure to ionising radiation with the risks caused by a uniform exposure of the whole body.

effective resistance The resistance of a single resistor which could replace a combination of several resistors combined in series and/or parallel.

efficiency (of a device) The percentage of energy that is transformed to a useful form.

elastic medium A medium/substance that allows energy to be conducted through it without loss of energy.

elastic potential energy When work is done by an applied force on an elastic material, energy is stored within the atomic bonds of the material (as it is compressed or stretched). The amount of elastic potential energy stored (in joules) is given by the area under the force–extension graph for the material.

electric charge The inherent property of electrons (negative) and protons (positive) which results in an electric force between them. Large objects become charged as a result of an imbalance in the number of positive and negative particles.

electric circuit A continuous path of conductors and circuit elements, and at least one source of EMF, around which an electric current can flow. (See circuit elements.)

electric current The rate of flow of electric charge through a conductor. The direction of an electric current is the direction of transfer of positive charge. This is often the result of the movement of electrons in the opposite direction.

electric field A region in which an electric charge experiences an electric force. The direction is that of the force on a positive charge.

electric force The force between electrically charged particles or objects due to their charge. Repulsion between like charges and attraction between unlike charges. See also Coulomb’s law.

electric potential The potential energy, per unit of charge, of a system of electric charges due to the mutual forces between them. Often called the voltage.

electrical potential energy Potential energy due to the concentration of charge in part of an electric circuit.

electric power Rate at which electrical energy is converted to some other form of energy in a circuit element.

electromagnetic radiation A wide range of frequencies (or wavelengths) that can be created by accelerating charges, which result in a rapidly changing magnetic field and electric field travelling out from the source.

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601

electromagnetic spectrum Family of electromagnetic radiations consisting of radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays. All electromagnetic radiations travel at 3.0 × 108 m s–1 in a vacuum.

electron current The flow of electrons around an electric circuit. Because electrons are negative charges, the flow is in the opposite direction to that of the electric current.

electronegativity The degree to which atoms attract electrons. Materials of different electronegativity are used in cells.

electrons The negative particles in the outer part of atoms. They can move from one object to another creating an electrostatic charge. When moving in a conductor they constitute an electric current.

electronvolt A small unit of energy. One electronvolt (1 eV) is the energy an electron would gain when accelerated across a potential difference of one volt in a vacuum; 1 eV = 1.6 × 10–19 J.

electrostatic charge An electric charge gained by an object as the result of the loss or gain of electrons from close contact with another object.

electrostatic force A force that acts between charged particles and can act over relatively large distances.

electrostatic induction The process of producing an electric charge on a conductor by placing it near an electrostatic charge and then earthing one end.

elementary charge The magnitude of the charge on an electron or proton: e = 1.6 × 10–19 C.

EMF or ElectroMotive Force The potential difference created by a device such as a battery, generator or other source of electric potential. Note that it is not a force—the name is misleading, hence only the abbreviation should be used.

EMF generator A device that produces an EMF by rotating coils relative to a magnetic field. Generators which produce AC current are often called alternators.

endoscope (fibrescope) A long, flexible tube of thin optic fibres with a telescope at one end and a lens at the other, which is inserted into the patient and used to produce images.

602

Glossary

energy An object possesses energy if it has the ability to do work. Energy takes many forms, for example gravitational potential energy, kinetic energy, light and heat.

enriched nuclear fuel Uranium ore in which the concentration of the fissile uranium-235 is increased.

equilibrium The condition whereby the sum of the forces acting on a body add to zero; the forces are balanced.

equinox The two points through the year when the Sun crosses the celestial equator (approximately March 21 (vernal equinox) and September 21 (autumnal equinox)) .

fast breeder reactor A nuclear power plant that uses plutonium as fuel.

fast neutron A high-speed neutron that is capable of initiating nuclear fission in a plutonium-239 nucleus.

fertile Nuclide that is capable of absorbing a neutron into its nucleus and transforming into a fissile nuclide.

fissile Capable of undergoing nuclear fission. fission (nuclear fission) When a nucleus splits into two or more pieces usually after bombardment by neutrons.

fission fragments Nuclides formed during nuclear fission; these are usually radioactive.

focal length The distance from a lens or mirror at which a parallel beam of light is brought to a focus.

frequency The number of vibrations or cycles that are completed per second or the number of complete waves that pass a given point per second, measured in hertz (Hz).

friction When two surfaces move over one another the force opposing the motion is called the force of friction. There are a range of frictional forces. Air resistance is also a type of frictional force.

fuel rods Long thin rods of enriched uranium used in a nuclear reactor.

fusion (nuclear fusion) A process taking place inside stars in which small nuclei are forced together to make larger nuclei. Energy is released in the process.

gamma ray (γ) High-energy electromagnetic radiation ejected from the nucleus of a radioactive nuclide.

Geiger counter A device for measuring radioactive emissions.

genetic Relating to genes found in DNA of body cells and the ways in which physical characteristics are passed from parent to offspring.

geometrical optics The branch of optics that models the path of light using straight rays.

gravitational field strength The size of the gravitational force acting on a 1 kg mass. Near the surface of Earth g = 9.8 N kg–1 and objects in free fall will accelerate at 9.8 m s–2.

half-life Time taken for half of the nuclei of a radioactive isotope to decay.

Hertzsprun–Russell (H-R) diagram A plot of the luminosity of stars against surface temperature and classifies stars by types.

hypertmetropia Long-sightedness or a difficulty in seeing near objects clearly.

impulse The product of an applied force (in newtons) and the time interval for which it is applied (in seconds). Impulse is equivalent to the change in momentum of the affected body.

in phase Two points within a wave that are simultaneously experiencing the same displacement and velocity, and are, therefore, moving in the same direction.

inertia The tendency of a body to maintain its state of motion as described by Newton in his first law.

infrasound Sound of very low frequencies below our hearing range (<20 Hz).

insulator A material which conducts an insignificant electric current when subjected to an electric field.

intensity (of a wave) The amount of energy passing through a square metre area each second.

internal resistance A battery will have a resistance to the current flowing through it which will result in a loss of energy inside the battery, and hence a loss of EMF at the terminals. Resistance increases as a battery goes flat.

inverse square law A mathematical pattern expressing a relationship in which one quantity decreases with the square of another; e.g. if the first doubles, the second decreases to one-quarter, and so on. See Coulomb’s law.

ionising ability A measure of how strongly radiation interacts with matter and causes molecules and atoms to become ions.

ionising radiation Any radiation that has enough energy to free electrons from atoms to produce ions. Ionising radiation, which includes alpha, beta and gamma radiation, as well as X-rays and gamma rays, is considered to be dangerous.

isolated system A group of bodies interacting with no influence from any other body nor input of energy.

isotopes Atoms with the same number of protons but with different numbers of neutrons.

I–V characteristic The relationship between the current through, and the voltage across, a circuit element. Often given as an I–V graph, or the resistance in the case of an ohmic conductor.

kinematics The study of moving objects without reference to mass or force.

kinetic energy The energy of a moving body, measured in joules.

Kirchhoff’s laws 1. The sum of all currents at a point is zero. 2. The total potential drops around a closed circuit equal the total EMF.

lift The lift force, lifting force or simply lift is a mechanical force generated by solid objects as they move through a fluid such as air.

load When work is done on an object by the action of a force or forces, the object is often referred to as the load. Load can also refer to the object being supported by another item or the object causing the extension or compression of another item.

longitudinal waves Waves that involve particles of the medium vibrating parallel to the direction of travel of energy.

luminosity The absolute brightness of a star measured in watts of total energy output.

magnification The factor by which the size of an image has changed from the size of the original.

Glossary

603

magnitude (of stars) Apparent magnitude is an arbitrary scale based on the brightest stars in the northern sky being magnitude +1. A change of –1 corresponds to a brightening by about 2.5 times. The absolute magnitude of a star corresponds to the apparent magnitude of the star if it was 10 pc from us.

mass The ability of a body to resist an acceleration when acted upon by an unbalanced force. Mass is an indication of the inertia of a body.

mass number The number of nucleons (protons and neutrons) in a nucleus.

mechanical energy The energy that a body possesses due to its position or motion. Kinetic energy, gravitational potential energy and elastic potential energy are all forms of mechanical energy.

mechanical wave Energy transferred from one location to another by the passing of the energy from one particle to the next within a substance.

medium The substance carrying a mechanical wave. model A system of some type that is well understood and which is used to build a mental picture or analogy for an observed phenomenon that is not well understood.

moderator A material, usually graphite or water, that slows neutrons in a nuclear reactor.

modulation To impress a (digital) signal onto a carrier frequency that has been produced for the purpose of carrying the signal.

momentum The product of the mass of an object and its velocity—a vector quantity measured in kg m s–1.

multimode fibre (step-index or graded-index) Optical fibre with a relatively large core made of uniform or graded-index glass.

myopia Short-sightedness or a difficulty in seeing distant objects clearly.

neutral No electric charge. neutrons Uncharged sub-atomic particles. newton SI unit of force. One newton (1 N) is defined as the force required to make a mass of 1 kg accelerate at 1 m s–2.

non-contact forces Forces that act at a distance and do not require the bodies to actually touch each other. Gravitational, magnetic and electric forces are noncontact forces.

non-ionising radiation Any radiation that does not have enough energy to ionise atoms. Radio waves, microwaves, visible light and UV-A radiation are non-ionising.

normal reaction force The force N or FN that a surface

exerts on a body, which acts perpendicular to the surface.

nuclear reactor A nuclear power plant that uses nuclear fuel to generate electricity.

nucleons Particles located in the nucleus of an atom. nucleus The central part of an atom. nuclide An atom as characterised by its atomic and mass number.

ohmic conductor/resistor A conductor that obeys Ohm’s law; i.e. has a constant resistance.

Ohm’s law For metals at a constant temperature, the current flowing is directly proportional to the voltage across them; i.e. their resistance is constant.

optical density A measure of the speed at which light is able to travel through a substance.

optical fibre Thin fibre of transparent material, encased in plastic and used to transmit light signals.

parallax angle (of stars) Half the change in angular position of a star as seen from one side of the Earth’s orbit to the other. Stellar parallax uses this angle to find the distance (see parsec).

parallel circuit A circuit in which the elements are connected such that the voltage across them all is the same and the current is split up between them. Compare series circuit.

parsec (pc) The distance to a star that has a parallax angle of one arc second; equal to 206 265 AU.

penetrating ability A measure of how well radiation passes through matter.

period The time interval for one vibration or cycle to be completed.

photovoltaic cell A semiconductor device that produces a low voltage current when light falls on it.

604

Glossary

piezoelectric effect The production of sound by a disc of special crystals in a transducer in response to a varying electric signal of particular frequency.

polarised light Light waves that have their electric fields aligned with one another; i.e. varying in the same plane as one another.

polarising filters Filters that are able to convert unpolarised light into polarised light.

position The location of an object with respect to a reference point. Position is a vector quantity.

potential difference (p.d.) The difference in electric potential between two points in a circuit. What is measured by a voltmeter when placed across a circuit. A battery creates the p.d. across a circuit which drives the current.

potential energy Energy that can be considered to be ‘stored’ within a body due to its position, composition or molecular arrangement.

power The rate at which work is done—a scalar quantity measured in watts (W).

propagation The motion of waves through or along a medium.

protons Positively charged sub-atomic particles. radiation Rays or particles that carry energy. radioactive substance A substance that spontaneously emits radiation in the form of alpha particles, beta particles and gamma rays.

radioisotope An isotope of an element that is radioactive.

radius of curvature The radius of the sphere from which a mirror has been cut.

rarefaction An area of lower pressure in the transmission of a sound wave.

ray tracing Using known pathways of light to model image formation on a scaled, two-dimensional diagram.

real image An image that is formed at the point of intersection of light from a common point of origin.

redshift (Doppler shift) The change (lowering) of frequency that occurs in any wave phenomenon as the source of the waves moves away from the observer. Likewise, a blueshift occurs if the source is moving towards the observer.

reflection The rebound of energy or particles incident on a surface.

reflection coefficient The proportion of the incident sound wave energy that is reflected at a boundary. It is determined by the difference in the acoustic impedance values for the two media.

refraction The bending of the direction of travel of light as it enters a medium of different optical density.

regular reflection Parallel rays of light incident on a plane mirror or a flat polished metal surface remain parallel to each other on reflection.

relative refractive index An index or number that is allocated to each pair of substances indicating their relative optical densities.

renewable energy An energy source that is largely inexhaustible and is constantly replenished by natural processes.

residual current device (RCD) A device used in a household circuit to cut off the circuit if the current in the neutral is not equal to the current in the active—a situation that arises if current goes to earth through a person.

resistance The ratio of the voltage across, to the current through, a circuit element. Measured in ohms (1 Ω = 1 V A–1). Resistance depends on the size, shape, temperature and material of a conductor.

resistor A circuit element which usually has a straight I–V characteristic; that is, a constant resistance.

retrograde motion The apparent backwards (westwards) motion of the planets in comparison to their normal eastward motion through the stars.

right ascension (RA) One of the two coordinates describing the position of a star. It is the angular distance from the vernal equinox.

scalar A physical quantity that can be represented by a magnitude only. Mass, time, speed and refractive index are scalar quantities.

semiconductors Materials such as silicon which normally have a moderately high resistance but which can be ‘doped’ with other elements to considerably increase their conductivity, by adding either mobile electrons (n-type) or ‘holes’ (p-type) to the lattice structure of the crystal.

Glossary

605

series circuit A circuit in which the elements are connected such that the same current flows through one after the other. Compare parallel circuit.

short circuit The situation in which a good conductor is inadvertently placed across a battery and an excessive current flows, which may cause damage.

sidereal time Time as measured from the rotation of the stars rather than the Sun (solar time); 24 hours of sidereal time (one full revolution of the celestial sphere) pass in approximately 23 hours 56 minutes of solar time.

single-mode fibre A fibre that is so thin it allows only one path on which light can travel.

solstice (summer or winter) The point when the Sun is at its highest (about December 21) or lowest (about June 21) elevation in our sky.

somatic Relating to the human body. spectroscopic parallax Misleading term describing the process of finding the distance to stars that are beyond the range of stellar parallax.

speed The ratio of distance travelled to time taken. Speed is a scalar quantity.

spherical mirror A mirror that has a specially shaped surface such that it could have formed part of the surface of a sphere.

superconductor A conductor with zero resistance. Once a current is established it does not die out. Only certain materials at very low temperatures are superconductors.

superposition When two or more waves travel in a medium, the resulting wave at any moment is the sum of the displacements associated with the individual waves.

sustainable energy An energy source that can be used in a way in which human needs are met without diminishing the ability of other people, wild species, or future generations to survive.

tension Force acting in a stretched material such as a rope or cable.

thermal neutron (slow neutron) A neutron released during fission that is travelling slowly enough to be absorbed into a uranium-235 nucleus.

thrust The force of flight that pushes a plane forward.

606

Glossary

torque The tendency of a force to cause an object to rotate equal to the force times the radius.

total internal reflection Light or waves are only reflected when striking a boundary between two media; there is no transmission.

transducer A small handheld probe used in diagnostic ultrasound to produce and detect sound waves.

transformer A device that uses the electromagnetic effects of a current to transform the voltage up or down from that supplied.

transmutation The changing of one element into another.

transverse waves Waves in which the direction of the vibration of the particles of the medium is at 90° to the direction of travel of the wave energy itself.

ultrasound Sound of frequencies beyond our hearing range (> 20 000 Hz).

unpolarised light Light waves that have electric field variations that are not aligned with one another.

Van de Graaff generator Machine that produces a large electrostatic charge on its dome by the transfer of charge on a rubber belt in contact with a plastic roller.

vector A variable that requires both a magnitude and a direction in order to be to fully defined. Velocity, acceleration and force are vector quantities.

velocity The ratio of displacement to time taken. Velocity is a vector quantity.

virtual image Light paths that originate from a common point on an object are diverged by a mirror or lens and appear to have come from a common image point.

visible light The range of electromagnetic radiation frequencies to which the cells in human eyes can respond.

volt The unit of electrical potential. One volt is equal to one joule of potential energy given to one coulomb of charge in a source of EMF. The voltage (or the number of volts) is another name for potential difference.

voltmeter A meter that displays the potential difference across part of a circuit. It must be connected in parallel with the appropriate circuit elements. Compare ammeter.

wave pulse A single disturbance passed through a medium.

wavelength The distance between successive points in a wave that have the same displacement and are moving in the same direction; that is, points that are in phase.

weight The gravitational force, W or Fg, acting on a

body, measured in newtons, N. Near the surface of the Earth, the acceleration due to gravity is approximately constant, so an object’s weight is roughly proportional to its mass.

wind tunnel A tunnel-like passage through which air is blown at a known velocity to investigate air flow around an object (as an airplane part or model) placed in the passage.

work Work is done on a body when a force acts on the body and the energy of the body is altered. One joule of work is done on an object when the application of a net force of one newton moves an object through a distance of one metre in the direction of the net force.

Glossary

607

INDEX A-scan (ultrasound device) 528–9 aberration 317

ionising ability 16

Becquerel, Henri 440

alpha radiation 8, 15–16, 17, 551

Bernoulli’s equation 472–3, 479

chromatic 317–18

alternating current (AC) 70, 94, 97

Bernoulli’s principle 472, 474

spherical 291, 318

beta decay 9

absolute magnitude 394–5

alternative energy sources investigation 498–512

absolute refractive index 263

altitude–azimuth mount 373

energy 16, 17

absorbed dose (radiation) 26–7

americium-241 18

industrial applications 18

absorption

ammeters 59

beta particles 15–16

ionising ability 16

of light 254–5

ampere 51

beta radiation 8, 15–16, 17, 551

of sound 514

amplitude 236, 518, 523

‘big bang’ theory 432–3, 434–5

acceleration 115–16

Andromeda Nebula 382

‘big crunch’ 435, 436

average 116, 125

angiogram 546

binary stars 392, 416

due to gravity 159–61

angle of incidence 252, 259

binding energy 463–4

uniform 130–1

angle of reflection 252

binoculars 271

acceleration–time graphs 126

angle or refraction 259

biomass fuels 504

accommodation (eye) 324

antineutrino 9

black body ‘spectrum’ 399, 411, 433

accuracy 568

aperture 320

black dwarfs 418

acoustic impedance 525–7

apparent brightness 395, 396, 398

black holes 429–30

action–reaction pairs 164–7

apparent magnitude 393–4, 395

blind spot 324

activity, radioisotopes 21–2

apparent weightlessness 476

Bohr, Niels 440

Adams, John Couch 379

Aristotle 135, 150, 191, 335, 357

Brahe, Tycho 401, 419

adaptive optics 376

asteroids 380–1, 382

adding vector quantities 561, 562

astronomy

astronomical observations 360–3 geocentric universe 362, 367

aerodynamic forces 471–80

galaxies 383, 424–5

brightness of stars 393–5

aerofoils 473, 475–6

nebulae 382–3

brown dwarfs 414

agricultural waste and bioenergy 503–4

scientific advances 382, 383–6

air bags 183

see also planets; stars; telescopes; universe

air resistance 136, 211

athletics, timing and false starts 126–7

aircraft

atmosphere

centre of gravity 482

refraction in 266

centre of mass 481

Sun 406

camera, simple 318 aperture (f-stop) 320 focusing 319 shutter speed 319–20 camera, SLR 320

drag 478–9

atomic charge 38

cancer detection, with radioactive tracers 30–1

equilibrium 482–3

atomic models 440–1

cancer treatment

flight forces 471–9

atomic number 3

modelling 481–4

atoms 3, 38, 440

flight investigation 486–91

attenuation 272

radiation for liver cancer 556–7 using ionising radiation 550, 556 carbon dioxide emissions 72–3

lift 471–3

ultrasound waves 524

cars, determining speed of 132–3

thrust 477–8

X-rays 543

celestial coordinates 344

torque 482, 483–4

average acceleration 116, 125

weight 476

average speed 114, 119

position of 343

average velocity 114–15, 130

see also north celestial pole; south celestial pole

aircraft engines 477–8 alpha decay 8 alpha particles 15, 18 energy 16, 17

B-scan (ultrasound device) 529–31

celestial sphere 340–3, 357, 387

bandwidth 272

cells 51, 53, 54

batteries see cells

608

Index

celestial poles

chemical 54, 89–90

electrochemical 513–14

conservation of energy 206, 211–13, 493

destructive interference 244

internal resistance 90–1 photovoltaic 53, 93–4

conservation of momentum 187–8

diagnostic imaging

rechargeable 90

constructive interference 243, 244

with PET 555–6

in series and parallel 90–1

contact forces 143

with radioisotopes 552–4

solar 93–4

continuous spectrum of X-rays 540

diagnostic ultrasound 516, 528–33

continuous waves 227, 232, 236

diffraction 245, 287

centre of gravity 482 centre of mass 111

superposition 243–4

deuterium 465, 466

diffraction gratings 409–10

Cepheid variables 423–4

contrast (X-rays) 543, 545–6

diffuse reflection 252–3

Ceres 380

control rods (nuclear reactors) 455

diodes 60

Chadwick, James 440, 442

conventional current 52

dioptric power of a lens 326

chain reactions 445, 448

conversion factors 567–8

direct current (DC) 70, 94, 97

charge concentration 53

convex lenses 306, 308–10

dispersion 280–2

charge-coupled device (CCD) 385–6

convex mirrors 291, 299–303

displacement 112–13, 124

chemical cells 54, 89–90

Copernican heliocentric universe 358–60, 367

displacement–distance graphs 232–4

Chernobyl nuclear accident 439, 460 chromatic aberration 317–18 chromosphere 406 circuit breakers 98 circuit device symbols 78 circuit elements 54, 78 circumpolar stars 343 closed fuel cycle 459

corona 406 coronal mass ejection 407 cosmic microwave background radiation 433 coulomb 40 Coulomb, Charles 48 Coulomb’s law 45–6

closed systems, energy in 493

critical angle, and total internal reflection 270–1

collisions

displacement–time graphs 236 distance travelled 112 Doppler effect 427, 532–3 Doppler imaging 533 Doppler shift 426 dose equivalent (radiation) 27–8 double insulation 99 drag (aircraft) 478–9 drag coefficient 479

critical mass 449–50

dwarf planets 379

forces during 182–4

CT scans 546–7

dynamics 143

and law of conservation of momentum 187–8

Curie, Marie 8

and pedestrians 189 colour addition 277–8 colour television 278 coloured light, and wavelengths 276–7 colours of stars 336, 397–8 complementary colours 278

current see electric current current–voltage graphs 59–60 curved mirrors

Earth circumference 357 modern creation story 335–6

applications 291–3

Earth-centred universe 357–8

see also concave mirrors; convex mirrors

earth leakage system 99

cyclists, and air resistance 211

compound microscope 323

earth wire 98–9 eclipses 253–4 lunar 253, 353, 354, 357

compression 517

‘dark energy’ 436

computerised tomography scans see CT scans

dark matter 428

ecliptic 348, 349, 351, 354, 355

dark side of the Moon 352–3

effective dose (radiation) 29

data 568

effective resistance 81

concave lenses 306, 312–14, 315 concave mirrors 291–2, 296–7 graphical ray tracing 293 information formation 292–3 ray diagrams 294–6 summary of images formed 296 conductors 41–2 cones 324 conservation of charge 37, 38, 78

accuracy and precision 568–9

solar 253, 254, 353, 354

Einstein, Albert 444

graphical analysis 571–5

general theory of relativity 224

significant figures 571

mass and energy 404, 439, 443–4, 463, 464

uncertainty in a result 569–71 daughter nucleus 8 decay series 22 decibel scale 523 derived units 564, 565

special theory of relativity 158, 403, 432 elastic medium 226 elastic potential energy 203 calculating 204–5

Index

609

and elastic materials 203

in closed systems 493

and ideal springs obeying Hooke’s law 203–4

conservation of 206, 211–14

force(s) 144 action–reaction pairs 164–7

heat 206–7

aerodynamic 471–80

electric blankets, power settings 87

kinetic 198–200, 494

on an inclined plane 169–71

electric charge 38

and mass 404, 439, 443–4, 463, 464

constant, and work 192–3

electric circuits 54, 78

mechanical 198–207

contact 143

flowing water analogy 56–7

potential 53, 67, 201–5, 494

during collisions 182–4, 187–8

parallel circuits 80, 84–6

and power 214–15, 494, 495

electrical 45–6, 48

and potential difference 54–5

radiation 16–17

electrostatic 12, 440, 441, 464

rules 78–80

review 494–5

in equilibrium 152–3

series circuits 80–1

and waves 225–6

frictional 173, 193–4

electric current 51, 57, 59

energy consumption, electrical appliances 71–2

and impulse 181

direction 52–3 see also alternating current; direct current

energy transformations 209–10, 225, 493–4

and motion 143, 144–8, 150–3, 156–61, 164–73

and momentum 179–80, 187–9

electric eels 93

and conservation of energy 211–14

non-contact 143

electric fields 38, 43–4, 46–8, 55

efficiency 210, 495–6

normal reaction 67–9

electric potential, and EMF 54

enriched uranium 454

strong nuclear 12, 441, 464

electric power 67–9, 495

entropy 496–7

tension 171–2

and greenhouse gases 72–3

equations of motion 130–3

upward, and work 195

production and transmission 70–1

equatorial mount 373–4

as vector quantity 144–8

electric shock 99–100, 101

equilibrium (aircraft) 482–3

electrical appliances, energy consumption 71–2

equilibrium, forces in 152–3

energy values 504

equinoxes 347, 350–1

electrical energy 53, 67

and greenhouse gases 499

Eratosthenes 357

Franklin, Benjamin 37, 38, 41

ethanol, as fuel 503

electrical force 45–6, 48

frequency 235, 275, 517–18

Euclid 255

electrical safety 98–9

frictional forces 173

experimental error 570

electrochemical cells 513–14

eye, human 323–4

electromagnetic radiation 274–5, 538

fuel cells 511–12

eye defects, correcting 325

fuel rods 454, 458

units of 67–9

electromagnetic spectrum 26, 274, 275, 384, 537, 538 electromagnetic waves 273–6, 383, 537 electron current 52 electron degenerate matter 419 electronegativity 89 electrons 3, 38, 39, 440 mass and size 40 electronvolt 16, 541 electrostatic charge 38–9, 49 electrostatic forces 12, 440, 441, 464 electrostatic induction 41 elementary charge 39, 40, 42–3 EMF (electromotive force) 53 and electric potential 54 sources of 89, 90, 92–3 endoscopes 534 energy 191, 198

610

Index

fossil fuels

and work 193–4

fuses 98 f-stop 320 far side of the Moon 352

galaxies 383, 424–5, 429, 431

fast breeder reactors 456–7

distance of 426–7

fast neutrons 448, 457

early 420

Fermi, Enrico 442, 445

total mass 427–8

fibrescopes 534–5

types of 424

film badges 13

Galilean telescope 369–70, 371

fissile nuclides 442

Galileo 135–6, 150, 151, 154, 191, 367

fission reactions see nuclear fission

telescope observations 364–7

fixed speed cameras 133

gamma decay 9–10

flight forces 471–9

gamma radiation 8, 9, 16–17, 551

modelling 481–4

gamma rays 16, 274, 276

flight investigation 486–91

energy 16, 17

focal length 292, 307, 326

ionising ability 16

focal point 291 focusing 319 force–displacement graphs 195–6

for radiotherapy 551 Geiger counter 12

generators 53

hydrogen isotopes 464

kilowatt hours 68, 495

genetic effects, ionising radiation 29–30

hypermetropia 325

kinematics 143

geocentric universe 362

I–V graphs 59–60

geometrical optics 287 glossary 576–81

kinetic energy 198–200, 494 Kirchhoff’s laws 79, 80

parallel circuit elements 84–5 ideal springs obeying Hooke’s law 203–6

laser light 535

graphical analysis of data 571–5 graphical ray tracing see ray tracing

image distance 294, 308

graphing motion 122–7

images

law of conservation of energy 206, 211–13, 493

acceleration 126

in multiple mirrors 289

position 122–3

in a plane mirror 288–9

velocity 123–5

impulse 181

laser speed guns 132

law of conservation of momentum 187–8 law of refraction 262–3

gravitational field strength 160–1

and momentum 181–2

Le Verrier, Urbain 379

gravitational potential energy 201–3, 494

and work 196

length, standards 563–4

incident ray 259

lens equation 314

gravity, vertical motion under 136–8

inclined plane 169–71

lenses

gravity waves 224

index of refraction 263–5

concave 306, 312–14

greenhouse gases 72–3, 499

inertia 150–1, 156, 158

convex 306, 308–19

inferior mirages 267

power of 326

hailstones 44

infrared radiation 276, 383

spherical 306–7

half-life 20–1

infrasound 516

leptons 6

half-value thickness (HVT) 543–4

instantaneous speed 113, 119

lift (aircraft) 471–3, 474

hard X-rays 540

instantaneous velocity 113, 114

light

heat death 496

insulators 42

absorption 254–5

heat energy 206–7

intensity (sound waves) 523, 526

coloured 276–8

heating effects of ultrasound 533

interference 245

dispersion 280–2

heliocentric universe 358–60, 362

constructive 243, 244

Herschel, William 379, 383

destructive 244

Hertzsprung–Russell diagram 413–14

particle/ray model 250–2, 256, 259, 260, 261 polarisation 282–3

interpreting 417–19

intermediate-level radioactive waste 458

placing stars on 414–15

internal resistance 91

refraction 258–63

high-grade energy 493

intrinsic brightness (stars) 394, 395

transmission 254, 255

high-level radioactive waste 458

inverse square laws 45

visible 274, 276, 277

Hipparchus 358

ionising ability, radiation types 16

Hooke’s law 204–6

ionising radiation 26–30, 550, 556

wave model 225, 246, 250–1, 254, 259–60, 282–3, 287

house wiring 97–8

isotopes 3–4, 11

household electricity 97–101

reflection 251–2, 254, 255

light rays see rays of light light-year 392–3 lightning 43–4

Hoyle, Fred 432

jet engines 477

Hubble, Edwin 423, 424, 425–6

Joule, James 206–7

Hubble Space Telescope 376, 385, 386, 409, 429

Jupiter, moons around 366

Hubble’s constant 426, 434 human eye, structure 323–4

Kepler, Johannes 362, 363–4, 381, 401, 419, 431

Huygen’s principle 237–8

Keplerian telescope 370

longitudinal waves 228, 230, 516–17

Huygen’s wavelets, and refraction 265

Kepler’s first law 363

low-grade energy 493

hydroelectric power 507

Kepler’s second law 363

low-level radioactive waste 458

hydrogen 3

Kepler’s third law 364

luminosity (stars) 395–6, 398–9

hydrogen fuel cells 511–12

kilogram 563

lunar eclipses 253, 353, 354, 357

‘lightning rod’ 37, 41 line spectra 540 linear relationships 571–3 liver cancer, radiation treatment 556–7 long-sightedness 325

Index

611

phases 351–2 magnetic field, Sun 407

motion

magnetic resonance imaging (MRI) 548–9

acceleration 115–16

magnification 295–6, 308, 314

and centre of mass 111

of a telescope 369–70

acceleration–time graphs 126 and changes in velocity 116–17

main sequence stars 413, 417, 418

displacement 112–13

mass

equations of 130–3

of a body 158

explanation of 166–7

and energy 404, 439, 443–4, 463, 464 standard 563

and forces 143, 144–8, 150–3, 156–61, 164–73

of stars 416–17

graphing 122–7

Newton’s second law of motion 156–61, 179, 199, 476 Newton’s third law of motion 164–72, 187, 477 night sky, viewing the 337, 339–40 non-contact forces 143 non-ionising radiation 26 non-linear data 573–5 non-ohmic conductors 62 normal reaction force 167–9 north celestial pole (NCP) 338, 342, 343 nuclear equations 8, 9

mass number 3, 9

and mechanical energy 198–206

mass–luminosity relationship (stars) 417

Newton’s laws 150–3, 156–61, 164–72, 179, 187, 199, 471, 476, 477

mechanical energy 198–207 mechanical waves 226–7, 241, 516–17

position and distance travelled 111–12

medical diagnostics

position–time graphs 122–3

nuclear fission reactors 453–7, 506

PET 555–6

speed and velocity 113–15, 118–19

nuclear fission weapons 447, 449–51

radioisotopes 552–4

of the stars 338, 340–3

nuclear force 12, 440, 441, 464

ultrasound 516, 528–33

straight line 111–19

nuclear fuel 449–50, 454

X-rays 546

theories of 135–6

nuclear fuel rods 454, 458

medical radioisotopes 551–5, 556

velocity–time graphs 123–5, 130

nuclear fusion 404, 405, 464–5

medical therapies

vertical 136–8

nuclear fusion reactors 465–6

nuclear explosions 447, 449, 450–1 nuclear fission 442, 445 energy released 443–4 release of neutrons during 442–3

radiotherapy 550–1

MRI scans 548–9

nuclear medicine 552

ultrasound 533, 534

multimode fibres 272, 273

nuclear power 506

X-rays 550–1

multiple mirrors, images in 288

nuclear power plants 453, 457

medium 226, 241

mutations 30

nuclear reactors 453–7, 465–6

micro hydro installations 507

myopia 325

accidents 439, 460 nuclear transmutations 8

microscope, compound 323 microwaves 276

n-type semiconductors 94

Milky Way galaxy 383, 424–5

nebulae 382–3, 423, 428

Millikan, Robert 42–3

negatively charged 39

nuclear weapons 447, 449–51

mirages 267

Neptune 379

nucleons 440

mirror formula 304

net force 145

nucleus 440–1

mirrors

neutral atom 3

concave 291–7

neutrinos 12

convex 291, 300–3

neutron stars 420, 421

moderators (nuclear reactors) 454–5 modulation 272

neutrons 3, 442 release during fission 442–3

nuclear waste 443, 506 disposal 457–9

forces within 12, 440–1 nuclides 4, 442 object distance 294, 308 Occam’s razor 359

newton 144, 156

ocean energy 508

conservation of 187–9

Newton, Isaac 142, 191

ohmic conductors 60, 61

and impulse 181–2

Newtonian reflector telescope 372–3

ohmic resistors, power in 86–7

Newton’s first law of motion 150–3, 471

Ohm’s law 60

Newton’s gravitation law 402

one-dimensional waves 228

Newton’s law of cooling 509

opaque materials 254

momentum 179–80

Moon 351 dark side 352–3 eclipses 253, 353, 354 mountains on 365

Olbers’s paradox 431

open fuel cycle 458, 459

612

Index

optical centre 306

polarisation 282–3

non-ionising 26

optical density, and speed of light 263

polarised light 282

penetrating ability 15, 16, 540, 543–4

optical fibres 272–3, 534–5

polarising filters 282–3

see also X-rays

optical instruments 321–3

Polaroid sunglasses 283

radiation exposure

optical systems 317–26

position 111–12

absorbed dose 26–7

position–time graphs 122–3

dose equivalent 27–8

p-type semiconductors 94

positively charged 39

effective dose 29

parabolic mirrors 292

positron emission tomography (PET) scans 555–6

radiation shield 456

potential difference 54–5

radio telescopes 384–5

potential energy 201, 494

radio waves 274, 276

parallax 303, 361, 391, 392–3 parallel circuit elements, I–V graph 84–5

radiative diffusion 405

parallel circuits 80, 84–6

elastic 203–5

radioactive fission fragments 443

parent nucleus 8

electrical 53, 67

radioactive nuclei, instability 11, 12

parsec 392, 393

gravitational 201–3, 494

radioactive tracers 30–1

particle/ray model of light

power

radioactive waste 443, 457–9

reflection 250–2, 256

electric 67–9

radiocarbon dating 23–4

refraction 259, 260, 261

and energy 214–15, 494, 495

radioisotopes 4, 11

penetrating ability radiation 15, 16 X-rays 540, 543–4

in ohmic resistors 86–7

activity 21–2

and work 214–15

artificial 4–5

precession of the equinoxes 350–1

choice of for diagnostic imaging 554

percentage uncertainty 568–9

precision 568–9

decay series 22

period 235, 518

prefixes 567

diagnostic imaging with 552–4

periodic table 5

pressurised water reactors 455–6

half-life 20–1

periodic waves 227

primary colours 277–8

in medicine 551–5

periscopes 271

principal axis (lens) 306

PET scans 555–6

principal focus 307

radiopharmaceuticals 553, 554, 555

photocopiers 49

prisms, dispersion of white light 280–1

radiotherapy 550–1

photographic film (X-rays) 544–5

propellers (aircraft) 477, 483

radius of curvature 292

photosphere 406

proper motion 392

rainbows 280–1

photovoltaic cells 53, 93–4

protons 3, 38, 39, 440

rarefaction 517

piezoelectric effect 519, 520–1

mass and size 40

therapeutic treatments 556

ray diagrams 251–2, 256, 261, 294–6, 371

piezoelectric materials 520

protostars 418–19

piezoelectric transducers 520–1 pinhole camera 256

Ptolemy’s system of planetary motion 358, 359, 360

plane mirrors

pulsars 421

concave mirrors 293

Pythagoras 336

convex lenses 308–10

images in 288–9 reflection 252 planetary motion 355, 358, 363–4, 381 planetary nebulae 420

quasars 429

planets 337–8 distance to 401 elliptical orbits 363 search for 379–81 Pluto 379 plutonium-239 443, 448, 456, 457 point-to-point cameras 133

concave lenses 312–14

convex mirrors 300–3 quarks 5

average distance from the Sun 360

ray tracing 256, 260, 288, 289

spherical lenses 307 rays of light 250–2, 256, 273 real image 289

radiation 26 alpha, beta and gamma 8, 9, 15–16, 17, 551 detection 12–13 effects of 29–30 energy 16–17 ionising 26–30, 550, 556 ionising abilities 16

rechargeable batteries 90 red giants 148, 414, 419 redshift 426, 429 reflecting telescopes 321–2 Newtonian 372–3 reflection diffuse 252–3

Index

613

light 251–2, 254, 255

short circuit 91

laboratory measurement 118–19

modelling 251–2

short-sightedness 325

of waves 234

plane mirror 252

shutter speed 319–20

speed camera radar 132

regular 252, 253

SI units 563–4

speed of light, and optical density 263

sound waves 526–7

sidereal month 352, 353

speed measuring devices 132–3

ultrasound waves 516, 524

significant figures 571

spherical aberration 291, 318

waves 241–2, 244

single-mode fibres 272

spherical lenses 306

reflection coefficients 526

skydivers 162

ray paths 307

refracted ray 259

slinky springs 228, 230

terminology 306–7

refracting telescopes 321, 369, 371–2

slow neutrons 447, 453

spherical mirrors 292–3

refraction 258–62

SLR camera 320

sports shoe design 182

apparent depth 260–1

smoke detectors 18

stable isotopes 4, 11

in the atmosphere 266

Snell’s law 262, 264, 270

standard units of measurement 563–4

bent straw 260

soft X-rays 540

star clusters 418

and Huygen’s wavelets 265

solar cells 53, 93–4

star time 344–5

index of 263–5

solar eclipses 253, 254, 353, 354

starlight, analysis 409–12

law of 262–3

solar electricity 95

stars 337, 338

and ray/particle approach 259, 260, 261

solar energy (heat) 508–9

brightness 393–5, 396, 398

solar flares 407

celestial coordinates 344

and wave approach 259–60

solar power 510–11

circumpolar 343

solar tower 509

colours and temperature 336, 397–8

solar wind 406, 407

distance to 391–3

regular reflection 252, 253

somatic effects, ionising radiation 29, 30

Hertzsprung–Russell diagram 413–15, 417–19

relativistic mass 158

sound energy, measuring 523

luminosity 395–6, 398–9

renewable energy sources 498

sound waves 229, 516

mass of 416–17

refractive index absolute 263–4 relative 262–3

reprocessed nuclear fuel 459

acoustic impedance 525–7

mass–luminosity relationship 417

residual current device (RCD) 99

attenuation 524

motion of 338, 340–3

resistance 60, 62–3, 64

detected by piezoelectric transducers 520

sizes of 398–9

resistivity 63 resistors 60

intensity 523, 526

spectrum 409–12

as longitudinal mechanical waves 516–17

types of 413–14

in parallel 85–6 in series 81–2 resolution (telescopes) 374–5 retina 324 right ascension (RA) 344 rock dating 22, 24 rods 324

produced by piezoelectric transducers 521 properties 517–19 reflection 526–7 see also ultrasound waves

spectral classification 411–12

static equilibrium 483 ‘steady state’ theory 432–3 Stefan–Boltzmann law 399, 412 stellar evolution 417–21 stellar parallax 392, 412 straight line motion 111–19

Roentgen, Wilhelm 537

south celestial pole (SCP) 338, 341, 342, 343

rotational equilibrium 482, 483

spacetime, explosion of 432

subatomic particles 6

Rutherford, Ernest 8, 440

spectrographs 409

subcritical mass 450

spectrum

summer solstice 347

scalars 113, 560

stars 409–10, 411–12

scientific notation 566

Sun 410, 411

seismic waves 228

speed 113, 240

strong nuclear force 12, 441, 464

Sun 347, 401–2 atmosphere 406 eclipses 253, 254, 353, 354

semiconductor diodes 94

average 114, 119

magnetic field 407

semiconductors 42

of cars 132–3

mass of 402, 416

series circuits 80–2

instantaneous 113, 119

614

Index

modelling of processes occurring in 404–5

thrust (aircraft) 477–8

unit symbols, correct use 565–6

thunderclouds 43–4

units and symbols 565

movement along the ecliptic 349–50

time, standards 564

universe

nuclear fusion 464–5

Titius–Bode Law 380, 381

age of 434–5

physical properties 402

tokamaks 466

Aristotle’s view 357

source of energy 403–4

torque (aircraft) 482, 483–4

‘big bang’ theory 432–4

spectrum 411

total internal reflection 270–1, 272, 273

‘big crunch’ 435, 436

through the year 348–9

towering 267

Brahe’s model 362

transducers 519, 520, 528, 529

Copernican model 358–60

SunRace (solar car challenge) 510–11 sunspots 365, 407

piezoelectric 520–1

early models 357–8

superconductors 78

transformers 70–1

Earth-centred 357–8

supercritical mass 450

translational equilibrium 482, 483

expanding 431–6

supergiants 414, 419

translucent materials 254

future of 435–6, 496

supernovae 419, 421, 465

transmission

geocentric 362

superposition 243–4

light 254, 255

heliocentric 358–60

sustainable energy sources 498

ultrasound waves 524

Ptolemy’s view 358

investigation 499–502 starting points 503–12 switches 98

transmutation

‘steady state’ theory 432–3

artificial 4–5

unpolarised light 282

nuclear 8

upper gastrointestinal tract, X-ray 546

synchronous rotation 353

transparent materials 254

upward force and work 195

synchrotrons 6

transuranic elements 445

uranium decay series 22

synodic months 352

transverse waves 228, 232, 516

uranium-235 449

tritium 465

chain reaction 448

technetium-99m 13, 554

tsunamis 224, 226

fission reaction 443, 444, 447

telescopes

two-dimensional waves 229

Galilean 369–70, 371 Galileo’s observations 365–7

ultrasound 516

Keplerian 370

medical diagnostic 516, 528–33

light-gathering power 371

producing 519–20

magnification 369–70

ultrasound devices

mounts 372–3

A-scan 528–9

new techniques 382

B-scan 529–31

Newtonian reflector 372–3

Doppler imaging 533

radio telescopes 384–5

phase scanning 531–2

reflecting 321–2

real time images 531

refracting 321, 369, 371–2

three-dimensional images 532

resolution 374–6

ultrasound therapies

very big 377

heat treatments 533

tellurium-131 552

to destroy cells 534

tension 171–2

ultrasound waves

terminal velocity 162, 479

attenuation 524

thermal neutrons 447, 453

reflection and transmission 516, 524

thermal nuclear reactors 453–6

ultraviolet radiation 276

thermocouple 92

uncertainty 568–9

thermoluminescent dosimeters (TLDs) 13 thermopiles 93 three-dimensional waves 229

estimating in a result 570–1 uncontrolled chain reaction 448, 455 uniform acceleration, equations for 130–1

as nuclear fuel 454 uranium-238 3, 8, 448, 449, 456, 457 Uranus 379 Van de Graaff generator 39 vector components 147–8 vector quantities 113, 144–8, 484, 560 adding 561, 562 multiplying by a scalar 561–2 vehicle safety design 183–4 velocity 113, 122 average 114–15 changes in 116–17 instantaneous 113, 114 terminal 162 velocity–time graphs 123–4, 130 Venus, phases of 366–7 vertical motion, analysing 136–8 virtual focus 207 virtual images 28, 303 virtual rays 288 visible light 274, 276, 277 visible spectrum 276–7 volt 54 voltage 54

Index

615

voltmeters 58

wind power 505 wind tunnels, model 486, 490–1

water waves 228, 229, 237–8

wings (aircraft)

watt 67

lift 473, 474

wave equation 240–1

testing 475–6

wave model of light 225, 246, 287 dispersion 280–2

winter solstice 347 work 191–2, 494

polarisation 282–3

done by a constant force 192–3


and friction 193–4

refraction 259–60 wave pulses 227

from force–displacement graphs 195–6

wavelengths 236–7, 275, 518

and impulse 196

and coloured light 276–7

and power 214–15 and upward force 195

waves amplitude 236, 518, 523

world record speeds 117–18

continuous 227, 232, 236 diffraction 245, 287

X-ray beams

displacement–distance graphs 232–4

filtering 542

displacement–time graphs 236

heterogeneous and homogeneous 542–3

electromagnetic 273–6, 383, 537 and energy 225–6

half-value thickness 543–4

X-ray image

frequency 235, 275, 517–18

beam cross-sectional area 544

gravity 224

contrast enhancement 545–6

Huygen’s principle 237–8

diagnostic procedures 546

interference effects 243–4, 245

film cassettes 544–5

longitudinal 228, 230, 516–17

X-ray photon energy 541

mechanical 226–7

X-ray radiotherapy 550–1

meeting barriers 241–2

X-rays 276

one-, two-, and three-dimensional 228–9

attenuation and contrast 543

period 235, 518

for CT scans 546–7

periodic 227

discovery 537


as electromagnetic waves 537

seismic 229

energy of 540–3

sound 229, 516, 517–19, 524–7

penetrating ability 540, 543–4

speed of 233–4

production 539–40

superposition 243–4 transverse 228, 232, 516 ultrasound 516, 524 water 228, 229, 237–8 wavelength 236–7, 275, 276–7, 518 weight 159–60, 476 white dwarfs 414, 418 white light 277 dispersion 280–1

616

Index

continuous and line spectra 540

ePhysics 11

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