Space Mechanics for Engineers

Space Mechanics for Engineers

2nd Edition

Space Mechanics for Engineers

2nd Edition

Raed I. Kafafy Department of Mechanical Engineering, Engineering, International Islamic University University Malaysia, Malaysia

Ossama O. Abdelkhalik Department of Mechanical Engineering Engineering - Engineering Mechanics, Michigan Tech University, USA

Copyright ©2013, 2009 by Raed I. Kafafy and Ossama O. Abdelkhalik.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without any prior written permission of the authors.

Perpustakaan Negara Malaysia Cataloguing-in-Publication Data Kafafy. Raed I., 1973- and Abdelkhalik, Ossama O., 1973Space Mechanics for Engineers ISBN 978-983-2599-98-2

1. Space Mechanics – Orbital Mechanics – Astrodynamics. I. Title.

Printed and bound in Malaysia by Kaci Trading Sdn Bhd.

1.1

What is Space Mechanics? ................................................................................................... 1

1.2

Historical Background ......................................................................................................... 2

1.2.1 Ancient Astronomy ................................................................................................................................ 2 1.2.2 Islamic Astronomy ................................................................................................................................. 3 1.2.3 Medieval European Astronomy ............................................................................................................. 5 1.2.4 Astrodynamics ....................................................................................................................................... 7

1.3

Our Celestial Residence ....................................................................................................... 7

1.3.1 What is Space?....................................................................................................................................... 7 1.3.2 Earth’s Atmosphere ............................................................................................................................... 7 1.3.3 Space Thermal Environment .................................................................................................................. 9 1.3.4 Microgravity ........................................................................................................................................ 10 1.3.5 Near Earth Radiation Environment ...................................................................................................... 11 1.3.6 Solar Activity and Emissions ............................................................................................................... 11 1.3.7 Galactic Cosmic Rays .......................................................................................................................... 13 1.3.8 The Magnetosphere.............................................................................................................................. 13 1.3.1 The Van Allen Belts ............................................................................................................................. 15 1.3.2 Environmental Impact on Spacecraft Design....................................................................................... 15 1.3.3 Spacecraft Electrostatic Charging ........................................................................................................ 16 1.3.4 Metroids and Micrometroids................................................................................................................ 17 1.3.5 Space Debris ........................................................................................................................................ 17 1.3.6 The Solar System ................................................................................................................................. 18

References ...................................................................................................................................... 18

2.1

Background ......................................................................................................................... 21

2.1.1 Particle Kinematics .............................................................................................................................. 21 2.1.2 Particle Dynamics ................................................................................................................................ 23

2.2

The Two-Body Problem ..................................................................................................... 24

2.2.1 Problem Description ............................................................................................................................ 24

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2.2.2 Problem Formulation ........................................................................................................................... 24 2.2.3 Equation of Motion .............................................................................................................................. 25 2.2.4 Constants of Motion ............................................................................................................................. 26 2.2.5 Trajectory Equation ............................................................................................................................. 28 2.2.6 Orbital Elements .................................................................................................................................. 30 2.2.1 Orbit Equations .................................................................................................................................... 31 2.2.2 Characteristics of the Two-Body Problem ........................................................................................... 32

2.3

Keplerian Orbits ................................................................................................................. 32

2.3.1 Circular Orbits ..................................................................................................................................... 32 2.3.2 Elliptical Orbits .................................................................................................................................... 33 2.3.3 Parabolic Orbits ................................................................................................................................... 35 2.3.4 Hyperbolic Orbits................................................................................................................................. 36 2.3.5 Summary of Keplerian Orbits .............................................................................................................. 39

References ...................................................................................................................................... 39 Problems ......................................................................................................................................... 40

3.1

Introduction......................................................................................................................... 43

3.2

Classical Coordinate Systems ............................................................................................ 43

3.2.1 Geocentric-Equatorial Coordinate System........................................................................................... 43 3.2.2 Heliocentric-ecliptic Coordinate System ............................................................................................. 44 3.2.3 Right Ascension-declination Coordinate System................................................................................. 44

3.3

Classical Orbital Elements ................................................................................................. 45

3.3.1 Classical Orbital Elements ................................................................................................................... 45 3.3.2 Alternative Orbital Elements ............................................................................................................... 46 3.3.3 Ephemerides......................................................................................................................................... 47

3.4

Perifocal Coordinate System ............................................................................................. 47

3.5

Coordinate Transformations ............................................................................................. 48

3.5.1 Transformation from orbital elements to r and v ................................................................................. 48 3.5.2 Transformation from r and v to orbital elements ................................................................................. 49


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4.1

Solar Time ........................................................................................................................... 56

4.1.1 Julian Calendar..................................................................................................................................... 56 4.1.2 Gregorian Calendar .............................................................................................................................. 57

4.2

Terrestrial time ................................................................................................................... 57

4.3

International Atomic Time ................................................................................................ 57

4.4

Sidereal time........................................................................................................................58

4.4.1 Universal Time..................................................................................................................................... 59 4.4.2 Coordinated Universal Time (UTC) .................................................................................................... 60 4.4.3 Global Positioning System Time ......................................................................................................... 60

4.5

Summary of Time Systems ................................................................................................ 61


5.1

Initial Orbit Determination ............................................................................................... 64

5.2

Two Position Vectors (Lambert’s Problem)..................................................................... 64

5.2.1 Minimum Energy Solution................................................................................................................... 66

5.3

Three Position Vectors (Gibbs Method) ........................................................................... 69


6.1

Atmospheric Drag...............................................................................................................77

6.2

Non-symmetric Earth’s Gravitational Field .................................................................... 78

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6.2.1 The J2 Perturbation ............................................................................................................................... 78 6.2.2 Regression of Line of Nodes................................................................................................................ 79 6.2.3 Precession of line of Apsides ............................................................................................................... 79 6.2.4 Triaxiality............................................................................................................................................. 80

6.3

Solar and Lunar Gravitation ............................................................................................. 80

6.4

Solar Radiation Pressure ................................................................................................... 80

6.5

Modeling Perturbations ..................................................................................................... 82

6.5.1 Special Perturbation Techniques:......................................................................................................... 82 6.5.2 General Perturbation Techniques: ........................................................................................................ 82


7.1

Spacecraft Coverage ........................................................................................................... 85

7.1.1 Ground Station Visibility ..................................................................................................................... 85 7.1.2 Elevation and Azimuth of Satellite ...................................................................................................... 86

7.2

Ground Track ..................................................................................................................... 87

7.2.1 Ground Track on a Non-rotating Earth ................................................................................................ 87 7.2.2 Effect of Earth Rotation on Ground Track........................................................................................... 87

7.3

Effect of Launch on Orbit Inclination .............................................................................. 88

7.4

Special Earth Orbits ........................................................................................................... 88

7.4.1 Polar Orbits .......................................................................................................................................... 89 7.4.2 Earth-Synchronous Orbits .................................................................................................................... 90 7.4.3 Geosynchronous Orbits ........................................................................................................................ 90 7.4.4 Geostationary Orbits ............................................................................................................................ 90 7.4.5 Molniya Orbit....................................................................................................................................... 91 7.4.1 Sun-Synchronous Orbit ........................................................................................................................ 91


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8.1

Basics of Orbital Maneuver ............................................................................................... 95

8.1.1 Orbital Energy...................................................................................................................................... 95

8.2

Principles of Orbital Maneuvers ....................................................................................... 96

8.2.1 Satellite Launch ................................................................................................................................... 96 8.2.2 The Delta–V Budget ............................................................................................................................ 96

8.3

Coplanar Maneuvers .......................................................................................................... 97

8.3.1 Tangential-Orbit Maneuver.................................................................................................................. 98 8.3.2 Non-Tangential Coplanar Maneuver ................................................................................................... 98 8.3.3 Hohmann Transfer ............................................................................................................................... 98 8.3.4 Bi-elliptic Transfer............................................................................................................................. 102 8.3.5 General Coplanar Transfer between Circular Orbits ......................................................................... 105 8.3.6 Phasing Maneuver.............................................................................................................................. 106

8.4

Out-of-Plane Orbit Maneuvers ..................................................................................... 107

8.4.1 Simple Plane Change ......................................................................................................................... 107 8.4.2 General Plane Change Maneuver ....................................................................................................... 108 8.4.3 Combined Maneuver.......................................................................................................................... 110

References .................................................................................................................................... 112 Problems ....................................................................................................................................... 113

9.1

Trajectories ....................................................................................................................... 117

9.2

The Solar System .............................................................................................................. 118

9.2.1 Planets ................................................................................................................................................ 118 9.2.2 Asteroids ............................................................................................................................................ 119 9.2.3 Comets ............................................................................................................................................... 120

9.3

The Patched-Conics Method ............................................................................................ 121

9.3.1 Description of Method ....................................................................................................................... 121

9.4

Lagrange Liberation Points ............................................................................................. 126

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9.4.1 L1 (stable) .......................................................................................................................................... 126 9.4.2 L2 (stable) .......................................................................................................................................... 126 9.4.3 L3 (stable) .......................................................................................................................................... 127 9.4.4 L4 and L5 (unstable) .......................................................................................................................... 127

9.5

Aerobraking ...................................................................................................................... 127

9.6

Gravity Assist .................................................................................................................... 128

9.7

Aerogravity Assist ............................................................................................................. 128


10.1

General Relative Motion ............................................................................................... 133

10.2

What does an Observer on one Spacecraft see? .......................................................... 136

10.3

Linearized Equations of Relative Motion .................................................................... 136

10.4

Clohessy-Wiltshire (CW) Equation .............................................................................. 139

10.4.1 Closed-form solution for near circular orbits ................................................................................... 140


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Space Mechanics for Engineers 1st Edition

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1

Introduction

The Sun, with all the planets revolving around it, and depending on it, can still ripen a bunch of grapes as though it had nothing else in the Universe to do. Galileo Galilei (1564-1642)

Getting into space is tremendously expensive and associated with many challenges and even risks 1. So why do humans bother the cost and take the risk? The answer to this question is several folds. Space offers human kind several compelling advantages which have been exploited in modern societies. It provides a global perspective - Space provides the highest post for surveillance and resource monitoring. Space also provides a universal perspective - un-obscured view of the heavens above all metropolitan, weather and pollution effects which obscure Earth telescopes. In addition, space offers a unique environment - a free-fall environment which can be used to perform unprecedented material and biological experiments. Space has an abundant storage of resources which can be used to replace depleting Earth resources. Human eminent curiosity and intrinsic longing for knowledge has always been a strong drive for exploration in land, sea and space. Many uses of space technology are inherently dual use capabilities having civil and military uses. Therefore, space technology is inevitably a significant element of national security which throws many obstacles in the international collaboration road.

Galileo Galilei (1564-1642). Italian natural Philosopher, Astronomer and Mathematician who made fundamental contributions to the development of the scientific method and to the sciences of motion, astronomy and strength of materials.

1.1 What is Space Mechanics? Space mechanics is the branch of mechanics, which studies the motion of objects traveling into space – subject to all possible forces. These objects may be celestial or heavenly bodies created by God such as the Earth, the Sun, and

1 Recall,

for instance, the tragedy of the most recent space shuttle accidents, in which the entire crews of Challenger and Columbia were lost in 1998 and 2003 respectively.

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CHAPTER 1│INTRODUCTION

the Moon; or vehicles made by man such as satellites, space shuttles and space probes. The branch of the subject studying the former objects is usually referred to as celestial mechanics , whereas the branch studying the latter is usually referred to as astrodynamics (Vallado, 2007). Orbital mechanics may be considered as a synonym for space mechanics.

The Hubble Space Telescope's launched in 1990. Its position above the atmosphere, which distorts and blocks the light reaching Earth, gives it a view of the universe that far surpasses that of ground-based telescopes.

Astronomy is the scientific study of celestial objects (such as stars, planets, comets, and galaxies) and phenomena that originate outside the Earth's atmosphere (such as the cosmic background radiation). It is concerned with the evolution, physics, chemistry, meteorology, and motion of celestial objects, as well as the formation and development of the universe. Therefore, in a correct sense, space mechanics may also be considered as a branch of astronomy. Old or even ancient astronomy is not to be confused with astrology , the belief system that claims that human affairs are correlated with the positions of celestial objects. Although the two fields share a common origin and a part of their methods (namely, the use of ephemerides), they are clearly distinct2.

1.2 Historical Background As a child of astronomy, the historical roots of space mechanics may be traced back in the history of astronomy and even tangled in some ancient periods with the false science of astrology. In the following historical overview, we will focus on the achievements made to astronomy with no mention to astrology even if such achievements were in fact intended to serve the need for better astrological prediction.

Claudius Ptolemaeus (83–168), known in English as Ptolemy. Greek-Egyptian mathematician, astronomer, geographer and astrologer. He was born, lived and died in Roman Egypt. He was the author of several scientific treatises, which had great importance to later Islamic and European science. His most famous treastise on astronomy is now known as the Almagest, or “The Great Treatise. The second is the Geography, which is a thorough discussion of the geographic knowledge of the Greco-Roman world.

1.2.1 Ancient Astronomy Nearly all early cultures identified celestial objects with gods and spirits. Ancient peoples related these objects (and their movements) to climate and weather changes either favorable or uninvited such as rain, drought, seasons, tides, cyclones, and storms. Beside the belief in mighty heavens, ancient civilizations also developed calendar systems based on the apparent motion of the Sun or the Moon (the most brightest and noticeable celestial objects). Many Egyptian gods were seen in the constellations or represented by actual astronomical bodies. For instance, the Milky Way represented the sky goddess Nut giving birth to the sun god Re. The sun itself was represented by several gods, depending on its position within the sky. Other than mythology, the Egyptian astronomers, who were actually priests, recognized the coincidence of flooding of the Nile and the rise of the bright star Sirius before the sun. They were subsequently able to predict the annual flooding, a considerable powerful skill. They divided the year into 12 months of 30 days each in addition to a five-day long feast. The Egyptians knew of the almost quarterday discrepancy between the Egyptian calendar and the actual tropical year, but they maintained the 365 day calendar for ceremonial reasons. Many 2 Albert

et al., 2001: through www.wikipedia.com

CHAPTER 1│ INTRODUCTION

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Egyptian buildings, such as the Abu Simbel Temple and the infamous Pyramids of Giza, were built with an astronomical orientation (Astronomy in ancient Egypt). Babylonians invented a sexagesimal (base 60) number system which is still used in the modern practice of dividing a circle into 360 degrees, of 60 minutes each, began with the Sumerians. Egyptian and Babylonian astronomy was the basis for much of the astronomical traditions that later developed in Greek and Hellenistic astronomy, in classical Indian astronomy, in Sassanid, Byzantine and Syrian astronomy, in medieval Islamic astronomy, and in Western European astronomy. Some historians, even, goes to the extreme that most of the scientific and philosophical body of knowledge unanimously accredited to the great Greek philosophers such as Democritus, Aristotle and Socrates are in fact the mere Greek translations of the original unpublished knowledge of the Egyptian priests (James, 1954)! Beginning around 600 BC, Greek philosophers and scientists developed a number of important astronomical ideas. The early Greek astronomers knew many of the geometrical relationships of the heavenly bodies. Pythagoras, who lived during the 500s BC, argued that the earth was round. He also tried to explain the nature and structure for the universe as a whole. He developed an early system of cosmology. In about 370 BC, Euxodus of Cnidus had developed a mechanical system to explain the motion s of the planets. Euduxos taught that the planets, sun, the moon, and the stars revolved around the earth. In 300s BC, Aristotle incorporated this earth centered or geocentric, theory into his philosophic system. Ptolemy was the author of several scientific treatises including the astronomical treatise, Almagest. His Planetary Hypotheses went beyond the mathematical model of the Almagest to present a physical realization of the universe as a set of nested spheres, in which he used the epicycles of his planetary model to compute the dimensions of the universe. He estimated the Sun was at an average distance of 1210 Earth radii while the radius of the sphere of the fixed stars was 20,000 times the radius of the Earth.

Astrolabe is a sophisticated tool for observing the position of the stars which was invented in ancient Greece and vastly improved in early Islam. Through Islamic Spain, the new astrolabe was introduced to Europe (Astrolabe).

1.2.2 Islamic Astronomy In the history of astronomy, Islamic astronomy or Arabic astronomy refers to the astronomical developments made in the Islamic world, particularly during the Islamic Golden Age (8th-16th centuries), and mostly written in the Arabic language. These developments mostly took place in the Middle East, Central Asia, Islamic Spain, North Africa, and later in China and India. It closely parallels the genesis of other Islamic sciences in its assimilation of foreign material and the amalgamation of the disparate elements of that material to create a science that was essentially Islamic. These included Indian, Sassanid and Hellenistic works in particular, which were translated and built upon. In turn, Islamic astronomy later had a significant influence on Indian and European astronomy (see Latin translations of the 12th century) as well as Chinese astronomy (Saliba).

Muhammad ibn Jabir al-Batani (858– 929) Latinized as Albategnius, Albategni or Albatenius. Arab astronomer, astrologer, and mathematician. One of his best-known achievements in astronomy was the determination of the solar year as being 365 days, 5 hours, 46 minutes and 24 seconds.

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It is really interesting to note that Muslims were in fact the first to differentiate and separate the science of Astronomy from the pseudo science of astrology. A significant number of stars in the sky, such as Aldebaran and Altair , and astronomical terms such as alhidade, azimuth, zenith, and almucantar , are still today recognized with their Arabic names. A large corpus of literature from Islamic astronomy remains today, numbering approximately 10,000 manuscripts scattered throughout the world, many of which have not been read or cataloged. Even so, a reasonably accurate picture of Islamic activity in the field of astronomy can be reconstructed. Naṣ īr al-Dīn al-Ṭūsī, or Tusi (12011274). Persian astronomer, philosopher, physician, mathematician, physician, physicist, and theologian. Tusi made very accurate tables of planetary movements as depicted in his book Ziji ilkhani. Tusi invented a geometrical technique called Tusi-couple, which generates linear motion from the sum of two circular motions. He used this technique to replace Ptolemy's problematic equant.

An important area in Astronomy is optics. It is very relevant in the development of tools for observation like telescopes that employ lenses or mirrors. Ibn al-Haytham (the Latin Alhazen) studied the property of lenses, discovered the camera obscura, explained correctly the process of vision, studied the structure of the eye, and explained for the first time why the sun and the moon appear larger on the horizon (very simply put, it is because the thicker layer of atmosphere at the horizon acts as magnifying lens compared to overhead). Other Muslim mathematicians such as Khayyam and al-Tusi examined Euclidean geometry that is the geometry of flat surfaces. The Muslim mathematicians, especially al-Battani, Abu'l-Wafa', Ibn Yunus and Ibn al-Haytham, also developed spherical Astronomy. Euclidean and spherical geometry are particularly useful in studying the overall geometry of the Universe in the study of cosmology. The Muslims also applied their astronomical knowledge to questions of timekeeping and the calendar in making almanacs, this word too being Arabic in origin. The most exact solar calendar existing to this day is the Jalali calendar which was developed under the direction of Omar Khayyam in the 12th century. This is still in use in Persia and Afghanistan.

Omar Khayyam (1048-1131). Persian poet, mathematician, and astronomer. Khayyam measured the length of the year as 365.24219858156 days, which shows an incredible confidence. For comparison the length of the year at the end of the 19th century was 365.242196 days, while today it is 365.242190 days.

Many people think that “the Arabs gave us (the) zero”. It is more appropriate to look at the phenomenon of Islamic science within its own cultural context. Then, if we want to compare, we discover that actually, as far as astronomy is concerned, little was achieved in Europe until ca. 1550 that had not been achieved previously by Muslim scholars at some time between the 9th and the 15th century.


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1.2.3 Medieval European Astronomy Nicholas Copernicus (1473 – 1543) Copernicus was the first astronomer to formulate a scientifically-based heliocentric cosmology that displaced the Earth from the center of the universe. His epochal book, De revolutionibus orbium coelestium (On the Revolutions of the Celestial Spheres), is often regarded as the starting point of modern astronomy and the defining epiphany that began the Scientific Revolution. The book was published the year of Copernicus’ death, 1543, though he had arrived at his theory several decades earlier. Although Greek, Indian and Muslim savants had published heliocentric hypotheses centuries before Copernicus, his publication of a scientific theory of heliocentrism, demonstrating that the motions of celestial objects can be explained without putting the Earth at rest in the center of the universe, stimulated further scientific investigations and became a landmark in the history of modern science that is known as the Copernican Revolution.

Nicholas Copernicus (1473-1543). Polish astronomer and mathematician. He began to believe that the earth went round the sun about 1507 and from that time until his death worked, more or less intermittently, on his exposition of his theory. He delayed the publication of this exposition because of fear of being accused of heresy.

Galileo Galelei (1564-1642) Galileo Galelei provided the crucial observations that proved the Copernican hypothesis, and also laid the foundations for a correct understanding of how objects moved on the surface of the earth (dynamics) and of gravity. (Valado, 2007) Johanns Kepler (1571-1630) The greatest achievement of Kepler was his discovery of the laws of planetary motion. There were such three laws, but here we shall deal only with the first two - those that govern the motion of an individual planet. These are found in Astronomia Nova (New Astronomy, 1609), underpinned by important work in Epitome (of Copernican Astronomy) Book V (1621). The laws are: The orbit of a planet/comet about the Sun is an ellipse First Law: with the Sun's center of mass at one focus. A line joining a planet/comet and the Sun sweeps out Second Law: equal areas in equal intervals of time. The squares of the periods of the planets are Third Law: proportional to the cubes of their semimajor axes. Isaac Newton (1643 – 1727) According to the well-known story, it was on seeing an apple fall in his orchard at some time during 1665 or 1666 that Newton conceived that the same force governed the motion of the Moon and the apple. He calculated the force needed to hold the Moon in its orbit, as compared with the force pulling an object to the ground. He also calculated the centripetal force needed to hold a stone in a sling, and the relation between the length of a pendulum and the

Johannes Kepler (1571 - 1630). German mathematician and astronomer who discovered that the Earth and planets travel about the sun in elliptical orbits. He gave three fundamental laws of planetary motion. He also did important works in optics.

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time of its swing. These early explorations were not soon exploited by Newton, though he studied astronomy and the problems of planetary motion. Correspondence with Hooke (1679-1680) redirected Newton to the problem of the path of a body subjected to a centrally directed force that varies as the inverse square of the distance; he determined it to be an ellipse, so informing Edmond Halley in August 1684. Halley's interest led Newton to demonstrate the relationship afresh, to compose a brief tract on mechanics, and finally to write the Principia . Sir Isaac Newton (1642-1727). British mathematician and physicist, one of the foremost scientific intellects of all time. He was elected a Fellow of Trinity College in 1667 and Lucasian Professor of Mathematics in 1669. During two to three years of intense mental effort he prepared Philosophiae Naturalis Principia Mathem Mathematic atica a ( Mathema Mathematic tical al Principl Principles es of Natural Natural Phil Philoso osophy phy ) common commonly ly known known as the Principia, although this was not published until 1687.

Book I of the Principia states the foundations of the science of mechanics, developing upon them the mathematics of orbital motion round centers of force. Newton identified gravitation as the fundamental force controlling the motions of the celestial bodies. He never found its cause. To contemporaries who found the idea of attractions across empty space unintelligible, he conceded that they might prove to be caused by the impacts of unseen particles. The three laws of motion are stated below. Every object remains in its state of rest or First Law of Motion: motion in a straight line unless a force is exerted upon which. Second Law of Motion: The rate of change of (linear) momentum of a body is proportional to the force exerted on which and is in the same direction. Third Law of Motion: opposite reaction.

To every action there is always an equal and

Book III shows the law of gravitation at work in the universe: Newton demonstrates it from the revolutions of the six known planets, including the Earth, and their satellites. However, he could never quite perfect the difficult theory of the Moon's motion. Comets were shown to obey the same law; in later editions, Newton added conjectures on the possibility of their return. He calculated the relative masses of heavenly bodies from their gravitational forces, and the oblateness of Earth and Jupiter, already observed. He explained tidal ebb and flow and the precession of the equinoxes from the forces exerted by the Sun and Moon. All this was done by exact computation. Universal Gravitational Law: Any two bodies attract one another with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.


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1.2.4 As 1.2.4 Astr trod odyn ynam amic icss The birth date of astrodynamics may be marked with the successful launch of the first man-made Earth satellite known as Sputnik I by the former Soviet Union on October 4, 1957. Only few months later, to be followed by the launch of the second man-made satellite Explorer I on January 31, 1958 by the United States. Thousands of satellites have been launched since then to serve numerous applications ranging from purely civil to purely military. In addition to Earth satellites missions, many interplanetary space exploration missions have been successfully accomplished to the Moon, Mercury, Mars, Saturn and other planets in the solar system. From Sputnik to Ulysses (the joint ESA/NASA solar orbiter mission and one of the longest and most successful space missions ever conducted, which has finally ended on June 30, 2009), space utilization has become an integral and crucial part of the modern human life. Due to the highly ambitious programs in NASA and ESA, in recent years, it has become evident that there is piling interest innovative space missions which require collaborative scientific and engineering efforts from different disciplines. Recent advances in modern orbital mechanics, space propulsion systems, and orbital and attitude control techniques paved the way to realize missions which used to be unfeasible. New space mission concepts such as formation flying have gained much of the space community interest. Such progresses were supported by peer advances in orbital determination methods and maneuver techniques which constitute what is referred as modern astrodynamics (Gurfil, 2006).

1.3 1. 3 Our Celestial Residence Studying the space environment is important to understand how its characteristics will affect the design and operation of spacecraft and the possible crew inside. What constitutes the space environment? Where space begins? How does the space environment differ from that of Earth?

1.3.1 What is Space? The question “Where does space begin” is complex. In fact, it depends not only on physical properties, but also on international politics! A loose answer which is serves our purposes is that space occurs beyond the Earth’s protective atmosphere. However, higher segments of the atmosphere are usually considered as part of the space environment where Low Earth Satellites (LOE) and space shuttles are normally operating (Newman, 2002).

1.3.2 Earth’s Atmosphere Earth’s atmosphere protects us from the hostile environment of space and provides our life support. The atmosphere is mainly composed of molecular nitrogen and oxygen with trace elements in the following proportions: Nitrogen (78%), Oxygen (21%), Argon (0.97%), and Carbon dioxide/water vapor (0.03%).

Sputnik I (which means traveling companion) is the Earth's first artificial satellite, which was launched by the former Soviet Union on October 4, 1957. This launch of that 18-lb, 22inch-diameter sphere sent a dramatic shockwave through the west, which resulted in a series of ground-breaking events including the establishment of NASA itself.

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The atmosphere itself is divided into five segments: the troposphere, stratosphere, mesosphere, thermosphere, and exosphere. The thickness of each segment is determined mainly by temperature gradients. Troposphere [sea level – 18 km] The temperature in this region varies with altitude from about 290 K (17°C) at sea level to 220 K (− ( −53°C) at 11 km. Nearly, all the weather effects we experience on the surface of Earth, such as rain and snow, occur within the troposphere. The tropopause is an extension of the troposphere where the temperature remains relatively constant, and it extends from 11 to 18 km. Stratosphere [18 km – 50 km]

Fig. 1-1 The standa standard rd atmo atmosph sphere. ere.

Here, the temperature gradient reverses, and the air actually gets warmer. At 18 km, the temperature is about 220 K and rises to about 270 K ( −3°C) at 50 km. At an altitude of 22 km, cabin pressurization is not economic due to low atmospheric density and the threat of ozone poisoning (So, human flight above this height requires a sealed environment with independent oxygen and pressure supplies). For humans, 22 km might be a good altitude to denote the beginning of the space environment. At about 45 km, aircraft propulsion requires an independent supply of fuel and oxidizer. So, essentially, for aircraft the space environment begins at 45 km. Astronauts who fly above the stratosphere receive “wings” for their achievements, so we might define the beginning of space here. The stratosphere is also the house of the ozone layer (Ozone = O 3) which screens out ultraviolet radiation (a radiation which would kill off all forms of life) but transmits visible light (which is essential for life and photosynthesis). High-altitude carbon dioxide allows visible energy from the sun to pass through, and Earth absorbs it. When Earth reradiates that energy as longwavelength infrared energy, the CO2 will not let it back out. Mesosphere [50 km – 85 km] The temperature in this region decreases with altitude from about 270 K at 50 km to 190 K (-83°C) at 85 km (this is the coldest region of the atmosphere). The mesosphere is essential for human survival on earth as it absorbs primary cosmic radiation and deadly solar ultraviolet and X-ray radiation, and vaporizes incoming meteorites entering from interplanetary space. At 60 km altitude no atmosphere is present to scatter sunlight, making the sky appear black and allowing the curvature of the earth to be discerned. This lack of atmosphere also prevents sound and shock waves from permeating. It is too low for spacecraft and can only be reached by very large high-altitude balloons. The mesosphere is often thought of as a “no-person’s land.” Thermosphere [85 km – 300 km] Beyond the mesosphere the temperature gradient rapidly reverses and the temperature rises from 190 K (− ( −83°C) to 1,000 K (773°C) at higher altitudes


where ultraviolet radiation from the Sun is absorbed. If the solar cycle is at a maximum (many solar flares), the temperature can exceed 1,000 K due to the higher number of energetic particles in the region, which results in an inflation of the upper atmosphere and an increase in the density at a given altitude The same effects—inflation and variation of density—are observed to a lesser extent in a day–night cycle. One way to think about it is to imagine the atmosphere breathing on a day–night basis, inflating during the day (or taking an extremely large breath in) during solar maximum (7 years) and exhaling during the night (or a large sigh during solar minimum) (4 years). At 100 km, the Von Karmen line denotes the altitude where aerodynamic forces (drag, viscosity, etc.) are minimal. (100 km is the height that the U.S. Air Force Office of Aerospace Research defines as the beginning of space.) Expososphere [300 km and beyond] The temperature remains constant at 1,000 K with the exc eption of solar cycle variations. Here, atomic oxygen is more abundant than molecular oxygen or nitrogen due to UV radiation. Atomic oxygen is highly reactive and may cause damage to satellite surfaces. Upper atmosphere aerodynamic drag affects spacecraft design and operations. This drag is created by the impact of atmospheric particles on the spacecraft surface. Although its magnitude is in no way comparable to aerodynamic drag encountered by aircraft, it is nevertheless present and below ~ 600 km should be considered. (A satellite at 200 km cannot remain there for more than several days based on the ballistic coefficient of the spacecraft). More than an order of magnitude of variation in atmospheric density is observed between solar minimum and solar maximum which should be considered when designing a space system to be operated below 600 km. Above the exosphere is the region known as hard space. While the idea of space is usually accompanied by a thought of emptiness, hard space is by no means empty. At 2,000 km altitude, the density is 108 particles per cubic meter (the density at the surface of Earth is 1025 particles/m3) So, space is not empty but, rather, is filled with electromagnetic radiation and particles.

1.3.3 Space Thermal Environment There are limits to the temperature range that humans and equipment can endure. The extreme thermal conditions in space require not only shielding and insulation, but heat rejection capabilities as well. On Earth, heat can be transferred by one or more of three modes: conduction, convection, and radiation. In the vacuum of space, an isolated body, such as a space-suited astronaut during a spacewalk or a planet or a satellite, can transfer heat to or from another body solely by radiation, provided the body is in a state of thermal equilibrium.

9

1 0


The extent to which a body absorbs solar radiation is determined by the solar absorption coefficient α. Likewise, the amount of power a body emits also depends on the emissivity. The physical phenomenon of radiation is governed by Kirchhoff’s law. Kirchhoff’s law is satisfied when the amount of power absorbed is equal to the amount of power emitted.

    emitted εσ 4π  =

(1-1)

(1-2) P = T r (1-3) P =P Combining the above equations results in a solution for the equilibrium temperature for any object at a distance d from the Sun of

emitted asborbed





=

1√   αε  

Where is the average surface temperature of the sun and radius of the sun.



(1-4) is the average

1.3.4 Microgravity When we hear astronauts describe the feeling of “weightlessness” while in orbit, what they are actually referring to is the effect of microgravity. Microgravity can be simulated either by placing an object in an environment where the force of gravity is naturally small (i.e., placing an object between two gravitationally equal massive bodies) or by placing an object in free fall, such as in low Earth orbit. In a low Earth orbit, a spacecraft (and the astronauts inside) experiences a radial gravity effect that is only one-tenth less than the standard 9.81 m/s2 (1 G) environment. While orbiting Earth, the spacecraft and astronauts experience a constant state of free fall; and thus, they are considered to be in a microgravity environment where the centripetal acceleration of the spacecraft (acting tangentially) is responsible for the resulting microgravity environment (10-6 G) onboard. The term microgravity (or mG) is used to describe this very low-acceleration environment. Working in a microgravity environment allows researchers to investigate essential questions of fundamental physics, life science, materials science, space science, earth observation, medicine, gravitational biology, and engineering technology. Microgravity allows scientists to observe phenomena usually overshadowed by the effect of gravity on the surface of Earth. Engineers explore new technologies and develop devices specifically designed to function in microgravity, but the big payoff might be back down on Earth. Research efforts in both aeronautics and space contribute immensely to the benefit of everyone on Earth. Many disciplines have benefited from studies conducted in microgravity by investigators all over the world. Medical professionals today use medicines developed in orbit; materials scientists and engineers have a better understanding of how substances interact to form various materials; and countless other areas of research have been developed from our ventures into the microgravity environment. A major purpose of ISS


11

is to provide a laboratory for long-duration microgravity experiments in the life and physical sciences.

EXAMPLE 1-1 If the acceleration of gravity at the surface of Earth is 9.8 m/s2, what will be the acceleration of gravity at an altitude of 600 km. Assume the average radius of Earth = 6,378 km.

SOLUTION The acceleration of gravity can be found from

    ℎ  0      0  ℎ  0 ℎ   ⁄ ≈   =

=

(

+ )

At sea level, the acceleration of gravity becomes

=

Sun spot

Then, we can write

=

(

+ )

Fig. 1-2 Sunspots are places where very intense magnetic lines of force break through the Sun’s surface. They are used to measure the intensity of solar activities.

1

=

(1 + /

)

At an altitude of 600 km, the acceleration of gravity will be

= 9.8 ×

1

(1 + 600/6378)

= 8.187 m s

83% g

0

Solar flare

1.3.5 Near Earth Radiation Environment The near Earth environment looks like a cavity in the interplanetary scene under which the Earth surface is more or less protected from the hostile space environment. The blue planet offers to its inhabitants a fragile shield that consists of both its atmosphere and its magnetic field, without which, life on Earth would be impossible. Outside this double protection, various types of radiation are encountered, which are highly variable in nature, energy, origin, and distribution.

1.3.6 Solar Activity and Emissions The Sun is a modest star (one out of more than 100 billion stars in our galaxy). It provides all the heat input to the solar system and dominates the gravitational field. The Sun contains 99.85% of the solar system mass. The gravity of the Sun creates extreme pressures and temperatures within itself, which makes it fundamentally a giant thermonuclear fusion reactor, fusing hydrogen nuclei and producing helium. So, it produces a tremendous amount of energy. The Sun has no distinct surface or discrete physical boundary (the apparent surface is merely optical). The radius of the Sun is 5-96 105 km, or about 109

Fig. 1-3 Solar flares are associated with a violent release of energy for a short time, from an hour to a few days.

1 2


times the radius of Earth. The distance from Earth to the Sun is referred to as an astronomical unit or AU (1 AU ~ 1.5x106 km). Temperature, pressure, and density are highest at the core, or center, of the Sun. The temperature reaches as high as 16 million Kelvin at the core then decreases farther away from the sun to abo ut 106 K at the highest point of the atmosphere. Solar Activity The activity of the Sun is measured by the number of visible sunspots (organized in groups). This activity presents a periodicity of approximately 11 years, with essentially 7 years of maximums (i.e., high solar activity levels related to an increase in the number of sunspots, and associated with violent emissions of particles), and 4 years of minimums. Solar Wind The external gaseous envelope of the Sun, the corona, has an extremely high temperature, and thus it continuously ejects particles, mainly electrons and protons. This continuous flux of charged particles constitutes the solar wind which streams off the Sun in all directions at average speed of charged particles ranges from 400 to 1,000 km/s (1.44 to 3.6 million km/hr). The solar wind–charged particles, under the influence of the solar magnetic field, diffuse in the entire interplanetary space. These particles originate from two regions, the Sun’s equatorial and polar regions. The equatorial area emits ions at ~400 km/s, which continuously affect the near-Earth environment. The Sun’s polar region spits particles out at 1,000 km/s, but these only occasionally affect our neighborhood (when these regions extend to lower latitudes). Now we might ask, what happens when these energetic particles hit Earth’s magnetic field (or shield)? Sunspots and Solar Flares The main features of our active star are best understood through the Sun’s magnetism. The Sun’s magnetism, or magnetic field, is produced by the flow of electrically charged ions and electrons. Sunspots are places where very intense magnetic lines of force break through the Sun’s surface. The sunspot cycle results from the recycling of magnetic fields by the flow of material in the interior. Magnetic fields are at the root of virtually all the features we see on and above the Sun without them the Sun would be a rather boring star. Sunspots are the most obvious dynamic phenomenon on the Sun. Large ones can be seen from Earth with the unaided eye, and they may look like dark objects passing in front of the Sun. Galileo was the first to show that sunspots are actually on the Sun’s surface. In 1851, a German amateur astronomer, Heinrich Schwabe, published a paper in which he concluded that the number of sunspots was not constant, but went from a minimum to a maximum about every 10 years (not too far off the actual 11-year cycle). The magnetic field of


the Sun forms vertical bands from the north magnetic pole to the south magnetic pole. From these active regions stem the solar flares that are associated with a violent release of energy for a short time, from an hour to a few days. This burst of energy produces various types of radiation, mainly X-rays and gamma rays, and ejects particles that can have extremely high energy into the interplanetary environment.

1.3.7 Galactic Cosmic Rays Galactic cosmic rays are mainly protons, heavy ions, and particles with extremely high energies. As their name indicates, they are of galactic and/or extragalactic origin emitted by distant stars and even more distant galaxies, GCRs diffuse through space and arrive at Earth from all directions. It has been shown that outside the Earth magnetosphere, the GCR fluxes are isotropic. Moreover, during solar maximums, the GCR fluxes are lower than during solar minimums. One way to imagine this effect is to think about solar maximum periods as a strengthening of the heliosphere, the Sun equivalent of the Earth magnetosphere, that is, the magnetic shield of the Sun that extends throughout the entire solar system, and hence protects the planets from these particles. GCRs are a serious danger to spacecraft, and the humans inside, because a single particle, since it is highly energetic, can damage any common electronic component onboard. When a single particle causes such a malfunction, its effect is called a single event effect (SEE).

1.3.8 The Magnetosphere The magnetosphere is defined by the interaction of Earth’s magnetic field and the solar wind. It is in the magnetosphere that Earth’s space environment meets head on the effects of the Sun. Up to 4 or 5 Earth radii, the Earth’s magnetic field resembles that of a simple magnetic dipole (a bar magnet), with field lines originating from the magnetic North pole, curving around in a symmetric arc, and entering at the magnetic South pole. A clarification of the North–South axis is needed because what is referred to as the North–South direction is the axis of the magnetic dipole that is tilted ~11° relative to the axis of rotation of Earth. Moreover, this dipole is offset 500 km toward the west Pacific (it is not at the exact center of Earth), and hence is not truly symmetrical around Earth, even below 4 Earth radii.

13

1 4


Fig. 1-4 Earth’s magnetosphere and its structure. Cosmic charged particles will be either repelled (orange trajectories) or trapped by the magnetosphere lines as illustrated by the (yellow trajectories).

One noticeable effect of this offset is a weaker magnetic field over the south Atlantic, which is called the south Atlantic anomaly (SAA). In the SAA, radiation particles interfere with satellite, aircraft, and spacecraft communications. Usually, trapped particles are repelled by the strong magnetic field of Earth. However, within the SAA the weaker field strength allows more particles to reach lower altitudes, causing potential damage to spacecraft or communications blackouts (i.e., no radio signals are received from a spacecraft) when passing through the SAA because of the high concentration of charged particles in the region. These blackouts last about 15 to 30 min per orbit, and typical spacecraft pass through the SAA approximately 9 to 15 times per day. The SAA is also responsible for exposing astronauts to high doses of radiation in space. The SAA traps high-energy radiation from solar flares, solar wind, and GCRs. To avoid unnecessary exposure to this radiation, extravehicular activity (EVA), or spacewalks, is planned, as much as possible, for orbits that do not pass though the SAA. At an altitude of 500 km above Earth the SAA ranges from 90° to 40° in geographic longitude and 50° to 0° in geographic latitude. The field lines deviate substantially from those of a simple dipole. As the solar wind hits Earth’s magnetosphere, the magnetic field is compressed on the day side of the Earth. The region where the solar wind is stopped (but not exactly) is called the magnetopause. The magnetopause is situated around 10 Earth radii on the day side in the equatorial plane. However, during violent solar flares, this boundary could be as low as 6 Earth radii [i.e., approximately geostationary earth orbit Charged particles of solar or cosmic origin are deviated by Earth’s magnetic field when they are not trapped in the Van Allen belts. Therefore, in order for


15

a charged particle to reach a given depth in the magnetosphere, it needs to have certain energy. This energy is attenuated; hence so is its damaging potential, as it passes through the magnetosphere. This effect is called magnetosphere filtering, and it implies the shielding function of the magnetosphere. One implication of shielding is that an orbiting spacecraft in LEO has an extremely small exposure to solar flares and GCRs, if its orbital inclination is below 50°. This is below the polar regions, which are areas that offer little resistance to radiation since this is where magnetic field lines converge. The following section further expands on the consequences of designing spacecraft for the harsh environment of space.

1.3.1 The Van Allen Belts In addition to SCRs and GCRs, there is a radiation effect closer to home. The third type of ionizing radiation is contained in the Van Allen belts, which are doughnut-shaped (toroid) regions of trapped particles, mainly protons and electrons, around Earth discovered on the Explorer 1 satellite mission on January 31, 1958. Where do these charged particles come from and how do they get trapped? The sources of these high-energy particles are basically the same as those previously described. As the particles hit Earth’s magnetosphere, they undergo a velocity modification (charged particle moving in a magnetic field). Under certain conditions of incident angles and energy, the velocity modification results in a relatively stable trajectory of the particle around Earth, hence the particle gets trapped.

James Van Allen (September 7, 1914– August 9, 2006) was an American space scientist at the University of Iowa. The Van Allen radiation belts were named after him, following the 1958 satellite missions (Explorer 1 and Explorer 3) in which Van Allen had argued that a Geiger counter should be used to detect charged particles.

As observations have shown, after detonation of nuclear weapons in space (i.e., the Starfish program in 1962), particles may reside in the Van Allen belts for periods of weeks to several years. Naturally, the particles’ energy, structure, and temporal dimensions of the Van Allen belts and radiative environment are functions of the solar activity. Two concentric regions of trapped electrons have been identified: The first extends to about 2.4 Earth radii, and its radiation particle population is rather stable; The second is an outer belt from 4 to 6 Earth radii whose electron density is highly variable and very sensitive to solar activity. The doughnutshaped Van Allen belts are distributed nonuniformly within the magnetosphere, and extended stays in either could be fatal to humans.

1.3.2 Environmental Impact on Spacecraft Design Energetic radiation can severely degrade the optical, mechanical, and electrical properties of a spacecraft. Specifically, satellite degradation results from ionization of atoms encountered the breakup of chemical liaison, and displacement of atoms from crystal lattice sites.

Inner belt Outer belt

Fig. 1-5 Van Allen belts.

1 6


The important parameters that aerospace engineers design for include cumulative dose of radiation, transient effects that depend on the instantaneous flux of radiation, and electrostatic arcing due to the accumulation of electric charges encountered Cumulative Dose Effects The absorbed dose is defined as the ratio of the average energy transferred to a given volume of a material by the radiation, per mass of this volume The SI unit for the absorbed dose is the Gray (Gy), defined as 1 joule absorbed in 1 kilogram of matter. An older unit that is still often used is the rad, where 100 rad = 1 Gy. Note that the absorbed dose is a macroscopic cumulative parameter, and therefore it cannot pretend to capture the entire range of radiation effects on a spacecraft or material. Single Event Effects SEEs are radiation events caused by a single energetic particle (galactic cosmic rays, solar protons, trapped particles in the Van Allen belts), which are most damaging to electronic components.

Ulysses (named after the Latin translation of Dantie’s Odysseus) was a robotic space probe designed to study the Sun, launched on October 6, 1990 as a joint mission of NASA and ESA to study the Sun at all latitudes. As one of the longest and most successful space missions ever, it has finally ended, with ground shut-down commands on June 30, 2009.

As the particle plows through a chip, it creates along its path a localized ionization. This ionization in turn can result in the following: If the electronic component is a memory device, local ionization can result in a change of the data point or state of the device (a change of 0 to 1 or vice versa). This phenomenon is called a single event upset (SEU) and is often nondestructive. Some electronic components undergo a single event latch-up (SEL). In this case, a conduction path is created between the power input and a substrate of the component power source, which can drag down the satellite bus voltage. If the device is not being monitored (by a watchdog) and is not disconnected when a latch-up is detected, the excessive current may lead to the destruction of the component, or burnout (single event burnout, SEB). SELs and SEBs tend to be quite destructive SEEs. The SEE severity depends on the specific type of event and the system criticality of the component. Shielding the satellite and components has little effect against SEEs.

1.3.3 Spacecraft Electrostatic Charging Electrostatic charging of a spacecraft on orbit always occurs. Some materials accumulate charges on the day side of the orbit but discharge during eclipse, maintaining a relatively low voltage (~1 kV for kapton or 3 kV for quartz), depending on the duration of eclipse and Sun lighting. Other materials such as Teflon continuously accumulate charges, regardless of the position of the spacecraft on orbit; hence high voltage, 6,000 - more than 20 kV, can be measured. These materials should not be used on the surface of a satellite. However, the high voltage encountered on a satellite surfaces is not the main danger of surface charging, but rather the differential charging (that is, if a


17

surface material is at 6,000 V while the next to it is at 1,000 V) which might result in arc discharges. Arcing induces severe perturbations to the onboard electronics, ranging from clock resets and mode changes in instruments to complete loss of the payload. The simplest way to prevent this is to use conductive surfaces (same potential everywhere) whenever possible on the spacecraft. Highly energetic particles can induce internal charging in a spacecraft as well as external, and internal arcing is potentially more dangerous, but it is less probable.

1.3.4 Metroids and Micrometroids The background interplanetary meteoroids are solid objects whose size and mass vary over many orders of magnitude from very small to extremely large (10-15 to 1013 kg). They are most likely of cometry or asteroidal origin. Micrometeoroids are usually solid particles with a mass range from 10-15 to 0.1 kg. To the aerospace engineer, the primary concern with meteoroids is their probability of collision with the spacecraft. Micrometeoroids near Earth often have sufficient energy to penetrate protective satellite coatings and to degrade surface thermal properties. Technical solutions exist to harden satellites (e.g., through coatings, thermal conditioning, extra outside material layers). Regardless, a meteoroid can be fatal to a spacecraft, and nothing can be done to prevent it except to spot the object prior to impact and maneuver the satellite out of its orbital path.

1.3.5 Space Debris The space debris picture shown raises our concern and consciousness of the problem, and the dangers posed by this human-made debris to operational spacecraft (whether manned or unmanned). A dramatic illustration of this fact is the recent impact of the mini satellite Cerise with an Ariane 4 third stage. Ground controllers lost contact with Cerise, a small British-built satellite, only to discover weeks later that the satellite was struck by a fragment from an Ariane rocket which had exploded several years previous. During a recent Hubble Space Telescope repair mission, astronauts found a small hole in the dish of one of the telescope’s antennas. The likely cause is space junk. To make matters worse, during the same repair mission while Hubble was docked to the Shuttle payload bay, a fragment from a Pegasus rocket that exploded several years ago came dangerously close, causing an immediate action to fire the Shuttle maneuvering jets to avoid this close encounter. More recently, a US commercial Iridium communications satellite and a defunct Russian satellite (Cosmos 2251, a communications relay station launched in June 1993 and was non-operational for almost 10 years before) ran into each other on 11 February 2009, creating a cloud of debris. The two satellites collided at an altitude of 790 km over northern. US radar tracking detected some 600 pieces of debris.

Fig. 1-6 This is how human-made spacecraft objects occupying the near-Earth space environment look like to an external observer.

1 8


Fig. 1-7 The solar system. The well-known nine planets in addition to the largest known asteroids to-date. The planets shown are of the correct scale and order from the sun. Distances are not to scale.

1.3.6 The Solar System Fig. 1-7 illustrates the relative sizes of the planets belonging to our solar system. Jupiter is the largest and Ceres is the smallest.

Hubble Telescope. (n.d.). Retrieved 2008, from www.hubblesite.org/the_telescope/hubble_essentials/

Hubble

Site:

Ptolemy. (n.d.). Retrieved 2008, from History of Mathematicians: wwwgroups.dcs.st-and.ac.uk/~history/Mathematicians/Ptolemy.html Astrolabe. (n.d.). Retrieved 2008, from www.metmuseum.org Astronomy in ancient Egypt. (n.d.). Retrieved February 12, 2010, from Egyptology Online: http://www.egyptologyonline.com/astronomy.htm Doyle, S. E. (1994). Civil Space Systems: Implications for International Security. United Nations Institute for Diarmament Research (UNDIR). James, G. G. (1954). Stolen Legacy: Greek Philosophy is Stolen Egyptian Philosophy. New York: Philosophical Library. King, D. (n.d.). Astronomy in the Baghdad of the Caliphs. Institute of the History of Science, Johann Wolfgang Goethe University. Newman, D. (2002). Interactive Aerospace Engineering and Design (1st ed.). New York: McGraw-Hill. Newton’s Life. (n.d.). Retrieved 2008, from www.newton.ac.uk/newtlife.html


19

Pisacane, V. (2005). Fundamentals of Space Systems (2 ed.). Oxford University Press. Role of Islam. (n.d.). Retrieved www.moonsighting.com/articles/roleofislam.html

2008,

from

Saliba, G. (n.d.). Whose Science is Arabic Science in Renaissance Europe? Retrieved 2008, from www.columbia.edu/~gas1/project/visions/case1/sci.1.html Stephen Schneider and Thomas Arny. (2007). Pathways to Astronomy. New York: McGraw-Hill Higher Education. Unsöld, A., & Baschek, B. (2001). The New Cosmos: An Introduction to Astronomy and Astrophysics. Berlin Springler. Vallado, D. A. (2007). Fundamentals of Astrodynamcis and Applications. Microcosm. Walter, U. (2008). Astronautics: The Physics of Space Flight (2 ed.). Wiley-VCH.

2 0


2

Two Body Problem and Keplerian Motion

I can calculate the motion of heavenly bodies, but not the madness of people! Sir Isaac Newton (1642-1727)

Earth orbiting spacecraft are attracted by the Earth as the main gravitational force. The Earth gravitational force is not the only force affecting the spacecraft; however, it is the dominating force on the spacecraft. In order to obtain a good approximate description of spacecraft motion, we assume that the spacecraft is attracted only by the central body (Earth for Earth orbiting satellites.) If the central body is spherical with a homogenous mass distribution, then we can assume the central body is a point mass. In this case, the spacecraft motion is called two-body motion. The motion of a spacecraft is usually described by its position and velocity vectors, as functions of time, in a defined reference frame. In order to understand the motion of a spacecraft, we will start by reviewing fundamental kinematic and dynamic relations, in section 2.1. Section 2.2 derives the governing equations of motion of a spacecraft, starting from Newton’s gravitational law. The geometrical solutions for the spacecraft equations of motion, in a two-body problem frame, are discussed in section 2.3.

2.1 Background

V

This section provides a brief review of necessary concepts and definitions from kinematics and dynamics of particles. We strongly recommend the reader to go through this review material before proceeding to the two-body problem.

a C Z

2.1.1 Particle Kinematics The motion of any particle P (as shown in Fig. 2-1) can be tracked in a Euclidian space with the help of a Cartesian coordinate system and a clock! In the frame of reference XYZ, we can define the particle position r(t ) as

P

O

r

s

Y

X

o

Fig. 2-1 Particle kinematics.

21

C H A P T E R 2 │ T W O B O D Y P R O B L E M A N D K E P L E R I A N M O T I O N

2 2

r(t ) = x(t )i + y (t) j + z(t)k where i ,

(2-1)

j , and k are unit vectors in the X, Y, and Z directions. r = r = (r ⋅ r )1/ 2 =

x 2 + y 2 + z 2

(2-2)

Then, the particle velocity is

v (t ) =

d r

= r = xi + y  j + zk dt = vx i + v y j + vz k

(2-3)

and its acceleration is

d v = v = r dt = v x i + v y j + v z k

a(t ) =

(2-4)

= a x i + a y j + a z k Particle Trajectory The trajectory or path of a particle is the locus of points the particle occupies as it moves through space. Since a velocity vector describes the direction of motion (or the future position of the particle), it is always tangent to the trajectory. The velocity vector of a particle is always tangent to its trajectory. Let us introduce the unit vectors

V

ut

a un

C Z

O X

P

are the tangent and normal to the particle trajectory at its local position respectively. Since the velocity is always tangent to the trajectory, then we can write it as

r

s

Y

v = vu t

(2-5)

v = v = v ⋅ v

(2-6)

Where

o

Fig. 2-2 Particle trajectory and osculating plane.

u t and u n (as illustrated in Fig. 2-2) which

The distance traveled by the particle along its trajectory, s is related to the particle speed (magnitude of velocity) through

ds = v.dt v = s

 , or Note that s = v ≠ r

(2-7)

CHAPTER 2│ TWO BODY PROBLEM AND KEPLERIAN MOTION

d

r ≠

dt

(r )

23

(2-8)

See Example 2-1 for illustration. The acceleration of the particle can be expressed in the osculating plane (the plane of motion) in terms of the unit vectors

u t and u n as follows a = at u t

+ anu n

(2-9)

where V

= v = s, an =

at and

ρ is

v2

(2-10)

ρ

Fnet

the radius of curvature of the trajectory at the particle position,

Z

which is the distance from the particle position to the center of curvature C as illustrated in Fig. 2-2.

EXAMPLE 2-1 Take

P

   =3

+ 2 . Calculate

SOLUTION

dtd ‖‖

) and compare it to

(

ddt

O X Fig. 2-3 Particle kinetics.

=3

=3

=

=

Obviously, for all

>

9

9

9

Y

.

̇   ‖ ̇‖ ‖‖   ‖‖ √    ∞  ‖‖ ≠‖ ̇‖ =

m

r

+4

+4

=

3

1+

m

Fg

M

Fg r

4 9

> 0,

2.1.2 Particle Dynamics Angular Momentum The angular momentum of a particle about a point is the moment of momentum (or more specifically, linear momentum) of the particle about that point. For the particle P shown in Fig. 2-3, which has mass (m), the angular momentum H about O is given by

Fig. 2-4 Earth and a rotating satellite is a good approximation of two-body system.


2 4

(H)O = r × (mv)

(2-11)

Then, for constant m, we can find the rate of change of angular momentum

(H )O = d [r × (m v)]

dt = r × (m v ) + r × (m a)

(2-12)

The first term on the right hand side will cancel by vector identity (Note that

r × v = v × v = 0 ). If the net force acting on the particle is Fnet , then from Fg

Z

Newton’s second law (conservation of linear momentum), we can write, for constant m,

m

M O

Fnet = m a

Y

X

Therefore, equation (2-12) can be written as

Z’



H O = r × Fnet

Y’ X’

(2-14)

Now, r × Fnet is exactly the moment of the net force Fnet about O, or (M net )O .

Fig. 2-5 Formulation of the two-body problem.

Then, we can write

(H )O = (M net )O

Z

O

(2-13)

(2-15)

The above equation is analogues to Newton’s second law for linear motion, and is called Newton’s second law for angular motion or the conservation of angular momentum.

r

Moving frame

2.2 The Two-Body Problem

Y

X Inertial frame Fig. 2-6 Inertial frame and moving frame.

2.2.1 Problem Description The two-body problem is the dynamic problem to find the trajectory of motion of a system composed of two body masses M and m (see figure Fig. 2-5), for instance, in the absence of any effect other than the mutual gravitational force – given some initial condition on the positions and velocities of these body masses. From this description, we notice that an actual two-body system does not exist in reality, but as we will see later, the trajectory of motion of many body pairs in space can be approximated, to a sufficiently high-degree of accuracy, by a two-body motion.

2.2.2 Problem Formulation In order to mathematically formulate the problem, let us consider the system of two body masses M and m (as shown in Fig. 2-5). Assume X’Y’Z’ is an inertial frame of reference (frame of reference which is neither accelerating


25

nor rotating as illustrated in Fig. 2-6). Let XYZ be a non-rotating frame of reference parallel to X’Y’Z’ with its origin O coincident with the center of mass M.

F g = −G

Mm r r 2 r

(2-16)

2.2.3 Equation of Motion The position vectors of M and m with respect to X’Y’Z’ are rM and rm respectively. Then, the position of m relative to M will be

r = rm − rM

(2-17)

Applying Newton’s second law of motion to m and M, we get

Mm r r 2 r

(2-18)

Mm r r 2 r

(2-19)

M r r 2 r

(2-20)

mrm = −G

M rM = G Or

rm

= −G

rM

= G

m r

(2-21)

r 2 r

If we subtract (2-21) from (2-20), we get

rm

− rM = r = −G

(M + m)

r 3

r

(2-22)

Equation (2-22) is the vector differential equation of the relative motion of t he two-body system. Now, if we assume one of the masses is much smaller than the other, specifically, m << M, then G (M + m) ≈ GM and equation (2-22) becomes

r =

−

GM r 3

r

(2-23)

If we compare equation (2-23) and equation (2-20), we notice that r and r will measure the same magnitude and direction whether in XYZ or X’Y’Z’ (Recall that the frame XYZ is non-rotating and parallel to the inertial frame X’Y’Z’). The quantity GM is called the gravitational parameter which can be

2 6


found from the astronomical data of most known celestial objects. Hence, the equation of motion will become r +

µ

r 3

r=0

(2-24)

We notice that the equation of motion (2-24) is a 2nd order differential equation of position in time which needs to be integrated twice to obtain the trajectory of a spacecraft for example. Then, two constant vectors are needed for the solution of the equation of motion, which may be taken as the initial position and velocity vectors.

2.2.4 Constants of Motion If we take the dot product of the equation of motion (2-24) with r we get

r ⋅ r +

µ

r 3

r ⋅ r = 0

(2-25)

The first term can be mathematically manipulated as

r ⋅ r =

d  v 2  1 d 1 d (r ⋅ r ) = ( v ⋅ v ) =   dt  2  2 dt 2 dt

(2-26)

We know that

d dt

(r ⋅ r) = 2r ⋅ r

(2-27)

But

d dt

(r ⋅ r ) =

d dt

(r 2 ) = 2r r 

(2-28)

Then, we get the important relation

r ⋅ r = r r 

(2-29)

 µ  , we get   r 

We notice that, if we take the time derivative of  −

µ  µ µ µ  −  = 2 r  = 3 (r r ) = 3 r ⋅ r dt  r  r r r d 

Now, if we substitute from (2-26) and (2-30) into (2-25) to get

(2-30)


d  v2  dt  2

µ  −  = 0 r 

27

(2-31)

By integration, the quantity in brackets will be a constant. This quantity is also the total mechanical energy per unit mass of spacecraft, ε. The total mechanical energy of spacecraft is the summation of the kinetic energy, 1 mv 2 and 2

potential energy, − m

µ r

h

γ

. We know from dynamics that the

potential energy of an object depends upon the selection of the datum at which the potential energy vanishes. Here, we opt to have the datum for gravitational potential at infinity. For other selections of datum, at the center of M, for example, a constant should be added to − m

µ r

. Therefore, for the

Y

X

two-body motion, we can write Fig. 2-7 Angular momentum.

ε=

v2 2

µ − = const

(2-32)

r

Equation (2-32) is known as the energy equation which is also referred to as the vis viva equation (vis viva, in Latin, means the living-force). This is the first constant of motion which complies with the conservation of energy principle. According to which, a body moving in a conservative field, such as the gravitational field, has constant mechanical energy. Now, let us take the cross product of the equation of motion (2-24) with r from left to get

r × r +

µ

r 3

r×r = 0

(2-33)

The second term on left hand side vanishes by vector identity, since r × r So, we have

r ×  r = 0

=0.

(2-34)

Now, if we take the time derivative of r × r , we get

d (r × r) = r × r + r × r = r × r = 0 dt

(2-35)

Note that r × r = 0 . By integration, the quantity in brackets will be a constant. This quantity is nothing but the angular momentum of spacecraft per unit mass h about the center of the central body (O). So, we can write

h = r × r = r × v = constant

(2-36)

φ m

Z M O

V


2 8

This is the second constant of motion which complies with the conservation of angular momentum principle. According to equation (2-15), since the gravitational force acting on spacecraft always passes through the center of the central body (O), it will have zero moment about O. Hence, the angular momentum of the particle about point O will be invariant. We also notice that since h = r × v = const , the plane of motion which contains r and v (the osculating plane) will be fixed in space. For convenience, we will usually refer to angular momentum per unit mass as angular momentum and mechanical energy per unit mass as mechanical energy.

2.2.5 Trajectory Equation To find the trajectory, we need to integrate the equation of motion. The nonlinear equation of motion cannot be integrated directly. Instead, we will use the following procedure to integrate the equation of motion indirectly. First, let us take the cross product of equation (2-24) with h from right, we get

r × h

V

ν

µ

r 3

r×h = 0

(2-37)

We can show that

m M O

+

d

B

dt

(r × h ) = r × h + r × h = r× h

(2-38)

The term r × h will vanish since h is constant. Also since

r × h = r × ( r × v ) = r( r ⋅ v ) − v ( r ⋅ r)

Fig. 2-8 True anomaly.

(2-39)

= r r  r − r 2 v We also notice that

d  r  µ  r r  2  µ  = µ − 2 r   = − 3 r r  r − r v dt  r   r r  r

(

) = − µ3 r × h r

(2-40)

Now, if we substitute from (2-40) and (2-38) into (2-37), we get

d  r × − µ =0 r  h dt  r 

(2-41)

r r × h − µ = B r

(2-42)

By integration we get

Where B is a constant vector which can be determined from initial conditions of the motion. Take the dot product of (2-42) with r , to get


r ⋅ (r × h ) =

µ

r

r⋅r + r⋅B

29

(2-43)

By vector identity, we can write the left hand side as

r ⋅ (r × h ) =

µ

r

r⋅r + r⋅B

(2-44)

Then, h2

= µr + rB cos ν

(2-45)

Where is the angle between B and r (as shown in Error! Reference source not found.The angle ν is called true anomaly . The true anomaly of spacecraft may be considered as an angular coordinate of the spacecraft position measured from a fixed direction in the plane of motion. Now, we can solve (245) for r

r =

h2 / µ 1 + (B / µ) cos ν

(2-46)

So, the two-body motion is a planar motion which is described by the above equation. But how does the trajectory look like? Let us have a look on the equation of a conic section in polar coordinates, r and ν.

r =

p 1 + e cos ν

(2-47)

p and e are called the parameter 3 and eccentricity of the conic section,

respectively. If we compare the two-body trajectory equation and the equation of a conic section in polar coordinates, we find that they will match exactly, if we let p = h 2 / µ

(2-48)

e = B/µ

(2-49)

and

So, let us define

e=

B

µ

(2-50)

Then, we can write the trajectory equation as

3 In

many texts, it is also referred to as the semi-latus rectum. In this textbook, we will use the name parameter throughout.

Fig. 2-9 A conic section is a curve obtained by intersecting a circular cone with a plane. The conic sections were named and studied as early as 200 BC, when Apollonius of Prego undertook a systematic study of their properties.

3 0


h2 / µ r = 1 + e cos ν

(2-51)



Since cosine is an even function, therefore the trajectory described by the orbit equation is symmetric about the apse line (the line defined be vector ).

2.2.6 Orbital Elements

The position of a spacecraft can be specified using six parameters; for example, the six scalar components of the position and velocity vectors r and v at epoch t . Another set of six parameters which are commonly used in specifying spacecraft positions in space are the classical orbital elements. In order to specify a spacecraft position in space completely, we need to specify: The Orbit Plane In general a plane can be defined in space using two parameters. These two parameters could be two components of a unit vector normal to the plane, or two angles measured from a reference frame. In the standard orbital elements, the orbit plane is defined using two angles:

≡ ≡

inclination of the orbit plane to the fundamental plane (the equatorial plane in an Earth orbit),

right ascension of ascending node, which is the angle measured from a reference direction to the ascending node in the fundamental plane.

Ω

The Shape of Orbit The shape of orbit is specified by any two of the following four parameters.

≡ ≡ ≡ ≡

semi-major axis

semi-minor axis

eccentricity parameter

 

In standard orbital elements, and are used to describe the shape of orbit. Orientation of the Orbit in Plane

Once we have specified the orbit shape and plane, then we need to describe how to place the ellipse in the plane. This is done by specifying the angle ω ω

≡

argument of perigee, which is the angle measured from the ascending node to the periapsis.

The Position of Spacecraft on the Orbit



Finally, the spacecraft position in orbit is l by the true anomaly angle .


31

≡

true anomaly, which is the angle measured from periapsis to spacecraft position.

2.2.1 Orbit Equations

  

Equation (2-51) is known as the orbit equation, which defines the path of a spacecraft with respect to the central body. It gives the spacecraft position as a function of true anomaly . Velocity Vector Components

⊥





The spacecraft velocity components along and perpendicular to are and , respectively.

⊥ ̇ ℎ ℎℎ⁄  ℎ     ⁄  ℎ  ℎ  3̇       −    ℎ ℎ      ℎ⁄  ℎ   =

=

=

=

(1 + cos )

(2-52)

1 + cos

=

=

sin

=

1 + cos

(1 + cos )

sin

(1 + cos )

=

1 + cos

= (2-53)

sin

Flight Path Angle The spacecraft flight path angle can be calculated from

tan Energy of Orbital Motion

 ⊥    ν =

=

sin

1 + cos



(2-54)



From the orbital equation, is only minimum when = 0. This point is called the periapsis of the orbit (which means the closest point in Greek). At periapsis, we have

 ℎ      ⊥ ℎ          ⊥ ℎ  ℎ     1

=

(2-55)

1+

=0;

=

(1 + )

(2-56)

At periapsis, the energy function becomes

=

But at periapsis,

= 0. Therefore,

=

but

2

=

1 2

=

(2-57)

. Then,

(2-58)


3 2

 ℎ       ℎ  =

1 2

(1

(2-59)

1+

Substituting into the energy, we get

=

   

1

)=

2

(1

(2-60)

)

2.2.2 Characteristics of the Two-Body Problem The only possible path for an orbiting satellite in a two-body system is a conic section (circle, ellipse, parabola, or hyperbola). The focus of the conic orbit is located at the center of the central body. The mechanical energy (ε) of a satellite (sum of kinetic and potential energies) does not change as the satellite moves along its conic orbit. However, the kinetic and potential energies may exchange. The orbital motion takes place in a plane which is fixed in inertial space. The specific angular momentum (h) of a satellite about the central body remains constant. Hence, as r and v changes, the flight path angle (γ) must change as well to keep h = r v constant. V

2.3 Keplerian Orbits ν

Fig. 2-10 Circular orbit.

We have in the previous section that the solution of the two-body system results in a planar trajectory which has the shape of a c onic section. Since such orbits agrees with the Kepler’s laws of planetary motion, the two-body orbits are usually referred to as Keplerian orbits. A Keplerian orbit can be circular, elliptic, parabolic, or hyperbolic.

2.3.1 Circular Orbits If we let



= 0 in the orbit equation,

 ℎ ∴  ≡ ̇ ℎ ⊥  =

⊥

constant

This proves that the orbit is circular. Since .

=

= 0, therefore

= 0 and



=

=

 ℎ      =



=

(

=

)

(2-61)

(2-62)

The time, T required for one orbit is known as the period. Since speed is constant,




=

circumference speed

 √  3      ℎ =

2

2

=

=

;

1

=

2

2

(2-63)

(2-64)

=

The specific energy becomes

=

     3  

33

(2-65)

2.3.2 Elliptical Orbits







If 0 < < 1, then the orbit equation is bounded. The initial orbit equation describes an ellipse if it is bounded from 0 < < 1. The minimum value of is

 ℎ     ℎ         ℎ        ℎ   ⇒ ℎ                              1

=

1+

(2-66)

and its maximum value is

1

=

1

2 =

1

2 =

=

1

=

=

cos( cos(180

1+

= (1

1

on apse line is

1

+

1

=

Projection of

+

(1

)

)

1 + cos (1

)

1+

(2-67)

(2-68)

(2-69) apoapsis

(2-70)

(2-71)

(2-72)

.

)=

cos

=

(1

1 + cos

)

cos

V

(2-73)

Fig. 2-11 Elliptic orbit.

periapsis periapsi s

3 4


a

a

r B

b

C

F’

F ae

Fig. 2-12 Geometry of an ellipse.

r a

r p



Solving for ,

                        ℎ   ℎ         =

(2-74)

cos

(2-75)

=

=

(

) =

=

1

=

But for an elliptic orbit,

=

1

(1

(1

2

(1

(2-76)

)

(2-77)

(2-78)

)

)

=

=

2

(2-79)

2

(2-80)

2

The area swept by the position vector during time dt can can be found from the area of the triangle which side are r and r + d r as shown in in Fig. 2-13. 2-13. The differential area is then

Fig. 2-13 Kepler’s law.

   ⊥ ≈ ⊥ ℎ   ℎ  ℎ   =

1 2

( +

1

)

2

=

2

(2-81)

This gives the well known Kepler’s law

=

2

(2-82)

If we integrate over one full orbit, we get

=

The area of ellipse =

2 . Then, we can write

(2-83)


 ℎ   ℎ ℎ     ℎ    3    2π µ    3                   ̅  0        0  ̅       =

or

2

=

=

But

=

2

2

1

35

(2-84)

(2-85) (2-86)

), then we can write

(1

=

=

2

(2-87)

where

=

=

(2-88)

1

1+

Solving for

=

Since + for is:

+

= 2 , therefore is the average of

=

1

2

( )

=

=

1

1

(1

2

=

)

=

and

1+

(2-89)

. The average value

(2-90)

(2-91)

2.3.3 Parabolic Orbits If



V

= 1,

 ℎ        ℎ     →  → ∞       =

=

As

180°,

1 2

1

1+ (1

)=0

(2-92)

v

r

(2-93)

.

=0

2

=

2

periapsis

(2-94)

(2-95) Fig. 2-14 Parabolic orbit.

3 6


∞

If a spacecraft is on a parabolic orbit, it will reach with zero velocity. Parabolic orbits are called escape trajectories. At a distance from Earth, the escape velocity is

   =

. If spacecraft is in a circular orbit,

 √ circ γ            

(2-96)

= 2

This indicates a required velocity boost of 41.4%

tan

=

sin

(2-97)

1+cos

From trigonometric identities,

sin

cos

= cos

(2-98)

= 2 sin cos 2 2 sin

2

tan

2

= 2 c os

= tan

=

2

1

2

2

(2-99)

(2-100)

(2-101)

2.3.4 Hyperbolic Orbits If



> 1, then the orbit equation

 ℎ    1

=

(2-102)

1+

describes a hyperbola. Two symmetric curves, one is occupied by the spacecraft and the other one is empty. The true anomaly of asymptotes,



= cos

−1   1

;

90° <

periapsis empty focus

C

f’

r p

a |r a|

Fig. 2-15 Hyperbolic orbit.

a



< 180°

(2-103)


Where



corresponds to

→∞

. From trigonometry,

sin

       

 √   =

1

(2-104)

< < For , spacecraft is in hyperbola I while for ), vacant orbit in hyperbola II is traced. For

For

37

  <

< (360°



= 0,

= ,

 ℎ     ℎ           ℎ                      ℎ       →    ∞                  esc   3   1

=

1+

1

=

;

1

2 =| |

=

1

=

=

= (

1

(

1+

=

1

(1

2

=

Hyperbolic excess speed,

2

1)

1);

=

=

2

=

(2-107)

(2-108)

(2-109) (2-110)

)

(2-111)

occurs when spacecraft is at

=

(2-106)

( + 1)

2

=

Recall that

<0

(2-105)

2

(2-112)

, given that:

+

(2-113)

,

=

+

(2-114)

The characteristic energy , where represents excess kinetic energy = over that which is required to simply escape from the center of attraction.

3 8


EXAMPLE 2-2

  3 ℎ ν         ℎ ℎ ⊥ ⊥    ℎ   ℎ  ⁄       ℎ            ℎ     ℎ    

  

At a given point on a spacecraft geocentric trajectory, = , , = . / and = °, show that the path is hyperbolic. Then, calculate , , , , , and .

SOLUTION

2

=

Since

>

= 7.389 km/s

, therefore it is hyperbolic.

a) Compute

=

=

cos

= 8.6 cos 50° = 5.528 km/s

= 80708.4 km /s

b) Compute

=

= sin

sin

cos

1 + cos

cos

= 0.1193

=

sin

sin

= 1.334

= tan

= 6.588 km/s

= 11.18

= 84.89°

c) Compute

= 1.339

d) Compute

=

1

= 6986 km

1+

e) Compute

=

1

1

= 20609 km


2.3.5 Summary of Keplerian Orbits The following table summarizes all the important relations for Keplerian orbits. We notice that the parameters of a parabolic orbit can be obtained from the parameters of an elliptic orbit if we let . The parameters of hyperbolic orbit can be obtained from those of elliptic orbit if we replace each with – .

→∞

  e

r p , r a

Circular

Elliptic

Parabolic

Hyperbolic

e =0

0 < e < 1

e =1

e > 1

r p = r a = R

v

T

2π µ

−

r p

=

a(1 − e)

r a

=

a(1 + e)

µ

2µ

R

r

R3

µ 2R

/2

−

µ a

2π 3 / 2 a

µ

−

µ 2a

r p = p/2

2µ r

∞

0

r p

= −a(1 − e)

r a

= −a(1 + e)

2µ µ + r a

∞

µ 2a

Bate, R., Mueller, D., & White, J. (1971). Fundamentals of Astrodynamics. New Work: Dover Publications. Curtis, H. (2005). Orbital Mechanics for Engineers (1 ed.). ButterworthHeinemann. Gurfil, P. (2006). Modern Astrodynamics. Academic Press. Pisacane, V. (2005). Fundamentals of Space Systems (2 ed.). Oxford University Press. Vallado, D. A. (2007). Fundamentals of Astrodynamcis and Applications. Microcosm.

39

4 0


P2-1

Search the web for the orbit radii and eccentricities of the following satellites: (a) Global Positioning System (GPS) (b) MEASat - 5 (c) RazakSat (d) Ofeq-5 (e) Molniya-1

P2-2

The position and velocity of a satellite at a given instant are described respectively by

       

= 2 + 2 + 2 DU ,

=– 0.5 + 0.3 + 0.4 DU/DT

where i,j,k is a non-rotating geocentric coordinate system. (a) Find the specific angular momentum ( h) and (b) total specific mechanical energy (ε) of the satellite.

P2-3

Show that the two-body motion is confined to a plane, which is fixed in space. [Hint: Start with the two-body equation of motion. Then, show that the osculating plane (the plane of the position and velocity vectors) is fixed in space].

P2-4

A certain Earth satellite orbit has a periapsis radius = 1,507 km and apoapsis radius = 39,305 km. (a) Find the orbit eccentricity. (b) Find the specific energy of the trajectory. (c) Find the semi-latus rectum or parameter (p) of the orbit. (d) Find the height of the satellite at a true anomaly 100 °. Take Earth’s radius as 6,378 km.

P2-5

Repeat problem P2-4, if the Earth satellite orbit has a semi-major axis, a, of 10,000 km and eccentricity, e, of 0.3.

P2-6

A spacecraft has the following position vector as a function of time (t ) in canonical units = 1.5 cos cos + 1.5 sin 1.5 cos sin



 

  

where i,j,k are the unit vectors of a non-rotating geocentricequatorial coordinate system. (a) Find the velocity vector of the spacecraft as a function of time and its magnitude. (b) Find the angular momentum vector of the spacecraft as a function of time and its magnitude.


(c) Find the total mechanical energy of the spacecraft. (d) Find the inclination of the spacecraft orbit.

P2-7

 ℎ    

Show that for any Keplerian orbit, the velocity is given by

=

1 + 2 cos +

P2-8

The United States BMEWS (Ballistic Missile Early Warning System) detects a UFO (Unidentified Flying Object) with the following parameters: Altitude = 0.5 DU, Speed = 0.8164965809 DU/TU, and Flight-path angle = 30°. May this object be a space probe intended to escape the Earth, an Earth satellite or a ballistic missile?

P2-9

Determine the true anomaly ν of the point(s) on an elliptic orbit at which the speed is equal to the speed of a circular orbit with the same radius, i.e., . =

 circ

vellip

ν

vcirc

[Answer: cos P2-10

−1  (

Fig. P2-9

)]

A flying object has been observed from the Earth. The object was determined to be travelling at an altitude of 300 km and velocity of 7.5 km/s with a flight path angle of 20°. (a) Calculate the angular momentum of the object about the center of Earth. (b) Calculate the total mechanical energy of the object. (c) Determine the eccentricity of the object trajectory and specify its shape. (d) Calculate the true anomaly of the object at epoch.

41

4 2


3

Coordinate Systems

I can calculate the motion of heavenly bodies, but not the madness of people! Sir Isaac Newton (1642-1727)

3.1 Introduction

Z

(North pole)

Coordinate system is a reference system that is used to represent the trajectories of artificial objects or natural celestial bodies in three-dimensional space. It is classified by: •

Being inertial or non-inertial,

•

Location of the origin, and

•

Orientation of two axes or the fundamental plane.

Inertial systems are used for the integration of the equation of motion whereas non-inertial systems are used for observation of space objects which is needed for orbit determination. Orbit determination will be discussed in details in a later chapter.

Y

X

Equatorial plane

(Vernal equinox)

Fig. 3-1. Geocentric-equatorial system.

coordinate

3.2 Classical Coordinate Systems 3.2.1 Geocentric-Equatorial Coordinate System Vernal equinox is the line of intersection between the Earth’s equatorial plane and the ecliptic plane. On the first day of spring in the northern hemisphere, when the noontime sun crosses the equator from South to North, the position of the Sun at that time defines the direction of vernal equinox, . This coordinate system has these characteristics:

γ

Origin: Center of the Earth Fundamental Plane: The equatorial plane X-axis: Vernal equinox direction from the center of the Earth Usage: Earth orbiting objects (satellites)

43

C H A P T E R 3 │ C O O R D I N A T E S Y S T E M S

4 4

3.2.2 Heliocentric-ecliptic Coordinate System The heliocentric-ecliptic coordinate system has these characteristics: Origin: Center of the sun

Z (Celestial north pole) Y Ecliptic plane

X

(Vernal equinox)

Fundamental Plane: Coincides with the ecliptic (plane of the earth’s revolution around the sun).

γ

X-axis: Vernal equinox direction , pointing from the sun towards center of Earth on the first day of autumn Usage: Planets, asteroids, comets, some space probes

3.2.3 Right Ascension-declination Coordinate System The right ascension-declination coordinate system has these characteristics:

Fig. 3-2. Heliocentric-equatorial coordinate system.

Fundamental Plane: Celestial equator (extension of earth’s equator to a fictitious infinite radius sphere called the celestial sphere) Right ascension, α: Measured eastward in the plane of the celestial equator from the vernal equinox direction Declination, δ: Measured northward from the celestial equator to the line-ofsight.

Origin: Center of the earth (geocenteric), or a point on the surface of the earth (topocenteric), or anywhere else. Usage: Observation of celestial bodies and space vehicles



We can write as:

 r

(3-1)

=r

Where Fig. 3-3. Right ascension-declination system.

r δ α δ α δ = cos cos

+ cos sin

+ sin

EXAMPLE 3-1 Given

    =

5368 + 1784 + 3691

(3-2)

CHAPTER 3│ COORDINATE SYSTEMS

45

 

Find and .

SOLUTION

r ‖‖    ∴ δ −1 δ α  δ α ∴ α ∴α  ∴    =

=

0.7947

= sin

0.5464 = 33.12°

cos cos

=

0.7947

cos sin

=

0.2642

cos



0.2642 + 0.5464

=

Earth’s north polar axis

0.9489

S/C

= 161.6° or 198.4°

Since

< 0,

=

e

Z

i h

Perigee

v r

. °

Equatorial plane

3.3 Classical Orbital Elements

Y

Ascending node X

Node line

Solution of the two-body problem requires the knowledge of two vector constants (position and velocity) at some instant, such that the spacecraft position and velocity can be predicted at any subsequent instant. Vector description of motion is suitable for computational work, but helps a little to understand the orbit itself.

Fig. 3-4. Orbit orientation in space.

Long time ago, astronomers developed orbital elements which easily express the orbit size, shape and orientation. Orbital elements (sometimes referred to as Keplerian elements) are five scalar quantities to describe orbit size, shape and orientation and a six scalar quantity to locate the spacecraft in its orbit. In a two-body motion (Keplerian orbit), the five elements determining the shape, size and orientation of the orbit remain fixed with time, whereas the sixth scalar quantity progresses with time. Fig. 3-5. Alternative orbital elements.

3.3.1 Classical Orbital Elements 1. a, semi-major axis Semi-major axis is a constant defining the size of the conic orbit. 2. e, eccentricity Eccentricity is a constant defining the shape of the conic orbit. 3. i , inclination

4 6


Inclination is the angle between the angular momentum vector, h and the K unit vector. 4. Ω, longitude of the ascending node Longitude of the ascending node is the angle between the I unit vector and the satellite crossing through fundamental plane northward (ascending node) measured counterclockwise in fundamental plane when viewed from north of fundamental plane. 5. ω, argument of periapsis Argument of periapsis is the angle in the plane of the satellite’s orbit, between the ascending node and the periapsis point, measured in the direction of the satellite’s motion. 6.

0, true anomaly at epoch

True anomaly is the angle in orbital plane, from periapsis to satellite position at a particular time, t o, called epoch.

3.3.2 Alternative Orbital Elements 1. p, parameter The orbit parameter can be substituted for semi-major axis, a. 2. Π, longitude of periapsis Longitude of periapsis is the angle from I to periapsis measured eastward to the ascending node, and then in the orbital plane to periapsis to replace ω and Ω.

ΠωΩ =

+

(3-3)

3. u0, argument of latitude at epoch Argument of latitude at epoch is the angle in the orbital plane, from the ascending node (if exists) to the position of the satellite at t 0. 4. ℓ0, true longitude at epoch True longitude at epoch is the angle between I and ro (radius vector of the satellite at t 0) measured eastward to the


47

ascending node, then, in the orbital plane, to satellite position at t 0.

o ω νo ℓo Ω ω νo Π νo Ω o =

=

+

+

+

=

+

=

(3-4)

+

(3-5)

5. T, time of periapsis passage Time of periapsis passage is the time when the satellite passed through the periapsis.

3.3.3 Ephemerides An ephemeris4 (plural: ephemerides) was, traditionally, a table giving the coordinates (in a Cartesian coordinate system, or in right ascensiondeclination system), of the Sun, the Moon, and the planets in the sky at a given time. A modern planetary ephemeris is software which generates the coordinates of the planets and often of their satellites, or of asteroids or comets at virtually any time provided by the user. Typically, such ephemerides cover very long period of time (several centuries) past and future. The biggest uncertainties are due to the perturbations on the planets of numerous asteroids, most of whose masses are poorly known, rendering their effect a bit uncertain. Solar system ephemerides are essential for the navigation of spacecraft and for all kinds of space observations of the planets, their natural satellites, stars and galaxies. 5 Fig. 3-6 Perifocal coordinate system

3.4 Perifocal Coordinate System



The perifocal frame is shown in Fig. 3-6. frame centered at focus of orbit.

̅ �

plane is the orbit plane

Origin: Focus of orbit Center of the central body (Earth for example), which is the focus of the Fundamental Plane: The orbital plane



-axis: Direction of periapsis from focus with unit vector

• •

� 



-axis: is at = 90° -axis: is normal to orbit plane in direction of

=



Usage: Earth orbiting objects (satellites)

4 From 5

the Greek word ephemeros which means daily.

http://www.statemaster.com/encyclopedia/Ephemerides

̂

=





, with a unit vector


4 8

3.5 Coordinate Transformations The position of a spacecraft may be described in many coordinate systems. The selection of the appropriate coordinate system depends on the source of position measurements, i.e. the sensor used for determination of the spacecraft position. The position of the sensor itself is usually given in another coordinate system. Neither of these coordinate systems is inertial; hence not suitable for integration of the spacecraft equations of motion. Transformation from one coordinate system to the other is necessary such as the satellite position data will be clearly specified in an inertial coordinate system for orbit determination or propagation. In the following sections, we will discuss some of the most important coordinate transformations used.

3.5.1 Transformation from orbital elements to r and v Q V

P

Transformation from r and v to orbital elements is an interesting and practical astrodynamic problem. It can be used to update the position and velocity of a spacecraft to some future time. If and are known at some time , we can find , , , , , and . Under the two-body assumptions, the elements: , , , , and are fixed. Only true anomaly changes with time. Having the new set of orbital elements , , , , , and at a subsequent instant of time , we can find and in perifocal coordinates; then, transform and to geocentric-equatorial coordinates.

   0 0 0    Ω      Ω  0    Ω         ̅ �      ∴  ℎ     ̇ ̇ ̅  ̇� ̇ ̇�̅ ̇̇  ̇̇  ̇  ℎ  ̇ ⊥ ℎ  =

Fig. 3-7. Transformation from elements to r and v.

orbital

=

=

cos 1

(3-6)

+ sin

(3-7)

(

1+

=

+

=

+

+

)

(3-8)

(3-9)

where

= cos = sin +

sin cos

(3-10)

But

=

=

(3-11)

and

=

=

(1 +

)

(3-12)


∴ ̇̅ ∴

=

49

 ℎ  ̇� ℎ   µ ν ν ;

v=

h

=

( +

)

[ sin p + (e + cos )q]

(3-13)

(3-14)

3.5.2 Transformation from r and v to orbital elements Assume a radar site on the Earth provides the position vector, r and the velocity vector, v of a satellite relative to geocentric-equatorial coordinate system at a particular time (t 0). How can we find the six orbital elements describing the satellite motion? First, we form the three vectors: angular momentum h, nodal vector n, and eccentricity vector e. The angular momentum vector is simply formed as



(3-15)



(3-16)

=

×

The node vector or (nodal vector) is the vector pointing from the center of earth to the ascending node, which is calculated from (n is perpendicular to both K (normal to equatorial plane) and h (normal to orbital plane)).

=

×

The eccentricity vector e is the vector pointing from the center of the Earth (focus of orbit) toward perigee with magnitude equal to orbit eccentricity, e. We can calculate e from

 µ   µ    =

1

r

( . )

(3-17)

K I n

Fig. 3-8 Transformation r and v to orbital elements.

5 0


Now we solve for the orbital elements according to the following procedure:

 ℎ   ‖‖  ∙ℎ Ω ∙  ∙ 0 ∙ =

/

=

(3-18) (3-19)

cos =

(3-20)

cos

=

(3-21)

cos

=

(3-22)

cos

=

(3-23)

The inverse of the cosine function gives an angle between 0° and 180° (See Fig. 3-9.). For inclination, this is not a problem since the correct range of inclination is already between 0° and 180°. However, attention should be given to the angles Ω, ω, ν, and u0 which all range from 0° to 360°. Since cos θ = cos (360 − θ), the value of the cosine is not enough to determine the correct value of Ω, ω, ν, and u0. By inspection of figure (orbit figure below), we notice that the ascending node will be east of vernal equinox direction, if 0 < Ω < 180° and west of the vernal equinox direction otherwise. Therefore, if n.J > 0 (or ny > 0), the ascending node will be east of the vernal equinox direction, hence 0 < Ω < 180°. Otherwise, 180° < Ω < 360°. Also, by inspection of the same figure, we notice that the periapsis of the orbit will be north to the fundamental plane (the equatorial plane in case of an Earth satellite), if 0 < ω < 180° and south to the fundamental plane otherwise. The eccentricity vector e can be considered as the position vector of the periapsis in the XYZ coordinate system. Therefore, if e.K > 0 (or ez > 0), the periapsis will be north to the fundamental plane, hence 0 < ω < 180°. Otherwise, 180° < ω < 360°. Similarly, we can show that if r.K > 0 (or r z > 0), the satellite will be north to the fundamental plane, hence 0 < u0 < 180°. Otherwise, 180° < u0 < 360°.

Fig. 3-9. Cosine function.


Bate, R., Mueller, D., & White, J. (1971). Fundamentals of Astrodynamics. New Work: Dover Publications. Curtis, H. (2005). Orbital Mechanics for Engineers (1 ed.). ButterworthHeinemann. Gurfil, P. (2006). Modern Astrodynamics. Academic Press. Pisacane, V. (2005). Fundamentals of Space Systems (2 ed.). Oxford University Press. Vallado, D. A. (2007). Fundamentals of Astrodynamcis and Applications. Microcosm. Walter, U. (2008). Astronautics: The Physics of Space Flight (2 ed.). Wiley-VCH.

51

5 2


P3-1

Given the orbital elements for objects A, B, C and D, fill in the blanks to correctly complete the following statements: ℓ

Object

i

A

0°

undefined

210°

B

114°

180°

260°

C

63.4°

90°

110°

D

23°

60°

260°

30° 90° 140° 160°

(e) (f) (g) (h)

Object … is in retrograde motion. Object … has a true anomaly at epoch of 180°. Object … has its apogee north of the equatorial plane. Object … has a line of nodes coinciding with the vernal equinox direction. (i) Object … has an argument of perigee of 20°. P3-2

The position and velocity of a satellite at a given instant are given, respectively, in canonical units by r = 2.0 i + 3.0 j + 4.0 k DU, v = 0.2 i – 0.2 k DU/TU where i, j, and k are the unit vectors of non-rotating geocentricequatorial coordinate system. (a) Find the satellite specific angular momentum, . (b) What is the eccentricity, and type of the orbit? (c) Determine orbit inclination, . (d) Calculate the orbit semi-parameter, . (e) Find the true anomaly of the satellite,



P3-3



ℎ

 ℎ    

Show that for any Keplerian orbit, the velocity is given by

=

1 + 2 cos +

P3-4

The United States BMEWS (Ballistic Missile Early Warning System) detects a UFO (Unidentified Flying Object) with the following parameters: Altitude = 0.5 DU, Speed = 0.8164965809 DU/TU, and Flight-path angle = 30°. May this object be a space probe intended to escape the Earth, an Earth satellite or a ballistic missile?

P3-5

Determine the true anomaly ν of the point(s) on an elliptic orbit at which the speed is equal to the speed of a circular orbit with the same


radius, i.e.,

 circ =

53

.

vellip

ν

vcirc

[Answer: cos P3-6

−1  (

Fig. P2-9

)]

A flying object has been observed from the Earth. The object was determined to be travelling at an altitude of 300 km and velocity of 7.5 km/s with a flight path angle of 20°. (e) Calculate the angular momentum of the object about the center of Earth. (f) Calculate the total mechanical energy of the object. (g) Determine the eccentricity of the object trajectory and specify its shape. (h) Calculate the true anomaly of the object at epoch.

5 4


4

Time Systems

Time and space are modes by which we think and not conditions in which we live. Albert Einstein (1879-1955)

Time may be defined as the measurement of the interval between two successive events. It is a fundamental dimension or, at least, a vital element in every science. When a scientist or layman uses the terms “hours, minutes or seconds” he is understood to mean units of mean solar time, which is the time kept by ordinary clocks. Details of this time scale will be elaborated later. Many time systems have been utilized to increase the accuracy of time measurements. There has to be a standard primary epoch 6 from which we count intervals. In engineering and science, four time systems are employed to keep time. They are the sidereal time, solar (universal time), dynamical time, and atomic time. The foundation for the sidereal time and solar time is the Earth’s rotation whereas the dynamical and atomic times are not; they are independent. In astrodynamics, the precision of time measurements is severely crucial. This can be observed in several space applications. For example, Earth satellites have velocities as high as 7 km/s. Therefore, to determine a satellite position, maybe in a satellite formation, to 5 - 10 cm precision, we need time measurement to be as precise as 7 - 15 µs. In another example, spacecraft observations information, which are essentially electromagnetic waves, travels at the speed of light (~ 3×105 km/s). So, if we seek a precision in position of 10 cm, we need a precision of 0.3 ns in. In the following, we will introduce some of the time systems that were used or are being used for calculation of time.

6

Epoch: an instant in time th at is arbitrarily selected as a point of reference.

Albert Einstein (1879-1955). GermanSwiss-American theoretical physicist. Einstein realized the inadequacies of Newtonian mechanics and his special theory of relativity stemmed from an attempt to reconcile the laws of mechanics with the laws of the electromagnetic field. He dealt with classical problems of statistical mechanics and problems in which they were merged with quantum theory. He investigated the thermal properties of light with a low radiation density and his observations laid the foundation of the photon theory of light. Einstein received Nobel prize in physics in 1921.

55

5 6

CHAPTER 4│TIME SYSTEMS

4.1 Solar Time Solar time is the ordinary time which is reckoned by the sun. The combination of the Earth’s rotation about its axis and the motion about the Sun leads to the Sun’s apparent motion. Apparent solar time is the interval between successive transits which is observed from a certain longitude (So, apparent solar day will be time between two successive upper transits of the sun across the local meridian, i.e. from noon to noon). The projection of the solar motion along the ecliptic onto the celestial equator leads to varying motion along the celestial equator. So, the apparent solar day varies in length over the year. The average of the apparent solar day is called the mean solar day .

4.1.1 Julian Calendar The Julian Date, or JD is defined as the interval of time measured in days from the epoch at 12:00:00 on January 1, 4713 B.C. This was determined by Joseph Scalinger7 in 1582. He used the combination of the solar cycle (28 years), the Metonic cycle (19 years), and the Roman indication (15 years) to produce a Julian period which is made of 7980 Julian years (365.25 days). In Julian calendar, each year consists of 12 months and includes 365 days, unless it is divisible by 4 (e.g. 2008/4 = 502), then it becomes a leap year and includes 366 days. Therefore, the average Julian year = 365.2500 days. Tropical year is the time taken by the Earth to travel from equinox back to equinox which is 365.2422 days. Therefore, Julian year > Tropical year.

Fig. 4-1. Tropical year.

Celestial observations are measured in Julian Day Numbers which are the number of Julian days (noon to noon) from epoch of noon at Greenwich on 1 January 4713 BC (e.g. 1 Jan 2004 = 2,453,006 JD). The JD enables astronomers to conduct all their observations in one day . Julian day numbers for any given day is provided in the Astronomical Almanac which is published yearly. Table 4-1 can be used to find the Julian date equivalent of Gregorian dates. There are also various conversion programs which are available over the internet. The Julian date is an integer and decimal that accounts for the fraction of day since last noon. A fraction of 0.1 is equivalent to 2.4 hours. Astronomical observations when specified by JD are denoted by prefix J. Therefore, J2000 = 12:00:00 Universal Time (UT) on 0 Jan 2000 (31 Dec 1999).

7 The

Julian period was named after Scalinger’s father and not Julius Caesar as commonly perceived!


57

Year

Jan 0

Feb 0 Mar 0 Apr 0 May 0 Jun 0

Jul 0

Aug 0

Sep 0

Oct 0

Nov 0

Dec 0

2005

2453371

3402

3430

3461

3491

3522

3552

3583

3614

3644

3675

3705

2006

3736

3767

3795

3826

3856

3887

3917

3948

3979

4009

4040

4070

2007

4101

4132

4160

4191

4221

4252

4282

4313

4344

4374

4405

4435

2008

4466

4497

4526

4557

4587

4618

4648

4679

4710

4740

4771

4801

2009

4832

4863

4891

4922

4952

4983

5013

5044

5075

5105

5136

5166

2010

5197

5228

5256

5287

5317

5348

5378

5409

5440

5470

5501

5531

2011

5562

5593

5621

5652

5682

5713

5743

5774

5805

5835

5866

5896

2012

5927

5958

5987

6018

6048

6079

6109

6140

6171

6201

6232

6262

2013

6293

6324

6352

6383

6413

6444

6474

6505

6536

6566

6597

6627

2014

6658

6689

6717

6748

6778

6809

6839

6870

6901

6931

6962

6992

2015

7023

7054

7082

7113

7143

7174

7204

7235

7266

7296

7327

7357

2016

7388

7419

7448

7479

7509

7540

7570

7601

7632

7662

7693

7723

4.1.2 Gregorian Calendar The average Julian year is longer than the Tropical year by approximately 11 min per year. And by the 16th century, Easter (a feast which occurs in spring) had already occurred in the summer! To resolve this, Pope Gregory XIII issued the Gregorian calendar in 1582 in which 10 days were omitted from the calendar and the leap rule was modified. A leap year is divisible by 4, unless it is a century year then is should be divisible by 400. For example, 2000 is a leap year, but 2200 is not. Then, the average year becomes 365.2425 days, which differs from the tropical year by one day every 3300 years! Gregorian calendar is relevant in that it forms the time basis for western civilization which is also adopted worldwide. Besides the Gregorian calendar, other calendars are also adopted in non-western nations such as the Hijri calendar in Muslim countries and the Chinese calendar in China and south-east Asia. Julian Date Converter To convert from Calendar date to Julian date and vice versa, you can use the following link to the Julian Date Converter maintained by the US Navy.

4.2 Terrestrial time According to Siedelmann (1992), terrestrial time TT is the theoretical timescale of apparent geocentric ephemerides of bodies in the solar system. It is the independent argument in the apparent geocentric ephemerides and uses the SI second as the primary interval. It is the uniform time that would be measured by an ideal clock on the surface of the Earth and measures in 86,400 SI seconds per day. TT is independent of the variations in the rotation of the Earth, polar motion, and nutation and precession. It is also more uniform than astronomical times.

Precession

Nutation

Earth’s rotation

4.3 International Atomic Time Table 4-1 Conversion from Gregorian dates to ulian dates.

It is a time system which is independent of the average rotation of the Earth. It is also known as Temps Atomique International or TAI and is defined in terms of vibrations of a cesium–133 atom and is measured in SI seconds. One SI second equals the duration of

Fig. 4-2. Precession and nutation of the Earth's Polar Axis


5 8

9,192,631,770 cycles of radiation emitted between two hyperfine ground states of cesium–133 at 0°K. TAI was developed by the Bureau International des Poids et Mesures in Sèvres, France. TAI achieves a precision that permits the observation of relativistic effects for clocks in motion or accelerated by a local gravitational field. TAI is not explicitly tied to the Earth's rotation, although that was precisely the motivation for the original definition of the second. The epoch of TAI was arbitrarily defined to be an offset of 32.184 s from Terrestrial Time (TT) for 1 January 1997 (Pisacane, 2005). Therefore, Fig. 4-3. Mean solar day versus sidereal day.

TT = TAI + 32.184 s

(4-1)

4.4 Sidereal time Sidereal time is a direct measure of the Earth’s rotation about the celestial pole with respect to the equinox. Vernal equinox is always on the equator. Hence, sidereal time is defined as the hour angle of the vernal equinox relative to the local meridian. During one sidereal day, the Earth has to turn slightly more than one complete rotation on its axis relative to the fixed stars as shown in Fig. 4-3. This is because the Earth travels about 1/365.24 of the way around its orbit in one mean solar day. Since the vernal equinox is the reference point, the sidereal time related to the Greenwich meridian is termed Greenwich Mean Sidereal Time, GMST . The sidereal time at a particular longitude is called Local Sidereal Time, LST . LMST stands for Local Mean Sidereal Time. Thus, the following relationships can be obtained: 1 mean solar day

= = =

1.0027379093 mean sidereal day 24:03:56.55536 of sidereal time 86636.55536 mean sidereal seconds

Greenwich Apparent Sidereal Time (GAST) is the hour angle of the apparent or true vernal equinox of date with the hour angle measured west from Greenwich meridian to the equinox. Greenwich mean sidereal day begins when the Greenwich meridian coincides with the vernal equinox. Greenwich Mean Sidereal Time (GMST) is the hour angle of the mean vernal equinox of date with the hour angle measured west from Greenwich meridian to the equinox. The difference between GAST and GMST lies in the nutation of the equinox. Illustrations of GMST, LMST, GAST and LAST are shown in Fig. 4-4.

Fig. 4-4. Difference between mean apparent sidereal times.

and


59

4.4.1 Universal Time Universal Time (UT) is the number of hours, minutes, and seconds which have elapsed since midnight (when the Sun is at a longitude of 180°) in the Greenwich time zone. The Mean Solar Time equals the Universal Time. There are three distinct categories of UT. UT0, UT1, UT2 need to be distinguished for precise applications. UT0 is found by reducing the observations of stars form many ground stations. UT0 is also known as the Greenwich Mean Time GMT . It is a local approximation of the universal time which is often used and not corrected for polar motion. In UT0, 24 universal hours are defined as mean solar day . To obtain UT1 or UT, the UTO is corrected for polar motion so time is independent of station location. The actual universal time (UT) or (UT1) is tied to the rotation of the Earth. Because the Earth's rotation rate is rather irregular and unpredictable to 0.1 s, Universal Time can only be deduced from observations of star transits. The expressions for GMST (0h UT1) at a desired time in seconds and degrees are:





(0h UT1) = 24110.54841 + 8640184.812866 ( + 0.093×10 ( ) –6.2×10 ( )

  

      λ =

= [

– 2 451545.0

−6  3

(4-2)

(4-3)

36525

(0h UT1) + UT1]

)

360° 86400

s

(4-4)

We can convert between GMST and LMST at a particular lo ngitude, λ using

=

+

(4-5)

The above formula needs a designation for east and west longitudes. The designation that is used here is positive for east longitudes and negative for west longitudes.

EXAMPLE 4-1

6 0


Determine the GMST and right ascension of Greenwich at 0h UT on 1 January 2010.

SOLUTION At 0h UT on 1 January 2010, JD = 2 455 166 + 31 + 0.5 = 2 455 197.5

JD of 12h UT 31 days of on 0 Dec 2009 December (30 Nov 2009)

Shift from 12h UT to 0h UT

JD of 0h UT on 1 Jan 2010

TU = (2 455 197.5 – 2 451 545.0) / 36 525 = 0.1 GMST (0h UT on 1 Jan 2010) = 24,110.548 41 + 8,640,184.812866 (0.1) + 0.093104 (0.1) 2 – 6.2 ×10 −6 (0.1)3 = 8.881290306276 × 10 5 s GMST (0h UT) = 8.881290306276 × 10 5 (360°/86400s) = 100.537 627615 deg

30 Dec 2009 30 Dec 2009

t r h a g e i n Y d i w e 31 Dec 2009 M N 1 Jan 2010

31 Dec 2009

1 Jan 2010

0 1 0 n 2 o n a o J N 0

2 Jan 2010

2 Jan 2010

Gregorian Julian Date

4.4.2 Coordinated Universal Time (UTC)

Fig. 4-5. Comparison between various time systems adapted from (Pisacane, 2005).

The coordinated universal time (UTC) is an approximation of the UT which differs from TAI by an integer of seconds. It is the basis of most radio time systems and worldwide legal time systems. UTC is the time standard provided by the world-wide web and other time broadcast services. A step adjustment (leap seconds) is usually inserted after the 60 th second of the last minute of December 31 or June 30 such that the following inequality is always satisfied.

|UT1 – UTC| < 0.9 s

(4-6)

4.4.3 Global Positioning System Time The Global Positioning System GPS is a constellation of 18 operational satellites which are used to provide worldwide navigation information such as longitude, latitude, altitude, and time. The GPS time is an estimate of the UTC. The GPS time has a fixed shift from the TAI.


GPS = TAI – 19 s

61

(4-7)

4.5 Summary of Time Systems We have introduced many time systems which are involved in the analysis and design of a space mission. To better understand the relation among all these time systems we refer to the time chart in Fig. 4-5.

From the time systems chart, we recognize the regularity of the GPS, TAI and TT time systems which all differ by fixed time shifts. The irregularity of the UT1 is also evident. We also notice from the chart the coordination of the UTC to satisfy the condition on difference with UT1.

Pisacane, V. (2005). Fundamentals of Space Systems (2 ed.). Oxford University Press. Bate, R., Mueller, D., & White, J. (1971). Fundamentals of Astrodynamics. New Work: Dover Publications. Vallado, D. A. (2007). Fundamentals of Astrodynamcis and Applications. Microcosm. Walter, U. (2008). Astronautics: The Physics of Space Flight (2 ed.). Wiley-VCH.

6 2


P4-1

Which takes longer, a solar day or a sidereal day? Explain.

P4-2

Why the universal time (UT) does not measure the same seconds as terrestrial time (TT)?

P4-3

What is the local sidereal time (degrees) of Greenwich, England (GMST), at 02:00 AM on 15 August 2009?

P4-4

What is the local sidereal time (degrees) of Kuala Lumpur (101°42’ E longitude) at 03:30 PM (local time) on 15 August 2009?

P4-5

How many significant digits of decimal degrees are required to match the accuracy of 12h 41m 37.4567s? Convert your answer for units of degrees, arc-minutes and arc-seconds.

P4-6

What calendar date corresponds to the following Julian dates? 2,363,592.5, 2,391,598.5, 2,418,781.5, 2,446,470.5, 2,474,033.5. Are these dates significant?

5

Orbit Determination

We must admit with humility that, while number is purely a product of our minds, space has a reality outside our minds, so that we cannot completely prescribe its properties a priori. Johann C. F. Gauss (1777 –1855)

The determination of the orbits of celestial bodies has been a long lasting scientific interest which possessed most of the greatest minds in mathematics and physics over the centuries. Tens of most accurate mathematical techniques have been used and even invented purposely for the sake of orbit determination. Below is a list of the mathematicians who primarily contributed to the problem of orbit determination and their major contributions. •

•

•

• • • •

Newton (1642 – 1727) introduced the first method of orbit determination from observations, which was a semi-graphical method, in his Principa Halley (1656 – 1742) used the method to calculate the orbits of 24 comets, discover a new comet (Halley comet) in 1705 and foresee its return in 1758. Euler (1707 – 1783) introduced the first completely analytical method to solve the same problem in his Theory of the Motion of Planets and Comets Lambert (1728 – 1777) gave a general formulation of the problem for the case of elliptic and hyperbolic orbits Lagrange (1736 – 1813) put all incomplete theories in a general mathematical form. Laplace (1749 – 1827) introduced a new method for orbit determination. Gauss (1777 – 1855) invented the method of Least Squares to fit the best orbit to large number of observations.

Johann Carl Friedrich Gauss (1777 – 1855). German mathematician and scientist who contributed significantly to many fields, including number theory, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy and optics.

63

6 4

CHAPTER 5│ORBIT DETERMINATION

5.1 Initial Orbit Determination The orbit determination means finding the orbit of a spacecraft based on some observations. Usually the observations are collected on the ground using equipments like telescopes or radars. An observation is a measurement for the spacecraft position or direction at a certain time. With few observations collected over a period of time, it is possible to determine the orbit of the spacecraft. The determination of the orbit means calculating the six orbital elements at any point in the orbit, or equivalently calculating the position and velocity vectors of the spacecraft at any point on the orbit. When a telescope detects a spacecraft in its field of view, it provides the direction to the spacecraft at the time of observation. A measurement for the direction is composed of two angles (e.g. azimuth and elevation). A telescope cannot provide the distance to the spacecraft (sometimes called the range to the spacecraft). Radar on the other hand provides the range to the spacecraft as well as the two direction angles to the spacecraft at each observation time. One observation (either from radar or a telescope) is not enough to determine the orbit of the spacecraft. Few observations collected over a period of time can be processed together to determine the spacecraft orbit. Several methods have been developed over time for orbit determination. The selection of the most suitable method to use depends on the available set of observations and the times at which these observations were collected. This chapter is not meant to detail all available orbit determination methods, but rather the a sample of the most common methods will be presented to acquaint the reader about the type of problem and the methods of orbit determination. Fig. 5-1. Short way and long way.

The observations that have only direction are called “Angles Only Observations”. One of the effective methods for orbit determination given angular measurements is Laplace’s method, which requires 3 sets of angular measurements. It can process data collected from different sites at different times. Gauss’ technique is another method for orbit determination from angular measurements. It is primarily useful if the measurements are separated by less than 10 deg. If we have observations for the positions of the spacecraft at two different locations, then we can implement a Lambert problem solution to determine the orbit as will be detailed in the next section. If we have three position vectors available, then Gibbs method can be implemented.

5.2 Two Position Vectors (Lambert’s Problem) In this section we discuss the problem of orbit determination given that the observations are two position vectors for the spacecraft at two different times. This problem is known in the literature as Lambert Problem and it has


65

several applications beyond the orbit determination problem including orbit design and optimal maneuver planning. We will discuss Lambert’s original geometrical formulation developed in 1761. However, it worth mentioning that there have been several other methods for solving the Lambert problem developed since then, the latest of them was developed by Giulio Avanzini and is published in the AIAA Journal of Guidance, Control, and Dynamics in 2008. In this section, we assume that the spacecraft is moving on an ellipse. We can measure the position of the spacecraft at different locations. The position vectors are and . As can be seen from Figure 5-2, for an orbit that includes the two measured position vectors, the spacecraft could be moving in two possible directions. We distinguish between the two directions using the difference in true anomaly between the two positions of the spacecraft. If is the true anomaly of the spacecraft at and is the true anomaly of the = . If > 180 then the spacecraft is moving spacecraft at , then along the long way and the discrete variable = 1. If < 180 then the spacecraft is moving along the short way and the discrete variable = 1. Now, how do we know the value of in the first place? The answer to this question is simple since we know and

 

     ∆   ∆   ∆ ∆   ∆ ∆Ѳ  ∆    ∆ cos(

sin(

∆ ∆ 

) = cos(

)=

1

)=

cos (

.

)







(5-1)

(5-2)

The two vectors, in fact, identify the orbit plane (any two vectors define a = 180 , then there are infinite orbit planes. The special case of plane). If = 180 will not be considered here.



Fig. 5-2. Minimum energy solution.

6 6






There are infinite number of orbits that include the two known vectors and . All of them are in the same plane. For an ellipse, the sum of the distances from any point on the ellipse to the two fuci is always constant and is equal to twice the semi major axis of the orbit. Hence, the semi major axis of any orbit that includes the two vectors and should satisfy the following relation:

        

2 =

+ (2

)=

+ (2

)

(5-3)

5.2.1 Minimum Energy Solution As eluded to in section 5.2, there is infinite number of orbits that include the two measured vectors. In order to fix one of these orbits, we need an additional piece of information. For instance, we can select the time of spacecraft flight between the two vectors. This way we will be able to determine a finite number of orbits. In this section, we will show how to compute the particular orbit that has the minimum possible value of semi major axis (the minimum energy orbit). Lambert’s theorem: “The orbit transfer time depends only upon the semi major axis, the sum of the distances of the initial and final points of the arc from the centre of force and the length of the chord joining these points.” The length of the cord can be geometrically computed as follows:

       ∆ =

+

2

(

)

(5-4)

Define the semi perimeter as:

    =

 

1 2

( +

(5-5)

+ )

Recall that sum of the distances from any point on the ellipse to foci is constant and is equal to 2 . So, the secondary focus F’ is the intersection of the two circles, as shown in Figure 5-3. The first circle is centered at the first point and has a radius of 2 . The second circle is centered at the second point and has a radius of 2 . So, in order for a solution to exist (in order for F’ to exist) the two circles must intersect. On the other hand, the radius of each of them is proportional to . Since we are interested in the minimum energy solution, then the circles touch. Show that the touch point is located on the cord. Let’s denote the semi major axis of the minimum energy orbit by . Then we can write:





∴      (2

) + (2

)=

(5-6)


67

∴      +

=

+

4

=

(5-7)

2

Note: If we are looking for the orbit that has minimum eccentricity, we can show that in that case the major axis of the ellipse will be parallel to the chord. This particular ellipse is called the fundamental ellipse. To find the eccentricity of the minimum energy orbit, note that:

          

2

=

+

+

=

2

+

(5-8)

=

2

Consider the blue triangle, and recall that the distance between the two foci is 2a e , then we can write the following identity:

min min

                ∴          (2

) = {(

4

=(

)

} +{

) (1

)+

2 (

But,

(

+(

)

}

)

)

     α       0 ∴                =

+

2

cos

+

=

2

(5-9)

(5-10)

(5-11)

(5-12)

And note that:

2 (

)

=2

+

+

+

2

2

+

=

+

+

2

(5-13)

(5-14)

Compare Equation 5-12 and Equation 5-13:

∴     =



2 ( 1

Substitute for cos in Equation 5-10,

)

1

(5-15)

6 8


∴        0   4

4

=

We can then solve for

(

)(

(5-16)

)

.

Exercise: show that

      ∴        ∆ 4

=

And show that:

2

=

(

(5-17)

2

)(

)=

(1

)

∴     =

2

1

(5-18)

(5-19)

To calculate the time, the following parameters are introduced:

              ℎ        ℎ       ∆        ∓      − ℎ  ∓ ℎ  2

2

(

(

2

2

=

=

)=

)=

+ + 4

+ + 4

+

+

4

+

=

=

(5-21)

2

=

(5-22)

2

=

4

(5-20)

2

(5-23)

2

Kaplan shows that the general time of flight is:

=

=

+

[2

[

(

(

)]

)]

(5-24)

(5-25)


69

5.3 Three Position Vectors (Gibbs Method) If we have three position vectors for the spacecraft then we can use Gibbs method to determine the orbit. Although the mathematical derivation for the method may seem lengthy, the Gibbs algorithm is very simple and easy to implement.



Given three position vectors for the spacecraft

vector in the direction of

:

   , and

,

. Define the unit

 1 3 ‖ ‖ =

(5-26)

Also define a unit vector perpendicular to the orbit plane: ×

=

r 3

(5-27)

×

Both U and C are perpendicular to each other.

∴ r 3 U .C

(5-28)

=0

The three position vectors have to be in the same plane (the orbit plane), so we can write the following identity which is valid for any three coplanar vectors:

 1 3 =



(5-29)

+

From the kinematics of the orbit motion described in Chapter two, It is easy to show that for any :

        µ    ×

=

(5-30)

+

Left cross both sides of the equation with , we get:

×( × )=

×

r

+

(5-31)

×

Using the mathematical identity below, we get:

         ℎ ℎ ×( × )= ( . )

( . )=

0=

(5-32)

7 0


∴  ℎ           ∴  µ    ×

=



+

(5-33)

×

Recall the pqw coordinate system, which is defined as:

= ep ;

(5-34)

= hw; w × p = q

=

w×

h

(5-35)

+ eq

r

The above identity is valid at each of the three given positions of the spacecraft. From Equation 5-29, we can write:

   1 3    ℎ ℎ ℎ    1        3    ℎ ℎ ℎ ∴   1   1 3   3 .

= . (C

+C

)

(5-36)

Also, From the orbit equation, we can write for each position vector:

.

=

;

.

=

;

.

=

(5-37)

Substitute into equation 5-36:

=

Recall the orbit equation:

+

 ℎ 1    ℎ 1 1Ѳ   ℎ ∴ 1  1   1 Ѳ =

It can be rearranged to be:

1

1+

+

=

=

But

. e = r ecos

(5-38)

(5-39)

(5-40)

(5-41)

(5-42)


71

 1

∴   µ  1 3  ⇒   3    1 3  ⇒   1   3 1 ℎ             h

.e =

(5-43)

r

Now, note the following identities:

(

=

+

)×

(

=

+

)×

Eliminate the constants

and

(

×

×

+

×

=

×

(

×

)

(5-44)

(

×

)

(5-45)

=

from Equation 5-38, we get:

+

)

×

1     3   (

=

)+

×

(

)+

×

(

×

(5-46)

)

Define the following vectors:

 1      3          =

(

×

=(

)+

×

(

×

+

×

)+

+

(

×

×

)

)

(5-47)

(5-48)

Note that both vectors, N and D, are functions only in the known three position vectors. From equation (5-46) we can write:

Note:



 ∴ℎ



µ  

h

=

=

(5-49)

(5-50)

is perpendicular to the orbit plane.

∴ 

‖‖

(5-51)

  

(5-52)

=

Exercise: show that

=

Where,

1

7 2


 1      3   ∴  ℎ      ∴         =

(

)+

(

)+

(

)

(5-53)

Recall Equation (5-35):

=

×

+

(5-54)

Substitute for w, q, and h

=

×

+

(5-55)

Despite the lengthy derivation of the method, the algorithm becomes very simple and short as follows:

Gibbs Algorithm: Given

   1  3 3  3 ∶     ,

,

Calculate

,

Verify:

Calculate

.

,

,

=0

, , ,





The computation of V along with the known r completes the orbit determination!

Bate, R., Mueller, D., & White, J. (1971). Fundamentals of Astrodynamics. New Work: Dover Publications. Curtis, H. (2005). Orbital Mechanics for Engineers (1 ed.). ButterworthHeinemann. Gurfil, P. (2006). Modern Astrodynamics. Academic Press. Pisacane, V. (2005). Fundamentals of Space Systems (2 ed.). Oxford University Press. Vallado, D. A. (2007). Fundamentals of Astrodynamcis and Applications. Microcosm.


P5-1

    1 Δ1   

Given the following position vectors of a satellite

= 0.51 + 0.93 + 0.4 DU, = 0.2 = 0.37 + 1.0 + 0.35 DU

where I,J,K are unit vectors of geocentric-equatorial coordinate system. Find the minimum energy orbit.

P5-2

    1 Δ1    Δ33   

Given the following position vectors of a satellite

= 0.51 + 0.93 + 0.4 DU, = 0.1 = 0.43 + 0.98 + 0.38 DU, = 0.1 = 0.37 + 1.0 + 0.35 DU



where I,J,K are unit vectors of geocentric-equatorial coordinate system. Find using the Gibbs Method

73

7 4


Orbit Perturbations

6

Absolute true, and mathematical time, in and of itself and of its own nature, without reference to anything external, flows uniformly… Absolute space, of its own nature, without reference to anything external, always remains homogeneous and immovable. Sir Isaac Newton (1643-1727)

Perturbations are movements away from a normal or undisturbed motion. It is common to view the universe as being regular and predictable. However, accurate observational data often shows and clarifies the irregularities of motion superimposed on the more regular motions of the celestial bodies. The actual motion will vary from the theoretical two-body path due to perturbations caused by other bodies (such as the Sun and Moon) and additional forces not considered in Keplerian motion (such as a non-spherical central body and drag). Perturbations can occur due to predictable or unpredictable effects. The usual perturbing forces we consider for the orbit problem are both random and predictable. The spherically symmetric gravitational field of the central body (e.g. Earth) is not the only force field acting on the spacecraft. Other forces are mainly due to asphericity (not being a perfect sphere) of the central body, atmospheric drag and lift, third body effects, solar-radiation pressure, thrust, magnetic fields, solid-Earth and ocean tides, Earth reradiation (albedo), relativistic effects, and others. Examples are atmospheric drag during reentry and the third body in the restricted three-body problem. Perturbating effects need to be considered to ensure that interplanetary missions hit their target and to accurately predict the orbit of the Moon about the Earth. Ignoring the effects of the central body's oblateness on any satellite keeps us from accurately predicting its position over a long time. These secondary effects, or perturbations, were not considered in the two-body problem formulation. To be considered, we may write the equation of motion of spacecraft as

r

= ∇U +

b

U = gravitatio nal potential b = force vector per unit mass of spacecraf t

(6-1)

75

7 6

CHAPTER 6│ORBIT PERTURBATIONS

Source

Table 6-1 Relative magnitudes of perturbing accelerations.

Acceleration (m/s²) 500 km

Geostationary orbit (35786 km)

Air drag*

6E-5 A/m

1.8E-13 A/m

Radiation pressure

4.7E-6 A/m

4.7E-6 A/m

Sun(mean)

5.6E-7

3.5E-6 A/m

Moon(mean)

1.2E-6

7.3E-6

Jupiter(max)

8.5E-12

5.2E-11

* Dependent on the level of solar activity A = spacecraft surface area or frontal area, m = spacecraft mass Perturbations are usually classified by astrodynamicists into secular perturbations and periodic perturbations. Secular perturbations cause steady divergence of orbital elements over time. Periodic perturbations cause sinusoidal variation of orbital elements over time. The main perturbations on orbits will be considered in further details.

Long term effects on orbital elements for various types of perturbations 9

short-term priodic

8

secular

7 n o i t 6 a ) i r s t i a v n t u 5 n y e r a m r t e i 4 l b e r l a a ( t i 3 b r

long-term periodic

O

2 1 0 0

1

2

3

4

5

Orbit Periods

Fig. 6-1. Secular and periodic perturbations.

6

7

8

CHAPTER 6│ ORBIT PERTURBATIONS

77

6.1 Atmospheric Drag Next to the oblateness of the Earth, atmospheric drag most strongly influences the motion of a satellite near Earth and thus cannot be neglected; in fact during the last few revolutions of the satellite's life, drag effects can be more dominant than those from the Earth's oblateness. Atmospheric acts with a force which may be decomposed as drag and lift. Drag can be simply expressed as

F D =

1 ρAC DV 2 2

ρ

= local atmospheric density

A

= spacecraft frontal area

C

= drag coefficient

D

V

(6-2) Successive orbits ∆V

= spacecraft velocity relative to the atmosphere

Original orbit

drag

For more distant satellites, third-body effects and solar-radiation pressure dominate more than oblateness and drag. Investigations of aerobraking and satellite tethers require accurate atmospheric models. Many references describe the effects and various analytical procedures to model the drag effect, but models are still incomplete.

Earth’s atmosphere

The study of drag usually is diveived into three areas: Fig. 6-2. Geocentric-equatorial system.

(1) determining orbits under the influence of drag (2) estimating a satellite's lifetime (3) determining physical properties of the upper atmosphere Lift is usually neglected unless high precision of orbit determination is needed. The dominant influence of drag is to shrink and circularize orbits leading eventually to re-entry. For a small change of orbit radius, the increment of orbit period due to atmospheric drag is approximated by

∆τ τ B

≈ −3πρ r / B = m/AC

D

= ballistic coefficient of the spacecraft

m = spacecraft mass

( 6-3 )

coordinate


7 8

6.2 Non-symmetric Earth’s Gravitational Field It is convenient to describe the Earth’s gravitational field using spherical harmonic expansion

   ∞  RE n  − 1 + ∑   J n P n0 cos Φ r   2 n =  µ   (r, Φ, Λ) =   n r   n  RE  ∑ ( ) C m S m P + Λ + Λ Φ cos sin cos     nm nm n0  r    m =1   U(r, Φ, Λ)

= gravitatio nal potential at radius r, latitude Φ and longitude Λ.

P mn

= Legendre polynomials

J ,C ,S n nm nm

= coefficien ts dependent on mass distribution

J

n

( 6-4 )

= zonal harmonic coefficien ts which are independen t of longitude.

This expression describes the gravitational attraction resulting from the irregular distribution of the Earth’s mass using a potential function. Table 6-2 Magnitude of the lowest order J, C, S coefficients.

J2

C21

S21

1082.6E-6

0

0

J3

C22

S22

-2.53E-6

1.57E-6

-0.90E-6

J4

C31

S31

-1.62E-6

2.19E-6

0.27E-6

J5

C32

S32

-0.23E-6

0.31E-6

-0.21E-6

J6

C33

S33

-0.54E-6

0.10E-6

0.20E-6

6.2.1 The J2 Perturbation

E W

Line of nodes

Fig. 6-3. Regression of line of nodes.

Zonal harmonics are defined by zeroth order (m=0), where the dependence of the potential on longitude vanishes and the field is symmetrical about the polar axis. These are simply bands of latitude. It is apparent that the term J2 is much dominating compared to others. This term represents the Earth’s polar flattening (or equatorial bulge). J2 is almost 1000 times larger than the next largest coefficient ( J3). Excess gravitational mass in the equator region affects spacecraft orbits in two ways: (1) Regression of the line of nodes, and (2) Precession of the line of apsides.


79

4 3 2

) y a d / ° (

1 0

˙ Ω

0

20

40

60

80

100

120

140

160

180

-1 100 x 200km, e = 0.128 100 x 180km, e = 0.09 100 x 160km, e = 0.037 100 x 140km, e = 0.023 100 x 120km, e = 0.009 100 x 100km, e = 0

-2 -3

Fig. 6-4. Regression of line of nodes. [Nodal regression in degree/day].

-4 -5

Inclination (°)

6.2.2 Regression of Line of Nodes

Line of apsides

The equatorial bulge produces a torque which rotates the angular momentum vector. For direct orbits (i < 90°), the orbit rotates westward (regression of line of nodes). Whereas for indirect orbits (i > 90°), the orbit rotates eastward. Nodal regression can be calculated from solution of the equation of motion of the satellite taking into account the J2 term in the earth’s potential function. The perturbed solution gives the regression of nodes per orbit as

3 J 2 R 2 E n t cos i + O( J 2 2 ) Ω = Ω0 − 2 2 p

( 6-5 )

We notice that orbits with zero and 180-degress inclination (i.e. equatorial orbits) will experience no regression of nodes. This is expected since such types of orbits have no defined nodes. Fig. 6-3. shows the variation of regression of nodes in degrees per day as a function of orbit inclination, eccentricity and altitude.

6.2.3 Precession of line of Apsides The excess mass at the equator over the mean mass tends to curve the orbit more rapidly at spacecraft crossings with the equator resulting in orbit rotation. At inclination i ≈ 63.4°, the precession is zero. This is the inclination of Molniya orbits.

ω

2   = ω 0 + 3 J 2 R2 E n  2 − 5 sin 2 i t + O ( J 22 )

2

p



2



( 6-6 )

Fig. 6-5. Precession of line of apsides.


8 0

100 x 100km, e = 0 4

100 x 200km, e = 0.03 100 x 140km, e = 0.065

3

100 x 160km, e = 0.128 100 x 180km, e = 0.183

) s / ° (

2

˙ ω

100 x 200km, e = 0.231

1

0 0

20

40

60

80

100

120

140

160

180

-1

Fig. 6-6. Daily apsidal degree/day.

regression

in -2

Inclination(°)

105°W

6.2.4 Triaxiality 165°E

15°W 0° Greenwich 75°E

Equator

Fig. 6-7. Triaxiality.

The terms representing longitudinal variation of the Earth’s gravitational field affects most geostationary (GEO) satellites at an altitude ~ 35,786km, since for non-synchronous orbits this effect will be averaged out. The term J22 = [(C 22)2 + (S22)2]1/2 has the greatest influence and it represents the slight ellipticity of the Earth’s equator with major axis along the line 15°W165°E.

 = k 2 sin 2( Λ − Λ ) Λ 0

rsd

( 6-7 )

A synchronous satellite will be stable at either Λ0 = 75°E or 105°W.

rs

6.3 Solar and Lunar Gravitation

β

rd

Disturbing body

Fig. 6-8. Solar and lunar gravitation.

IIncoming photons

Other bodies in the solar system impose additional gravitational force on spacecraft. The Moon provides the most significant influence due to its mass and proximity. The Sun provides an influence of the same order due to its enormous mass. Both perturbations are termed luni solar perturbations. Their most significant effect is to change the inclination ( i ) of orbit. Solution can be obtained by solving the restricted three-body problem.

6.4 Solar Radiation Pressure Specularly reflected photons

Fig. 6-9. Solar radiation pressure.

Like drag, solar-radiation pressure is a non-conservative perturbation, but it becomes more pronounced at somewhat higher altitudes. One of the more difficult aspects of analyzing solar radiation is accurately modeling and predicting the solar cycles and variations.


81

During periods of intense solar storms, this effect may be much larger than all the other perturbations (depending on the altitude); at times of low activity, the effect may be negligible. A spacecraft moving within the solar system will experience a perturbation owing to the incidence of solar radiation on its illuminated surface. To arrive at the solar-radiation pressure, begin with the intensity of the energy of the incoming radiation from the Sun. A solar-radiation constant—often called the intensity, irradiance or the solar radiation flux—is W ~ 1400 W/m2. Electromagnetic radiation (photons) carries momentum and the reflection of incident radiation on a surface represents exchange of momentum, thus exerting a tiny, but measurable, pressure on the surface which is given by

= W Θ / c P = solar radiation pressure P

c

( 6-8 )

= speed of light in vacuum ~ 3x108 km/s at Earth, P E ~ 4.7x10 -6 N/m 2

0 0 -1

) n o i t a r e l e c c a d e z i l a m r o n ( 0 1 g o l

500

1000

1500

drag

2000 primary gravity

-2 J2 -3

-4

-5

J3 J4

-6 J5 -7 solar gravity -8

-9

-10

Spacecraft altitude(km)

Fig. 6-10. Perturbing Accelerations on Earth’s Satellite vs. Altitude (normalized by standard acceleration of gravity at Earth’s surface).

8 2


6.5 Modeling Perturbations Modeling orbit perturbations is primarily important for astrodynamists. Perturbations may be classified due to their relative effect on orbital elements into secular and periodic perturbations.

6.5.1 Special Perturbation Techniques: Those dealing with the direct numerical integration of the equations of motion including all necessary perturbing accelerations.

6.5.2 General Perturbation Techniques: Those involving analytic integration of series expansions of the perturbing accelerations. These techniques are more difficult and lengthy than special techniques but lead to better understanding of the source of the perturbation

Bate, R., Mueller, D., & White, J. (1971). Fundamentals of Astrodynamics. New Work: Dover Publications. Curtis, H. (2005). Orbital Mechanics for Engineers (1 ed.). ButterworthHeinemann. Pisacane, V. (2005). Fundamentals of Space Systems (2 ed.). Oxford University Press. Vallado, D. A. (2007). Fundamentals of Astrodynamcis and Applications. Microcosm.


83

P6-1

Describe the regression of nodes perturbation. What is the main cause for this perturbation?

P6-2

Describe three distinct perturbations to Earth satellite orbits. Explain how each of them will affect the classical orbital elements.

P6-3

Write a MatLAB script (You may also use Simulink) to integrate the equations of motion of an Earth’s satellite. Assume the satellite is inserted into orbit with initial velocity at the point = = where i,j,k are the unit vectors of geocentric-equatorial coordinate system xyz. Plot the satellite trajectory in the geocentricequatorial coordinate axes and compare it to Keplerian orbit. Consider the motion of the satellite under the influence of a homogenous and spherically symmetric Earth which is described by the equation

 0

0



̈ ∇  3     ‖‖        0 ⁄ 03⁄     −6  =

⊕

=

where = , = + + = + + = 7.5 km s , = ⊕ + 1000 km , ⊕ = 398600.4415 km s , = 1802.627 × 10 , 6378.1363 km P6-4

,

⊕

=

Study the effect of considering non-spehericity of the Earth on the satellite orbit. Repeat problem P6-3 considering the motion of the satellite under the influence of a non-spherically symmetric Earth which is approximated by the equation

 ∇ ̈ −  3                                 =

where

=

+

=

+

=

=

=

and 3 ⊕ 5

2 3

⊕

5

2 3

⊕

5

2

⊕

( + )

1

,

1

,

3

8 4


7

Mission Analysis

I can The diversity of the phenomena of nature is so great, and the treasures hidden in the heavens so rich, precisely on order that the human mind shall never be lacking in fresh nourishment. Johannes Kepler (1571-1630)

The utmost goal of learning space mechanics for an aerospace engineer is to be able to analyze and design a space mission; whether this mission is to send a communication satellite, a weather satellite, a navigation satellite, or even scientific space probe to help us learn more about the environment and the outer space. There are many aspects of mission analysis which cannot be fully covered in single chapter. However, we are going to introduce here the main concepts and outline the many of the practical aspects of mission analysis.

7.1 Spacecraft Coverage Spacecraft coverage is so important for orbit selection (design) for a variety of space missions including communication, navigation, weather, surveillance, rescue, Earth resource monitoring, and space environment monitoring. Many criteria may be used to specify coverage such as rate of sampling the Earth, size of coverage area, time intervals between revisits of specific locations, duration of satellite visibility from some location, and the fraction of a day that specific locations will be covered (within the coverage area). As can be seen, spacecraft coverage specification greatly depends on the mission details. In the following analysis, Earth will be assumed a perfect sphere (the maximum difference between Earth’s radii ~ 0.003 of mean radius).

7.1.1 Ground Station Visibility At any instant in time, the point on a ground track is defined as the point of intersection between the surface of the Earth and the line connecting the Earth center and the satellite. This point is called the sub-satellite point 85

CHAPTER 7│MISSION ANALYSIS

8 6

(SSP).A satellite is visible at all points on the Earth’s surface within a circle centered on the SSP, whose diameter increases with altitude. Signals from satellites at the horizon limit are attenuated by the atmosphere, so for practical purposes the surface coverage is restricted to the region in which satellite elevation is greater than ~ 5°. The geocentric semi-angle over which the satellite is visible can be calculated from

 R  φ = −ε + cos−1 E cos ε   R E + h 

S

Earth

ε

φ E

s

( 7-1 )

The slant range, range from ground station to satellite, is given by

G

s

= ( R E + h) sin φ / cos ε

( 7-2 )

The duration of over-head pass is given by Fig. 7-1. Geometry visibility.

τ = 2φ / ωES for

ground

station

( 7-3 ) where

ω2 ES

=ω

2

+ ω − 2 ωE ωcos i 2

E

7.1.2 Elevation and Azimuth of Satellite Elevation and azimuth are required to point an antenna or telescope to a satellite. Elevation (El) is the angle from the horizon of the observer in the plane containing the satellite, observer, and the center of Earth. Azimuth (Az) is angle measured in the horizontal plane of the observer from local north to the intersection with the plane containing the satellite, observer, and the center of Earth. The geographic longitude and latitude of the subsatellite point can be calculated from

ωes ωE ω

φ s

= sin −1[sin i sin(ω + µ )]

λ s

 cos i sin(ω + ν )   + C = Ω − λ g + tan − 1   ω ν cos( ) +  

i

Fig. 7-2. The angular velocity of the satellite with respect to Earth can be found from the vector subtraction of the satellite inertial angular velocity from the Earth’s inertial angular velocity about its axis.

where n = true anomaly i = inclinatio n f = latitude of subsateli te point l s = longitude of subsatell ite point ω W lg

= argument of peria psis = longitude of ascending node = hour angle of Greenwich

C

= 0 if cos( ω + n) ≥ 0 = p if cos( ω + n) < 0 and cos i sin( ω + n) > 0 = p if cos( ω + n) < 0 and cos i sin( ω + n) < 0

( 7-4 )

CHAPTER 7│ MISSION ANALYSIS

87

The central angle or great circle angle between the observer and subsatellite point is given by

cos α









= cos π − φ 0  cos  π − φ s  +

 2  π sin   2

 2   π φ 0  sin    2 

−



−

 φ s  cos(λ s 

( 7-5 )

− λ 0 )

Then, the elevation and azimuth will be

 cos α − RE   r  El = tan − 1   sin α      sin φs − sin φ0 cos α   Az = ± cos − 1    cos φ sin α 0  

( 7-6 )

Where r = h + RE

7.2 Ground Track 7.2.1 Ground Track on a Non-rotating Earth The orbit of Earth’ satellite lies in a plane passing through the center of Earth. The track of this plane on the surface of a non-rotating spherical Earth will become a great circle. If the Earth did not rotate the satellite would retrace the same ground track over and over. On a Mercator projection, the ground track would look like as shown:

Fig. 7-3. Ground track on a non-rotating Earth.

The maximum latitude north or south of the equator that the satellite passes over is just equal to the orbit inclination of orbit (i ), if the orbit is direct, and to 180 – i , if the orbit is retrograde.

7.2.2 Effect of Earth Rotation on Ground Track The orbital plane of a satellite remains fixed in space while the Earth turns under the orbit. The net effect of Earth rotation is to displace the ground track westward on each successive revolution of the satellite by the number of degrees the earth turns during one orbital period. Instead of retracing the same ground track over and over, a satellite eventually covers a swath around the Earth between latitudes north and south of the equator equal to the inclination.

Fig. 7-4. Effect of Earth rotation on ground track.


8 8

A global surveillance satellite would have to be in a polar orbit to fly over the Earth’s entire surface. If the time required for one complete rotation of the earth on its axis (23 hr 56 m) is an exact multiple of the satellite’s period then eventually the satellite will retrace exactly the same path over the earth as it did on its initial revolution (repeating orbit ). This is a desirable property for a reconnaissance satellite where you wish to have it fly over a specific target once each day.

7.3 Effect of Launch on Orbit Inclination We can determine the effect of launch site latitude and launch azimuth on orbit inclination from spherical geometry. Suppose a satellite is launched from point C on the earth whose latitude and longitude are ΦL and λL, respectively with a launch azimuth, AzL. Since we know two angles and the included side of the spherical triangle ABC we can solve for the third angle i as

cos i = sin AzL cos Φ L

From the above equation, we notice that for a direct orbit (0 < i <90°), cos i is positive. Since -90° < AzL < 90°, for any launch site, then cos ΦL is always positive. We also notice that a direct orbit requires, that the launch azimuth, AzL, be easterly (0 < AzL < 180°). To find the minimum orbital inclination that can be achieved from a launch site at ΦL, cos i must be maximized which implies that AzL should be 90°.

β

C A

Fig. 7-5. Launch site inclination.

i

L0

effect

B

on

( 7-7 )

orbit

For a due east launch, the orbital inclination will be the minimum possible from a launch site at latitude, ΦL and i will be exactly equal to L0. A satellite cannot be put directly into an equatorial orbit (i = 0°) from a launch site which is not on the equator. The previous Soviet Union had no launch site closer than 45° to the equator. So, it cannot launch a satellite whose inclination is less than 45°. To insert a satellite into an equatorial orbit, we should perform a plane change maneuver of at least 45° (a maneuver which requires a lot of fuel) after the satellite is inserted into orbit. For example, The previous Soviet Union had no launch site closer than 45° to the equator. So, it cannot launch a satellite whose inclination is less than 45°. To insert a satellite into an equatorial orbit, a plane change maneuver of at least 45° should be performed (which requires a lot of fuel) after the satellite is inserted into orbit.

7.4 Special Earth Orbits For analysis of spacecraft mission it is convenient to arrange these missions into the following categories such as: Low Earth Orbits (LEO), Medium-Height Earth Orbits (MEO), Geostationary Earth Orbits (GEO), Highly Elliptic Orbits (HEO), Non-geocentric Orbits (lunar and interplanetary). Some Earth’s satellite orbits are commonly used to perform specific missions. We will consider the following Earth’s satellite orbits:


1. 2. 3. 4. 5.

89

Polar Orbit Geo-synchronous Orbit Molniya Orbit Sun-synchronous Orbit Repeating Ground Track Orbit

7.4.1 Polar Orbits Polar orbits (i = 90º) have no regression of nodes. So, a polar orbit will remain fixed in inertial space.

Ω = Ω0 −

3 J 2 R 2 E 2

p

2

nt cos i + O( J 22 )

( 7-8 )

Polar orbits provide global coverage (from pole to pole). A near-polar orbit at 1000 km and with (68.4°< i <111.6°) provides global coverage with min elevation (εmin = 10°) at poles and better coverage at higher altitudes. Therefore, polar LEO is usually used for Earth’s remote sensing and surveillance. The altitude of satellite orbit is determined by trade-off between instrument resolution (the lower, the better) and fuel required to maintain orbit (the higher, the better). Civil remote sensing satellites are typically placed between 600 – 900 km. Military surveillance missions usually have more complex mission scenarios. In addition, the polar orbit may also be Geosynchronous, Sun-synchronous or sometimes both.

Fig. 7-6. Ground track and swath width of a polar satellite (i = 90°, h = 1000 km, Ω0 = 104.5°, elevation = 10°).

Fig. 7-7. Ground track and swath width of a near-polar satellite (i = 68.4°, h = 1000 km, Ω0 = 104.5°, elevation = 10°).


9 0

7.4.2 Earth-Synchronous Orbits Earth-synchronous orbit results when the sub-satellite point follows a ground track identical to some previous orbit after a certain period of time such that payload requirement to revisit ground target sites or regions can be satisfied. Repetition occurs on a regular basis and can be achieved in many ways. The drift of the sub-satellite point on the equator between two successive orbits (∆φ) is determined by two factors: (1) The Earth’s rotation

∆φ E = −2π

Fig. 7-8. Earth-synchronous orbit.

τ

rad/orbit

τ

(7-9)

E

(2) The regression of line of nodes

∆φ R = −

3 π J R 2 E cos i 2

2

2 2

rad/orbit

(7-10)

a (1 − e )

So, ( i ≠ 0)

∆φ = ∆φ E + ∆φ R rad/orbit

(7-11)

For an Earth-synchronous orbit, we need

n ∆φ

= m 2π

where n and m are integers

(7-12)

7.4.3 Geosynchronous Orbits (e ≠ 0)

A geosynchronous orbit is special case of Earth-synchronous orbit, in which

Fig. 7-9. Geosynchronous orbit.

the satellite period is identical to the sidereal period of the Earth. This can be achieved if the semimajor axis of the orbit is equal to 42,156 km. It has the advantage of large coverage area (~ 42% of Earth's surface). If eccentricity is non-zero, the satellite appears to oscillate in longitude. If inclination is non-zero, the satellite traces out a figure eight.

7.4.4 Geostationary Orbits Fig. 7-10 A geostationary satellite.

A geostationary orbit is special case of geo-synchronous orbit, in which the orbit has zero eccentricity and zero inclination (i.e. equatorial circular orbit). A geostationary satellite can provide fixed-point communication. At least, three geostationary satellites are required to provide global communication network.


91

7.4.5 Molniya Orbit

 ∆

Orbits with inclinations precision of apsides (

ω - ω 0

−1

= sin = 0).

=0=

(±4/5) = 63.4°, 116.6° have zero

3 J 2 R 2

p

2 E

2

 n 2  

−

5 2

sin

2

 i t  

(7-13)

The Russian Molniya satellite takes advantage of this property to maintain apogee over the northern hemisphere. The orbit is a highly elliptic (HEO), with e ≈ 0.75 and period of one-half sidereal day ≈ 11h 58m. It allows long observation time (up to 8 hrs) when satellite is in around apogee. Communication can be provided to high latitude regions inaccessible from geostationary satellites. Three satellites are required to provide 24-hr communication.

7.4.1 Sun-Synchronous Orbit Sun-synchronous orbit results when the orbit rotates in space at the same rate as the Earth’s rotation around the Sun (one revolution per year),that is

∆φ s = 2 πτ / τ ES

rad/orbit

Fig. 7-11 Molniya orbit in space.

(7-14)

Fig. 7-12. Ground track of a Molniya satellite ( i = 63.4°, h p = 1000 km, Ω0 = -0.7174°, e = 0.722)

Sun

Sun

Dawn/dusk orbit Mid-afternoon/night Noon/midnight -along terminator orbit orbit (short shadows)

Fig. 7-13.

Illustration of synchronous orbit.

Sun-

9 2


Sun-synchronism is needed for some remote-sensing missions in which interest is focused on a particular geographical region. The satellite will pass over these particular regions approximately at the same local solar times each day (once in daylight and once at night). Free sun-synchronization can be achieved using regression of line of nodes.

Bate, R., Mueller, D., & White, J. (1971). Fundamentals of Astrodynamics. New Work: Dover Publications. Curtis, H. (2005). Orbital Mechanics for Engineers (1 ed.). ButterworthHeinemann. Gurfil, P. (2006). Modern Astrodynamics. Academic Press. Pisacane, V. (2005). Fundamentals of Space Systems (2 ed.). Oxford University Press. Unsöld, A., & Baschek, B. (2001). The New Cosmos: An Introduction to Astronomy and Astrophysics. Berlin Springler. Vallado, D. A. (2007). Fundamentals of Astrodynamcis and Applications. Microcosm. Walter, U. (2008). Astronautics: The Physics of Space Flight (2 ed.). Wiley-VCH.


P7-1

What are polar orbits? What are the missions they are more suitable for?

P7-2

What is a sun-synchronous orbit? How can the orbit of an Earth satellite, which does not have an on-board propulsion system, be made sun-synchronous?

P7-3

An Earth satellite has the following orbital parameters: period = 180 min, eccentricity = 0.005, inclination = 10 º, argument of perigee = 90 º, right ascension (longitude) of ascending node = 100 º. Determine the geographic longitude and latitude of the subsatellite point when the true anomaly of the satellite = 180 º and the hour angle of Greenwich = 10,000 s sidereal time (GMST).

P7-4

Find the geographic longitude and latitude of the subsatellite point of the satellite in P7-3 after 2 hours. If you want to use a telescope to observe the satellite, where shall you point your telescope (i.e. what is the azimuth and elevation of the satellite)? Will you be able to observe the satellite? [Hint: You need to update the GMST after the 12 hours passage. Use eccentric anomaly to convert between time and true anomaly].

P7-5

An Earth satellite has a perigee altitude = 500 km and eccentricity = 0.74. The satellite has been detected by a ground station to be at a true anomaly of 20 º. (a) Find the mean motion, n of the satellite in rad/s. (b) What is the time taken by the satellite since it last passed through perigee? (c) Find the satellite position (true anomaly) after 2:00:00 hours.

P7-6

The two line element (TLE) data of the international space station (ISS) retrieved from (www.celestrak.com/NORAD/elements) at some past epoch are as follows:

12345678901234567890123456789012345678901234567890123456789012345678 1 25544U 98067A 10063. 37915313 . 00014094 00000- 0 10260- 3 0 6988 2 25544 51. 6465 48. 6647 0007661 11. 8053 136. 8643 15. 73611041646963

Find the classical orbital elements of ISS at epoch from the TLE data. Use the Appendix for explanation of the TLE designations), specifically find the following: eccentricity (e), inclination (i ), argument of periapsis (ω), longitude of ascending node (Ω), mean motion in rad/s (n), semi-major axis (a).

93

9 4


P7-7

At the given epoch (as indicated by the TLE data given in P7-6), calculate the following: (a) The time past since last perifocal passage ( ) of the ISS. (b) The true anomaly at epoch ( ) (c) The Greenwich mean sidereal time (GMST) at epoch (d) The geographic longitude and latitude of the ISS subsatellite point at epoch (e) The elevation and azimuth of the ISS with respect to a ground station at 8°N, 101°42’E. (f) Assuming the ISS orbit is a Keplerian orbit; find the geographic longitude and latitude of the subsatellite point 30 minutes after epoch.

0

0 

8

Orbital Maneuvers

I Philosophy is such an impertinently litigious lady that a man had as good be engaged in lawsuits as to have to do with her. Isaac Newton in a letter to his friend Edmund Halley, June 20, 1687

Spacecraft is not inserted in an orbit to stay forever! A spacecraft may need to change its orbit once or more during its life time due to many reasons. A launch vehicle may insert a geostationary (GEO) satellite into an initial low Earth orbit (LEO) which is much lower than the final operational orbit. Then, the satellite should transfer from the initial orbit to its final orbit. Another need may arise if a surveillance satellite has to change its orbit in order to track a new target. Interplanetary missions usually require many orbit transfers until the spacecraft is inserted into the operational orbit or to use the same spacecraft to accomplish more than one mission. At the satellite end of life (EOF), the satellite may be kicked out of its orbit whether to reenter the Earth’s atmosphere or to rest in a graveyard orbit.

8.1 Basics of Orbital Maneuver 8.1.1 Orbital Energy Any analysis of orbital maneuvers, i.e., the transfer of a satellite from one orbit to another by means of a change in velocity, logically begins with the energy as

     = (

2

1

)

(8-1)

95

CHAPTER 8│ORBITAL MANEUVERS

9 6



Where V is the magnitude of the orbital velocity at some point, r the magnitude of the radius from the focus to that point, the semimajor axis of the orbit, and μ the gravitational constant of the attracting body. Equation (91) can be rearranged as

V

µ

2

r

=

µ

(8-2)

2a

Where it is evident that

2nd burn

Kinetic Energy

1st burn

Satellite Mass

+

Potential Energy Satellite Mass

=

Total Energy Satellite Mass

Note that total energy/satellite mass is dependent only on a. As a increases, energy increases.

8.2 Principles of Orbital Maneuvers Fig. 8-1. Satellite launch.

V

∆V

+

V

=

Impulse

(1)

(2)

Orbital maneuvers are based on the principle that an orbit is uniquely determined by the position and velocity vectors at any point. Conversely, changing the velocity vector at any point instantly transforms the trajectory to a new one corresponding to the new velocity vector. Any conic orbit can be transformed into another conic orbit by changing the spacecraft velocity vector instantaneously.

(3)

8.2.1 Satellite Launch Fig. 8-2. Delta-V Budget.

v e

V

High-altitudes (above 200 km) may be achieved through two burns separated by coasting phase. The first burn is nearly vertical and places the satellite into an elliptic orbit with apogee at the final orbit radius. The satellite then coast (no burn) until it reaches the apogee. A second burn can be used to insert the satellite into its final LEO orbit.

8.2.2 The Delta–V Budget

Fig. 8-3. Impulsive thrust produced based on rocket theory.

Orbital transfers are usually achieved using the propulsion system onboard the spacecraft. Since the propellant mass on board is limited, it is very crucial for mission planning to estimate the propellant required for every transfer. The overall need for propulsion is usually expressed in terms of spacecraft total velocity change, or DV (Delta-V) budget. We assume the propulsion is applied impulsively , i.e. the velocity change will be acquired instantaneously. This assumption is reasonably valid for high-thrust propulsion. From rocket theory, shown in Fig. 8-3. we can express the force produced by the impulsive thrust as:

CHAPTER 8│ ORBITAL MANEUVERS

97

  ̇  ̇0             ∆       =

=

=

=

=

=



ln

=

ln

(8-3)

(8-4)

(8-5)

(8-6)

where =specific impulse = thrust/rate of fuel consumption. The spacecraft‘s initial and final mass, and the propellant mass are:

 0

= spacecraft initial mass = spacecraft final mass =propellant mass used =9.81m/s²

∆ 0     ∆0     ∆0 =

=

exp

ln(1 +

1 =

)

1

(8-7)

exp

8.3 Coplanar Maneuvers Orbit maneuver had its roots in the classical formulas and dynamics of Astrodynamics from several centuries ago. However, the application of orbit maneuver did not occur until after the launch of Sputnik in 1957. Orbit maneuver is based on the fundamental principle that an orbit is uniquely determined by the position and velocity vectors at any point. Therefore, changing the velocity vector at any point instantly transforms the trajectory to correspond to the new velocity vector. Thus, the orbit of a satellite is changed. Coplanar maneuver only involves the change of semimajor axis and eccentricity of the orbit without changing the orbit plane. In this section, four kind of coplanar maneuvers are introduced: 1)

Tangential-Orbit Maneuver,

2)

Non-tangential Orbit Maneuver,

3)

Hohmann Transfer,

4)

Bielliptic Orbit Transfer.

Fig. 8-4. Single coplanar maneuver.


9 8

8.3.1 Tangential-Orbit Maneuver Tangential-orbit maneuver occurs at the point where the velocity vector of spacecraft is tangent to its position vector, typically at perigee point.

EXAMPLE 8-1 Determine the ∆V required to transfer from a circular orbit into elliptic orbit.

SOLUTION

∆V

V1

α

V2

The ∆V between two orbits can be shown as follow: V circ

V

V p

        ∆       =

,

=

2

Error! Reference source not found. 8-4

shows a typical tangential orbit maneuver at perigee point. Using equation 9-5, the ∆V required is,

Fig. 8-5. Basic orbital maneuver.

2

=

8.3.2 Non-Tangential Coplanar Maneuver ∆V1

∆V2

r2 Fig. 8-6. Hohmann transfer.

r1

The orbit maneuver is not limited to apogee and perigee points. If needed, the satellite may perform the orbit maneuver at any point. When the spacecraft performs a burn from one orbit to another, the burn has to be performed at the intersection point of the old and the new orbits. Fig. 8-5 shows the ∆V vector required for a non-tangential orbit maneuver, where α is the angle difference between the flight path angles before and after the maneuver, at the intersection point.

∆  α1 ∅ ∅ 1  α 1  V=

v +v

2v v cos

(8-8)

=

8.3.3 Hohmann Transfer The Hohmann’s transfer is the minimum cost two-impulse transfer between coplanar circular orbits. It can be used to transfer a satellite between two nonintersecting orbits (Walters Hohmann 1925). The Hohmann transfer employs an intermediate elliptic orbit that is tangent to both initial and final


99

orbits at their apsides. To accomplish the transfer, two burns are needed. The first burn will insert the spacecraft into the transfer orbit, where it will coast from periapsis to apoapsis. At apoapsis, the second burn is applied to insert the spacecraft into final orbit. Fig. 8-6 represents a Hohmann transfer from a circular orbit into another circular orbit. A tangential ΔV1 is applied to the circular orbit velocity. The magnitude of ΔV1 is determined by the requirement that the apogee radius of the resulting transfer ellipse must equal the radius of the final circular orbit. When the satellite reaches apogee of the transfer orbit, another ΔV must be added or the satellite will remain in the transfer ellipse. This ΔV is the difference between the apogee velocity on the transfer orbit and the circular orbit velocity in the final orbit. The first impulse magnitude can be calculated as:

∆ 1 p t  1  1    1 V =V

pt

V =

,

2μ

μ

r

a

μ

r

(8-9)

Where, V , is the velocity of the spacecraft on the transfer orbit at perigee. The second impulse magnitude can be calculated as:

∆    a t  µ  µ  µ V =V

V

,

=

2

r

r

a

(8-10)

Where,

at

V , is the velocity of the spacecraft on the transfer orbit at apogee. The time of flight on the transfer orbit is:

3        1    =

1 2

=

(8-11)

The perigee and apogee radii of the transfer orbit are:

,

=

,

,

=

(8-12)

EXAMPLE 8-2 Determine the total ∆V required for Hohmann transfer to transfer from a LEO with hinitial = 191 km into GEO.

SOLUTION The initial and final radii are: rinitial = 191 + 6378.145 = 6569.145 km

1 0 0


rfinal = 42164.215 km At the first impulse, the delta-v required is:

∆ 1  in2µitial  µ   inµitial V =

where,

∆1

a=

r

r

a

initial final +r

2

r

= 24366.796 km

Thus, V = 2.457 km/sec For the second impulse, the delta-v required is:

∆   µfinal  f2µinal  µ V =

∆ TOTAL

r

r

a

= 1.478 km/sec

The total delta-v require is,

V

= 2.457 + 1.478 = 3.935 km/sec

EXAMPLE 8-3 Two geocentric elliptical orbits have common apse lines and their perigees are on the same side of the Earth. The first orbit has a perigee radius of = = km and = . , whereas for the second orbit km and = . . a. Find the minimum total delta-v and the time of flight for a t ransfer

      

 

from the perigee of the inner orbit to the apogee of the outer orbit. b. Do part (a) for a transfer from the apogee of the inner orbit to the perigee of the outer orbit.

SOLUTION a.

1    1 1 1−        + ⇒              

The initial orbit:

(1 +

=

=

) = 9100 km

= 10000 km

=

The final orbit:

= 48000 km

= 8.6038 km/sec

= 64000 km

&

=

=

= 96000 km

= 1.4408 km/sec

The transient orbit:

=

= 7000 km &

=

= 96000 km


101

∴             △△1      ∴△      5  

= 51500 km

=

= 10.3027 km/sec

=

= 0.7512 km/sec

=

= 1.6989 km/sec

=

= 0.6896 km/sec

= 2.3885 km/sec

=

=

Time of flight:

=

−5 −1 5

= 5.402 × 10

sec

= 1.1631 × 10 sec

= 2.3262 × 10 sec = 16.1544 hr

b.

 1  1             ∴              △△1      ∴△    

For initial orbit:

=

(1 +

) = 13000 km

=

= 4.6328 km/sec

=

= 4.3225 km/sec

For final orbit:

For transient orbit:

=

= 13000 km &

=

= 32000 km

= 22500 km

=

= 6.6036 km/sec

=

= 2.6827 km/sec

=

= 1.9708 km/sec

=

= 1.6398 km/sec

= 3.6106 km/sec = 3.3588 × 10 sec

Time of flight:

= 4.665 hr


1 0 2

EXAMPLE 8-4 A spacecraft is in a 300 km circular earth orbit. Calculate the transfer time for a Hohmann transfer to a 3000 km coplanar circular Earth orbit.

SOLUTION

1 3

For initial orbit, 1:

= 6678.145 km

For final orbit, 3:

= 9378.145 km For elliptical transient orbit, 2: = = 6678.145 km

   +      ∴   =

= 7.7258 km/sec = 6.5194 km/sec

= 9378.145 km = 8028.1 km

=

Transfer orbit time

=

⇒ 1 ⇒ 3

− −1

= 8.7771 × 10

=

sec

= 0.9943 hr

8.3.4 Bi-elliptic Transfer The bi-elliptic transfer requires a total of three impulses with two transfer orbits. The first burn inserts the spacecraft into first transfer orbit at its periapsis. When the spacecraft coasts to the apoapsis of the first transfer orbit, the second impulse is fired to insert the spacecraft into second transfer orbit. The spacecraft orbits along the second transfer orbit to its apoapsis point. Finally, a third impulse inserts the spacecraft into the destination orbit. Fig. 8-7 illustrates a bi-elliptic transfer between two c ircular orbits. ∆V1 ∆V2

Fig. 8-7. Bielliptic Transfer.

∆V3

The bi-elliptic transfer requires much longer transfer time compared to the Hohmann Transfer. However, bi-elliptic transfers are more efficient in terms of total delta-V needed for the mission, in some cases. Fig. 8-8 shows the cost comparison between Hohmann and Bi-elliptic Transfers. R is the ratio of final to initial radii for both orbits, where R* is the ratio of apogee radius of transfer orbit to initial orbit in bi-elliptic orbit. For R < 11.94, Hohmann transfer is more cost effective than the bi-elliptic transfer. For R > 15.58, the bi-elliptic transfer is more cost effective than the Hohmann transfer.


Bi-elliptic

103

Hohmann

) c e s / m k ( , V

R* Increases R* ≈ 50 R* ≈ 60

∆

f o t s o C l a t o T

R* ≈ 100 R* ≈ 200 R* = ∞

R* ≈ 11.94

R* ≈ 15.58

Final radius to initial radius ratio, R

EXAMPLE 8-5 Determine the total ∆V required and time of flight for a bi-elliptic transfer with given orbit properties: Initial orbit, hinitial = 191.344 km Apogee altitude of transfer orbit, h apog = 503873 km Final orbit, h final = 376310 km

SOLUTION The initial, transfer orbit apogee and final radius are, rinitial = 191.344 + 6378.145 = 6569.489 km rtrans = 503873 + 6378.145 = 510251.145 km rfinal = 376310 + 6378.145 = 382688.145 km And the semimajor axis for both transfer orbits are, r +r a = = 258410.317 km 2 r +r a = = 446469.645 km 2 At first impulse, the delta-v required is,

1 initial trans  final trans

∆ 1  in2µitial  µ1   inµitial ∆   trans2µ  µ  trans2µ  µ1 V =

r

a

= 3.156 km/sec

r

At the second impulse, the delta-v required is,

V =

r

a

r

At the third impulse, the delta-v required is,

a

= 0.677 km/sec

Fig. 8-8. Delta-v cost comparison between for Hohmann and bieleptic orbit transfers.

1 0 4


∆ 3  µfinal  f2µinal  µ  V =

r

r

a

∆ TOTAL 3 3 1   π  µ µ

=

0.0705 km/sec

The total delta-v require is,

V

= 3.156 + 0.677 + 0.0705 = 3.9035 km/sec

The time of flight is,

TOF =

a

×

+

a

= 2138113.26 sec = 593.92 hr

EXAMPLE 8-6 A spacecraft is in a 300 km circular Earth orbit. Calculate a. The total delta-v required for the bi-elliptical transfer to a 3000 km altitude coplanar circular orbit shown, and b. Compare the total transfer time with the Hohmann’s transfer time in Example 9-4.

SOLUTION a. For initial orbit, 1:

1 

r = 6678.145 km For final orbit, 4: r = 9378.145 km For first elliptical transient orbit, 2:

⇒1 ⇒

v = 7.7258 km/sec v = 6.5194 km/sec

1 

=0.3

 r  1−  B      r  a   r  a r

= r = 6678.145 km = 9540.21 km

a =

= r = a (1 +

r

v

=

v

=

μ



μ

) = 12402.26 km

= 8.8087 km/sec

= 4.74316 km/sec

For second elliptical transient orbit, 3:


3  3 r+rB  3  3  r  a 3  r  a r

= r = 9378.145 km

r

= r = 12402.26 km

a =

v

105

=

μ

μ

= 10890.2 km = 6.9573 km/sec

 = 5.26088 km/sec vA = v v1 = 1.0829 km/sec vB = v3 v = 0.51772 km/sec vC = v3 v = 0.43793 km/sec

v

=

△△  △  ∴△ △A △B △C   3 3     3    ∴  =|

|+|

| = 2.03855 km/sec

|+|

b. Total transfer time,

=

where,

2

+

2

=

= 9.2736 × 10 sec

=

= 1.131 × 10 sec

= 1.02918 × 10 sec = 2.859 hr From Example 9-4, Hohmann Transfer only requires 0.9943 hr to transfer the spacecraft into another orbit. However, the bi-elliptic requires 3 times longer of transfer time to transfer the spacecraft into another orbit.

8.3.5 General Coplanar Transfer between Circular Orbits In the Hohamnn and Bi-elliptic transfers, the first and second impulses are applied at either the periapsis or apoapsis; however, this is not a requirement. For a general coplanar orbit transfer, the periapsis radius of the transfer orbit must be equal to or less than the radius of the inner orbit; and the apoapsis radius of the transfer orbit must be equal to or exceed the radius of the outer orbit. This condition is illustrated in Fig. 8-9. . Transfer Orbit

Transfer Orbit

r2

r1

Possible because rp < r 1 and ra > r2

r2

r1

Impossible because rp > r1

Transfer Orbit

r2

r1

Impossible because ra < r2

Fig. 8-9. General coplanar transfer between circular orbits.

1 0 6


The amount of the first and second impulses in this case are not optimal, in general. To calculate the amount of Delta-V needed, first we compute the points of the first and second impulses. At each point, we compute the velocity of the spacecraft before and after applying the impulse. The difference is the required Delta-V at that point.

8.3.6 Phasing Maneuver Coplanar maneuvers usually involve a change in orbit size and shape. However, in some cases, a spacecraft is required to change its position in the orbit, but not the orbit. One example is the case of two spacecraft, in the same orbit, in a rendezvous mission. The interceptor spacecraft is required to intercept (or meet) the target spacecraft, which could be behind or ahead of the intercept spacecraft, in the same orbit. Fig. 8-10 shows an illustration of the phasing maneuver. The maneuvering spacecraft first transfers to a phasing orbit, stays there for one (or more) orbit periods, and then transfers back to its original orbit. If the interceptor is behind the target spacecraft, then the phasing orbit is smaller than the original orbit, and vice versa. Given that the phase angle (the difference in true anomaly) between the two spacecraft is ∆θ. Then, the period of the phasing orbit is:

τphase 2π∆θ tgt =

tgt

(8-13)

n

Where n n is the mean motion of the target spacecraft (the original orbit). The semimajor axis for the phasing orbit is then:

3 phase phase  phase µ 3 phase  phase√  =

τ

a

2π

n

=

= 2π

τ

Original Orbit

a

μ

/

2π

Target

Phasing Orbit ∆θ ∆V

Interceptor Phasing Orbit

Fig. 8-10. Phasing orbit maneuver.

(8-14)


107

EXAMPLE 8-7 Determine the semimajor axis of the phasing orbit, given that the position of target and interceptor spacecraft are:

⃗⃗  √    =

=

+

SOLUTION

⃗ ⃗  ∙   −1 ∆θ ⃗‖⃗‖ µ  tgt 3 τphase 2π∆θ tgt 3 √ phase phase  

First, the phase angle between spacecraft is,

= cos

= 60°

The mean motion of the original orbit is,

n

=

= 0.125 rad/TU r Then, the one orbit period required for the phasing orbit is, =

Fig. 8-11. Simple plane change.

= 41.888 TU

n

The semimajor axis for the phasing orbit is,

a

=

τ

μ

/

2π

= 3.5422 DU

8.4 Out-of-Plane Orbit Maneuvers A velocity change which lies in the plane of the orbit can change its size or shape, or rotate the line of apsides. To change the orientation of the orbit plane in space, the ∆V impulse-vector inserted to the spacecraft should not parallel to the spacecraft velocity vector.

∆V

V θ

8.4.1 Simple Plane Change V

Orbital maneuvers are characterized by a change in orbital velocity. If a velocity vector increment, ΔV that is perpendicular to a satellite velocity vector, V1 is added, then its results a new satellite velocity vector, V 2. The perpendicular ΔV does not change the speed and flight-path angle of the satellite, but only the inclination of the orbit. The maneuver is called simple plane change (see Fig. 8-11). For a circular orbit spacecraft that performs the simple plane change through an angle θ, the semimajor axis, a and eccentricity, e remain unchanged. Thus, the velocity of spacecraft at before and after the plane change are equal,

V 1

= V 2 . Using the velocity vector triangle illustration in Fig.

required is,

∆   = 2 sin

2

8-12., the delta-v (8-15)

Fig. 8-12. Velocity vector triangle for circular orbit plane change.

1 0 8


EXAMPLE 8-8 Determine the ∆V required for a satellite to change its orbit plane from inclination 10° to inclination 25° at altitude 600km.

SOLUTION The radius of the orbit is, r = 600 + 6378.145 = 6978.145 km

∆    

The delta-v required for the plane change i s, μ 25 10 V=2× × sin = 1.973 km/sec r 2

8.4.2 General Plane Change Maneuver In general, plane change maneuver involves the inclination and RAAN change while the size and shape of orbit remain the same. The change of the RAAN in plane change maneuver results that both orbit do not intersect at the original RAAN location. Fig. 8-13 shows the example of general plane change maneuver. Nodes 1 and 2 are the direction of RAAN for both initial and final orbits respectively, where else the e is the eccentricity vector (also known as argument of perigee) for the initial orbit.



The delta-v required for the general plane change maneuver is given by equation Fig. 8-13 . ALa is the argument of latitude of intersection point.

∆    ∆Ω

α   ΔΩ La α  θω

V = 2Vsin

cos = cos

cos

A Z

+ sin

)=

sin(

La

2

sin

sin A sin

=

Final Orbit

+

X Ωinitial

ω

Fig. 8-13. General Plane Change Maneuver.

node 1

ˆ

e

)

(8-17) (8-18)

sin

Initial Orbit

θ

cos(

(8-16)

node 2

(8-19)


109

Z Final Orbit Initial Orbit

α ALa

i final

Equatorial Plane

Fig. 8-14. Argument of latitude of intersection point.

i initial ∆Ω

node 1, Ωi

node 2, Ω f

EXAMPLE 8-9 Compute ΔV required to change the right ascension of the ascending node

of the following orbit to 100 o West: rp1 = 1.1DU, e1 = 0.1, i = 45˚ , Ω = 40˚ West, ω = 10˚

SOLUTION The orbit of satellite is transfer to Ω = 100 ˚ West.

∆ α  ΔΩ    ΔΩ     

The ΔV required is given by,

V = 2Vsin

where α is,

cos = cos

cos

2

+ sin

sin

=

40° + 100° = 60° = 41.4° The speed of satellite at that particular point is, 2

V=

where,

a = 1.22DU And r can be obtain through,

sin A

La

=

sin

 α ∆Ω sin(

)

sin

ALa = 67.8˚

θ = ALa – ω = 57.8˚

r= Thus,

p1

r (1 + e) 1 + ecos

θ

= 1.14878 DU

V = 0.96 DU/TU ΔV = 0.679 DU/TU

cos(

)

1 1 0


8.4.3 Combined Maneuver Frequently, the spacecraft orbit needs to be raised as well as titled. Two orbital transfers may then be applied: 1) A coplanar maneuver to raise the orbit (change radius), then 2) A plane change to tilt the orbit. However, performing two separates orbit maneuvers is fuel inefficient because number of burns increased. Also, the time required for spacecraft to arrive at final orbit is much longer. Therefore, as an alternative, these two maneuvers can be combined in one maneuver to perform both tasks in one burn which is more economic (require less fuel) and faster. There are a few type of combined maneuver available in study. In this section, we will introduce the minimum inclination maneuver. Fig. 8-15 shows the minimum inclination maneuver for a spacecraft. Both initial and final velocity of plane change maneuver contains the Hohmann transfer’s contribution. The change of inclination between initial, transfer and final orbit is chosen in the way such that the required cost is minimum. Here, a scaling term, s is introduced to determine change of inclination required between orbits.

∆ ∆ ∆  ∆ =

= (1

(8-20)

)

The total delta-v that required for the combined maneuver is,

∆       ∆          ∆ =

+

+

_

2

+

2

_

ˆ K Transfer Orbit Final Orbit

∆Vb

∆Va

Initial Orbit

Fig. 8-15. Orbit transfer of a spacecraft using combined maneuver.

ˆ J

ˆ I

)

cos(

cos (1

)

(8-21)


111

Now, the optimum scaling, s is required to determine to produce minimum cost. Then, we take

∆

=0

 ≈ ∆ −1 ∆ ∆      3   1

)

sin(

tan

_

cos(

_

(8-22)

)

For circular initial and final orbits:

_

=

,

=

(8-23)

_

EXAMPLE 9-10 Calculate the total delta-v required for a spacecraft to transfer from an orbit, r 1 = 1.02 DU to r 2 = 2.33 DU with the change of inclination ∆i = 10°.

SOLUTION We have the initial and final radius, r 1 and r2. Then semimajor axis for the transfer orbit is,

 1   initial  µ1 trans   2µ1  µ trans   2µ  µ final  µ  Δ −1 3  Δ Δ  ∆ +

=

2

= 1.675

The velocities at each location are:

=

= 0.9901 DU/TU

_

=

= 1.1678 DU/TU

_

=

= 0.5512 DU/TU

=

= 0.6551 DU/TU

Then, we need to determine the scaling, s, that is: 1 sin( ) = tan = 0.224105 / + cos( ) Therefore, the total delta-V is,

= 0.3464 DU/TU

1 1 2


Bate, R., Mueller, D., & White, J. (1971). Fundamentals of Astrodynamics. New Work: Dover Publications. Chobotov, V. (2002). Orbital Mechanics (3 ed.). AIAA. Curtis, H. (2005). Orbital Mechanics for Engineers (1 ed.). ButterworthHeinemann. Gurfil, P. (2006). Modern Astrodynamics. Academic Press. Pisacane, V. (2005). Fundamentals of Space Systems (2 ed.). Oxford University Press. Vallado, D. A. (2007). Fundamentals of Astrodynamcis and Applications. Microcosm. Walter, U. (2008). Astronautics: The Physics of Space Flight (2 ed.). Wiley-VCH.


113

P8-1

A space vehicle in a circular orbit at an altitude of 0.0784 DU above the Earth executes a Hohmann transfer to a 0.1568 DU circular orbit. Calculate the total delta-v requirement.

P8-2

Calculate the total delta-v required for a Hohmann transfer from a circular orbit of radius 4 to a circular orbit of radius 16 .

P8-3

Determine the total time of flight for Hohmann and bi-elliptic transfer with given orbit properties: Initial orbit, hinitial = 200 km, Apogee altitude of transfer orbit for bi-elliptic transfer, hapog = 35,000 km, Final orbit, h final = = 30,000 km.

P8-4

Determine the total ∆V required for a bi-elliptic transfer with given orbit properties: Initial orbit, r1 = 1.013 DU, Apogee altitude of transfer orbit, r 2 = 3.021 DU, Final orbit, r3 = 2.601 DU.

P8-5

Determine the total ∆V required and time-of-flight for a bi-elliptic transfer to place a spacecraft from r1 = 6578.145 km into GEO. Given that, the apogee radius of transfer orbit is, r2 = 46378.145 km.

P8-6

Given that the target and interceptor spacecraft are orbiting around the Earth in equatorial orbit. Determine the semimajor axis of the phasing orbit. Both spacecrafts’ positions at that time are: r = 12546.38 12546.387I 7I + 10527.66 10527.667J 7J km km



⃗⃗tgtint r







= 16129 16129.3 .324I 24I + 2844.03 2844.035J 5J km km

P8-7

Given that the target and interceptor spacecraft are in an equatorial orbit with semimajor axis of 15,000 km and eccentricity 0.1. Determine the semimajor axis of the phasing orbit if the distances of target and interceptor spacecraft to the Earth center are 13,574.4216 13,574.4216 km and 13,615.9737 km respectively at the time. (Assume that the true anomaly of both spacecraft are in first quadrant.)

P8-8

Determine the ∆V required for a satellite to change its orbit plane from equatorial orbit to an orbit with inclination 10° at altitude 400km.

P8-9

Determine the total ∆V required for a satellite to change its orbit plane at inclination 5° to GEO.

P8-10

Compute ΔV required to change from following orbit to the right ascension of the ascending node at 35˚ and inclination at 15˚: rp1 = 1.08 DU, e1 = 0.05, i = 20˚ , Ω = 20˚, ω = 5˚

1 1 4


P8-11

Calculate the ∆V required required at point C in in orbit 1 for a single impulsive maneuver to rotate the apse line by 180° 180 ° counterclockwise (to become orbit 2), but keep the eccentricity, e and the angular momentum, h unchanged.

C

1

2 r

 ℎ

[Answer: 2

Fig. P8-11

/ ]

P8-12

Compute ΔV required to change from following orbit to the right ascension of the ascending node at 30˚ and inclination at 25˚: Altitude, h = 300 km, e1 = 0, i = = 10˚ , Ω = 25˚

P8-13

A satellite in orbit 1 undergoes a delta-v maneuver at perigee P1 such that the new orbit 2 has the same eccentricity e, but its apse line is rotated 90° 90° clockwise from the original one. Calculate the specific angular momentum of orbit 2 in terms of that of orbit 1 and the eccentricity e.

2 1

Fig. P8-13 P8-14

Calculate the total delta-V required for a spacecraft to transfer from an orbit with altitude, h1 = 400 km to geosynchronous orbit with the change of inclination ∆i = 25°.


P8-15

Calculate the total delta-v required for a spacecraft to transfer from an orbit, r1 = 1.157 DU to r 2 = 4.136 DU with the change of inclination ∆i = 20° using: (a) Hohmann transfer followed by simple plane change. (b) Combined Maneuver.

P8-16

A satellite is in a circular earth orbit of altitude 400 km. Determine the new perigee and apogee altitudes if the satellite on-board propulsion system increases the speed of the satellite (a) in the flight direction by 240 m/s; (b) gives the satellite a radial (outward) component of velocity of 240 m/s.

115

1 1 6


9

Interplanetary Trajectories

The apparent retrograde and direct motion of the planets arises not from their motion but from the earth's. The motion of the earth alone, therefore, suffices to explain so many apparent inequalities in the heavens. Nicolaus Copernicus (1473-1543)

9.1 Trajectories Interplanetary missions involve spacecraft travels within the solar system. Interplanetary trajectories are different from satellite orbits around planets mainly because the spacecraft is exposed to more than one dominant force. A spacecraft may leave the Earth, pass close to the Moon, travel round a heliocentric trajectory dominated by the Sun, and have close encounters with planets or asteroids before reaching its target planet. Parking orbits around initial and final bodies also add intermediate goals to the mission but give more flexibility to the mission design. One of the major exploration projects in the United States was the Galileo mission to Jupiter in 1989. The trajectory to Jupiter employed a two stage inertial upper-stage (IUS) using one-gravity assist maneuver at Venus and two maneuvers at Earth, requiring more than 6 years of travel time. Galileo was launched in October 1989 from Earth on a Space Shuttle. Instead of heading towards Jupiter or the asteroid belt, Galileo took a flight path that carried to Venus. Galileo arrived there in February 1990. Venus’ gravity accelerated Galileo and sent it on a flight path back toward Earth. When Galileo passed Earth in December 1990, the Earth’s gravitational field added energy to send Galileo out to the asteroid belt. A propulsive maneuver, performed in December 1991, brought Galileo past Earth again in December 1992 for a last gravity assist before the spacecraft began its final path to Jupiter. Arrival at Jupiter occurred late in 1995.

117

1 1 8

CHAPTER 9│INTERPLANETARY TRAJECTORIES

9.2 The Solar System The Sun is attended by enormous number of smaller bodies. The most noticeable are the nine planets: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, and Pluto.

9.2.1 Planets Except for Mercury and Pluto, the planetary orbits are nearly circular and lie nearly in the ecliptic plane. Pluto’s orbit is so eccentric that the perihelion point lies inside orbit of Neptune. The size, shape and orientation of planetary orbits are given by five classical orbital elements which remain relatively fixed except for slight perturbations caused by the mutual attraction of the planets.

Fig. 9-1. Components of the Solar System.

Fig. 9-2.

Tilting of the rotation axes of the planets and the Sun

Fig. 9-3.

A side view of the planetary orbits. Orbits and planets are not to scale.


Orbital period (years)

Mean distance (106 km)

Orbital speed (km/s)

—

—

—

Mercury

0.241

57.9

47.87

Venus

0.615

108.1

Earth

1.000

Mars

Planet

Mass Relative to Earth

(km3/s2)

119

Equatorial radius (km)

Inclination of equator to orbit

333432 1.327×10 11 696,000

7° 15’

0.055

2.232×104

2,487

0

35.04

0.815

3.257×105

6,187

149.5

29.79

1.000

3.986×105

6,378

1.881

227.8

24.14

0.107

4.305×104

3,380

Jupiter

11.86

778

13.06

317.9

1.268×108

71,370

3.1°

Saturn

29.46

1426

9.65

95.18

3.795×107

60,400

26.7°

Uranus

84.01

2868

6.80

14.54

5.820×106

23,530

97.9°

Neptune

164.8

4494

5.49

17.13

6.896×106

22,320

28.3°

Pluto

247.7

5896

4.74

0.00216

860.976

1,154

122.5°

Sun

177.4 23.5° 25.2

Table 9-1

Physical Planets

Properties

of

the

9.2.2 Asteroids Asteroids (a Greek word for star-like), also called minor planets or planetoids, are a class of small celestial bodies in the solar system that orbit around the Sun. Asteroids are material left over from the formation of the solar system. One theory suggests that they are the remains of a planet that was destroyed in a massive collision long ago. More likely, asteroids are material that never coalesced into a planet. They have been found inside Earth's orbit to beyond Saturn's orbit. Most, however, are contained within a main belt that exists between the orbits of Mars and Jupiter. The first asteroid to be discovered, Ceres, is the largest asteroid known to date and is now classified as a dwarf planet. All others are currently classified as small solar system bodies. A newly discovered asteroid is given a provisional designation consisting of the year of discovery and an alphanumeric code (such as 2002 AT4). Once its orbit has been confirmed, it is given a number, and later may also be given a name (e.g. 433 Eros).

Sun Mercury Venus

Earth

Moon

Mars Jupiter Saturn Uranus Neptune

Pluto

Fig. 9-4. Astronomical symbols.

1 2 0


Fig. 9-5.

Asteroids belt.

9.2.3 Comets The word "comet" comes from the Greek word komē, meaning “hair of the head”. Comets are small Solar System bodies that orbit the Sun and, when close enough to the Sun, exhibit a visible coma (or atmosphere) and/or a tail — both primarily from the effects of solar radiation upon the comet's nucleus. Comet nuclei are themselves loose collections of ice, dust and small rocky particles, measuring a few kilometers or tens of kilometers across. Comets originate in the outer solar system; they are thrown inwards towards the Sun by gravitational perturbations from planets or nearby stars. They have a variety of different orbital periods, ranging from a few years, to 50 or 100 years, to thousands of years, while some are believed to pass through the inner Solar System only once before being thrown out into interstellar space. Dust tail coma Ion tail nucleus

Fig. 9-6.

Comet structure.


121

9.3 The Patched-Conics Method Interplanetary spacecraft spends most of the flight time moving under the gravitational influence of a single body – the Sun. Its path of motion is affected by the gravitational fields of the departure and arrival planets only for very brief intervals compared to total mission time. Perturbations caused by other planets while the spacecraft is moving along its heliocentric course are normally negligible. Precise computation of the mission trajectory, of course, requires numerical integration of the equations of motion where all perturbation effects are inserted. Though, for preliminary mission analysis, it is sufficient to use an approximate method to determine Delta-V requirements of the mission. The patched conics approximation is the best available method for that purpose. It involves partitioning the overall transfer into several two-body problems. In other words, only one celestial body’s influence is considered to be acting upon the spacecraft at all times. However, the patched-conic approximation is still limited in that it only considers the gravity of one celestial body at a time.

9.3.1 Description of Method Assume that at any particular time; only one central body is acting on the spacecraft. The region in which a particular body is dominant is called its sphere of influence, which is a concept introduced by Lagrange. In this region, the spacecraft is assumed to follow a Keplerian orbit. As it leaves one region, it enters the sphere of influence of another body and its trajectory becomes a new conic orbit with the second central body at its focus. Thus, the entire trajectory is formed by patching together the various conic sections. To determine the sphere of influence the equations of motion are first constructed with respect to the disturbing body, treating the central body as a perturbation. Then, constructing the equations of motion with respect to the central body, treating the disturbing body as a perturbation. Then, the radius of the sphere of influence, RSI can be approximated as

5    ≈  

( 9-1 )

Mc = Mass of the central body Md = Mass of the disturbing body r d = Distance from the central body to the disturbing body

For Earth, RSI ~ 0.93×106 km, assuming the Sun as a disturbing body. Compare this to the radius of Moon’s orbit around Earth ~ 0.38×106 km The following table gives the sphere of influence of planets with respect to the Sun.

1 2 2

Table 9-2 Sphere of influence of the Planets.


Surface Escape Velocity (m/s)

J2

Sphere of influence (106 km)

273.98

6.18×105

−

−

Mercury 2.232×104

3.70

4250

−

0.09 - 0.14

Venus

3.257×105

8.87

10360

2.7×10 -5

0.61 - 0.62

Earth

3.986×105

9.81

11180

Mars

4.305×104

3.71

5020

Jupiter

1.268×108

23.12

59530

0.01475

45.9 – 50.5

Saturn

3.795×107

9.05

35560

0.01645

51.6 – 57.5

Uranus

5.820×106

7.77

21250

0.012

49.4 – 54.1

Neptune 6.896×106

273.98

6.18×10 5

−

Pluto

3.70

4250

−

Planet

Sun

μ (km³/s²)

1.327×101 1

860.976

Equatorial Surface Gravity (m/s²)

0.00108 3 0.00196 4

0.91 – 0.94 0.52 – 0.63

−

0.09 - 0.14

Axial rotation period (sidereal) ~ 27 days 58.646 days 243.019 days 23h56m22.7s 24h37m22.6s ~ 9h50m ~ 10h15m ~ 17h50m ~ 27 days 58.646 days

The patched-conic approximation offers an efficient method for describing interplanetary orbits. By partitioning the overall orbit into a series of two-body orbits, it greatly simplifies mission analysis. For each of the portions of an orbit, one gravitational force is assumed to be acting upon the spacecraft at a time. To illustrate the efficiency of the patched-conic approximation, we partition the standard Hohmann transfer of a spacecraft traveling from Earth to Jupiter into three separate conic stages.

EXAMPLE 9-10 Calculate the total delta-v required to transfer of spacecraft from an orbit around the Earth to an orbit around Jupiter using a Hohmann minimumenergy transfer. Orbits of the Earth and Jupiter around the Sun may be assumed circular and coplanar.

SOLUTION The mission is defined by the following phases Phase 1: Geocentric –the boost from Earth orbit to a heliocentric escape trajectory. Phase 2: Heliocentric – the cruise in transfer orbit from Earth to Jupiter. Phase 3: Jupiter centered – the powered deceleration from heliocentric approach trajectory to Jupiter orbit.


Jupiter

Earth

1. Heliocentric Phase: The spacecraft speed relative to Earth as it exits the Earth’s sphere of influence is given by



=

 ⊙      2

1

+

= 8.792 km/s

Similarly, the spacecraft speed relative to Jupiter as it enters Jupiter’s sphere of influence is given by

  ⊙        3           ⊙ =

which means that

1

2

+

=

5.643 km/s

is in the opposite direction. The transfer time is half

orbital period of the transfer orbit

=

2

=

+

8

= 2.731 years

It is interesting to note here that the minimum time for the Earth-Jupiter mission is 2.73 years. This is by no means a long trip in the context of interplanetary missions as we will see later. 2. Geocentric Phase:

∆1

The spacecraft is assumed to be initially parking in a 300-km circular orbit. Impulsive velocity increment is provided at point (A) to inject the vehicle into hyperbolic escape trajectory having speed at a great distance from the Earth.



123

1 2 4


V E A

V ∞ E Asymptote

ν∞

(periapsis)

The spacecraft velocity in its parking orbit is calculated from

    ℎ      ℎ   ∆1      ℎ   ⊕

=

⊕

+

In order to reach (B) with velocity velocity

=

Therefore,

2

⊕

⊕

+

= 7.726 km/s

, the spacecraft should leave (A) with

+(

=

) = 14.024 km/s

= 6.298 km/s

The eccentricity of the hyperbolic trajectory can be calculated as

=1+

=1+

+

(

) = 2.295

The hyperbola asymptotes must be in the same direction as Earth’s orbital velocity. Then, the angular location of point (A) can be determined from

Arrival at Jupiter:



= cos

−1 



( 1/ ) = 115.8°

Within the sphere of influence of Jupiter, the transfer trajectory will appear as a hyperbola.


V J D

V ∞ J Asymptote

ν∞

(peria psis)

Assume that after arrival at Jupiter the spacecraft will be inserted into a circular capture orbit with radius = 6RJ, in which the velocity is given by

   

=

6

= 17.173 km/s

Assume (D) is the point of closest approach to Jupiter on the hyperbola. As the spacecraft approaches (C) with velocity



, it will arrive at (D) with velocity

     ∆   ∆ ∆ ∆1 ∆     2

=



6

+(

) = 24.934 km/s

Therefore, the velocity required velocity increment is

=

= 6.298km/s

The total mission Delta-V budget

=

+

can be calculated from

= 7.761 + 6.298 = 14.059 km/s

The eccentricity of the hyperbolic trajectory can be calculated from

=1+

=1+

6

= 1.1080



The hyperbola asymptotes must be in opposite direction to Jupiter’s orbital velocity, because the velocity of Jupiter with respect to the Sun is greater than the velocity of the spacecraft with respect to the Sun. Then, the angular location of point (D) can be determined from



= cos

−1 



( 1/ ) = 154.5°

125

1 2 6


9.4 Lagrange Liberation Points Lagrange liberation points (or Lagrange points) are the five positions where the combined gravitational field of two large masses will cause a smaller object to remain stationary relative to the two larger objects. The points are stationary solutions of the circular restricted three-body problem. Euler discovered three of these points and Lagrange two of them. There are three points that are unstable because the slightest disturbance to any object located at one of them causes the object to drift away permanently. Until recently, this meant that the unstable Lagrangian points seemed to have no practical application for spaceflight. Now, however, they are known to have immense significance and have become the basis for chaotic control.

9.4.1 L1 (stable) The L1 point Lies on the line defined by the two large masses between them. Example: A body which orbits the Sun closer than the Earth would have a shorter orbital period than the Earth. However, this is based on the assumption that the effect of the Earth’s own gravitational pull is negligible. If the object is directly between the Earth and the Sun, then the effect of the Earth's gravity is to weaken the force pulling the object towards the Sun, and therefore increase the orbital period of the object. The closer to Earth the object is, the greater this effect is. At the L1 point, the orbital period of the object becomes exactly equal to the Earth's orbital period. L1 is well-suited to solar observations. Objects here are never shadowed by the Earth or the Moon.

9.4.2 L2(stable) The L2 point Lies on the line defined by the two large masses, beyond the smaller of the two. Here, the gravitational forces of the two large masses balance the centrifugal force on the smaller mass.

Fig. 9-7. Liberation points


On the side of the Earth away from the Sun, the orbital period of an object would normally be greater than that of the Earth. The extra pull of the Earth's gravity decreases the orbital period of the object, and at the L2 point that orbital period becomes equal to the Earth's. As the Earth rotates around the Sun, these two points rotate around the Sun with it, always staying in the same place with respect to the planet. These stable orbits are being eyed as places to deploy satellites that observe the Earth-Moon system.

9.4.3 L3 (stable) The L3 point lies on the line defined by the two large masses, beyond the larger of the two. This Lagrange point is located the farthest away from the Earth, on the opposite side of the Sun. Until the development of interplanetary probes which could travel to areas of the solar system outside the Earth and Moon, L3 could never be observed due to its position. L3 in the Sun–Earth system exists on the opposite side of the Sun, a little outside the Earth's orbit but slightly closer to the Sun than the Earth is. At the L3 point, the combined pull of the Earth and Sun again causes the object to orbit with the same period as the Earth.

9.4.4 L4 and L5 (unstable) Two additional Lagrange points — L4 and 5 —The Earth-Moon system also has these Lagrange points, which may contain interplanetary dust L4 and 5 in the Sun-Earth system are also called the triangular Lagrange points or Trojan points. The latter name comes from the Trojan asteroids at the Sun–Jupiter L4 and L5 points. Every planet has its own Trojan points and the larger the planet, the more interstellar dust will be found trapped in them. They lie at the third corners of the two equilateral triangles in the plane of orbit whose common base is the line between the centers of the two large masses. They are located on either side of the Earth with respect to the Sun, lie 60° ahead of the Earth and 60° behind the Earth in its orbit around the Sun, and contain interplanetary dust grouped into what are called Kordylewski clouds.

9.5 Aerobraking A technique that can be employed to capture a spacecraft by the atmosphere of a planet in order to reduce the fuel required for the mission. When the spacecraft is at its closest point, the propulsion system is used to insert the spacecraft into an elliptic orbit. The periapsis radius will approximately remain fixed, while the apoapsis radius will decay at twice the rate of the semimajor axis decay

127

1 2 8


dr a/dt = 2 da/dt

( 9-2 )

When the apoapsis radius reaches the required altitude of the target orbit, the propulsion system is used to raise the periapsis out of the atmosphere to the desired altitude such that the target orbit will be sustained. Aerobraking was first successfully used in the Magellan mission to Venus Proper predictions of aerobraking requires good knowledge of the planet’s atmospheric density. Consider the example of the Mars Global Surveyor spacecraft.

9.6 Gravity Assist Significant trajectory changes require substantial propellant mass. An alternative approach is to make use of the gravitational field of an intermediate planet which can change the magnitude or direction of the spacecraft velocity. Such propellant-free maneuver is known as Gravity Assist , Swing-by or Fly-by maneuver. Gravity assist was essential for several missions including the Galileo mission to Jupiter and Cassini mission to Saturn.

Fig. 9-8. Gravity assist maneuver

9.7 Aerogravity Assist In the last years, a new concept in the field of aero-assisted maneuvers has been developed. This new technique, envisaged for interplanetary missions and named AGA (aero gravity assist), utilizes atmospheric flight in order to augment the gravitational bending of heliocentric velocity occurring during a planetary encounter (McRonald, 1992). The resultant ∆V is much larger than that obtained from traditional (exoatmospheric) gravity assist. This rather paradoxical result (an atmospheric pass, with its inevitable loss of energy, being made to increase the inertial velocity) is illustrated in Fig. 1. The spacecraft enters the planet sphere of influence at the hyperbolic excess velocity following an approach asymptote with pericenter within the atmosphere. The atmospheric flight is controlled by aerodynamic lift to maintain nearly constant altitude. This circular trajectory


129

needs very high values of lift, directed downwards, to counteract the centrifugal force. After the atmospheric turn through an angle, the spacecraft follows the leaving asymptote with a final velocity that, due to drag force, will be smaller than the incoming velocity. It is expected that this reduction will be minimized with appropriate vehicles, such as hypersonic waveriders.3'4 The triangles of velocities shown at the bottom of the figure demonstrate that, in spite of the reduction in V, the ∆V with such an AGA maneuver is greater than that obtained with gravity assist alone. It is well known that the maximum ∆V achieved in a gravity assist maneuver is equal to the circular velocity at the hyperbola pericenter. One may ask: given the aerodynamic performance L/D of a vehicle and an approach velocity V, what will be the maximum ∆V in an aerogravity assist maneuver? It is the goal of this Note to give the answer to this question.

Fig. 9-9. Aerogravity assist maneuver

1 3 0


Bate, R., Mueller, D., & White, J. (1971). Fundamentals of Astrodynamics. New Work: Dover Publications. Chobotov, V. (2002). Orbital Mechanics (3 ed.). AIAA. Curtis, H. (2005). Orbital Mechanics for Engineers (1 ed.). ButterworthHeinemann. Gurfil, P. (2006). Modern Astrodynamics. Academic Press. Hamilton, C. (1997). Asteroid Introduction. http://www.solarviews.com/eng/asteroid.htm

Retrieved

2009,

from

McRonald, A. D. (1992). Hypersonic Maneuvering for Augmenting Planetary Gravity Assist. Journal of Spacecraft and Rockets, 216-222. Pisacane, V. (2005). Fundamentals of Space Systems (2 ed.). Oxford University Press. Stephen Schneider and Thomas Arny. (2007). Pathways to Astronomy. New York: McGraw-Hill Higher Education. Unsöld, A., & Baschek, B. (2001). The New Cosmos: An Introduction to Astronomy and Astrophysics. Berlin Springler. Vallado, D. A. (2007). Fundamentals of Astrodynamcis and Applications. Microcosm. Walter, U. (2008). Astronautics: The Physics of Space Flight (2 ed.). Wiley-VCH.


P9-1

Describe aerobraking. How can it be used to reduce the total velocity change of an interplanetary mission? Consider a nuclear waste disposal mission from earth (1 A.U.) to a heliocentric circular orbit of radius 0.86 A.U.

P9-2

What is the characteristics energy required for a Hohmann transfer from earth to the circular orbit or radius 0.86 A.U.? What is the flight time via the Hohmann-transfer from earth to the 0.86 A.U. orbit about the sun?

P9-3

To accomplish certain measurements of phenomena associated with sunspot activity, it is necessary to establish a heliocentric orbit with a perihelion of 0.85 A.U. The departure from the earth’s orbit will be at apohelion. What must the burnout velocity be at an altitude of 1300km to accomplish this mission?

P9-4

A Venus probe departs from a 6378-km altitude circular parking orbit with a burnout speed of 8.69km/s. Find the hyperbolic excess speed at infinity.

P9-5

Calculate the sphere of influence for the nine planets in the solar system.

P9-6

When the Earth is 147.4×106 km away from the Sun, a spacecraft parked in a 200-km altitude circular Earth orbit is to be launched directly into an elliptical orbit around the Sun with perihelion of 120×106 km and aphelion equal to the Earth’s distance from the Sun on the launch date. (a) Calculate the ∆V required. (b) Calculate the velocity of the departure hyperbola.

P9-7

Compute the distance to L1, the Lagrangian liberation point, from the center of the moon along the Earth-moon line. To a first approximation, L1 is an equilibrium point between the gravitational and centrifugal accelerations of the attracting bodies.

P9-8

Compute the stationkeeping requirements (ΔV) to remain within 10 km of L1 in P9-7 for a year.

P9-9

Compute the velocity impulse (ΔV) for transfer from a 100-km circular orbit at Mars to a hyperbolic orbit Earth return trajectory with an eccentricity e=1.5.

P9-10

Write the equation of motion for a solar sail in the solar system. Discuss the type of trajectories possible.

131

1 3 2


P9-11

If the Earth were stopped in its orbit, what would be the elapsed time, in days, until collision with the sun? Assume point masses, and assume the Earth’s orbit to be circular, with r=1.0 A.U.

P9-12

Assuming the orbits of Earth and Mars are circular (with radius equal to semi-major axis) and coplanar, calculate (a) the total ∆V required to send a spacecraft from a 600-km Earth orbit into a Hohmann transfer from Earth to Mars, (b) the time required for the Hohmann transfer maneuver, and (c) the initial position of Mars (α) in its orbit relative to Earth for interception to occur.

Fig. P9-12 P9-13

Your space mission analysis team is assigned to design the first interplanetary mission to Mars. The spacecraft will be initially launched into a 500-km parking circular orbit around Earth. The spacecraft propulsion system will be used to escape the gravitation field of Earth and insert the spacecraft into a Hohmann transfer orbit along which the spacecraft will coast towards Mars. On arrival at Mars, the propulsion system will be used to decelerate the spacecraft and insert it into a 13,500-km altitude circular orbit around Mars. Assume the orbits of Earth and Mars around the Sun are both circular and coplanar (a) Calculate the velocity at which the spacecraft will be inserted into the Hohmann transfer orbit relative to the Sun, then relative to Earth. (b) Calculate the velocity at which the spacecraft will exit the Hohmann transfer orbit relative to the Sun, then relative to Mars. (c) Calculate the velocity change required to escape the Earth’s orbit assuming a simple coplanar maneuver is used. (d) Calculate the velocity change required to insert the spacecraft into the Mars’s circular orbit assuming a simple coplanar maneuver is used.

Relative Motion

10

The apparent retrograde and direct motion of the planets arises not from their motion but from the earth's. The motion of the earth alone, therefore, suffices to explain so many apparent inequalities in the heavens. Nicolaus Copernicus (1473-1543)

10.1 General Relative Motion Relative motion usually refers to the motion of one spacecraft with respect to another close spacecraft. Relative motion (relative position and relative velocity) can be described by the difference between the absolute positions and absolute velocities, in an inertial frame. However, an easier way of modeling the relative motion can be developed if we take into consideration that the relative distance between the two spacecraft is much smaller than the size of the spacecraft orbits. In fact, we will see in this chapter that when we take advantage of the later fact, we can describe the relative motion between the two spacecraft with a system of linear equations. This linear system facilitates the design of several tasks like spacecraft docking, debris avoidance, and position correction tasks. Consider two spacecraft, A and B. The relative position between the two spacecraft, in the inertial frame, is defined as: Z

⃗rNel ⃗BN  ⃗AN

r

=r

r

k

⃗rAel ⃗rNel = [Q]r

rrel

j i

(10-1)

Where, the superscript N denotes the inertial reference frame. The relative position vector can be expressed spacecraft A’s RSW-reference frame as:

r

Chaser

Target X

Y

(10-2) Fig. 10-1. Relative position of Chaser to Target in RSW frame..

133

1 3 4

CHAPTER 10│RELATIVE MOTION

Where, the superscript A denotes the RSW reference frame of spacecraft A, and [Q] is the rotation matrix of the RSW reference frame from the inertial frame. From Kinematics (see chapter 2) and using Eq. (10-1), the velocity of spacecraft B is:

 B  A ΩA ⃗rel  rel v =v +

×r

+v

ΩA  B  A ̇ΩA ⃗rel ΩA ΩA ⃗rel ΩA  rel  rel ̇ΩA ΩA  A ⃗A  A A A   ⃗ A A ΩA A  A AΩA ΩA ⃗A A A ΩA

(10-3)

where is the angular velocity of the spacecraft A with respect to the Earth. Also, the acceleration of spacecraft B is:

a =a +

×r

+

×

×r

+2

is the rate of change of the angular velocity,

×v

+a

(10-4)

.

Recall that the specific angular momentum of spacecraft A is:

h =r ×v =r v

Where v

k

(10-5)

is the component of the velocity vector perpendicular to r . Recall

also the relationship between the angular velocity,

v

and v

=r

: is,

(10-6)

Substituting Eq. (10-6) into (10-5), we get:

=

r ×v r

=

k

(10-7)

Taking the time derivative for Eq. (10-7), we obtain:

̇ΩA  A ̇A ⃗A  A  ⃗AA∙  A ΩA =

2

r

r (r × v ) =

v )

2(r

r

(10-8)

We have the angular velocities and accelerations derived in Eqs. (10-7) and (10-8), then we can determine the relative velocities and accelerations in Inertial frame using Eqs. (10-3) and (10-4). In spacecraft A’s RSW reference frame are, we have:

 ArelA  relNN  rel  rel v a

= [Q]v = [Q]a

(10-9)


135

EXAMPLE 10-1 At a given time, the orbital elements of spacecraft A are: °, = °, = °, = ° = , = . , = and the orbital elements of spacecraft B are: °, = °, = °, = ° = , = . , = Determine the relative position, velocity and acceleration of spacecraft B with repect to spacecraft A, as expressed in spacecraft A reference frame.

                         

SOLUTION

⃗⃗  ⃗⃗ 

First, the absolute positions and velocities for both spacecraft are,

T T T T

= [6086.2086 4941.2610 1176.8355] km = [ 4.5384 5.2771 1.5146] km/sec = [6159.0316 4871.8140 1571.0703] km = [ 4.5076 5.1031 1.9933] km/sec

Next, we determine the angular velocity and rate of angular velocity of spacecraft A, expressed in Spacecraft A’s RSW frame:

−  ωA  µ3A  ̇ωA  ⃗ ∙ ⃗ ω  −8  A Ωω θ  A   ω ̇ωNN TATA ̇ωωAA T ⃗ ⃗ ⃗  N  N ⃗     ω   B A  rel T  relN   Nrel  B   A ω ̇ N ⃗ ωN ωN ⃗  ωN  relN =

=

= 8.82334 × 10

a

2(rA

vA )

A

=

rad/sec

0.6641 × 10

rad

sec Next, we determine the Directional Cosine Matrix of spacecraft A at that point, which is, [Q] = C( , i, + ) 0.76775 0.62332 0.14845 [Q] = 0.64035 0.73824 0.21201 0.02256 0.25783 0.96593 r2A

Then, the angular velocity and rate of angular velocity can be expressed in Inertial frame as: = [Q] = [Q]

The relative position in Inertial frame is, = = [72.8230 69.4471 394.2348] km The relative velocity in Inertial frame is, v =v v × v = [0.06122 0.22816 0.46354] km/sec The relative acceleration in Inertial frame is, ) 2 a =a a × ×( × ×v


1 3 6

 Nrel ⃗⃗ a



= [ 0.03850

0.11369

 

T −3

0.29040] × 10

km/sec



Thus, the relative position, velocity and acceleration expressed in spacecraft A’s RSW frame are, = [Q] = [71.1472 14.3185 400.3501] km = [Q] = [ 0.02639 0.10937 0.50795] km/sec = [Q] = [ 0.001802 0.047013 0.310685] km/sec

⃗

⃗A⃗  A A⃗ 



TT T



10.2 What does an Observer on one

X 3 X

Spacecraft see?

X 4

2 Y 1

5

X

X

Y 6 8 X

Spacecraft A Spacecraft B

Consider two spacecraft in two orbits with the same semimajor axis. That is, both spacecraft have same orbit period. Let the first orbit be circular and the second orbit be elliptic with small eccentricity.

X 7 X

Fig. 10-2. Relative direction observation in Inertial frame.

The spacecraft A’s reference frame is always rotating such that one axis is pointing to the earth center. Consider an observer on spacecraft A and is observing spacecraft B location at all times (see figure 10-2.) Because spacecraft B is in elliptic orbit, its velocity is not uniform. If we plot the observations on an XY coordinate system attached to spacecraft A, we get the bean-shape shown in figure 10-3. Spacecraft B travels faster when it is close to its perigee (when its altitude is below that of spacecraft A,) and travels slower when it is close to its apogee (when its altitude is above that of spacecraft A). The following section shows how to compute the position vectors shown in figure 10-3 as a function of time.

Y 3 2 4 5

1 8

X

10.3 Linearized Equations of Relative Motion

⃗A

6 7

Fig. 10-3. As viewed from co-moving frame by spacecraft A

⃗B

Consider two spacecraft A and B, and let r be the position vector of chaser spacecraft A and r be the position vector of target spacecraft B, both expressed in the inertial frame. The relative position between the two spacecraft can is:

⃗ ⃗B  ⃗A

r=r

r

(10-10)


137

Assume that the relative distance between the spacecraft is small, compared to the absolute distance to the Earth’s center:

⃗B ⃗A δ δA ≪

r =r + r r

1

r

Target

The assumption in Eq. (10-11) is reasonable, especially for spacecraft rendezvous cases. Recall the orbit equation for spacecraft B:

̈⃗B µ ⃗BB3 r

r =

(10-12)

r

Substituting Eq. (10-12) into the second order time derivation of Eq. (10-10) yields,

δ ̈⃗  ̈⃗A µ δ⃗ B3 ⃗A  B δ⃗ ⃗A ∙ δ⃗ ⃗A ⃗A ∙ ⃗A δ⃗∙ ⃗A δ⃗∙ δ ⃗ B A  δ⃗∙ ⃗A δ  ≈ A  δ⃗∙ ⃗A A A A r=

r+r

r

r

The range of the target spacecraft, as:

(10-13)

, to the Earth’s center can be expressed

r =( r+r ) ( r+r )=r

r +2 r r + r

r

(10-14)

If we expend the Eq. (10-14), and using the assumption in Eq. (10-11), we get:

r =r

Note that r

1+

−3  −3  = (r )

/

2 r r r

+

r

r

r

2 r r

1+

r

(10-15)

;Eq. (10-15) can be replaced by:

r

B−3 A−3  =r

1+

3 − δ⃗∙A ⃗A 

2 r r

(10-16)

r

To linearize Eq. (10-16), we use the binomial series:

       ⋯

(1 + ) = 1 + where,



we have,

=

rr∙r

+

(

1)

2!

+

(10-17)

. By neglecting the higher order terms in the binomial series,



1+

3 − δ⃗∙A ⃗A  ≈

2 r r r

Z

(10-11)

1+

  δ⃗∙A ⃗A 3

2 r r

2

r

(10-18)

rB X

δr

rA

Chaser

Y

Fig. 10-4. Relative Position of Target Spacecraft to Chaser Spacecraft.

1 3 8


Then, substituting Eq. (10-18) into (10-16), we obtain,

−3B A−3   δ⃗∙ ⃗A 3  5 δ⃗∙ ⃗A A A A

r

=r

1

3 r r

=

r

1

3

r

r

(10-19)

r r

Substituting Eq. (10-19) into Eq. (10-13):

δ ̈⃗  ̈⃗A  µ  A3  A5 δ⃗∙ ⃗A δ⃗ ⃗A  ̈⃗A  µ rr rr  r3 δ⃗∙ ⃗A ⃗A  r3 δ⃗∙ ⃗A δ⃗ r=

=

r

r

1

3

r

r

r r

( r+r )

( r r )r

+

(10-20)

( r r ) r

Since the relative dstance between the two spacecraft is small, then

3r δ⃗∙ ⃗A δ⃗≪ ( r r ) r

1. Eq. (10-17) becomes:

δ ̈⃗  ̈⃗A  µ rr rr  r3 δ⃗∙ ⃗A ⃗A ̈⃗A µ rr  δ ̈⃗ µµ δA3⃗  A5 δ⃗∙ ⃗A ⃗A r=

Recall that r =

r

( r r )r

+

(10-21)

and substitute that into Eq. (10-18).The linearized

equation of relative motion is:

r=

r

r

3

r

( r r )r

(10-22)

EXAMPLE 10-2 Given the initial positions and velocities of two spacecraft: ] =[ . . . ] =[ . . . ] =[ . / . . ] =[ . / . . Using the linearized method, determine the relative position after 30mins. Then compare the results results with the relative relative position obtained through the determination of the absolute position of each spacecraft.

⃗⃗  

                            

SOLUTION

⃗⃗11⃗⃗⃗1⃗1 

At initial time, the relative positions and velocities are,

=

=





T

= [ 22.0744 57.6290 14.0791] km = [ 0.054772 0.058184 0.017981] km/sec

T


139

⃗A ⃗1 δ⃗ ⃗1 ⃗1     ⃗⃗1   ⃗1    T

Using the Eq. (10-22), where r =

and r =

,. the relative

position after 30 mins is,

T

( = 30mi 30min n) = [ 52.2169 30.0041 5.4692] km The absolute position of each spacecraft after 30mins are: (t = 30min) = [ 7414.0065 376.3 6.3305 245.66 .6600] km (t = 30min) = [ 7465.1619 407.2 7.2900 239.8 9.8874] km Then, the relative position is,

= [ 51.1554

T T

5.7726] km

30.9595

10.4 Clohessy-Wiltshire (CW) Equation Another form of the equations of relative motion between two close spacecraft in near circular orbits o rbits is the Clohessy-Wiltshire (CW) equations8.

δ

⃗A

Define the position vector r for spacecraft A, and r is the relative position vector of spacecraft B with respect to spacecraft A, both vectors in the RSW frame as:

⃗A δ 

r =

r=

+

  

(10-23)

+

Based on the assumption that the relative distance between the two spacecraft is small compared to the absolute position, Clohessy and Wiltshire showed that the acceleration of the relative position vector is:

δ̈   3   δ  ⃗ r =

⃗

3

+ r +

(10-24)

The term represents all external forces, like thrust forces or aerodynamic drag forces, affecting spacecraft B. Another way of looking at Equation (10-24) is to consider spacecraft A as a target spacecraft and spacecraft B as an interceptor spacecraft. The interceptor spacecraft may have thrusters onboard and use them to intercept the target after a given amount of time. Consider the equation,

̈ ⃗ ̈ ⃗  ̇  ⃗A    ⃗̇ A      ⃗A r

8 Clohessy

= r

×r

2

×

×(

×r )

(10-24)

and Whiltshire are both engineers who first developed this form of the equations of relative motion, in the middle of the twentieth century

1 4 0


The vector

 

is the angular velocity of spacecraft A:

      =

=

̇    δ̈        ⃗ ̇    ̇  ̈  ̇    ̈  ̇  ̈  

Note that

is constant, and hence

= 0.

.

Substitute for the angular velocity vector into equation (10-25), we get:

r =

+ +

+

3

+

+2

2

+

(10-25)

Which can be written in terms of its three components as:

2

3

2

+

=

=

(10-26)

=

Eq.(10-27) is also known as Hill’s equation.

10.4.1 Closed-form solution for near circular orbits It is possible to obtain a closed form solution for the Hill’s equation. Assume that both spacecraft are in near circular orbits; with no external forces, i.e.

⃗

= 0. Note that using impulsive thrust does not violate the assumption of zero external forces since impulsive thrust affects only the initial conditions of the problem. Taking time derivatives derivatives for the x -component -component equation, we get:

̈ ⃛

̇  ̈  ̇ ̈ ̇ ⃛ ̇  3   0  ̇ 0  ̇ ̈0   ̈  0   0  0  0  0 ̇0 ̈0

Substitute for

=2

=2

+3

=2

(10-27)

+3

into equation above, we obtain: into

=

(10-28)

Then, taking the Laplace transform of equation (10-41), we get:

( )

+

( )=

+

(

+

( )

)

+

(

+

=0 (10-29)

)

Where , and are the x -components -components of the initial relative position, velocity and acceleration vectors, respectively. Take the inverse Laplace transform to obtain: o btain:


141

  0 ̈0 ̇0   ̈0  ̈0 ̇0 0   0 ̇0 ̇0   0 ̇0   ̇  ̇0   0 ̇0  ̈ ̈   ̇0    0 ̇0  ̇  ̇  ̇  ̇0  0 ̇0    0   0 ̇0        ̇0  ̇0   0 0   ̇0 0  0 ̇0   0   ̇0 ̇   ̇0  0 ̇0   0 ( )=

+

+

sin(

From Eq. (10-28), we can see that becomes,

( )=4

+

2

+

)

cos(

=2

sin(

3

+

(10-30)

. Then Eq. (10-31)

+3

)

)

2

cos(

)

(10-31)

Eq. (10-32) shows the closed form solution of Hill’s equation for the xcomponent of the relative position vector. The relative velocity is obtained by taking of the time derivative of ( ):

( )=

)+ 3

cos(

+

2

sin(

)

(10-32)

Substitute Eq. (10-33) into the equation,

( )=

2

)

cos(

2 (3

+2

)sin(

)

(10-33)

Integrate the Eq. (10-34) twice, we obtain t he expression of and ,

( )=

( )=

2

2

cos(

The constants and

) + 2(3

sin(

)+

2

(3

+2

)cos(

)sin(

+2

)+C

)+

(10-34)

+

can be determined from initial conditions: =

6

3

2

=

(10-35)

+

Therefore,

( )=

2

)+

cos(

2

(3

)sin(

+2

)

(6

+)

2

+

( )=

2

sin(

) + 2(3

+2

)cos(

)

(6

(10-36)

+)

The solution of z-component in Eq. (10-27) is straight forward, since it is uncoupled from the other two equations. The equation of motion in zcomponent is:

̈  =

Taking the Laplace transform, we obtain:

(10-37)

1 4 2


   0  ̇0       0 ̇ ̇0   0  0   ̇0  0  ̇  ̇0  0  ( )

=

( )(

( )=

)=

+

(

( )

)

+

+

+

(

(10-38)

)

+

Then, through inverse Laplace transform, the z-component equations are:

( )=

( )=

)+

sin(

cos(

)

cos(

)

sin(

(10-39)

)

From these results for the equation of relative motion, we can see that the closed form solutions are linear functions in the initial relative positions and velocities. Thus, they can be expressed in m atrix form as follows:

⃗ Φrr⃗0 Φrv ̇⃗0 ⃗̇  Φvr⃗0 Φvv ̇⃗0   Φrr           Φrv        ⎣     Φvr          Φvv       =

r +

=

where,

r +

4 3cos( = 6(sin( ) 0

1

=

2

sin(

(cos(

2

)

)

1)

1

)

=

EXAMPLE 10-3

0 1 0

(1

cos(

(4sin(

)

0 0 cos(

)

))

3

0

)

0 0 0

2sin( ) 4 cos( ) 3 0

0

1

0

3 sin( ) = 6 (cos( ) 1) 0

(10-40)

r

)

0

cos( ) 2sin( ) 0

r

0 0 sin(

0 0 cos(

sin(

)

)

) (10-41)


143



Due to an impulsive external disturbance (debris hit), a GEO satellite = is found an hour afterwards to have drifted to the position [ ] relative to its original position. At that time the satellite initiates a two-impulse maneuver to return to its original location in 4 hours. Find the total Delta-v for this maneuver. SOLUTION

 

 

For GEO satellite, the height and mean motion are, h = 36000km

ω = 7.292 × 10 −5 rad / sec The closed form solutions of relative position for GEO satellite are:

⃗1 Φrr⃗0 Φrv ̇⃗0 =

r +

r

⃗ ̇T 1 ⃗Φ1vr⃗0 Φvv ̇⃗0 T ⃗1

⃗0

= r + r with = [0 0 0] km and = [ 5 5 0] km Given that the satellite is drifted to position after one hour, then:

1

= 3600 secs.

⃗̇̇ ⃗0  1

The relative velocities at initial time and at t 1 are,

T T

r = [ 0.001669

0.000999

0] km/sec

= [ 0.001093

0.001728

0] km/sec

Time required to return to designate position is 4hrs. Then,

14400 secs.



=

⃗ Φrr⃗1 Φrv ⃗̇ 1+

Using the CW closed form solution,

⃗

where

= [0

0

T

=

r +

=

r +

⃗̇  Φvr⃗1 Φvv ⃗̇ 1+

0] km and,

The required initial and final relative velocity to return to designate

⃗̇ ⃗̇ 1+    ∆  ⃗̇ 1  ⃗̇ 0  ⃗̇ 1+  ⃗̇ 

position are,

= [0.0008235

= [ 0.0001941

0.0001988 0.0005303

T

0] km/sec

T

0] km/sec

Thus, the total Delta-v is,

V=

+

= [0.001593

0.001458

T

0] km/sec

1 4 4


Chobotov, V. (2002). Orbital Mechanics (3 ed.). AIAA. Curtis, H. (2005). Orbital Mechanics for Engineers (1 ed.). ButterworthHeinemann. Gurfil, P. (2006). Modern Astrodynamics. Academic Press. Vallado, D. A. (2007). Fundamentals of Astrodynamcis and Applications. Microcosm.


145

P10-1

A spacecraft is about to be released from the space shuttle, which is in circular orbit at 600 km altitude. The relative velocity (from the shuttle bay) of the ejection is 0.9 m/s down, 0.05 m/s backwards, and 0.01 m/s to the right. Find the position and velocity of the spacecraft after 4 minutes.

P10-2

Spacecraft A is in a circular orbit of radius 6000 km. An approaching spacecraft B executes a delta-v burn when its position vector relative to spacecraft A is

δr 0 = [1 0.9 

T 0.95] km.

Just before the burn the relative velocity of the spacecraft was 

δ v0 = [0. 01 0.01 5] m / s. T

Calculate the total delta-v required for the spacecraft B to rendezvous with spacecraft A in one-third period of spacecraft A orbit.

Space Mechanics for Engineers

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