British Astronomy and Astrophysics Olympiad 2015 -2017
PROBLEMS Solutions
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Science Olympiad Blog
British Astronom y and Astrophysics Olympiad 2016-2017 Astronomy & Astrophysics Competition Paper Monday 23 rd January 2017 Instructions Time: 3 hours plus 15 minutes reading time (no writing permitt ed). Approx 35 minutes per question. Questions: All five question s should be attempted. Marks: The questions carr y similar marks. Solutions: Answers and calcu lations are to be written on loose paper or in examination booklets. Students should ensure their name and school is clearly written on all answer sheets and pages are numbered. A standard formula booklet with standard physical constants should be supplied. Instructions: To accommodate students sitting the paper at different times, please do not discuss any aspect of the paper on the internet until 8 am Saturday 28 th January. Clarity: Solutions must be written legibly, in black pen (the papers are photocopied), and worki ng down the page. Scribble will not be marked and overall clari ty is an important aspec t of this exam paper. Eligibility: The International Olympiad will be held during November 2017; all sixth form students are eligible to participate, even if they will be attending university in November.
Training Dates and the International Astronomy and Astrophysics Olympiad (IOAA)
The IOAA this year will be held in Phuket, Thailand, from 12 th to 21st November 2017. The team will be selected from sixth form students taking this paper and Y12 students taking the AS Challenge in March. The best students eligible to represent the UK at the IOAA will be invited to attend the Training Camp to be held in the Physics Department at the University of Oxford, (Tuesday 4th April to Friday 7 th April 2017). Astronomy material will be cover ed; problem solving skills and observational skills (telescope and naked eye observations) will be developed. At the Training Camp a data analysis exam and a short theor y paper will be sat. Five stud ents (plus one reserve) will be selected for further trainin g. From May there will be mentoring by email to cover some topics and problems, followed by a training camp in the summer and also one in the autumn.
Important Constants Constant
Symbol Value
c
Speed of light
3.00
× 108 m s
Earth’s rotation period
1 day
24 hours
Earth’s orbital period
1 year
365.25 days
parsec
pc
AstronomicalUnit
AU
Radius of the Earth
−1
× 1016 m 1.49 × 1011 m 6.37 × 106 m 3.09
R⊕
Semi-major axis of the Earth’s orbit
1 AU
Radius of the Sun
R
Mass of the Sun
M
Mass of the Earth
M⊕
Luminosity of the Sun
L
Gravitational constant
G
× 108 m 1.99 × 1030 kg 5.97 × 1024 kg 3.85 × 1026 W 6.67 × 10 11 m3 kg 6.96
−
−1
s− 2
You might find the diagram of an elliptical orbit below useful in solving some of the questions:
Elements of an elliptic orbit: a =
OA
a=
OB
e=
1 −
F P A
(= OP) semi-major axis (= OC) semi-minor axis b2 a2
eccentricity focus periapsis (point nearest to F) apoapsis (point furthest from F)
Keplers Third Law: For an elliptical orbit, the square of the period, T , of orbit of an object about the focus is proportional to the cube of the semi-major axis, a (the average of the minimum and maximum distances from the Sun). The const ant of proportionality is 4π 2 /GM , where M is the mass of the central object and G is the universal gravitational constant. Magnitudes: The apparent magnitudes of two objects, brightnesses, b1 and b0 , via the formula:
m1
b1 = 10−0.4(m1 −m0 ) b0
and
m0 ,
are related to their apparent
Qu 1. Martian GPS On Earth the Global Positioning System (GPS) requires a minimum of 24 satellites in orbit at any one time (there are typically more than that to allow for redundancies, with the current constellation having more than 30) so that at least 4 are visible above the horizon from anywhere on Earth (necessary for an x, y, z and time co-ordinate). This is achieved by having 6 different orbital planes, separated by 60 ◦ , and each orbital plane has 4 satellites.
Figure 1: The current set up of the GPS system used on Earth. Credits: Left: Peter H. Dana, University of Colorado; Right: GPS Standard Positioning Service Specification, 4th edition
The orbits are essentially circular with an eccentricity < 0.02, an inclination of 55◦ , and an orbital period of exactly half a sidereal day (called a semi-synchronous orbit). The receiving angle of each satellite’s antenna needs to be about 27.8 ◦ , and hence about 38% of the Earth’s surface is within each satellite’s footprint (see Figure 1), allowing the excellent coverage required. a. Given that the Earth’s sidereal day is 23h 56 mins, calculate the orbita l radius of a GPS satellite. Express your answer in units of R⊕ . b. How long would it take a radio signal to travel directly between a satellite and its closest neighbour in its orbital plane (assuming they’re evenly spaced )? How far would a car on a motorway (with a speed of 3 0m s−1 ) travel in that time? [This can be taken to be a very crude estimate of the positional accuracy of the system for that car.] In the future we hope to colonise Mars, and so for navigation purposes it is likely that a type of GPS system will eventually be establis hed on Mars too. Mars has a mass of 6.42 1023 kg, a mean radius of 3390 km, a sidereal day of 24h 37 mins, and two (low mass) moons with essentially circular orbits and semi-major axes of 9377 km (Phobos) and 23 46 0 km (Deimos).
×
c. Using suitable calculations, explore the viabili ty of a 24-satellite GPS constellati on similar to the one used on Earth, in a semi-synchronous Martian orbit, by considering: (i) Would the moons prev ent such an orbit? (ii) How would the GPS positiona l accuracy compare to Earth? (iii) What would the recei ving angle of each satellite’ s antenna need to be, and what would be the associated satellite footprint? By comparing these with the ones utilised by Earth’s GPS, make a final comment on the viability of future Martian GPS.
1
Qu 2. Hohmann Transfer Orbits In order to move a spacecraft between orbits we must apply a thrust using rockets, which changes the velocity of the spacecraft by ∆v . In this question we will ignore changes in the mass of the spacecraft due to the burning of fuel. For an object of mass m in a circular orbit of radius r around an object with mass M (where m the orbital velocity, vorb , is given by the formula vorb =
GM . r
a. Show that vorb in low Earth orbit (LEO; about 200km above the surface), is about 8kms is an estimate of the ∆v the rockets need to provide for the spacecraft to reach LEO.
M)
−1
. This
An economical route to take when travelling between planets is called a Hohmann transfer orbit. This is an ellipse for which the perihelion coincides with the inner planetary orbit (with radius r A ) and the aphelion coincides with the outer planetary orbit (with radius rB ). It is achie ved by incr easing the velocity of the spacecraft at point A by ∆vA before then increas ing it again at point B by ∆vB .
Figure 2: A diagram of a Hohmann transfer orbit between an inner and outer planet
For an ellipse with semi-major axis a it can be shown that the velocity v , at a distance r from mass M , can be written as:
v 2 = GM
2 1 r
−a
b. Derive expressions for ∆vA and ∆vB by comparing their circular orbital speeds with their transfer orbit speeds. Simplify your final expressions to include G , M , rA and rB only. c. Approximating Mars’ orbit as circular with a radius of 1.52 AU, calculate the ∆v to go from Earth LEO to Mars i.e. ∆v = ∆vA + ∆vB . Compare your answer to the ∆v to reach Earth LEO.
|
| |
|
d. Derive an expression for the total time spent on the transfer orbit , t H , and calculate it for an Earth to Mars transfe r. Give your answer in months. (Use 1 month = 30 days.) e. Hence calculate the direct distance betwe en Earth and Mars at the moment the spacecra ft reaches Mars. How long would it take a radio message from the spacecraft to reach Earth? f. How long woul d any astr onauts on board the space craft need to wait until they could use a Hohmann transfer orbit to return to Earth? Hence calculate the total duration of the mission.
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Qu 3. Starkiller Base As part of their plan to rule the galaxy the First Order has created the Starkiller Base. Built within an ice planet and with a superweapon capable of destroying entire star systems, it is charged using the power of stars. The Starkiller Base has moved into the solar system and seeks to use the Sun to power its weapon to destroy the Earth.
Figure 3: The Starkiller Base charging its superweapon by draining energy from the local star. Credit: Star Wars: The Force Awakens, Lucasfilm.
For this question you will need that the gravitational binding energy, object with mass M and radius R is given by
U=
U , of a uniform density spherical
3GM 2 5R
and that the mass-luminosity relation of low-mass main sequence stars is given by L
∝ M 4.
a. Assume the Sun was initia lly made of pure hydrogen, car ries out nuclea r fusion at a constant rate and will continue to do so until the hydrogen in its core is used up. If the mass of the core is 10% of the mass of the star, and 0.7% of the mass in each fusion reaction is converted into energy, show that the Sun’s lifespan on the main sequence is approximately 10 billion years. b. The Starkiller Base is able to stop nuclear fus ion in the Sun’s core . (i) At its current luminosity, how long would it take the Sun to radiate away all of its gravitational bindin g energy? (This is an estim ate of how long it would take to drain a whole star when radiatively charging the superweapon.) (ii) How does your value compare to the main sequence lifetime of the Sun calculated in part a.? (iii) Comment on whether there were (or will be) any events in the life of the Sun with a timescale of this order of magnitude. c. In practice, the gravitational binding energy of the Earth is much lower than that of the Sun, and so the First Order would not need to drain the whole star to get enough energy to destroy the Earth. Assuming the weapon is able to channel towards it all the energy being radiated from the Sun’s entire surface, how long would it take them to charge the superweapon sufficiently to do this?
3
The First Order find that radiative charging of the weapon is too slow to satisfy their plans for galactic domination, and so instead the weapon charging process compresses and stores part of the star within the Starkiller Base (as shown in Figure 3). To avoid creating a black hole, the First Order cannot compre ss stellar matter below its Schwarzschild radius, R S = 2GM/c 2 . d. Taking the Starkil ler Base’s ice planet to have a diameter of 660 km, show that the Sun can be safely contained, even if it was fully drained. e. The Starkiller Base wants to destroy all the plane ts in a stellar system on the far side of the galaxy and so drains 0.10 M from the Sun to charge its weapon. Assuming that the U per unit volume of the Sun stays approximately constant during this process, calculate: (i) The new luminosity of the Sun. (ii) The new radius of the Sun. (iii) The new temperature of the surfa ce of the Sun (current T = 5780 K), and suggest (with a suitable calculation) what change will be seen in terms of its colour. The Resistance defeat the First Order and destroy the Starkiller Base when it was almost fully charged. Upon releasing the energy stored in the base it causes the planet to turn into a small star.
Figure 4: The Resistance fighters escaping the Starkiller Base as it turns into a star. Credit: Star Wars: The Force Awakens, Lucasfilm.
f. Assume that at the moment of destr uction of the Starkil ler Base the mass of the new star formed is equal to the mass drained from the Sun (0.10 M ). Derive an expression for the main sequence lifetime in terms of stellar mass, and hence calculate the main sequence lifetime of this new star.
4
Qu 4. Hanny’s Voorwerp Hanny’s Voorwerp (Dutch for ‘object’) is a rare type of astronomical object discovered in 2007 by the school teacher Hanny van Arkel whilst part icipating as a volunteer in the Galaxy Zoo project. When inspecting the image of the galaxy IC 2497 in the constellation Leo Minor, she observed a bright green blob close to the galaxy.
Figure 5: HST image of galaxy IC 2497 and the glowing Voorwerp below it. Credit: Keel et al. (2012) & Galaxy Zoo.
Subsequent observations have shown that the galaxy IC 2497 is at a redshift of z = 0.05, with the Voorwerp at a similar distance and with a projected angular separation of 20 arcseconds from the centre of the galaxy (3600 arcseconds = 1 ◦ ). Radio observations suggest that the Voorwerp is a massive cloud of gas, made of ionized hydrogen, with a size of 10 kpc and a mass of 10 11 M . It is probably a cloud of gas that was stripped from the galaxy during a merger with another nearby galaxy. In this question you will explore the cause of the ‘glow’ of the Voorwerp and will learn about a new type of an astronomical object; a quasar. a. Given that Hubble’s constant is measured as H 0 = 70 km s−1 Mpc−1 , calculate the distance to the galaxy (in Mpc). The rate of ionizing photons from a source (in photons per second) can be expressed as:
S∗ = V n2 α where V is the volume of the ionized region, n is the number density of the ionized gas and α is the ionization coefficient, α = 2.6 10−13 cm3 s−1 .
×
b. Calculate the power (luminosity) of the source required to completely ionize the Voorwerp (assumed to be spherical), given that the mass of a hydrogen atom is 1.67 10−27 kg and the ionization energy of hydrogen is 13.6 eV, where 1 eV = 1.60 10−19 J.
×
5
×
One possible source of ionizing radiation is the jet arising from the accretion of material onto the supermassive black hole (SMBH ) situated in the centr e of the galax y. This produces an enormous amount of energy, greatly brightening the galaxy; a galaxy shining due to this process is known as a quasar. c. The gravitational potential ener gy of the material falling to radius R , which in this case is a black hole with radius equal to the Schwarzschild radius, RS = 2GM/c 2 , at a mass accretion rate m ˙ δm/δt, is converted into radiation with an efficiency of η . Show that the power (luminosity) output of the SMBH is given by:
≡
L=
1 η mc ˙ 2. 2
∼
d. The typical mass accretion rate onto an activ e SMBH is 2 M yr −1 and the typical efficiency is η = 0.1. Calculate the typical luminosit y of a quasar. Compare the luminosity of the quasar with the power needed to ionize the Voorwerp. Detailed astronomic al observations have shown than the nucleus of the galaxy has a modest luminosit y of L < 1033 W, thus the black hole in IC 2497 is not currently active (i.e. the accretion rate is very low). Quasars are thought to ignite every time the black hole starts accret ing a fresh source of matter , and switch off once that supply is exhaust ed. Therefore, this might be the first evidence of a quasar switching off recently (by astronomical standards), with the Voorwerp reflecting the light emitted by the quasar whilst it was still active. This would make the Voorwerp a ‘quasar ionization echo’ and IC 2497 the nearest galaxy to us to host a quasar. e. Calculate the projecte d physical separa tion, rp , between the galaxy and the Voorwerp. f. Derive an expression for the difference in the light travel time between photons travelling directly to Earth from the galaxy and photons reflected off the Voorwerp first. Give your formula as a function of rp andasθ measured , where θ is between lines of of sight to the Earth and to theθcentre of the Voorwerp bythe anangle observer at thethe centre IC 2497. (For example = 90◦ would correspond to the galaxy and Voorwerp both being the exact same distance from the Earth, and so the projected distanc e rp is therefore also the true distance between them.) g. High precision measurements showed that the Voorwerp is slightly further away than the galaxy, and so θ = 125 ◦ . Use this with your expression from the previous part of the question to estimate an upper limit for the number of years that have passed since the quasar was last active.
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Qu 5. Imaging an Exoplanet Recently a group of researchers announced that they had discovered an Earth-sized exoplanet around our nearest star, Proxima Centauri. Its closeness raises an intriguing possibility about whether or not we might be able to image it directly using telescopes. The difficulty comes from the small angular scales that need to be resolved and the extreme differences in brightness between the reflected light from the planet and the light given out by the star.
Figure 6: Artist’s impression of the view from the surface of Proxima Centauri b. Credit: ESO / M. Kornmesser
Data about the star and the planet are summarised below: Proxima Centauri (star)
Proxima Centauri b (planet)
Distance
1.295pc
Mass
0.123M Mass (min)
Radius
0.141R
Surface temperature
3042 K
Apparent magnitude
11.13
Orbital period
Radius (min)
11.186 days
≈ 1.27 M ≈ 1.1 R
⊕
⊕
The following formulae may also be helpful: m
where
− M = 5 log
m
d
M
10
is the apparent magnitude,
M
−M
=
−2.5log
L L
∆m = 2.5log CR
is the absolute magnitude, d is the distance in parsecs, and the fstar
contrast ratio (CR) is defined as the ratio of fluxes from the star and planet,
CR =
fplanet .
a. Calculate the maximum angular separ ation between the star and the planet, assumin g a circular orbit. Give your answer in arcseconds (3600 arcseconds = 1 ◦ ). b. Determine the luminosity of the star and hence calculate the flux received on the Earth (in W m−2 ) from both the star and the planet. Use them to work out the contrast ratio and thus the apparent magnitude of the planet. Assume the planet refle cts half of the incident light and that M = 4.83.
7
The resolving power of a diffraction limited telescope is given by
θmin = 1.22
λ D
where λ is the wavelength being observed at, D is the diameter of the telescope aperture, and θ min is the smallest angular separation (in radians) the telescope can distinguish. Data about some current and planned telescopes are summarised below: Telescope
Diameter (m)
Faintest
m
detectable
Hubble Space Telescope (HST) Keck II (based in Hawaii)
2.4 10.0
31 (variable)
James Webb Space Telescope (JWST)
6.5
34
c. Verify that the HST (which is diffraction limited since it’s in space) would be sensitive enough to image the planet in the visible, but is unable to resolve it from its host star (take λ = 550 nm). Ground-based telescopes have bigger mirrors than the HST, but are not diffraction limited due to movements in the atmosphere and so need to be fitted with ‘adaptive optics’ (AO) to compensate for this effect. However, even with perfect AO the faintest object the telescope can detect is limited by the brightness of the atmosphere. The signal-to-noise ratio (SN R) can be approximated as:
SN R
fAt ≈ √fAt + bt
where f is the flux from the object (in photons m −2 s−1 ), A is the area of the telescope mirror, is the overall efficiency of the telescope and detector, b is the flux from the sky (in photons s −1 ), and t is the length of the exposure. d. Calculate the exposur e time needed for a Keck II image of the exoplanet to have an S N R of 3 (i.e. barely detectable). Assume that the telesco pe has perfect AO (so it is diffraction limited), is observed at the longest wavelength for which the planet can still be resolved from the star, all the received flux from the planet consists of photons of that longest wavelength, = 0.1 and b = 109 photons s −1 (so b f ). Comment on your answer.
The James Webb Space Telescope (JWST) is the successor to the HST and is due to launch in 2018. It should be able to both resolve the system and cope with the contrast ratio. Since it is in space it is diffraction limited, and the SN R should be dominated by the flux from the planet (i.e. f b).
e. How long an exposure would JWST nee d in order to get the same SN R as Keck II, again if observed at the longest wavelength for which the planet can still be resolved from the star by the telescope? (Make similar assumptions about the received flux and use the same value of .)
END OF PAPER Questions proposed by: Dr Alex Calverley (Bedford School) Dr Emile Doran (The Langley Academy) Sandor Kruk (University of Oxford)
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BAAO 2016/17 Solutions and Marking Guidelines
Note for markers:
Answers to two or three significant figures are generally acceptable. The solution may give more in order to make the calculation clear. There are multiple ways to solve some of the questions; please accept all good solutions that arrive at the correct answer. If a candidate gets the final (numerical) answer then allow them all the marks for that part of the question (as indicated in red), so long as there are no unphysical / nonsensical steps or assumptions made.
Q1 - Martian GPS
[Total = 20]
a. Given that the Earth’s sidereal day is 23h 56 mins, calculate the orbital radius of a GPS satellite.
⊕
Express your answer in units of R .
⊕ ⊕
[1] [1]
[1]
[3]
b. How long would it take a radio signal to travel directly between a satellite and its closest neighbour in its orbital plane (assuming they’re evenly spaced)? How far would a car on a motorway (with a -1 speed of 30 m s ) travel in that time? [This can be taken to be a very crude estimate of the positional accuracy of the system for that car.].
Closest neighbour in orbital plane should be 90° away (since evenly spaced) so can use Pythagoras [1]
Time for the signal to travel that distance: [1]
[2]
[1]
[1]
Distance travelled by a car on a motorway in that time:
[In practice the positional accuracy of a GPS system is much harder to calculate; at low speeds it is typically a function of fluctuations and reflections of the signal within the atmosphere, as well as the presence of objects that might block the signal]
c.
Using suitable calculations, explore the viability of a 24-satellite GPS constellation similar to the one used on Earth, in a semi-synchronous Martian orbit, by considering: i.
Would the moons prevent such an orbit?
[1] [1] [1]
[3]
This is about ½ the distance to Diemos and about 3500 km (about RM) away from Phobos so the moons should not provide a problem for any GPS satellite constellation ii.
[1]
[1]
How would the GPS positional accuracy compare to Earth?
(Using similar reasoning to part b.) (or
[1] )
The positional accuracy is about twice as good as on Earth iii.
[1]
[2]
[1]
[1]
What would the receiving angle of each satellite’s antenna need to be, and what would be
the associated satellite footprint? By comparing these with the ones utilised by Earth’s GPS, make a final comment on the viability of future Martian GPS. Receiving angle:
[1] [1]
[2]
The area of a 'zone' of a sphere is 2π Rh where h is the radial
height of the zone. From the geometry of the situation: [1] [1]
Fraction of surface area:
[1]
[3]
Receiving angle similar to Earth's so can use current GPS satellite technology
[0.5]
Satellite footprint similar to Earth's so should get sufficient coverage
[0.5]
Martian GPS system is viable
[1]
[2]
Q2 - Hohmann Transfer
[Total = 20]
-1
a. Show that vorb in low Earth orbit (LEO; about 200 km above the surface) is about 8 km s . This is an estimate of the Δv the rockets need to provide for the spacecraft to reach LEO.
⊕ ⊕
[1] [1]
[2]
-1
[In practice a Δv of 8 km s assumes no external forces, but atmospheric drag can increase the -1 necessary Δv by 1.3 – 1.8 km s . When travelling between objects in space, however, such drag forces are absent and so the Δv calculated is much more accurate]
b. Derive expressions for ΔvA and ΔvB by comparing their circular orbital speeds with their transfer orbit speeds. Simplify your final expressions to include G, M⨀, rA and rB only.
[1]
But 2a = rA + rB
[1]
[2]
[1]
[1]
Similarly:
[These equations have been written so that the change in speed is positive, however give full credit for reversed signs (so long as they are consistent)] c.
Approximating Mars' orbit as circular with a radius of 1.52 AU, calculate the Δv to go from Earth LEO to Mars i.e. Δv = |ΔvA|+|ΔvB|. Compare your answer to the Δv to reach Earth LEO.
Δv LEO This is less than to get theinto [1] (So most of the effort needed in going to Mars simply comes from leaving Earth)
[1]
[1]
[1]
[3] [1]
[The Δv calculated here would be for the spacecraft to enter a circular orbit around the Sun at the same distance as Mars, but this would not constitute landing – the extra Δv to get to the Martian -1 surface increases the total for the whole transfer to roughly 8 km s (same as to get into LEO). Other, more complicated routes can be taken, some of which offer substantial efficiencies so the Δv for the trip can be much lower (although the time taken to complete the manoeuvre will be longer, and the Δv to reach LEO will still be the biggest single step)] d. Derive an expression for the total time spent on the transfer orbit, tH, and calculate it for an Earth to Mars transfer. Give your answer in months. (Use 1 month = 30 days).
From Kepler's third law: Since the spacecraft only covers half of the ellipse the time on the journey is half the period, and given that 2a = rA + rB then:
e.
[1]
[1]
[1] [1]
[2]
Hence calculate the direct distance between Earth and Mars at the moment the spacecraft reaches Mars. How long would it take a radio message from the spacecraft to reach Earth? Initially Earth is at A. When the spacecraft reaches B (after 8.56 months), the Earth has moved round the Sun in its orbit and is
x A'
θ
now at A'.
Angle between A' and B:
[1]
Using the cosine rule:
[1]
[2]
[1]
[1]
Since radio waves travel at the speed of light the time taken by the message is:
[In practice, due to the eccentricity of Mars' orbit, the signal transmission time varies depending on the year the spacecraft was launched. When the rover Curiosity arrived at Mars the engineers described the landing as 7 minutes of terror, since the signal from the spacecraft would take 14 minutes to reach Earth but the time to transverse the Martian atmosphere was only 7 minutes (hence the process had to be completely automated)]
f.
How long would any astronauts on board the spacecraft need to wait until they could use a Hohmann transfer orbit to return to Earth? Hence calculate the total duration of the mission.
A"
θ
During the transfer the Earth moves by , so the spacecraft should launch from Mars when Earth is at position A" (exhibiting symmetry with when it arrived). Since the planets move anti-clockwise in this diagram the angle covered by Earth from A' to A" is [1]
From Kepler's third law, the period of Mars
[1]
Therefore, the relative angular velocity of Earth if Mars' motion is subtracted out is: -1
-8
[1]
-1
(allow any equivalent units e.g. 168° year , 9.38 × 10 rad s etc.)
Consequently, the time the astronauts need to wait for Earth to get from position A' to A" is:
[1]
[4]
[1]
[1]
Thus the total duration of a return mission to Mars is:
[Shorter missions are possible, but would require a greater Δv and hence need much more fuel - any future mission will have to balance the cost (and mass) of more fuel on a fast trip with the cost (and mass) of more supplies on a slow trip]
Q3 - Starkiller Base
[Total = 20]
a. Assume the Sun was initially made of pure hydrogen, carries out nuclear fusion at a constant rate and will continue to do so until the hydrogen in its core is used up. If the mass of the core is 10% of the star, and 0.7% of the mass in each fusion reaction is converted into energy, show that the Sun’s lifespan on the main sequence is approximately 10 billion years. Time on main sequence = total nuclear energy available / luminosity
[1] [1]
[2]
b. The Starkiller Base is able to stop nuclear fusion in the Sun’s core i. At its current luminosity, how long would it take the Sun to radiate away all of its gravitational binding energy? (This is an estimate of how long it would take to drain a whole star when radiatively charging the superweapon.) Time radiating energy = total gravitational binding energy / luminosity
ii.
[1]
[2]
How does your value compare to the main sequence lifetime of the Sun calculated in part a.? This is much shorter than tMS
iii.
[1]
[1]
[1]
Comment on whether there were (or will be) any events in the life of the Sun with a timescale of this order of magnitude. An event in the Sun's life that happened on a timescale of this order of magnitude is the gravitational collapse of the protostar before it joined the main sequence [1] [1] [The Sun will also be on the asymptotic giant branch (AGB) for a similar order of magnitude of time - this is when the core is completely carbon / oxygen (but no longer undergoing fusion) and there is a spherical shell of helium burning happening just outside the core (with a shell outside that of hydrogen burning). Credit this answer too if a student mentions it.]
c.
In practice, the gravitational binding energy of the Earth is much lower than that of the Sun, and so the First Order would not need to drain the whole star to get enough energy to destroy the Earth. Assuming the weapon is able to channel towards it all the energy being radiated from the Sun’s
entire surface, how long would it take them to charge the superweapon sufficiently to do this? Time charging the weapon = total energy needed / rate of energy transfer
⊕ ⊕⊕
[1] [1]
(So it would only take a week to absorb enough energy from the Sun to destroy the Earth!)
[2]
d. Taking the Starkiller Base’s ice planet to have a diameter of 660 km, show that the Sun can be safely contained, even if it was fully drained.
Need to work out the Schwarzschild radius for the Sun, and compare it to the size of the base
(About 100 times) Smaller than the radius of the base e.
[1]
[1]
the Sun can be safely contained [1]
[1]
The Starkiller Base wants to destroy all the planets in a stellar system on the far side of the galaxy and so drains 0.10 M⨀ from the Sun to charge its weapon. Assuming that the U per unit volume of the Sun stays approximately constant during this process, calculate: i.
The new luminosity of the Sun. Need to use the mass-luminosity relation for main sequence stars (L
⨀ ∝
luminosity of a 0.9 M star (since that is the new mass of the Sun)
ii.
∝
4
M ) to work out the
[1]
[1]
The new radius of the Sun. Energy density = U / V = constant
iii.
∝
[1] [1]
[2]
The new temperature of the surface of the Sun (current T ⨀ = 5780 K), and suggest (with a suitable calculation) what change will be seen in terms of its colour. Need to use the Stephan-Boltzmann Law to get the new temperature and then Wien's Law
to determine the effect on the peak wavelength (and hence the colour)
[1]
[1] [1]
[2] [1]
This is a longer wavelength than the current peak (500 nm) so the Sun is redder [1]
[1]
f.
Assume that at the moment of destruction of the Starkiller Base the mass of the new star formed is equal to the mass drained from the Sun (0.10 M ⨀). Derive an expression for the main sequence lifetime in terms of stellar mass, and hence calculate the main sequence lifetime of this new star.
∝ ∝ ∝ ∝
We can combine the mass-luminosity relation with the expression we used in part a. [1]
[1]
[1] [1]
[2]
Q4 - Hanny's Voorwerp
[Total = 20]
-1 -1 a. Given that Hubble’s constant is measured as H 0 = 70 km s Mpc , calculate the distance to the galaxy
(in Mpc).
Need to turn the redshift into a recessional velocity and then combine with Hubble's Law [1] [1]
[2]
b. Calculate the power (luminosity) of the source required to completely ionize the Voorwerp (assumed -27 to be spherical), given that the mass of a hydrogen atom is 1.67 × 10 kg and the ionization energy -19 of hydrogen is 13.6 eV, where 1 eV = 1.60 × 10 J. Given we know the radius of the cloud (10 kpc) and the mass (10
11
⨀
M ) we can work out the number
density of hydrogen atoms
[1] [1] [1]
(watch that the units of α are converted correctly to SI)
The luminosity can then be calculated as we know the energy of each photon [1] [1]
[5]
(Allow full credit for interpreting the 10 kpc 'size' of the cloud to mean its diameter rather than its 56 -1 38 radius, giving S* = 2.39 × 10 photons s and L = 5.20 × 10 W) 2 [Working out S* directly may prove difficult for some calculators as (M/m H) may exceed their largest power of ten, in which case students should work out √S* and then square it later.] c.
The gravitational potential energy of the material falling to radius R, which in this case is a black hole 2
with radius equal to the Schwarzschild radius, RS = 2GM/c , at a mass accretion rate
, is
converted into radiation with an efficiency of . Show that the power (luminosity) output of the SMBH is given by
.
We know the gravitational potential energy of a particle of mass m at the Schwarzschild radius is the same as the kinetic energy it has gained moving from infinity to that point, so [1]
Given that a fraction η is converted into radiation and the given mass accretion rate then
[1]
[2]
(So the maximum energy you can get from a black hole is half the rest mass energy of the material falling in – this is a much more efficient process for generating energy than the 0.7% you get from nuclear fusion in stars, which in themselves are much more efficient than chemical reactions!) -1
d. The typical mass accretion rate onto an active SMBH is ∼2 M⨀ yr and the typical efficiency is η = 0.1. Calculate the typical luminosity of a quasar. Compare the luminosity of the quasar with the power needed to ionize the Voorwerp. -1
Need to convert the mass accretion rate into kg s and then put into the formula
[1]
The luminosity of the quasar is high enough to ionize the Voorwerp e.
[1]
[2]
[1]
[1]
Calculate the projected physical separation, r p, between the galaxy and the Voorwerp.
Since the angle is so small, we can use the small angle approximation for tan θ ≈ θ
f.
[0.5] [0.5] [1]
[2]
Derive an expression for the difference in the light travel time between photons travelling directly to Earth from the galaxy and photons reflected off the Voorwerp first. Give your formula as a function of rp and θ, where θ is the angle between the lines of sight to the Earth and to the centre of the Voorwerp as measured by an observer at the centre of IC 2497. (For example θ = 90° would correspond to the galaxy and Voorwerp both being the exact same distance from the Earth, and so the projected distance r p is therefore also the true distance between them.) Given the small angular separation we can treat the light rays from the galaxy to Earth and from the Voorwerp to Earth as essentially parallel, and so the difference in light travel time comes from the extra distance travelled in being reflected off the Voorwerp Relevant diagram, suitably labelled
[2]
Extra distance = x + y
[1] [1] [1]
[5]
(alternative form)
[Allow any equivalent formula, for example expressing it in terms of csc θ and cot θ, so long as some attempt has been made to simplify it. It is quicker and simpler to derive if θ is assumed to be acute – we show it this way in case students see that the angle is obtuse from the next part of the question and want to have a consistent picture throughout]
g. High precision measurements showed that the Voorwerp is slightly further away than the galaxy, and so θ = 125°. Use this with your expression from the previous part of the question to estimate an upper limit for the number of years that have passed since the quasar was last active.
(This is remarkably recent on astronomical timescales!)
[0.5] [0.5]
[1]
Q5 - Imaging an Exoplanet
[Total = 20]
a. Calculate the maximum angular separation between the star and the planet, assuming a circular orbit. Give your answer in arcseconds (where 3600 arcseconds = 1°).
[1] [1] [0.5] [0.5]
[3] -2
b. Determine the luminosity of the star and hence calculate the flux received on the Earth (in W m ) from both the star and the planet. Use them to work out the contrast ratio and thus the apparent magnitude of the planet. Assume the planet reflects half of the incident light and that ⨀ = 4.83.
[1] [1]
[2]
[1]
[1]
[1] [1]
[2]
[1]
[1]
[1] [1]
[2]
[Since we only know the minimum radius of the exoplanet it could be larger and hence brighter, however it may also reflect less than half the incident light from the star and so be fainter – in practice the numbers used here are an optimistic estimate and it is more likely to be fainter.] (Accepted alternative methods: if they use Stephan-Boltzmann's Law then L = 5.88 × 10 23 W, fstar = 2.92 × 10-11 W m-2, and fplanet = 3.39 × 10-18 W m-2, though the contrast ratio and magnitude should be the same. Also accept if they assume only the day side is able to reflect and hence the apparent magnitude brightens to 27.9)
c.
Verify that the HST (which is diffraction limited since it's in space) would be sensitive enough to image the planet in the visible, but is unable to resolve it from its host star (take λ = 550 nm).
Since θHST >then θmaxthe HST can't resolve it
[0.5]
[0.5]
Apparent magnitude of planet (28.5) is brighter (greater) than limiting magnitude (31)
[1] [1]
[1]
d. Calculate the exposure time needed for a Keck II image of the exoplanet to have an SNR of 3. Assume that the telescope has perfect AO, is observed at the longest wavelength for which the planet can still be resolved from the star, all the received flux from the planet consists of photons of that longest wavelength , ε = 0.1 and b = 10 9 photons s-1 (so b >> f). Comment on your answer.
[1]
Photon flux:
-2
[1]
-1
(SB-Law fplanet gives f = 25.4 photons m s , and day side reflection increases f by a factor of 2)
Since b >> f we can simplify the denominator of the SNR formula by ignoring the first term, so [1]
[3]
This is really long (> 50 years!) so it is unlikely Keck will ever be able to directly image it [1]
[1]
(alternative f gives 22.7 ks, and hence the conclusion is that it e.
is feasible)
How long an exposure would JWST need in order to get the same SNR as Keck II, again if observed at the longest wavelength for which the planet can still be resolved from the star by the telescope? (Make similar assumptions about the received flux and use the same value of ε).
(using similar reasoning to the previous part of the question)
-2
[1] [1] -1
(alternative method gives f = 16.5 photons m s ) Since b << f we can simplify the denominator of the SNR formula by ignoring the second term, so [1]
[3]
(alternative method gives t = 0.16 s, and day side reflection decreases t by a factor of 2) [This is much more reasonable – the optimistic assumptions we have made throughout this question mean this is just a lower limit and so the actual exposure time will be longer, perhaps several minutes to get a much higher SNR and to compensate for the fact for that the sensitivity is lower at 970 nm than we have quoted since it varies with wavelength. The main thing that would prevent direct observation of the exoplanet by JWST would be if the contrast ratio was much higher than calculated here as that would mean the star's light would need to be blocked out by the Near Infrared Camera's coronagraph – unfortunately this can only be used at wavelengths too long to resolve the system.]
END OF PAPER
Astronomy & Astrophysics A2 Challenge September - December 2016 Solutions and marking guidelines •
•
•
•
•
The total mark for each question is in bold on the right hand side of the table. The breakdown of the mark is below it. There is an explanation for each correct answer for the multiple-choice questions. However, the students are only required to write the letter corresponding to the right answer. n Section !, students should attempt either "u #$ or "u #%. f both are attempted, consider the question with the higher mark. &nswers to one or two significant figures are generally acceptable. The solution may give more in order to make the calculation clear. There are multiple ways to solve some of the questions, please accept all good solutions that arrive at the correct answer.
"uestion &nswer Section A #. (
'ark 10 #
The !oriolis force is an inertial force that acts on ob)ects that are in motion relative to a rotating reference frame. &s a result, hurricanes rotate anticlockwise in the northern hemisphere and clockwise in the southern hemisphere. *this is also explained through cons. of angular mom.+
.
#
There are two tides every day one caused by the gravitational attraction of the 'oon on the side facing it, and the other due to the centrifugal force experienced by the side furthest from the centre of mass.
$.
(
#
The number of photons collected is proportional to the area of the aperture, which is proportional to the diameter. Therefore if the diameter is % times bigger it will receive % / #0 times more photons.
%.
&
#
&quila is not a 1odiacal constellation. &ccording to the astronomical definition, there are #$ 1odiacal constellations, with 2phiucus being the least known of them.
3.
(
#
The orbit of the 'oon is inclined 34 to the ecliptic. (uring winter, the Sun has its lowest declination. Thus the 5ull 'oon, which is on the opposite side of the sky, will have its highest declination and will be visible highest in the sky.
0.
#
6ach fold doubles it, so new thickness after n folds is #7 μm 8 n. Setting this equal to # &9 gives n / 3$.: so closest is 37.
1
:.
!
#
The altitude of ;olaris *
=.
°N
&
°.
#
The rising and setting sun alignments will happen an equal number of days before and after the winter and summer solstices respectively *roughly #st >une and #st (ecember+. Since ##th >uly is # days after the summer solstice, you were looking for a date about # days before *in order to get a setting Sun+.
?.
!
#
The first clue is that you are not at the
16 Miles. #7.
.° ° 0.0011 1.9 10 %.1 &' The distance the image was taken at is . !" #$
#
The angular si1e of a pixel is
Section B
##.
10 a. i. Answer:
2
764 ' s
Sun will appear stationary if the 6urofighter moves at the 6arthAs rotational speed, so
4510 764 ' s ()*+,-/ %3 2 %%76.6060 ii. Answer: %>? ' s
#
2xford is less far from the 6arthAs rotational axis, so the speed needed will be less by a factor of cos θ *where θ is the latitude of 2xford+
(8/: % 32 ;
?
m s-#
#
b. 2
()*+,-/
&t the 6quator a day is # hours *due to the value of +, and during a day the Sun travels #=74, so since it is 7.34 across it covers that angular distance in
. @ ×1% = 0.044 A
?00F764 = 45 G H = 0.0> ()*+,-/
So time to rise the sun by 7.34 becomes minutes
.@ #.
=%? 1% 900 &'
# # #
minutes
a. Answer:
IJ M KL = :
#
d / # ?77 km
#
This is a much higher altitude than the SS.
#
B
b. Answer
5.71 A
2
9sing DeplerAs Third Eaw
3 = OPN QR S3=T
N N OPU V2 WXYR = T.Z×[[×.Z×N\ V6.45×10 W1%.9×10 YR T / .0: 8 #7% s */ :.%# hours+
B
3
#
#
Section C !ither "u #$ or " #% a. #$.
10 "
&t Foche limit 5grav / 5tidal, so
Oa+N = OP+/ / :fgh !orrect rearrangement and cancelling
[
XI] =^% aPjh
#
'ass of planet and satellite in terms of density
ut
#
k = R `PR and G = R `a^R
Substitution of densities into equation
S XI] ^_% ``Pa^RRbR
#
!ancelling of r to leave required expression
S XI] =c% ``aP dR
#
b.
2
?.6>×10 = 619 &n ' o4 `l,-+/m = 7kl,-+/m = 7 ×V6. R 4 l,-+/m 4 0%5×10 ZYR R 619 R XI] = l,-+/m c% `l,-+/m &' `pq) d = 60 %50c% 940d V== 1.661400 0 l,-+/m Y c.
#
#
R XI]ra,8 = %.77×60 %50c 619 d 940 = 1%> 700 &' V= %.14 l,-+/mY
d. The limits *roughly+ agree with the observed extent of the ring system
1 # 1 #
B&llow students to say the inner edge agrees well with the simple model, but the outer edge is a poorer fitC e. 5inding the density of a moon that reached the fluid Foche limit at
`a = V.tYh = V.Yh =11%7 kg m 4
-$
l,-+/m
2
#
&ssuming Geritas was spherical then the radius is
R
R
[z ^ _P\huvwxy b _\hR b 1.>?×10 m V= 1>? kmY
#
Biven that this is similar in si1e to several other moons of Saturn, the idea that the rings came from the tidal destruction of a moon is not completely outlandishC
#%
a.
1
The period is the time interval between two consecutive peaks of the blue curve. 5rom the radial velocity curve, the period is ## days.
#
B5ull marks for the period within I daysC b. 9sing DeplerAs Third Eaw
1
~Nh = N , OP The semi-ma)or axis of the planetJs orbit is
R Q = _{k73 b =5.16×10 ' = 0.07> |}
#
B
1
(qp/q = %Q3 (qp/q = 75.40 &' s
#
B
S •€-,/ W•€‚ƒ,m)- = 0
•€…P = 0
-,/ FG((‚ƒ,m)- = 0 S k(SG=k -,/ (‚ƒ,m)9sing (-,/ = ? &' A
#
#
the planetJs orbit is not known. &stronomers are only able to measure the radial velocity of the star, not the tangential one. n the calculations above we assumed that the plane of the orbit is in the line of sight , thus the radial velocities are the maximum, total velocities. Therefore, the mass we determined is a minimum mass for the planetK it is in fact , where is the inclination of the orbit.
GsiD†
#
†
e. Le are given that the orbit is an ellipse, hence using the diagram on page , we can determine the minimum and maximum distance from the planet to the star. The minimum distance *also known as periapsis+ is
^apm = QV1F‡Y ^apm =0.07> V1F0.4? Y = 0.041 |} The maximum distance *also known as apoapsis+ is
^a,8 = QV1W‡Y ^apm =0.07> V1W0.4? Y = 0.06? |} Thus, the distance from the planet to ;roxima !entauri ranges between 7.7$#&9 and 7.703 &9.
#
Hence the maximum equilibrium temperature of the planet is
3ˆƒ,m)-ra,8 = 3l-,/ c l-,/ d 3ˆƒ,m)-ra,8 = 405%^apm‰ The minimum equilibrium temperature of the planet is
3ˆƒ,m)-rapm = 3l-,/ c%^l-,/ d a,8 3ˆƒ,m)-rapm = %1% ‰ The temperature of the planet ranges between # D and $7: D, or -0# 4! and $%4!.
#
The habitable 1one is the band around a star where a planet can have water on its surface in liquid form, at normal pressure. Hence, the equilibrium temperature of the planet must be between 74! and #774!. The temperature on ;roxima !entauri reaches values above 74!, thus it is in the habitable
#
1one of its host star. f. The distance to ;roxima !entauri, the nearest star to the Sun, is %. light years. &ssuming that the lifetime of a human being is =7 years, the robotic space probe would need to travel at a speed of #3 =77 to reach the star in a lifetime. Such high velocities have not been achieved yet, currently the fastest man-made ob)ect is the >uno 'ission, travelling at %7 . Thus, the prospects of reaching ;roxima !entauri in our lifetime are low, unless new technology is developed.
&' s &' s
6
1
#
Astronomy & Astrophysics A2 Challenge September – December 2016 Instructions
Time Allowed: One hour In Section C, you can choose to answer either Q13 on the Saturn’s rings, or Q14 on the discovery o the nearest e!o"lanet# $ar%s allocated or each &uestion are shown in 'rac%ets on the right# (ou will need to use a ruler# (ou may use any calculator# (ou may use any standard ormula sheet# This is the irst "a"er o the )ritish Astronomy and Astro"hysics *lym"iad in the +1-.+1/ academic year# To "rogress to the ne!t stage o the )AA*, you must ta%e )0h* ound 1 in 2ovem'er, which is a general "hysics "ro'lem "a"er# To 'e awarded the highest grade istinction5 in this "a"er, it should 'e sat under test conditions and mar%ed "a"ers achieving -6 or a'ove should 'e sent in to the )0h* *ice at *!ord 'y 7ednesday 8th 2ov +1-# To solve some o the &uestions, you will need to write e&uations, draw diagrams and, in general, show your wor%ing# (ou are also encouraged to loo% at the clear s%y and identiy the 'rightest stars, a ew days 'eore sitting the "a"er# This "a"er has more than an hour’s worth o &uestions# (ou are encouraged to have a go at as many as you can and to ollow u" on those that you do not com"lete in the time allocated#
British Physics Olympiad Sponsors
Worshipful Company of Scientific Instrument Makers
1
Useful constants
S"eed o light 9ravitational constant Solar mass Solar radius Astronomical nit ;ight year
0×10 6.1.3.6907×10 9×10 6.1.4996×10 5×10 9.4365.6×1024125 5.6.937×10 7×10 23.4
c G M R A ly
or a!is
1 year 1 day M
N mmskg %g m m m
A days hours %g m
°
(ou might ind the diagram o an elli"tical or'it 'elow useul in solving some o the &uestions: B
b P F
A
a
O
C
= 1! $"##
? *A ?0*5 = semi=ma>or a!is ? *) ?C*5 = semi=minor a!is = eccentricity
@ = ocus 0 "eria"sis "oint nearest to @5 A a"oa"sis "oint urthest rom @5
Be"ler’s Third ;aw:
%&'%'
@or an elli"tical or'it, the s&uare o the "eriod o or'it o an o'>ect a'out the ocus is "ro"ortional to the cu'e o the semi=ma>or a!is the average o the minimum and ma!imum distances rom the Sun5# The constant o "ro"ortionality is , where M is the mass o the central o'>ect#
4(+)*
The &uestions were "ro"osed 'y: r Ale! Calverley )edord School5 Sandor Bru% niversity o *!ord5
2
Section A: ultiple Choice Write the correct answer to each question. Each question is worth 1 mark. There is only one correct answer to each question. Total: 10 marks.
1# 7hy do hurricanes rotate anti=cloc%wise in the northern hemis"here and cloc%wise in the southern hemis"here A# rotating
+ 4 H 1-
4# 7hich o the ollowing is not a odiacal constellation, according to the astronomical deinition A# A&uila
)# A&uarius
C# *"hiucus
3
# 0isces
E# 7hen o'serving rom the B, during which season is the @ull $oon visi'le highest in the s%y A# )# C# #
S"ring Summer Autumn 7inter
-# 9iven an overwhelmingly large "iece o "a"er, with a thic%ness o 1 μm, a""ro!imately how many times do you need to old it in hal theoreticallyJ5 or the thic%ness o the inal stac% to reach rom
4 E - /
°° ° N °°
/# At a latitude o E+ A# )# C# #
what is the altitude o 0olaris a'ove the horion
3H 4H E+ 8
H# K$anhattanhengeK is the name given to when, >ust 'eore sunset or >ust ater sunrise 4 times a year twice or setting, and twice or rising5, the Sun aligns with the east=west streets o the 2ew (or% grid system# *ne o the setting dates this year was on 11th Luly# 7hich o these is another date you are li%ely to see the K$anhattanhengeK sunset A# )# C# #
3th $ay 3th Lune 11th ecem'er 11th Lanuary
8# (ou travel 1 miles South, 1 miles
South 0ole 1 $iles rom the 2orth 0ole 11- $iles rom the South 0ole + $iles rom the South 0ole
××
1# *n Eth Se"tem'er +1-, the osetta mission has inally ound the 0hilae lander on Comet -/0.Churyumov9erasimen%o# Considering that 0hilae 1 1 1 m5 a""eared in an image rom the high=resolution camera with +4H +4H "i!els and ield o view +#+ +#+ 5 as +E +E "i!els, rom what distance did osetta manage to image 0hilae
×
°× ° × A# )# C# #
1#1 %m +#1 %m 1+#+ %m +-#H %m 4
<#
Section ): Short Answer
Each short question is worth 5 marks. Total:
10 marks.
!uestion 11 "orces o# $ature In the ))C "rogramme Forces of Nature, )rian Co! uses a
a# )y considering the circumerence o the etN M+ mar%sN '# A ust as the to" edge o the Sun has gone 'elow the horion, and ra"idly accelerates due west u" to a s"eed o E m s=1# 9iven that the Sun has an angular diameter o #EF as viewed rom et needs to ly or in order to see the whole o the Sun a'ove the horion M3 mar%sN
5
!uestion 12 Star %ars ogue One The new Star 7ars ilm Roue !ne concentrates on the creation o the irst eath Star, which in one scene causes a total ecli"se on the "lanet Scari, where it is 'eing 'uilt#
a# Assume the eath Star is 'eing 'uilt in or'it around the
6
Section C: 'ong Ans(er Each lon question is worth 1" marks. #nswer either $u 1% or $u 1&. Total: 10 marks.
!uestion 1) Saturn*s ings *ne "ossi'le theory or why the gas giants have ring systems is that a small moon got too close to the "arent "lanet# 7hen the gravitational tidal orces due to the dierence 'etween the strength o the "lanetDs "ull on the near and ar sides o the moon5 'ecame greater than the gravitational orces holding the moon together, it was ri""ed a"art# This minimum distance is called the Koche limitK, named ater the @rench astronomer
D
- * )- $ =
,
Consider a s"herical "lanet with mass and radius , and a "erectly rigid s"herical moon with mass and radius , or'iting the "lanet in a circular or'it o radius # @or a small "article o mass on the surace o the moon, the gravitational and tidal orces it e!"eriences will 'e 1567$8
/
2)*0.
/
a# )y ma%ing these two e!"ressions e&ual, derive an e!"ression or the oche limit, /9: , "urely in terms o , and the uniorm densities o the "lanet and the moon ;< and ;= res"ectively5 M4 mar%sN
7
'# se your ormula to calculate the oche limit o Saturn or a moon made o water ice ;= ? 83 %g m =35, given that *>$5?3@ ? E#-H O 1+- %g and ,>$5?3@ ? - +/ %m M+ mar%sN c# In "ractice, as a moon a""roaches the oche limit it will start to deorm and 'ecome more o an elli"soid than a s"here, causing the tidal orces to increase, and so the oche limit rom our sim"le model is really a minimum radius# The o""osite e!treme would 'e to assume that 'oth the "lanet and moon are made o a luid, and so can deorm without resistance this wor%s well when loo%ing at things li%e stars in close 'inary systems5# In that situation it can 'e shown that the e&uivalent ormula or the oche limit 'ecomes
/9: A
2.44,B ;;CD
7or% out this new ma!imum value or the oche limit or water ice around Saturn# M1 mar%N
,>$?@,>$?@
d# The inner edge o SaturnDs rings ring5 occurs at 1#11 , and the outer edge o the A ring the last main visi'le ring5 is at +#+/ # o the rings o Saturn all roughly5 'etween the two oche limits calculated or the e!treme cases o a "erectly rigid and a luid moon made o water ice M1 mar%N e#
,>$?@
M+ mar%sN
8
!uestion 1+ $earest e,oplanet -isco.ere*n +4th August +1-, astronomers discovered a "lanet or'iting the closest star to the Sun, 0ro!ima Centauri, situated 4#++ light years away, which ulils a long=standing dream o science=iction writers: a world that is close enough or humans to send their irst interstellar s"acecrat# Astronomers have noted how the motion o 0ro!ima Centauri changed in the irst months o +1-, with the star moving towards and away rom the
igure 'elow# 0ro!ima is a""roaching
a# @rom the radial velocity curve a'ove, determine the "eriod o the "lanet around 0ro!ima Centauri# M1 mar%N '# 0ro!ima Centauri is a red dwar star, unli%e our Sun, with a mass o only #1+ $ # 7hat is the semi=ma>or a!is o the "lanet’s or'it in A M1 mar%N
c# Assuming that the or'it o 0ro!ima Centauri ) is circular, what is the "lanet’s or'ital velocity M1 mar%N d# )y considering that the total linear momentum o the star="lanet system in the centre o mass rame is ero, estimate the minimum mass o the "lanet in terms o
9
e# sing a sim"le a""ro!imation, the e&uili'rium tem"erature o a "lanet can 'e calculated as
/
&<8$@E =&>$F,2/>$
where is the distance 'etween the star and the "lanet# 9iven that the astronomers discovered that the or'it o the "lanet is in act an elli"se with an eccentricity o #3E, and that the star has a surace tem"erature o 3 B and a radius o #14 R , what are the minimum and ma!imum e&uili'rium tem"eratures o 0ro!ima Centauri )
Comment on whether or not the "lanet is in the ha'ita'le one o 0ro!ima Centauri# MThe ha'ita'le one is the 'and around a star where a "lanet can have water on its surace in li&uid orm, at normal "ressure#N M3 mar%sN # Comment on the "ros"ects o studying the "lanet directly, during your lietime, using ro'otic s"ace "ro'es# M1 mar%N
/$D O" A/
10
British Astronomy and Astrophysics Olympiad 2015-2016 Astronomy & Astrophysics Competition Paper th
Monday 18 Janary 2016 !nstrctions "ime: 3 hours (approximately 35 minutes per question). #estions: All five questions should be attempted. Mar$s: The questions carry similar marks. %oltions Ansers and calculations are to be ritten on loose paper or in examination booklets. !tudents should ensure their name and school is clearly ritten on all anser sheets and pa"es are numbered. A standard formula booklet may be supplied. To accommodate students sittin" the paper at different rd times# please do not !nstrctions discss any aspect of the paper on the internet until $ am !aturday %3 &anuary.
Clarity !olutions must be ritten le"ibly# in black pen (the papers are photocopied)# and orkin" don the pa"e. !cribble ill not be marked and overall clarity is an important aspect of this exam paper.
'li(i)ility The 'nternational lympiad ill be held durin" ecember %*+,- all A evel students are eli"ible to participate# even if they ill be attendin" university in ecember. ////////////////////////////////////////////////////////////////////////////////////////////////---
"rainin( *ates and the !nternational Astronomy and Astrophysics Olympiad Following this round the best students eligible to represent the UK at the International Olympiad in Astronomy and Astrophysics (IOAA) will be invited to attend the Training Camp to be held in the Physics Department at the University o O!ord" (Monday 4th April –Thursday 7th April 2016 )# Astronomy material will be covered$ problem solving s%ills and observational s%ills (telescope and na%ed eye observations) will be developed# At the &raining 'amp a practical e!am and a short theory paper will be sat# Five will be selected or urther training# From ay there will be mentoring by email to cover some topics and problems" ollowed by a training camp at the beginning of uly and a wee%end training camp in autumn# &he IOAA this year will be held in hubaneswar" India" rom !th to 1!th "ecember 2016#
!mportant constants c
!peed of li"ht in free space 0arth1s rotation period
+ day
0arth1s orbital period
+ year
2arsec
pc
Astronomical nit
A
4adius of the 0arth
⨀ % ( +
4adius of the 0arth1s orbit 4adius of the !un ass of the !un ass of the 0arth uminosity of the !un 6ravitational constant
3.00×10 m s 24 hours 365.25 days 3.09×10 m 1.6.439×10 7×10 1 AU m 6.96×10 !" 1.99×10 &' #$ 5.97×10 & #$ 3.)5×10 * 6.67×10 m! #$ s& m
7ou mi"ht find the dia"ram of an elliptical orbit belo useful in solvin" some of the questions: B
b P
A
a
O
C
'lements o+ an elliptic or)it:
a 8 A (82) b8 (8;)
semi/ma9or axis semi/minor axis
e8
eccentricity
,1- //
< = focus 2 = periapsis (point nearest to <) A = apoapsis (point furthest from <) ,eplers "hird .a/:
"
4&+ " " 10".'8:
Ma(nitdes: The apparent ma"nitudes of to ob9ects# bri"htnesses# and # via the formula:
%
and
# are related to their apparent
!upported by:
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Random House Publishers
Cambridge University
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3
# 1 Asteroid Belt 'n science fiction films the asteroid belt is typically portrayed as a re"ion of the !olar !ystem here the spacecraft needs to dod"e and eave its ay throu"h many lar"e asteroids that are rather close to"ether. ?oever# if this ima"e ere true then very fe probes ould be able to pass throu"h the belt into the outer !olar !ystem.
i(re 1 Artist conceptal illstration o+ the asteroid )elt le+t3 %chematic o+ the %olar %ystem /ith the asteroid )elt )et/een Mars and Jpiter ri(ht3
This question ill look at the real distances beteen asteroids. a.
1 .) 10? ⨀ calculate the
6iven that the total mass of the asteroid belt is approximately
;<=>
radius of the ob9ect that could be formed# assumin" it has a density typical of rock @ 3.0 $ Am! ). ;ompare this to the radius of the lar"est member of the asteroid belt# ;eres. (B@3 km) b. The main part of the asteroid belt extends from %.+ A to 3.3 A# and has an avera"e an"ular idth of +,.*# as vieed from the !un. ;alculate the avera"e thickness of the belt# and hence its total volume# E;<=> . c.
Assumin" this volume is uniformly filled by spherical rocky asteroids of avera"e radius F # derive a relationship beteen the avera"e distance beteen asteroids# GF # and their radius F # rememberin" to keep the total mass equal to ;<=> .
d. 'f F 8 %.* km# calculate GF . ?o does this compare to the 0arth/oon distanceB (GHI 3$>#*** km) An ob9ect ith an apparent ma"nitude " 0 has an apparent bri"htness " 2 .52 10 * m& . e.
sin" the luminosity of the !un# calculate the total poer incident on an asteroid in the middle of
f.
the asteroid belt. Assumin" only 3*C of that is reflected by its rocky surface# calculate the apparent ma"nitude of the asteroid hen vieed from its nearest nei"hbour. 6iven that ob9ects ith J 6 are too faint for the naked eye# ould it be visible to an astronaut stood on the asteroid surfaceB
>
# 2 %permoons
i(re 2 %permoons at Peri(ee and Apo(ee !ma(e credit John a(han7Pete .ardia)al7 9J.A
A DsupermoonE is a ne or full moon that occurs ith the oon at or near its closest approach to 0arth in a "iven orbit (peri"ee). The media commonly associates supermoons ith extreme bri"htness and siFe# sometimes implyin" that the oon itself ill become lar"er and have an impact on human behaviour# but 9ust ho different is a supermoon compared to the Gnormal1 oon e see each monthB unar ata: !ynodic 2eriod Anomalistic 2eriod !emi/ma9or axis rbit eccentricity 4adius of the oon ass of the oon
8 %H.53*5$H days (time beteen same phases e.". full moon to full moon) 8 %@.55>55* days (time beteen peri"ees i.e. peri"ee to peri"ee) 8 3.)44 10K #m 8 *.*5>H 8 +@3$.+ km 8 7.342 10&& #$
!n this 4estion /e /ill only consider a +ll moon that is at peri(ee t o )e a spermoon a. ;alculate ho many days separate to supermoons. b. !ho that the difference in distance beteen the apo"ee and peri"ee is 4.22 10 ' #m. (The data "iven in this question allos the mean orbital parameters to be calculated. Iote that perturbations in the lunar orbit mean that the peri"ee and apo"ee continually chan"e over the course of the year.) c. etermine the difference in the an"ular diameter of a supermoon and a full moon observed at apo"ee. Thus# determine the percenta"e difference in the bri"htness of a supermoon and a full moon observed at apo"ee. ('"nore the effects of the oon1s orbital tilt ith respect to the 0arth.) d. Jhat chan"e in ma"nitude does this bri"htness difference correspond toB e. !u""est hy it can be difficult to detect any differences in the bri"htness of supermoon compared to a Gnormal1 full moon hen observin" ith the naked eyeB f. ;alculate the "ravitational field of the supermoon at the 0arth. Jhat fractional mass increase ould a oon at apo"ee need in order to create the same "ravitational fieldB
5
# : !nterstellar 'n the science fiction movie Interstellar# the crops on 0arth are failin"# makin" farmin" difficult# and the existence of humanity is threatened. To save the human race# a cre of astronauts travelled throu"h a ormhole in search of a ne home and they sent encoura"in" data from planets near 6ar"antua# a supermassive black hole: iller1s planet is the first planet in the system orbitin" 6ar"antua. 't is a ater orld ith a similar composition to the 0arth# covered in an endless shallo ocean. The planetKs "ravity is +3*C of the 0arthKs# forcin" human astronauts to move sloly and ith some difficulty hile on its surface. ein" ell ithin the tremendous "ravitational field of 6ar"antua# time on the surface of illerKs planet passes very sloly relative to the rest of the universe: a sin"le hour on iller ould equate to seven years back on 0arth. ecause of the planetKs proximity to 6ar"antua# the immense "ravitational pull from the black hole causes the planet to be afflicted by massive tidal aves as tall as + km. There is no si"n of dry land on iller# hich may not exist due to the enormous erosive poer of the planetKs aves.
i(re : C! model o+ a spermassi;e )lac$ hole and Millers Planet Cre dit !nterstellar
The "ravitational time dilation is "iven by:
L" LMN
1-
2+ O &
here L" is the time for a slo/tickin" observer ithin the "ravitational field# LP # is the time for a fast/tickin" observer at an arbitrarily lar"e distance from the massive ob9ect# is the mass of the massive body# and is the distance of the observer from the centre of the body. 7ou are asked to estimate the folloin": a. b. c. d.
The characteristics of the planet (mass and radius). The mass of the supermassive black hole. The orbital parameters of the planet (orbital radius and period). The planet orbits the black hole# but the ?ollyood director seems not to have checked his numbers carefully. 'n hat ay is this apparent from the values "iven and the results you have calculatedB (the radius of the black hole is the event horiFon# the value of hen LP H Q.)
,
# < *yson strctres !ince its first li"ht in %**H# the Lepler Telescope has been scannin" the universe in search of habitable orlds beyond our !olar !ystem. Lepler is desi"ned to observe stars and look for tiny dips in their bri"htness. These dips# especially if they repeat# can be a si"n the star has planets orbitin" it. y measurin" the timin" and the siFe of the dips# scientists can learn a lot about the transitin" planet. urin" its routine observations of the star L'; $>,%$5%# similar to our !un (same radius and mass)# the telescope observed somethin" very unusual. A "roup of citiFen scientists noticed that this star appeared to have to small dips in %**H# folloed by a lar"e dip lastin" almost a eek in %*++# and finally a series of multiple dips si"nificantly dimmin" the star1s li"ht in %*+3. The pattern of the dips indicates that a lar"e# irre"ular/shaped ob9ect orbits the star. !ome people have speculated that the star mi"ht be orbited by a "iant alien me"astructure# called a Dyson structureE. 't is a structure that harnesses a star1s ener"y to be used by a civilisation# like solar poer# but on a massive scale. 't ould be composed of thousands of spacecraft that ould be theoretically lar"e enou"h to block out a si"nificant portion of a starKs li"ht.
i(re < "he li(ht cr;es o+ ,!C 8<62852 sho/in( t/o transits "he time is in days since a re+erence point Credit Boya=ian et al 2015 Planet >nters ? ,!C 8<62852 - 9here@s the l
a. 0xplain hy the scientists believe that the ob9ect is unlikely to be a star or a planet. b. &ud"in" from the li"ht curves in the plots above# hat ould be the area of the yson structureB c. ased on the lar"est dips# hat is the avera"e distance of the yson structure from the starB Assume its mass is much smaller than the mass of the central star.
The problem of creatin" a yson structure is that it cannot be free floatin" in space. ne possible solution is creatin" a cloud of solar sails. These ob9ects ould be in perfect balance beteen the "ravity pullin" them
(
inards# and the li"ht pressure pushin" them outards. The luminosity (poer output) of the star is ⨀ . d. Assumin" that the sails are made of a reflective material ith reflectivity *# hat is the pressure on the sail due to photon bombardmentB (momentum of a photon is +,c) e. Jhat is the force exerted by the photons on the yson structureB Assume * 8 +. f. Assumin" that the net acceleration of the solar sails is Fero# hat ould be the mass of the structureB ". 0xplain hy the scenario of buildin" such a structure is unrealistic.
@
# 5 ra;itational lensin( The deflection of li"ht by a "ravitational field as first predicted by Albert 0instein a century a"o# su""estin" that massive ob9ects can bend li"ht like a classical lens. This prediction as confirmed by !ir Arthur 0din"ton in +H+H# hile observin" a solar eclipse. ;onsider a spherically symmetric ob9ect ith mass . This ob9ect ill act like a lens# ith an impact parameter b measured from the centre of the ob9ect. The an"le of deflection due to the massive lens# "iven by 6eneral 4elativity# is calculated as:
R
4+ O &
'n a simplified model# the impact parameter may be seen as the shortest separation beteen the centre of the lens and the path of a particular li"ht ray. The dia"ram belo shos the "eometric model of a "ravitational lens (
-.
-
-%
i(re 5 %chematic dia(ram o+ a ( ra;itational lens
a.
0xplain ho Arthur 0ddin"ton mi"ht have used the "ravitational lens effect to confirm the predictions of 6eneral 4elativity.
b. !ho that the source an"le ST is related to SU # # VT W VX via the expression:
ST Y SU c.
VXT 4+ VX VT O & SU
$
d. !ho that the time delay for a photon in the presence of an 0instein 4in" is "iven by:
\L
1 VT VX 2 VXT O
S%&
]^ou may us_ `h_ arobcma`code
1 Aosf
1g
R& 2
0xtendin" the example of "ravitational lensin" into 3 dimensions# instead of to ima"es of the source# sometimes multiple ima"es of the source can be seen# arran"ed in a cross (called an D0instein crossE). ast year# astronomers discovered a "alaxy that is "ravitationally lensed by a "iant elliptical "alaxy situated in a "alaxy cluster in the fore"round. !urprisin"ly# they discovered a supernova explosion (called the 4efsdal !upernova) in the ima"e of the lensed "alaxy# arran"ed in an 0instein cross.
i(re 6 Mltiply-lensed e+sdal sperno;a )y a massi;e (alay in the (alay clster MAC% J11))le7FA%A7'%A7%"%ci7GC.A
e.
y knoin" that the distance to the "alaxy is >.> 6pc# and to the cluster is %.* 6pc# find the time delay caused by the lens for a photon from the supernova explosion. 7ou can take the +%
f.
H
mass of the elliptical "alaxy to be 8 +* ⨀ . A 6pc is +* parsecs. 2erhaps even more surprisin"ly# the astronomers realised that they ere seein" the four ima"es of the supernova at different time instances. 'n some of the pictures they took# ima"es of the supernova ere missin". 0xplain ho this is possible.
'nd o+ #estions
H
Asteroid Belt (solutions)
⨀
a. Given that the total mass of the asteroid belt is approximately Mbelt = 1.8 × 10-9 M , calculate the radius of the object that could be formed, assuming it has a density typical of rock (ρ = 3.0 g cm-3). Compare this to the radius of the largest member of the asteroid belt, Ceres. (RCeres = 473 km)
Mbelt = 1.8 × 10-9 M⨀ = 3.6 × 1021 kg density = 3.0 g cm-3 = 3.0 × 103 kg m-3
=
×.× × ∴ = = ×.
[1]
= 660 km = 1.4
[1] [1]
Ceres
[This means Ceres contains a sizeable fraction of the material in the whole belt, although it has a lower density than we have assumed, so less mass than implied here]
b. The main part of the asteroid belt extends from 2.1 AU to 3.3 AU, and has an average angular width of 16.0°, as viewed from the Sun. Calculate the average thickness of the belt, and hence its total volume, Vbelt. h
8° 2.1 AU
3.3 AU
Average height above the orbital plane, h
= .. tan8 = 0.38 AU
[0.5]
So total thickness = 2h = 2 × 0.38 = 0.76 AU
[0.5]
Volume of belt = area of disk × total thickness
∴ = 3.3 −2.1 ×0.76 =15.45 =5.1×10 =5.1×10
[1]
AU
m
[1]
km
[This is a huge volume, which explains why each asteroid gets so much space to itself]
c.
Assuming this volume is uniformly filled by spherical rocky asteroids of average radius Rav, derive a relationship between the average distance between asteroids, d av, and their radius Rav, remembering to keep the total mass equal to M belt. Assuming there are N asteroids, each filling a cube with side length dav then
=
belt
[1]
av
Conserving the volume of rock in the belt means
=
belt
av
[1]
so we can cancel N to give
= ∴ = belt
belt
av
av
av av
belt
[1]
belt
(a spherical approximation for the volume of belt allocated to each asteroid can also gain full credit so long as it is clear that dav in that case is equal to double the radius of the spherical volume used) i.e.
= belt
av
leading to
av av
=
belt belt
d. If Rav = 2.0 km, calculate d av. How does this compare to the Earth-Moon distance? (d E→M = 384,000 km)
= = av
av
belt
belt
.×
×.× ×.××.×
=1.1×10 = 2.9
m ( =1.km) 1×10
[1] [1]
E→M
(using a spherical approximation in the previous question yields dav = 1.4 × 10 9 m = 3.6 dE→M)
[An average separation of about 1 million km is close to the real value for our asteroid belt and emphasises the vast space between asteroids, which is why many probes can travel through unharmed, although the real value of Rav is somewhat smaller – this is because the real asteroid belt contains far more small asteroids than big ones] e.
Using the luminosity of the Sun, calculate the total power incident on an asteroid in the middle of the asteroid belt. Distance to middle of the belt,
= .. = 2.7 mid
AU
Apparent brightness of the Sun in the middle of the belt:
= ×..×.×× =190
= ⊙
mid
W m-2
[1] [1]
Incident power, Pi = apparent brightness × cross-sectional area of an asteroid
∴ = =190× 2.0×10 =2.4×10 i
av
W
[1] [1]
f.
Assuming only 30% of that is reflected by its rocky surface, calculate the apparent magnitude of the asteroid when viewed from its nearest neighbour. Given that objects with m > 6 are too faint for the naked eye, would it be visible to an astronaut stood on the asteroid surface? Apparent brightness of one asteroid as viewed from another with only 30% reflectivity:
= .
.×.× = ×. × =4.5×10
∴=− . log =−2. =6.9 5log
[1] -2
Wm
..× ×
[1]
This means that the astronaut could not see the nearest asteroid with their naked eye
[1] [1]
[1]
(Using a spherical approximation gives b = 2.9 × 10 -11 W m-2 and m = 7.3, so the same conclusion) [In practice even though asteroids come in a wide variety of sizes, and the spacing can vary quite far from our values of dav, most asteroids only reflect about 10% of their light and so despite all the simplifying approximations we have made we still get the same result – that without binoculars you would be unable to tell you were flying through an asteroid belt, which is rather different to the picture painted by science fiction films!]
1
Figure 1: Image credit: John Gaughan / Pete Lardizabal / WJLA
Supermoons The term“s upermoon” was coined by astrologer Richard Nolle in 1979. He defined a supermoon as a new or full moon that o ccurs with the Moon at or near (within 90% of) its closest approach to Earth in a given orbit (perigee). The value of 90% was arbitrarily chosen, and other definitions are often used. The media commonly associates supermoons with extreme brightness and size, sometimes implying that the Moon itself will become larger and have an impact on human behaviour, but just how different is a supermoon compared to the ‘normal’ Moon we see each month? Lunar Data: •
Synodic Period = 29.530589 days - time between same phases e.g. full moon to full moon, new moon to new moon.
•
Anomalistic Period = 27.554550 days - time between perigees e.g. perigee to perigee.
•
Semimajor axis ( a)= 3.844 × 105 km
2 •
Orbit eccentricity ( e)= 0.0549
•
Radius of the Moon ( R )= 1738. 1 km
•
Mass of the Moon ( M )= 7.342 × 1022 kg
In this question, we will only consider a full moon that is at perigee to be a supermoon. a) Calculate how many days separate a supermoon. b). Show that the mean diffe rence in the distance between the apogee and perigee is 4 .22 × 104 km. (The data given in this question allows the mean orbital parameters to be calculated. Note that in reality, perturbations in the lunar orbit mean that the perigee and apogee continually change over the course of the year) . c). Determine the differe nce in the angular diameter of a supermoon and a full moon observ ed at apogee. Thus, determ ine the percentage differ ence in the brightness of a supermoon and a full moon observed at apogee. (Ignore the effects of the Moon’s orbital tilt with respect to the Earth) . d). What change in magnitude does this brightness difference correspond to? e). Suggest why it can be difficult to detect any differences in the brightness of a supermoon when observing with the naked eye? f). Calculate the grav itational force of the superm oon on the Earth. What mass increase would a Moon at apogee need, to create the same gravitational force? Solutions
a). 27 .554550/(29.530589 each supermoon,
−
27.554550) = 13 .9443 synodic periods between
so 13 .9443 × 29.530589 = 411 .78 days between each supermoon. −
b). R From equation of ellipse we have that: Ra = a (1 + e) and Rp = a (1 e), where a is the lunar apogee, Rp is the lunar perigee, a is the semi-major axis and e is the eccentricity. Therefore the differen ce in ap ogee and perigee is: Ra −Rp = a (1+e)−a(1−e) = 2ae = 4.22 × 104 km.
3 c). Angular diameter of a full moon at apogee:
R /Ra = 4.286 × 10−3
Angular diameter of a full moon at perigee: R /Rp = 4.784 × 10−3 So the Moon appears 4 .784 × 10−3 /4.286 × 10−3 = 1.116 times or 11.6% larger at perigee. Since light will follow an inverse square law, the brightness will decrease according to the distance 2 . Hence, in ter ms of brightness, a Moon appears 1.112 = 1.246 times or 25% brighter at perigee. d). Difference in magnitude = log 10 (difference in brightness) /0.4 = log 10 (1.246)/0.4 = 0.239 magnitudes. e). Typically, a difference of 0.239 magnitudes should be detectable with the naked eye. However, several months separate these two extremes of the brightest and dimmest full moons. In additi on, the Moon is passing throug h different phases (e.g. crescent, quarter, gibbous) as it reaches full Moon, all of which have different brightnesses. Therefore, it is unlikely that a naked eye observer will notice, and furthermore remember, the difference in brightness between a supermoon and a full moon at apogee. f). Gravitational force of Moon at perigee:
Fgrav,p =
and gravitational force of Moon at apogee: Fgrav,a = or write as: Fgrav,a =
G M M
2 Rp
,
G M M
2 Ra
G M M
+m . (Rp +r )2
But from part b), know that r = 4.22 × 104 km. If Fgrav,p = Fgrav,a , the increase in lunar mass, M + m = 1.246M , or m = 0.246M . (since gravity also follows the inverse square law, this is analagous to the answer in part c).
British Olympiad in Astronomy and Astrophysics Competition Paper Name School Total Mark/50
24th April 2015 Time Allowed: One hour Attempt as many questions as you can. Write your answers on this question paper. Marks allocated for each question are shown in brackets on the right. You may use any calculator. You may use any standard formula sheet. This is the first competition paper of the British Olympiad in Astronomy and Astrophysics. To solve some of the questions, you will need to write equations, draw diagrams and, in general, show your working.
This paper has real problems and is not like an A level paper. The questions are more difficult because you are not told how to proceed. If you cannot do many, do not be disheartened. If you can do some then you should be delighted. A good mark is from a few questions solved. There are two optional parts that you may attempt after the exam. These are more difficult questions that follow up on the questions to indicate how much information can be deduced from the data by a keen astrophysicist.
Useful constants
Speed of light Gravitational constant Solar mass Astronomical Unit Parsec Earth’s orbit semi-major axis Earth’s rotation period Earth’s mass Earth’s axial tilt
3.00 × 10 6.67 × 10 1.99 × 10 1.496 3.086 ×× 10 10
c G Msolar AU pc
ms N m kg kg m m AU hours kg
1 24
1 day MEarth
5.97 × 10 23.4
°
You might find the diagram of an elliptical orbit below useful in solving some of the questions:
b P F
Elements of an elliptic orbit:
c
O
a
A
– semi-major axis axis – semi-minor = ! –eccentricity, where " = #$
F – Sun/Earth - focus P – perihelion/perigee (point nearest to F) A – aphelion/apogee (point furthest from F)
Kepler’s Third Law: For an elliptical orbit, the square of the period of orbit of a planet about the Sun is proportional to the cube of the semi-major axis ( a) (the average of the minimum and maximum distances from the Sun). List of symbols used in the paper: φ - geographic latitude L – geographic longitude UT – Universal Time
1
Section A: Multiple Choice Circle the correct answer to each question. Each question is worth 2 marks. There is only one correct answer to each question. Total: 20 marks.
1. Why is the Moon heavily cratered, but not the Earth? A. The Moon has stronger gravity, so it attracts more space debris B. The Moon formed earlier than the Earth, so it had more time to be bombarded by asteroids C. The craters on Earth were eroded by the oceans and atmosphere over a long period of time D. The Moon orbits around the Earth in addition to orbiting around the Sun, so it collects more space debris 2. We do not expect to find life on planets orbiting around high-mass stars because: A. B. C. D.
High-mass stars are far too luminous The lifetime of a high-mass star is too short High-mass stars are too hot to allow for life to form Planets cannot have stable orbits around high-mass stars
3. What would happen to the Earth’s orbit if the Sun suddenly became a black hole with the same mass? A. It would spiral inwards because of the strong gravitational forces B. It would fall on a straight line into the black hole C. It would become an open orbit and the Earth would escape from the Solar System D. Nothing 4. A 10-inch refracting telescope with focal ratio (defined as the ratio of the focal length and aperture) of 10 is used with a 25 mm focal length eyepiece. What is the magnifying power of the telescope?
%1 &'() = 2.54 (m*
A. B. C. D.
10x 50x 100x 200x
5. Which of the following planets has the longest day, defined as the period of a complete rotation about its axis?
A. Venus
B. Earth
C. Mars
2
D. Jupiter
6. Which of the following is not a zodiacal constellation?
A. Virgo
B. Cancer
C. Aquila
D. Gemini
7. The Sun is seen setting from London (+ = 51°30 , N- = 0°8/ * at 21:00 UT. At what time UT will it be seen setting in Cardiff (+ = 51°30, N- = 3°11/ * on the same day? A. B. C. D.
21:12 21:00 20:48 20:58
8. How far away must your friend be standing from you such that the attractive force exerted on you is similar to the maximum gravitational force exerted on you by Mars? Assume that your friend’s mass is 65 kg. The mass of Mars is 6.4 x 10 23 kg and the minimum distance from Earth to Mars is 0.52 AU. A. B. C. D.
2.3 m 0.8 mm 0.8 m 2.3 mm
9. A comet follows an elliptical orbit that is 31.5 AU at aphelion and 0.5 AU at perihelion. What is the period of the comet? A. B. C. D.
181 years 16 years 64 years 6.3 years
10. In which of the following places is the length of the shortest day of the year equal to half the length of the longest night?
+ = 25°N* London (+ = 52°N* Rio de Janeiro (+ = 23°* Tromsø (+ = 70°N*
A. Dubai ( B. C. D.
[HINT: Do not attempt to calculate the latitude, but rather look at the answers and consider how the length of the day varies with latitude and time of year.]
/20
3
Section B: Short Answer Write your answers to the following questions. Each question is worth 5 marks. You should show your working in the spaces provided. Total: 10 marks.
Question 11 A geostationary satellite is one that orbits in the equatorial plane of the Earth with the same period and in the same direction as the Earth’s rotation. These orbits are important for communication and weather observation because the satellite always remains above the same point on Earth. The orbits of geostationary satellites are circular.
a) Calculate the radius of the orbit of a geostationary satellite. Ans:
[3]
b) Imagine now that the satellite was orbiting the Earth at the same orbital radius and same period, but in the opposite direction. For approximately how many hours a day would a satellite be above the horizon for an observer at ground level, situated on the Equator? Assume that the radius of the Earth can be neglected. Ans: [2]
/5
4
Question 12 The light from distant galaxies has distinct spectral features characteristic of the gas which makes them up. The astronomer Edwin Hubble noticed that the lines in the spectra of most galaxies are shifted towards the red end of the spectrum. This lead to his famous discovery that the recessional velocity of a galaxy is proportional to the distance to the galaxy, the constant of proportionality being , implying that the Universe is expanding.
To measure the redshift of a galaxy, astronomers usually use the parameter. Suppose that we observe a galaxy with a redshift of and find that one of the lines in the hydrogen spectrum has been redshifted, compared to its rest wavelength of 486.1 nm. Assume that the Universe is undergoing a uniform expansion, with the rate given by the Hubble constant, .
= 0.30
= 72 km s (
is a reasonable approximation, what is the redshifted wavelength λ of the receding galaxy, in 'm?
a) Assuming that the classical Doppler effect
Ans:
[3]
b) When we observe the galaxy, how far into its past are we looking? Ans:
[2]
[A Mpc (abbreviation for megaparsec) is one million parsecs. It is a useful unit used by astronomers to measure the large distances to galaxies]
/5
5
Section C: Long Answer Write complete answers to the following questions. Total: 20 marks.
Question 13
Solar Eclipse
A major astronomical event happened on the morning of Friday 20 th March 2015: a partial solar eclipse visible from the whole of the UK (at least from the parts not fully covered by clouds). The next partial solar eclipse of the same totality will happen in 2026 and the next total solar eclipse visible from the UK will be in 2090. In the image below you can see a time lapse of the eclipse, as seen from Sheffield, UK.
Figure 1 The partial solar eclipse visibility as seen from Sheffield, UK.
a) From the images in Figure 1 identify the time corresponding to the maximum of the partial solar eclipse.
[1] b) The apparent magnitude of an object is a measure of its brightness as seen by an observer on Earth. Note that the brighter the object appears, the lower its magnitude. From Figure 1, the maximum coverage of the solar eclipse, as seen from Sheffield, was 90%. Using the relation between the difference in apparent magnitudes and the %;< ; * , variation in brightness, , also known as Pogson’s formula,
# ># 2.512
#
:
estimate the magnitude of the Sun at the maximum of the eclipse, if the apparent magnitude of the Sun is -26.74. Assume that the brightness of the solar disc is uniform, therefore being proportional to the surface area.
6
[3] c) The only two populated places where the totality could be seen were the Faroe and Svalbard Islands. Explain if it would ever be possible solar eclipse from , to see ,a total the Capital of Svalbard, Longyearbyen ( during December.
+ = 78°13 N 15°33 ?*
[1] d) The tidal interaction between the Earth and the Moon causes the Moon to move away from the Earth (increase its semi-major axis) by 3.82 cm/year, and the Earth to spin down very slowly. Considering the most favourable case and using the data below estimate in how many years a total solar eclipse will not be visible from anywhere on Earth. Assume that the eccentricity of the Moon’s orbit does not change. The radius of the Moon is
km km @385-000 = 695-800
@A = 1737.5 km- the mean distance to the Moon is A = = 0.055 = 149.6×10 km = 0.0167
and the eccentricity of the Moon’s orbit is A . The radius of the Sun is , the mean distance to the Sun is and the B C eccentricity of Earth’s orbit is C . [HINT: Make use of the ellipse on page 1 and identify where the most favourable case lies, considering the angular diameters of the Sun and of the Moon, respectively]
7
[7]
/12 Question 14
Transiting extrasolar planet
One method of detecting extrasolar planets is to observe their transit across the disc of their host star. During the transit, the observed brightness of the star drops by a small amount, depending on the size of the planet. In 1999, following the spectroscopic detection of a planet around star HD 209458, astronomers David Charbonneau and Gregory Henry were able to observe a transit of the planet across the disc of the star, making it the first detection of a transiting extrasolar planet. The planet, named HD 209458b was found to be orbiting the star with a mass of 1.15 Msolar on a circular orbit every 3.525 days, much faster than the Earth is orbiting the Sun. Hundreds of extrasolar planets have since been detected using the transit method by the Kepler mission. However, the main disadvantage of this method is that the orbit of the planet has to be very close to edge-on, for the transit to occur from our vantage point. In this question, assume that the planet’s orbit is perfectly edge-on, such that the transit is central. The figures below are the plot of the light curve of star HD 209458, showing the drop in brightness during the transit, and a schematic of the transit. Because the surface brightness of the star’s disc is not uniform (an effect called limb darkening), the real light curve in Figure 2 does not fully resemble the idealised case in Figure 3.
Figure 2. The light curve of the star HD 209458.
Figure 3. Schematic of the transit.
a) From Figure 2, what percentage of the star’s disc is covered by the planet in the middle of the transit? Estimate an error in your determination.
[1] 8
b) From your answer in a) determine the ratio of the radius of the planet and the radius of the star.
[2]
c) Estimate the radius of the planet’s orbit in AU, assuming that the mass of the planet is much smaller than the mass of the star.
[2] d) From Figure 2 estimate the total transit time, from first to last contact (as shown in the Figure 3). Assuming that the speed at which the transit occurs is equal to the circular speed of the planet around the star, calculate the radii of the star and of the planet. Express them in units of solar radii and Jupiter radii, respectively ( DEF
@ = 6.96 ×
10G km - @HEIJKLM = 7.0 × 10 km*.
____[3]
/8 END OF PAPER 9
British Olympiad in Astronomy and Astrophysics April 2015 Solutions and marking guidelines for the BOAA Competition Paper The total mark for each question is in bold on the right hand side of the table. The breakdown of the mark is below it. There are multiple ways to solve some of the questions, so please accept all the good solutions that arrive at the correct answer.
Question
Answer
Mark
Section A
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
20
C B D C A C A C C B
2 2 2 2 2 2 2 2 2 2
Section B
11.
a.
Answer:
10
42!200 "m
3
On a circular orbit, the centripetal force is due to the gravitational force:
= = = 4 = 4.22 × 10 m 42!200 "m
= 24 '()*+ & '()*+
1 1
, therefore the radius of the geostationary orbit is:
1 2
b. Answer:
# = 24 '()*+
In the case when the satellite is orbiting the Earth with a period of the opposite direction to Earth’s rotation, the relative angular velocity is:
, but in
= 2 # = 4#
1
The observer can be considered as being stationary on Earth and the satellite revolving with . Because the radius of the geostationary orbit is larger than the radius of the Earth, the satellite will be visible above the horizon for the observer for half of its (relative) orbit, covering an angle of (or π). The time the satellite is visible for the observer is:
1,0-
$% = = 4# = & '()*+ &
[Since the radius of the orbit is only , the visibility of the satellite is less than 6 hours. In fact it is 5.42 hours.] This occurs twice in a 24 hour period, so strictly it is 12 hours (or 10.8 h). Either answer of 6 h or 12 h gains the mark.
1
1
12.
a.
&70 9m H
Answer:
3
From the classical Doppler effect: long as
3 0.7
/ /#/# = 3
Doppler shift formula:
Where
. This is a reasonable approximation as
, otherwise the full relativistic Doppler effect has to be used.
/# = 4,&.1 9m
1
the rest wavelength. Hence:
#
1
/ = &71./ =8/9m1 536&70 9m 4 BCD C(9 EFG*+ $% = = :#
The observed (redshifted) wavelength is:
b.
Answer:
1 2
The time it takes the light (travelling at speed ) from the galaxy to reach us is:
1
According to Hubble’s law the distance to the galaxy is:
Hence,
1
$% :3# ;2 "m +0.<7 >?@:<# AI4<BCD C(9 EFG*+
In the equation above notice that the unit of conversion from Mpc to km.
is
and take care when doing
the
Section C
13.
20
a. From Figure 1 identify that the maximum of the eclipse occurs between 09:24 and 09:34, therefore in the image taken at 09:31.
1
b. The brightness of the solar disc is proportional to the visible surface area. The apparent magnitude of the Sun is corresponding to a brightness of . During the eclipse, the Moon covers 90% of the solar disc, thus the visible area of the Sun is only 10%. The brightness of the solar disc during the maximum of the eclipse, is:
3
= 2&.;4 = 0.1 = D(DJ(#J#2.K12 = 2.KD(J # = 2&.;4 2.KD(J # 0.1 = 24.24
1
This corresponds to an apparent magnitude . Inverting the Pogson’s formula, the apparent magnitude of the Sun during the eclipse is:
2
1
1
The Sun was 2.5 magnitudes less bright during the eclipse. It appeared to be slightly dimmer outside, indicating that only 10% of the Sun was able to provide sufficient light to continue our daily activities. Even with a small percentage of the Sun being visible, it is still 40,000 brighter than the full Moon, whose apparent magnitude is -12.74. c. Longyearbyen (latitude ) is situated above the Arctic circle (latitude ) and it experiences polar night during the whole month of December (from November to February, more specifically). Hence, the Sun is not visible above the horizon during these months and a solar eclipse would not be observable. Luckily, the total solar eclipse occurred in March. d. Total solar eclipses would no longer be visible from Earth when the angular size of the Moon will be smaller than the angular size of the Sun. To calculate the angular sizes of the
80- 27-2;N
;,-17L M
&&-74L M =
1
7
Sun and of the Moon, respectively, make use of a diagram such as the one below: R ]
d
From the diagram, the apparent diameter of the object is:
For small angles,
O = 2 +C9 < PQRS +C9 O \G9O O C9 *GRCG9+6
1
, so any of these is acceptable.
The most favourable case for the total eclipse not to occur is for the Moon to have the largest possible angular size, and the Sun the smallest possible angular size. This happens when the Moon is at perigee (nearest point to the Earth) and the Earth at aphelion (furthest point from the Sun). First calculate the distance to the Sun, and its angular diameter.
TU TU = V 5 = V 1 5 W6 = 1K2.1 × 10 X "m O =Y0.T[K24VZ
1
From the diagram of the ellipse on page 1, the distance to the aphelion is:
1
The angular diameter of the Sun (and of the Moon for total eclipses not to occur) is:
VZ
1
Now calculate where the Moon must be, and hence obtain . Then calculate the change of from current value and, knowing the rate, calculate the time taken. The distance the Moon needs to be situated at is:
Y = +C9ZO2 7,0!000 "m YT[ YT[ = V = V Z1 WZ6 VZ = 1YT[WZ 402!120 "m $% 4K0 mCD C(9 EFG*+
This will be the perigee of the new orbit, page 1, the distance to the perigee is:
1
. Again, from the diagram of the ellipse on
Hence, the new semi-major axis of the Moon will be:
At a rate of 3.82 cm/year, the Moon will move from the current semi-major axis of 385,000 km to 402,120 km in:
In 450 million years we will only be able to observe annular solar eclipses from Earth.
3
1
1
14
a. The percentage of the star’s disc covered by the planet is 1.65%. The light-curve has high quality data, with an error of only ~ 0.05%
1
0.5 0.5
1.&^ 0.1_ 1.; ^0.1_
[Accept or and the errors propagated in the following calculations] b. The covered area of the star, during the transit is:
`ab = ab c = 0.01&K d c = ``ab d = ab ab d = ec = 0.12, Vf = hijk4g5 ljnoi6
The percentage determined in a) of
2
corresponds to:
1
Therefore, the ratio of the radius of the planet and of the star is:
1
c. Use Newton’s third law (which gives the proportionality of Kepler III but with constants):
Neglect
mljnoi H >hijk
2
and find the radius of the planet’s orbit:
Using the values in the question, radius of the planet’s orbit is:
V = 4d >hijk = 1.1K > hrjk V = 0.04; pq
and the period of
1
s = 7.K2K RGE+
d. From the light curve (Figure 2), the total transit time (from the beginning of the drop, to its end) is:
$%tb 0.17 RGE+ 7.12 '()*+
1 3
1
The transit time is equivalent to the duration of an eclipse from first to last contact, as seen in Figure 3. During the eclipse, the centre of the planet’s disc moves a distance equal to . Because the star and the planet are practically at the distance from the observer, the speed at which the transit occurs is equal to the circular speed of the planet around the star,
2d 5 ab6 aby
1 Hence,
ab = 2d$%5tbab6 = 2V d 5 ab = $%tb V A ,.1K × 10 u "m 0.12,! d = ;.27 × 10 u "m 1.04 vw ab = 0.82 × 10 u "m 1.72 xwa[
And using the ratio in b),
the radius of the star and of the planet are:
4
0.5 0.5
British Olympiad in Astronomy and Astrophysics Trial Paper Name School
Total
March 2015
Mark/50
Time Allowed: One hour Attempt as many questions as you can. Write your answers on this question paper. Marks allocated for each question are shown in brackets on the right. You may use any calculator. You may use any standard formula sheet.
This is a trial paper for the British Olympiad in Astronomy and Astrophysics. The first competition paper of the British Olympiad in Astronomy and Astrophysics will take place in April 2015 and will have a similar format and questions to this trial paper. To solve some of the questions, you will need to write equations, draw diagrams and, in general, show your working. There are two optional parts that you should attempt in extra time. These are more difficult questions that will not be marked, but they are useful for your training.
Useful constants
Speed of light Gravitational constant Solar mass Astronomical Unit Earth’s orbit semi-major axis
c G Msolar AU
3.00 × 10 6.67 × 10 1.99 × 10 1.496 × 10 1
kg m AU
Earth’s orbital period
1 year
365.25
days
1
m s N m kg
Section A: Multiple Choice Circle the correct answer to each question. Each question is worth 2 marks. There is only one correct answer to each question. Total: 20 marks.
1. Which of the following types of stars is the hottest? A. B. C. D.
Red giant Brown dwarf O-type blue giant Yellow main sequence star
2. The majority of the mass in the Universe is contained in: A. B. C. D.
The most massive stars Gas and dust Dark matter Supermassive black holes
3. How much more or less light can an 8-metre aperture telescope collect compared to a 4metre aperture telescope (in the same amount of time)? A. Half the amount B. The same amount C. Twice as much D. Four times as much 4. Why aren’t solar/lunar eclipses observed at every new and full moon? A. Because the orbital plane of the Moon is tilted compared to the Earth’s orbit around the Sun B. Eclipses happen every month, but they can be seen from different places on Earth C. Most eclipses happen during the day, so they are not visible D. Because the Earth-Moon distance changes in time 5. Which planet would you not be able to see on the night sky at midnight from the UK?
A. Jupiter
B. Venus
C. Mars
2
D. Saturn
6. Which of the following constellations is not visible from the UK?
A. Canis Major
B. Cygnus
C. Crux
D. Gemini
7. What is Earth’s mean orbital speed around the Sun? A. B. C. D.
150 km s 30 km s 15 km s 0.3 km s
8. If your mass is 65 kg, what is the maximum value of the attractive force exerted on you by Jupiter? The mass of Jupiter is 1.9 x 1027 kg and its semi-major axis is 5.2 AU. -5
A. B. C. D.
1.4 x 10-11N 2.1 x 10 N 9.6 x 10-6 N -5 2.1 x 10 N
9. The famous comet, Halley’s comet, appeared in the night sky in 1986. The semi-major axis of its orbit is 17.8 AU. When is it going to return next? A. B. C. D.
2023 2061 2064 2093
10. Suppose a colony is established on Mars. How long would it take for a Martian doctor to send a question to a colleague on Earth and receive a response, when Mars is closest to Earth? Assume that the colleague replies instantly. The radius of Mars’s orbit is 1.524 AU. A. 8.7 minutes B. 25.3 minutes C. 4.3 minutes D. 12.7 minutes
/20
3
Section B: Short Answer Write your answers to the following questions. Each question is worth 5 marks. You should show your working in the spaces provided. Total: 10 marks.
Question 11 In the image below you can see the projection of the shadow of a child onto the wall he faces. The height of the child from head to toes is 1.8 m, the length of the shadow on the wall is 0.8 m and on the ground is 1.6 m.
[HINT: It is useful to draw a diagram of the child, the wall and the position of the Sun] a) What is the altitude of the Sun above the horizon?
Ans:
[3]
b) What would be the length of the shadow in the absence of the wall? Ans:
[2]
/5
4
Question 12 An astronomer observes a galaxy and finds that one of the lines in the hydrogen spectrum has been redshifted to 669.4 nm, compared to the rest wavelength of 656.3 nm. Assume that the Universe is undergoing a uniform expansion, with the rate given by the Hubble constant, = 72 km s Mpc . a) The shift of the line is due to the Doppler effect. What is the velocity v of the receding galaxy, in km s ? Ans: [3]
b) Edwin Hubble discovered his famous law that the recession velocity of a galaxy is proportional to the distance to the galaxy, the constant of proportionality being . What is the distance r to the galaxy in Mpc? Ans: [2]
[A Mpc (abbreviation for megaparsec) is one million parsecs. It is a useful unit used by astronomers to measure the large distances to galaxies]
/5
5
Section C: Long Answer Write complete answers to the following questions. Each question is worth 10 marks. The last point of each question is optional – for your training only. Total: 20 marks.
Question 13
Solar Eclipse
On the morning of Friday 20 th March 2015 a partial solar eclipse will be visible from the whole of the UK. Solar eclipses are quite rare and this will be a major event, with the Moon passing in front of the Sun and covering a large portion of the solar disc. This will be an event you will remember for the rest of your life, but remember you shouldn’t watch the Sun without a suitable filter! The radius of the Moon is 1737.5 km and the distance to the Moon will be 365,100 km on that day. The radius of the Sun is 695,800 km and the distance to the Sun is 149.6 million km. The Moon orbits the Earth, in an anticlockwise direction (viewed from the above the North Pole), the same direction as the Earth rotates about its axis. The period of the Moon’s orbit around the Earth, relative to the Sun (the synodic period – the period between when the Sun, Moon and Earth are in line), is 29.5 days. Using this information: a) Calculate the angular diameters in degrees (how large they appear) of the Sun and of the Moon, respectively, as they will be seen in the sky on that day.
[3]
b) To observe the eclipse you will be using a telescope with a focal length of 200 cm and an eyepiece with a focal length of 25 mm and field of view (FOV) of 52°. Is it possible to see the entire image of the solar disc in the eyepiece of this telescope?
[2]
6
c) Suppose that you are observing the eclipse from a place near the North Pole where the Earth’s spin can be neglected. Calculate the duration of the eclipse from the first to last contact, assuming that the eclipse is central.
[3]
d) Explain if the image below (Figure 1) shows the beginning or end of the solar eclipse.
Figure 1. Binocular view of the solar eclipse. The N and E directions for the observer are shown.
[2]
/10 7
e) Optional Estimate the duration of the eclipse from the first to last contact, assuming that the eclipse is central, for an observer situated in London (latitude = 52.5°) . In this case, Earth’s spin cannot be neglected. The radius of the Earth is !"#$% = 6370 km and the spin period is &!"#$% = 24 '(*s. (the distance to the Moon can be taken to be the same).
8
Question 14
Extrasolar planet discovery
In 1995 a team of Swiss astronomers from the Geneva Observatory announced that they had discovered the first planet outside our solar system around the star 51 Pegasi. They found it by looking at the spectrum of the star and observing the slight change in its velocity, as the star and planet move around their common centre of mass. The planet was found to be orbiting the star on a near circular orbit every 4.23 days, much faster than the Earth is orbiting the Sun. Since then, the radial velocity method that relies on the Doppler effect has been used to discover hundreds of extrasolar planets. The figure below is the srcinal plot of the radial velocity of star 51 Pegasi as it varies with the time.
D
A
C
B
Figure radial velocity of the of star Pegasi. The phase of2.1 The is equivalent to onecurve full period the51 planet around the star.
a) In the figure below (Figure 3), mark in the boxes the letters (A, B, C and D) corresponding to the positions of the star 51 Pegasi around the star-planet centre of mass, as inferred from the radial velocities in Figure 1. Assume that the star is orbiting in a clockwise direction. The figure is not drawn to scale.
Direction to observer
Centre of mass
Figure 3.
[2]
9
b) From Figure 2, determine the velocity of the star around the star-planet centre of mass and estimate an error in your determination.
[1]
c) The star 51 Pegasi has a similar mass to the Sun. Estimate the distance from the planet to the star in AU, assuming that the mass of the planet is much less than that of the star.
[3]
d) Estimate the mass of the planet and express it in terms of Jupiter masses, +,-/$# = . Why is this a lower limit for the real mass of the planet?
1.9 × 10
kg
[HINT: For a binary system = , where is the distance from the object with mass to the centre of mass of the system]
[4]
/10
10
e) Optional Without making the approximation that the mass of the planet is much less than the mass of 51 Pegasi, calculate the mass of the planet (expressed in terms of Jupiter masses +,-/$# =
1.9 × 10 kg) by obtaining an expression involving the ratio
END OF PAPER 11
8: ;:<
.
Solutions for the BOAA trial paper Question Section A 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Section B 11.
Answer
Mark 20 2 2 2 2 2 2 2 2 2 2 10 3 2
C C D A B C B D B A a. b.
32° 2.9 m
Solution:
α
H
h l
a. Let H = 1.8 m, h = 0.8 m, l = 1.6 m. From the information in the question and the diagram above, the altitude of the Sun is:
tan 32° b. The length of the shadow in the absence of the wall is:
12.
a. b.
6."300#$% 10 m !
tan 2.9 m 3 2
Solution: a. Doppler shift:
6&6.3 nm &9"" m ! (6000 669.m' !nm "3.1) #$% ( "3 #$% .
Where the rest wavelength and observed wavelength. Hence, b. Hubble law:
Using v and the given H0,
1
.
the
Section C 13.
20 3
a. Diagram
T
R 1
d
From the diagram, the apparent diameter of the object is:
For small angles, acceptable. Numerically:
2 !*n +, !*n tan *n 4a,*an!5 -n/ 0.0.&&33°'&° #n/
1 , so any of these is
1
b.
2
The magnification of the telescope is:
200 %m 778:;<=>?; ;@;A>;<; 2.& %m "0
1
The field of view of the telescope is thus:
&2° BCD=;E;F<8A; BCD;@;A>;<; "0 0.6&°
1
The FOV is larger than the apparent diameter of the Sun (0.533 , so it is possible to see the entire image of the Sun in the eyepiece.
°5
c.
3
At the North Pole the Earth is static. The reason why the eclipse occurs is that the Moon has its own motion around the Earth with a period of 29.5 days, relative to the Earth’s motion around the Sun. T he angular velocity of the Moon is thus:
G 2HI 29.360° & ,aJ! 12.2°K,aJ Last contact
1
First contact
As seen in the diagram above, the centre of the Moon covers an angular distance of from first to last contact.
L OPQ R S88Q 1.0)"°
1
The time needed for the Moon to cover this distance is:
LM LG 1.12.0)"2 ,aJ! 0.0"" ,aJ! 2.12 N4! Thus, the duration of the eclipse seen from the North Pole is 2.12 hours.
2
1
d.
2
Earth rotates about its axis from West to East (anticlockwise direction), so the Sun and Moon appear to move in the sky from East to West (clockwise). The Moon orbits the Earth, in an anticlockwise direction, from W to E, so the eclipse will begin on the W side of the Sun and will end in the E (as seen by an observer on Earth). Judging from the coordinates given in the image, this is the beginning of the solar eclipse.
e. Optional In part (c) we calculated the duration of the eclipse in case of a static
LM 2.12 N4!
Earth, . This is a special case that only occurs at the Poles of the Earth. At any other latitudes, we need to consider Earth’s spin in the calculations, as the observer will be moving along Earth’s surface. This will extend the duration of the eclipse, since the observer and the Moon will rotate in the same direction (anticlockwise). During the eclipse, the shadow of the Moon moves on the surface of the Earth with a linear velocity:
UVVW 3Z2'0 m N UVVW 2XYIUVVW The distance the shadow of the Moon travels on the static Earth (assuming that the Earth had a flat surface) during the of eclipse is:
LM 2.12 N4! UVVW LM 6Z")0 m
Now, consider the rotating Earth. Earth’s rotational velocity at the Equator is:
[\]^_ 2X`I[\]^_[\]^_ 1Z6)0 m N b &2.&° [\]^_Z [\]^_ %!b 1Z020 m N
At London’s latitude
), Earth’s rotational velocity is:
The Earth-Moon relative velocity is thus:
]cd UVVW [\]^_Z 2220 m N In this case, the duration of the eclipse will increase to:
LM e ]cd LM UVVW UVVW [\]^_ %!b 3.1 N4! In reality, the duration of the eclipse, as viewed from London, will be less than 3.1 hours as we need to consider that the Earth’s surface is curved. So far, we considered that the shadow of the Moon moves on a projection of Earth’s curved surface on a flat surface, but an accurate calculation is complicated. Also, in the question we considered the case of a total eclipse (“ the eclipse th is central”), while from London the eclipse on 20 March will be a partial one, with an obscuration of 85%. Therefore, the angular distance the Moon covers from first to last contact is smaller than the one we calculated. The partial eclipse on 20th March will last for 2h16 min, with first contact at 08:45 UT and last contact at 10:41 UT.
3
14
a.
2 D
C
A
0.5 marks each
B
b.
1
From Figure 2 the velocity of the star around the centre of mass is:
f^\] 60 g 10 m ! &" 60 m ! 10 1' m !
Accepted values for the velocity between
Accepted values for the uncertainty in velocity between
c.
0.5 marks each 3
Use Kepler’s third law:
Ih 'Hh mkd\Wc^ #f^\]Y if^\] Rjkd\Wc^5 Ih 1 Y '.23 ,aJ! Y 0.0& l Neglect of the Sun use:
1
, and because the mass of 51 Pegasi is the same as
1
I
Where T is in years and a in AU. The period of the planet is therefore the semi-major axis (same with radius in this case as the orbit is nearly circular) of the planet’s orbit is:
d.
1 4
The position of the centre of mass, in the CM frame is zero. I.e.
Therefore:
qrsuv R f^\]pwvsx 0 o jkd\Wc^jpkd\Wc^ R f^\] jkd\Wc^ kd\Wc^ f^\]f^\] F^\] 2XIf^\] F^\] I jkd\Wc^ fVd\] 2Xkd\Wc^ jkd\Wc^ 9.3 10 hyz 0.& {|k}^c]
The star-CM distance:
1
From where we get:
4
1 1
Different method For a closed system (like this planet-star system) the total linear momentum, in the centre of mass frame, is 0. Therefore, the momenta of the star and of the planet are equal. Conservation of momentum:
jkd\Wc^ kd\Wc^ f^\]F^\] kd\Wc^ 2Xkd\Wc^ I f^\~ fVd\] f^\] I 0.& {|k}^c] jkd\Wc^ fVd\] 2Hkd\Wc^
Circular velocity:
Mass of the planet is:
This is a lower estimate of the mass of the planet because we are not given any information about the inclination of the orbit, which we assume to be edge-on. If the orbit were tilted by an angle i to the line of sight, the measured radial velocity would be , and hence the true mass of the planet is: . Unfortunately, using the radial velocity method we are not able to determine the inclination of the system, so all the masses we measure are lower estimates. [Any explanation about the tilt of the orbit is acceptable]
=~P; €>Q„‚ƒ‚ •!*n*
e. Optional In part (c) we neglected the mass of the planet as If we don’t neglect it Kepler’s 3 rd law becomes:
jkd\Wc^ f^\]
Ihi …f^\] Rh jkd\Wc^† …kd\Wc^ R f^\]† 'H jA F AF
Using the hint in the question:
Replacing in Kepler’s 3 rd law:
Ihi hF ‡A R Fˆ …A R F† 'H A Ihih F A 1 R Ah 'H F F F
Rearranging,
From Figure 3 we can get the radius of the orbit of 51 Pegasi:
F f^\]2H I 3.'9 10 y m
1.0&) 10 1 R 5h A‰F 219& j 0.'"
Replacing numerically,
where
Hence
and
kd\Wc^
{|k}^c]
5
.
1
