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BRITISH OLYMPIADIN ASTRONOMY & ASTROPHYSICS 2015

British Olympiad in Astronomy and Astrophysics

Competition Paper

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.

Name School

Total Mark/50

1

Useful constants

Speed of light c 3.00 × 10� ms� Gravitational constant G 6.67 × 10�� Nm�kg� Solar mass Msolar 1.99 × 10�� kg Astronomical Unit AU 1.496 × 10�� m Parsec pc 3.086 × 10�� m Earth’s orbit semi-major axis 1 AU Earth’s rotation period 1 day 24 hours Earth’s mass MEarth 5.97 × 10�� kg Earth’s axial tilt 23.4 ° You might find the diagram of an elliptical orbit below useful in solving some of the questions:

Elements of an elliptic orbit:

� – semi-major axis � – semi-minor axis � = ! –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

F

b

O a

c

P A

2

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. High-mass stars are far too luminous B. The lifetime of a high-mass star is too short C. High-mass stars are too hot to allow for life to form D. 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? %1inch = 2.54cm*

A. 10x B. 50x C. 100x D. 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 D. Jupiter

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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′W* at 21:00 UT. At what time UT will it be seen setting in Cardiff (+ = 51°30,N, . = 3°11′W* on the same day?

A. 21:12 B. 21:00 C. 20:48 D. 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 1023 kg and the minimum distance from Earth to Mars is 0.52 AU.

A. 2.3 m B. 0.8 mm C. 0.8 m D. 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. 181 years B. 16 years C. 64 years D. 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?

A. Dubai (+ = 25°N* B. London (+ = 52°N* C. Rio de Janeiro (+ = 23°S* D. Tromsø (+ = 70°N*

[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.]

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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]

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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 2�, implying that the Universe is expanding. To measure the redshift of a galaxy, astronomers usually use the 3 parameter. Suppose that we observe a galaxy with a redshift of 3 = 0.30 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, 2� = 72kms�Mpc�.

a) Assuming that the classical Doppler effect 63 ≈ 8 9is a reasonable approximation,

what is the redshifted wavelength λ of the receding galaxy, in nm? 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]

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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 20th 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.

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.

Figure 1 The partial solar eclipse visibility as seen from Sheffield, UK.

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[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 to see a total solar eclipse from the Capital of Svalbard, Longyearbyen (+ = 78°13,N15°33,E* during December.

[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 @A = 1737.5km, the mean distance to the Moon is �A =385,000km and the eccentricity of the Moon’s orbit is �A = 0.055. The radius of the Sun is @B = 695,800km, the mean distance to the Sun is �C = 149.6 × 10�km and the eccentricity of Earth’s orbit is �C = 0.0167. [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]

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[7]

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.

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]

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Figure 2. The light curve of the star HD 209458. Figure 3. Schematic of the transit.

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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 ×10Gkm, @HEIJKLM = 7.0 × 10�km*.

____[3]

END OF PAPER

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1

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 20

1. C 2 2. B 2 3. D 2 4. C 2 5. A 2 6. C 2 7. A 2 8. C 2 9. C 2 10. B 2

Section B 10 11.

���� = ���� �� = ������� �

= ����������4���

= 4.22 × 10�m ≈ 42,200km

� = 2�# = 4��#

$% = �� = �#4 = 6hours

a. Answer: 42,200km

On a circular orbit, the centripetal force is due to the gravitational force:

� = 24hours, therefore the radius of the geostationary orbit is:

b. Answer: 6hours In the case when the satellite is orbiting the Earth with a period of �# = 24hours, but in the opposite direction to Earth’s rotation, the relative angular velocity is:

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 180° (or π). The time the satellite is visible for the observer is:

[Since the radius of the orbit is only 6.�, 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.

3

1

1

1 2

1

1

2

12.

/ −/#/# =12 ≈ 3

/ = /#(1 + 3)

/ = 631.9nm ≈ 630nm

$% = 2

= 1:#

$% ≈ 3:# ≈ 0.372kms<=Mpc<= ~4billionyears

a. Answer: 630nm From the classical Doppler effect: � ≈ 3 ≈ 0.3. This is a reasonable approximation as

long as 1 ≪ 2, otherwise the full relativistic Doppler effect has to be used. Doppler shift formula:

Where /# = 486.1nm − the rest wavelength. Hence:

The observed (redshifted) wavelength is:

b. Answer: 4billionyears The time it takes the light (travelling at speed 2) from the galaxy to reach us is:

According to Hubble’s law the distance to the galaxy is:

Hence,

In the equation above notice that the unit of :# is I<= and take care when doing the conversion from Mpc to km.

3

1 1 1 2

1 1

Section C 20 13. 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

�� = 0.1�=

� −= = − log=# ���=log=# 2.512

� = = − 2.5 log=# ���= � = −26.74 − 2.5 log=# 0.1

� = −24.24

b. The brightness of the solar disc is proportional to the visible surface area. The apparent magnitude of the Sun is = = −26.74corresponding 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:

This corresponds to an apparent magnitude � . Inverting the Pogson’s formula, the apparent magnitude of the Sun during the eclipse is:

3

1

1

1

3

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 78°13LN) is situated above the Arctic circle (latitude 66°34LN =90° − 23°27′) 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.

1

O = 2sin<= PRdS

T�U = V + 2 = V�(1 + W�) = 152.1 × 10Xkm

O = 0.524°

Y = .Zsin O2 ≈ 380,000km

YT[� = V − 2 = VZ(1 − WZ)

VZ = YT[�1 − WZ ≈ 402,120km

$% ≈ 450millionyears

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 Sun and of the Moon, respectively, make use of a diagram such as the one below:

From the diagram, the apparent diameter of the object is:

For small angles, sinO ≈ tanO ≈ O(inradians), 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, T�U and its angular diameter. From the diagram of the ellipse on page 1, the distance to the aphelion is:

The angular diameter of the Sun (and of the Moon for total eclipses not to occur) is:

Now calculate where the Moon must be, YT[� and hence obtain VZ. Then calculate the change of VZ from current value and, knowing the rate, calculate the time taken. The distance the Moon needs to be situated at is:

This will be the perigee of the new orbit, YT[�. Again, from the diagram of the ellipse on page 1, the distance to the perigee is:

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.

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1

1

1 1

1

1

1

R

d α

4

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% [Accept 1.6 ± 0.1% or 1.7 ± 0.1% and the errors propagated in the following calculations]

1 0.5 0.5

`ab���� = �.ab�����

c = `ab����`d��� = .ab�����.d����

.ab����.d��� = ec = 0.128

b. The covered area of the star, during the transit is:

The percentage determined in a) of c = 0.0165corresponds to:

Therefore, the ratio of the radius of the planet and of the star is:

2 1

1

��Vf = 4π��(�hijk +lmjnoi)

V = ���d�����4���

V = 0.047AU

c. Use Newton’s third law (which gives the proportionality of Kepler III but with constants):

Neglect mlmjnoi ≪ Mhijk and find the radius of the planet’s orbit:

Using the values in the question, Mhijk = 1.15Mhrmjk and the period of T = 3.525days radius of the planet’s orbit is:

2 1 1

$%�t��b ≈ 0.13days ≈ 3.12hours

1ab���� =2(.d��� + .ab����)$%�t��b = 2�V�

.d��� + .ab���� = � $%�t��b� V~8.15 ×10ukm

.d��� = 7.23 × 10ukm ≈ 1.04.vw�

.ab���� = 0.92 × 10ukm ≈ 1.32.xwa[���

d. From the light curve (Figure 2), the total transit time (from the beginning of the drop, to its end) is:

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 2(.d��� + .ab����). 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, 1ab����:

Hence,

And using the ratio in b), 0.128, the radius of the star and of the planet are:

3 1

1

0.5 0.5

British Olympiad in Astronomy and Astrophysics

Trial Paper

March 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 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.

Name School

Total Mark/50

1

Useful constants

Speed of light c 3.00 × 10� ms� Gravitational constant G 6.67 × 10�� Nm�kg� Solar mass Msolar 1.99 × 10�� kg Astronomical Unit AU 1.496 × 10�� m Earth’s orbit semi-major axis 1 AU Earth’s orbital period 1 year 365.25 days

2

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. Red giant B. Brown dwarf C. O-type blue giant D. Yellow main sequence star

2. The majority of the mass in the Universe is contained in:

A. The most massive stars B. Gas and dust C. Dark matter D. Supermassive black holes

3. How much more or less light can an 8-metre aperture telescope collect compared to a 4-metre 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 D. Saturn

3

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. 150 kms� B. 30kms� C. 15kms� D. 0.3kms�

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.

A. 1.4 x 10-5 N B. 2.1 x 10-11 N C. 9.6 x 10-6 N D. 2.1 x 10-5 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. 2023 B. 2061 C. 2064 D. 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

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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]

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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, �� = 72kms�Mpc�.

a) The shift of the line is due to the Doppler effect. What is the velocity v of the receding galaxy, in kms�? 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]

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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 20th 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]

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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.

[2]

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Figure 1. Binocular view of the solar eclipse. The N and E directions for the observer are shown.

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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 !"#$% = 6370kmand the spin period is &!"#$% = 24hours. (the distance to the Moon can be taken to be the same).

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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 original plot of the radial velocity of star 51 Pegasi as it varies with the time.

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.

[2]

A

B

C

D

Figure 2. The radial velocity curve of the star 51 Pegasi. The

phase of 1 is equivalent to one full period of the planet around

the star.

Direction to observer Centre of mass

Figure 3.

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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, +,-./$0# =

1.9 × 10�1kg. Why is this a lower limit for the real mass of the planet? [HINT: For a binary system 2�3� =2�3�, where 3 is the distance from the object with mass 2 to the centre of mass of the system] [4]

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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 +,-./$0# =

1.9 × 10�1kg) by obtaining an expression involving the ratio 456789:

4;:7<.

END OF PAPER

1

Solutions for the BOAA trial paper

Question Answer Mark Section A 20

1. C 2 2. C 2 3. D 2 4. A 2 5. B 2 6. C 2 7. B 2 8. D 2 9. B 2 10. A 2

Section B 10 11.

� � tan�� � � � � 32°

� � tan� � 2.9m

a. 32° b. 2.9m Solution: 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:

b. The length of the shadow in the absence of the wall is:

3 2

12.

� ����� ���

� � ��

a. 6.00 � 10�kms��

b. 83Mpc Solution: a. Doppler shift:

Where �� � 656.3nm � the rest wavelength and � � 669.4nm � the observed wavelength. Hence, � � 5988kms��~6000kms��. b. Hubble law:

Using v and the given H0, � � 83.17Mpc~83Mpc.

3 2

α H

h

l

2

Section C 20 13.

� � 2sin�� Rd�

Sun:� � 0.533° Moon:� � 0.545°

a. Diagram

From the diagram, the apparent diameter of the object is:

For small angles, sin 1 2 tan 1 2 13inradians5, so any of these is acceptable. Numerically:

3

1 1

1

6 � 789:;<=>?;7;@;A>;<; � 200cm2.5cm � 80

BCD=;E;F<8A; � BCD;@;A>;<;6 � 52°80 � 0.65°

b. The magnification of the telescope is:

The field of view of the telescope is thus:

The FOV is larger than the apparent diameter of the Sun (0.533°5, so it is possible to see the entire image of the Sun in the eyepiece.

2 1

1

G � 2πI � 360°

29.5days � 12.2°/day

LM � L�G � 1.07812.2 days � 0.088days � 2.12hours

c. 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. The angular velocity of the Moon is thus:

As seen in the diagram above, the centre of the Moon covers an angular distance of L� � �OPQ R �S88Q � 1.078° from first to last contact. The time needed for the Moon to cover this distance is:

Thus, the duration of the eclipse seen from the North Pole is 2.12 hours.

3 1 1 1

R

d α

First contact Last contact

3

d. 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.

2

�UVVW �2XYUVVWIUVVW 2 3,240kmh��

� � �UVVWLM � 6,870km

�[\]^_ �2X`[\]^_I[\]^_ 2 1,670kmh��

�[\]^_,a ��[\]^_ cosb 2 1,020kmh��

�]cd � �UVVW � �[\]^_,a 2 2220kmh��

LMe � ��]cd � LM � �UVVW�UVVW � �[\]^_ cosb 2 3.1hours

e. Optional In part (c) we calculated the duration of the eclipse in case of a static Earth, LM � 2.12hours. 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:

The distance the shadow of the Moon travels on the static Earth (assuming that the Earth had a flat surface) during the LM � 2.12hours of eclipse is:

Now, consider the rotating Earth. Earth’s rotational velocity at the Equator is:

At London’s latitude 3b � 52.5°), Earth’s rotational velocity is:

The Earth-Moon relative velocity is thus:

In this case, the duration of the eclipse will increase to:

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 is central”), while from London the eclipse on 20th 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 2h16min, with first contact at 08:45 UT and last contact at 10:41 UT.

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14 a.

2

0.5 marks each

�f^\] � 60 ± 10ms��

b. From Figure 2 the velocity of the star around the centre of mass is:

Accepted values for the velocity between 58 � 60ms�� Accepted values for the uncertainty in velocity between 10 � 14ms��

1

0.5 marks each

IhY� � 4πh

i36f^\] Rjkd\Wc^5

IhY� � 1

Y � 0.05AU

c. Use Kepler’s third law:

Neglect mkd\Wc^ ≪ Mf^\], and because the mass of 51 Pegasi is the same as of the Sun use:

Where T is in years and a in AU. The period of the planet is I �4.23days therefore the semi-major axis (same with radius in this case as the orbit is nearly circular) of the planet’s orbit is:

3 1 1

1

o � jkd\Wc^pqrstuv R6f^\]pwvsxjkd\Wc^ R6f^\] � 0

jkd\Wc^�kd\Wc^ �6f^\]�f^\]

�F^\] � 2X�f^\]I

jkd\Wc^ � 6fVd\]�F^\]I2X�kd\Wc^

jkd\Wc^ � 9.3 � 10hykg 2 0.56{|k}^c]

d. The position of the centre of mass, in the CM frame is zero. I.e.

Therefore:

The star-CM distance:

From where we get:

4

1

1

1

A

B

C

D

5

jkd\Wc^�kd\Wc^ � 6f^\]�F^\]

�kd\Wc^ � 2X�kd\Wc^I

6f^\~ � 6fVd\]

jkd\Wc^ � 6fVd\]�f^\]I2π�kd\Wc^ 2 0.56{|k}^c]

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:

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 v sin i, and hence the true mass of the planet is: 6=~P; � 6�>Q ��� �⁄ . 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]

1

Ihi�6f^\] Rjkd\Wc^�4πh � ��kd\Wc^ R �f^\]��

jA � 6F�F�A

Ihi6F4πh ��A R �F�A � � ��A R �F��

Ihi6F4πh�F� � �A�F 1 R�A�F�

h

�F � �f^\]I2π � 3.49 � 10ym

e. Optional In part (c) we neglected the mass of the planet as jkd\Wc^ ≪ 6f^\]. If we don’t neglect it Kepler’s 3rd law becomes:

Using the hint in the question:

Replacing in Kepler’s 3rd law:

Rearranging,

From Figure 3 we can get the radius of the orbit of 51 Pegasi:

Replacing numerically,

1.057 � 10�� � 131 R 15h where 1 � �A �F� Hence 1 � 2195 and jkd\Wc^ � 0.486{|k}^c]