BAAO 2020/21 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. Units should be present on final answers when appropriate.
• There are multiple ways to solve some of the questions; please accept all good solutions that arrive
at the correct answer. Students getting the answer in a box will get all the marks available for that
calculation / part of the question (as indicated in red), so long as there are no unphysical /
nonsensical steps or assumptions made (students may not explicitly calculate the intermediate
stages and should not be penalised for this so long as their argument is clear).

Q1 – Mercury Rotation Period with Aricebo [25 marks]

a. Calculate the power of each echo received by the Aricebo telescope and hence determine the total
number of photons in each echo, given the echo was detected 579.3 s after being transmitted and
Mercury’s surface only reflects 6.5% of the incident radio photons. Assume θ = 0.16° and the
reflected photons from Mercury are scattered randomly within only the hemisphere facing Earth.

We can use the light travel time to get the distance to Mercury
𝑡𝑀𝑒𝑟𝑐 579.3
𝑑𝑀𝑒𝑟𝑐 = 𝑐 × = 3.00 × 108 × = 8.69 × 1010 m [1]
2 2
From this we can work out the radius of the beam
𝑟𝑏𝑒𝑎𝑚 = 𝑑𝑀𝑒𝑟𝑐 tan 𝜃 = 8.69 × 1010 × tan 0.16° = 2.43 × 108 m [1]
[Accept 𝑟𝑏𝑒𝑎𝑚 = 𝑑𝑀𝑒𝑟𝑐 tan 𝜃 + 𝑅𝐴𝑟𝑖𝑐𝑒𝑏𝑜 although the effect of 𝑅𝐴𝑟𝑖𝑐𝑒𝑏𝑜 is negligible as Mercury is so
far away. Also accept small angle approximation if 𝜃 is correctly converted into radians]

Consequently the intensity of the beam at Mercury is then

𝑃 2.0×106
𝑏𝑀𝑒𝑟𝑐 = 2 = = 1.08 × 10−11 W m−2 [1]
𝜋𝑟𝑏𝑒𝑎𝑚 𝜋(2.43×108 )2
This means the total power reflected is
2
𝑃𝑟𝑒𝑓𝑙 = 𝑏𝑀𝑒𝑟𝑐 × 𝜋𝑅𝑀𝑒𝑟𝑐 × 0.065
−11
= 1.08 × 10 × 𝜋(2440 × 103 )2 × 0.065 = 13.1 W [1]
Treating it as a point source radiating in just the hemisphere that faces Earth
𝑃 13.1
𝑏𝐸𝑎𝑟𝑡ℎ = 2𝜋𝑑𝑟𝑒𝑓𝑙
2 = 2𝜋(8.69×1010 )2 = 2.77 × 10−22 W m−2 [1]
𝑀𝑒𝑟𝑐
Consequently, the power received by Aricebo is
2
𝑃𝑟𝑒𝑐𝑒𝑖𝑣𝑒𝑑 = 𝑏𝐸𝑎𝑟𝑡ℎ × 𝜋𝑅𝐴𝑟𝑖𝑐𝑒𝑏𝑜
304.8 2
= 2.77 × 10−22 × 𝜋 ( ) = 2.02 × 10−17 W [1] [6]
2

Turning this into a number of photons:

𝐸𝑝ℎ𝑜𝑡 = ℎ𝑓 = 6.63 × 10−34 × 430 × 106 = 2.85 × 10−25 J [1]
𝐸 𝑃𝑟𝑒𝑐𝑖𝑒𝑣𝑒𝑑 𝑡𝑝𝑢𝑙𝑠𝑒 2.02×10−17 ×500×10−6
𝑛 = 𝐸 𝑡𝑜𝑡 = 𝐸𝑝ℎ𝑜𝑡
= 2.85×10−25
= 35500 photons [1] [2]
𝑝ℎ𝑜𝑡

[This drop in power of 1023 is why you need such a powerful transmitter and such a large receiving
dish to pick up the echo, and hence why it was such a hard measurement to make at the time]
b. Averaging over a series of pulses from August 1965, after correcting for the relative motion of the
Earth and Mercury and the rotation rate of the Earth during the observations, the difference
between the frequencies of photons from the extreme left and right parts of an annulus received
500 μs after the initial echo was Δftotal = 4.27 Hz.
i. Given that the pulse was Doppler shifted twice (once when reflected, once again when
received back at Aricebo) show that the rotational period of Mercury is ≈ 60 days. Assume
that the axis of rotation is normal to the plane of observations. [1 day = 24 hours]

Helpful diagram of the situation [1]

Calculating x from the path difference between a beam that reflects of the sub-radar point
and the edge of the annulus:
1 1
𝑥 = 𝑐 × 2 𝑡 = 3.00 × 108 × 2 × 500 × 10−6 = 7.50 × 104 m [1]
We can now calculate α,
𝑅−𝑥 2440−75
cos 𝛼 = = = 0.969 ∴ 𝛼 = 14.2° [1]
𝑅 2440
Using the Doppler effect equation given, with Δ𝑓 halved twice (once for the two journeys,
once to compare from sub-radar point to edge rather than edge to edge),
1 1
Δ𝑓𝑡𝑜𝑡𝑎𝑙 ×4.27
8 −1
𝑣𝑜𝑏𝑠 = 4 𝑓
4
𝑐 = 430×106 × 3.00 × 10 = 0.745 m s [1]
Considering the geometry of the situation to get the tangential (rotational) velocity,
𝑣 0.745
𝑜𝑏𝑠
𝑣𝑟𝑜𝑡 = sin 𝛼
= sin 14.2° = 3.027 m s −1 [1]
2𝜋𝑅𝑀𝑒𝑟𝑐 2𝜋×2440×103
∴ 𝑇𝑟𝑜𝑡 = 𝑣𝑟𝑜𝑡
= 3.027
= 5.06 × 106 𝑠 = 58.6 days [1] [6]
Δ𝑓
[Students using 𝑣𝑟𝑜𝑡 = 𝑡𝑜𝑡𝑎𝑙
𝑓
𝑐 = 2.98 𝑚 𝑠 −1 to get 𝑇𝑟𝑜𝑡 = 59.56 days only get the final
mark for ecf on their vrot (i.e. 1 out of 6). Trot can be given in seconds or days, but if in the
former a comparison to 60 days must be made]

[We are fortunate that the rotational axis of mercury is almost perfectly normal to the plane
of observations, so this is almost identical to the real rotational period]

ii. Mercury has a semi-major axis of 0.387 au. Rounding slightly if necessary, express the ratio
orbital period : rotational period in a simple integer form (the integers should be < 10).

From Kepler’s third law,

4𝜋2 𝑎 3 4𝜋2
𝑇𝑜𝑟𝑏 = √ =√ × (0.387 × 1.50 × 1011 )3 [1]
𝐺𝑀⊙ 6.67×10−11 ×1.99×1030
= 7.63 × 106 s = 88.28 days [1]
We can now work out the ratio
𝑇𝑜𝑟𝑏 88.2836
∴ ratio = 𝑇𝑟𝑜𝑡
= 58.617
= 1.506 ≈ 1.5 [allow ecf with Trot = 60 days] [0.5]
∴ 3: 2 [0.5] [3]
c. The eccentricity of the planet’s orbit became the prime suspect as to why its actual ratio would be
stable over long time periods.
i. Calculate how many times larger the tidal torque is when Mercury is at perihelion than when
it is at aphelion, given the eccentricity of the orbit is 0.206.

Given 𝑟𝑝𝑒𝑟𝑖 = 𝑎(1 − 𝑒) and 𝑟𝑎𝑝ℎ = 𝑎(1 + 𝑒) and 𝜏 ∝ 𝑟 −6

−6
𝜏𝑝𝑒𝑟𝑖 𝑟 1−𝑒 −6 1−0.206 −6
∴ 𝜏𝑎𝑝ℎ
= ( 𝑟𝑝𝑒𝑟𝑖 ) = (1+𝑒) = (1+0.206) = 12.3 [1] [1]
𝑎𝑝ℎ

ii. Assuming the tidal torque at perihelion is the dominating factor in setting Mercury’s rotation
rate, predict the rotational period of Mercury if it were to behave as though it was tidally
locked when passing through perihelion. Compare this to the measured value and comment
on validity of the assumption.

Finding the perihelion distance first

𝑟𝑝𝑒𝑟𝑖 = 𝑎(1 − 𝑒)
= (0.387 × 1.50 × 1011 )(1 − 0.206) = 4.61 × 1010 m [1]
[Give this mark if they have already calculated rperi in c) i) instead]

Using the vis-viva equation given at the beginning of the paper,

2 1
𝑣𝑝𝑒𝑟𝑖 = √𝐺𝑀⊙ ( − )
𝑟𝑝𝑒𝑟𝑖 𝑎

2 1
= √6.67 × 10−11 × 1.99 × 1030 (4.61×1010 − 0.387×1.50×1011 )
= 5.89 × 104 m s−1 [1]
Working out the orbital angular velocity,
𝑣𝑝𝑒𝑟𝑖 5.89×104
𝜔𝑜𝑟𝑏 = = = 1.28 × 10−6 rad s −1 [1]
𝑟𝑝𝑒𝑟𝑖 4.61×1010
If 𝜔𝑜𝑟𝑏 = 𝜔𝑟𝑜𝑡 [which gives 𝑣𝑟𝑜𝑡 = 3.12 m s−1]
2𝜋 2𝜋
∴ 𝑇𝑟𝑜𝑡 = 𝜔 = 1.28×10−6 = 4.91 × 106 s = 56.9 days [1] [4]
𝑟𝑜𝑡

This is similar to the measured value of Trot ∴ valid assumption [1] [1]

iii. Fig 3 shows the orientation of Mercury’s axis of minimum moment of inertia (the axis the
tidal torque acts upon) if the ratio had been 1 : 1. Redraw this diagram but for the ratio
found in part b (ii). Assume the initial orientation at perihelion looks the same in both cases
(so only the other five positions are needed, separated equally in time) and that the planet
both orbits and rotates in an anticlockwise direction. [Note: the orientation when it returns
to perihelion may not be the same as it was initially.]

New axes are alternating vertical and horizontal lines

[1] [1]
Ends for a & b correctly labelled [1] [1]

[This does mean that it needs two Mercurian years

before the perihelion orientation is the same – as such
one (solar) day on Mercury lasts two years!]
Q2 – Great Conjunctions and the Star of Bethlehem [35 marks]
a. In ideal observing conditions the two planets are far enough apart that they should be (just about)
distinguishable to the naked eye, however to some observers in imperfect conditions they would
appear as a single bright dot, brighter than either planet on its own.
i. During the conjunction, the apparent magnitudes of Jupiter and Saturn were 𝓂𝐽 = −1.97
and 𝓂𝑆 = 0.63, respectively (ignoring dimming by the atmosphere). What would be the
apparent magnitude of the two planets if they appeared to an observer as a single point?
[Hint: it is not simply −1.97 – 0.63 = −2.60]

Given that brightness adds linearly, given the brightness of Jupiter (bJ) and of Saturn (bS), the
joint brightness of Jupiter and Saturn bJ+S is
𝑏𝐽+𝑆 𝑏𝐽 +𝑏𝑆 𝑏
𝑏𝑆
= 𝑏𝑆
= 𝑏𝐽 + 1 [1]
𝑆
Using the formula connecting magnitudes and brightness given at the start of the paper,
𝑏𝐽+𝑆 𝑏
𝑏𝑆
= 10−0.4(𝓂𝐽+𝑆−𝓂𝑆) ∴ 𝓂𝐽+𝑆 = 𝓂𝑆 − 2.5 log (𝑏𝐽 + 1)
𝑆
−0.4(𝓂𝐽 −𝓂𝑆 )
= 𝓂𝑆 − 2.5 log(10 + 1) [1]
−0.4(−1.97−0.63)
= 0.63 − 2.5 log(10 + 1)
= −2.06 [1] [3]
[Coincidentally, −√(−1.97)2 + (0.63)2 = −2.068 ; this approach receives 0 marks]

ii. Although they appeared close in angle, there was a very considerable distance between the
two planets. At conjunction, Jupiter was 5.926 au from Earth whilst Saturn was 10.827 au
(see Fig 5). If they were actually next to each other in space such that they could be treated
as a single object, how far from the Earth (in au) would they need to be to have the same
apparent magnitude as calculated in the previous part? For simplicity, assume that both
planets can be modelled as (very low luminosity) stars so that the change in brightness is
only due to changing the distance from the Earth (i.e. ignore the complications from the
changing distance from the Sun affecting the number of reflected photons and the changing
geometry affecting the illuminated fraction of the planet’s surface).

To balance the magnitudes for being at the same distance, we can use absolute magnitudes
given the formula at the beginning of the paper (with distances converted into pc)
𝑑 5.926×1.50×1011
ℳ𝐽 = 𝓂𝐽 − 5 log (10𝐽 ) = −1.97 − 5 log ( 10×3.09×1016
) = 25.7
Similarly, ℳ𝑆 = 27.0 [need both absolute magnitudes for the mark] [1]

Using a similar approach to the previous part of the question to get the combined absolute
magnitude,
ℳ𝐽+𝑆 = ℳ𝑆 − 2.5 log(10−0.4(ℳ𝐽 −ℳ𝑆 ) + 1)
= 27.0 − 2.5 log(10−0.4(25.7−27.0) + 1) = 25.4 [1]
(𝓂𝐽+𝑆 −ℳ𝐽+𝑆 )/5
∴ 𝑑𝐽+𝑆 = 10 pc × 10
= 10 pc × 10(−2.06−25.4)/5
= 3.15 × 10−15 pc = 9.72 × 1011 m = 6.48 au [1] [3]
[Must be in au for final mark. Make some allowance for high sensitivity to rounding errors]

[Even at a standardised distance, Jupiter is still considerably brighter. If we had kept it as

reflected photons and thus had to take into account the change in incident radiation, we
would get 𝑑𝐽+𝑆 = 6.91 au, showing that our simplifying approximation is reasonable]
b. Jupiter has a period of 4332.589 days and Saturn has a period of 10759.22 days (where 1 day = 24
hours). Note: be careful as your calculations will be very sensitive to rounding errors.
i. Calculate the time between great conjunctions as viewed from the centre of the Solar
System (this will be equal to the average synodic period). Give your answer in years (where 1
year = 365.25 days). [Hint: consider a reference frame rotating at the same rate as Jupiter.]

Moving into a frame of reference where Jupiter is stationary,

1 1 1
𝜔𝐽+𝑆 = 𝜔𝐽 − 𝜔𝑆 ∴ 𝑇 =𝑇 −𝑇 [1]
𝐽+𝑆 𝐽 𝑆
1 1 −1
∴ 𝑇𝐽+𝑆 = (4332.589 − 10759.22) = 7253.455 days = 19.86 years [1] [2]
[Must be in years for the final mark. First mark is for any sensible approach in terms of ω]

ii. Use your answer to predict the date (to the nearest day) of the next great conjunction.

Rewriting 𝑇𝐽+𝑆 into something more convenient

𝑇𝐽+𝑆 = 20 years − 51.545 days
∴ next conjunction 52 days before 21st Dec [1]
52 days = 21 in Dec + 30 in Nov + 1 in Oct ∴ 30th Oct 2040 [2] [3]
[One mark for the year (2040), one for the date. Allow ±1 day. Alternatively, some students
may look for the 304th day in 2040]

[The real date is 31st Oct 2040 so this model has done remarkably well – due to the
dependence on the Earth’s position in its orbit the gap between conjunctions can vary up to
∼ 320 days above or below the average value]

iii. Some astronomers have suggested that the ‘Star of Bethlehem’ seen by the magi (‘wise
men’) on their way to Jesus’ birth was in fact a great conjunction. Use your average synodic
period to find the date of the great conjunction in the first decade BC and give your answer
to the nearest month. [Note: be careful with BC years as year 0 in your calculation is
equivalent to 1 BC, since 31st December 1 BC is followed by 1st January 1 AD]

2020
In 2020 years there have been 19.86 = 101.7 cycles
∴ need 102 cycles before 2020 conjunction [1]
356
Start date = 10 days before the end of the year = 2020 + 365.25
= 2020.975 [1]
[Give this mark if they have already calculated a decimalised start date in b) ii) instead. The
top of the fraction must be 356 as 2020 is a leap year – lose this mark for other values]

∴ end date = 2020.975 − (102 × 19.86) = −4.63 [1]

st
0.63 years = 7.57 months before 31 Dec ∴ May [1]
Since a number between 0 and 1 corresponds to 1 BC, between -1 and 0 corresponds to 2BC,
and so on, ∴ a number between -5 and -4 corresponds to 6 BC ∴ May 6 BC [1] [5]
[One mark for the year (6BC), one for the month. Allow ±1 month]

[From our data a more precise prediction is 230.4 days before 31st Dec and so 14th May 6 BC.
The real great conjunction in that decade occurred on 3rd June 6 BC, so is rather close to our
predicted date. This is before the death of King Herod in 4 BC, as required to fit in with the
story, but it would have been seen in the constellation of Pisces in the NNE, so is unlikely to
have been a ‘star in the East’]
c. By empirically fitting a sinusoidal function (which is assumed to be the same for each track, just with
a fixed phase difference between them) and assuming all conjunctions are separated by the average
synodic period, we can give rough estimations for the separations of any given great conjunction.
Note: be careful as your calculations will be very sensitive to rounding errors.
2𝜋𝑡
i. By reading off the graph, give an equation for Track A of the form 𝜃 = |𝐷 sin ( 𝜆
+ 𝜙𝐴 )|,
where 𝑡 is the (decimalised) date in years, and 𝐷, 𝜆, and −𝜋/2 < 𝜙𝐴 ≤ 𝜋/2 are values that
need to be determined. [Hint: ensure your function passes through the 2020 data point, and
the function is decreasing as it does.]

Reading off the graph the amplitude (D) and the period (λ) of the sine wave:
𝐷 = 1.20° [allow 1.10° ≤ 𝐷 ≤ 1.30°] [1] [1]
𝜆 = 2500 years [allow ±300 years] [1] [1]
[Throughout the rest of this question, propagate the student’s values accordingly]
[Note: in reality D is set by the difference in the inclinations of Jupiter and Saturn’s orbits]

For 𝜙𝐴 the required properties are:

Correct behaviour (i.e. function is decreasing as it passes through 2020) [1]
In the specified range (i.e. −𝜋/2 < 𝜙𝐴 ≤ 𝜋/2) [1] [2]

Guidance on awarding the marks for 𝜙𝐴 :

0.102 2𝜋×2020.975
𝜙𝐴 = sin−1 ( 1.2
)− 2500
= −4.99 rad (= −286.1°)
Not in right range, or right behaviour so 0 marks
𝜙𝐴 = −4.99 + 𝜋 = −1.85 rad (= −106.1°)
Not in right range, or right behaviour so 0 marks
𝜙𝐴 = −4.99 + 2𝜋 = 1.29 rad (= 73.8°)
In right range, but not right behaviour so 1 mark
0.102 2𝜋×2020.975
𝜙𝐴 = (𝜋 − sin−1 ( 1.2
)) − 2500
= −2.02 rad (= −115.9°)
Not in right range, but right behaviour so 1 mark
𝜙𝐴 = −2.02 + 𝜋 = 1.12 rad (= 64.1°)
In right range and right behaviour so 2 marks
2𝜋𝑡
[So final formula is 𝜃 = |1.20 sin (2500 + 1.12)|, but do not penalise students for not writing
this so long as their values of D, λ and 𝜙𝐴 are clear]

ii. Without having to read anything else off the graph, write down the equations for Tracks B
and C, given the same restrictions on 𝜙𝐵 and 𝜙𝐶 .

D and λ remain the same [this can be implied by calculations in later parts] [1] [1]
The phase difference is 𝜋/3, the only extra consideration being staying in the required range
2𝜋
𝜙𝐵 = 𝜙𝐴 − 3
= −0.98 rad (= −55.9°) [1] [1]
2𝜋𝑡
𝜃 = |1.20 sin (2500 − 0.98)|
𝜋
𝜙𝐶 = 𝜙𝐴 − 3 = 0.07 rad (= 4.1°) [1] [1]
2𝜋𝑡
𝜃 = |1.20 sin ( + 0.07)|
2500
𝜋
[If they go outside the required range e.g. 𝜙𝐵 = 𝜙𝐴 + = 2.17 rad (= 124.1°) then lose
3
that mark, unless they have already been penalised in part i. for being outside the range (in
which case allow full ecf) . As in the previous part, do not penalise students for not writing
the equations in the form 𝜃 = ⋯ so long as their values of D, λ and 𝜙𝐴 are clear]
iii. State which track the ‘Star of Bethlehem’ great conjunction is on, and hence use the relevant
equation to predict its separation. How does it compare to the 2020 great conjunction?

Want 102 cycles before 2020 great conjunction, and 102 is divisible by 3
∴ Track A [1] [1]
2𝜋𝑡
∴ 𝜃 = |1.20 sin (2500 + 1.12)|
2𝜋×−4.63
= |1.20 sin ( + 1.12)| = 1.07° [1] [1]
2500
This is much further apart than the 2020 great conjunction [1] [1]
[Note: other tracks coincidentally give similar values of θ; award 0 marks if wrong track used]

[Since this separation is over 10 times larger than the 2020 great conjunction (the real
separation was almost the same as in this model), this adds evidence to suggest it would not
have been as spectacular in the sky (and certainly not close enough to look like a single star),
so that particular great conjunction is probably not an explanation for the Star of Bethlehem]

iv. When is the next great conjunction at least as close as the 2020 one (i.e. 𝜃 ≤ 0.102°)? Give
its year and the value of 𝜃.

Rather than using brute force to find it, the graph suggests a good one to try is the next
point after 2020 on Track A
∴ 𝑡 = 2020.975 + (3 × 19.86) = 2080.55 ∴ year = 2080 [1] [1]
2𝜋×2080.55
∴ 𝜃 = |1.20 sin ( 2500
+ 1.12)| = 0.078° (so good guess) [1] [1]

[Using intuition here saved us a lot of time. The real date is 15th March 2080, and the real
separation will actually be almost identical to the 2020 great conjunction]

v. Similarly, when was the last great conjunction at least as close as the 2020 one? Give its year
and the value of 𝜃.

Again, to avoid brute force, the graph suggests we see whether there is anything suitable
around Track B’s last x-intecept
2𝜋𝑡 2500×0.98
0 = 1.20 sin (2500 − 0.98) ∴ 𝑡 = 2𝜋
= 388
2500
These repeat every ½ λ so the last one was 388 + 2
= 1638 [1]
Looking at the number of track intervals you need to go back,
(2020.975+19.86)−1638
𝑛= 3×19.86
= 6.76 ∴ try 7 intervals back from 2040 [1]
∴ 𝑡 = (2020.975 + 19.86) − (7 × 3 × 19.86) = 1623.80
∴ year = 1623 [1] [3]
2𝜋×(2020.975+19.86)
∴𝜃= |1.20 sin ( − 0.98)| = 0.043° (good guess) [1] [1]
2500

[Another example of intuition saving us lots of time. We have added the 19.86 to the
2020.975 to move from Track A to Track B. The real one was on 16th July 1623 (our model
predicts 19th October) and had a separation of 0.086°. Some closer separations do happen,
including transits (where the disc of Jupiter overlaps with the disc of Saturn), the next of
which are due to happen in 7541 and 8674, and occultations (where Jupiter completely
blocks out Saturn) in 7541, 13340 and 13738. (Note: 7541 sees part of a sequence called a
triple conjunction, where retrograde motion means that Jupiter and Saturn meet three times
in a short period. The transit is in February and the occultation is in June)]
vi. Using your equations, calculate the probability that a great conjunction has 𝜃 ≤ 0.102°.

We need to know for how long in each cycle the curve has a value below 0.102° and
compare that to the length of each cycle to get the overall probability

Using the notation of the diagram above,

𝑥
probability = [1]
𝜆/4
We can ignore phase information as that will not affect the size of 𝑥, just the co-ordinates of
its start and end points, and hence we can simplify the expression for 𝜃 to find 𝑥
2𝜋𝑥 2500 0.102
0.102 = 1.20 sin (2500) ∴ 𝑥 = 2𝜋
× sin−1 ( 1.20 ) = 33.86 years [1]
33.86
∴ prob = = 0.0542 = 5.42% [1] [3]
2500/4
[Accept it given as either a decimal or a percentage. Note that λ cancels in the expression for
sin−1(0.102/𝐷)
probability so it is only dependent on their value of D, giving prob = 𝜋/2
; this
approach gains all the marks and gives values ranging from 5.00% to 5.91% in the allowed
range of D. Allow any other sensible approaches that give results within this range]

[The real value (taken from the number of conjunctions satisfying this condition over 16000
years of conjunctions) is 5.34% (38 instances out of 711) showing how well this model does.
The orbits of the planets change slightly over time due to their gravitational interactions, so
extending this pattern arbitrarily far into the future or past is unlikely to give precise results]
Q3 – High Resolution Stellar Photos [40 marks]
a. When it reached first perihelion, radio signals from the probe took 446.58 s to reach Earth.
i. Show that the spacecraft’s perihelion is ≈ 0.5 au, giving your answer to 4 s.f., and hence
estimate the launch date, assuming the Earth’s orbit is circular. Note that 2020 is a leap year
and take 1 year = 365.25 days. [Hint: You may wish to use a numerical method.]

This is a diagram of the situation (note: Earth and probe are orbiting anticlockwise)

Given the light travel time,

𝑥 = 446.58 × 3.00 × 108 = 1.3397 × 1011 m = 0.8932 au [1]
Considering the geometry of the triangle Earth-Sun-Probe after travelling for time t:
2
𝑥 2 = 𝑟𝑝𝑒𝑟𝑖 + (1 au)2 − 2𝑟𝑝𝑒𝑟𝑖 (1 au) cos 𝜃 [cosine rule] [0.5]
2𝜋𝑡
𝜃 = 𝜋 − 1 year [0.5]
Keeping 𝑟𝑝𝑒𝑟𝑖 in au, and given t is half the period of the probe’s orbit, T, we can use Kepler’s
third law [= K3] to work out t in SI units:
1 4𝜋2
𝑎 = 2 (1 + 𝑟𝑝𝑒𝑟𝑖 )(1 au) and (2𝑡)2 = 𝐺𝑀 𝑎3
⊙

𝜋2 3
∴ 𝑡 = √8𝐺𝑀 ((1.50 × 1011 )(1 + 𝑟𝑝𝑒𝑟𝑖 )) [2]
⊙
Substituting this into our equations above to form one where 𝑟𝑝𝑒𝑟𝑖 is the only unknown:
𝜋2 3
√8𝐺𝑀 ((1.50×1011 )(1+𝑟𝑝𝑒𝑟𝑖 ))
⊙
2
𝑟𝑝𝑒𝑟𝑖 + 1 − 2𝑟𝑝𝑒𝑟𝑖 cos 𝜋 − 2𝜋 365.25×24×60×60
− 𝑥 2 = 0 [1]

( )
Using an appropriate numerical method, 𝑟𝑝𝑒𝑟𝑖 = 0.4956 au [2] [7]
[Allow any valid method to find the root of the equation such as Newton-Raphson starting
with rperi = 0.5 au, using a graphical calculator to find the roots, or even using trial and error
(starting from 0.5 au). The final answer must be 4 s.f. otherwise a mark is dropped. If given in
metres, a comparison must be made to 0.5 au (also in metres) to show they are
approximately the same]

[If students use the ‘simple’ version of K3, 𝑇 2 = 𝑎3 , where 𝑇 is already in years and 𝑎 is
1 3
already in au, then they will get 𝑡 = √32 (1 + 𝑟𝑝𝑒𝑟𝑖 ) so the equation to be solved is

2 1 3
𝑟𝑝𝑒𝑟𝑖 + 1 − 2𝑟𝑝𝑒𝑟𝑖 cos (𝜋 − 2𝜋√32 (1 + 𝑟𝑝𝑒𝑟𝑖 ) ) − 𝑥 2 = 0, which gives 𝑟𝑝𝑒𝑟𝑖 = 0.4998 au .

This approach receives full marks (and is somewhat simpler to type into a calculator)]
 Given we know 𝑟𝑝𝑒𝑟𝑖 , we can find 𝑎, 𝑡 and therefore the launch date:
𝑟𝑝𝑒𝑟𝑖 +1 0.4956+1
𝑎= = = 0.7478 au [1]
2 2
[This mark may be given in part (iii) instead if not awarded here]
1 4𝜋2 1 4𝜋2
∴ 𝑡 = 2 √𝐺𝑀 𝑎3 = 2 √6.67×10−11 ×1.99×1030 (0.7478 × 1.50 × 1011 )3 [1]
⊙

= 1.0244 × 107 s = 118.567 days [1]

th
So we need 15 June – 118.567 days
118.567 days = 15 in June + 31 in May + 30 in April + 31 in March + 11.567 in Feb
so will be 29 – 11.567 = 17.433 ∴ 17th Feb [allow 18th Feb] [1] [4]

[Using the ‘simple’ version of K3 gives a = 0.7499 au, t = 118.592 days, and the same launch
date. This is effectively the same as if students use the rperi = 0.5 au given, although doing
that with the SI version of K3 gives t = 119.090 days and so 16th Feb. Using the given rperi and
the simple version of K3 gives t = 118.618 days. A launch date of 17th Feb corresponds to
arriving at perihelion at 00:00 on 15th June, whilst 18th Feb allows for students presuming a
later time of arrival (up to 23:59 on 15th June). Allow ecf on the launch date from their own
values of t]

[In reality, the probe reached a perihelion distance of 0.514 au and was launched on 10th
Feb. The reason for the discrepancy is that the Earth’s orbit is elliptical and reaches
perihelion on 4th Jan so the probe was not actually at aphelion at launch (although it did
reach it on 17th Feb), and the aphelion distance was 0.986 au]

ii. Use the Ramanujan approximation to work out the distance travelled by the probe between
launch and perihelion to 4 s.f.

Using the equations given at the beginning of the paper,

𝑟𝑝𝑒𝑟𝑖 0.4956
𝑟𝑝𝑒𝑟𝑖 = 𝑎(1 − 𝑒) ∴ 𝑒 = 1 − 𝑎
= 1 − 0.7478 = 0.3372 [1]
𝑏2
𝑒 = √1 − 𝑎2 ∴ 𝑏 = 𝑎√1 − 𝑒 2 = 0.4956√1 − 0.33722 = 0.7040 au [1]
𝑎−𝑏 0.7478−0.7040
∴ 𝑗 = 𝑎+𝑏 = 0.7478+0.7040 = 0.030176 [1]
3𝑗 2
∴ 𝐶 = 𝜋(𝑎 + 𝑏) [1 + ]
10+√4−3𝑗 2
3(0.030176)2
= 𝜋(0.7478 + 0.7040) [1 + ] = 4.562 au [0.5]
10+√4−3(0.030176)2
But only half an ellipse has been covered, so distance travelled = 2.281 au [0.5] [4]

[Must be 4 s.f. for the final mark. Carrying forward values from the ‘simple’ version of K3
gives e = 0.3335, b = 0.7070 au, j = 0.02947, C = 4.578 au and so distance travelled is 2.289
au (essentially the same as if using the given rperi)]
iii. Find the values of D and E, given as fractions in their simplest terms, and hence calculate a
new value for the distance travelled by the probe (also to 4 s.f.). Compare this to the
approximation in the previous part and comment on your answer.

Using the given formula for a binomial expansion,

𝜋/2
𝐼 = ∫0 (1 − 𝑒 2 sin2 𝑡)1/2 d𝑡
1 1
𝜋/2 1 (− )
= ∫0 1 + 2 (−𝑒 2 sin2 𝑡) + 2 2 2 (−𝑒 2 sin2 𝑡)2 + ⋯ d𝑡
𝜋/2 1 1
= ∫0 1 − 2 𝑒 2 sin2 𝑡 − 8 𝑒 4 sin4 𝑡 d𝑡 [correctly simplified expansion] [1]
𝜋/2 1 𝜋⁄2 1 𝜋⁄2
= ∫0 1 d𝑡 − 2 𝑒 2 ∫0 sin2 𝑡 d𝑡 − 8 𝑒 4 ∫0 sin4 𝑡 d𝑡
The first term is very easy to evaluate (and is effectively already given),
𝜋/2 𝜋/2 𝜋
∫0 1 d𝑡 = [𝑡]0 = 2
To do the other integrals we need to use the trig identities:
1 1 1 1
sin2 𝑥 = 2 − 2 cos 2𝑥 and cos2 𝑥 = 2 + 2 cos 2𝑥
The second term,
𝜋/2 𝜋/2 1 1 1 1 𝜋/2 𝜋
∫0 sin2 𝑡 d𝑡 = ∫0 (2 − 2 cos 2𝑡) d𝑡 = [2 𝑡 − 4 sin 2𝑡] =4 [1]
0
The third term,
𝜋/2
∫0 sin4 𝑡 𝑑𝑡
𝜋/2 1 1 2
= ∫0 (2 − 2 cos 2𝑡) 𝑑𝑡
𝜋/2 1 1 1
= ∫0 4 − 2 cos 2𝑡 + 4 cos 2 2𝑡 𝑑𝑡
𝜋/2 1 1 1 1 1
= ∫0 4 − 2 cos 2𝑡 + 4 (2 + 2 cos 4𝑡) 𝑑𝑡
𝜋/2 3 1 1
= ∫0 8 − 2 cos 2𝑡 + 8 cos 4𝑡 𝑑𝑡
3 1 1 𝜋/2 3𝜋
= [8 𝑡 − 4 sin 2𝑡 + 32 sin 4𝑡] = 16 [3]
0
So putting back into our original formulae,
𝜋 1 𝜋 1 3𝜋 1 3
𝐶 = 4𝑎𝐼 = 4𝑎 ( 2 − 2 𝑒 2 [ 4 ] − 8 𝑒 4 [ 16 ] − ⋯ ) = 2𝜋𝑎 [1 − 4 𝑒 2 − 64 𝑒 4 − ⋯ ]
1 3
∴ 𝐷 = 4 and 𝐸 = 64 [one mark for each correct fraction] [2] [7]
We can now evaluate C numerically,
1 3 5
𝐶 ≈ 2𝜋𝑎 [1 − 4 𝑒 2 − 64 𝑒 4 − 256 𝑒 6 ]
1 3 5
= 2𝜋(0.7478) [1 − 4 (0.3372)2 − 64 (0.3372)4 − 256 (0.3372)6 ]
= 4.562 au [0.5]
∴ distance travelled = 2.281 au [0.5] [1]
To 4 s.f., this is the same as the Ramanujan approximation (so very good!) [1] [1]
6
[Must be 4 s.f. for the final mark. If students forget the e term, they will still get the same
answer to 4 s.f. – do not penalise them for that mistake. Using the ‘simple’ version of K3
gives 2.289 au, again the same as before. The real difference between the two values is
4.5 × 10−6 au = 680 km, which is remarkably close]

[In reality, the distance the probe had travelled from launch to perihelion was 2.393 au,
however the main reason for the discrepancy is the extra week of travel between launch and
aphelion]
b. The energy density of black-body radiation, u, and number density, n, at temperature T are given.
𝑢
i. The average energy per photon is given as 𝐸̅ = = 𝜀𝑘𝐵 𝑇 . Find the numerical value of 𝜀.
𝑛

To get the integral to look like the standard one, we need to use a substitution:
ℎ𝜈 𝑘𝐵 𝑇 𝑘𝐵 𝑇
Use 𝑥 = 𝑘 ∴𝜈= 𝑥 ∴ d𝜈 = d𝑥 [need correct dν for mark] [1]
𝐵𝑇 ℎ ℎ
Substituting this into the original integral,
∞ 8𝜋ℎ𝜈 3 1 ∞ 8𝜋ℎ 𝑘𝐵 𝑇 3 𝑥3 𝑘𝐵 𝑇
𝑢 = ∫0 𝑐3 ℎ𝜈
𝑑𝜈 = ∫0 𝑐 3 ( ℎ
) exp(𝑥)−1 ℎ
𝑑𝑥 [1]
exp( )−1
𝑘𝐵 𝑇
8𝜋ℎ 𝑘𝐵 𝑇 4 𝜋4 8 𝑘 𝑇 3
= 𝑐3
( ℎ ) [15] = 15 𝜋 5 ( ℎ𝑐
𝐵
) 𝑘𝐵 𝑇 [1]
8 5 𝑘𝐵 𝑇 3
𝑢 𝜋 ( ) 𝑘𝐵 𝑇 𝜋4
∴ 𝐸̅ = 𝑛 = 15 ℎ𝑐
𝑘𝐵 𝑇 3
= 30𝜁(3) 𝑘𝐵 𝑇 [1]
16𝜋( ) ζ(3)
ℎ𝑐
𝜋4
∴𝜀= = 2.701 [1] [5]
30𝜁(3)
[Third mark is for some attempt at simplification of the algebraic expression. No s.f.
requirement in this question but expect at least 2 s.f. (and should really use 4 s.f. in the next
part of the question). If get dν = dx and hence a value of ε that still has a value of T in it this is
considered a major error and there is no ecf]

ii. Assuming the plasma of Fe ions is in thermal equilibrium with the photons, and that the
average energy of the photons is equal to the ionisation energy of FeX (which is 22 540 kJ
mol-1), calculate the temperature of the plasma. Give your answer to 4 s.f.

Converting the ionisation energy into an energy of each photon

22540×10 3
22540×10 3
𝐸̅ = 𝑁
= 6.02×1023 = 3.744 × 10−17 J [1]
𝐴
𝐸̅ 3.744×10−17
∴𝑇= = = 1.004 × 106 K [must be 4 s.f.] [1] [2]
𝜀𝑘𝐵 2.701×1.38×10−23

[Despite the considerable simplifications we have made to our model, this is remarkably
close to the real value of temperature for which Fe X is the dominant fraction. In reality, it
forms an equilibrium over a range of temperatures (contributing a detectable fraction from
about 105.4 to 106.4 K) with various other ions in ionisation equilibrium – all you can compute
is the ionic fraction at a given temperature (i.e. the plasma is never purely one ion)]

c. The Rayleigh criterion and speed of sound in a plasma are given.

i. Determine the theoretical minimum angular diameter of an element resolvable by this
optical system. Give your answer in arcseconds (“).

First, we can work out the wavelength given the photon energy,
ℎ𝑐 ℎ𝑐 6.63×10−34 ×3.00×108
𝐸= 𝜆
∴𝜆= 𝐸
= 71.0372×1.60×10−19
= 1.75 × 10−8 m [1]
𝜆 1.75×10−8
∴ 𝜃 = 1.22 = 1.22 × = 4.50 × 10−7 rad = 0.0929" [1] [2]
𝐷 47.4×10−3
[Must be in arcseconds for the final mark]

[Given the wavelength investigated is 17.5 nm, it is clearly in the extreme UV]
ii. In practice, this is not achieved as the pixels are not small enough. Given that each picture
element is spread across two pixels (in 1D) to allow adequate sampling, what is the actual
minimum angle resolved on the CCD? Give your answer in arcseconds (“). [Hint: consider the
geometry of the optical system and note that the angles are small enough that the small
angle approximation can be used.]

Considering the geometry of the situation:

where s is the distance on the CCD and f is the focal length. Using the small angle approx:
𝑠
tan 𝜃 ≈ 𝜃 = 𝑓 [1]
2×10×10−6
∴𝜃= = 4.78 × 10−6 rad = 0.985" [1] [2]
4187×10−3
[Must be in arcseconds for the final mark. If only done for s = 1 pixel (i.e. 10 μm), lose 1
mark. Another valid approach for the first mark is to use the formula for the plate scale,
𝜃 1
𝑠
= 𝑓 = 0.239 rad m−1 = 2.39 × 10−6 rad pixel−1 = 0.493" pixel−1 ]

iii. Later on in its mission, Solar Orbiter will have a perihelion of 0.284 au. Calculate the physical
size on the Sun (in km) of each picture element in an image taken with HRIEUV as well as the
FOV in units of R⊙.

Using similar geometry to the previous part of the question,

where d is the distance on the Sun and r is the distance to the Sun. Using the small angle
approximation again,
𝑑 ≈ 𝑟𝜃 = (0.284 × 1.50 × 1011 ) × 4.78 × 10−6 = 203 km [1] [1]
Considering the field of view (FOV) of the telescope, we need to convert 1000” into rad
1000 𝜋
𝐹𝑂𝑉 = 𝑟𝜃 = (0.284 × 1.50 × 1011 ) × (3600 × 180) = 2.07 × 108 m
= 0.297 𝑅⊙ [1] [1]
[So the field of view is 0.297 R⊙ by 0.297 R⊙. First answer must be in km and second in R⊙
for the marks]

[Originally the mission plan was to go even closer, however it was adjusted to have a
minimum perihelion of > 0.28 au so that the solar panel technology could be reused from
the BepiColombo mission, which will enter an orbit around Mercury in December 2025.
Mercury’s perihelion is ≈ 0.30 au so Solar Orbiter will pass fully inside the orbit of Mercury.
The minimum perihelion for Solar Orbiter will be on 7th Feb 2027]
iv. If the mass fractions of the surface of the Sun are X = 0.7381, Y = 0.2485, and Z = 0.0134, and
treating the plasma as an ideal monatomic gas so that γ = 5/3, determine the speed of sound
of the plasma (in km s-1) at the base of the corona. Hence, by comparison to your answer
from the previous part, estimate the upper limit on the length of an exposure to avoid
motion blur in the plasma. You should ignore any motion blur from the relative motion of
the spacecraft or the rotation of the Sun.

Using the given mass fractions and formulae, plus the temperature from b. (ii)
𝑚𝑝 𝑚𝑝
𝜇= 3𝑌 𝑍 = = 0.599𝑚𝑝 = 1.00 × 10−27 kg
3(0.2485) 0.0134 [1]
2𝑋+ + 2(0.7381)+ +
4 2 4 2
5
𝛾𝑘𝐵 𝑇 ×1.38×10−23 ×1.004×106
∴ 𝑣𝑠 = √ =√ 3
= 152 km s −1 [1] [2]
𝜇 1.00×10−27
[Must be in km s-1 for the final mark. If students did not get a value for T from part b. they
should have seen in the opening paragraph of the question that typical corona temperatures
are 106 K and used that – penalise those that use the temperate of the surface of the Sun]

By comparing to the physical size of each picture element,

𝑑 203
𝑡 = 𝑣 = 152 = 1.34 s [1] [1]
𝑠

[The HRIEUV will take images of 1 second exposure (with shorter exposure times in subfields
of the FOV) every second in the ‘Discovery’ program, however most images will have a 2
second exposure with the interval time between images varying from seconds to minutes
depending on the different observation programs]