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BAAO 2017, Теория — Задача 2. Hohmann Transfer Orbits

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British Astronomy and Astrophysics Olympiad 2016-2017 — Astronomy & Astrophysics Competition Paper — Monday 23rd January 2017 · 23 януари 2017 г. · 20 т.

Указания

Time: 3 hours plus 15 minutes reading time (no writing permitted). Approx 35 minutes per question.

Questions: All five questions should be attempted.

Marks: The questions carry similar marks.

Solutions: Answers and calculations 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 28th January.

Clarity: Solutions must be written legibly, in black pen (the papers are photocopied), and working down the page. Scribble will not be marked and overall clarity is an important aspect 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 12th 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 7th April 2017). Astronomy material will be covered; 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 theory paper will be sat. Five students (plus one reserve) will be selected for further training. 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
Speed of light	$c$	$3.00\times10^{8}\ \mathrm{m\,s^{-1}}$
Earth’s rotation period	1 day	24 hours
Earth’s orbital period	1 year	365.25 days
parsec	pc	$3.09\times10^{16}\ \mathrm{m}$
Astronomical Unit	AU	$1.49\times10^{11}\ \mathrm{m}$
Radius of the Earth	$\mathrm{R}_{\oplus}$	$6.37\times10^{6}\ \mathrm{m}$
Semi-major axis of the Earth’s orbit		1 AU
Radius of the Sun	$\mathrm{R}_{\odot}$	$6.96\times10^{8}\ \mathrm{m}$
Mass of the Sun	$\mathrm{M}_{\odot}$	$1.99\times10^{30}\ \mathrm{kg}$
Mass of the Earth	$\mathrm{M}_{\oplus}$	$5.97\times10^{24}\ \mathrm{kg}$
Luminosity of the Sun	$\mathrm{L}_{\odot}$	$3.85\times10^{26}\ \mathrm{W}$
Gravitational constant	$G$	$6.67\times10^{-11}\ \mathrm{m^{3}\,kg^{-1}\,s^{-2}}$

You might find the diagram of an elliptical orbit below useful in solving some of the questions:

Elements of an elliptic orbit:

$a = \mathbf{OA}\ (= \mathbf{OP})$ semi-major axis

$a = \mathbf{OB}\ (= \mathbf{OC})$ semi-minor axis

$e = \sqrt{1-\frac{b^2}{a^2}}$ eccentricity

F focus

P periapsis (point nearest to F)

A 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 constant of proportionality is $4\pi^2/GM$ , where $M$ is the mass of the central object and $G$ is the universal gravitational constant.

Magnitudes: The apparent magnitudes of two objects, $m_1$ and $m_0$ , are related to their apparent brightnesses, $b_1$ and $b_0$ , via the formula:

\frac{b_1}{b_0} = 10^{-0.4(m_1-m_0)}

Note for markers

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.

Условие

In order to move a spacecraft between orbits we must apply a thrust using rockets, which changes the velocity of the spacecraft by $\Delta v$ . 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 $m \ll M$ ) the orbital velocity, $v_{\mathrm{orb}}$ , is given by the formula $v_{\mathrm{orb}} = \sqrt{\frac{GM}{r}}$ .

An ellipse with dashed major axis from P (left) to A (right) and dashed minor axis from B (top) to C (bottom), centre O, focus F marked with a blue dot between P and O, semi-minor axis labelled b and semi-major axis labelled a. — Diagram of an elliptical orbit (Important Constants page)

a. Show that $v_{\mathrm{orb}}$ in low Earth orbit (LEO; about $200\ \mathrm{km}$ above the surface), is about $8\ \mathrm{km\,s^{-1}}$ . This is an estimate of the $\Delta v$ the rockets need to provide for the spacecraft to reach LEO. [2 т.]

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 $r_{\mathrm{A}}$ ) and the aphelion coincides with the outer planetary orbit (with radius $r_{\mathrm{B}}$ ). It is achieved by increasing the velocity of the spacecraft at point A by $\Delta v_{\mathrm{A}}$ before then increasing it again at point B by $\Delta v_{\mathrm{B}}$ .

Two concentric circles: the inner labelled “Initial orbit” with radius rA, the outer labelled “Final orbit” with radius rB. A dashed half-ellipse labelled “Transfer orbit” runs from point A at the bottom of the inner circle to point B at the top of the outer circle. Arrows ΔvA at A (pointing right) and ΔvB at B (pointing left); a dotted vertical line through A and B. — 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\left(\frac{2}{r}-\frac{1}{a}\right)

b. Derive expressions for $\Delta v_{\mathrm{A}}$ and $\Delta v_{\mathrm{B}}$ by comparing their circular orbital speeds with their transfer orbit speeds. Simplify your final expressions to include $G$ , $\mathrm{M}_{\odot}$ , $r_{\mathrm{A}}$ and $r_{\mathrm{B}}$ only. [3 т.]

c. Approximating Mars’ orbit as circular with a radius of $1.52\ \mathrm{AU}$ , calculate the $\Delta v$ to go from Earth LEO to Mars i.e. $\Delta v = |\Delta v_{\mathrm{A}}| + |\Delta v_{\mathrm{B}}|$ . Compare your answer to the $\Delta v$ to reach Earth LEO. [4 т.]

d. Derive an expression for the total time spent on the transfer orbit, $t_{\mathrm{H}}$ , and calculate it for an Earth to Mars transfer. Give your answer in months. (Use 1 month = 30 days.) [3 т.]

e. 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? [3 т.]

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. [5 т.]

Отговори

Решение

Q2 - Hohmann Transfer [Total = 20]

a. Show that $v_{orb}$ in low Earth orbit (LEO; about 200 km above the surface) is about 8 km s⁻¹. This is an estimate of the Δv the rockets need to provide for the spacecraft to reach LEO.

$v_{\mathrm{orb}} = \sqrt{\frac{G\mathrm{M}_{\oplus}}{\mathrm{R}_{\oplus}+r}} = \sqrt{\frac{6.67\times10^{-11}\times5.97\times10^{24}}{6.37\times10^{6}+200\times10^{3}}}$ [1]

$= 7790\ \mathrm{m\,s^{-1}} \quad (= 7.79\ \mathrm{km\,s^{-1}} \approx 8\ \mathrm{km\,s^{-1}})$ [1] [2]

[In practice a Δv of 8 km s⁻¹ assumes no external forces, but atmospheric drag can increase the necessary Δv by 1.3 – 1.8 km s⁻¹. 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 $\Delta v_A$ and $\Delta v_B$ by comparing their circular orbital speeds with their transfer orbit speeds. Simplify your final expressions to include G, $\mathrm{M}_{\odot}$ , $r_A$ and $r_B$ only.

$\Delta v_{\mathrm{A}} = \sqrt{G\mathrm{M}_{\odot}\left(\frac{2}{r_{\mathrm{A}}}-\frac{1}{a}\right)} - \sqrt{G\mathrm{M}_{\odot}\left(\frac{1}{r_{\mathrm{A}}}\right)} = \sqrt{\frac{G\mathrm{M}_{\odot}}{r_{\mathrm{A}}}}\left(\sqrt{2-\frac{r_{\mathrm{A}}}{a}}-1\right)$ [1]

But $2a = r_{\mathrm{A}} + r_{\mathrm{B}}$

$\therefore \Delta v_{\mathrm{A}} = \sqrt{\frac{G\mathrm{M}_{\odot}}{r_{\mathrm{A}}}}\left(\sqrt{2-\frac{2r_{\mathrm{A}}}{r_{\mathrm{A}}+r_{\mathrm{B}}}}-1\right) = \sqrt{\frac{G\mathrm{M}_{\odot}}{r_{\mathrm{A}}}}\left(\sqrt{\frac{2r_{\mathrm{B}}}{r_{\mathrm{A}}+r_{\mathrm{B}}}}-1\right)$ [1] [2]

Similarly:

$\Delta v_{\mathrm{B}} = \sqrt{\frac{G\mathrm{M}_{\odot}}{r_{\mathrm{B}}}}\left(1-\sqrt{2-\frac{2r_{\mathrm{B}}}{r_{\mathrm{A}}+r_{\mathrm{B}}}}\right) = \sqrt{\frac{G\mathrm{M}_{\odot}}{r_{\mathrm{B}}}}\left(1-\sqrt{\frac{2r_{\mathrm{A}}}{r_{\mathrm{A}}+r_{\mathrm{B}}}}\right)$ [1] [1]

[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. $\Delta v = |\Delta v_A| + |\Delta v_B|$ . Compare your answer to the Δv to reach Earth LEO.

$\Delta v_{\mathrm{A}} = \sqrt{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{1.49\times10^{11}}}\left(\sqrt{\frac{2\times1.52}{1+1.52}}-1\right)$

$= 2935\ \mathrm{m\,s^{-1}}\ (= 2.94\ \mathrm{km\,s^{-1}})$ [1]

$\Delta v_{\mathrm{B}} = \sqrt{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{1.52\times1.49\times10^{11}}}\left(1-\sqrt{\frac{2\times1}{1+1.52}}\right)$

$= 2642\ \mathrm{m\,s^{-1}}\ (= 2.64\ \mathrm{km\,s^{-1}})$ [1]

$\Delta v = 2.94 + 2.64$

$= 5.58\ \mathrm{km\,s^{-1}}$ [1] [3]

This is less than the Δv to get into LEO [1] [1]

(So most of the effort needed in going to Mars simply comes from leaving Earth)

[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 surface increases the total for the whole transfer to roughly 8 km s⁻¹ (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, $t_H$ , and calculate it for an Earth to Mars transfer. Give your answer in months. (Use 1 month = 30 days).

From Kepler's third law: $T^2 = \frac{4\pi^2}{GM}a^3$

Since the spacecraft only covers half of the ellipse the time on the journey is half the period, and given that $2a = r_{\mathrm{A}} + r_{\mathrm{B}}$ then:

$t_{\mathrm{H}} = \pi\sqrt{\frac{(r_{\mathrm{A}}+r_{\mathrm{B}})^3}{8G\mathrm{M}_{\odot}}}$ [1] [1]

$= \pi\sqrt{\frac{\left((1+1.52)\times1.49\times10^{11}\right)^3}{8\times6.67\times10^{-11}\times1.99\times10^{30}}} = 2.22\times10^{7}\ \mathrm{s}$ [1]

$= 8.56\ \mathrm{months}$ [1] [2]

e. 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 now at A'.

Hohmann transfer diagram (Initial orbit, Final orbit, Transfer ellipse, ΔVA at A, ΔVB at B, radii rA and rB) with the Earth’s later position A' marked in red on the inner orbit, the angle θ between A' and the vertical line to B, and a red dotted line x from A' to B.

Angle between A' and B:

$\theta = \left(\frac{8.56-6}{12}\right)\times360^{\circ} = 76.7^{\circ}$ [1]

Using the cosine rule:

$x^2 = r_A^2 + r_B^2 - 2r_Ar_B\cos\theta$

$\therefore x = 2.41\times10^{11}\ \mathrm{m} \quad (= 1.62\ \mathrm{AU})$ [1] [2]

Since radio waves travel at the speed of light the time taken by the message is:

$t = \frac{2.41\times10^{11}}{3.00\times10^{8}} = 803\ \mathrm{s} \quad (= 13.4\ \mathrm{minutes})$ [1] [1]

[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.

During the transfer the Earth moves by $180^{\circ} + \theta = 257^{\circ}$ , so the spacecraft should launch from Mars when Earth is at position A" (exhibiting symmetry with when it arrived).

The same Hohmann transfer diagram with the Earth’s launch position A' marked in red on the inner orbit at angle θ to the right of the vertical line through B.

Since the planets move anti-clockwise in this diagram the angle covered by Earth from A' to A" is $360^{\circ} - 2\theta = 206.5^{\circ}$ [1]

From Kepler's third law, the period of Mars $= \sqrt{1.52^3} = 1.87\ \mathrm{years}$ [1]

Therefore, the relative angular velocity of Earth if Mars' motion is subtracted out is:

$\omega_{\mathrm{rel}} = \frac{360^{\circ}}{12} - \frac{360^{\circ}}{1.87\times12} = 14.0^{\circ}\ \mathrm{month^{-1}}$ [1]

(allow any equivalent units e.g. 168° year⁻¹, 9.38 × 10⁻⁸ rad s⁻¹ etc.)

Consequently, the time the astronauts need to wait for Earth to get from position A' to A" is:

$\frac{206}{14.0} = 14.8\ \mathrm{months} \quad (= 443\ \mathrm{days})$ [1] [4]

Thus the total duration of a return mission to Mars is:

$t = 14.8 + (2\times8.56) = 31.9\ \mathrm{months} \quad (= 956\ \mathrm{days} = 2.66\ \mathrm{years})$ [1] [1]

[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]

Questions proposed by

END OF PAPER

Questions proposed by:

Dr Alex Calverley (Bedford School)

Dr Emile Doran (The Langley Academy)

Sandor Kruk (University of Oxford)

Оригинал в Архива: BAAO_2017.pdf · официални решения: BAAO_2017_solutions.pdf