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IOAA 2016, Отборен тур — Задача 1. Group Examination

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Group Examination · 50 т.

Условие

(G1) A spacecraft of mass $m$ and velocity $\vec{v}$ approaches a massive planet of mass $M$ and orbital velocity $\vec{u}$ , as measured by an inertial observer. We consider a special case, where the incoming trajectory of the spacecraft is designed in a way such that velocity vector of the planet does not change direction due to the gravitational boost given to the spacecraft. In this case, the gravitational boost to the velocity the spacecraft can be estimated using conservation laws by measuring asymptotic velocity of the spacecraft before and after the interaction and angle of approach of the spacecraft.

Schematic of a flyby in which the planet moves with velocity u and the spacecraft with incoming velocity v parallel to u; the outgoing velocity vector points to the right, also parallel to u. — Figure 1

Schematic of a flyby with incoming velocity v making an angle θ with the direction opposite to the planet's velocity u; the outgoing asymptote points upward to the right. — Figure 2

Orbit diagram of Voyager 2 around the Sun showing encounters with Jupiter (9 Jul 79), Saturn (25 Aug 81), Uranus (24 Jan 86) and Neptune (25 Aug 89). — Figure 3

G1.1 What will be the final velocity $(\vec{v_f})$ of the spacecraft, if $\vec{v}$ and $\vec{u}$ are exactly anti-parallel (see Figure 1). [3 т.]

G1.2 Simplify the expression for the case where $m \ll M$ . [1 т.]

G1.3 If angle between $\vec{v}$ and $-\vec{u}$ is $\theta$ and $m \ll M$ (see Figure 2), use results above to write expression for the magnitude of final velocity $(v_f)$ . [3 т.]

G1.4 Table on the last page gives data of Voyager-2 spacecraft for a few months in the year 1979 as it passed close to Jupiter. Assume that the observer is located at the centre of the Sun. The distance from the observer is given in AU and $\lambda$ is heliocentric ecliptic longitude in degrees. Assume all objects to be in the ecliptic plane. Assume that the orbit of the Earth to be circular. Plot appropriate column against the date of observation to find the date at which the spacecraft was closest to the Jupiter, and label the graph as G1.4. [8 т.]

G1.5 Find the Earth-Jupiter distance, $(d_{\mathrm{E-J}})$ on the day of the encounter. [4 т.]

G1.6 On the day of the encounter, around what standard time $(t_{\mathrm{std}})$ had the Jupiter transited the meridian in the sky of Bhubaneswar (20.27° N; 85.84°E; UT + 05:30)? [6 т.]

G1.7 Speed of the spacecraft (in km s⁻¹) as measured by the same observer on some dates before the encounter and some dates after the encounter are given below. Here day n is the date of encounter. Use these data to find the orbital speed of Jupiter $(u)$ on the date of encounter and angle $\theta$ .

date	n-45	n-35	n-25	n-15	n-5	n
vtot	10.1408	10.0187	9.9078	9.8389	10.2516	25.5150
date	n+5	n+15	n+25	n+35	n+45
vtot	21.8636	21.7022	21.5580	21.3812	21.2365

[12 т.]

G1.8 Find eccentricity, $e_J$ , of Jupiter's orbit. [8 т.]

G1.9 Find heliocentric ecliptic longitude, $\lambda_p$ , of Jupiter's perihelion point. [5 т.]

Month	Date	$\lambda$ (°)	Distance (AU)
June	1	135.8870	5.1589731906
June	2	135.9339	5.1629499712
June	3	135.9806	5.1669246607
June	4	136.0272	5.1708975373
June	5	136.0736	5.1748689006
June	6	136.1200	5.1788390741
June	7	136.1662	5.1828084082
June	8	136.2122	5.1867772826
June	9	136.2582	5.1907461105
June	10	136.3040	5.1947153428
June	11	136.3496	5.1986854723
June	12	136.3951	5.2026570402
June	13	136.4405	5.2066306418
June	14	136.4857	5.2106069354
June	15	136.5307	5.2145866506
June	16	136.5756	5.2185705999
June	17	136.6202	5.2225596924
June	18	136.6647	5.2265549493
June	19	136.7090	5.2305575243
June	20	136.7532	5.2345687280
June	21	136.7970	5.2385900582
June	22	136.8407	5.2426232385
June	23	136.8841	5.2466702671
June	24	136.9273	5.2507334797
June	25	136.9702	5.2548156324
June	26	137.0127	5.2589200110
June	27	137.0550	5.2630505798
June	28	137.0969	5.2672121872
June	29	137.1384	5.2714108557
June	30	137.1795	5.2756542053
July	1	137.2200	5.2799520895
July	2	137.2600	5.2843175880
July	3	137.2993	5.2887686308
July	4	137.3378	5.2933308160
July	5	137.3754	5.2980426654
July	6	137.4118	5.3029664212
July	7	137.4467	5.3082133835
July	8	137.4798	5.3140161793
July	9	137.5116	5.3210070441
July	10	137.5628	5.3312091210
July	11	137.6898	5.3405592121
July	12	137.8266	5.3466522674
July	13	137.9599	5.3516661563
July	14	138.0903	5.3561848203
July	15	138.2186	5.3604205657
July	16	138.3453	5.3644742164

Month	Date	$\lambda$ (°)	Distance (AU)
July	17	138.4707	5.3684017790
July	18	138.5949	5.3722377051
July	19	138.7183	5.3760047603
July	20	138.8409	5.3797188059
July	21	138.9628	5.3833913528
July	22	139.0841	5.3870310297
July	23	139.2048	5.3906444770
July	24	139.3250	5.3942369174
July	25	139.4448	5.3978125344
July	26	139.5641	5.4013747321
July	27	139.6831	5.4049263181
July	28	139.8016	5.4084696349
July	29	139.9198	5.4120066575
July	30	140.0377	5.4155390662
July	31	140.1553	5.4190683021
August	1	140.2725	5.4225956100
August	2	140.3895	5.4261220723
August	3	140.5062	5.4296486357
August	4	140.6225	5.4331761326
August	5	140.7387	5.4367052982
August	6	140.8546	5.4402367851
August	7	140.9702	5.4437711745
August	8	141.0856	5.4473089863
August	9	141.2007	5.4508506867
August	10	141.3157	5.4543966955
August	11	141.4303	5.4579473912
August	12	141.5448	5.4615031166
August	13	141.6591	5.4650641822
August	14	141.7731	5.4686308707
August	15	141.8869	5.4722034391
August	16	142.0006	5.4757821220
August	17	142.1140	5.4793671340
August	18	142.2272	5.4829586711
August	19	142.3402	5.4865569133
August	20	142.4530	5.4901620256
August	21	142.5657	5.4937741595
August	22	142.6781	5.4973934544
August	23	142.7904	5.5010200385
August	24	142.9024	5.5046540300
August	25	143.0143	5.5082955377
August	26	143.1260	5.5119446617
August	27	143.2375	5.5156014948
August	28	143.3488	5.5192661222
August	29	143.4599	5.5229386226
August	30	143.5709	5.5266190687
August	31	143.6817	5.5303075275

Решение

Внимание: Непълно решение

The supplied key gives complete primary solutions for G1.1–G1.9, but prints only the heading “Alternative solution in COM frame” without that alternative derivation.

Anti-parallel case: planet moving right with velocity u, spacecraft approaching from below with velocity v, making a loop around the planet and leaving in the direction of u. — Figure 1

General case: planet moving right with velocity u, spacecraft approaches with velocity v at angle θ to the horizontal and leaves deflected upward. — Figure 2

Voyager 2 trajectory diagram with the printed planetary labels and encounter dates. — Figure 3

G1.1

Solution:

Let $\vec{v_f}$ and $\vec{u_f}$ be the final velocity of the spacecraft and the planet respectively. As the planet For anti-parallel case, using conservation of linear momentum,

M\vec{u} + m\vec{v} = M\vec{u_f} + m\vec{v_f}

\therefore\ Mu - mv = Mu_f + mv_f

u_f = u - \frac{m}{M}(v_f+v)

Now, using conservation of energy,

Mu^2 + mv^2 = Mu_f^2 + mv_f^2

u^2 + \frac{m}{M}v^2 = \left(u - \frac{m}{M}(v_f+v)\right)^2 + \frac{m}{M}v_f^2

\cancel{u^2} + \cancel{\frac{m}{M}}v^2 = \cancel{u^2} + \frac{m}{M}\cdot\cancel{\frac{m}{M}}(v_f+v)^2 - 2u\cancel{\frac{m}{M}}(v_f+v) + \cancel{\frac{m}{M}}v_f^2

0 = \frac{m}{M}(v_f+v)^2 - 2u(v_f+v) + (v_f^2 - v^2)

0 = \frac{m}{M}(v_f+v)\cancel{(v_f+v)} - 2u\cancel{(v_f+v)} + \cancel{(v_f+v)}(v_f-v)

0 = \frac{m}{M}(v_f+v) - 2u + v_f - v

\therefore\ v_f\left(1+\frac{m}{M}\right) = 2u + \left(1-\frac{m}{M}\right)v

v_f = \frac{2u + \left(1-\frac{m}{M}\right)v}{\left(1+\frac{m}{M}\right)}

Alternative solution in COM frame

G1.2

Solution:

If $m \ll M$ ,

v_f \approx 2u+v

G1.3

Solution:

As velocity vector of the planet is not changing direction, there is no momentum transfer in direction perpendicular to $\vec{u}$ . We will resolve $\vec{v}$ and $\vec{v_f}$ into components parallel and perpendicular to $\vec{u}$ .

v_x = -v\cos\theta \qquad\qquad v_y = v\sin\theta

v_{f_x} = 2u + v\cos\theta \qquad\quad v_{f_y} = v\sin\theta

v_f^2 = v_{f_x}^2 + v_{f_y}^2 = (2u + v\cos\theta)^2 + (v\sin\theta)^2

= 4u^2 + 4uv\cos\theta + v^2\cos^2\theta + v^2\sin^2\theta

= 4u^2 + 4uv\cos\theta + v^2

\therefore\ \boxed{v_f = \sqrt{4u^2 + v^2 + 4uv\cos\theta}}

G1.4

Solution:

Graph labelled G1.4: ecliptic longitude (135–145 degrees) plotted against date (Jun-01 to Aug-30 1979); curve rises slowly to about 137.5° by Jul-10, then steeply to about 143.6° by Aug-30.

From the graph, it can be inferred that the encounter with Jupiter occured on 10th July (day 191) and its distance from the Sun on that day is 5.33121 AU

G1.5

Solution:

The day number of Vernal Equinox is 80. Thus, ecliptic longitude of the Sun as seen from the Earth on the day of encounter will be,

\lambda_\odot = (191 - 80)*360^\circ/365.25 = 109.4045^\circ

Thus, the ecliptic longitude of the Earth as seen from the Sun on the day of encounter will be,

\lambda_\oplus = 180^\circ + 109.4045^\circ = 289.4045^\circ

Applying cosine rule,

d_{\oplus-J} = \sqrt{d_\oplus^2 + d_J^2 - 2d_\oplus d_J\cos\Delta\lambda}

= \sqrt{1^2 + 5.3312^2 - 2\times 1 \times 5.3312 \times \cos(289.4045^\circ - 137.5628^\circ)}

= 6.2308\ \mathrm{AU}

i.e. the Earth is 6.2308 AU from Jupiter on that day.

G1.6

Solution:

Thus, the angle of eastern elongation for Jupiter ( $\angle SEJ$ ) on that day would be,

\xi = \sin^{-1}\left(\frac{5.3312 \times \sin(289.4045^\circ - 137.5628^\circ)}{6.2308}\right)

= 23.8146^\circ

It would rise 95 minutes after the Sun rise, i.e. around 7:35am. It would transit the meridian after around 6 hours i.e. around 13:35 local time or 13:22 IST.

For more precise answer, R.A. of Jupiter on the day of encounter is approximately,

\lambda_{J_\mathrm{geocentric}} = 109.4045^\circ + 23.8146^\circ = 133.2191^\circ

\tan\alpha_J = \tan\lambda_{J_\mathrm{geocentric}}\,\cos\epsilon

= \tan 133.2191^\circ\,\cos 23^\circ 26'

\therefore\ \alpha_J = 135.68^\circ = 9^h\ 3^m

Thus, it will culminate at that sidereal time. On that day, sidereal time at noon is 07:24 (111 days from V.E. times 4 minutes). Thus, it will culminate 1 hour 39 minutes after the local noon i.e. at 13:39 local time or at about 13:26 IST.

Sun S at centre with Earth's orbit and Jupiter J on an outer orbit; arrow labelled 'To V.E.' upward from S; lines SE and SJ drawn. — Figure 4

Geometry of the encounter: Sun S at top right with angle δλ between directions to A and C; spacecraft path A–B–C near Jupiter J with angles φ1 at A, φ2 at C, and velocity direction JD labelled u.

G1.7

Solution:

In Figure 5, path of Voyager-2 is shown as A-B-C. The Sun is shown as S and the Jupiter is shown as J. From the data we note that $r$ is increasing continuously. The same should be reflected in the diagram. For practical purpose, J and B are the same points. The direction of velocity vector of Jupiter is given by $JD$ . In the figure,

\angle ASB = \delta\lambda_1 \qquad\qquad \angle ASB = \delta\lambda_2

\angle ASC = \delta\lambda \qquad\qquad \angle ABD = \theta

\angle ABC = \theta_1 \qquad\qquad \angle DBC = \angle ABC - \angle ABD = \theta_1 - \theta

\angle SAB = \phi_1 \qquad\qquad \angle SCB = \phi_2

Now the lines originating from the Sun indicate radial direction on the respective dates. Let us take speed of the spacecraft sufficiently far from the day 190, to avoid any influence of Jupiter in initial and final velocity estimation. We can choose dates 35 days on either side of July 10 i.e. June 5 and August 14.

\delta\lambda_1 = 137.5628^\circ - 136.0736^\circ = 1.4892^\circ

l(AB) = \sqrt{l(SA)^2 + l(SB)^2 - 2 \times l(SA) \times l(SB) \times \cos\delta\lambda_1}

= \sqrt{5.17487^2 + 5.33121^2 - 2 \times 5.17487 \times 5.33121 \times \cos 1.4892^\circ}

= 0.20755\ \mathrm{au}

\phi_1 = \sin^{-1}\left(\frac{l(SB)\sin\delta\lambda_1}{l(AB)}\right) = \sin^{-1}\left(\frac{5.33121 \times \sin 1.4892^\circ}{0.20755}\right)

= \sin^{-1}(0.66755)

\phi_1 = 41.8783^\circ \quad\text{or}\quad 138.1217^\circ

\delta\lambda_2 = 141.2007^\circ - 137.5628^\circ = 3.6379^\circ

l(BC) = \sqrt{l(SC)^2 + l(SB)^2 - 2 \times l(SC) \times l(SB) \times \cos\delta\lambda_2}

= \sqrt{5.45085^2 + 5.33121^2 - 2 \times 5.45085 \times 5.33121 \times \cos 3.6379^\circ}

= 0.36253\ \mathrm{au}

\phi_2 = \sin^{-1}\left(\frac{l(SB)\sin\delta\lambda_2}{l(BC)}\right) = \sin^{-1}\left(\frac{5.33121 \times \sin 3.6379^\circ}{0.36253}\right)

= \sin^{-1}(0.66755)

\phi_2 = 68.9199^\circ \quad\text{or}\quad 111.0801^\circ

from the figure, $\phi_1$ should be obtuse and $\phi_2$ may be acute. In $\square SABC$

\delta\lambda = \lambda_2 - \lambda_1 = 141.2007^\circ - 136.0736^\circ

= 5.1271^\circ

\therefore\ \theta_1 = 360^\circ - \delta\lambda - \phi_1 - \phi_2

= 360^\circ - 5.1271^\circ - 138.1217^\circ - 68.9199^\circ

\theta_1 = 147.8313^\circ

In $\triangle SBC$ , we notice

\angle SBC = 180^\circ - \phi_2 - \delta\lambda_2

= 180^\circ - 68.9199^\circ - 3.6379^\circ

= 107.4422^\circ

\tan\angle DBC = \frac{v_{y_f}}{v_{x_f}} = \frac{v\sin\theta}{v\cos\theta + 2u}

\tan(\theta_1 - \theta) = \frac{\sin\theta}{\cos\theta + 2\frac{u}{v}}

\therefore\ \frac{2u}{v} = \frac{\sin\theta}{\tan(\theta_1 - \theta)} - \cos\theta

We use this expression to find $|\vec{u}|$ .

v_f^2 = 4u^2 + v^2 + 4uv\cos\theta

\frac{v_f^2}{v^2} = \frac{4u^2}{v^2} + 1 + \frac{4u}{v}\cos\theta

\left(\frac{v_f}{v}\right)^2 = \left(\frac{2u}{v}\right)^2 + 1 + 2\left(\frac{2u}{v}\right)\cos\theta

= \left(\frac{\sin\theta}{\tan(\theta_1 - \theta)} - \cos\theta\right)^2 + 1 + 2\left(\frac{\sin\theta}{\tan(\theta_1 - \theta)} - \cos\theta\right)\cos\theta

= \left(\frac{\sin\theta}{\tan(\theta_1 - \theta)}\right)^2 - \cancel{\frac{2\sin\theta\cos\theta}{\tan(\theta_1 - \theta)}} + \cos^2\theta + 1 + \cancel{\frac{2\sin\theta\cos\theta}{\tan(\theta_1 - \theta)}} - 2\cos^2\theta

= \frac{\sin^2\theta}{\tan^2(\theta_1 - \theta)} + 1 - \cos^2\theta

= \frac{\sin^2\theta}{\tan^2(\theta_1 - \theta)} + \sin^2\theta = \sin^2\theta\left(\cot^2(\theta_1 - \theta) + 1\right)

= \frac{\sin^2\theta}{\sin^2(\theta_1 - \theta)}

\therefore\ \frac{v_f}{v} = \frac{\sin\theta}{\sin(\theta_1 - \theta)} = \frac{\sin\theta}{\sin\theta_1\cos\theta - \cos\theta_1\sin\theta}

\therefore\ \frac{v}{v_f} = \sin\theta_1\cot\theta - \cos\theta_1

\tan\theta = \frac{\sin\theta_1}{\frac{v}{v_f} + \cos\theta_1}

= \frac{\sin 147.8313^\circ}{\frac{10.0187}{21.3812} + \cos 147.8313^\circ} = -1.4088

\therefore\ \theta = 180^\circ - 54.6328^\circ = 125.3672^\circ

v_f^2 = 4u^2 + v^2 + 4uv\cos\theta

21.3812^2 = 4u^2 + 10.0187^2 + 4u \times 10.0187\cos 125.3672^\circ

0 = u^2 - 5.7990u - \frac{(457.1557 - 100.3743)}{4}

0 = u^2 - 5.7990u - 89.1953

\therefore\ u = \frac{5.7990 + \sqrt{5.7990^2 + 4 \times 89.1953}}{2}

= 12.7789\ \mathrm{km\ s^{-1}}

Jupiter's orbital velocity on the day of encounter is $12.779\ \mathrm{km\ s^{-1}}$ and the angle between the initial velocity of the spacecraft and Jupiter's velocity vectors is 125°22′.

G1.8

Solution:

The angle between $\vec{r}$ and $\vec{u}$ on the day of encounter will be,

\psi = \angle SBC - (\theta_1 - \theta)

= 107.4422^\circ - 147.8313^\circ + 125.3672^\circ

= 84.9781^\circ

Now we use angular momentum conservation to estimate eccentricity. If $u_p$ and $r_p$ represent perihelion velocity and perihelion distance of Jupiter,

r_pu_p = a_J(1-e)\sqrt{\frac{GM_\odot}{a_J}\left(\frac{1+e}{1-e}\right)}

= \sqrt{GM_\odot a_J (1-e^2)}

r_pu_p = ru\sin\psi

\therefore\ 1 - e^2 = \frac{r^2u^2\sin^2\psi}{GM_\odot a_J}

= \frac{5.33121^2 \times 1.496 \times 10^{11} \times (12.7789 \times 10^3)^2 \sin^2 84.9781^\circ}{6.6741 \times 10^{-11} \times 1.9891 \times 10^{30} \times 5.20260}

= 0.99761

\therefore\ e = \sqrt{1 - 0.99761} = 0.0489

The eccentricity of Jupiter's orbit is $0.0489$ .

G1.9

Solution:

To estimate longitude of perihelion, one should estimate true anomaly of Jupiter on that day.

r = \frac{a(1-e^2)}{1+e\cos\Theta}

\therefore\ 0.0489\cos\Theta = \frac{a(1-e^2)}{r} - 1 = \frac{5.20260 \times 0.99761}{5.33121} - 1

= -0.02646

\Theta = 122.754^\circ

Thus, the longitude of perihelion of Jupiter is,

\lambda_p = \lambda_J - \Theta

= 137.5628^\circ - 122.754^\circ

\lambda_p = 14.809^\circ

Data

Month	Date	λ (°)	Distance (AU)	Month	Date	λ (°)	Distance (AU)
June	1	135.8870	5.1589731906	July	17	138.4707	5.3684017790
June	2	135.9339	5.1629499712	July	18	138.5949	5.3722377051
June	3	135.9806	5.1669246607	July	19	138.7183	5.3760047603
June	4	136.0272	5.1708975373	July	20	138.8409	5.3797188059
June	5	136.0736	5.1748689006	July	21	138.9628	5.3833913528
June	6	136.1200	5.1788390741	July	22	139.0841	5.3870310297
June	7	136.1662	5.1828084082	July	23	139.2048	5.390644477
June	8	136.2122	5.1867772826	July	24	139.3250	5.3942369174
June	9	136.2582	5.1907461105	July	25	139.4448	5.3978125344
June	10	136.3040	5.1947153428	July	26	139.5641	5.4013747321
June	11	136.3496	5.1986854723	July	27	139.6831	5.4049263181
June	12	136.3951	5.2026570402	July	28	139.8016	5.4084696349
June	13	136.4405	5.2066306418	July	29	139.9198	5.4120066575
June	14	136.4857	5.2106069354	July	30	140.0377	5.4155390662
June	15	136.5307	5.2145866506	July	31	140.1553	5.4190683021
June	16	136.5756	5.2185705999	August	1	140.2725	5.4225956100
June	17	136.6202	5.2225596924	August	2	140.3895	5.4261220723
June	18	136.6647	5.2265549493	August	3	140.5062	5.4296486357
June	19	136.7090	5.2305575243	August	4	140.6225	5.4331761326
June	20	136.7532	5.2345687280	August	5	140.7387	5.4367052982
June	21	136.7970	5.2385900582	August	6	140.8546	5.4402367851
June	22	136.8407	5.2426232385	August	7	140.9702	5.4437711745
June	23	136.8841	5.2466702671	August	8	141.0856	5.4473089863
June	24	136.9273	5.2507334797	August	9	141.2007	5.4508506867
June	25	136.9702	5.2548156324	August	10	141.3157	5.4543966955
June	26	137.0127	5.2589200110	August	11	141.4303	5.4579473912
June	27	137.0550	5.2630505798	August	12	141.5448	5.4615031166
June	28	137.0969	5.2672121872	August	13	141.6591	5.4650641822
June	29	137.1384	5.2714108557	August	14	141.7731	5.4686308707
June	30	137.1795	5.2756542053	August	15	141.8869	5.4722034391
July	1	137.2200	5.2799520895	August	16	142.0006	5.4757821220
July	2	137.2600	5.2843175880	August	17	142.1140	5.4793671340
July	3	137.2993	5.2887686308	August	18	142.2272	5.4829586711
July	4	137.3378	5.2933308160	August	19	142.3402	5.4865569133
July	5	137.3754	5.2980426654	August	20	142.4530	5.4901620256
July	6	137.4118	5.3029664212	August	21	142.5657	5.4937741595
July	7	137.4467	5.3082133835	August	22	142.6781	5.4973934544
July	8	137.4798	5.3140161793	August	23	142.7904	5.5010200385
July	9	137.5116	5.3210070441	August	24	142.9024	5.5046540300
July	10	137.5628	5.3312091210	August	25	143.0143	5.5082955377
July	11	137.6898	5.3405592121	August	26	143.1260	5.5119446617
July	12	137.8266	5.3466522674	August	27	143.2375	5.5156014948
July	13	137.9599	5.3516661563	August	28	143.3488	5.5192661222
July	14	138.0903	5.3561848203	August	29	143.4599	5.5229386226
July	15	138.2186	5.3604205657	August	30	143.5709	5.5266190687
July	16	138.3453	5.3644742164	August	31	143.6817	5.5303075275

Оригинал в Архива: ioaa-qp-team.pdf · официални решения: team_sol.pdf · допълнителни данни: team.pdf