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IOAA 2016, Теория, английски — Задача 9. Mars Orbiter Mission

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Всички материали от IOAA 2016

Theoretical Examination · 20 т.

Внимание: Бележка към темата

Both archived English question editions print T12 AstroSat, but both 27-page English worked keys substitute T12 Exoplanets. No matching AstroSat solution was found in the inspected full year inventory. Summary Answer Sheets and Cover Sheet named by the instructions were not found in this archived source pack. Instructions specify 13 questions totalling 300 points; both question editions contain 12 questions totalling 250 points. Printed numerical/formula discrepancies are retained: T8 uses c/lambda_max for nu_max and prints 1233 pi then 3813; T9 numerical velocity denominator lacks a power of ten and the stated semimajor axis is 20933 km; T11.4 prints a direct rather than inverse mass ratio in the radius-cubed relation; T11.7 prose says 166.67 Hz while its mass calculation uses 142.86 Hz. No inferred corrections are substituted.

Instructions to the Contestants

(1) Total time duration of this examination is five hours.

(2) This examination consists of

Questions	Marks each	Total
Question 1 to 5	10	10 x 5 = 50 marks
Question 6 to 10	20	20 x 5 = 100 marks
Question 11 to 13	50	50 x 3 = 150 marks
	Total	300 marks

(3) Inside the envelope you will find

a. Question paper in English

b. Question paper in your native language (if applicable)

c. Table of Constants in English

d. Table of Constants in your native language (if applicable)

e. A set of Summary Answersheets

f. Cover Sheet

(4) Use only black or blue pen for writing. For figures, you may use pencils.

(5) There is no negative marking.

(6) Some marks will be deducted if the final answer is given without detailed solution.

(7) Some marks will be deducted if the final answers have inappropriate number of significant digits, no units or wrong units.

(8) The necessary values of fundamental and astronomical constants should be taken from the Table of Constants provided to you.

(9) Instructions regarding using the answersheets

a. For each question a separate Summary Answersheet has been provided. Final answer(s) for each question / each part of the question must be written in the corresponding box in the Summary Answersheet. Write your contestant code and page number on each Summary Answersheet.

b. You should also show detailed solution in the blank space provided on each Summary Answersheet. If necessary, you can ask for extra blank sheets from the invigilators.

c. Use separate blank sheets for each question.

d. Write your contestant code, question number and page number on FRONT side of each blank sheet. Write your contestant code and page number on the BACK side of each blank sheet.

e. The page numbers should be continuous, i.e., if you use 20 sheets for the entire examination (including the Summary Answersheets), the page numbers should run from 1 to 40.

f. Write only inside the boxed area.

g. For the work that you do not want to be evaluated, cross that part out.

(10) At the end of the examination

a. On the Cover Sheet, clearly write page numbers for each question.

b. Check that you have written your contestant code on all pages.

c. Put the Cover Sheet, all Summary Answersheets, blank sheets and rough sheets inside the envelope. You may keep question paper and Table of Constants with you.

Table of Constants

Fundamental Constants

Name	Symbol	Value
Speed of light in vacuum	$c$	$2.998\times10^8\ \mathrm{m\,s^{-1}}$
Planck constant	$h$	$6.626\times10^{-34}\ \mathrm{J\,s}$
Boltzmann constant	$k_B$	$1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma$	$5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Elementary charge	$e$	$1.602\times10^{-19}\ \mathrm{C}$
Gravitational constant	$G$	$6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Gas constant	$R$	$8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Avogadro constant	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Wien's displacement constant	$\lambda_mT$	$2.898\times10^{-3}\ \mathrm{m\,K}$
Mass of electron	$m_e$	$9.109\times10^{-31}\ \mathrm{kg}$
Mass of proton	$m_p$	$1.673\times10^{-27}\ \mathrm{kg}$
Mass of neutron	$m_n$	$1.675\times10^{-27}\ \mathrm{kg}$
Atomic mass unit	$u$	$1.661\times10^{-27}\ \mathrm{kg}$

Astronomical Constants

Name	Symbol	Value
Parsec	pc	$3.086\times10^{16}\ \mathrm{m}$
Astronomical unit	$a_\oplus$	$1.496\times10^{11}\ \mathrm{m}$
Mass of the Sun	$M_\odot$	$1.989\times10^{30}\ \mathrm{kg}$
Radius of the Sun	$R_\odot$	$6.955\times10^8\ \mathrm{m}$
Luminosity of the Sun	$L_\odot$	$3.826\times10^{26}\ \mathrm{W}$
Apparent magnitude of the Sun at midday		$-26.72\ \mathrm{mag}$
Solar constant		$1366\ \mathrm{W\,m^{-2}}$
Angular diameter of the Sun		$30'$
Mass of the Earth	$M_\oplus$	$5.972\times10^{24}\ \mathrm{kg}$
Radius of the Earth	$R_\oplus$	$6.371\times10^6\ \mathrm{m}$
Tropical year		$365.242$ solar days $=3.156\times10^7\ \mathrm{s}$

Условие

India's Mars Orbiter Mission (MOM) was launched using the Polar Satellite Launch Vehicle (PSLV) on 5 November 2013. The dry mass of MOM (body + instruments) was 500 kg and it carried fuel of mass 852 kg. It was initially placed in an elliptical orbit around the Earth with perigee at a height of 264.1 km and apogee at a height of 23903.6 km, above the surface of the Earth. After raising the orbit six times, MOM was transferred to a trans-Mars injection orbit (Hohmann orbit).

The first such orbit-raising was performed by firing the engines for a very short time near the perigee. The engines were fired to change the orbit without changing the plane of the orbit and without changing its perigee. This gave a net impulse of $1.73 \times 10^{5}\ \mathrm{kg\ m\ s^{-1}}$ to the satellite. Ignore the change in mass due to burning of fuel.

T9.1 What is the height of the new apogee, $h_{\mathrm{a}}$ above the surface of the Earth, after this engine burn? [14 т.]

T9.2 Find the eccentricity ( $e$ ) of the new orbit after the burn and the new orbital period ( $P$ ) of MOM in hours. [6 т.]

Решение

T9.1

What is the height of the new apogee, $h_{\mathrm{a}}$ , above the surface of the Earth, after this engine burn?
Solution:

Let apogee and perigee distances be $r_{\mathrm{a}}$ and $r_{\mathrm{p}}$ respectively.

r_{\mathrm{p}} = R_{\oplus} + h_{\mathrm{p}}^{i} = (6371 + 264)\ \mathrm{km} = 6635\ \mathrm{km}

[1.0]

r_{\mathrm{a}} = R_{\oplus} + h_{\mathrm{a}}^{i} = (6371 + 23904)\ \mathrm{km} = 30275\ \mathrm{km}

[1.0]

Conservation of energy and angular momentum gives

E = -\frac{GmM_{\oplus}}{r_{\mathrm{p}} + r_{\mathrm{a}}}

Total energy at perigee

\frac{1}{2}m{v_{\mathrm{p}}}^2 - \frac{GmM_{\oplus}}{r_{\mathrm{p}}} = E = -\frac{GmM_{\oplus}}{r_{\mathrm{p}} + r_{\mathrm{a}}}

\frac{1}{2}{v_{\mathrm{p}}}^2 = \frac{GM_{\oplus}}{r_{\mathrm{p}}}\left(1 - \frac{r_{\mathrm{p}}}{r_{\mathrm{p}} + r_{\mathrm{a}}}\right)

\therefore\ v_{\mathrm{p}} = \sqrt{2\frac{GM_{\oplus}}{r_{\mathrm{a}} + r_{\mathrm{p}}}\frac{r_{\mathrm{a}}}{r_{\mathrm{p}}}}

[2.0]

= \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times 3.0275 \times 10^7}{6.635 \times 10^6 \times (3.0275 + 6.635 \times 10^6)}}

= 9.929\ \mathrm{km\ s^{-1}}

[1.0]

As the engine burn is just 41.6 s, we assume that the entire impulse is applied instantaneously at perigee. The impulse is $J = 1.73 \times 10^5\ \mathrm{kg\ m\ s^{-1}}$ . Note that the total mass of MOM must include the fuel, so we have to use $m = 500 + 852 = 1352\ \mathrm{kg}$ . [1.0]

Change in velocity due to impulse at perigee is

\Delta v = \frac{J}{m} = \frac{1.73 \times 10^5}{1352} = 128.0\ \mathrm{m\ s^{-1}}

[1.0]

The new velocity will be given by (we use $'$ symbol to denote quantities after the first orbit-raising maneuvre)

v_{\mathrm{p}}' = v_{\mathrm{p}} + \Delta v = 10.06\ \mathrm{km\ s^{-1}}

[1.0]

The perigee remains unchanged. So we get $r_{\mathrm{p}}' = r_{\mathrm{p}}$ . [1.0]

Since the satellite is moving faster, the new apogee will be higher.

v_{\mathrm{p}}' = \sqrt{2GM_{\oplus}\frac{r_{\mathrm{a}}'}{r_{\mathrm{p}}'(r_{\mathrm{a}}' + r_{\mathrm{p}}')}}

\therefore\ 1 + \frac{r_{\mathrm{p}}'}{r_{\mathrm{a}}'} = \frac{2GM_{\oplus}}{(v_{\mathrm{p}}')^2 \times r_{\mathrm{p}}'}

[2.0]

= \frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{(10.06 \times 10^3)^2 \times 6.635 \times 10^6} = 1.188

r_{\mathrm{a}}' = \frac{6635}{0.188} = 35\,380\ \mathrm{km}

[2.0]

h_{\mathrm{a}} = 35380 - 6371 = \boxed{29\,009\ \mathrm{km}}

[1.0]

Acceptable range: ±150 km

T9.2

Find eccentricity ( $e$ ) of the new orbit after the burn and new orbital period ( $P$ ) of MOM in hours.
Solution:

As seen above,

\frac{r_{\mathrm{p}}'}{r_{\mathrm{a}}'} = 0.188 = \frac{1 - e}{1 + e}

[1.0]

\therefore\ e = \frac{1 - 0.188}{1 + 0.188} = \boxed{0.683}

[1.0]

Acceptable range: ±0.002

The new orbital semi-major axis and orbital period will be,

a' = \frac{r_{\mathrm{a}}' + r_{\mathrm{p}}'}{2}

[1.0]

= \frac{35380 + 6635}{2} = 20\,933\ \mathrm{km}

[1.0]

P = 2\pi\sqrt{\frac{a'^3}{GM_{\oplus}}}

[1.0]

= 2\pi\sqrt{\frac{(2.0933 \times 10^7)^3}{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}} = \boxed{30\,136\ \mathrm{s} = 8.37\ \mathrm{h}}

[1.0]

Acceptable range: ±0.1 h

Оригинал в Архива: theory-qp.pdf · официални решения: theory_sol.pdf · допълнителни данни: tableofconstants.pdf · допълнителни данни: tableofconstants.pdf · допълнителни данни: Instructions_theory.pdf · допълнителни данни: Instructions_theory.pdf · допълнителни данни: theory.pdf · допълнителни данни: theory.pdf