Съдържание

Fundamental constants Astronomical data Approximations Gauss’s formulae The Sun The Earth and Moon Условие Решение

IOAA 2023, Теория — Задача 12. DART

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Всички материали от IOAA 2023

IOAA 2023 — Theoretical competition · 40 т.

Указания

Do not touch envelopes until the start of the examination.
The theoretical examination lasts for 5 hours and is worth a total of 250 marks.
There are Answer Sheets for carrying out detailed work and Working Sheets for rough work, which are already marked with your student code and question number.
Use only the answer sheets for a particular question for your answer. Please write only on the printed side of the sheet. Do not use the reverse side. If you have written something on any sheet which you do not want to be evaluated, cross it out.
Use as many mathematical expressions as you think may help the evaluator to better understand your solutions. The evaluator may not understand your language. If it is necessary to explain something in words, please use short phrases (if possible in English).
You are not allowed to leave your work desk without permission. If you need any assistance (malfunctioning calculator, need to visit a restroom, etc.), please draw the attention of the supervisor.
The beginning and end of the examination will be indicated by the supervisor. The remaining time will be displayed on a clock.
At the end of the examination you must stop writing immediately. Put everything back in the envelope and leave it on the table.
Once all envelopes are collected, your student guide will escort you out of the examination room.
A list of constants and useful relations are included in the envelope.

Table of Constants

Fundamental constants

Constant	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	$e$	$1.602\times10^{-19}\,\mathrm C$
Universal gravitational constant	$G$	$6.674\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$
Universal electric constant	$\epsilon_0$	$8.854\times10^{-12}\,\mathrm{m^{-3}\,kg^{-1}\,s^4\,A^2}$
Universal 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	$b=\lambda_mT$	$2.898\times10^{-3}\,\mathrm{m\,K}$
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	$m_n$	$1.675\times10^{-27}\,\mathrm{kg}$
Mass of Helium nucleus	$m_{\mathrm{He}}$	$6.645\times10^{-27}\,\mathrm{kg}$
Atomic mass unit (a.m.u., Dalton)		$1.661\times10^{-27}\,\mathrm{kg}$

Astronomical data

Data	Symbol	Value
Hubble constant	$H_0$	$70\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$
North Ecliptic Pole ( $J2000.0$ )	$(\alpha_E,\delta_E)$	$(18\mathrm h\,00\mathrm m\,00\mathrm s,+66^\circ33'39'')$
North Galactic Pole ( $J2000.0$ )	$(\alpha_G,\delta_G)$	$(12\mathrm h\,51\mathrm m\,26\mathrm s,+27^\circ07'42'')$
1 jansky	$1\,\mathrm{Jy}$	$10^{-26}\,\mathrm{W\,m^{-2}\,Hz^{-1}}$
1 parsec	$1\,\mathrm{pc}$	$3.086\times10^{16}\,\mathrm m=206\,265\,\mathrm{au}=3.262\,\mathrm{ly}$
1 astronomical unit (au)	$1\,\mathrm{au}$	$1.496\times10^{11}\,\mathrm m$
1 sidereal day	$T_{\mathrm{SD}}$	$23.93444\,\mathrm h=23\mathrm h\,56\mathrm m\,04\mathrm s$
1 tropical year		$365.2422$ solar days
1 sidereal year		$365.2564$ solar days

Approximations and Gauss’s formulae

Approximations

(1+x)^n\approx1+nx

(1+x)(1+y)\approx1+x+y\quad\text{if }x\ll1\text{ and }y\ll1

Gauss’s formulae

Spherical law of cosines: $\cos a=\cos b\cos c+\sin b\sin c\cos A$

Spherical law of sines: $\dfrac{\sin A}{\sin a}=\dfrac{\sin B}{\sin b}=\dfrac{\sin C}{\sin c}$ .

The Sun, Earth and Moon

The Sun

Data	Symbol	Value
Solar luminosity	$L_\odot$	$3.826\times10^{26}\,\mathrm W$
Apparent angular diameter of Sun	$\theta_\odot$	$32'$
Effective temperature of Sun	$T_{\mathrm{eff},\odot}$	$5778\,\mathrm K$
Apparent visual magnitude		$-26.75$
Absolute visual magnitude		$+4.82$
Apparent bolometric magnitude		$-26.83$
Absolute bolometric magnitude		$+4.74$
Distance of the Sun from the Galactic centre		$\approx8\,\mathrm{kpc}$

The Earth and Moon

Data	Symbol	Value
Obliquity of the ecliptic (Earth)	$\epsilon$	$23.5'$
Platonic year (period of precession of Earth’s axis)		$25\,765$ years
Apparent visual magnitude of full Moon		$-12.74$
Apparent angular diameter of Moon	$\theta_L$	$31'$
Inclination of the lunar orbit to the ecliptic		$05^\circ08'43''$
Inclination of the lunar equator to its orbital plane		$6.687^\circ$
Lunar sidereal month	$T_{\mathrm{SL}}$	$27.321661\,\mathrm d=655.71986\,\mathrm h$
Synodic month		$29.530589\,\mathrm d$
Tropical month		$27.321582\,\mathrm d$
Anomalistic month (the average time between two successive Moon’s perigees)		$27.554550\,\mathrm d$
Draconic month (time between two successive transits of the Moon through the same node)		$27.212221\,\mathrm d$

The Solar System

Object	Mean radius [km]	Mass [kg]	Semimajor axis [au]	Eccentricity
Sun	$695\,700$	$1.988\times10^{30}$	—	—
Mercury	$2\,440$	$3.301\times10^{23}$	$0.387$	$0.206$
Venus	$6\,052$	$4.867\times10^{24}$	$0.723$	$0.007$
Earth	$6\,378$	$5.972\times10^{24}$	$1.000$	$0.016\,710$
Moon	$1\,737$	$7.346\times10^{22}$	$3.844\times10^5\,\mathrm{km}$	$0.054\,900$ (range $0.026$ – $0.077$ )
Mars	$3\,390$	$6.417\times10^{23}$	$1.524$	$0.093$
Jupiter	$69\,911$	$1.898\times10^{27}$	$5.203$	$0.048$
Saturn	$58\,232$	$5.683\times10^{26}$	$9.537$	$0.054$
Uranus	$25\,362$	$8.681\times10^{25}$	$19.189$	$0.047$
Neptune	$24\,622$	$1.024\times10^{26}$	$30.070$	$0.009$

General Marking Scheme

Case	Marking
Using incorrect physical concept (despite correct answers)	No points given
Giving correct answer without detailed calculation	Deduct 50% of the marks for that part
Minor mistakes in the calculations, e.g., wrong signs, symbols, substitutions	Deduct 20% of the marks for that part
Units missing from final answers	Deduct 0.5 pts
Too few or too many significant figures in the final answer	Deduct 0.5 pts
Error resulting from another error in an earlier part for which the student already lost marks, if the answer is physically reasonable.	Full points (i.e., no deductions)
Error resulting from another error in an earlier part, where the student should have realised the answer was physically unreasonable.	Deduct 20% of the marks for that part

For example, if due to an error in an earlier part, the student calculates the mass of a star as $2.5\times10^{30}\,\mathrm{kg}$ instead of $2\times10^{30}\,\mathrm{kg}$ , they will only lose marks for the earlier part. However, if, for the same reason, a student calculates the mass as $2\times10^{25}\,\mathrm{kg}$ , they should realize this is wrong (a few times the Earth’s mass) and thus should lose some marks for this part as well.

Условие

The Double Asteroid Redirection Test (DART) was a NASA mission to evaluate a method of planetary defense against near-Earth objects. The spacecraft hit Dimorphos, a moon of the asteroid Didymos, to study how the impact affected its orbit.

a. Calculate the expected orbital period change (in minutes), assuming that the collision was head-on, central, and perfectly inelastic.

Assume that before the impact Dimorphos orbited Didymos on a circular orbit with a period of $P = 11.92\,\mathrm{h}$ . The masses of Dimorphos and Didymos are $m = 4.3\times 10^9\,\mathrm{kg}$ and $M = 5.6\times 10^{11}\,\mathrm{kg}$ , respectively. The mass and speed of the DART spacecraft relative to Dimorphos at a moment of impact were $m_{\mathrm{s}} = 580\,\mathrm{kg}$ and $v_{\mathrm{s}} = 6.1\,\mathrm{km\,s^{-1}}$ . Neglect the gravitational influence of other bodies. (20 points)

b. In reality, the orbital period of Dimorphos was observed to be changed by $\Delta P_0 = -33\,\mathrm{min}$ . This is due to the momentum transfer associated with the recoil of the ejected debris: the spacecraft was absorbed by the asteroid, but the impact excavated some material from the asteroid and ejected it into space. Calculate the momentum of the ejected debris and express it as a fraction of the momentum of Dimorphos before the collision. You can assume that the mass of the ejected material is much smaller than the mass of Dimorphos. (15 points)

c. Calculate the velocity change (in $\mathrm{mm\,s^{-1}}$ ) of Dimorphos as a result of the impact, taking into account the effect of the ejected debris. (5 points)

Решение

Part (a)

Didymos’ mass is much larger than Dimorphos’ mass. Therefore, the radius of the orbit of Dimorphos before impact, $a$ , can be calculated from Kepler’s 3rd law:

\frac{GM}{4\pi^2}=\frac{a^3}{P^2},

and therefore:

a=\left(\frac{GMP^2}{4\pi^2}\right)^{1/3}=1.2\,\mathrm{km}.

The orbital velocity of Dimorphos before impact was:

v_0=\frac{2\pi a}{P}=0.176\,\mathrm{m/s}.

(2 points)

Let $v'$ be the Dimorphos velocity right after the collision. Using the law of conservation of momentum, we have:

mv_0-m_s v_s=(m+m_s)v',

and:

v'=\frac{mv_0-m_s v_s}{m+m_s}\approx v_0-\frac{m_s}{m}v_s,

where we used the fact that the mass of the spacecraft mass is much smaller than the mass of Dimorphos.

We then use the vis-viva equation to calculate the semi-major axis of the orbit after collision $a'$ :

(v')^2=GM\left(\frac2a-\frac1{a'}\right)=\frac{GM}{a}\left(2-\frac a{a'}\right)=v_0^2\left(2-\frac a{a'}\right),

2-\frac a{a'}=\left(\frac{v'}{v_0}\right)^2=\left(1-\frac{m_s}{m}\frac{v_s}{v_0}\right)^2\approx1-\frac{2m_s}{m}\frac{v_s}{v_0}.

Thus, the semi-major axis changed by:

\frac{\Delta a}{a}=\frac{a'-a}{a}=-\frac{2m_s}{m}\frac{v_s}{v_0}.

(8 points)

If the semi-major axis changes from $a$ to $a+\Delta a$ , then the orbital period changes from $P$ to $P+\Delta P$ , and the mass of the spacecraft can be neglected. Then:

\frac{a^3}{P^2}=\frac{(a+\Delta a)^3}{(P+\Delta P)^2}=\frac{a^3(1+\Delta a/a)^3}{P^2(1+\Delta P/P)^2}=\frac{a^3}{P^2}\left(1+\frac{3\Delta a}{a}-\frac{2\Delta P}{P}\right),

hence:

\frac{\Delta P}{P}=\frac32\frac{\Delta a}{a}.

Thus, the orbital period changes by:

\frac{\Delta P}{P}=\frac32\frac{\Delta a}{a}=-\frac{3m_s}{m}\frac{v_s}{v_0}

(8 points)

We therefore expect that the orbital period of Dimorphos should decrease by 1.4%, that is, 10 minutes. (2 points)

Alternative solution (requires better numerical precision)

\begin{aligned} v_0&=0.17622\,\mathrm{m/s}\\ v'&=0.17662-0.00082=0.17540\,\mathrm{m/s}\\ a'&=0.99080a=1192.5\,\mathrm{m}\\ P'&=705.4\,\mathrm{min} \end{aligned}

Hence $\Delta P=705.4-715.2=-9.8\approx-10\,\mathrm{min}$ . I propose to grant full points ONLY if the first four significant figures match the solution. If the first three significant figures match the solution, grant 80% of the points. Otherwise (if the method is correct), grant HALF of the points.

Part (b)

Let $\Delta p$ be the momentum of the ejected debris. Then, the momentum conservation equation becomes:

mv_0-m_s v_s=(m+m_s)v'+\Delta p,

(4 points)

so:

v'=\frac{mv_0-m_s v_s-\Delta p}{m+m_s}\approx v_0-\frac{m_s}{m}v_s-\frac{\Delta p}{m}.

Using similar calculations as in point a), we get:

\frac{\Delta a}{a}=-2\left(\frac{m_s}{m}\frac{v_s}{v_0}+\frac{\Delta p}{mv_0}\right),

(8 points)

hence:

\frac{\Delta P_0}{P}=\frac32\frac{\Delta a}{a}=-3\left(\frac{m_s}{m}\frac{v_s}{v_0}+\frac{\Delta p}{mv_0}\right).

Thus:

\frac{\Delta p}{mv_0}=-\frac{\Delta P_0}{3P}-\frac{m_s}{m}\frac{v_s}{v_0}=0.011.

(3 points)

Alternative solution (requires better numerical precision)

The orbital period after the collision is $P'=11.92-33/60=11.37\,\mathrm{h}$ . Thus, the semi-major axis of the orbit (after the collision) is

a'=\left(\frac{GMP'^2}{4\pi^2}\right)^{1/3}=1166\,\mathrm{m}.

The velocity of Dimorphos right after the collision is $v'=v_0\sqrt{2-a/a'}=0.1734\,\mathrm{m\,s^{-1}}$ , so $\Delta p/mv_0=0.011$ . I propose to grant full points ONLY if the first four significant figures match the solution. If the first three significant figures match the solution, grant 80% of the points. Otherwise (if the method is correct), grant HALF of the points.

Part (c)

\Delta v=v'-v_0=-\frac{m_s}{m}v_s-\frac{\Delta p}{m}=-\frac{m_s}{m}v_s+\frac{\Delta P_0}{3P}v_0+\frac{m_s}{m}v_s=\frac{\Delta P_0}{3P}v_0=-2.7\,\mathrm{mm/s}.

(5 points)

Original PDF comments

The original solution PDF contains two comments attributed to IOAA on this page:

“In previuos erratic version it was "... - 1".”

“In previous erratic version it was L_2/L_1 = ...”

Бележки за изданието

The English question packet includes a Ukrainian answer form for problem 9. The constants table prints Earth’s obliquity as 23.5′ and the Moon’s semimajor axis in km under an au heading. In the eclipse key, the magnitude addition and luminosity-ratio labels differ from its derivation and results; the original PDF correction comments are retained. Aldebaran’s key prints 4412 km/h and then uses 4416 in the angular-speed numerator. The X-ray question and key allocate points to parts (c) and (d) differently. DART’s alternative calculation prints both 0.17622 and 0.17662 m/s. The LISA key labels the component-mass calculation Part (b), repeats one algebra line and prints the period factor with a positive square. These printed discrepancies remain literal.

Оригинал в Архива: theory.pdf · официални решения: th_solutions.pdf