Съдържание

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

IOAA 2023, Теория — Задача 1. Neptune

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Всички материали от IOAA 2023

IOAA 2023 — Theoretical competition · 5 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

Given that Neptune will be at opposition on 21 September 2024, calculate in which year Neptune was last at opposition near the time of the northern-hemisphere spring equinox. Assume that the orbits of Earth and Neptune are circular.

Решение

Using Kepler’s Third law and the semi-major axis ( $a = 30.070\,\mathrm{au}$ ) of the orbit from the table of constants, the sidereal period of Neptune’s orbit is:

P_{\mathrm{N}} = (a^3)^{\frac12} = \sqrt{27189.4} = 164.89\,\mathrm{years}

(1 point)

The rest of the calculation can be done in (at least) two ways:

(1) ‘day counting’/‘date drift’ solution:

Since Neptune is an outer planet (relative to Earth), the synodic period $S$ in years is given by:

\frac1S = \frac{1}{1\,\mathrm{year}}-\frac{1}{P_{\mathrm{N}}}

(1 point)

\begin{aligned} &=1-\frac{1}{164.89}=1-0.006065=0.99394\\ \Longrightarrow\quad S&=1/0.99394=1.006102\,\mathrm{years}\\ \Longrightarrow\quad 1.006102\times365.2422&=367.4708\,\mathrm{solar\ days} \end{aligned}

(1 point for $S$ )

Therefore the date of opposition drifts by:

D=367.4708-365.2422=2.2286\,\mathrm{days/year}

(1 point)

The approximate date of the northern spring equinox is 20 March. The number of days between 21 September and 20 March $=185$ days, therefore year of desired opposition is:

2024-(185/D)=2024-83=1941

(1 point)

If the student takes 21 March as the spring equinox, they get $184/D=82.5\,\mathrm{yr}\Longrightarrow1942$ .

(2) ‘angular drift’ solution (more accurate)

As before, the synodic period of Neptune can be derived as:

\begin{aligned} \frac1S&=\frac{1}{1\,\mathrm{year}}-\frac{1}{P_{\mathrm{N}}}\\ &=1-\frac{1}{164.89}=\frac{163.89}{164.89}\quad\Longrightarrow\quad S=\frac{164.89}{163.89}\,\mathrm{years} \end{aligned}

(1 point)

Therefore the ecliptic longitude of Neptune at opposition drifts by $(360^\circ/163.89)/\mathrm{year}$ . We want to find how many years it takes for the longitude of opposition to move by $180^\circ$ , i.e. for what $t$ :

t\times\frac{360^\circ}{163.89}=180^\circ.

(2 points)

\Longrightarrow\quad t=163.89/2=81.95\,\mathrm{years}\quad\Longrightarrow\quad2024-82=1942

(1 point)

We accept 1941 or 1942 for full points for calculations using the assumptions in the question. If the student uses some other method which is conceptually correct and results in 1943 (the true answer) they should also get full points.

Table of spring equinoxes and oppositions of Neptune:

Year	Equinox (UT)	Opposition (UT)	Coordinates of Neptune	$\Delta T$ [days]
1940	Mar 20 18:42	Mar 14 21:08	$11\mathrm{h}40\mathrm{m}\ +3^\circ28'$	5.9
1941	Mar 21 00:20	Mar 17 07:40	$11\mathrm{h}49\mathrm{m}\ +2^\circ39'$	3.7
1942	Mar 21 06:11	Mar 19 18:12	$11\mathrm{h}57\mathrm{m}\ +1^\circ50'$	1.5
1943	Mar 21 12:03	Mar 22 04:51	$12\mathrm{h}04\mathrm{m}\ +1^\circ00'$	−0.7
1944	Mar 20 17:49	Mar 23 15:29	$12\mathrm{h}13\mathrm{m}\ +0^\circ11'$	−2.9
2024	Mar 20 03:06	Sep 21 00:16	$23\mathrm{h}55\mathrm{m}\ +1^\circ56'$	−184.9

Taking into account all effects, the last opposition closest to the spring equinox was actually in 1943. 1942 results from the assumptions made in the question.

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