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GeCAA 2020, Теория — Задача 2. Flat Earth

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Всички материали от GeCAA 2020

GeCAA-2020 — Theory · 10 т.

Внимание: Бележка към темата

Eight complete native whole tasks and 27 requests, with all 25 official key pages, alternatives, marking criteria, and the complete two-page GeCAA-2020 constants companion. This bilingual Russian/English question edition is transcribed from its printed English column; task titles use the matching English key. Russian-only figure captions remain native. Printed scientific and source-edition discrepancies remain with attributed local notes. Task points total 150 by addition; no aggregate paper total or overall time limit is printed, so those fields remain unspecified. Referenced answer sheets/boxes are not in the obtained packet or complete 2020 archive-folder inventory.

Native source editions and gaps

Question masthead: GeCAA-2020 [Russian], Теоретический тур / Theory, with Russian and English columns. Official English key masthead: GeCAA - Theory Solutions, October 3, 2020. That is the printed key edition date; no competition held date/place is inferred. The designated answer sheets/boxes mentioned for Light Curves and Meteor Radiant were not obtained. The primary question/official key wording variants are recorded inside their original tasks.

Table of constants

The dated GeCAA-2020 constants companion prints Russian and English labels alongside the following native values.

Fundamental constants

Quantity	Symbol	Value
Speed of light in vacuum	$c$	$2.998\times10^8\ \mathrm{m/s}$
Planck constant	$h$	$6.626\times10^{-34}\ \mathrm{J\cdot s}$
Boltzmann constant	$k_B$	$1.381\times10^{-23}\ \mathrm{J/K}$
Stefan-Boltzmann constant	$\sigma$	$5.670\times10^{-8}\ \mathrm{W/(m^2\cdot K^4)}$
Elementary charge	$e$	$1.602\times10^{-19}\ \mathrm C$
Universal gravitational constant	$G$	$6.674\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
Universal gas constant	$R$	$8.315\ \mathrm{J/(mol\cdot K)}$
Avogadro constant	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Wien’s displacement law	$\lambda_{\max}T$	$2.898\times10^{-3}\ \mathrm{m\cdot K}$
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 (a. m. u.)		$1.661\times10^{-27}\ \mathrm{kg}$

Astronomical data

Quantity	Symbol	Value
1 parsec	1 pc	$3.086\times10^{16}\ \mathrm m=206265\ \mathrm{au}=3.262\ \mathrm{ly}$
1 astronomical unit (au)	1 au	$1.496\times10^{11}\ \mathrm m$
1 jansky	1 Jy	$10^{-26}\ \mathrm{W/(m^2\cdot Hz)}$
Hubble constant	$H_0$	$70\ \mathrm{(km/s)/Mpc}$
Solar luminosity	$L_\odot$	$3.826\times10^{26}\ \mathrm W$
Apparent angular diameter of Sun	$\theta_\odot$	$32'$
Effective temperature of Sun	$T_{\mathrm{eff}\,\odot}$	$5778\ \mathrm K$
Obliquity of the ecliptic (Earth)	$\epsilon$	$23.5^\circ$
Inclination of the lunar orbit w. r. t. ecliptic		$5^\circ08'43''$
Apparent visual magnitude of full moon		$-12.74^{\mathrm m}$
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 sidereal day		$23^{\mathrm h}56^{\mathrm m}04^{\mathrm s}$
1 tropical year		365.2422 solar days
1 sidereal year		365.2564 solar days

Solar magnitudes

Quantity	Value
Apparent visual magnitude	$-26.75^{\mathrm m}$
Absolute visual magnitude	$+4.82^{\mathrm m}$
Apparent bolometric magnitude	$-26.83^{\mathrm m}$
Absolute bolometric magnitude	$+4.74^{\mathrm m}$

Solar system

Object	Mean radius (km)	Mass (kg)	Semi-major axis (au)	Eccentricity
Sun	695500	$1.988\times10^{30}$	—	—
Mercury	2440	$3.301\times10^{23}$	0.387	0.206
Venus	6052	$4.867\times10^{24}$	0.723	0.007
Earth	6378	$5.972\times10^{24}$	1.000000	0.016710
Moon	1737	$7.346\times10^{22}$	0.002572	0.054900
Mars	3390	$6.417\times10^{23}$	1.524	0.093
Jupiter	69911	$1.898\times10^{27}$	5.203	0.048
Saturn	58232	$5.683\times10^{26}$	9.537	0.054
Uranus	25362	$8.681\times10^{25}$	19.189	0.047
Neptune	24622	$1.024\times10^{26}$	30.070	0.009

Gauss’s formulae

Cosine theorem:

\cos a=\cos b\cos c+\sin b\sin c\cos A.

Sine theorem:

\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C}.

Source note: The English native heading is “Gauss’s formulae”; its Russian counterpart is “Сферическая тригонометрия”. Blank source symbol cells remain blank. This is the dated same-competition constants PDF obtained from the GeCAA 2020 archive Constants folder.

Условие

A new model of the world is gaining in popularity among some people. These people believe in the “Flat Earth” view of the world, where the Earth is not a spheroid, but rather a circle with radius $R_\oplus$ . The central axis of the Earth (normal to the circle passing through its centre $C$ ) passes through the observer’s zenith.

This model must at least remain consistent with the observed phenomena, as listed below:

The value of the solar constant is $S_\oplus=1366\ \mathrm{W/m^2}$ .
The Earth’s central axis precesses in a circle with a period 25800 years.
The radius of the precession circle is $23.5^\circ$ .

Assume that (1) the Earth is a perfect blackbody radiator with uniform temperature and the Sun is sufficiently far away such that all solar rays are parallel and (2) the Sun’s current (initial) location is at the zenith.

Determine how many years it will take for the Earth’s equilibrium temperature to decrease by $1\ ^\circ\mathrm C$ .

Решение

Assume the surface area of one side of the flat Earth is $A$ . Let the angle between the Sun and the flat Earth’s center axis be $\theta$ , where $\theta$ is initially $0^\circ$ . As the Sun’s rays are parallel, the power delivered to the Earth by the Sun will be $S_\odot A\cos\theta$ at any given point in time.

At equilibrium, this is the energy radiated away via blackbody radiation, so the equilibrium temperature $T$ satisfies

S_\odot A\cos\theta=\sigma(2A)T^4.\qquad[2.0]

The factor of 2 comes from the fact that the flat Earth would radiate energy from both sides. This yields

T(\theta)=\sqrt[4]{\frac{S_\odot\cos\theta}{2\sigma}},

and we wish to find the value of $\theta_1$ such that $T(\theta_1)=T(0)-\Delta T$ . Thus,

\sqrt[4]{\frac{S_\odot\cos\theta_1}{2\sigma}}=\sqrt[4]{\frac{S_\odot}{2\sigma}}-\Delta T.\qquad[2.0]

\cos\theta_1=\left(1-\Delta T\sqrt[4]{\frac{2\sigma}{S_\odot}}\right)^4.\qquad[1.0]

=\left(1-\sqrt[4]{\frac{2\times5.67\times10^{-8}}{1366}}\right)^4,

\cos\theta_1=0.9880.\qquad[1.0]

Now, we find the time it takes for the axis to make such an angle with the Sun. On the celestial sphere, let $O$ be the center of precession, $Z$ be the current direction of the axis, and $X$ be direction of the axis when it makes an angle of $\theta_1$ with the sun, so $\angle ZCX=\theta_1$ . If $\epsilon$ is the radius of precession, then $\angle OCZ=\angle OCX=\epsilon$ . [1.0]

By the spherical Law of Cosines on angle $O$ of spherical triangle $OXZ$ , we have

\cos\theta_1=\cos\epsilon\cos\epsilon+\sin\epsilon\sin\epsilon\cos(\widehat O)=\cos^2\epsilon+\sin^2\epsilon\cos(\widehat O),

\widehat O=\cos^{-1}\left(\frac{\cos\theta_1-\cos^2\epsilon}{\sin^2\epsilon}\right).\qquad[1.0]

\therefore\Delta t=\frac{\widehat O}{2\pi}\times P=\frac{P}{2\pi}\times\cos^{-1}\left(\frac{\cos\theta_1-\cos^2\epsilon}{\sin^2\epsilon}\right).\qquad[1.0]

=\frac{25800}{2\pi}\times\cos^{-1}\left(\frac{0.9880-\cos^2 23.5^\circ}{\sin^2 23.5^\circ}\right)\approx1606\ \mathrm{yr}.

Thus, the average temperature of the earth will go down by $1\ ^\circ\mathrm C$ in just over 1600 years. [1.0]

Source note: The primary question denotes the solar constant $S_\oplus$ , while the English key’s reprint and derivation use $S_\odot$ . The value is 1366 W/m² in both editions.

Теми и трудност

Оценена трудност: 3/5.

Светлина и астрономически измервания · Небесна сфера и сферична астрономия · Абсолютно черно тяло · Прецесия и нутация

Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка

Оригинал в Архива: GeCAA-2020_Theory.pdf · официални решения: Theory Solutions.pdf · допълнителни данни: GeCAA-2020_Constants.pdf