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GeCAA 2020, Теория — Задача 3. Mirror

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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 bored cosmologist comes up with a thought experiment to determine the Hubble constant ( $H_0$ ) for his model of a Steady-State-Universe. In this experiment, a large, fully reflecting flat mirror — carrying several gyroscopes that would maintain its spatial orientation in the same plane — would be placed at a distance $D$ from the Solar System in a region without gravitational influences. From the Earth, a laser beam would be directed towards that region for a long period of time. After a time $T$ , the radiation would return and be detected, allowing the determination of the fixed constant $H_0$ .

a) Find an expression for $H_0$ as a function of $D$ , $c$ (speed of light) and $T$ . Consider that the separation $S$ between the Solar System and the mirror increases only due to the expansion of the universe according to the law $S=s\cdot e^{H_0t}$ , where $s$ is the initial separation. You may use $e^x\approx1+x$ for $x\ll1$ , if necessary and assume that light travels a distance $ct$ in time $t$ in this expanding universe. [7 т.]

b) Imagine that such a mirror is located in the vicinity of the star Vega (which also features on the logo of the 1st GeCAA). Vega was the first star outside the Solar System to be photographed and one of the first stars whose parallax ( $p=0.125''$ ) was accurately measured in 1840 by G. W. von Struve. Estimate the total duration of this $H_0$ measurement experiment. Give the answer in years. [3 т.]

Отговори

Решение

(a) Let $t_1$ be the time taken by the light beam from the Solar System to the mirror, let $t_2$ be the time taken by the beam from the mirror to the Solar System and $T$ the total time to go back and forth. As a first order approximation, we will take distance travelled by the photon in each part as an average of the initial and final distance. Therefore, equating the kinematics of the situation, we have:

S_1=\frac{D+De^{H_0t_1}}2=\frac{D(1+e^{H_0t_1})}2=ct_1,

S_2=\frac{S_1+S_1e^{H_0t_2}}2=\frac{S_1(1+e^{H_0t_2})}2=ct_2.\qquad[1.0]

=\frac D4(1+e^{H_0t_1})(1+e^{H_0t_2})\approx\frac D4(2+H_0t_1)(2+H_0t_2)\approx\frac D4[4+2H_0(t_1+t_2)],

S_2=D\left(1+\frac12H_0T\right).\qquad[1.5]

From the first equation, we also find:

S_1=ct_1=\frac{D(1+e^{H_0t_1})}2,

ct_1=D\left(1+\frac12H_0t_1\right),\qquad\therefore t_1=\frac D{c-\frac12DH_0},

S_1=ct_1=\frac{2Dc}{2c-DH_0}.\qquad[1.5]

Similarly, $S_2=2S_1c/(2c-S_1H_0)$ . [0.5] Joining the expressions found, we obtain:

S_2=D\left(1+\frac12H_0T\right)=\frac{2S_1c}{2c-S_1H_0},

D\left(1+\frac12H_0T\right)=\frac{\frac{4Dc^2}{2c-DH_0}}{2c-\frac{2Dc}{2c-DH_0}\cdot H_0},

1+\frac12H_0T=\frac{2c}{2c-DH_0-DH_0}=\frac c{c-DH_0},

c=\left(1+\frac12H_0T\right)(c-DH_0)=c-DH_0+\frac12cH_0T-\frac12DH_0^2T,

0=\frac{H_0}2(cT-2D-DH_0T),\qquad H_0=\frac{cT-2D}{DT}.\qquad[2.5]

Alternative solution

First note that the time taken for the laser beam to travel to the mirror and back again is equal to the time taken for the laser beam to travel a distance of $2D$ (measured at $t=0$ ) in a straight line. [2.0] The co-moving distance the beam has to cover is the same in both scenarios. We then equate the distance travelled by the light with the amount by which the space has expanded in that time.

In time $t$ , the space has expanded by a factor of $\exp(H_0t)$ , to first order we can linearise this to $1+H_0t$ . This means that the beam, on average, travels through space that’s stretched out by a factor of $1+H_0T/2$ . Thus, [2.5]

cT=2D\left(1+\frac{H_0T}2\right),\qquad H_0=\frac{cT-2D}{DT}.\qquad[2.5]

Better approximation (not required in the exam)

In reality, the light does not travel a distance of $ct$ in this expanding universe. To find the exact expression, consider a time differential $dt$ time $t$ after the beam was emitted. In this small time interval, the beam travels a distance of $c\,dt$ . Because the space is stretched out, the travelled distance corresponds to a smaller segment of space at $t=0$ , smaller by a factor of $\exp(H_0t)$ . The distance spanned at $t=0$ is then

dr=\exp(-H_0t)c\,dt.

We integrate this from $t=0$ to $t=T$ :

\int_0^{2D}dr=c\int_0^T\exp(-H_0t)\,dt,\qquad2D=\frac c{H_0}(1-\exp(-H_0T)).

The result so far is accurate within the constraints of the model, but it’s not analytically solvable for $H_0$ . To get an estimate, we can approximate the right hand side to

2D=\frac c{H_0}\left(1-1+H_0T-\frac{(H_0T)^2}2\right)=cT-\frac{cH_0T^2}2.

Expressing $H_0$ , we get

H_0=\frac2{cT^2}(cT-2D).

The difference between this answer and the initial estimate is $2D/cT$ which is almost unitary.

(b) From the $H_0$ expression found in the previous item, we find the travel time

T=\frac{2D}{c-DH_0}.\qquad[1.0]

=\left(\frac c{2D}-\frac{H_0}2\right)^{-1}\approx\left(\frac c{2D}\right)^{-1}=\frac{2D}c.\qquad[1.0]

T=\frac{2\times8\times3.086\times10^{16}}{3\times10^8}=1.65\times10^9\ \mathrm s,\qquad T\approx52.2\ \mathrm{yr}.\qquad[1.0]

Source note: The primary bilingual question explicitly assumes light travels $ct$ in time $t$ ; this sentence is omitted from the English key’s reprinted question. The key itself labels the complete final derivation “Better approximation (not required in the exam)”. Both the requested model and this native supplementary approximation are retained.

Теми и трудност

Оценена трудност: 4/5.

Галактики и космология · Светлина и астрономически измервания · Разширение на Вселената · Космология · Определяне на разстояния

Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка

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