Съдържание

GeCAA 2020, Теория — Задача 1. Astrophotography

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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

An astrophotographer, based on the Equator, uses a good digital camera on a tripod, but with no tracking. The camera has a telescopic lens with focal length of 150 mm and aperture (objective diameter) of 84 mm. The sensor has an effective light collecting diameter of 22.5 mm. The photographic target is a star field at the observer’s Zenith.

a) Calculate the field of view (the angular width of the image) which can be captured on the sensor using this equipment. Give the answer in degrees. [2 т.]

b) The pixels in the camera’s sensor are separated by a distance of $3.23\ \mu\mathrm m$ . What is the maximum possible exposure time for a single frame, so that no star trails appear on the exposed image? Give the answer in seconds. [5 т.]

c) For a better-quality image of the star field, the photographer decides to take multiple shots at the exposure time calculated in b) and then to stack them together. The total time for all these shots is 30 seconds (ignore any time taken to write data to the memory card). What fraction of the total field of view is always visible throughout the shots? Give the answer as a number between 0 and 1. [3 т.]

Отговори

Решение

(a) By simple right angle triangle,

\mathrm{FOV}=2\times\tan^{-1}\left(\frac{\text{sensor width}}{2\times\text{Focal Length}}\right)=2\times\tan^{-1}\left(\frac{22.5}{2\times150}\right),

\alpha\approx8.58^\circ.\qquad[2.0]

(b) Number of pixels on the sensor will be given by,

N=\frac{\text{Sensor width}}{\text{pixel width}}=\frac{22.5}{3.23\times10^{-3}}=6965.94.\qquad[1.0]

As the number of pixels have to be integer, we take $N=6966$ . [1.0] Angular coverage of each pixel will be,

\text{Pixel view}=\frac{8.578^\circ}{6966}=0.00123^\circ/\text{pixel}\approx4.43''/\text{pixel}.\qquad[1.0]

As the stars complete one full circle in 23.9344 hours, [1.0]

Marking note: student loses this one mark if $15^\circ$ per hour is used.

t_1=1.23\times10^{-3}\frac{23.9344^{\mathrm h}}{360}=0.294\ \mathrm s\approx0.3\ \mathrm s.\qquad[1.0]

(in reality this is probably a factor of 10 smaller than the eye would detect)

(c) In 30 seconds, the sky will move by,

\Delta\alpha=\frac{360^\circ}{23.9344\times120}=0.125342^\circ.\qquad[1.0]

\therefore\frac{\alpha-\Delta\alpha}{\alpha}=\frac{8.578-0.125342}{8.578}=98.5\%.\qquad[2.0]

(or $8.453^\circ$ is total field of view in high resolution images from the stack)

Marking note: Only penalise $15^\circ/\mathrm{hr}$ once in the whole solution.

Source note: The English key’s reprinted part (c) asks “What proportion of the total field of view is possible at the higher signal to noise ratio?”, while the primary bilingual question asks the fraction always visible and specifies a number between 0 and 1. The key prints 98.5%. Its exposure-time line prints hours in the numerator and seconds in the result without an explicit hour-to-second conversion. Both printed forms and both marking notes are retained.

Теми и трудност

Оценена трудност: 2/5.

Уреди и наблюдения · Небесна сфера и сферична астрономия · Детектори и изображения · Телескопи · Денонощно движение и видимост

Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка

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