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IAO 2015, Теория, Група α — Задача 2. Eclipse on the Poles

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Всички материали от IAO 2015

XX International Astronomy Olympiad. Theoretical round. Group α · 15 октомври 2015 г.

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General note. Maybe not all problems have correct questions. Some questions (maybe the main question of the problem, maybe one of the subquestions) may make no real sense. In this case you have to write in your answer (in English or Russian): «impossible situation – ситуация невозможна». Of course, this answer has to be explained numerically or logically.

Data from the tables (Planetary data, stars, constants, etc.) may be used for solving every problem.

The answers «Да-Yes» or «Нет-No» have to be written in English or Russian.

Ещё раз прочитайте перед началом работы.

Общее замечание. Не исключено, что не во всех задачах вопросы поставлены корректно. Некоторые вопросы (возможно, главный вопрос задачи, возможно – подвопрос) могут не иметь смысла. В этом случае следует написать в ответе (по-русски или по-английски): «ситуация невозможна – impossible situation». Естественно, ответ должен быть подкреплён вычислениями или логическими рассуждениями.

Данные из таблиц (Солнечная система, звёзды, константы) могут быть использованы в любой задаче.

Ответы «Да-Yes» или «Нет-No» должны быть написаны по-русски или по-английски.

Условие

The White Bear and the Penguin from the previous International Astronomy Olympiads returned to their poles (North and South respectively), and decided to observe an annular solar eclipse. The Penguin was lucky to see an amazing picture: at the maximum phase of the eclipse the centres of both discs, solar and lunar, appeared just on the visible horizon. And what did the Bear observe at this time? Draw what the White Bear saw at that moment, and also contour by dotted line the true positions of the Sun and the Moon. Assume that the Earth is spherical. The drawing should include an artistic picture with an image of the Bear on North Pole; necessary sizes or angular sizes should be pointed out in the picture. Recollect for yourself the necessary information about the animals.

Отговори

Решение

αβ-2. Eclipse on the Poles.

It seems that since the points of observations are opposite, the situation should be absolutely symmetric: the White Bear should also observe half of the eclipsed solar disk on the horizon at this time. But this quite incorrect.

It is necessary to take into account three important circumstances.

First, for any observer, the physical horizon is slightly lowered. For example, if a person stands on a flat surface, the depression of physical horizon she sees is about 2.5′. The depression of the physical horizon is easy to calculate: if R is the radius of the Earth, and h is the height of eye level above the surface of the Earth, the horizon dip is equal to

\arccos(R/(R+h)) \approx ((R+H)^2 - R^2)^{1/2}/R \approx (2h/R)^{1/2}\ \text{(in radians)}.

Penguins may be different. For an estimation let us take the sizes of an Imperial Penguin, whose eyes' height above the surface of the Earth is about 1 m. The horizon depression is then about 2′. The White Bear, as it is mentioned, is sitting, so height of his eyes above the surface of the Earth will be about 1 m too, and so the horizon depression is about 2′ too (If it stands up, this height would be more than 2 m and about 3′ accordingly).

This means that if, as seen by the Penguin, the centre of the solar disk is on the horizon, this centre already is 2′ below the mathematical horizon. So at the opposite point of the Earth – at the North Pole – it is 2′ above the mathematical horizon. Accordingly, the sitting Bear-observer sees it already 4′ above physical horizon, i.e. a quarter of the radius of the Sun.

But horizon depression is actually the smallest effect in this problem.

Secondly, there is refraction of light beams in the atmosphere. The value of a refraction depends slightly on weather conditions, but on average contributes with about 35′ at the horizon. So at the moment when the centre of the solar disk is visible on the horizon at the South Pole, at the opposite point of the Earth – at the North Pole – the Sun may be seen quite above the horizon. In fact the refraction at the South Pole yields 35′, and at North Pole about 25′ (not 35′, since the Sun not at the horizon but higher).

But this effect is also not the main one in the given problem.

The parallax of the Moon should be taken into account. It is no accident the annular eclipse may be observed only in a narrow band on the Earth surface. The distance to the Moon is 384 thous. km. The polar radius of the Earth is 12730 km. So the parallactic displacement of the Moon is 12730 / 38400 ≈ 0,0332 rad ≈ 1,9°.

That is, the Bear do not observe any solar eclipse this time, neither annular nor even partial.

Deduction: the White Bear sees the centre of the Sun above the horizon, at approximately 4′+35′+25′ = 64′ or about one degree. And the lunar disc is at approximately 1°54′ below the solar one, i.e. totally under the horizon. (And it should be noted, 1°54′ below the true position of the solar disk, while taking into account the refraction the true position of the Moon is 2°19′ below the apparent position of the Sun).

Parameters of orbits and physical characteristics of planets, Sun and Moon

Body, planet	Average distance to central body, in astr. units	Average distance to central body, in mln. km	Sidereal (or analogous) period, in tropical years	Sidereal (or analogous) period, in days	Eccentricity, e	Equat. diameter, km	Mass, $10^{24}$ kg	Av. density, g/cm³	Grav. acceler. at surf., m/s²	Max. magn. from Earth **)	Albedo
Sun	$1{,}6\cdot10^9$	$2{,}5\cdot10^{11}$	$2{,}2\cdot10^8$	$8\cdot10^{10}$		1392000	1989000	1,409		$-26{,}8^{\mathrm{m}}$
Mercury	0,387	57,9	0,241	87,969	0 ,206	4 879	0,3302	5,43	3,70	$-2{,}2^{\mathrm{m}}$	0,06
Venus	0,723	108,2	0,615	224,701	0,007	12 104	4,8690	5,24	8,87	$-4{,}7^{\mathrm{m}}$	0,78
Earth	1,000	149,6	1,000	365,256	0,017	12 756	5,9742	5,515	9,81		0,36
Moon	0,00257	0,38440	0,0748	27,3217	0,055	3 475	0,0735	3,34	1,62	$-12{,}7^{\mathrm{m}}$	0,07
Mars	1,524	227,9	1,880	686,980	0,093	6 794	0,6419	3,94	3,71	$-2{,}0^{\mathrm{m}}$	0,15
Jupiter	5,204	778,6	11,862	4 332,59	0,048	142 984	1899,8	1,33	24,86	$-2{,}7^{\mathrm{m}}$	0,66
Saturn	9,584	1433,7	29,458	10 759,20	0,054	120 536	568,50	0,70	10,41	$0{,}7^{\mathrm{m}}$	0,68
Uranus	19,191	2871,0	84,015	30 685,93	0,046	51 118	86,625	1,30	8,44	$5{,}5^{\mathrm{m}}$	0,74
Neptune	30,071	4498,6	164,778	60 187,64	0,008	49 532	102,78	1,76	11,20	$7{,}8^{\mathrm{m}}$	0,58

**) For outer planets and Moon – in mean opposition.

Coordinates	Observatory	Kazan kremlin	Petrovskoye
λ ( E / в.д. )	48° 49′	49° 06′	49° 06′
φ ( N / с.ш. )	55° 50′	55° 48′	55° 41′
Timezone	UT+3	UT+3	UT+3

Some constants and formulae

Constant or formula	Value
Speed of light in vacuum c (m/s)	299792458
Constant of gravitation G (N·m²/kg²)	$6.674\cdot10^{-11}$
Solar constant A (W/m²)	1367
Hubble parameter H₀ (km/s/Mpc), mean	68
Hubble parameter, range	50–100
Planck constant h (J·s)	$6.626\cdot10^{-34}$
Charge of electron e (C)	$1.602\cdot10^{-19}$
Mass of electron mₑ (kg)	$9.109\cdot10^{-31}$
Proton-to-electron ratio	1836.15
Faraday constant F (C/mol)	96485
Magnetic constant μ₀ (H/m)	$1.257\cdot10^{-6}$
Universal gas constant R (J/mol/K)	8.314
Boltzmann constant k (J/K)	$1.381\cdot10^{-23}$
Stefan–Boltzmann constant σ (W/m²/K⁴)	$5.670\cdot10^{-8}$
Wien displacement constant b (m·K)	0.002897
Laboratory wavelength Hα (Å)	6562.81
Tropical year length T (days)	365.242199
Nodal period of lunar orbit (years)	−18.6
Standard atmosphere (Pa)	101325
Visible light extinction by terrestrial atmosphere in zenith (minimum)	19%, 0.23 magnitudes
Height of homogeneous atmosphere (m)	7991
Refractive index of water for 20°C n	1.334
Moment of inertia of a solid ball	$I=\frac25MR^2$
Volume of a ball	$V=\frac43\pi R^3$
Area of sphere	$S=4\pi R^2$
π	3.14159265
e	2.71828183
Golden ratio φ	1.61803399

20 brightest stars, other stars and Greek alphabet

Name	Designation	RA	Dec	Parallax (arcsec)	Magnitude	Spectral class
Altair	α Aql	19h50m47s	08°52′06″	0.195	0.77	A7
Capella	α Aur	05h16m41s	45°59′53″	0.073	0.08	G5+G0
Arcturus	α Boo	14h15m38s	19°10′57″	0.089	−0.04ᵛ	K1
Canopus	α Car	06h23m57s	−52°41′45″	0.028	−0.72	F0
Toliman (Rigel Kent)	α Cen A; B	14h39m36s	−60°50′07″	0.747	−0.01; 1.33	G2; K1
Hadar	β Cen	14h03m49s	−60°22′23″	0.009	0.61	B1
Sirius	α CMa	06h45m09s	−16°42′58″	0.375	−1.46	A1
Procyon	α CMi	07h39m18s	05°13′30″	0.288	0.38	F5
Acrux	α Cru	12h26m36s	−63°05′57″	0.010	0.77	B0
Becrux	β Cru	12h47m43s	−59°41′20″	0.009	1.30	B0
Deneb	α Cyg	20h41m26s	45°16′49″	0.002	1.25	A2
Achernar	α Eri	01h37m43s	−57°14′12″	0.026	0.46	B3
Pollux	β Gem	07h45m19s	28°01′35″	0.097	1.14	K0
Vega	α Lyr	18h36m56s	38°47′01″	0.123	0.03	A0
Betelgeuse	α Ori	05h55m10s	07°24′25″	0.005	0.5ᵛ	M2
Rigel	β Ori	05h14m32s	−08°12′06″	0.013	0.12	B8
Fomalhaut	α PsA	22h57m39s	−29°37′20″	0.130	1.16	A3
Antares	α Sco	16h29m24s	−26°25′55″	0.024	0.96	M1+B4
Aldebaran	α Tau	04h35m55s	16°30′33″	0.048	0.85ᵛ	K5
Spica	α Vir	13h25m12s	−11°09′41″	0.023	0.98	B1

Some other stars

Name	Designation	RA	Dec	Parallax (arcsec)	Magnitude	Spectral class
Hamal	α Ari	02h07m10s	23°27′45″	0.050	2.01	K2
Polaris	α UMi	02h31m49s	89°15′51″	0.007	1.97ᵛ	F7
Kochab	β UMi	14h50m42s	74°09′20″	0.025	2.07	K4
Proxima Centauri	V645 Cen, α Cen C	14h29m43s	−62°40′46″	0.769	11.05	M5.5

Greek alphabet

Uppercase	Lowercase	Name
Α	α	alpha
Β	β	beta
Γ	γ	gamma
Δ	δ	delta
Ε	ε	epsilon
Ζ	ζ	zeta
Η	η	eta
Θ	θ	theta
Ι	ί	iota
Κ	κ	kappa
Λ	λ	lambda
Μ	μ	mu
Ν	ν	nu
Ξ	ξ	xi
Ο	ο	omicron
Π	π	pi
Ρ	ρ	rho
Σ	σ	sigma
Τ	τ	tau
Υ	υ	upsilon
Φ	φ	phi
Χ	χ	chi
Ψ	ψ	psi
Ω	ω	omega

Theoretical round. Sketches for solutions. For jury job ONLY

Theoretical round. Sketches for solutions. For jury job ONLY

Note. The given sketches are not full; the team leaders have to give more detailed explanations to students. But the correct solutions in the students' papers (enough for 8 pts) may be shorter.

Note. Jury members should evaluate the student's solutions in essence, and not by looking on formal existence the mentioned sentences or formulae. The formal presence of the mentioned positions in the text is not necessary to give the respective points. Points should be done if the following steps de facto using these positions.

Theoretical round. Basic criteria. For work of Jury

Note. The given sketches are not full; the team leaders have to give more detailed explanations to students. But the correct solutions in the students' papers (enough for 8 pts) may be shorter.

Теми и трудност

Трудност: неоценена.

Слънчева система · Небесна сфера и сферична астрономия · Затъмнения и покрития · Атмосферни ефекти

Оригинал в Архива: theo_alpha.pdf · официални решения: theosol1.jpg · допълнителни данни: theosol2.jpg · допълнителни данни: theosol3.jpg · допълнителни данни: theosol4.jpg · допълнителни данни: theosol5.jpg · допълнителни данни: theo_instr.pdf · допълнителни данни: data_1.pdf · допълнителни данни: data_2.pdf · допълнителни данни: data_3.pdf · допълнителни данни: hrd.pdf · допълнителни данни: sky_map.pdf