Съдържание

Условие 1.1. (4 т.)1.2. (3 т.)Picture. (1 т.)Решение

IAO 2010, Теория, Група α — Задача 1. Circumpolar stars.

Оригинал на условието (изображение)Съобщи за грешка

Всички материали от IAO 2010

Theoretical round. Problems to solve · 16 октомври 2010 г. · 8 т.

Бележка

Maybe not all problems have correct questions. Some questions (maybe the main question of the problem, maybe one of the subquestions) may have 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 table of planetary data may be used for solving every problem.

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

Note for jury and team leaders.

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

Some constants and formulae

Constant	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 constant, $H_0$ (km/s/Mpc), mean value	70
Hubble constant, diapason of values	50–100
Plank constant, $h$ (J·s)	$6.626\cdot10^{-34}$
Charge of electron, $e$ (C)	$1.602\cdot10^{-19}$
Mass of electron, $m_e$ (kg)	$9.109\cdot10^{-31}$
Proton-to-electron ratio	1836.15
Faraday constant, $F$ (C/mol)	96485
Magnetic constant, $\mu_0$ (H/m)	$1.257\cdot10^{-6}$
Universal gas constant, $R$ (J/mol/K)	8.314
Boltzmann constant, $k$ (J/K)	$1.381\cdot10^{-23}$
Standard atmosphere (Pa)	101325
Stefan-Boltzmann constant, $\sigma$ (W/m²/K⁴)	$5.670\cdot10^{-8}$
Wien’s displacement constant, $b$ (m·K)	0.002897
Laboratory wavelength of H $\alpha$ (Å)	6563
Refractive index of water for 20°C, $n$	1.334
Area of sphere	$S=4\pi R^2$
$\pi$	3.14159265

Data of some stars

Star	Designation	RA	DEC	$p$	m	SC
Arcturus	$\alpha$ Boo	14h 15m 40s	19° 10′ 57″	0″.089	−0.05	K1
Vega	$\alpha$ Lyr	18h 36m 56s	38° 47′ 01″	0″.129	0.03	A0
Deneb	$\alpha$ Cyg	20h 41m 26s	45° 16′ 49″	0″.001	1.25	A2
Polaris	$\alpha$ UMi	02h 31m 51s	89° 15′ 51″	0″.007	2.02	F7
Sirius	$\alpha$ CMa	06h 45m 09s	−16° 42′ 58″	0″.379	−1.46	A1

Parameters of orbits. Physical characteristics of some planets, Moon, Sun and Eris

Body, planet	Average distance to central body (astr. units)	Average distance (mln. km)	Sidereal (or analogous) period (troph. years)	Period (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		$-26.8^m$
Mercury	0.387	57.9	0.241	87.97	0.206	4879	0.3302	5.43	3.70		0.06
Venus	0.723	108.2	0.615	224.70	0.007	12104	4.8690	5.24	8.87		0.78
Earth	1.000	149.6	1.000	365.26	0.017	12756	5.9742	5.515	9.81		0.36
Moon	0.00257	0.38440	0.0748	27.3217	0.055	3475	0.0735	3.34	1.62	$-12.7^m$	0.07
Mars	1.524	227.9	1.880	686.98	0.093	6794	0.6419	3.94	3.71	$-2.0^m$	0.15
Jupiter	5.204	778.6	11.862	4332.59	0.048	142984	1899.8	1.33	24.86	$-2.7^m$	0.66
Saturn	9.584	1433.7	29.458	10759.20	0.054	120536	568.50	0.70	10.41	$0.7^m$	0.68
Eris	68.01			204862	0.434	2600	0.0167		0.8		0.86

**) For Moon – in mean opposition.

Source discrepancies

Printed data and formulae are retained. The official key uses several ungrouped slash expressions. The Mercury parallax key prints approximately 40.9 despite its preceding numerical ratio. The climate key uses 272.8 and 294.4 K for the statement’s −0.4 and +21.2°C.

Условие

Usually we consider that there are about 6000 stars in the whole sky which are visible to the naked eye. The refraction near the horizon is $35'$ . Find, how many more stars visible to the naked eye become circumpolar (which means that they never set) due to refraction:

1.1. (4 т.)

for observer on the North Pole at zero height position (like Polar Bear in a position with its eyes at zero height).

1.2. (3 т.)

for observer on the Equator at zero height position (like Giraffe with the eyes lowered to zero level).

Picture. (1 т.)

In your solution include a picture with an image of the Bear on the ices of North Pole and the Giraffe on the Equator, and necessary dimensions and/or angular dimensions used.

Решение

Obviously, the problem is of estimation. So we will not take into account the effect of diminishing the number of visible stars due to weak transparency of the atmosphere near the horizon. Also, at first we will consider that stars are distributed more or less regularly in the sky.

1.1.

The number of circumpolar stars at the North pole equals to a half of 6000, i.e. 3000. It is those stars whose declination is above $0^\circ$ . Due to refraction the number of circumpolar stars is increased by those ones whose declination is below $0^\circ$ but above $-35'$ . Let us estimate the ratio between the number of stars which are located in the range of declinations from $-35'$ to $0'$ to the number of stars of all the sphere. It is the same as the ratio between the solid angle of the belt between the declinations $-35'$ and $0'$ and the solid angle of whole sphere ( $4\pi$ ). [It is good to draw a figure.]

It is more visual not to use solid angles but areas on an imagined sphere of radius $R_0$ , and calculate the above ratio as the ratio between the area of the belt near the celestial equator with $35'$ width and the area of the whole sphere ( $4\pi R_0^2$ ). The area of the belt is its length along the celestial equator ( $2\pi R_0$ ), multiplied by its width $R_0\cdot35\cdot\pi/(180\cdot60)$ (where $\pi/(180\cdot60)$ is the value of an arc minute expressed in radians). The ratio between these areas is

2\pi R_0\cdot R_0\cdot35\cdot\pi/(180\cdot60)/4\pi R_0^2=35\pi/21600.

The number of stars in this belt may be estimated as

6000\cdot35\pi/21600\approx30.5,

but the accuracy even to one star here is obviously excessive, therefore it is necessary to say that about 30 stars become circumpolar due to refraction.

1.2.

One may use a similar procedure to find the answer for the question 1.2. of the problem. Due to refraction the stars placed closer $35'$ to the poles (both Northern and Southern), that is, with declinations above $+89^\circ25'$ or below $-89^\circ25'$ , become circumpolar. The ratio between the area of these slices of imagined heavenly sphere (area of each slice is $\pi\cdot(35')^2=\pi\cdot(35\pi/(180\cdot60))^2$ ) and the area of the whole sphere $4\pi$ is $5.2\cdot10^{-5}$ . By multiplying this value by 6000 we may get approximately 0.31 star. [And some participants just have written: there will be about 0.3 new circumpolar stars.] But what does this value mean? It means that the model of uniform distribution of stars on heavenly sphere does not work any more. Any stars ("one star", "two stars" etc.) may be present in these areas of the sky, or may not be present, but it can not be “half present”. And the right answer may be given at once (without any calculations!): no star, as we know that there are no stars visible with the naked eye star in radius $35'$ near to poles. The nearest one is Polaris, it is situated approximately $45'$ from the North Pole.

Basic criteria. For work of Jury

1.1. 4 pt.

1.2. 3 pt total, including:

Formula and calculation – 2 pt.
“We know that there are no any visible star in radius 35′ near to poles” – 1 pt.

Picture. Good picture of Bear and Giraffe (should be better than points or schematic objects) – 1 pt.

Оригинал в Архива: alphatheo.jpg · допълнителни данни: theosol1.jpg · допълнителни данни: theosol2.jpg · допълнителни данни: theosol3.jpg · допълнителни данни: theosol4.jpg · допълнителни данни: theosol5.jpg · допълнителни данни: theocriteria1.jpg · допълнителни данни: theocriteria2.jpg · допълнителни данни: constants.jpg · допълнителни данни: orbital-elements.jpg