Съдържание

IAO 2013, Теория — Задача 4. β-1. Радиоастрон

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IAO 2013

XVIII Международная астрономическая олимпиада. XVIII International Astronomy Olympiad. Литва, Вильнюс, 6 – 14. IX. 2013, Vilnius, Lithuania. Round Theo · 6 септември 2013 г. · 8 т.

Внимание: Бележка към темата

Ръкописни български условия за групи α, β и общи αβ; решенията и критериите са на английски. В αβ-6 ръкописът дава 3400 km, а английският оригинал и ключът — 3400 pc. Източниковите разминавания са запазени и отбелязани.

Общо указание

Общо указание. Възможно е не във всички задачи въпросите да са поставени коректно. Някои въпроси (възможно е това да е основният въпрос в задачата, а може и да е някой подвъпрос) може да нямат смисъл. В такива случаи в отговора трябва да се напише (на руски или на английски): „ситуация невозможна - impossible situation“. Разбира се, отговорът трябва да е подкрепен с пресмятания или логически разсъждения.

Данните от таблиците (Слънчева система, звезди, константи) могат да се използват във всички задачи.

Отговорите „Да - Yes“ или „Нет - No“ трябва да бъдат написани на руски или на английски.

Общо указание

Данните от таблиците (Слънчева система, звезди, константи) могат да се използват във всички задачи.

Отговорите „Да - Yes“ или „Нет - No“ трябва да бъдат написани на руски или на английски.

Parameters of orbits and physical characteristics

The bilingual source sheet gives the following values. Blank cells reproduce blank source cells.

Body / planet	Average distance (AU)	Average distance ( $10^6$ km)	Sidereal period (tropical years)	Sidereal period (days)	Orbital inclination $i$	Eccentricity $e$	Equatorial diameter (km)	Mass ( $10^{24}$ kg)	Average density (g/cm³)	Gravity at surface (m/s²)	Axial tilt	Maximum magnitude from Earth	Albedo
Sun	$1.6\cdot10^9$	$2.5\cdot10^{11}$	$2.2\cdot10^8$	$8\cdot10^{10}$			1392000	1989000	1.409			−26.74
Mercury	0.387	57.9	0.241	87.969	7.00°	0.206	4879	0.3302	5.43	3.70	0.01°		0.06
Venus	0.723	108.2	0.615	224.7007	3.40	0.007	12104	4.8690	5.24	8.87	177.36		0.78
Earth	1.000	149.6	1.000	365.2564	0.00	0.017	12756	5.9742	5.515	9.81	23.44		0.36
Moon	0.00257	0.38440	0.0748	27.3217	5.15	0.055	3475	0.0735	3.34	1.62	6.7	−12.7	0.07
Mars	1.524	227.9	1.880	686.98	1.85	0.093	6794	0.6419	3.94	3.71	25.19	−2.0	0.15
Jupiter	5.204	778.6	11.862	4332.59	1.30	0.048	142984	1899.8	1.33	24.86	3.13	−2.7	0.66
Saturn	9.584	1433.7	29.458	10759.20	2.48	0.054	120536	568.50	0.70	10.41	26.73	0.7	0.68

For outer planets and Moon — in mean opposition.

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, $H_0$ (km/s/Mpc), mean value	71
Hubble parameter, diapason of values	50–100
Planck 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)	$1.381\cdot10^{-23}$
Stefan–Boltzmann constant, $\sigma$ (W/m²/K⁴)	$5.670\cdot10^{-8}$
Wien’s displacement constant, b (m·K)	0.002897
Laboratory wavelength of Hα (Å)	6562.81
Tropical year length, T (days)	365.242199
Standard atmosphere (Pa)	101325
Visible light extinction by the terrestrial atmosphere in zenith (minimum)	19%, 0.23 mag
Refractive index of water for 20°C, n	1.334
Moment of inertia of a solid ball	$I=\frac25MR^2$
Volume of ball	$V=\frac43\pi R^3$
Area of sphere	$S=4\pi R^2$
$\pi$	3.14159265
e	2.71828183
Golden ratio, $\varphi$	1.61803399

Data of some stars

Star	Designation	RA	DEC	Parallax $p$	Apparent magnitude $m$	Spectral class	Mass ( $\mathcal M_{\odot}$ )
Sun	⊙	0h–24h	−23°26′ to +23°26′	8.794″	vis −26.74; bol −26.82	G2	1
Proxima Centauri	V645 Cen, α Cen C	14h 29m 43s	−62°40′46″	0.769″	11.05	M5.5	0.123
Alpha Centauri A	α Cen A	14h 39m 37s	−60°50′02″	0.747″	−0.01	G2	1.1
Alpha Centauri B	α Cen B	14h 39m 35s	−60°50′14″		1.34	K1	0.9
Beta Centauri	β Cen	14h 03m 49s	−60°22′23″	0.009″	0.61	B1	21
Epsilon Eridani	ε Eri	03h 32m 56s	−09°27′30″	0.311″	3.74	K2	0.82
Gliese 581	HO Lib	15h 19m 27s	−07°43′20″	0.16″	vis 10.57; bol 8.0	M3V	0.31

The Alpha Centauri parallax is printed once beside the pair; the second cell is left blank here as in the source.

Historical units

Units of power

1 horse-power (hp) = 735.49875 W.

Local units of length in the end of XVIII century

1 arshin (arsh) = 0.711187 m
1 span (sp) = 1/4 arshin
1 vershok (vrsh) = 1/4 span
1 sajene (sj) = 3 arshin
1 verst (vrst) = 500 sajene

Условие

Радиоастрон е международен научен проект, възглавяван от Астрокосмическия център на Руската академия на науките. На 18 юли 2011 г. в елиптична орбита около Земята е бил изведен спътникът „Спектър-Р“ с 10-метров (в диаметър) космически радиотелескоп. Заедно с наземни радиотелескопи „Спектър-Р“ работи като интерферометър. Радиоастрон работи на стандартните радиоастрономически дължини на вълните 1.19-1.63 cm (K-диапазон), 6.2 cm (C-диапазон), 18 cm (L-диапазон) и 92 cm (P-диапазон). Понастоящем „Спектър-Р“ се движи по силно изтеглена елиптична орбита с период $\tau = 8.3$ денонощия и височина на перигея $h = 600\ \mathrm{km}$ над земната повърхност.

1.1. Оценете най-добрата разделителна способност (в ъглови секунди) на проекта Радиоастрон. С помощта на схематична рисунка обяснете в каква ситуация се реализира тя. [3,5 т.]

1.2. Оценете разделителната способност на Радиоастрон при наблюдение на обекти, намиращи се по направление на голямата ос на орбитата на „Спектър-Р“. Решението също придружете със схема. [3 т.]

Отговори

Решение

Официалното решение и критериите по-долу са на английски, както в източника.

Official English solution sketch. 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.

Resolution and base

The resolution of the instrument is determined by the working wavelength $\lambda$ , and the base $D$ . For telescopes the base is just the diameter of the front lens or mirrors, and for interferometers it is the distance between its elements, or, rather, the projection of this distance on the plane perpendicular to the direction to the object under research.

\delta=\lambda/D.

Thus, the maximum resolution is achieved in the case of the minimum wavelength and the maximum base.

\delta_{\min}=\lambda_{\min}/D_{\max}.

1.1. Maximum resolving power

In our case it is achieved at the wavelength $\lambda_{\min}=1.2\ \mathrm{cm}$ and the base of the Earth–“Spektr-R”, in which “Spektr-R” is at apogee, and the ground-based telescope is at the opposite side of the globe; and while the observations are held in the directions perpendicular to the base of Earth–“Spektr-R”, that is perpendicular to the major axis of the orbit:

D_{\max}=A+R,

where A is the distance at apogee, and R is radius of the Earth. The apogee distance can be found as

A=2a-(R+h),

where a is semi-major axis of the orbit, which, in turn, follows Kepler’s third law. Comparing the motion of the satellite with the motion of the Moon we get:

a^3/a_{\mathrm{Moon}}^3=\tau^2/T_{\mathrm{Moon}}^2;\qquad a=a_{\mathrm{Moon}}(\tau/T_{\mathrm{Moon}})^{2/3},

a=384\,400\ \mathrm{km}\times(8.3\ \mathrm{days}/27.32\ \mathrm{days})^{2/3}=173\,700\ \mathrm{km}.

D_{\max}=2a-(R+h)+R=2a-h,

D_{\max}=2\times173\,700\ \mathrm{km}-600\ \mathrm{km}\approx347\,000\ \mathrm{km}.

\delta_{\min}=1.2\ \mathrm{cm}/347\,000\ \mathrm{km}\approx3.5\cdot10^{-11}\ \mathrm{rad},

or in angular seconds:

\delta_{\min}=3.5\cdot10^{-11}\ \mathrm{rad}\times206265^{\prime\prime}/\mathrm{rad}\approx7\cdot10^{-6}\,{}^{\prime\prime}=7\ \mu\mathrm{as}.

Orbit and maximum interferometer base perpendicular to the major axis

1.2. Observation along the major axis

In observations of objects located in the direction of the major axis of the orbit of “Spektr-R” the maximum base D will be equal to the sum of the semi-minor axis of the orbit and the radius of Earth:

D_{\max}=b+R.

Semi-minor axis can be found from the geometric properties of the ellipse:

a^2=b^2+(a-P)^2,\qquad b^2=[a^2-(a-P)^2]^{1/2},

b\approx48\,900\ \mathrm{km}.

D_{\max}=48\,900\ \mathrm{km}+6\,400\ \mathrm{km}=55\,300\ \mathrm{km}.

\delta_{\min}=\lambda_{\min}/D_{\max}=1.2\ \mathrm{cm}/55\,300\ \mathrm{km}\approx2.2\cdot10^{-10}\ \mathrm{rad},

or in angular seconds:

\delta_{\min}=2.2\cdot10^{-11}\ \mathrm{rad}\times206265^{\prime\prime}/\mathrm{rad}\approx4.5\cdot10^{-5}\,{}^{\prime\prime}=45\ \mu\mathrm{as}.

Source note: the key prints $b^2$ on the left of the square-root formula, and $10^{-11}$ in the final angular conversion after $10^{-10}$ in the preceding line. These printed discrepancies are retained; its stated result is 45 μas.

Semi-minor axis and interferometer base for observation along the major axis

Official basic criteria

1.common. Totally 1.5 pt, including:

General approach, formula $\delta=\lambda/D$ — 0.5 pt.
The base for interferometers it is the distance between its elements — 0.5 pt.
Understanding $\delta_{\min}=\lambda_{\min}/D_{\max}$ — 0.5 pt.

1.1. Totally 3.5 pt, including:

Understanding $\lambda_{\min}=1.19$ cm — 0.5 pt.
The maximum base of the Earth–“Spektr-R”, in which “Spektr-R” is at apogee, and the ground-based telescope is at the opposite side of the globe $D_{\max}=A+R$ — 0.5 pt.
The directions perpendicular to the base of Earth–“Spektr-R” — perpendicular to the major axis — 0.5 pt.
The apogee distance can be found as $A=2a-(R+h)$ — 0.5 pt.
Using Kepler’s third law to find a, $a^3/a_{\mathrm{Moon}}^3=\tau^2/T_{\mathrm{Moon}}^2$ — 0.5 pt.
Final calculations and correct result — 1 pt.

1.2. Totally 3 pt, including:

The maximum base $D_{\max}=b+R$ — 1 pt.
Calculation semi-minor axis using the geometric properties of the ellipse — 1 pt.
Final calculations and correct result — 1 pt.

Official solution sketches: 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.

Basic criteria: note for jury

Jury members should evaluate the students’ 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.

Оригинал в Архива: bgtheo.pdf · официални решения: theosol.pdf · допълнителни данни: theodata.pdf · допълнителни данни: theo_cr_1.pdf · допълнителни данни: theo_cr_2.pdf · допълнителни данни: theo_cr_3.pdf · допълнителни данни: constants.pdf · допълнителни данни: hrd.pdf · допълнителни данни: alphatheo.pdf · допълнителни данни: betatheo.pdf · допълнителни данни: theo_cr_1.png · допълнителни данни: theo_cr_2.png · допълнителни данни: theo_cr_3.png