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IAO 2017, Теория, α+β — β-1. Double star.

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Всички материали от IAO 2017

XXII International Astronomy Olympiad — Theoretical round, α and β · 27 октомври 2017 г. · 8 т.

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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.

Do not forget that in clean copy you should write solutions only on even (left) pages. The odd (right) pages will be used by your team leader to translate solutions into English and also to emphasize your thoughts in solutions.

EVEN PAGE / ЧЁТНАЯ СТРАНИЦА	ODD PAGE / НЕЧЁТНАЯ СТРАНИЦА
Текст решения. Text of solution. Текст решения. Text of solution. Текст решения. Text of solution.	×

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.

Не забудьте, что в чистовике вы должны писать решения только на чётных (левых) страницах. Нечётные (правые) страницы предназначены для того, чтобы руководитель вашей команды перевёл решения на английский язык и подчеркнул основные мысли решения.

Некоторые константы и формулы / Some constants and formulae

Величина / Quantity	Значение
Скорость света в вакууме, c (м/с) / Speed of light in vacuum	299 792 458
Гравитационная постоянная, G (Н·м²/кг²) / Constant of gravitation	6.674·10⁻¹¹
Солнечная постоянная, A (Вт/м²) / Solar constant	1367
Параметр Хаббла, H₀ (км/с/МПк) / Hubble parameter, mean; range	68; 50–100
Постоянная Планка, h (Дж·с) / Planck constant	6.626·10⁻³⁴
Заряд электрона, e (Кл) / Charge of electron	1.602·10⁻¹⁹
Масса электрона, mₑ (кг) / Mass of electron	9.109·10⁻³¹
Соотношение масс протона и электрона / Proton-to-electron mass ratio	1836.15
Постоянная Фарадея, F (Кл/моль) / Faraday constant	96 485
Магнитная постоянная, μ₀ (Гн/м) / Magnetic constant	1.257·10⁻⁶
Универсальная газовая постоянная, R (Дж/моль/К) / Universal gas constant	8.314
Постоянная Больцмана, k (Дж/К) / Boltzmann constant	1.381·10⁻²³
Постоянная Стефана-Больцмана, σ (Вт/м²/К⁴) / Stefan-Boltzmann constant	5.670·10⁻⁸
Константа смещения Вина, b (м·К) / Wien’s displacement constant	0.002897
Лабораторная длина волны Hα (Å) / Laboratory wavelength	6562.81
Длина тропического года, T (сут) / Tropical year length	365.242199
Длина сидерического года, T (сут) / Sidereal year length	365.25636
Длина аномалистического года, T (сут) / Anomalistic year length	365.259636
Период обращения узлов лунной орбиты (лет) / Nodal period of lunar orbit	-18.6
Стандартная атмосфера (Па) / Standard atmosphere	101 325
Ослабление видимого света слоем 1 атмосферы (минимально) / Visible light extinction in zenith	19%, 0.23ᵐ
Высота однородной атмосферы (м) / Height of homogeneous atmosphere	7991
Показатель преломления воды при 20°C, n / Refractive index of water	1.334
Момент инерции шара / Moment of inertia of a solid ball	I=²⁄₅MR²
Момент инерции сферы / Moment of inertia of sphere	I=²⁄₃MR²
Объём шара / Volume of a ball	V=⁴⁄₃πR³
Площадь сферы / Area of sphere	S=4πR²
π	3.14159265
e	2.71828183
Золотое сечение, φ / Golden ratio	1.61803399

Данные о некоторых звёздах / Data of some stars

Звезда / Star	Designation	R.A.	DEC	p	Mag.	Temperature	Mass
Солнце / Sun	☉	0ʰ–24ʰ	−23°26′–+23°26′	8″.794	−26ᵐ.74	5777 K	1 M☉
Бетельгейзе / Betelgeuse	α Ori	05ʰ55ᵐ10ˢ	07°24′25″	0″.005	0ᵐ.5	3590 K	11.6 M☉
Вега / Vega	α Lyr	18ʰ36ᵐ56ˢ	38°47′01″	0″.130	0ᵐ.03	9600 K	2.14 M☉
Наос / Naos	ζ Pup	08ʰ03ᵐ35ˢ	−40°00′12″	0″.003	2ᵐ.21	42000 K	40 M☉
Толиман / Toliman (Rigel Kent)	α Cen A	14ʰ39ᵐ36ˢ	−60°50′07″	0″.747	−0ᵐ.01	5810 K	1.11 M☉
Толиман / Toliman (Rigel Kent)	α Cen B	14ʰ39ᵐ36ˢ	−60°50′07″	0″.747	1ᵐ.33	5260 K	0.93 M☉
Проксима Центавра / Proxima Centauri	V645 Cen, α Cen C	14ʰ29ᵐ43ˢ	−62°40′46″	0″.769	11ᵐ.06	3040 K	0.123 M☉

Координаты / Coordinates	WHO Observatory	Xinglong Observatory
φ (N / с.ш.)	37°32′	40°24′
λ (E / в.д.)	122°03′	117°35′
Часовой пояс / Timezone	UT+08	UT+08

Theoretical round. Sketches for solutions

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. Jury members should elaborate more detailed criteria, and also create criteria for other correct ways of the student’s solutions.

Source limitations

The original PDF and its embedded scans clip the opening text on solution pages 2, 4 and 6. Ellipses mark these unrecoverable passages; complete surviving derivations, answers and grading rows are retained. The double-star solution refers to a figure that is absent from the supplied packet.

Printed discrepancies

The answer key uses 1.34ᵐ for α Centauri B while its data table gives 1.33ᵐ. The summit key invokes brown dwarfs despite the stated A–M main-sequence restriction, and calls an isosceles triangle equilateral. Its printed sin β/2 expressions are preserved. The mirror key says losing K% on transmission despite the statement saying transmitting K%. The eclipse key prints V₃ = V₂ − V_E and L_NS for the along-track width, while nearby prose uses Vφ and L_WE; these are not silently changed. The printed β-2 grading allocations and subtotals are retained.

Условие

A double star consists of components by all physical characteristics equal to α Centauri A and α Centauri B, which are rotating around the common center of gravity by circular orbits. As observed from the Earth the angular distance between the components varies from $0.17''$ till $2.2''$ with a period of $\tau=39.6$ years. Find the apparent magnitude of every component and also their combined magnitude.

Решение

Внимание: Непълно решение

The original solution scan clips text at the top of page 2. The surviving answer and derivation are retained; missing text is not invented.

As the components move in circular orbits around their common center of mass, the distance between them does not change. Therefore observed from Earth, the angular distance between them can be changed only due to the fact that we see the orbits of the stars inclined.

The angular distance between the stars becomes maximum ( $2.2''$ ) when the line connecting them is perpendicular to the direction to the Earth (see figure). This situation take place twice during one period $T_\Sigma$ of rotating the stars, so $T_\Sigma=2\tau=79.2$ years. The total mass of the system $M_\Sigma$ (in solar masses) is equal to $M_\Sigma=1.11M_\odot+0.93M_\odot=2.04M_\odot$ .

In the system of units $M_\odot$ – year – au the Kepler's third law may be expressed as

a^3/T^2\mathbin{\cdot}M=1.

Therefore, in this binary system the semimajor axis of the relative orbit (in this case, the distance between the components) will be equal to

a_\Sigma=(T_\Sigma^2\mathbin{\cdot}M_\Sigma)^{1/3}\approx23.4\ \mathrm{au}.

As follows from the problem situation these 23.4 au are visible from the Earth as an angle of $2.2''$ , therefore,…

… parallax of the double star equals to

p=2.2''/23.4\approx0.094''.

(You can also find the distance at which the double star is located, in parsecs

L=23.4/2.2\approx10.63\ \mathrm{pc}\approx10.6\ \mathrm{pc}.)

By comparing the components with the real α Centauri A and α Centauri B, which have parallax $0.747''$ and magnitudes $-0.01^{\mathrm m}$ and $1.34^{\mathrm m}$ , we get:

m_1=m_{\alpha\mathrm{CenA}}-5^{\mathrm m}\lg(p/p_{\alpha\mathrm{CenA}})=-0.01^{\mathrm m}+5^{\mathrm m}\lg(0.747/0.094)=-0.01^{\mathrm m}+4.48^{\mathrm m}\approx4.5^{\mathrm m},

m_2=m_{\alpha\mathrm{CenB}}-5^{\mathrm m}\lg(p/p_{\alpha\mathrm{CenB}})=1.34^{\mathrm m}+5^{\mathrm m}\lg(0.747/0.094)=1.34^{\mathrm m}+4.48^{\mathrm m}\approx5.8^{\mathrm m}.

…

\Delta m=+5^{\mathrm m}\lg(L_1/L_{\alpha\mathrm{Cen}})=+5^{\mathrm m}\lg(10.6/1.34)\approx4.49^{\mathrm m}).

The total magnitude of the two components will be equal to

m_{1+2}=-2.5^{\mathrm m}\lg(10^{-m_1/2.5}+10^{-m_2/2.5})\approx4.2^{\mathrm m}.

(One can also first find the total magnitude of both components of α Centauri as

m_{\alpha\mathrm{Cen}(A+B)}=-2.5^{\mathrm m}\lg(10^{-m_{\alpha\mathrm{CenA}}/2.5}+10^{-m_{\alpha\mathrm{CenB}}/2.5})\approx-0.28^{\mathrm m},

and then add to it the difference of $+4.48^{\mathrm m}$ and get the same $4.2^{\mathrm m}$ ).

Answers: $m_1\approx4.5^{\mathrm m}$ , $m_2\approx5.8^{\mathrm m}$ , $m_{1+2}\approx4.2^{\mathrm m}$ .

Basic criteria

Basic criteria	Points
Understanding effect is due to the orbits of stars inclined	1.5
$T_\Sigma=2\tau=79.2$ years	1
$a_\Sigma=(T_\Sigma^2M_\Sigma)^{1/3}\approx23.4$ au	2
Understanding relations between using parallaxes and distances	1
Correct formula or analogous relations to calculate magnitudes	1
Correct calculation of $m_1\approx4.5^{\mathrm m}$ , $m_2\approx5.8^{\mathrm m}$ , $m_{1+2}\approx4.2^{\mathrm m}$	1.5 (0.5 pt each)

Оригинал в Архива: theo.pdf · официални решения: theosol.pdf · допълнителни данни: instructions.pdf · допълнителни данни: data.pdf · допълнителни данни: hrd.pdf