Съдържание

IAO 2008, Теория, Група α — Задача 3. Great opposition

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Всички материали от IAO 2008

XIII International Astronomy Olympiad · Theoretical round. Problems to solve · 13 октомври 2008 г.

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.

Элементы орбит. Физические характеристики некоторых планет, Луны и Солнца

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

Небесное тело, планета / Body, planet	Среднее расстояние от центрального тела / Average distance to central body, астр. ед. / astr. units	Среднее расстояние от центрального тела / Average distance to central body, млн. км / mln. km	Сидерический (или аналогичный) период обращения / Sidereal (or analogous) period, тропич. годах / troph. years	Сидерический (или аналогичный) период обращения / Sidereal (or analogous) period, средних сутках / days	Эксцентриситет / Eccentricity, e	Экваториальн. диаметр / Equat. diameter, km	Масса / Mass, $10^{24}$ kg	Средняя плотность / Av. density, g/cm³	Ускор. своб. пад. у пов. / Grav. acceler. at surf., m/s²	Макс. блеск, вид. с Земли / Max. magn. from Earth **)	Альбедо / Albedo
Солнце / Sun	1,6·10⁹	2,5·10¹¹	2,2·10⁸	8·10¹⁰		1392000	1989000	1,409		−26,8ᵐ
Меркурий / Mercury	0,387	57,9	0,241	87,97	0,206	4 879	0,3302	5,43	3,70		0,06
Венера / Venus	0,723	108,2	0,615	224,70	0,007	12 104	4,8690	5,24	8,87		0,78
Земля / Earth	1,000	149,6	1,000	365,26	0,017	12 756	5,9742	5,515	9,81		0,36
Луна / Moon	0,00257	0,38440	0,0748	27,3217	0,055	3 475	0,0735	3,34	1,62	−12,7ᵐ	0,07
Марс / Mars	1,524	227,9	1,880	686,98	0,093	6 794	0,6419	3,94	3,71		0,15
Юпитер / Jupiter	5,204	778,6	11,862	4 332,59	0,048	142 984	1899,8	1,33	24,86		0,66
Сатурн / Saturn	9,584	1433,7	29,458	10 759,20	0,054	120 536	568,50	0,70	10,41		0,68

**) Для Луны – в среднем противостоянии.
**) For Moon – in mean opposition.

Координаты на эпоху 2000
Epoch 2000 coordinates

	α	δ
Полярная звезда / Polar star	2ʰ 31ᵐ 48ˢ.7	89° 15′ 51″

Условие

You know that there are Great oppositions of Mars for citizens of Earth. The magnitude of Mars were even $-2^{\mathrm m}.9$ during some of the very famous of these oppositions (like in August 27, 2003). But there are also Great oppositions of Venus for citizens of some planet(s). At what planet(s) is this possible? Find the apparent stellar magnitude of Venus visible from every such planet (or this planet) at such Great oppositions.

Отговори

Решение

3αβ. Great opposition. Oppositions (including Great oppositions) of a planet may be only for observers at the inner planets. There is only one inner planet –Mercury – for the Venus. So any opposition of Venus may be observed only from Mercury.

The flux to Earth from Mars

F_M\sim\alpha_{Ma}D_{Ma}^2(1/R_{E-Ma})^2(1/R_{S-Ma})^2,

(here and below $\alpha$ is albedo, $D$ are diameters of the bodies and $R$ are distances, indexes Ma, E, V, Me and S correspond to Mars, Earth, Venus, Mercury and Sun).

The flux to Mercury from Venus

F_V\sim\alpha_VD_V^2(1/R_{Me-V})^2(1/R_{S-V})^2.

The ratio of fluxes

F_V/F_M=(\alpha_V/\alpha_{Ma})(D_V^2/D_{Ma}^2)(R_{E-Ma}R_{S-Ma})^2/(R_{Me-V}R_{S-V})^2.

And what does it mean the word “Great”? The Great oppositions appear for the situation when the distance to opposition planet is smaller than during usual (mean) oppositions. For Mars observing from Earth it is situation when the Mars is in the perihelion of its orbit, as the eccentricity of Mars orbit is quite larger than the eccentricity of Earth.

For Venus observing from Mercury the situation is opposite, the eccentricity of Mercury (observer’s planet) orbit is quite larger than the eccentricity of Venus (the planet that observed). So the Great oppositions appear when the Mercury is in the aphelion of its orbit.

It means that for the comparison magnitudes (i.e. fluxes) for these two Great oppositions we have to use values $R_{S-Ma}=(1-e)A_{Ma}$ and $R_{S-Me}=(1+e)A_{Me}$ for positions of Mars and Mercury and may consider the orbits of Earth and Venus circular, $R_{S-E}=A_E$ , $R_{S-V}=A_V$ .

So the ratio of fluxes

F_V/F_M=(\alpha_V/\alpha_{Ma})(D_V^2/D_{Ma}^2)((1-e)A_{Ma}-A_E)^2((1-e)A_{Ma})^2/((A_V-(1+e)A_{Me})A_V)^2.

Taking the necessary values from the data of Solar system we may calculate

F_V/F_M=(0.78/0.15)(12104/6794)^2(0.382\cdot1.382)^2/(0.256\cdot0.723)^2.

F_V/F_M\approx134.

So Venus visible from Mercury on Great opposition is brighter than Mars visible from Earth on Great opposition, and the difference in stellar magnitudes is equal

\Delta m=-2^m.5\lg(F_V/F_M)\approx-5^m.3.

Stellar magnitude of Venus

m=-2^m.9+\Delta m\approx-8^m.2.

Of course, it is not the only possible correct way for solution.

Теми и трудност

Оценена трудност: 4/5.

Слънчева система · Светлина и астрономически измервания · Планетни конфигурации · Звездни величини

Предпоставки: Разширена олимпиадна подготовка

Оригинал в Архива: alphatheo.pdf · официални решения: theosol.pdf · допълнителни данни: IAO2008-data.pdf