Съдържание

Условие Русский English Примечание Note Справочные данные / Reference data Решение

IAO 2023, Теория — Задача 6 (βγ). Сфера Дайсона / Dyson sphere

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IAO 2023

XXVII International Astronomy Olympiad — Theoretical round / Задачи теоретического тура · 8 т.

Только для перевода. / For translation only.

Напоминание. Большинство задач составлены таким образом, что при написании решений от участника требуется минимальное использование какого-либо языка. Таким образом, допускается минимум текстов в решениях задач (ни на родном языке, ни на одном из официальных языков IAO). Уделяйте больше внимания формулам, рисункам и графикам. Разрешается употреблять несколько ключевых слов на английском языке (например, “Yes”, “No”, “Assume”, “Answer”, “Average”, “Minimum”, “Maximum”, “Impossible situation” и т.д.), а также термины, содержащиеся в прилагаемых таблицах, и названия объектов.

Reminding. Most of the problems are compiled in such a way that when writing solutions, minimal use of any language is required from the participant. So minimum texts is allowed in solutions of the problems (neither in the native language, nor in one of the official languages of the IAO). Pay more attention to formulas, drawings and graphs. It is permitted to write a few keywords in English (for example, “Yes”, “No”, “Assume”, “Answer”, “Average”, “Minimum”, “Maximum”, “Impossible situation”, etc.), as well as terms contained in the accompanying tables and names of objects.

Source notes / Примечания к источнику

The scan repeats the Russian second statement page. Russian and English versions are retained as sections of the same seven tasks. The note after 6.6 prints 8.3/8.5/8.6; the English scan has a handwritten correction. The solution’s final magnitude for task 2 prints +5 rather than −5, the eclipse expression prints L₁−L₂, and the redshift relation prints a(tₑ)/a(tₒ); these source expressions are preserved. The criteria for 6.5/6.6 have unrelated white-dwarf text crossed out without replacement. Reference data are also present within the question PDF and are not extra tasks.

Условие

Русский

Далёкое будущее. Земляне построили сферу Дайсона. Её размер таков, что, что температурные условия для её жителей аналогичны нынешним наиболее благоприятным климатическим условиям на Земле. Для строительства использовали материал планеты Земля. Построенная сфера не вращается.

6.1. Найдите радиус этой сферы Дайсона $R_D$ .

6.2. Найдите получившуюся толщину сферы Дайсона $d_D$ .

6.3. Найдите ускорение свободного падения на сфере Дайсона $g_D$ . С ускорением свободного падения на какой планете или объекте Солнечной системы можно сравнить это ускорение?

6.4. Для того, чтобы у жителей внутренней стороны сферы сменялись бы светлое и тёмное время суток с привычной для нас периодичностью, внутри сферы Дайсона вокруг Солнца была создана непрозрачная полусфера, вращающаяся вокруг Солнца. (Пояснение для понимания, как вращается полусфера: её вращение можно сравнить с вращением поверхности восточного полушария Земли вокруг оси вращения Земли). Вычислите, каков должен быть радиус этой полусферы $R_F$ , чтобы она могла обращаться вокруг Солнца только за счёт гравитационного взаимодействия с ним.

6.5. Климатические условия на каких планетах значительно изменятся после строительства сферы Дайсона?

6.6. Климатические условия на каких планетах после строительства сферы Дайсона изменятся, но незначительно?

English

The distant future. The Earthlings built a Dyson sphere. Its size is such that the temperature conditions for its inhabitants are similar to the current most favorable climatic conditions on Earth. The material of the planet Earth was used for its construction. The constructed sphere does not rotate.

6.1. Find the radius of this Dyson sphere $R_D$ .

6.2. Find the resulting thickness of the Dyson sphere $d_D$ .

6.3. Find the acceleration of gravity on the Dyson sphere $g_D$ . With the acceleration of gravity on which planet or object of the Solar system can this acceleration be compared?

6.4. In order for the inhabitants of the inner side of the sphere to change the light and dark time of day with the frequency familiar to us, an opaque hemisphere rotating around the Sun was created inside the Dyson sphere around the Sun. (Explanation for understanding how the hemisphere rotates: its rotation can be compared with the rotation of the surface of the eastern hemisphere of the Earth around the axis of rotation of the Earth). Calculate what the radius $R_F$ of this hemisphere should be so that it can revolve around the Sun only due to gravitational interaction with it.

6.5. Which planets’ climatic conditions will change significantly after the construction of the Dyson sphere?

6.6. Which planets’ climatic conditions will change, but only slightly after the construction of the Dyson sphere?

Примечание

В п. 8.3, 8.5, 8.6. напишите название или категорию объекта латиницей или по-английски.

Note

In paragraphs 8.3, 8.5, 8.6. write the name or category of the object in Latin or English.

[Handwritten above this line: 6.3 6.5 6.6]

Справочные данные / Reference data

Complete bilingual planetary orbital and physical data table

Complete bilingual constants and formulae reference sheet

**) Для внешних планет и Луны – в среднем противостоянии.

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

Решение

6.1.

The radius of the Dyson sphere $R_D$ is found from the thermal equilibrium condition. All the energy coming from the Sun ( $R_S=696\,000$ km, $T_S=5777$ K) will eventually be re-emitted by the sphere.

The temperature on the sphere should correspond to comfortable conditions for earthlings. Take $T_D=293$ K.

\sigma T_S^4\cdot4\pi R_S^2=\sigma T_D^4\cdot4\pi R_D^2\quad\Rightarrow\quad R_D=R_S\frac{T_S^2}{T_D^2}=271\ \text{млн. км}\approx1{,}8\ \text{а.е.}

6.2.

If the construction of the sphere took all the substance of the Earth, then equate their volumes:

4\pi R_D^2d_D=\frac43\pi R_E^3\quad\Rightarrow\quad d_D=\frac{R_E^3}{3R_D^2}=1.2\ \text{мм.}

A very thin sphere.

6.3.

g_D=GM/R_D^2\approx1.8\ \text{мм}/\text{с}^2

Asteroid (minor planet).

6.4.

In order for day and night to alternate, as on Earth, the hemisphere must orbit the Sun with a period $T=1$ сутки $=24$ часа $=86400$ с. Let's find the force of attraction of the hemisphere to the Sun. Let's denote the plane bounding the hemisphere as XZ, and the axis perpendicular to it as Y. Obviously, the total force of attraction will be directed along the Y axis. Consider the attraction of a single section of the hemisphere with mass $\Delta m=\rho_Fd_F\Delta S$ . In the projection on the Y axis, the force of attraction will be equal to

\Delta F=\frac{GM\Delta m}{R_F^2}\cos\varphi,

where φ is the angle from the Sun between the direction to this unit area and the Y axis. At the same time, the projection of this section onto the XZ plane is also proportional to cos φ. It turns out that the sections having the same projections on the XZ axis give the same contribution to the force of attraction. As a result, if the entire mass of the hemisphere is proportional to the area of the hemisphere $2\pi R_F^2$ and is equal to $m=\rho_Fd_F\cdot2\pi R_F^2$ , then the total force of attraction is proportional to the projection area of $\pi R_F^2$ and is equals to

F=\frac{GM\rho_Fd_F\cdot\pi R_F^2}{R_F^2}=\frac{GMm}{2R_F^2}.

The equation of the circular motion of the hemisphere

\sum m_i\omega^2R_i=\frac{GMm}{2R_F^2}.

Here, for each $m_i$ , you also need to count its contribution, and if you calculate the integral correctly

\sum m_i\omega^2R_i=\frac\pi4m\omega^2R.

\frac\pi4m\omega^2R_F=\frac{GMm}{2R_F^2},\quad\text{откуда}\quad R_F=\sqrt[3]{\frac{2GM}{\pi\omega^2}}=\sqrt[3]{\frac{GMT^2}{2\pi^3}}.

Calculations give 2.5 million km. (≈ 1/60 au).

But let's make a cylindrical approximation in this case and write

m\omega^2R_F=\frac{GMm}{2R_F^2},\quad\text{откуда}\quad R_F=\sqrt[3]{\frac{GM}{2\omega^2}}=\sqrt[3]{\frac{GMT^2}{8\pi^2}}.

Calculations give 2.3 million km. (≈ 1/65 au).

And if in the first case we considered not the projections of gravity, but the material point at a distance of $R_F$ , we would get a well-known expression

m\omega^2R_F=\frac{GMm}{R_F^2},\quad\text{откуда}\quad R_F=\sqrt[3]{\frac{GM}{\omega^2}}=\sqrt[3]{\frac{GMT^2}{4\pi^2}}.

Calculations give ~3 million km. (≈ 1/50 au), this is too rough.

6.5–6.

The orbits of Mercury, Venus and Mars (and the former orbit of the Earth, the substance of which was used in the construction of the sphere) will be completely inside the Dyson sphere. Due to the additional (compared to the current situation) energy flows circulating inside the Dyson sphere, climatic conditions will change on all planets inside the sphere. But if Mercury and Venus were already quite hot, then Mars will warm up more, because the temperature inside the sphere will not be lower than on its surface. Mars will probably become habitable.

6.5. Mars.

6.6. Mercury, Venus.

Basic criteria

Criterion	pts
6.1. Energy balance	0.5
Formulae and calculation for ratio $R_D$	1
6.2. Formulae and calculation for ratio $d_D$	1
6.3. Formulae and calculation for ratio $g_D$	1
6.4. Estimation formulae and calculations	2
Accurate formulae and calculations	+1
6.5. ~~Finding diameter of white dwarf from the mass-radius dependence~~	1
6.6. ~~Finding diameter of white dwarf from the mass-radius dependence~~	0.5
Total	8

The two printed white-dwarf rows are crossed out in the source; no replacement wording is supplied.

Theoretical round. Sketches for solutions. For jury job ONLY

These solutions are written for jury members, so they contain a lot of explanations in the form of text. Students are instructed to minimize text, mostly use formulas, calculations, drawings and graphs, writing extra tests is not encouraged in the solutions. It is permitted to write a few keywords in English (for example, “Yes”, “No”, “Assume”, “Answer”, “Average”, “Minimum”, “Maximum”, “Impossible situation”, etc.), as well as terms contained in the accompanying tables and names of objects.

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.

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 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 use these positions. (For example, except the final conclusion, all the points in solution of problem 2 may be in only one common formula.)

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

Note. Everywhere “formula” means “correct formula”, “calculation” means “correct calculation”, “picture” means “correct picture”, etc., and in all cases: based on correct facts and intermediate formulae or calculations.

Note. Everywhere final calculation means also using correct number of significant digits. Extra significant digits, especially in estimational problems, will lead to subtraction of the points.

Note. Solutions with absurd or almost absurd answers should be rated lower (maybe ignoring the scoring schemes below, and maybe down to zero).

Everywhere – excessive number of significant digits in the answer – minus 1–2 points.

Теми и трудност

Трудност: неоценена.

Светлина и астрономически измервания · Небесна механика · Абсолютно черно тяло · Гравитационни системи

Предпоставки: Разширена олимпиадна подготовка

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