IAO 2007, Теория, Група α, английски — Задача 2. Sidereal period.
Theoretical round. Problems to solve · 29 септември 2007 г.
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): «situation is impossible – ситуация невозможна». Of course, this answer has to be explained numerically or logically.
Source availability
The numerical table of planetary data is not present in the32-file2007 archive inventory. The file bgtheo6.jpg contains only a translation of column headings, without numerical planetary rows. No table from another year has been substituted. The separately archived airplane/Moon photograph is included in full with question4.
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 for8pts) may be shorter.
Source photograph damage
The second and fourth official solution photographs lose some initial letters or words at the left edge. Ellipses inside the affected solution text mark those source losses. Legible formulas, results and grading remarks are retained; missing letters have not been silently supplied. The solution for question4 contains no explanatory drawing, although the question requests one.
Printed source annotations
The supplied key’s Moon magnitude calculations and approximate values are retained. The source results and formulas are reproduced without scientific correction; punctuation and multiplication spacing are normalized only for mathematical rendering.
Условие
Estimate (at least roughly) the smallest and the largest possible orbital sidereal periods of bodies (any bodies, including artificial ones) of our Solar System.
Решение
Внимание: Непълно решение
The official solution photograph clips some initial letters/words at the left edge. All legible content is retained and ellipses mark the genuine source loss.
The smallest sidereal period is a period of rotation of a body around a body of the Solar System with the largest mean density inside the sphere of orbit,
It is known that the densest bodies in our Solar System are located not far from the Sun. They are planets of the Earth group. We may find from the table of Solar System that the densest bodies are Earth and Mercury. The Earth has a little larger density but the orbit cannot lie near the planet surface. And the orbit around Mercury can lie just near the planet surface. As a result, the mean density inside the sphere of orbit around the Earth is smaller than the mean density inside the sphere of orbit around Mercury that is equal to the density of substance, i.e. 5430 kg/m³.
…ote for jury. Some students will take into account only bodies rotating around the Sun. 1 pt difference recommended in the …valuation of a solutionusing such a way, the answer for the minimal T for bodies rotating around the Sun is about …0000 sec = 167 min.
…ote for jury. It seems in this part of the problem that some students may find solutions that are unexpected for the jury (and …orrect!) solutions. Keep attention.
The largest sidereal period is a period of rotation around the Sun of a body with the maximum possible semiaxis. We may propose roughly that the largest possible semiaxis is less than a half distance to the nearest stars, i.e. about 0.6 pc ≈ 120 000 a.u.
…ote for jury. The estimation is very rough. The semiaxis used in solutions may be in the interval 100 000 – 200 000 a.u., so the …orrect answers are 30 – 90 mln. years.
Data
Data from the table of planetary data may be used for solving every problem.
Оригинал в Архива: alphatheo.jpg · официални решения: theosol1.jpg · допълнителни данни: theosol2.jpg · допълнителни данни: theosol3.jpg · допълнителни данни: theosol4.jpg · допълнителни данни: photo.JPG