IAO 2023, Теория — Задача 4 (αβγ). Гибридное затмение / Hybrid eclipse
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
Гибридным называется такое солнечное затмение, которое в одних местах на Земле видно как кольцеобразное, а в других – как полное. 20 апреля этого года наблюдалось одно из таких затмений. Его полоса прошла через Индийский океан, Австралию, Восточный Тимор, Индонезию и Тихий океан. На отдельном листе приведена карта-схема полосы затмения (масштабы не соблюдены). Звёздочками отмечены: 1 – приблизительная точка перехода от кольцеобразного затмения к полному, высота Солнца над горизонтом в это время составляла 10°, 2 – точка наблюдения затмения с максимальной фазой, высота Солнца над горизонтом – 67°, 3 – приблизительная точка перехода от полного затмения к кольцеобразному, высота Солнца над горизонтом – 10°. Вычислите максимальную ширину полосы полной фазы затмения.
English
A hybrid eclipse is a solar eclipse that in some places on Earth is seen as annular, and in some others as total. On April 20 this year one of these eclipses was observed. Its band passed through the Indian Ocean, Australia, East Timor, Indonesia and the Pacific Ocean. A separate sheet shows a map-scheme of the path of the eclipse (not to scale). The asterisks are marked: 1 – the approximate transition point from annular eclipse to total, the altitude of the Sun above the horizon at that time was 10°, 2 – the observation point of the eclipse with the maximum phase, the altitude of the Sun was 67°, and 3 – the approximate transition point from total eclipse to an annular, the altitude of the Sun was 10°. Calculate the maximum width of the path of the totality.
Справочные данные / Reference data
**) Для внешних планет и Луны – в среднем противостоянии.
**) For outer planets and Moon – in mean opposition.
Map 4
Решение
(Drawing is desirable). At points 1 and 3, where the annular eclipse changed to the total and the total to the annular, the apparent angular dimensions of the Moon and the Sun coincide. These points are on the tip of the cone of total eclipse. Compared to the center of the Earth, these points are closer to the Moon by
(here we took into account the almost central passage of the cone around the globe, the cosine of the corresponding angle of deviation is almost equal to 1). Point 2, where a total eclipse with a maximum phase is observed, is closer to the Moon (also compared to the center of the Earth) by
The difference between these distances is
The width of the cone of the Moon's shadow at this distance is
The axis of the cone come to the Earth's surface at an angle of 67° to the normal and almost perpendicular to the trajectory of the shadow, so to get the width of the path, you just need to divide the diameter of the cone by the sine of the angle of the axis of the cone to the normal.
and, taking into account a low accuracy of our calculations,
Note. In reality, the maximum width of the path of this totality was 49 km.
Basic criteria
| Criterion | pts |
|---|---|
| General understanding | 1 |
| Calculation differences relative Earth center, its difference | 1.5 |
| Cone angle | 1.5 |
| Width of the cone | 1.5 |
| Width of the path | 1.5 |
| Final calculations and answer | 1 |
| Total | 8 |
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.
Слънчева система · Затъмнения и покрития
Предпоставки: Разширена олимпиадна подготовка
Оригинал в Архива: theo.pdf · официални решения: theosol.pdf · допълнителни данни: data.pdf