Съдържание

IAO 2025, Теория, Групи α, β и γ — Задача 5. Canopus and Sirius

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IAO 2025

XXIX IAO. IV International Remote Astronomy Olympiad. Dhaka, Bangladesh, 6-15. XI. 2025. Theoretical round. Problems to solve · 6 ноември 2025 г. · 8 т.

Внимание: Бележка към темата

The separate numbered exoplanet data table for problem 4, the table of constants and the Hertzsprung–Russell diagram referenced by the problems and solutions are not in the archive.

Reminding

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 are 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.

Условие

5.1. Calculate the most favorable time for observations of Canopus in Dhaka over the next 24 hours. Bangladesh time should be used. [3,5 т.]

5.2. Suppose that due to smog, Canopus will be visible in Dhaka only at the most favorable time for its observation, and at the limit of visibility to the naked eye. What magnitude will Sirius be visible at this time? [4,5 т.]

Отговори

Решение

5.1.

Canopus is visible in Dhaka, but not high above the horizon. Therefore, the best conditions for the observation should be at the time of its upper culmination. The upper culmination of a celestial body occurs when its sidereal time equals to its right ascension. It is $6^{\mathrm h}24^{\mathrm m}$ for Canopus. The sidereal time is equal to the mean solar time at the date of autumnal equinox, and then begins to outstrip it with speed $1^{\mathrm d}/365.2422$ per day (about $3^{\mathrm m}56^{\mathrm s}$ per day). 47–48 days passed since September 22 (autumnal equinox) till November 8–9 (the next 24 hours), let us use 47.5 days difference. The difference between the mean solar time and the sidereal time is now:

24^{\mathrm h}00^{\mathrm m}\times47.5/365.2422=3^{\mathrm h}07^{\mathrm m}.

Thus, now sidereal time $6^{\mathrm h}24^{\mathrm m}$ corresponds to

6^{\mathrm h}24^{\mathrm m}-3^{\mathrm h}07^{\mathrm m}=3^{\mathrm h}17^{\mathrm m}\quad\text{mean solar time}.

Dhaka is located about $0.5^\circ$ East from the $90^\circ$ meridian, so the mean solar time here is $0.5^\circ\times4^{\mathrm m}/{}^\circ=2^{\mathrm m}$ ahead of the local time UT+6 used in Bangladesh. So to get the answer for UT+6, you need to subtract these $2^{\mathrm m}$ , and we get

3^{\mathrm h}17^{\mathrm m}-0^{\mathrm h}02^{\mathrm m}=3^{\mathrm h}15^{\mathrm m}\quad\text{time of Bangladesh}.

Thus, there will be favorable conditions for observing Canopus tonight. Don't miss the chance!

5.2.

During the upper culmination, Canopus is at an altitude

h_c=90^\circ+\delta_c-\varphi=90^\circ-52^\circ42'-23^\circ48'=13^\circ30'.

At this time, Sirius is also located near the upper culmination (from the difference in right ascensions, it can be seen that its culmination will be only 21 minutes later), so its altitude is approximately equal to

h_s\approx90^\circ+\delta_s-\varphi=90^\circ-16^\circ43'-23^\circ48'=49^\circ29'.

If the light of celestial bodies from the zenith passes through a layer $d_0$ of polluted air, then the layer from the stars at an altitude of $h$ : $d_1=d_0/\sin h$ .

For Canopus, the layer is

d_c=d_0/\sin13^\circ30'=4.28h_0,

And for Sirius the layer is

d_s=d_0/\sin49^\circ29'=1.32h_0.

Under the most favorable conditions, a layer of one atmosphere weakens light by $0.23^{\mathrm m}$ , thus, the extratmospheric magnitude of Canopus is

-0.72^{\mathrm m}-0.23^{\mathrm m}=-0.95^{\mathrm m}.

We will assume that the limit of visibility by the naked eye is $6^{\mathrm m}$ . Thus, the layer $d_c=4.28h_0$ weakens the light by $6^{\mathrm m}-(-0.95^{\mathrm m})=6.95^{\mathrm m}$ . Accordingly, the layer $d_0$ weakens the light by $6.95^{\mathrm m}/4.28=1.62^{\mathrm m}$ , and a layer $d_s=1.32h_0$ weakens the light by $1.62^{\mathrm m}\cdot1.32=2.13^{\mathrm m}$ .

The extratmospheric magnitude of Sirius is

-1.46^{\mathrm m}-0.23^{\mathrm m}=-1.69^{\mathrm m}.

Thus, in the conditions of smog of the considered level, the apparent magnitude of Sirius will be

-1.69^{\mathrm m}+2.13^{\mathrm m}=0.44^{\mathrm m}.

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 use mostly formulas, calculations, drawings and graphs, writing texts is not allowed 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.

If you find typos, errors or missing points in the solution, please inform the organizing committee. Any suggestions for modifying the solutions are welcome, write justifications to gavrilovissp@list.ru and gavrilov@issp.ac.ru (to both addresses for reliability), if the suggestions are reasonable, the proposal can be approved.

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 and artistic picture, all the points in solution of problem 2 may be in only one common formula.) Of course, fractions of the mentioned points should be awarded if only part of the item was done.

Note. Jury members may 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. Excessive number of significant digits, especially in estimational problems, will lead to subtraction of the points (column “digits”, minus 1-2 points).

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

Any suggestions for modifying the criteria are welcome, write justifications to gavrilovissp@list.ru and gavrilov@issp.ac.ru (to both addresses for reliability), if the suggestions are reasonable, the proposal can be approved.

3. Comet Lemmon

		pts
3.1.	Actual semimajor axis of the orbit	1
	Actual orbital period	1
	Actual eccentricity of the orbit	1
3.2.	Steps to get new eccentricity of the orbit	2.5
	Actual eccentricity of the orbit	1
	Actual orbital period	1
	Final conclusion	0.5
	Total	8

4. On a remote exoplanet

		pts
	Thermal balance	0.8
	Formula for incoming power	0.5
	Formula for outgoing power	0.4
	Using albedo	0.4
	Using greenhouse effect	0.4
	Analyzing time from sunrise to sunset vs half a day, additional angular sizes	0.8
	Using dependence on the angular size of the central star	0.8
	Using any refraction	0.4
	Using refraction depending on atmospheric pressure	1.1
	Using spectral class of the Sun or temperature of the Sun	0.4
	Final formula and/or correct calculations	0.8
	Final correct result in a range K5-K9	0.8
	Postulating “Not depending on [#1, 2, 3]”	0.4
	Total	8

5. Canopus and Sirius

		pts
5.1.	Understanding about upper culmination	0.5
	Using right ascension of Canopus	1
	Using difference in days since autumnal equinox	0.5
	Using Dhaka is located about 0.5° East from the 90° meridian	0.5
	Final formula and calculations	1
5.2.	Altitude of Canopus at upper culmination	0.6
	Altitude of Sirius at the time of upper culmination of Canopus	0.6
	Extinction in atmosphere	1
	Using limit of visibility by the naked eye	0.4
	Calculation the layers of atmosphere for Canopus and Sirius	0.8
	Final formula and/or correct calculations, final answer	1.1
	Total	8

Оригинал в Архива: problems_t_345_en.pdf · официални решения: solutions_t_en.pdf · допълнителни данни: criteria_t_en.pdf