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IAO 2025, Теория, Групи α, β и γ — Задача 3. Comet Lemmon

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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

Today, exactly during the theoretical round (at the end of the round for participants from the Eastern Hemisphere and at the beginning for the Western Hemisphere), comet C/2025 A6 (Lemmon) passes the perihelion of its orbit at a distance of p = 0.5299 au from the Sun. In its previous approach to the Sun, the comet was observed in the second half of the 7th century AD.

3.1. Estimate the speed of the comet at the given moment. [3 т.]

3.2. During the passage of the inner regions of the Solar System, a slight deceleration of the comet occurred, equivalent to the fact that at the perihelion point, the comet instantly lost δ = 0,023 % of its velocity. In what century will the inhabitants of the Earth see the comet during its next visit to the inner regions of the Solar System? [5 т.]

Отговори

Решение

If the comet was observed in the second half of the 7th century AD during its previous approach to the Sun, then its orbital period in the previous period can be estimated as $1350\pm25$ years. According to Kepler's Third Law, such a period corresponds to the semimajor axis of the orbit

a=T^{2/3}=122\pm1.5\,\mathrm{au},

The eccentricity of the orbit:

e=(a-p)/a=0.99566\pm0.00005.

3.1.

Such a small difference in eccentricity from unity indicates that the comet is moving along an almost parabolic trajectory. The velocity of the comet at perihelion (i.e., the speed at the moment) is almost equal to the II cosmic one:

V_{II}=\sqrt{\frac{2GM_\odot}{p}}=\sqrt{\frac{2\cdot6.67\cdot10^{-11}\cdot1.99\cdot10^{30}}{0.53\cdot1.496\cdot10^{11}}}=5.79\cdot10^4\,\mathrm{m/s}=57.9\,\mathrm{km/s}.

To calculate, we converted astronomical units to meters. The result is a value in meters per second. Thus, the comet's perihelion velocity $V_p\approx V_{II}=57.9\,\mathrm{km/s}\approx58\,\mathrm{km/s}$ .

3.1.* A more accurate result can be obtained by applying the law of conservation of energy and Kepler's Second Law for the points of perihelion and aphelion of the orbit of the comet:

\frac{V_p^2}{2}-\frac{GM_\odot}{p}=\frac{V_a^2}{2}-\frac{GM_\odot}{a},\qquad V_p p=V_a a.

where $G$ is the gravitational constant, $M_\odot$ is the mass of the Sun, $p$ and $a$ the distances to the Sun at perihelion and aphelion (which are equal to the radii of the orbits of the Earth and Saturn) and $V_p$ and $V_a$ are the corresponding velocities.

Solving these two equations together, we get:

V_p=\sqrt{\frac{2GM_\odot}{p(p/a+1)}}=\sqrt{\frac{2GM_\odot}{p((1-e)+1)}}\approx57.8\,\mathrm{km/s}.

As you can see, the answer is only 0.2% different from the one obtained for the parabolic velocity and is completely unnecessary for the estimation solution. However, this solution is absolutely correct.

A note to the jury. The part of the text highlighted in blue (p.1.1*.) is not required for the solution of p.1.1., but may be useful for one of alternatives of the solutions to p.1.2.

3.2.

But in this part of the problem, high accuracy is required, which was not important in clause 1.1. Using the formula obtained earlier

V_p=\sqrt{\frac{2GM_\odot}{p(p/a+1)}}=\sqrt{\frac{2GM_\odot}{p(2-e)}},

we can get

2-e=\frac{2GM_\odot}{pV_p^2}.

for small variations

\frac{\Delta(2-e)}{2-e}=\Delta\left(\frac{2GM_\odot}{pV_p^2}\right)\bigg/\frac{2GM_\odot}{pV_p^2},

\frac{-\Delta e}{2-e}=\Delta\left(\frac{1}{V_p^2}\right)\bigg/\frac{1}{V_p^2}=\frac{-2\Delta V_p}{V_p}.

since $\Delta e$ is significantly less than $(2-e)$ , and $(2-e)\approx1$ , we obtain

\Delta e\approx\frac{2\Delta V_p}{V_p}=-2\delta\approx-0.046\%.

The new eccentricity is:

e_1=(a_1-p)/a_1=e\cdot(1-0{,}00046)=0{,}9952,\qquad1-e_1=0{,}0048.

The new semi-major axis of the orbit of the comet:

a_1=p/(1-e_1)=0{,}5299\,\mathrm{au}/0{,}0048\approx110\,\mathrm{au}.

The new orbital period:

T_1=a_1^{3/2}\approx1160\text{ years}.

$2025+1160=3185$ . Thus, the next time inhabitants of the Earth will see comet C/2025 A6 (Lemmon) will be at the end of the 32th century (*). This is two centuries earlier than if the comet had not lost 0.023% of its velocity at perihelion.

(*) Provided that there are no major changes in the orbit in the outer part of the trajectory of the comet. There, even a minor affect from massive objects can lead to a significant change in trajectory, up to ejection of the comet outside the Solar System.

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