Съдържание

IAO 2019, Теория, α+β — α-5. Interstellar comet.

Оригинал на условието (изображение)Съобщи за грешка

Всички материали от IAO 2019

XXIV International Astronomy Olympiad — Theoretical round, α and β · 19 октомври 2019 г. · 8 т.

For translation only.

Note for translators: in texts for Group β by the red color marked the text identical to that used for Group α.

Source availability

The three photographed sheets contain the combined English translator edition and the bilingual diagram for β-5. The archived data sheets contain Solar-System parameters, geographical coordinates, constants and the Hertzsprung–Russell diagram. The star-data table referred to in the key (including Ruchbah, Vega, Altair and UY Scuti) and the additional mid-round formula promised to Group α are not included in these archived theory/data sources. The surviving key supplies its own working values; no missing handout has been reconstructed.

Условие

2I/Borisov is the first observed interstellar comet. It was discovered on August 30 this year by the Crimean astronomer G. V. Borisov with a 65-cm telescope of his own development. The point on celestial sphere from where the comet moved before entering the Solar System is located near the star Ruchbah in the constellation Cassiopeia. On December 7, the comet will pass the perihelion of its orbit at a speed of 43 km/s relative to the Sun, being 2.01 au from it.

Original reference sheet: Solar-System parameters and geographical coordinates

Original reference sheet: constants and formulae

5.1.

Will it be possible to observe the star Ruchbah from Piatra Neamt tonight?

5.2.

Estimate how long ago the comet 2I/Borisov passed near the star Ruchbah.

5.3.

Calculate whether our Sun is visible with the naked eye from the vicinity of the Ruchbah star (write the answer in English, “Yes” or “No”).

By the way. The comet was discovered at the observational round place of the IV and VI IAO (1999, 2001), and G. V. Borisov was a member of the Organizing Committee of these IAOs.

Решение

5.1.

Sure, we can. The entire constellation of Cassiopeia is circumpolar in Piatra Neamt. To confirm, you can find the height of Ruchbah at the lower culmination:

h=\varphi-90°+\delta=17°10'.

Thus, the star is always above the horizon, it can be observed at any clearly night.

5.2.

According to the law of conservation of energy, the perihelion speed of the comet and its speed at infinity are related by the equation:

mV_P^2/2-GMm/R_P=mV_\infty^2/2,

V_\infty^2=V_P^2-2GM/R_P,

V_\infty=(V_P^2-2GM/R_P)^{1/2}=(43000^2-2\cdot6.67\cdot10^{-11}\cdot2\cdot10^{30}/(2.01\cdot1.5\cdot10^{11}))^{1/2}\approx31\ \mathrm{km/s}.

The comet makes almost its entire journey away from the stars. Therefore, most of the way (namely, relatively speaking, outside the Solar System) it traveled from the vicinity of the Ruchbah star to us with a speed $V_\infty\approx31\ \mathrm{km/s}$ relative to the Earth.

Let us find the distance to Ruchbah.

D=1/p=1/0.0328\approx30.5\ \text{Пк}.

It would seem that it is enough to calculate the speed $V$ in units of pc/year ( $3.2\cdot10^{-5}$ pc/year is obtained), divide 30.5 pc by this value and get an answer: approximately 960,000 years. Alas, we forgot that Ruchbah also moves relative to the Sun, as indicated by the radial velocity given in the table $V_r=-6.7\ \mathrm{km/s}$ (the star is approaching us). Thus, this star is approaching us, earlier it was at a farther distance and it took more time to overcome the path to the Solar system. Relative Ruchbah, the comet moves at a speed

V_1=V_\infty+V_r\approx24.3\ \mathrm{km/s},

Or in another units, about $3.2\cdot10^{-5}$ pc/year. Dividing 30.5 pc by this value, we get the answer: about 1.2 million years.

5.3.

As we previously found out, the distance to Ruchbah is:

D=1/p=1/0.0328\approx30.5\ \mathrm{pc}.

The sun at such a distance will have a magnitude

\begin{aligned}m_1&=m_0+5^{\mathrm m}\log(D/R_0)=-26.8^{\mathrm m}-5^{\mathrm m}\log(pR_0)\\ &=-26.8^{\mathrm m}+5^{\mathrm m}\log(206265/0.0328)\\ &=-26.8^{\mathrm m}+34.0^{\mathrm m}\approx7.2^{\mathrm m}.\end{aligned}

Невооружённым глазом можно видеть только звёзды $7.2^{\mathrm m}$ и ярче. Ответ: «No».

With the naked eye, only stars $7.2^{\mathrm m}$ and brighter can be seen. The answer is «No».

Basic criteria

5.1. 1 pt.

5.2.

Formula and calculation the velocity outside Solar System – 2 pt.
Formula and calculation the distance to Ruchbah – 1 pt.
Using the radial velocity of Ruchbah – 1 pt.
Calculation and the final answer – 1.5 pt.

Additionally:

Wrong number of significant digits in the answer – minus 0.5 pt.

5.3. 1.5 pt.

Theoretical round. Sketches for solutions. For jury job ONLY

Note. 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.

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 using these positions.

Theoretical round. Basic criteria. For work of Jury

Note. The given sketches of solutions are not full; the team leaders have to give more detailed explanations to students. The correct solutions in the students’ papers (enough for 8 pts) may be shorter.

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

Crossed-out source passage

A red handwritten cross on solution page 2 deletes the following duplicated block between the first estimate $\delta=0.5'$ and “The physical diameter of the central part of the cluster is”. It is recorded here rather than treated as an active step of the α solution:

“The physical diameter of the cluster is:

D=L\times\delta=50\ \text{кПк}\times0.5'/3438'\approx7.3\ \text{Пк}.

M_{\mathrm{A2}}-M=2.5^{\mathrm m}\times\lg N,\qquad N=10^{8.6/2.5}\approx2800.

By comparing the cluster size in the image with the total image size of $4'$ , we can find that the angular diameter of the central region of the ‘sphere’ is approximately $0.4'$ – $0.5'$ . We may take the value $\delta=0.5'$ for the solution.”

Printed-source discrepancies

The original key is retained literally, including its inconsistent numerical steps: the comet-radiation derivation prints $F=2F_p=2(E/c)\pi r^2$ ; the interstellar-comet key repeats $3.2\cdot10^{-5}$ pc/year for different speeds and ends with “only stars $7.2^{\mathrm m}$ and brighter” followed by “No”. The sunset key calls the Chukchi correction “sunrise” and the Eskimo correction “sunset”. The satellite question uses $\zeta$ , while the key uses $\xi$ , and its Fig. 3 is also called Figure 5.3. These are source discrepancies, not silently corrected calculations. The NGC rubric prints “The answer using the both criteria – pt.” without a numerical score; no value is invented.

Оригинал в Архива: theo1.JPG · официални решения: theosol.pdf · допълнителни данни: theo3.JPG · допълнителни данни: hrd.JPG · допълнителни данни: data_2.JPG · допълнителни данни: data_1.JPG · допълнителни данни: theo2.JPG