Съдържание

IAO 2019, Теория, α+β — β-5. Two satellites.

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

Two artificial satellites move around an unknown planet in circular orbits. The satellite $S_1$ moves in an equatorial orbit with the angular velocity $\omega_1$ , and the satellite $S_2$ moves in a polar orbit with the angular velocity $\omega_2$ , as shown in the Fig.5. The planet rotates around its axis with the angular velocity $\omega_0$ . At the initial time, the satellites are in zenith for observer O.

Fig.5. Equatorial and polar satellite orbits

5.1.

It is known that for observer O, the duration of the visibility (from the zenith to setting) of satellite $S_2$ is $t_2$ . Find the similar visibility duration $t_1$ of satellite $S_1$ .

5.2.

Find the azimuth $\zeta$ of the satellite $S_2$ setting point (the point of its “disappearance” for the observer). Consider the South direction as the zero azimuth point (0°).

Решение

5.1.

1) Satellite visibility $S_1$

According to the notations in Figure 5.2, results:

Fig.5.2. Equatorial satellite visibility

\alpha_1=\omega_1t_1;\quad\omega_1=\sqrt{\frac{GM_P}{R_1^3}};\quad R_1=\sqrt[3]{\frac{GM_P}{\omega_1^2}};

\alpha_{01}=\omega_0t_1;\quad\cos\beta_1=\frac{R_P}{R_1};\quad\beta_1=\arccos\left(\frac{R_P}{R_1}\right)=\arccos\left(R_P\sqrt[3]{\frac{\omega_1^2}{GM_P}}\right);

\alpha_1=\alpha_{01}+\beta_1;\qquad t_1(\omega_1-\omega_0)=\beta_1;

t_1=\frac{\beta_1}{\omega_1-\omega_0};\qquad t_1=\frac{\arccos\left(R_P\sqrt[3]{\omega_1^2/(GM_P)}\right)}{\omega_1-\omega_0},

being the duration of the satellite's visibility $S_1$ .

2) Satellite visibility $S_2$

According to the notations in Figure 3, at the beginning (initial moment), the observer, located in $O_0$ , has direct visibility to the satellite in $S_0$ . From this moment, due to Planet's rotation, the observer on Planet's equator has visual acces to the left space of the plane $\Pi_P$ , tangent plane on Planet's surface at the point on the equator where the observer is. If the satellite is in this space, of course it will be visible. While the $\Pi_P$ plane rotates, being solidary with the observer on the Planet's equator, the satellite „get down” into its orbit in $\Pi_2$ plane.

The two plans, $\Pi_P$ and $\Pi_2$ , intersects, the direction of this intersection (AD direction from Figure 3) are moving, being constantly parallel to the Planet's axis rotation. The dihedral angle between the two planes first decreases from the 90°, corresponding to the initial moment, when the two planes are perpendicular, to the final value 0°, when the two planes become parallel, so that this angle rises again until the two planes are again perpendicular, etc.

At some point in its evolution, the satellite disappears from the observer's field of view. This occurs when the observer is in $O_2$ position, and the satellite is in $S_2$ position, located on the intersection axis of the two planes.

The $t_2$ represents the time in which $S_2$ satellite are visible, when satellite comes out from the observer's field of view, which is in $O_2$ position, and passes behind the $\Pi_P$ plane, tangent to the Planet's surface in the point on the equator where the observer is located, but remaining in the plane of its orbit, $\Pi_2$ .

Under these circumstances, from rectangular triangles $ACS_2$ and $ACO_2$ , results:

AC=R_2\cos\alpha_2;\qquad AC=\frac{R_P}{\cos\alpha_{02}};

R_2\cos\alpha_2=\frac{R_P}{\cos\alpha_{02}};

\cos\alpha_2\cdot\cos\alpha_{02}=\frac{R_P}{R_2};\quad\alpha_2=\omega_2t_2;\quad\alpha_{02}=\omega_0t_2;

\omega_2=\sqrt{\frac{GM_P}{R_2^3}};\quad R_2=\sqrt[3]{\frac{GM_P}{\omega_2^2}};

\cos\omega_2t_2\cdot\cos\omega_0t_2=\frac{R_P}{R_2},

\cos\omega_2t_2\cdot\cos\omega_0t_2=R_P\sqrt[3]{\frac{\omega_2^2}{GM_P}},

where $t_2$ is the duration of the satellite's visibility $S_2$ ;

R_P=\sqrt[3]{\frac{GM_P}{\omega_2^2}}\cdot\cos\omega_2t_2\cdot\cos\omega_0t_2.

Results:

t_1=\frac{\arccos\left(R_P\sqrt[3]{\omega_1^2/(GM_P)}\right)}{\omega_1-\omega_0};

t_1=\frac{\arccos\left(\sqrt[3]{\frac{\omega_1^2}{GM_P}}\cdot\sqrt[3]{\frac{GM_P}{\omega_2^2}}\cdot\cos\omega_2t_2\cdot\cos\omega_0t_2\right)}{\omega_1-\omega_0};

t_1=\frac{\arccos\left(\sqrt[3]{\frac{\omega_1^2}{\omega_2^2}}\cdot\cos\omega_2t_2\cdot\cos\omega_0t_2\right)}{\omega_1-\omega_0},

being the duration of the satellite's visibility $S_1$ .

Fig.3. Polar satellite visibility, tangent plane and triangles

5.2.

From rectangular triangles $AO_2S_2$ , $CO_2S_2$ and $ACO_2$ , in Figure 5.3, results:

\cos\varepsilon=\frac{AO_2}{O_2S_2};\quad\sin\beta_2=\frac{O_2S_2}{R_2};\quad\tan\alpha_{02}=\frac{AO_2}{R_P};

O_2S_2=R_2\sin\beta_2;\qquad AO_2=R_P\tan\alpha_{02};

\cos\varepsilon=\frac{AO_2}{O_2S_2};\qquad\cos\varepsilon=\frac{R_P\tan\alpha_{02}}{R_2\sin\beta_2};

O_2S_2=\sqrt{(CS_2)^2-(CO_2)^2}=\sqrt{R_2^2-R_P^2}=R_2\sqrt{1-\left(\frac{R_P}{R_2}\right)^2};

\sin\beta_2=\frac{O_2S_2}{R_2};\quad\sin\beta_2=\sqrt{1-\left(\frac{R_P}{R_2}\right)^2};\quad\alpha_{02}=\omega_0t_2;

\cos\varepsilon=\frac{R_P\tan\alpha_{02}}{R_2\sin\beta_2};\quad\alpha_{02}=\omega_0t_2;

\cos\varepsilon=\frac{R_P}{R_2}\cdot\frac{\tan\omega_0t_2}{\sqrt{1-(R_P/R_2)^2}};

R_P=\sqrt[3]{\frac{GM_P}{\omega_2^2}}\cdot\cos\omega_2t_2\cdot\cos\omega_0t_2;

\frac{R_P}{\sqrt[3]{GM_P}}=\frac1{\sqrt[3]{\omega_2^2}}\cdot\cos\omega_2t_2\cdot\cos\omega_0t_2;

R_2=\sqrt[3]{\frac{GM_P}{\omega_2^2}};\quad\sqrt[3]{GM_P}=R_2\cdot\sqrt[3]{\omega_2^2};

\frac{R_P}{R_2\sqrt[3]{\omega_2^2}}=\frac1{\sqrt[3]{\omega_2^2}}\cdot\cos\omega_2t_2\cdot\cos\omega_0t_2;

\frac{R_P}{R_2}=\cos\omega_2t_2\cdot\cos\omega_0t_2;

\cos\varepsilon=\frac{R_P}{R_2}\cdot\frac{\tan\omega_0t_2}{\sqrt{1-(R_P/R_2)^2}};

\cos\varepsilon=\frac{\cos\omega_2t_2\cdot\cos\omega_0t_2\cdot\tan\omega_0t_2}{\sqrt{1-(\cos\omega_2t_2\cdot\cos\omega_0t_2)^2}};

\varepsilon=90°-\xi,

where $\xi$ is the azimuth of the satellite when it disappears from the field of view of the observer;

\cos(90°-\xi)=\frac{\cos\omega_2t_2\cdot\cos\omega_0t_2\cdot\tan\omega_0t_2}{\sqrt{1-(\cos\omega_2t_2\cdot\cos\omega_0t_2)^2}};

\sin\xi=\frac{\cos\omega_2t_2\cdot\cos\omega_0t_2\cdot\tan\omega_0t_2}{\sqrt{1-(\cos\omega_2t_2\cdot\cos\omega_0t_2)^2}}.

Basic criteria

5.1.

Formula for $t_1$ dependence on $\omega_1,\omega_0,R_P,G,M_P$ – 2 pt.
Formula $R_P$ dependence on $t_2,\omega_2,\omega_0,G,M_P$ – 2 pt.
Formula $t_1$ dependence on $t_2,\omega_2,\omega_1,\omega_0$ – 1 pt.

5.2. up to 3 pt (depending on progress in solution).

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