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IOAA 2013, Теория, дълги задачи — Задача 2

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$7^{th}$ International Olympiad on Astronomy & Astrophysics · 27 July – 5 August 2013, Volos Greece · Theoretical Exam - Long Questions · 27 юли 2013 г.

Условие

Question 2

A spacecraft is orbiting the Near Earth Asteroid (2608) Seneca (staying continuously very close to the asteroid), transmitting pulsed data to the Earth. Due to the relative motion of the two bodies (the asteroid and the Earth) around the Sun, the time it takes for a pulse to arrive at the ground station varies approximately between 2 and 39 minutes. The orbits of the Earth and Seneca are coplanar. Assuming that the Earth moves around the Sun on a circular orbit (with radius $a_{\mathrm{Earth}}=1\ \mathrm{AU}$ and period $T_{\mathrm{Earth}}=1\ \mathrm{yr}$ ) and that the orbit of Seneca does not intersect the orbit of the Earth, calculate:

(1) the semi-major axis, $a_{\mathrm{Sen}}$ the eccentricity, $e_{\mathrm{Sen}}$ of Seneca’s orbit around the Sun

(2) the period of Seneca’s orbit, $T_{\mathrm{Sen}}$ and the average period between two consecutive oppositions, $T_{\mathrm{syn}}$ of the Earth-Seneca couple

(3) an approximate value for the mass of the planet Jupiter, $M_{\mathrm{Jup}}$ (assuming this is the only planet of our Solar system with non-negligible mass compared to the Sun). Assume that the presence of Jupiter does not influence the orbit of Seneca.

Решение

Answer :

(1) For $\Delta t_b=2\ \mathrm{min}=120\ \mathrm{sec}$ , the distance travelled by a light pulse is $R_1=c\times\Delta t$ or $R_1=0.24\ \mathrm{AU}$ , while for $\Delta t_a=39\ \mathrm{min}$ the maximum distance is $R_2=c\times\Delta t$ or $R_2=4.67\ \mathrm{AU}$ .

(4 points)

Since the orbits do not intersect and the $R_2$ exceeds by far 1 AU, the orbit of Seneca is exterior to that of the Earth. $R_1$ corresponds to the minimum relative distance of the two bodies (i.e. at opposition), while $R_2$ corresponds to the maximum relative distance (i.e. at conjunction).

If $q=a\times(1-e)$ is the perihelion and $Q=a\times(1+e)$ the aphelion distance of Seneca, then $R_1=q-1\ \mathrm{AU}$ (the minimum distance of Seneca from the Sun minus the semi-major axis of the Earth’s orbit), while $R_2=Q+1\ \mathrm{AU}$ (the maximum distance of Seneca from the Sun plus the semi-major axis of the Earth’s orbit).

(6 points)

Thus,

a_{\mathrm{Sen}}(1-e_{\mathrm{Sen}})=1+R_1=1.24\ \mathrm{AU}

a_{\mathrm{Sen}}(1+e_{\mathrm{Sen}})=R_2-1=3.67\ \mathrm{AU}

from which one finds $a_{\mathrm{Sen}}\sim2.46\ \mathrm{AU}$

(2 point)

and $e_{\mathrm{Sen}}\sim0.49$

(2 points)

(figures rounded to 2 dec. digits)

(2) The period, $T_{\mathrm{Sen}}$ , of Seneca’s orbit can be found by using Kepler’s 3rd law for Seneca and the Earth (ignoring their small masses, compared to the Sun’s):

\frac{a_{\mathrm{Sen}}^3}{T_{\mathrm{Sen}}^2}=\frac{a_{\mathrm{Earth}}^3}{T_{\mathrm{Earth}}^2}=1

(6 points)

from which one finds $T_{\mathrm{Sen}}\sim3.87\ \mathrm{yr}$

(2 point)

(Alternatively one can use Kepler’s law for Seneca only and use natural units for the mass of the Sun and $G$ , i.e. $[\mathrm{mass}]=1\ M_{\mathrm{Sun}}$ , $[t]=1\ \mathrm{yr}$ and $[r]=1\ \mathrm{AU}$ , in which case $T_{\mathrm{Sen}}=a_{\mathrm{Sen}}^{3/2}=3.86\ \mathrm{yr}$ )

(8 points)

Assuming non-retrogate orbit, the synodic period, $T_{\mathrm{syn}}$ of Seneca and Earth is given by

1/T_{\mathrm{syn}} = 1/T_{\mathrm{Earh}} - 1/T_{\mathrm{Sen}} \quad (\textit{Seneca} \text{ is superior})

(6 points)

which gives $T_{\mathrm{syn}} = 1.35\ \mathrm{yr}$

(2 point)

(3) From Kepler's law, we can calculate the mass of the $\textit{Sun}$

4\pi^2 a_{\mathrm{Sen}}^3/T_{\mathrm{Sen}}^2 = G M_{\mathrm{Sun}},\ \text{i.e. } M_{\mathrm{Sun}} \sim 1.988\times10^{30}\ \mathrm{kg}

(depending on accuracy)

(4 points)

Then, using Kepler's $3^{\mathrm{rd}}$ law for $\textit{Seneca}$ and $\textit{Jupiter}$ , one gets

\frac{a_{\mathrm{Jup}}^3/T_{\mathrm{Jup}}^2}{a_{\mathrm{Sen}}^3/T_{\mathrm{Sen}}^2}=\frac{M_{\mathrm{Sun}}+M_{\mathrm{Jup}}}{M_{\mathrm{Sun}}}=1+x

(4 points)

where $x =$ is the mass ratio of $\textit{Jupiter}$ to the $\textit{Sun}$ . Then, solving for $x$ we get (depending on the accuracy used in determining the elements of $\textit{Seneca's}$ orbit)

x = 0.0016,\ \text{thus } M_{\mathrm{Jup}} = 3.2\times10^{27}\ \mathrm{kg}

(4 points)

Теми и трудност

Трудност: неоценена.

Небесна механика · Закони на Кеплер · Синодични периоди

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