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IOAA 2022, Теория — Задача 12. Co-orbital satellites

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International Olympiad on Astronomy and Astrophysics 2022 Georgia — Theoretical Competition · 16 август 2022 г. · 50 т.

Условие

This question applies a method of determining the masses of two approximately co-orbital satellites developed by Dermott and Murray in 1981.

Suppose that two small satellites of masses $m_1$ and $m_2$ are approximately co-orbital (moving on very similar orbits) around a large central body of mass $M$ , with $m_1,\ m_2 \ll M$ . At any instant, the orbits of the satellites may be approximated as circular Keplerian orbits with radii $r_1$ and $r_2$ respectively, although $r_1$ and $r_2$ will vary slightly over time due to the mutual gravitational interaction between the satellites.

Figure 4 depicts the shapes of the orbits in the rotating reference frame with zero angular momentum, centred on the central body. We denote by $\theta$ the angle $\measuredangle m_1 M m_2$ , while $R$ , $x_1$ , and $x_2$ denote the mean orbital radius and radial deviations of the satellites.

Throughout this problem, write all answers in an inertial reference frame.

Hint: $(1 + x)^{\alpha} \approx 1 + \alpha x$ for $\alpha x \ll 1$

First we will determine the value of $\frac{m_1}{m_2}$ .

(a) Write down the angular momentum $L_i$ of the satellite with mass $m_i$ when its circular orbit has radius $r_i$ . [3 т.]

(b) The satellites total angular momentum $L_1 + L_2$ is conserved. Let $x_1,\ x_2 \ll R$ be the distances as shown in Figure 4. Find a simple relation between the ratios $\frac{m_1}{m_2}$ and $\frac{x_1}{x_2}$ . [8 т.]

Now, we will try and determine the value of $m_1 + m_2$ . For next parts, we will use the actual barycenter of the system, which may not be exactly at the center of the planet.

(c) The individual angular momenta of the satellites $m_1$ and $m_2$ will vary over time due to their gravitational interactions. Show that the rate of change of the angular momentum of the second satellite is given by

\frac{\Delta L_2}{\Delta t} \approx -\frac{Gm_1 m_2}{R}h(\theta) \quad \text{where} \quad h(\theta) = \left[\frac{\cos\left(\frac{\theta}{2}\right)}{4\sin^2\left(\frac{\theta}{2}\right)} - \sin\theta\right]

[18 т.]

(d) Show that $s = r_2 - r_1$ satisfies

\frac{\Delta s}{\Delta t} \approx -2\sqrt{\frac{G}{MR}}(m_1 + m_2)h(\theta)

[8 т.]

(e) For the angle $\theta = \measuredangle m_1 M m_2$ as indicated on Figure 4, find an expression for $\frac{\Delta\theta}{\Delta t}$ [5 т.]

(f) Using the results above, find the relation between $\Delta s$ and $\Delta\theta$ . [2 т.]

(g) After integrating the expression above, we will obtain the result,

\overline{x}^2 \approx \frac{4R^2}{3}\frac{m_1 + m_2}{M}\left(\frac{1}{\sin\left(\frac{\theta_{\min}}{2}\right)} - 2\cos\theta_{\min} - 3\right)

where $\overline{x} = \frac{x_1 + x_2}{2}$ .

Epimetheus ( $m_1$ ) and Janus ( $m_2$ ) are two approximately co-orbital moons of Saturn. Detailed observations of their orbits have been performed by the Voyager 1 and Cassini spacecraft, which found that $R = 150\,000\,\mathrm{km}$ , $x_1 = 76\,\mathrm{km}$ and $x_2 = 21\,\mathrm{km}$ . The minimum distance between Janus and Epimetheus is $13\,000\,\mathrm{km}$ . The mass of Saturn is known to be $5.7 \times 10^{26}\,\mathrm{kg}$ . Estimate the masses of Epimetheus and Janus. [6 т.]

Central body M (grey disk) with dashed reference circles; a large horseshoe-shaped orbit of m1 around almost the whole circle and a small tadpole-like closed loop of m2 at the top, both with direction arrows; radii r1 to m1 and r2 to m2 with the angle θ between them; R from M to the mean orbit; radial widths x1 (bottom) and x2 (top); two dashed lines from M bound the angle θmin at the upper left. — Figure 4: This figure schematically depicts the shapes of the orbits in the rotating reference frame, selected such that in this frame the total angular momentum of the two satellites is zero.

Отговори

Решение

(a) By Kepler’s third law we have

\begin{aligned} T_i^2 &= \frac{4\pi^2}{GM}r_i^3 \\ \omega_i &= \frac{2\pi}{T_i} = \sqrt{\frac{GM}{r_i^3}} && \mathbf{1.5} \\ L_i &= m_i\omega_i r_i^2 \\ \therefore L_i &= m_i\sqrt{GMr_i} && \mathbf{1.5} \end{aligned}

(b) When the satellites are opposing positions (i.e. $\theta = \pi$ ), we have

r_1 = R \pm \frac{x_1}{2},\ \ r_2 = R \mp \frac{x_2}{2} \qquad \mathbf{2.0}

Thus, using the conservation of angular momentum

\begin{aligned} L_{tot} &= m_1\sqrt{GM\left(R + \frac{x_1}{2}\right)} + m_2\sqrt{GM\left(R - \frac{x_2}{2}\right)} \\ &= m_1\sqrt{GM\left(R - \frac{x_1}{2}\right)} + m_2\sqrt{GM\left(R + \frac{x_2}{2}\right)} && \mathbf{4.0} \\ \therefore m_1\left(1 + \frac{x_1}{4R}\right) + m_2\left(1 - \frac{x_2}{4R}\right) &= m_1\left(1 - \frac{x_1}{4R}\right) + m_2\left(1 + \frac{x_2}{4R}\right) \\ m_1\frac{x_1}{2R} &= m_2\frac{x_2}{2R} \\ \therefore \frac{m_1}{m_2} &= \frac{x_2}{x_1} && \mathbf{2.0} \end{aligned}

(c) Let the centre of mass of the ternary system be denoted by C. The total gravitational force on satellite $m_2$ due to the primary mass $M$ , and the satellite $m_1$ may be resolved into a radial component, parallel to the line between $m_2$ and C and a tangential component, perpendicular to the line between $m_2$ and C. Only the tangential component of the force will act to change the angular momentum of the satellite $m_2$ .

The tangential component of the force is given by

F_{2_{\perp}} = \frac{GMm_2}{r_2^2}\sin(\delta\beta) - \frac{Gm_1 m_2}{d^2}\sin\beta \qquad \mathbf{2.0}

where $d = dist(m_1, m_2)$ , $\beta = \measuredangle m_1 m_2 C$ , and $\delta\beta = \measuredangle M m_2 C$ . There are two ways to simplify this expression.

The first way straightforward. By immediately exploiting the fact that $m_1,\ m_2 \ll M$ and hence $\delta\beta \to 0$ , we can say,

\begin{aligned} \frac{d}{\sin\theta} &= \frac{r_1}{\sin(\beta + \delta\beta)} \\ \sin\beta &= \left(\frac{r_1}{d}\right)\sin\theta && \mathbf{3.0} \\ \text{Also, } \frac{\left(\frac{r_1 m_1}{M + m_1}\right)}{\sin(\delta\beta)} &= \frac{r_2}{\sin(180 - \theta - \delta\beta)} \\ \therefore \sin(\delta\beta) &= \left(\frac{m_1 r_1}{r_2(m_1 + M)}\right)\sin\theta && \mathbf{4.0} \end{aligned}

Another way is using the sine and cosine rules for triangles, we have the expressions

\begin{aligned} \frac{r_2}{\sin(180 - \gamma)} &= \frac{\left[\left(\frac{r_1 m_1}{M + m_1}\right)^2 + r_2^2 - \frac{2m_1 r_1 r_2}{M + m_1}\cos\theta\right]^{\frac{1}{2}}}{\sin\theta} \\ \frac{r_2\sin\theta}{\sin\gamma} &= \left[r_2^2 + \left(\frac{r_1 m_1}{M + m_1}\right)^2 - \frac{2m_1 r_1 r_2}{M + m_1}\cos\theta\right]^{\frac{1}{2}} \\ \frac{d}{\sin\gamma} &= \frac{\frac{Mr_1}{m_1 + M}}{\sin\beta} \\ \sin\beta &= \frac{Mr_1}{m_1 + M}\frac{r_2}{d}\sin\theta\left[r_2^2 + \left(\frac{m_1 r_1}{m_1 + M}\right)^2 - \frac{2m_1 r_1 r_2}{m_1 + M}\cos\theta\right]^{-\frac{1}{2}} \approx \frac{r_1}{d}\sin\theta \\ \sin(\delta\beta) &= \frac{m_1 r_1}{m_1 + M}\sin\theta\left[r_2^2 + \left(\frac{m_1 r_1}{m_1 + M}\right)^2 - \frac{2m_1 r_1 r_2}{m_1 + M}\cos\theta\right]^{-\frac{1}{2}} \approx \frac{m_1 r_1}{r_2(m_1 + M)}\sin\theta \end{aligned}

In both cases, finally we obtain

\begin{aligned} F_{2_{\perp}} &= \frac{GMm_2}{r_2^2}\frac{m_1 r_1}{r_2(m_1 + M)}\sin\theta - \frac{Gm_1 m_2}{d^2}\frac{r_1}{d}\sin\theta \\ \text{As, } M + m_1 &\approx M \\ \therefore F_{2_{\perp}} &= Gm_1 m_2 r_1\sin\theta(r_2^{-3} - d^{-3}) && \mathbf{3.0} \end{aligned}

Two circles: a large one with m2 at the upper left and m1 at the right on it, and a small inner circle around M; lines from M to m2 (r2) and from M to m1 (r1) with the angle θ at M; point C near M joined to m2 and m1 with the angle γ at C; side d between m2 and m1; at m2 the angles β (between m2m1 and m2C) and δβ (between m2C and m2M). — Figure 3: Configuration of the satellites and planet. Circle is centered at C(all bodies are rotating around this point).

The distance between satellites

d \approx 2R\sin\left(\frac{\theta}{2}\right) \qquad \mathbf{1.0}

The tangential component of the gravitational force then becomes

\begin{aligned} F_{2_{\perp}} &\approx \frac{Gm_1 m_2}{R^2}\sin\theta\left(1 - \frac{1}{8\sin^3\left(\frac{\theta}{2}\right)}\right) \\ F_{2_{\perp}} &= \frac{Gm_1 m_2}{R^2}\left(\sin\theta - \frac{\cos\left(\frac{\theta}{2}\right)}{4\sin^2\left(\frac{\theta}{2}\right)}\right) && \mathbf{2.0} \end{aligned}

Finally, the torque generated by the gravitational force is then

\frac{\Delta L_2}{\Delta t} = F_{2_{\perp}}r \approx -\frac{Gm_1 m_2}{R}\left(\frac{\cos\left(\frac{\theta}{2}\right)}{4\sin^2\left(\frac{\theta}{2}\right)} - \sin\theta\right) \qquad \mathbf{3.0}

(d) We can see that

\begin{aligned} \Delta L_i &= m_i\sqrt{GM(r_i + \Delta r_i)} - m_i\sqrt{GMr_i} \\ &= m_i\sqrt{GMr_i}\left[\left(1 + \frac{\Delta r_i}{r_i}\right)^{\frac{1}{2}} - 1\right] \\ &= m_i\sqrt{GMr_i}\left(\frac{\Delta r_i}{2r_i}\right) \\ \therefore \frac{\Delta L_i}{\Delta t} &= \frac{1}{2}m_i\sqrt{\frac{GM}{r_i}}\frac{\Delta r_i}{\Delta t} && \mathbf{2.0} \\ \frac{\Delta r_i}{\Delta t} &= \frac{2}{m_i}\sqrt{\frac{r_i}{GM}}\frac{\Delta L_i}{\Delta t} \\ \therefore \frac{\Delta r_2}{\Delta t} &\approx -\frac{2}{m_2}\sqrt{\frac{r_2}{GM}}\frac{Gm_1 m_2}{R}h(\theta) \\ \frac{\Delta r_2}{\Delta t} &\approx -2m_1\sqrt{\frac{G}{MR}}h(\theta) && \mathbf{1.0} \end{aligned}

Again, since $r1 \approx r2 \approx R$ , we can say by symmetry/the conservation of angular momentum

\begin{aligned} \frac{\Delta r_1}{\Delta t} &\approx 2m_2\sqrt{\frac{G}{MR}}h(\theta) && \mathbf{3.0} \\ \therefore \frac{\Delta s}{\Delta t} &= \frac{\Delta r_2}{\Delta t} - \frac{\Delta r_1}{\Delta t} \\ \frac{\Delta s}{\Delta t} &= -2\sqrt{\frac{G}{MR}}(m_1 + m_2)h(\theta) && \mathbf{2.0} \end{aligned}

(e) Since $\theta$ is the angle between the two satellites, we have

\begin{aligned} \frac{\Delta\theta}{\Delta t} &= \omega_2 - \omega_1 = \sqrt{\frac{GM}{r_2^3}} - \sqrt{\frac{GM}{r_1^3}} && \mathbf{2.0} \\ &= \sqrt{\frac{GM}{R^3}}\left[\left(\frac{r_2}{R}\right)^{-\frac{3}{2}} - \left(\frac{r_1}{R}\right)^{-\frac{3}{2}}\right] \\ &= \sqrt{\frac{GM}{R^3}}\left[\left(1 + \frac{r_2 - R}{R}\right)^{-\frac{3}{2}} - \left(1 + \frac{r_1 - R}{R}\right)^{-\frac{3}{2}}\right] \\ &= \sqrt{\frac{GM}{R^3}}\left[1 - \frac{3}{2}\frac{(r_2 - R)}{R} - 1 + \frac{3}{2}\frac{(r_1 - R)}{R}\right] \\ &\approx \sqrt{\frac{GM}{R^3}}\left[\frac{3}{2}\frac{r_1 - r_2}{R}\right] \\ \frac{\Delta\theta}{\Delta t} &\approx -\frac{3}{2}\sqrt{\frac{GM}{R^3}}\frac{s}{R} && \mathbf{3.0} \end{aligned}

(f) Using the expressions in the previous two parts,

\begin{aligned} \frac{\Delta s}{\Delta\theta} &= \frac{\Delta s}{\Delta t}\cdot\frac{\Delta t}{\Delta\theta} && \mathbf{1.0} \\ &= 2\sqrt{\frac{G}{MR}}(m_1 + m_2)h(\theta)\cdot\frac{2}{3}\sqrt{\frac{R^3}{GM}}\frac{R}{s} \\ &= \frac{4}{3}\frac{R^2}{M}(m_1 + m_2)h(\theta)\left(\frac{1}{s}\right) \\ \therefore s\Delta s &= \frac{4R^2}{3}\frac{(m_1 + m_2)}{M}h(\theta)\Delta\theta && \mathbf{1.0} \end{aligned}

(g) The minimum distance of 13 000 km corresponds to a minimum angle of

\theta_{min} \approx \frac{13000}{150000} \approx 0.0868\,\mathrm{rad} = 4^{\circ}58' \qquad \mathbf{2.0}

Then we substitute the given values the into given expression and the result of (b) which gives

\frac{m_2}{m_1} \approx 3.6 \qquad \mathbf{1.0}

and

m_1 + m_2 \approx 2.5 \times 10^{18}\,\mathrm{kg} \qquad \mathbf{2.0}

Finally

m_1 \approx 5.3 \times 10^{17}\,\mathrm{kg} \quad m_2 \approx 1.9 \times 10^{18}\,\mathrm{kg} \qquad \mathbf{1.0}

Оригинал в Архива: Theory Problems.pdf · официални решения: Theory Solutions.pdf