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Two stars, A of mass 2M and B of mass M, where M = 1.0 × 10³⁰ kg, move in circular orbits about…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Two stars, A of mass $2M$ and B of mass $M$ , where $M = 1.0 \times 10^{30}\ \text{kg}$ , move in circular orbits about their common centre of mass, always a distance $d = 1.5 \times 10^{11}\ \text{m}$ apart, as shown in the figure. Take $G = 6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2$ and treat the stars as points.

Two stars on a horizontal line, A of mass 2M on the left and B of mass M on the right, a distance d apart. A cross marked CM lies one third of the way from A to B, each star moves on a dashed circle about CM, and the velocity arrows point down at A and up at B.

Find the radius of each star's orbit, and the period in days.
Find the speed of each star.
Star A suddenly throws off part of its mass as a thin, round shell of gas that leaves evenly in all directions, so that at that instant the velocities of both stars are unchanged. The gas flies off fast, so ignore its pull once it has left. Above what mass must A stay for the two stars to remain bound together?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Radii. The centre of mass is fixed: $2M\,r_A = M\,r_B$ and $r_A + r_B = d$ . So $r_A = \dfrac{d}{3} = 5.0 \times 10^{10}\ \text{m}$ and $r_B = \dfrac{2d}{3} = 1.0 \times 10^{11}\ \text{m}$ .

Step 2. Period. For star B: $\dfrac{G(2M)M}{d^2} = M\omega^2\,\dfrac{2d}{3}$ , so $\omega^2 = \dfrac{3GM}{d^3}$ (star A gives the same). Then

\displaystyle T = 2\pi\sqrt{\frac{d^3}{3GM}} = 2.580 \times 10^{7}\ \text{s} = 299\ \text{days}

Step 3. Speeds. $v_A = \omega r_A = 12.2\ \text{km/s}$ and $v_B = \omega r_B = 24.3\ \text{km/s}$ , in opposite directions. Their relative speed is $v_{rel} = v_A + v_B = \sqrt{\dfrac{3GM}{d}} = 36.5\ \text{km/s}$ .

Step 4. After the shell leaves. A now has mass $m$ , B still has $M$ , they are still $d$ apart, and their relative speed is still $v_{rel}$ . The motion of the two about their centre of mass has energy $\tfrac12\mu v_{rel}^2 - \dfrac{GmM}{d}$ , with $\mu = \dfrac{mM}{m + M}$ . The pair stays bound if this is negative:

\displaystyle \tfrac12 v_{rel}^2 < \frac{G(m + M)}{d} \quad\Rightarrow\quad \frac{3GM}{2d} < \frac{G(m + M)}{d} \quad\Rightarrow\quad m > \frac{M}{2}

So A must keep more than $\dfrac{M}{2} = 5.0 \times 10^{29}\ \text{kg}$ . It can lose at most $1.5 \times 10^{30}\ \text{kg}$ , three quarters of its mass.

Answer: (a) $5.0 \times 10^{10}\ \text{m}$ and $1.0 \times 10^{11}\ \text{m}$ ; $299\ \text{days}$ . (b) $12.2\ \text{km/s}$ and $24.3\ \text{km/s}$ . (c) $5.0 \times 10^{29}\ \text{kg}$ .

Check. Before the loss, $\tfrac12 v_{rel}^2 = \dfrac{3GM}{2d}$ is half of $\dfrac{G(3M)}{d}$ , as it must be for a circular orbit: the kinetic energy of relative motion is half the size of the potential energy.

Watch out. After the loss the centre of mass of the pair moves, because the shell carried away momentum. So judge the binding by the energy of the relative motion, not by the total energy in the old frame.

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