Question

In a double star, two stars(one of mass m and the other of mass $2m$) distance d apart rotate about their common centre of mass. Deduce an expression for the period of revolution. Show that the ratio of their angular momenta about the centre of mass is the same as the ratio of their kinetic energies.

Answer 1

The centre of mass C will be at distance $d/3$ and $2d/3$ from masses $2m$ and m, respectively. Both the stars rotate round C in their respective orbits with the same angular velocity $\omega$. The gravitational force acting on each star due to the other supplies the necessary centripetal force.
The gravitational force on either star is $G\left(2m\right)m/d^2$. If we consider the rotation of the smaller star, the centripetal force $\left(mr{\omega }^2\right)$ is $\left[m\right(2d/3\left){\omega }^2\right]$ and $\left[\right(2md{\omega }^2\left)/3\right]$ for the bigger star, i.e., the same.
$\therefore \frac{G\left(2m\right)m}{d^2}=m\left(\frac{2d}{3}\right)$ ${\omega }^2$ or $\Omega =\sqrt{\left(\frac{3Gm}{d^3}\right)}$
Therefore, the period of revolution is given by
$T=\frac{2\pi }{\omega }=2\pi \sqrt{\left(\frac{d^3}{3Gm}\right)}$
The ratio of the angular momenta is
$\frac{(I\omega {)}_{big}}{(I\omega {)}_{small}}=\frac{I_{big}}{I_{smell}}=\frac{\left(2m\right){\left(\frac{d}{3}\right)}^2}{m{\left(\frac{2d}{3}\right)}^2}=\frac{1}{2}$
Since $\omega$ is the same for both.
The ratio of their kinetic energies is
$\frac{(\frac{1}{2}I{\omega }^2{)}_{big}}{(\frac{1}{2}I{\omega }^2{)}_{small}}$ $=\frac{I_{big}}{I_{small}}=\frac{1}{2}$
which is the same as the ratio of their angular momenta.