Gravitation Part 1

The given situation is shown below The gravitational force between these two stars provide the required centripetal force for rotation in a circle about their common centre. Assuming 2 mat origin, the centre of mass of the system lies at $x = \frac{2 m \times 0 + m \times d}{2 m + m} = \frac{d}{3}$ Hence, $F_{G}=F_{C}$ where, $F_{C}$ is gravitational force between them and $F_{C}$ is centripetal force. $\Rightarrow \; \frac{G m_{1} m_{2}}{r^{2}} = 2 m \omega^{2} x$ $\Rightarrow \; \frac{G(2 m)(m)}{d^{2}} = 2 m \omega^{2} \times \frac{d}{3} \Rightarrow \omega^{2} = \frac{3 G m}{d^{3}}$ $\omega = \sqrt{\frac{3 G m}{d^{3}}}$ We know that, $\omega = \frac{2 \pi}{T}$ $T = \frac{2 \pi}{\omega}$ $\Rightarrow$ $T = \frac{2 \pi}{\sqrt{\frac{3 G m}{d^{3}}}}$ $= 2 \pi \sqrt{\frac{d^{3}}{3 G m}}$ [using Eq. (i)]