JEE Main 2021 — Gravitation Question with Solution

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 m at origin, the centre of mass of the system lies at

$$x=\frac{2m\times 0+m\times d}{2m+m}=\frac{d}{3}$$

Hence, $$F_G=F_C$$

where, F_G is gravitational force between them and F_C is centripetal force.

$$\Rightarrow \frac{G{m}_1{m}_2}{r^2}=2m{\omega }^{2}x$$

$$\Rightarrow \frac{G(2m)(m)}{d^2}=2m{\omega }^2\times \frac{d}{3}\Rightarrow {\omega }^2=\frac{3Gm}{d^3}$$

$$\Rightarrow \omega =\sqrt{\frac{3Gm}{d^3}}$$

We know that,

$$\omega =\frac{2\pi }{T}$$

$$\therefore$$ $$T=\frac{2\pi }{\omega }$$

$$\sqrt{\frac{3Gm}{d^3}}$$

$$=2\pi \sqrt{\frac{d^3}{3Gm}}$$ [using Eq. (i)]