Q 11-07-134JEE MainJEE Main 2021 (27 Jul, Shift 2)Easy
Two identical particles of mass $1$ kg each go round a circle of radius $R$, under the action of their mutual gravitational attraction. The angular speed of each particle is:
Answer: (B) $\dfrac12\sqrt{\dfrac{G}{R^3}}$
The particles are at opposite ends of a diameter, $2R$ apart.
$$\frac{G(1)(1)}{(2R)^2} = (1)\omega^2R \Rightarrow \omega^2 = \frac{G}{4R^3} \Rightarrow \omega = \frac12\sqrt{\frac{G}{R^3}}$$
Solution by Sreeraj P, M.Sc Physics