Q 11-07-078JEE MainJEE Main 2023 (29 Jan, Shift 1)Easy
Two particles of equal mass $m$ move in a circle of radius $r$ under the action of their mutual gravitational attraction. The speed of each particle will be
Answer: (D) $\sqrt{\dfrac{Gm}{4r}}$
The particles are at opposite ends of a diameter, so their separation is $2r$. Gravity gives the centripetal force:
$$\frac{Gm^2}{(2r)^2}=\frac{mv^2}{r}\ \Rightarrow\ v=\sqrt{\frac{Gm}{4r}}$$
Solution by Sreeraj P, M.Sc Physics