Question 11:
difficult
Two particles of equal mass $m$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed $v$ of each particle is:
(1995)
The gravitational force provides the necessary centripetal force. $\frac{mv^2}{R} = \frac{Gmm}{(2R)^2} = \frac{Gm^2}{4R^2}$. Solving for $v$, we get $v^2 = \frac{Gm}{4R}$, which means $v = \frac{1}{2}\sqrt{\frac{Gm}{R}}$.