Q: Two massive particles of masses M & m (M > m) are separated by a distance *l* . They rotate with equal angular velocity under their gravitational attraction. The linear speed of the particle of mass m is

(A) $ \displaystyle \sqrt{\frac{G M m}{(M + m)l} } $

(B) $ \displaystyle \sqrt{\frac{G M^2}{(M + m)l} } $

(C) $ \displaystyle \sqrt{\frac{G m}{l} } $

(D) $ \displaystyle \sqrt{\frac{G m^2 }{(M + m)l} } $

**Click to See Answer : **

Solution : The system rotates about the centre of mass. The gravitational force acting on the particle m accelerates it towards the centre of the circular path, which has the radius

$ \displaystyle r = \frac{M l}{M+m}$

$\displaystyle F = \frac{m v^2}{r}$

$ \displaystyle \frac{G M m}{l^2} = \frac{m v^2}{\frac{M l}{M+m}}$

*By Solving , *

$ \displaystyle v = \sqrt{\frac{G M^2}{(M+m)l}}$