question_answer

Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is [CBSE PMT 1995; RPMT 2003]

A)                \[v=\frac{1}{2R}\sqrt{\frac{1}{Gm}}\]      

B)                \[v=\sqrt{\frac{Gm}{2R}}\]

C)               \[v=\frac{1}{2}\sqrt{\frac{Gm}{R}}\]

D)                             \[v=\sqrt{\frac{4Gm}{R}}\]

Correct Answer: C

Solution :

                         Centripetal force provided by the gravitational force of attraction between two particles i.e. \[\frac{m{{v}^{2}}}{R}=\frac{Gm\times m}{{{(2R)}^{2}}}\]               \[\Rightarrow v=\frac{1}{2}\sqrt{\frac{Gm}{R}}\]