question_answer

Two particles of equal mass 'm' go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is

A) \[\sqrt{\frac{Gm}{4R}}\]

B) \[\sqrt{\frac{Gm}{3R}}\]

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

D) \[\sqrt{\frac{Gm}{R}}\]

Correct Answer: A

Solution :

[a] Here, centripetal force will be given by the gravitational force between the two particles.                 \[\frac{G{{m}^{2}}}{{{(2R)}^{2}}}=m{{\omega }^{2}}R\] \[\Rightarrow \frac{Gm}{4{{R}^{3}}}={{\omega }^{2}}\Rightarrow \omega =\sqrt{\frac{GM}{4{{R}^{3}}}}\] If the velocity of the two particles with respect to the centre of gravity is v then \[v=\sqrt{\frac{Gm}{4{{R}^{3}}}}\times R=\sqrt{\frac{GM}{4R}}\]