question_answer

Two particles of equal mass [m] go round a circle of radius R under the action of their neutral gravitational attraction. The speed of each particle is

A)  \[v=\frac{1}{2R\sqrt{Gm}}\]

B)  \[v=\sqrt{\frac{4Gm}{R}}\]

C)  \[v=\sqrt{\frac{Gm}{2R}}\]

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

Correct Answer: D

Solution :

 For circular motion \[\frac{m{{v}^{2}}}{R}=\frac{G{{m}^{2}}}{{{(2R)}^{2}}}\] \[\Rightarrow \] \[{{v}^{2}}=\frac{Gm}{4R}\] \[\Rightarrow \] \[v=\frac{1}{2}\sqrt{\frac{Gm}{R}}\]