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 partial with respect to their centre of mass is:     AIEEE  Solved  Paper (Held On 11 May  2011)

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

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

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

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

Correct Answer: A

Solution :

                                \[\frac{G{{m}^{2}}}{{{(2R)}^{2}}}=m{{\omega }^{2}}R\]                 \[\frac{G{{m}^{2}}}{4{{R}^{3}}}={{\omega }^{2}}\]                 \[\omega =\sqrt{\frac{Gm}{4{{R}^{3}}}}\]                 \[v=\omega R\] \[v=\sqrt{\frac{Gm}{4{{R}^{3}}}}\times R=\sqrt{\frac{Gm}{4R}}\]