Question

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

• A. $\sqrt{\frac{Gm}{2R}}$
• B. $\sqrt{\frac{Gm}{R}}$
• C. $\sqrt{\frac{Gm}{4R}}$
• D. $\sqrt{\frac{Gm}{3R}}$

Answer 1

The correct option is C $\sqrt{\frac{Gm}{4R}}$

Their mutual attraction force will be equal to the centrifugal force..

$\therefore \frac{m{v}^2}{R}=\frac{G.m.m}{(2R{)}^2}=>v^2=\frac{Gm}{4R}=>v=\sqrt{\frac{Gm}{4R}}$