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

Options

A. `sqrt((Gm)/(R))`
B. `sqrt((Gm)/(4R))`
C. `sqrt((Gm)/(3R))`
D. `sqrt((Gm)/(2R))`

Answer

B

Explanation

`(Gm^(2))/(4R^(2))=(mV^(2))/(R)=V=sqrt((GM)/(4R))`