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

Options

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

Answer

C

Explanation

The gravitational force between the particles will give the required centripetal force
So, `(GM^(2))/((2R)^(2))=(Mv^(2))/(R)rArr(GM^(2))/(4R^(2))=(Mv^(2))/(R)rArrv=sqrt((GM)/(4R))`.