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. `v=1/(2R)sqrt(1/(Gm))`
B. `v=sqrt((Gm)/(2R))`
C. `v=1/2sqrt((Gm)/(R))`
D. `v=sqrt((4Gm)/R)`

Answer

C

Explanation

`(mv^(2))/R=(G.m.m)/((2R)^(2))`
`v=sqrt((Gm)/(4R))=1/2sqrt((Gm)/R)`