Two particles of equal mass go around a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

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

`F_(G)=(Gm^(2))/(4R^(2))implies(Mv^(2))/R(Gm^(2))/(4R^(2))`
`:. v=1/2sqrt((GM)/R`