Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is:

Explanation

Let a be the distance between any two particles,

The resultant gravitational force on any one of the particles is given by, `F_(R)=(GM^(2))/(a^(2))(sqrt(2)+1/2)`. This provided the necessary centripetal force for motion of mass `M` in circle, so
`(sqrt(2)+1/2) ((GM^(2))/(a^(2)))=(MV^(2))/(a/(sqrt(2)))`
`V^(2)=((2sqrt(2)+1)/(2sqrt(2)))(GM)/a=((2sqrt(2)+1)/(2sqrt(2)))(GM)/(sqrt(2)R)`
`rArr V= 1/2sqrt((GM)/R(1+2sqrt(2)))`