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:

Options

A. `sqrt((GM)/R (1+2sqrt2))`
B. `1/2sqrt((GM)/R (1+2sqrt2))`
C. `sqrt((GM)/2)`
D. `sqrt(2sqrt(2)(GM)/2)`

Answer

B

Explanation

Net force on any one particle
`=(GM^2)/(2R)^2+(GM)^2 (Rsqrt2)^2 cos 45^@ + (GM^2)/(Rsqrt2)^2 cos 45^@ =(GM^2)/R^2 [1/4+1/sqrt2]`
This force will be equal to centripetal force so
`(Mu^2)/R=(GM^2)/R^2 [(1+2sqrt2)/4] u =sqrt((GM)/R[(1+2sqrt2)/4])=1/2sqrt((GM)/R (2sqrt2+1))`