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

Answer

D

Explanation

Centripetal force on the particle at A Fig.1.31, i.e., the force along AO due to the particle at B,C and D, is
`F=(GM*M)/((sqrt(2)R)^2) cos 45^@+(GM*M)/((2R)^2)+(GM*M)/((sqrt(2)R)^2) cos45^@`
`=(GM^2)/(4R^2)+2*(GM^2)/(2R^2)*1/sqrt(2)`
`=(GM^2)/(4R^2)(1+4/sqrt(2))=(GM^2)/(4R^2)(1+2sqrt(2))`
Again ,`F=(Mv^2)/R`
`therefore (Mv^2)/R=(GM^2)/(4R^2)(1+2sqrt(2))`
or, `v=1/2sqrt((GM)/R(1+2sqrt(2)))`