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

Answer

C

Explanation

`(mv^(2))/(R )=(Gm^(2))/((sqrt(2)R)^(2))cos 45^(@)xx2+(Gm^(2))/((2R)^(2))`
`=(Gm)/(sqrt(2)R)+(Gm)/(4R)`
`=(Gm)/(4R)[1+2sqrt(2)]`