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

Answer

A

Explanation

Net force on any one particle

`=(GM^(2))/((2R)^(2))+(GM^(2))/((R sqrt(2))^(2))cos45^(@)+(GM^(2))/((R sqrt(2))^(2))cos45^(@)`
`=(GM^(2))/(R^(2))[(1)/(4 )+(1)/(sqrt(2))]`
This force will be equal to centripetal force so
`(Mu^(2))/(R )=(GM^(2))/(R^(2))[(1+2sqrt(2))/(4)]`
`u=sqrt((GM)/(4R)[1+2sqrt(2)])=(1)/(2)sqrt((GM)/(R )[2sqrt(2)+1])`
So correct choice is (a).