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

Answer

D

Explanation

Resultant force acting on mass M due to other three masses are `F=R_(1)=f_(2)=(MV^(2))/(R )`
`=F_(1)+F_(2)+F_(3)`
`rarr therefore =F_(1)sqrt(2)+F_(3)`

`A=Rsqrt(2)` and AC =2R
`(GM^(2))/(R )[1/4+(1)/sqrt(2)] = Mv^(2)`
`V=sqrt(GM)/(R )(sqrt(2)+4)/(4sqrt(2))=1/2sqrt((gm)/(R )(1+2sqrt(2))`