Four particles, each of mass `M`, move along a circle of radius `R` under the acting of their mutual graviational attraction. The speed of each particle is

Options

A. `(GM)/(R)`
B. `sqrt(2sqrt(2)(GM)/(R))`
C. `sqrt((GM)/(R)(2sqrt(2)+1))`
D. `sqrt((GM)/(R)((2sqrt(2)+1)/(4)))`

Answer

D

Explanation

Side of square : `r = 2R cos 45^(@) = sqrt(2)R`

Now, `2F_(1)cos 45^(@)+F_(2)=(Mv^(2))/(R)`
`rArr v=sqrt((R)/(M)(sqrt(2)F_(1)+F_(2)))`
`rArr v=sqrt((R)/(M)[sqrt(2)(GM.M)/((sqrt(2)R)^(2))+(GMM)/((2R)^(2))])`
`=sqrt((GM)/(R)((2sqrt(2+1))/(4)))`.