Four particles, each of mass M and equidistant from each other, move along a circleof radius R under the action of thir mutual gravitational attraction. The speed of each particle is :

Options

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

Answer

B

Explanation

`F_("net")=F_(1)+2F_(2)cos45^(@)=`Centripetal force
`implies(GM^(2))/((2R)^(2))+[(2GM^(2))/((sqrt(2)R)^(2))cos45^(@)]=(MV^(2))/(R)`
`V=(1)/(2)sqrt((GM)/(R)(1+2sqrt(2)))`