Four particles, each of mass M, move along a circle of radius R under the action of their mutual gravitational 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

`F_(12)=(GM^(2))/(2R^(2))`
`F_(14)=(GM^(2))/(2R^(2))`

The resultant of these two forces is `(sqrt(2)GM^(2))//2R^(2))`. Now `F_(12)(GM^(2))//(4R^(2))`.
The combined resultant of all the forces is
`(sqrt(2)GM^(2))/(2R^(2))+(GM^(2))/(4R^(2))` or `(GM^(2))/(4R^(2)) [(sqrt(2))/2+1/4]`
Equating this with centripetal force, we get
`(Mv^(2))/R=(GM^(2))/R[(2sqrt(2)+1)/4]`
or `v^(2)=(GM)/R[(2sqrt(2)+1)/4]` or `v=sqrt((GM)/R((2sqrt(2)+1)/4))`