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 + 2sqrt2) )`
B. `sqrt((GM)/(R ))`
C. `sqrt(2 sqrt2 (GM)/(R ))`
D. `sqrt((GM)/( R) (1 + 2sqrt2))`

Answer

A

Explanation

As shown in figure, each mass experiences three forces namely F, F and F.

Here, F = force between two masses at `sqrt2R` separation.
F. = force between two masses at 2R separation.
As all particles move in the circular path of radius R due to their mutual gravitational attraction. Then net force on mass M = mass x centripetal acceleration.
`therefore (F)/(sqrt2) + (F)/(sqrt2) + F. = M (v^2)/(R ) ` (from figure)
or ` F sqrt2 + F. = M (v^2)/( R) " or " (sqrt2 GM^2)/((sqrt2 R)^2) + (GM^2)/(4R^2) = (Mv^2)/(R )`
or `(GM)/(R ) (1/4 + (1)/(sqrt2) ) = v^2 " or " v =1/2 sqrt((GM)/( R) (1 + 2sqrt2) )`