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

Answer

A

Explanation

Masses situated force at point A, B, C and D are as shown in figure.
Total gravitational force on mass at C towards the centre of earth.

`F_("net") = (GMM)/((2R)^(2)) + (GMM)/((sqrt(2)R)^(2)) cos 45^(@) + (GMM)/((sqrt(2)R)^(2)) cos 45^(@)`
`= (GM^(2))/(4R^(2)) + (2GM^(2))/(2R^(2)) xx (1)/(sqrt(2))`
`= (GM^(2))/(R^(2))[(1)/(4) + (1)/(sqrt(2))]`
The total force acting on C is equal to the centripetal force acting on all particles of mass M revolving with velocity v
`:. (Mv^(2))/(R ) = (GM^(2))/(R^(2)) [(4+sqrt(2))/(4sqrt(2))]`
`v = sqrt((GM)/(4R)[1+2sqrt(2)]) = (1)/(2)sqrt((GM)/(R )[1+2sqrt(2)])`