Four identical particles of mass M are located at the corners of a square of side ‘a’. What should be their speed if each of them revolves under the influence of others’ gravitational field in circular orbit circumscribing the square ?

Options

A. `1.21sqrt((GM)/a)`
B. `1.41sqrt((GM)/a)`
C. `1.16sqrt((GM)/a)`
D. `1.35sqrt((GM)/a)`

Answer

C

Explanation

radius of the circle =`a/sqrt2=r`
FBD of any one particle

`(GM^2)/(2a^2)+(2GM^2)/(a^2)1/sqrt2=(Mv^2)/(a//sqrt2) rArr sqrt((GM)/a)sqrt((1)/(2sqrt2)+1)=1.16sqrt((GM)/a)`