Two particles of equal mass 'm' go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is :

Options

A. `sqrt((Gm)/(4R))`
B. `sqrt((Gm)/(3R))`
C. `sqrt((Gm)/(2R))`
D. `1/2sqrt((Gm)/(R ))`

Answer

A

Explanation

`(Gm^(2))/((2R)^(2))=m omega^(2)R`
`(Gm)/(4R^(3))=omega^(2)`
`omega=sqrt((Gm)/(4R^(3)))`
`v=omega R`
`v= sqrt((Gm)/(4R^(3)))xxR=sqrt((Gm)/(4R))`
So, correct choice is (a).