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. `sqrt((Gm)/(R ))`

Answer

A

Explanation

Let `.omega.` be the angular velocity. The mutual force of gravitational attraction provides the necessary centripetal force of rotation. Thus

`(Gm.m)/((R+R)^(2))=m.R.omega^(2)`
or `omega^(2)=(Gm)/(4R^(3))`
or `omega=sqrt((Gm)/(4R^(3))`
Now, `v=Romega=Rsqrt((Gm)/(4R^(3)))=sqrt((Gm)/(4R))`