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

Options

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

Answer

1

Explanation

`(Gm^(2))/((2R^(2)))=momega^(2)R`
`(Gm^(2))/(4R^(3))=omega^(2)`
`omega=sqrt((Gm)/(4R^(3)))`
`v=omegaR`
`v=sqrt((Gm)/(4R^(3)))xxR=sqrt((Gm)/(4R))`