Two particles of equal mass `m_(0)` are moving round a circle of radius r due to their mutual gravitational interaction. Find the time period of each particle.

Explanation

Two particle will always remain on diametrically opposite points so that the gravitational force is centripetal . Here mutual gravitational force is
`F = (Gmm)/((2r)^(2)) = (Gm^(2))/(4r^(2))……(i) `
If the speed of each particel is v, then the centripetal force
`F = (mv^(2))/r " "` ....(i)
If the speed of each particles is v , then the centripetal force
`F = (mv^(2))/r " " ....(ii)`
Equating (i) and (ii) , we get ` v = sqrt((Gm)/(4r))`
` :. ` The time period `T = (2pir)/v = (2pir)/(sqrt((Gm)/(4r))) = (4pi)/(sqrt(Gm)) r^(3//2)`