Two point masses `m` and `2m` initially at rest are at separation `R` apart and due to the mutual gravitational force they come togther and collide elastically and then separates and the process continues. The time period of the resulting oscillations is `pisqrt((R)/(xg))`, the value of `x` is__________(Take `g = (Gm)/(R^(2)))`

Answer

6

Explanation

keep a mass `M`' at the centre of mass for `2m` such that
`(GM'(2m))/((R//3)^(2)) = (G(2m)m)/((R )^(2)) rArr M' = (m)/(9)`
`rArr (2pi (R//3))/(sqrt((GM')/((R//3))))`
`rArr T = (2pi)/(sqrt(GM')) ((R )/(3))^(3//2)`
`T' = (2)/(4sqrt(2)) = (2pi)/(sqrt((GM)/(9))) = (R^(3//2))/(sqrt(27))`
`T' = (4piRsqrt(R))/(4sqrt(6)sqrt(gR^(2))) = (pi)/(sqrt(6)) sqrt((R )/(g))`