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)))

Two point masses m and 2m initially at rest are at separation R apart

1.	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» Solution : 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))`