Two bodies of masses M and m were initially at infinite distance from each other and they started approaching each other due to their mutural force of gravitation. Prove that their velocity of approach becomes `sqrt((2G)/r(M+m))` when they are at a distance of r from each other.

Explanation

At infinite distance from each other , both the bodies did not have any potential energy . Also, their velocities were zero.
Hence initial momentum of both the bodies=0 .
Total initial energy = kinetic energy + potential energy =0+0=0.
At a distance r from each other , let the velocity of the body of mass m be v and the that of mass M be V.
Hence, their velocity of approach c =v+V
As the velocities v and V are in opposite directions, the total momentum at this stage = mv-MV
Hence, from the law of conservation of momentum
0=mv-MV or , V=`m/Mv`
`therefore c=v+V=v+ m/Vv=v(1+m/M)`
Hence , the kinetic energy of the bodies when they are at a distance r from each other
`=1/2 mv^2+1/2MV^2=1/2mv^2+1/2M*(m^2)/(M^2)*v^2`
`=1/2mv^2(1+m/M) =1/2m*(v^2(1+m/M)^2)/(1+m/M) =1/2 m*(c^2)/((M+m)/M)`
`=1/2 *(Mmc^2)/(M+m)`
Also, potential energy at a distance r (due to gravitation)
`=-(GMm)/r`
Hence , from the law of conservation of energy ,
`0=1/2 (mMc^2)/(M+m)-(GMm)/r or c^2=(2G)/r(M+m)`
`therefore c= sqrt((2G)/r(M+m))`.