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.

Two bodies of masses M and m were initially at infinite distance from

1.	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.
Answer» Solution :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 energyat 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))`.