Question

Two masses `m_(1)` and `m_(2)` at an infinite distance from each other are initially at rest, start interacting gravitationally. Find their velocity of approach when they are at a distance `r` apart.

Answer 1

Since they move under mutual attraction and no external force acts on them, their momentum and energy are conserved. Therefore,
`:.0=1/2M_(1)v_(1)^(2)+1/2M_(2)v_(2)^(2)-(GM_(1)M_(2))/s`
It is zero because in the beginning, both kinetic energy and potential energy are zero.
`0=M_(1)v_(1)+M_(2)v_(2)`
Solving the equations
`v_(1)^(2)=(2GM_(2)^(2))/(s(M_(1)+M_(2))`
and `v_(2)^(2)=(2GM_(1)^(2))/(s(M_(1)+M_(2)))`
`V` (velocity of approach) `=v_(1)-(-v_(2))=v_(1)+v_(2)`
`=sqrt((2G(M_(1)+M_(2)))/s)`