Question

two bodies of masses m1 and m2 are initially at rest at infinite distance apart. They are then allowed to move towards each other under the mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is

Answer 1

Hello dear,
In this problem, we'll ignore effect of earth's gravity, only gravity between two bodies will be considered.

● Answer-
√[(2G/r)(m1+m2)]

● Explaination-
When two masses are at distance r, gravitational potential energy will be-
PE = -Gm1m2/r

When the masses move toward each other, potential energy is converted to kinetic energy.
PE = KE1 = KE2
-Gm1m2/r = 1/2 m1v1^2 = 1/2 m2v2^2

Solving this, we'll get
v1 = √(2Gm2/r)
v2 = √(2Gm1/r)

Relative velocity is-
v = v1 + v2
v = √(2Gm2/r) + √(2Gm1/r)
v = √[(2G/r)(m1+m2)]

Relative velocity of approach is √[(2G/r)(m1+m2)] .

Hope this is useful...

Answer 2

Answer:

[2g(m1-m2/r]^1/2

I hope that your answer