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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 mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is

a
2Gm1−m2r1/2
b
2Grm1+m21/2
c
r2Gm1m21/2
d
2Grm1m21/2

detailed solution

Correct option is B

Let velocities of these masses at r distance from each other be v1 and v2, respectively.By conservation of momentum.m1v1−m2v2=0⇒m1v1=m2v2  ……..(i)By conservation of energychange in PE = change in KEGm1m2r=12m1v12+12m2v22⇒m12v12m1+m22v22m2=2Gm1m2r  ……(ii)On solving equations (i) and (ii),v1=2Gm22rm1+m2 and v2=2Gm12rm1+m2∴vapp=v1+v2=2Grm1+m2

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