Question

# Two masses M1 & M2 are initially at rest and are separated by a very large distance. If the masses approach each other subsequently due to gravitational attraction between them, their relative velocity of approach at a separation d is

A
2G(M1+M2)d
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B
4G(M1+M2)d
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C
4G(M1M2)d
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D
G(M1+M2)d
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Solution

## The correct option is A √2G(M1+M2)d Relative velocity of M1 w.r.t. M2 or vice-versais, vrel=v1+v2........(1) Initial energy of system, Ei=0, Final energy of system, Ef=−GM1M2d+M1v212+M2v222 From conservation of energy, Ei=Ef −GM1M2d+M1v212+M2v222=0..........(2) By momentum conservation, M1v1−M2v2=0 v2=M1M2v1 Substituting the value of v2 in (2) (M1)2(v1)22M2+M1(v1)22=GM1M2d v1=√2G(M2)2d(M1+M2) v2=√2G(M1)2d(M1+M2) Substituting the value of v1,v2 in (1) we get vrel=√2G(M1+M2)d

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