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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

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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

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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