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Question

Two bodies of masses m1 and m2 are initially at rest at an 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

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Solution

The correct option is **B** [2Gr(m1+m2)]1/2

(When allowed to move towards each other)

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 energy

Decrease in P.E.=Increase in K.E.

Gm1m2r=12m1v21+12m2v22 ... (ii)

On solving equation (i) and (ii)

v1=√2Gm22r(m1+m2) and v2=√2Gm21r(m1+m2)

∴vapp=|v1|+|v2|=√2Gr(m1+m2)

(When allowed to move towards each other)

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 energy

Decrease in P.E.=Increase in K.E.

Gm1m2r=12m1v21+12m2v22 ... (ii)

On solving equation (i) and (ii)

v1=√2Gm22r(m1+m2) and v2=√2Gm21r(m1+m2)

∴vapp=|v1|+|v2|=√2Gr(m1+m2)

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