1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is

A
[2G(m1m2)r]1/2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
[2Gr(m1+m2)]1/2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
[r2G(m1m2)]1/2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
[2Grm1m2]1/2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is D [2Gr(m1+m2)]1/2By applying law of conservation of momentum, m1v1−m2v2=0⇒m1v1=m2v2........ (i)Where v1 and v2 are the velocities of masses m1 and m2 at a distance r from each other.By conservation of energy,Change in P.E= change in K.E.Gm1m2r=12m1v21+12m2v22........ (ii)Solving eqn. (i) and (ii) we getv1= ⎷2Gm22r(m1+m2) and v2= ⎷2Gm21r(m1+m2)Relative velocity of approach, vR=|v1|+|v2|=√2Gr(m1+m2)

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Gravitational Potential Energy
PHYSICS
Watch in App
Join BYJU'S Learning Program