Question

Two particles of masses m1
kgñ1 secñ1 kgñ1 secñ1 )
of mutual gravitational pull. Show that at any instant their relative velocity of approach is where R is their separation at that instant.

Answer 1

Hey mate,

From conservation of mechanical energy

We know,

Decrease in potential energy is equal to increase in kinetic energy

So,

Gm1m2/r = 1/2μvr^2

μ−reduced mass

=m1m2/m1+m2

vr - relative velocity of approach

vr= √(2Gm1m2/μr)me

=√(2Gm1m2/(m1m2/m1+m2)r)

= √(2G(m1+m2)/r)

Hope this helps you out!