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.
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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!
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!
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