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Two particles m(1) and m(2) are initiall...

Two particles `m_(1)` and `m_(2)` are initially at rest at infinite distance. Find their relative velocity of approach due to gravitational attraction when their separation is d.

Text Solution

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Initiallly when the separation was large there was no interaction energy and when they get closer the system gravitational energy decreases and the kinetic energy increases.
When separation between the two particles is d, then according to energy conservation we have
`(1)/(2)m_(1)v_(1)^(2) + (1)/(2)m_(2)v_(2)^(2) - (Gm_(1)m_(2))/(d) = 0`
As no other force is present we have according to momentum conservation
`m_(1)v_(1) = m_(2)v_(2)`
From equations written above
`(1)/(2)m_(1)v_(1)^(2) + (1)/(2)(m_(1)^(2))/(m_(2)) v_(1)^(2) = (Gm_(1)m_(2))/(d)`
or `v_(1) = sqrt((2Gm_(2)^(2))/(d(m_(1) + m_(2)))) = sqrt((2G)/(d(m_(1) + m_(2))))m_(2)`
And on firther solving we get
`v_(2) = sqrt((2G)/(d(m_(1) + m_(2))))m_(1)`
Thus approach velocity is given as
`v_(ap) = v_(1) + v_(2) = sqrt((2G(m_(1)+m_(2)))/(d))`
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