Two bodies of mass `m_(1) and m_(2)` are initially at rest placed infinite distance apart. They are then allowed to move towards each other under mutual gravitational attaction. Show that their relative velocity of approach at separation r betweeen them is
`v=sqrt(2G(m_(1)+m_(2)))/(r)`

Options

A. `[2G((m_(1)-m_(2)))/r]^(1//2)`
B. `[r/(2G(m_(1)m_(2)))]^(1//2)`
C. `[(2G)/r(m_(1)+m_(2))]^(1//2)`
D. `[(2G)/rm_(1)m_(2)]^(1//2)`

Answer

C

Explanation

Conservation of momentum
`m_(1)v_(1)=m_(2)v_(2)`…….i
Conservation of energy
`(Gm_(1)m_(2))/r=1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)`……ii
Solving equation (i) and (ii)
`v_(1)=sqrt((2Gm_(2)^(2))/(r(m_(1)+m_(2))))" "v_(2)=sqrt((2Gm_(1)^(2))/(r(m_(1)+m_(2))))`
`v_("app")=v_(1)+v_(2)=sqrt((2G)/r(m_(1)+m_(2)))`