A particle is projected from the mid-point of the line joining two fixed particles each of mass `m`. If the separation between the fixed particles is `l`, the minimum velocity of projection of the particle so as to escape is equal to

Options

A. `sqrt((GM)/(l))`
B. `sqrt((GM)/(2l))`
C. `sqrt((2GM)/(l))`
D. `2sqrt((2GM)/(l))`

Answer

D

Explanation

The gravitational potential at the mid-point `P`,
`V=V_(1)+V_(2)=(-Gm)/((l//2))-(Gm)/((l//2))=-(4Gm)/(l)`
`implies` The gravitational potential energy
`=U=-(4Gmm_(0))/(l)`, `m_(0)=` mass of particle
When it is projected with a speed `v`, it just escapes to infinity, and the potential & kinetic energy will become zero.
`impliesDeltaKE+DeltaPE=0`
`implies(0-(1)/(2)m_(0)v^(2))+{0-(-(4Gmm_(0))/(l))}=0impliesv=2sqrt((2Gm)/(l))`.