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A particle is projected from the mid - p...

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

A

`sqrt((GM)/(l))`

B

`sqrt((GM)/(2l))`

C

`sqrt((2GM)/(l))`

D

`2sqrt((2GM)/(l))`

Text Solution

Verified by Experts

The gravitational potential at the mid-point `P=V=V_(1)+V_(2)`
`=(-Gm)/((l//2))-(Gm)/((l//2))=-(4Gm)/(l)`
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 and kinetic energy will become zero.
`rArr DeltaK.E.+DeltaP.E.=0`
`rArr(0-(1)/(2)m_(0)v^(2))+{0-(-(4Gmm_(0))/(l))}=0`
`rArrv=2sqrt((2Gm)/(l))`
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