Find the escape velocity of particle of mass `m` which is situated at a radial distance `r` (from centre of earth) above the earth's surface. `M` is the mass earth.

Options

A. `sqrt((GM)/(2r))`
B. `sqrt((GM)/r)`
C. `sqrt((2GM)/r)`
D. none of these

Answer

C

Explanation

At the point `P` the gravitational potential energy of `m` is `U_(!)=-(GMm)/(r)`

If we give a velocity `v`, the KE of the particle's is `K_(1)=1/2mv^(2)`
Then, the total mechanical energy is
`E_(1)=U_(1)+K_(1)=-(GMm)/r+1/2mv^(2).......(i)`
Since, the particle escape to infinity, for minimum `v`, the total mechanical energy at infinity is
`E_(2)=U_(2)+K_(2)=0........(ii)`
Equating `E_(1)` and `E_(2)`, we have `v=sqrt((2GM)/r)`