A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is `V`. Due to the rotation of planet about its axis the acceleration due to gravity `g` at equator is `1//2` of `g` at poles. The escape velocity of a particle on the planet in terms of `V`.

Options

A. `V_(e )=2V`
B. `V_(e )=V`
C. `V_(e )=V//2`
D. `V_(e )=sqrt(3)V`

Answer

A

Explanation

`g_(e )=g_(p)-R omega^(2) rArr (g)/(2)=g-R omega^(2)`
`R omega^(2) = (g)/(2) rArr R^(2)omega^(2)=(gR)/(2) rArr V^(2)(gR)/(2)` ….(i)
`V_(e ) = sqrt(2gR)`
From (i) and (ii)
`V_(e ) = sqrt(2xx2V^(2))rArr V_(e )=2V`