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)=sqrt(3)v`
C. `v_(e)=v`
D. `v_(e)=v//2`

Answer

A

Explanation

`(GMm)/(R^(2))-(mv^(2))/R=(mg)/2=m/R.(GM)/(R^(2))rArr v=sqrt((GM)/(2R))`
As `v_(e)=sqrt((2GM)/R),v_(e)=2v`