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

`V = omega`R
`g = g_(0) - omega^(2)` R [ g = at equator, `g_(0) ` = at poles ]
`(g_(0))/(2) = g_(0) - omega^(2) R , omega^(2) R = (g_(0))/(2) , V^(2) = (g_(0)R)/(2)`
`V_(e) = sqrt(2g_(0) R) = sqrt(4V^(2)) = 2V`