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

Options

A. v
B. 2v
C. 3v
D. 4v

Explanation

According to question (At equator)
`Mg-(Mv^(2))/(R)=(Mg)/(2)impliesv^(2)=(Rg)/(2)=(GM)/(2R)`
using conservation of energy `-(GMm)/(R)+(1)/(2)mv_(e)^(2)=0impliesv_(e)^(2)=(2GM)/(R)=4v^(2)`