Question

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is $7.5km{s}^{-1}$ . Due to the rotation of the planet about its axis, the acceleration due to gravity $g$ at equator is $1/2$ of $g$ at poles. What is the escape velocity (in $km{s}^{-1}$ ) of a particle on the planet from the pole of the planet?

Answer 1

Answer: 15

$g_{\text{e}}=g_{\text{p}}-R{\omega }^2\Rightarrow \frac{g}{2}=g-R{\omega }^2$
$R{\omega }^2=\frac{g}{2}\Rightarrow R^2{\omega }^2=\frac{gR}{2}\Rightarrow V^2=\frac{gR}{2}$ ...(1)
The escape velocity, $V_{\text{e}}=\sqrt{2gR}$ ...(2)
From (1) and (2)
$V_{\text{e}}=\sqrt{2\times 2V^2}\Rightarrow V_{\text{e}}=2V$
$V_e=15km{s}^{-1}$