Question

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is $7.5 \, km \, s^{- 1}$ . 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

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