Discussion Forum

Consider a spherical planet which  is rotating about its axis such that the speed of a point on its equator is v and the effective acceleration due to gravity on the equator is $\frac{1}{3}$ of its value at the poles. What is the escape velocity for a particular at the pole of this planet

Answer:

Explanation:

We know that the escape velocity of a particle from the surface of a planet is given by

$v_{e}=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2gR^{2}}{R}}$            [ $\because GM= gR^{2}$]

     $=\sqrt{2gR}$

 where, g= acceleration due to gravity, and R= radius of the planet

 Given, the velocity at equator , $v_{E}= V$

 and  acceleration due to gravity at equator

 $g_{E}= \frac{1}{3}g_{p}$

$\therefore$     $g_{p}= 3 g_{E}$.........(i)

 The escape velocity of a particle at equator,

$V_{E}=\sqrt{2g_{E}R}$  ..........(ii)

 and at poles,

$V_{P}=\sqrt{2g_{P}R}$

 $=\sqrt{2\times 3g_{E}R}=\sqrt{3}\sqrt{2g_{E}R}$             [From Eq.(i)]

 $=\sqrt{3}V_{E}$                    [From Eq.(ii)]

 = $\sqrt {3}$ v                    [ $\because v_{E}=V$]

 Hence, $\sqrt{3}$ v  is the escape velocity for a particle  at the pole of this planet.