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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 of at equator is $\large\frac{1}{2}$ of acceleration due to gravity at poles. The escape velocity of a particle on the poles of planet in terms of v

\[(a)\;v_e=2V \quad (b)\;v_e=V \quad (c)\;v_e=v/2 \quad (d)\;v_e=\sqrt 3 v\]

1 Answer

Let $w$ be angular velocity of planet and 'R' its radius.
$v=w R$ at equator -----(1)
$g=g_0-w^2R \qquad g$ at equator $g_0$ at poles
$w^2 R=\large\frac{g_0}{2}$
$v^2=\large\frac{g_0\; R}{2}$ (Substituting from (1))
But we know escape velocity
$v_e=\sqrt {2g_0R}$
$\quad=\sqrt {4v^2}$
$ \therefore v_e=2v$
Hence a is the correct answer.


answered Aug 21, 2013 by meena.p
edited Mar 14, 2014 by thagee.vedartham

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