Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
0 votes

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

\[(a)\;v_e=2V \quad (b)\;v_e=V \quad (c)\;v_e=v/2 \quad (d)\;v_e=\sqrt 3 v\]
Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App