Q. The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $v$, then the escape velocity from the planet is
A) $\sqrt{3}v$
B) $\sqrt{2}v$
C) $v$
D) $\sqrt{5}v$
Explanation
(a) $Escape\ velocity,\ v_{e} = \sqrt{\frac{2GM_{e}}{R_{e}}}$
$Given,\ \ \ \ \ \ \ M_{p} = 6M_{e},\ R_{p} = 2R_{e}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{p} = \sqrt{\frac{2G{\bullet 6M}_{e}}{{(2R}_{e})}} = \sqrt{3}{\ v}_{e}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.