Q. A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is ${11kms}^{- 1}$, the escape velocity from the surface of the planet would be
A) ${1.1\ kms}^{- 1}$
B) ${11\ kms}^{- 1}$
C) ${110\ kms}^{- 1}$
D) ${0.11\ kms}^{- 1}$
Explanation
(c) Mass of planet, $M_{p} = 10M_{e},\ where\ M_{e}$ is mass of earth. Radius of planet,
$R_{p} = \frac{R_{e}}{10},\ where\ R_{e}\ is\ radius\ of\ earth.$
Escape speed is given by,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \sqrt{\frac{2GM}{R}}$
$So,\ for\ planet\ v_{p} = \sqrt{\frac{2\ G \times M_{p}}{R_{p}}} = \sqrt{\frac{100 \times 2\ {GM}_{e}}{R_{e}}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10 \times v_{e}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10 \times 11{kms}^{- 1} = 110{kms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.