Q. The mass of the moon is 1/81 of earth’s mass and its radius 1/4th that of the earth. If the escape velocity from the earth’s surface is $11.2\ kms^{- 1}$, its value for the moon will be
A) $0.15\ kms^{- 1}$
B) $5\ kms^{- 1}$
C) $2.5\ kms^{- 1}$
D) $0.5\ kms^{- 1}$
Explanation
(c) $\frac{g_{m}}{g_{e}} = \frac{M_{m}}{M_{e}} \times \left( \frac{R_{e}}{R_{m}} \right) = \frac{1}{81} \times (4)^{2} = \frac{16}{81}$
$g_{m} = \frac{16}{81}g_{e}$
$\therefore v_{e} = \sqrt{2g_{e}R_{e}} = \sqrt{2 \times 9.8 \times (6400 \times 1000)}$
$\approx 11.2\ kms^{- 1}$
$v_{m} = \sqrt{2g_{m}R_{m}} = \sqrt{2 \times \frac{16}{81}g_{e} \times \frac{1}{4}R_{e}}$
$= \frac{2}{9}\sqrt{2g_{e}R_{e}} = \frac{2}{9} \times 11 \approx 2.5\ kms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.