Question

Assuming $g_{(moon)} = \left( \frac{1}{6}\right) g_{earth}$ and $D_{(moon)} = \left( \frac{1}{4}\right) D_{earth}$ where g and D are. the acceleration due to gravity and diameter respectively, the escape velocity from the moon is

• A. $\frac{11.2}{24} km s^{-1}$
• B. $ 11.2 \times \sqrt{24} k m \, s^{-1}$
• C. $\frac{11.2}{\sqrt{24}} km s^{-1}$
• D. $ 11.2 \times 24\, k m \, s^{-1}$

Answer 1

The Correct Option is C

Escape velocity of any planet is given by, $v = \sqrt{2gR}$
where g = acceleration due to gravity
R = radius of the planet
Here, $g_{\left(moon\right)} = \frac{1}{6}g_{earth} $
$D_{\left(moon\right)} = \frac{1}{4}D_{earth}$
or, $ R_{\left(moon\right)}= \frac{1}{4} R_{earth}$
$\therefore \:\:\: v_{moon} = \sqrt{2 \times\frac{1}{6}g_{earth} \times \frac{1}{4}R_{earth}}$
$ = \frac{1}{\sqrt{24}} \sqrt{2_{g_{earth} } R_{earh} } = \frac{v_{earth}}{\sqrt{24}}$
$ = \frac{11.2}{\sqrt{24}} km s^{-1}$