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Physics Question on Gravitation

Assuming g(moon)=(16)gearthg_{(moon)} = \left( \frac{1}{6}\right) g_{earth} and D(moon)=(14)DearthD_{(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

11.224kms−1\frac{11.2}{24} km s^{-1}

B

11.2×24km s−1 11.2 \times \sqrt{24} k m \, s^{-1}

C

11.224kms−1\frac{11.2}{\sqrt{24}} km s^{-1}

D

11.2×24 km s−1 11.2 \times 24\, k m \, s^{-1}

Answer

11.224kms−1\frac{11.2}{\sqrt{24}} km s^{-1}

Explanation

Solution

Escape velocity of any planet is given by, v=2gRv = \sqrt{2gR}
where g = acceleration due to gravity
R = radius of the planet
Here, g(moon)=16gearthg_{\left(moon\right)} = \frac{1}{6}g_{earth}
D(moon)=14DearthD_{\left(moon\right)} = \frac{1}{4}D_{earth}
or, R(moon)=14Rearth R_{\left(moon\right)}= \frac{1}{4} R_{earth}
∴   vmoon=2×16gearth×14Rearth\therefore \:\:\: v_{moon} = \sqrt{2 \times\frac{1}{6}g_{earth} \times \frac{1}{4}R_{earth}}
=1242gearthRearh=vearth24= \frac{1}{\sqrt{24}} \sqrt{2_{g_{earth} } R_{earh} } = \frac{v_{earth}}{\sqrt{24}}
=11.224kms−1= \frac{11.2}{\sqrt{24}} km s^{-1}