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Acceleration due to gravity on the moon is ${1/6}^{th}$ of acceleration due to gravity on the earth. If the ratio of densities of earth $(ρ_{e})$ and moon $(ρ_{m}) is 5 : 3$ , then the radius of moon $R_{m}$ in terms of $R_{e}$ will be

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\frac{5}{18} R_{e}

\frac{1}{6} R_{e}

\frac{\sqrt{3}}{18} R_{e}

\frac{1}{2 \sqrt{3}} R_{e}

answer is A.

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Detailed Solution

$g = \frac{G M}{R^{2}}$

$M = ρ X V$ , where $ρ$ is the density and $V$ is the volume.

$g = \frac{G \frac{4}{3} {πR}^{3} ρ}{R^{2}}$

$g = \frac{4}{3} π G ρ R$

$\Rightarrow g \propto R ρ$

$g_{m} = \frac{g_{e}}{6}, \frac{ρ_{e}}{ρ_{m}} = \frac{5}{3}$

$\Rightarrow \frac{g_{m}}{g_{e}} = \frac{R_{m}}{R_{e}} \frac{ρ_{m}}{ρ_{e}}$

$\Rightarrow \frac{1}{6} = \frac{R_{m}}{R_{e}} \frac{3}{5}$

$\Rightarrow \frac{R_{m}}{R_{e}} = \frac{5}{18}$

$\Rightarrow R_{m} = \frac{5}{18} R_{e}$

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