The acceleration due to gravity on the moon i | vedclass

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(C) Given: $\frac{g_{m}}{g_{e}} = \frac{1}{6}$ and $\frac{\rho_{e}}{\rho_{m}} = \frac{5}{3}$.We know that the acceleration due to gravity is given by

Vedclass · Accepted Answer

$\left(\frac{5}{18}\right) R_{e}$

Vedclass · Answer

(C) Given: $\frac{g_{m}}{g_{e}} = \frac{1}{6}$ and $\frac{ho_{e}}{ho_{m}} = \frac{5}{3}$.We know that the acceleration due to gravity is given by $g = \frac{GM}{R^2}$.Since $M = ho 	imes V = ho 	imes \frac{4}{3} \pi R^3$,we have $g = \frac{G 	imes \frac{4}{3} \pi R^3 ho}{R^2} = \frac{4}{3} \pi G R ho$.Therefore,the ratio $\frac{g_{m}}{g_{e}} = \frac{R_{m} ho_{m}}{R_{e} ho_{e}}$.Substituting the given values: $\frac{1}{6} = \left(\frac{R_{m}}{R_{e}}ight) 	imes \left(\frac{ho_{m}}{ho_{e}}ight)$.Since $\frac{ho_{e}}{ho_{m}} = \frac{5}{3}$,then $\frac{ho_{m}}{ho_{e}} = \frac{3}{5}$.So,$\frac{1}{6} = \left(\frac{R_{m}}{R_{e}}ight) 	imes \left(\frac{3}{5}ight)$.$\frac{R_{m}}{R_{e}} = \frac{1}{6} 	imes \frac{5}{3} = \frac{5}{18}$.Thus,$R_{m} = \left(\frac{5}{18}ight) R_{e}$.

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