Acceleration due to gravity on the moon is fr | vedclass

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(A) The acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{GM}{R^2}$.Since mass $M = 	ext{density} (ho) 	imes 	ext

Vedclass · Accepted Answer

$\frac{5}{18}R_e$

Vedclass · Answer

(A) The acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{GM}{R^2}$.Since mass $M = 	ext{density} (ho) 	imes 	ext{volume} (V) = ho 	imes \frac{4}{3}\pi R^3$,we have $g = \frac{G(ho \cdot \frac{4}{3}\pi R^3)}{R^2} = \frac{4}{3}\pi G ho R$.This implies $g \propto ho R$,so $\frac{g_e}{g_m} = \frac{ho_e}{ho_m} 	imes \frac{R_e}{R_m}$.Given $\frac{g_e}{g_m} = 6$ and $\frac{ho_e}{ho_m} = \frac{5}{3}$,we substitute these values:$6 = \frac{5}{3} 	imes \frac{R_e}{R_m}$.Rearranging for $R_m$,we get $R_m = \frac{5}{3 	imes 6} R_e = \frac{5}{18} R_e$.

Acceleration due to gravity on the moon is $\frac{1}{6}$ of the acceleration due to gravity on the earth. If the ratio of densities of earth $(\rho_e)$ and moon $(\rho_m)$ is $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,then the radius of the moon $R_m$ in terms of $R_e$ will be:

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