The acceleration due to gravity on the moon i | vedclass

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Question

(C) We know that the acceleration due to gravity is given by $g = \frac{GM}{R^2}$.Since the mass of a planet is $M = \rho \left( \frac{4}{3} \pi R^3 \

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We know that the acceleration due to gravity is given by $g = \frac{GM}{R^2}$.Since the mass of a planet is $M = ho \left( \frac{4}{3} \pi R^3 ight)$,we can substitute this into the gravity formula:$g = \frac{G}{R^2} \left( ho \frac{4}{3} \pi R^3 ight) = \frac{4}{3} \pi G ho R$.This shows that $g \propto ho R$.Given that $g_m = \frac{1}{6} g_e$ and the ratio of densities $\frac{ho_e}{ho_m} = \frac{5}{3}$,which implies $\frac{ho_m}{ho_e} = \frac{3}{5}$.Using the proportionality $g \propto ho R$,we have:$\frac{g_m}{g_e} = \frac{ho_m R_m}{ho_e R_e}$$\frac{1}{6} = \left( \frac{3}{5} ight) \left( \frac{R_m}{R_e} ight)$$\frac{R_m}{R_e} = \frac{1}{6} 	imes \frac{5}{3} = \frac{5}{18}$Therefore,$R_m = \frac{5}{18} R_e$.

The acceleration due to gravity on the moon is $\frac{1}{6}$ times the acceleration due to gravity on the earth. If the ratio of the density of the earth $\rho_e$ to the density of the moon $\rho_m$ is $\frac{5}{3}$,then the radius of the moon $R_m$ in terms of the radius of the earth $R_e$ is:

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