The acceleration due to gravity on the Moon i | 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

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$\frac{5}{18}R_e$

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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 $M = 	ext{density} (ho) 	imes 	ext{volume} (V) = ho 	imes \frac{4}{3}\pi R^3$,we have $g = \frac{G}{R^2} 	imes ho 	imes \frac{4}{3}\pi R^3 = \frac{4}{3}\pi ho GR$.Therefore,$g \propto ho R$.Taking the ratio for the Moon and Earth: $\frac{g_m}{g_e} = \frac{ho_m}{ho_e} 	imes \frac{R_m}{R_e}$.Given $\frac{g_m}{g_e} = \frac{1}{6}$ and $\frac{ho_e}{ho_m} = \frac{5}{3}$,we have $\frac{ho_m}{ho_e} = \frac{3}{5}$.Substituting these values: $\frac{1}{6} = \frac{3}{5} 	imes \frac{R_m}{R_e}$.Solving for $R_m$: $\frac{R_m}{R_e} = \frac{1}{6} 	imes \frac{5}{3} = \frac{5}{18}$.Thus,$R_m = \frac{5}{18}R_e$.

The acceleration due to gravity on the Moon is $\frac{1}{6}$ times that of the Earth. If the ratio of the density of the Earth $(\rho_e)$ to the Moon $(\rho_m)$ is $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,what is the radius of the Moon $(R_m)$ in terms of the radius of the Earth $(R_e)$?

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