The mass of the moon is (1/8) of the earth,bu | vedclass

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(C) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.Let $M_e$ and $R_e$ be the mass and radius of the eart

Vedclass · Accepted Answer

the radius of the earth is $\sqrt{8/6}$ of the moon

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(C) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.Let $M_e$ and $R_e$ be the mass and radius of the earth,and $M_m$ and $R_m$ be the mass and radius of the moon.Given: $M_m = \frac{M_e}{8}$ and $g_m = \frac{g_e}{6}$.Taking the ratio: $\frac{g_m}{g_e} = \frac{G M_m / R_m^2}{G M_e / R_e^2} = \frac{M_m}{M_e} 	imes \left( \frac{R_e}{R_m} ight)^2$.Substituting the given values: $\frac{1}{6} = \frac{1}{8} 	imes \left( \frac{R_e}{R_m} ight)^2$.Rearranging gives: $\left( \frac{R_e}{R_m} ight)^2 = \frac{8}{6}$.Therefore,$R_e = \sqrt{\frac{8}{6}} R_m$.

The mass of the moon is $(1/8)$ of the earth,but the gravitational pull is $(1/6)$ of the earth. This is due to the fact that:

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