The mass of the moon is frac{1}{81} of the ea | vedclass

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(B) The acceleration due to gravity $g$ on the surface of a celestial body is given by $g = \frac{GM}{R^2}$,where $M$ is the mass and $R$ is the radiu

Vedclass · Accepted Answer

The radius of the earth is $\frac{9}{\sqrt{6}}$ of the moon

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(B) The acceleration due to gravity $g$ on the surface of a celestial body is given by $g = \frac{GM}{R^2}$,where $M$ is the mass and $R$ is the radius.Given: $M_m = \frac{1}{81} M_e$ and $g_m = \frac{1}{6} g_e$.We have the ratio: $\frac{g_m}{g_e} = \frac{M_m}{M_e} 	imes \left( \frac{R_e}{R_m} ight)^2$.Substituting the values: $\frac{1}{6} = \frac{1}{81} 	imes \left( \frac{R_e}{R_m} ight)^2$.Therefore,$\left( \frac{R_e}{R_m} ight)^2 = \frac{81}{6}$.Taking the square root on both sides: $\frac{R_e}{R_m} = \sqrt{\frac{81}{6}} = \frac{9}{\sqrt{6}}$.Thus,$R_e = \frac{9}{\sqrt{6}} R_m$.

The mass of the moon is $\frac{1}{81}$ of the earth,but the gravitational pull (acceleration due to gravity) is $\frac{1}{6}$ of the earth. This is due to the fact that:

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