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(A) The acceleration due to gravity on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M

Vedclass · Accepted Answer

$0.15$

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(A) The acceleration due to gravity on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass,and $R$ is the radius.For the moon,$g_m = \frac{GM_m}{R_m^2}$.For the earth,$g_e = \frac{GM_e}{R_e^2}$.Given that $M_e = 81M_m$ and $R_e = 3.5R_m$.The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is $\frac{g_m}{g_e} = \frac{GM_m}{R_m^2} 	imes \frac{R_e^2}{GM_e} = \frac{M_m}{M_e} 	imes \left(\frac{R_e}{R_m}ight)^2$.Substituting the given values: $\frac{g_m}{g_e} = \frac{M_m}{81M_m} 	imes (3.5)^2 = \frac{1}{81} 	imes 12.25 = \frac{12.25}{81} \approx 0.1512$.Rounding to two decimal places,the ratio is $0.15$.

The mass of the earth is $81$ times that of the moon and the radius of the earth is $3.5$ times that of the moon. The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is

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