The gravitational pull of the moon is (1/6)^{ | vedclass

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(B) The acceleration due to gravity $g$ is given by $g = \frac{GM}{R^2}$.Let $g_m, M_m, R_m$ be the gravity,mass,and radius of the moon,and $g_e, M_e,

Vedclass · Accepted Answer

radius of the earth is $(\sqrt{4/3})^{	ext{th}}$ of the moon's radius.

Vedclass · Answer

(B) The acceleration due to gravity $g$ is given by $g = \frac{GM}{R^2}$.Let $g_m, M_m, R_m$ be the gravity,mass,and radius of the moon,and $g_e, M_e, R_e$ be those of the earth.Given: $g_m = \frac{1}{6} g_e$ and $M_m = \frac{1}{8} M_e$.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 given values: $\frac{1}{6} = \frac{1}{8} 	imes \left(\frac{R_e}{R_m}ight)^2$.Rearranging gives: $\left(\frac{R_e}{R_m}ight)^2 = \frac{8}{6} = \frac{4}{3}$.Taking the square root: $\frac{R_e}{R_m} = \sqrt{\frac{4}{3}}$.Thus,the radius of the earth is $\sqrt{4/3}$ times the radius of the moon.

The gravitational pull of the moon is $(1/6)^{\text{th}}$ of the earth and the mass of the moon is $(1/8)^{\text{th}}$ of the earth. This implies that the:

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