The weight of 1 , kg becomes 1/6 on the moon. | vedclass

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(D) The acceleration due to gravity on the moon is given by $g_m = \frac{G M_m}{R_m^2}$.Given that the weight on the moon is $1/6$ of the weight on Ea

Vedclass · Accepted Answer

$7.65 	imes 10^{22} \, kg$

Vedclass · Answer

(D) The acceleration due to gravity on the moon is given by $g_m = \frac{G M_m}{R_m^2}$.Given that the weight on the moon is $1/6$ of the weight on Earth,we have $g_m = \frac{g_e}{6}$.Taking $g_e = 9.8 \, m/s^2$,we get $g_m = \frac{9.8}{6} \approx 1.633 \, m/s^2$.Using the formula $M_m = \frac{g_m R_m^2}{G}$,where $G = 6.67 	imes 10^{-11} \, N \cdot m^2/kg^2$ and $R_m = 1.768 	imes 10^6 \, m$:$M_m = \frac{1.633 	imes (1.768 	imes 10^6)^2}{6.67 	imes 10^{-11}}$$M_m = \frac{1.633 	imes 3.1258 	imes 10^{12}}{6.67 	imes 10^{-11}}$$M_m \approx 0.765 	imes 10^{23} \, kg = 7.65 	imes 10^{22} \, kg$.

The weight of $1 \, kg$ becomes $1/6$ on the moon. If the radius of the moon is $1.768 \times 10^6 \, m$,then the mass of the moon will be:

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