If the acceleration due to gravity at the Ear | vedclass

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(C) The acceleration due to gravity at the surface of a planet is given by $g = \frac{GM}{R^2}$.For Earth,$g_e = \frac{GM_e}{R_e^2} = g$.For the Moon,

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$g/5$

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(C) The acceleration due to gravity at the surface of a planet is given by $g = \frac{GM}{R^2}$.For Earth,$g_e = \frac{GM_e}{R_e^2} = g$.For the Moon,$g_m = \frac{GM_m}{R_m^2}$.Given that $M_m = \frac{M_e}{80}$ and $R_m = \frac{R_e}{4}$.Substituting these values into the expression for $g_m$:$g_m = \frac{G(M_e / 80)}{(R_e / 4)^2} = \frac{G M_e / 80}{R_e^2 / 16} = \frac{16}{80} \frac{GM_e}{R_e^2}$.Since $\frac{GM_e}{R_e^2} = g$,we get $g_m = \frac{1}{5} g = \frac{g}{5}$.

If the acceleration due to gravity at the Earth is $g$,the mass of the Earth is $80$ times that of the Moon,and the radius of the Earth is $4$ times that of the Moon,then the value of the acceleration due to gravity at the surface of the Moon will be:

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