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(D) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.Since $M = 	ext{density} (ho) 	imes 	ext{volu

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$g^{\prime} = 3g$

Vedclass · Answer

(D) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.Since $M = 	ext{density} (ho) 	imes 	ext{volume} (V) = ho 	imes \frac{4}{3} \pi R^3$,we can write $g = \frac{G}{R^2} 	imes ho 	imes \frac{4}{3} \pi R^3 = \frac{4}{3} \pi ho G R$.This shows that $g \propto R$ when the density $ho$ is constant.Given that the new planet has the same density as Earth but is $3$ times bigger in size (radius),let $R^{\prime} = 3R$.Therefore,$\frac{g^{\prime}}{g} = \frac{R^{\prime}}{R} = \frac{3R}{R} = 3$.Thus,$g^{\prime} = 3g$.

Imagine a new planet having the same density as that of Earth but it is $3$ times bigger than the Earth in size. If the acceleration due to gravity on the surface of Earth is $g$ and that on the surface of the new planet is $g^{\prime}$,then

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