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(C) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.Since density $\rho = \frac{M}{V} = \frac{M}{\frac

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$g^{\prime} = 18g$

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(C) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.Since density $ho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi R^3}$,we can express the radius $R$ as $R = \left( \frac{3M}{4\pi ho} ight)^{1/3}$.Substituting $R$ into the formula for $g$:$g = \frac{GM}{(\frac{3M}{4\pi ho})^{2/3}} = G M^{1/3} (\frac{4\pi ho}{3})^{2/3}$.This shows that $g \propto M^{1/3} ho^{2/3}$.Given for the planet: $M^{\prime} = 8M$ and $ho^{\prime} = 27ho$.Taking the ratio:$\frac{g^{\prime}}{g} = \left( \frac{M^{\prime}}{M} ight)^{1/3} \left( \frac{ho^{\prime}}{ho} ight)^{2/3}$.$\frac{g^{\prime}}{g} = (8)^{1/3} 	imes (27)^{2/3} = 2 	imes (3^3)^{2/3} = 2 	imes 3^2 = 2 	imes 9 = 18$.Therefore,$g^{\prime} = 18g$.

There is a planet which is $8$ times more massive and $27$ times denser than the Earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surfaces of the planet and the Earth respectively,then:

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