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(D) The acceleration due to gravity $g$ on the surface of a planet is given by the formula: $g = \frac{4}{3} \pi \rho G R$, where $\rho$ is the mean d

Vedclass · Accepted Answer

$19.6$

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(D) The acceleration due to gravity $g$ on the surface of a planet is given by the formula: $g = \frac{4}{3} \pi ho G R$, where $ho$ is the mean density and $R$ is the radius of the planet.From this relation, we see that $g \propto ho R$.Given for the planet $(1)$ and the earth $(2)$:Ratio of radii: $\frac{R_1}{R_2} = \frac{1}{2}$Ratio of densities: $\frac{ho_1}{ho_2} = \frac{4}{1}$Acceleration due to gravity on earth: $g_2 = 9.8 \,ms^{-2}$.Using the proportionality $g_1 / g_2 = (ho_1 / ho_2) 	imes (R_1 / R_2)$:$\frac{g_1}{g_2} = \frac{4}{1} 	imes \frac{1}{2} = 2$.Therefore, $g_1 = 2 	imes g_2 = 2 	imes 9.8 \,ms^{-2} = 19.6 \,ms^{-2}$.

The ratio of the radii of a planet and the earth is $1: 2$, and the ratio of their mean densities is $4: 1$. If the acceleration due to gravity on the surface of the earth is $9.8 \,ms^{-2}$, then the acceleration due to gravity on the surface of the planet is: (in $\,ms^{-2}$)

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