If the density of a small planet is the same | vedclass

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(A) The gravitational acceleration $g$ on the surface of a planet is given by the formula: $g = \frac{4}{3} \pi G R \rho$, where $G$ is the gravitatio

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$0.2$

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(A) The gravitational acceleration $g$ on the surface of a planet is given by the formula: $g = \frac{4}{3} \pi G R \rho$, where $G$ is the gravitational constant, $R$ is the radius, and $\rho$ is the density.Since the density $\rho$ is the same for both the planet and the earth, we have $g \propto R$.Let $g$ be the acceleration on earth and $g^{\prime}$ be the acceleration on the planet.Given $R^{\prime} = 0.2 R$, we have:$\frac{g^{\prime}}{g} = \frac{R^{\prime}}{R} = 0.2$Therefore, $g^{\prime} = 0.2 g$.

If the density of a small planet is the same as that of earth, while the radius of the planet is $0.2$ times that of the earth, the gravitational acceleration on the surface of that planet is (in $\,g$)

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