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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 the earth, the gravitational acceleration on the surface of that planet is
Options:
$0.2$ times that the earth, the gravitational acceleration on the surface of that planet is
Solution:
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Verified Answer
The correct answer is:
$0.2 \mathrm{~g}$
$g=\frac{4}{3} \pi G R \rho$
and
$$
g^{\prime}=\frac{4}{3} \pi G R^{\prime} \rho
$$
$\therefore$
$$
\begin{array}{l}
\frac{g^{\prime}}{g}=\frac{R^{\prime}}{R}=0.2 \\
g^{\prime}=0.2 g
\end{array}
$$
and
$$
g^{\prime}=\frac{4}{3} \pi G R^{\prime} \rho
$$
$\therefore$
$$
\begin{array}{l}
\frac{g^{\prime}}{g}=\frac{R^{\prime}}{R}=0.2 \\
g^{\prime}=0.2 g
\end{array}
$$
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