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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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The correct answer is:
$g^{\prime}=3 g$
The acceleration due to gravity on the new planet can be found using the relation $g=\frac{G M}{R^2}...(i)$
but $M=\frac{4}{3} \pi R^3 \rho, \rho$ being density.
Thus, Eq. (i) becomes
$\therefore g=\frac{G \times \frac{4}{3} \pi R^3 \rho}{R^2}$
$=G \times \frac{4}{3} \pi R \rho$
$\begin{array}{ll}\Rightarrow & g \propto R \\ \therefore & \frac{g^{\prime}}{g}=\frac{R^{\prime}}{R} \\ \Rightarrow & \frac{g^{\prime}}{g}=\frac{3 R}{R}=3 \\ \Rightarrow & g^{\prime}=3 g\end{array}$
but $M=\frac{4}{3} \pi R^3 \rho, \rho$ being density.
Thus, Eq. (i) becomes
$\therefore g=\frac{G \times \frac{4}{3} \pi R^3 \rho}{R^2}$
$=G \times \frac{4}{3} \pi R \rho$
$\begin{array}{ll}\Rightarrow & g \propto R \\ \therefore & \frac{g^{\prime}}{g}=\frac{R^{\prime}}{R} \\ \Rightarrow & \frac{g^{\prime}}{g}=\frac{3 R}{R}=3 \\ \Rightarrow & g^{\prime}=3 g\end{array}$
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