Consider a planet whose density is same as that of the earth but whose radius is three times the radius '…

Consider a planet whose density is same as that of the earth but whose radius is three times the radius ' $R$ ' of the earth. The acceleration due to gravity ' $\mathrm{g}_{\mathrm{n}}$ ' on the surface of planet is $g_n=x$. $g$ where $g$ is acceleration due to gravity on surface, of earth. The value of ' $\mathrm{x}$ ' is
  1. $9$
  2. $3$
  3. $\frac{1}{3}$
  4. $\frac{1}{9}$

Solution

$\begin{array}{ll} & \mathrm{As}, \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2} \text { and } \mathrm{M}=\rho \mathrm{V} \\ \therefore \quad & \mathrm{g}=\frac{\mathrm{G} \rho \mathrm{V}}{\mathrm{R}^2}=\frac{\mathrm{G} \rho \frac{4}{3} \pi \mathrm{R}^3}{\mathrm{R}^2} \\ \therefore \quad & \mathrm{g} \propto \mathrm{R} \end{array}$ For the planet: Radius $\mathrm{R}=3 \mathrm{R}$ $\begin{array}{ll} \therefore \quad & \mathrm{g}_{\text {planet }}=\frac{\mathrm{G} \mathrm{V}_{\text {planet }}}{(3 \mathrm{R})^2} \\ & \text { where } \mathrm{V}_{\text {planet }}=\frac{4}{3} \pi(3 \mathrm{R})^3 \\ \therefore \quad & \mathrm{g}_{\text {planet }}=\frac{\mathrm{G} \rho \frac{4}{3} \pi(3 \mathrm{R})^3}{(3 \mathrm{R})^2} \\ \therefore \quad & \mathrm{g}_{\text {planet }} \propto 3 \mathrm{R} \end{array}$ $\therefore \quad \frac{\mathrm{g}}{\mathrm{g}_{\text {planet }}}=\frac{\mathrm{R}}{3 \mathrm{R}} \quad$....(since $\rho$ is constant) $\begin{array}{ll}\therefore & \mathrm{g}_{\text {planet }}=3 \mathrm{~g} \\ \therefore & \mathrm{x}=3\end{array}$

Asked in: MHT CET 2023 (09 May Shift 2)

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