The acceleration due to gravity on moon is $\frac{1^{\text {th }}}{6}$ times the acceleration due to gravity…

The acceleration due to gravity on moon is $\frac{1^{\text {th }}}{6}$ times the acceleration due to gravity on earth. If the ratio of the density of earth ' $e_e$ ' to the density of moon ' $e_m$ ' is $\frac{5}{3}$ then the radius of moon ' $R_m$ ' in terms of the radius of earth ' $R_e$ ' is
  1. $\left(\frac{3}{18}\right) \mathrm{R}_{\mathrm{e}}$
  2. $\left(\frac{1}{2 \sqrt{3}}\right) \mathrm{R}_{\mathrm{e}}$
  3. $\left(\frac{5}{18}\right) \mathrm{R}_{\mathrm{e}}$
  4. $\left(\frac{7}{6}\right) \mathrm{R}_{\mathrm{e}}$

Solution

We know $g=\frac{\mathrm{GM}}{\mathrm{R}^2}$ And the relation between mass and density of the planet is given by $\begin{aligned} & M=\rho\left(\frac{4}{3} \pi R^3\right) \\ & \therefore g=\left(\frac{4 \pi G}{3}\right) \rho R=K \rho R\end{aligned}$ Given, $\mathrm{g}_{\mathrm{m}}=\frac{1}{6} \mathrm{~g}_{\mathrm{e}}$ and $\mathrm{e}_{\mathrm{m}}=\frac{3}{5} \mathrm{e}_{\mathrm{e}}$ $\begin{aligned} & \Rightarrow \frac{g_m}{g_e}=\frac{e_m R_m}{e_e R_e}=\frac{1}{6} \\ & \Rightarrow \frac{3 R_m}{5 R_e}=\frac{1}{6} \\ & \Rightarrow R_m=\frac{5}{18} R_e\end{aligned}$ ^

Asked in: MHT CET 2022 (05 Aug Shift 2)

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