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

The acceleration due to gravity on moon is $\left(\frac{1}{6}\right)^{\text {th }}$ times the acceleration due to gravity on earth. If the ratio of the density of earth ' $\varrho_{e}$ ' to the density of moon $\varrho_{m}$ ' is $\frac{5}{3}$, then the radius of moon ' $\mathrm{Rm}$ ' in terms of the radius of earth 'Re' is
  1. $\left(\frac{7}{6}\right) \operatorname{Re}$
  2. $\left(\frac{3}{18}\right) \mathrm{Re}$
  3. $\left(\frac{5}{18}\right) \mathrm{Re}$
  4. $\left(\frac{1}{2 \sqrt{3}}\right) \mathrm{Re}$

Solution

$\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{1}{6} \quad \frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{m}}}=\frac{5}{3} \quad \frac{\mathrm{R}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{e}}}=?$ $\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{\mathrm{M}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}^{2}} \times \frac{\mathrm{R}_{\mathrm{e}}^{2}}{\mathrm{M}_{\mathrm{e}}}=\frac{\frac{4}{3} \pi \mathrm{R}_{\mathrm{m}}^{3} \rho_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}^{2}} \times \frac{\mathrm{R}_{\mathrm{e}}^{2}}{\frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^{3} \rho_{\mathrm{e}}}=\frac{\mathrm{R}_{\mathrm{m}} \rho_{\mathrm{m}}}{\mathrm{R}_{\mathrm{e}} \rho_{\mathrm{e}}}=\frac{1}{6}$ $\therefore \frac{\mathrm{R}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{e}}}=\frac{1}{6} \times \frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{m}}}=\frac{1}{6} \times \frac{5}{3}=\frac{5}{18}$ ~

Asked in: MHT CET 2020 (15 Oct Shift 1)

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