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

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