Acceleration due to gravity on moon is $\frac 16$ of the acceleration due to gravity on earth. If the ratio of densities of earth $({\rho _e})$ and moon $({\rho _m})$ is $\left( {\frac{{{\rho _e}}}{{{\rho _m}}}} \right) = \frac{5}{3}$ then radius of moon $R_m$ in terms of $R_e$ will be
Acceleration due to gravity on moon is $\frac 16$ of the acceleration due to gravity on earth. If the ratio of densities of earth $({\rho _e})$ and moon $({\rho _m})$ is $\left( {\frac{{{\rho _e}}}{{{\rho _m}}}} \right) = \frac{5}{3}$ then radius of moon $R_m$ in terms of $R_e$ will be
$\frac{5}{{18}}{R_e}$
$\frac{1}{6}{R_e}$
$\frac{3}{{18}}{R_e}$
$\frac{1}{{2\sqrt 3 }}{R_e}$
Acceleration due to gravity on moon is $\frac 16$ of the acceleration due to gravity on earth. If the ratio of densities of earth $({\rho _e})$ and moon $({\rho _m})$ is $\left( {\frac{{{\rho _e}}}{{{\rho _m}}}} \right) = \frac{5}{3}$ then radius of moon $R_m$ in terms of $R_e$ will be
$g = \frac{4}{3}\pi G\rho R$ $ \Rightarrow \,\,g \propto \rho R\, \Rightarrow \,\frac{{{g_e}}}{{{g_m}}} = \frac{{{\rho _e}}}{{{\rho _m}}} \times \frac{{{R_e}}}{{{R_m}}}$
$ \Rightarrow \,\,\frac{6}{1} = \frac{5}{3} \times \frac{{{R_e}}}{{{R_m}}}\, \Rightarrow \,\,{R_m} = \frac{5}{{18}}{R_e}$
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