question_answer

The acceleration due to gravity on moon is only one sixth that on earth. Ratio of moon radius \[({{R}_{m}})\] to earth's radius \[({{R}_{e}})\]should be: (a) \[\frac{6}{1}\]if both are assumed to have same density (b) \[\frac{1}{6}\]if both are assumed to have same density (c) \[\frac{5}{18}\]lf\[\frac{{{\rho }_{m}}}{{{p}_{e}}}\]is given as\[\frac{5}{3}\]. (d) \[\frac{9}{5}\]if\[\frac{{{\rho }_{m}}}{{{p}_{e}}}\]is given as\[\frac{3}{10}\].

Answer:

\[{{g}_{m}}=\frac{g1}{6},\frac{{{R}_{m}}}{{{R}_{e}}}=?\] If\[{{e}_{m}}={{e}_{e}},\frac{{{g}_{m}}}{{{g}_{e}}}=\frac{{{e}_{m}}{{R}_{m}}}{{{e}_{e}}{{R}_{e}}}\Rightarrow \frac{{{R}_{m}}}{{{R}_{e}}}\frac{1}{6}\] If\[\frac{{{e}_{e}}}{{{e}_{m}}}=\frac{5}{3},\,\,\frac{{{g}_{m}}}{{{g}_{e}}}=\frac{{{e}_{m}}}{{{e}_{e}}}.\frac{{{R}_{m}}}{{{R}_{e}}}\Rightarrow \frac{{{R}_{m}}}{{{R}_{e}}}=\frac{1}{6}\times \frac{5}{3}=\frac{5}{18}\]