question_answer

Acceleration due to gravity at earth's surface is\[\text{1}0\text{ m}{{\text{s}}^{-\text{2}}}\]. The value of acceleration due to gravity at the surface of a planet of mass\[{}^{1}/{}_{5}\]th and radius\[{}^{1}/{}_{2}\]of the earth is (a) \[\text{4}\,\text{m}{{\text{s}}^{-\text{2}}}\]                               (b) \[~\text{6}\,\text{m}{{\text{s}}^{-\text{2}}}\] (c) \[\text{8 m}{{\text{s}}^{-\text{2}}}\]                               (d) \[\text{12 m}{{\text{s}}^{-\text{2}}}\]

Answer:

\[{{g}_{e}}=10\,m/{{s}^{2}},\,\,{{m}_{p}}=\frac{1}{5}{{m}_{e}},\,\,{{r}_{p}}=\frac{1}{2}{{r}_{e}};\,\,{{g}_{p}}=?\] \[{{g}_{e}}=\frac{GM}{{{R}^{2}}}\Rightarrow \frac{{{g}_{e}}}{{{g}_{p}}}=\frac{{{M}_{e}}}{{{M}_{p}}}\times \frac{R{{p}^{2}}}{{{\operatorname{Re}}^{2}}}=5\times {{\left( \frac{1}{2} \right)}^{2}}=\frac{5}{4}\] \[\Rightarrow {{g}_{p}}=\frac{4}{5}\times {{g}_{e}}=\frac{4}{3}\times 10=8m/{{s}^{2}}\]