Question

39. What will be the acceleration due to gravity on the surface of the moon, if its radius is1/4 the radius of the Earth and its mass is 1/80 times the mass of the Earth? (Take g as 9.8 m/s²)
40. The acceleration due to gravity at the surface of the moon is1.67m/s². The mass of the Earth is about 81 times more massivethan the moon. What is the ratio of the radius of the Earth to that ofthe moon?

Answer 1

$$\begin{gathered} \left(39\right) \\ {g}_e=G\frac{M_e}{{R_e}^2}=9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.---\left(1\right) \\ {g}_m=G\frac{M_m}{{R_m}^2} \\ {g}_m=G\frac{{\displaystyle \raisebox{1ex}{$M_e$}\!\left/ \!\raisebox{-1ex}{$80$}\right.}}{{\left({\displaystyle \raisebox{1ex}{$R_e$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}^2} \\ {g}_m=G\frac{16M_e}{80{R_e}^2} \\ {g}_m=\frac{1}{5}G\frac{M_e}{{R_e}^2} \\ {g}_m=\frac{1}{5}\times 9.8=1.96\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \\ \left(40\right) \\ {g}_m=1.67\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.²=G\frac{M_m}{{R_m}^2}or, \\ {R}_m=\sqrt{G\times \frac{M_m}{1.67}}=\sqrt{G\times \frac{M_e}{1.67\times 81}} \\ {g}_e=9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.²=G\frac{M_e}{{R_e}^2}or, \\ {R}_e=\sqrt{G\times \frac{M_e}{9.8}} \\ \frac{R_m}{R_e}=\frac{\sqrt{G\times \frac{M_e}{1.67\times 81}}}{\sqrt{G\times \frac{M_e}{9.8}}}=\sqrt{\frac{G\times \frac{M_e}{1.67\times 81}}{G\times \frac{M_e}{9.8}}} \\ \frac{R_m}{R_e}=\sqrt{\frac{9.8}{1.67\times 81}}=0.2691 \\ \frac{R_e}{R_m}=3.7152 \end{gathered}$$