Question

Q.10. If the density of two planets is the same, but their radii are in the ratio of 1 : 2, prove that the ratio of acceleration due to gravity onm their surfaces is also 1 : 2.

Answer 1

Dear Student,


$$\begin{gathered} \text{Weight(w) = mg} \\ \mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{r}}^2}---\left(\mathrm{A}\right) \\ \mathrm{M}=\frac{{\mathrm{gr}}^2}{\mathrm{G}} \\ \mathrm{r}1\mathrm{is}\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{object}\mathrm{d}=\mathrm{from}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{the}\mathrm{earth}. \\ \mathrm{M}=\mathrm{mass}\mathrm{of}\mathrm{an}\mathrm{object} \\ \mathrm{Density}\left(\mathrm{\rho }\right)=\frac{\mathrm{mass}}{\mathrm{volume}} \\ \mathrm{Mass}\left(\mathrm{M}\right)=\mathrm{Density}\times \mathrm{volume}=\mathrm{\rho }\times \frac{4}{3}{\mathrm{\pi r}}^3 \\ \mathrm{Put}\mathrm{in}\mathrm{A} \\ \mathrm{g}=\frac{\mathrm{G}\times \mathrm{\rho }\times 4{\mathrm{\pi r}}^3}{{\mathrm{r}}^2} \\ \mathrm{g}=\mathrm{G}\times \mathrm{\rho }\times 4\mathrm{\pi r} \\ \mathrm{For}\mathrm{planet}1 \\ \mathrm{g}=\mathrm{G}\times \mathrm{\rho }\times 4{\mathrm{\pi r}}_1 \end{gathered}$$