Question

the height at which the weight of a body becomes 1/ 16 its weight on the surface of earth is

Answer 1

$$\begin{gathered} \mathrm{Expression}\mathrm{for}\mathrm{weight}: \\ \mathrm{W}\mathrm{\alpha }\mathrm{g} \\ \frac{{\mathrm{W}}_2}{{\mathrm{W}}_1}=\frac{{\mathrm{g}}_2}{{\mathrm{g}}_1} \\ ={\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^2 \\ ={\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}\right)}^2 \\ \frac{{\mathrm{W}}_2}{{\mathrm{W}}_1}={\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}\right)}^2 \\ \frac{1}{16}={\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}\right)}^2 \\ \frac{1}{4}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}} \\ \mathrm{R}+\mathrm{h}=4\mathrm{R} \\ \mathrm{h}=\mathbf{3}\mathbf{R} \\ \mathrm{Substitute}\mathrm{radius}\mathrm{of}\mathrm{earth},\mathrm{R}=6400\mathrm{km} \\ \mathrm{h}=3\times 6400\mathrm{km} \\ =\mathbf{19200}\mathbf{}\mathbf{km} \end{gathered}$$