Physics - Gravitation Question with Solution | TestHub

$\Delta$ K.E. $=\Delta U$ Let mass of the particle be $\mathrm{M}$ and that of the Earth be ${\mathrm{M}}_{\mathrm{e}}$ $$\therefore \frac{1}{2}{\mathrm{Mv}}^2={\mathrm{GM}}_{\mathrm{e}}\mathrm{M}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right)$$ Also, $\mathrm{g}=\frac{{\mathrm{GM}}_e}{{\mathrm{R}}^2}$ Equation (i) can be written as, $$\begin{array}{ll}& \frac{1}{2}{\mathrm{v}}^2={\mathrm{GM}}_{\mathrm{e}}\left[\frac{\mathrm{R}+\mathrm{h}-\mathrm{R}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right]=\frac{{\mathrm{GM}}_{\mathrm{e}}}{{\mathrm{R}}^2}\left[\frac{\mathrm{Rh}}{(\mathrm{R}+\mathrm{h})}\right]\\ \therefore & \frac{1}{2}{\left(\frac{1}{3}{\mathrm{v}}_{\mathrm{e}}\right)}^2=\frac{\mathrm{gRh}}{\mathrm{R}+\mathrm{h}}\\ \therefore & \frac{1}{2}{\left(\frac{1}{3}\sqrt{2\mathrm{gR}}\right)}^2=\frac{\mathrm{gRh}}{\mathrm{R}+\mathrm{h}}\\ \therefore & \frac{1}{2}\times \frac{1}{9}(2\mathrm{gR})=\frac{\mathrm{gRh}}{\mathrm{R}+\mathrm{h}}\\ \therefore & \frac{\mathrm{h}}{\mathrm{R}+\mathrm{h}}=\frac{1}{9}\\ \therefore & 9\mathrm{h}=\mathrm{R}+\mathrm{h}\\ \therefore & 8\mathrm{h}=\mathrm{R}\\ \therefore & \mathrm{h}=\frac{\mathrm{R}}{8}\end{array}$$ $\begin{array}{rl}& \text{ Given }\mathrm{V}=\frac{\mathrm{Ve}}{3}\\ & \therefore \mathrm{h}=\frac{\mathrm{R}}{{\left[\frac{{\mathrm{V}}_{\mathrm{e}}}{{\mathrm{V}}_{\mathrm{e}}/3}\right]}^2-1}=\frac{\mathrm{R}}{9-1}=\frac{\mathrm{R}}{8}\end{array}$