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A projectile is projected with velocity $k{v_e}$ in vertically upward direction from the ground into the space. ($v_e$ is escape velocity and $k < 1$). If air resistance is considered to be negligible then the maximum height from the centre of earth to whichit can go, will be : ($R =$ radius of earth)
$\frac{R}{{{k^2} + 1}}$
$\frac{R}{{{k^2} - 1}}$
$\frac{R}{{1 - {k^2}}}$
$\frac{R}{{k + 1}}$
Solution
Kinetic energy $=$ Potential energy
$\frac{1}{2} \mathrm{m}\left(\mathrm{k} v_{\mathrm{e}}\right)^{2}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}} \Rightarrow \frac{1}{2} \mathrm{mk}^{2} 2 \mathrm{g} \mathrm{R}=$
$\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}} \Rightarrow \mathrm{h}=\frac{\mathrm{Rk}^{2}}{1-\mathrm{k}^{2}}$
Height of Projectile from the earth's surface $=\mathrm{h}$
Height from the centre $\mathrm{r}=\mathrm{R}+\mathrm{h}=\mathrm{R}+\frac{\mathrm{Rk}^{2}}{1-\mathrm{k}^{2}}$
By solving $\mathrm{r}=\frac{\mathrm{R}}{1-\mathrm{k}^{2}}$
