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A particle of mass $'\mathrm{m}'$ is projected with a velocity $v=\mathrm{kV}_{\mathrm{e}}(\mathrm{k}\,<\,1)$ from the surface of the earth.
$\left(\mathrm{V}_{\mathrm{e}}=\right.\text { escape velocity) }$
The maximum height above the surface reached by the particle is :
$\mathrm{R}\left(\frac{\mathrm{k}}{1-\mathrm{k}}\right)^{2}$
$\mathrm{R}\left(\frac{\mathrm{k}}{1+\mathrm{k}}\right)^{2}$
$\frac{\mathrm{R}^{2} \mathrm{k}}{1+\mathrm{k}}$
$\frac{\mathrm{Rk}^{2}}{1-\mathrm{k}^{2}}$
Solution
$-\frac{G M m}{R}+\frac{1}{2} m k^{2} v_{e}^{2}=-\frac{G M m}{r}$
$-\frac{G M m}{R}+\frac{1}{2} m k^{2} \frac{2 G M}{R}=-\frac{G M m}{r}$
$-\frac{1}{R}+\frac{k^{2}}{R}=-\frac{1}{r}$
$\frac{1}{r}=\frac{1}{R}-\frac{k^{2}}{R}$
$\frac{1}{r}=\frac{1-k^{2}}{R}$
$r=\frac{R}{1-k^{2}}$
$R+h=\frac{R}{1-k^{2}}$
$h=\frac{R}{1-k^{2}}-R=\frac{k^{2}}{1-k^{2}} R$
