The escape velocity of a body on the earth's surface is v_e . A body is thrown vertically up with a speed of ${\mathrm{kv}}_{\mathrm{e}}\left(\mathrm{k}<1\right)$. The maximum height reached by the body above the earth is

${\mathrm{V}}_{\mathrm{e}}=\sqrt{\frac{2\mathrm{GM}}{\mathrm{R}}}$

Apply law of conservation of energy at surface and at height h:

${\left(\mathrm{TE}\right)}_{surface}={\left(\mathrm{TE}\right)}_{max height=h}$

$\Rightarrow -\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2}{\mathrm{mv}}^2=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}+0$

$\Rightarrow -\frac{\mathrm{GM}}{\mathrm{R}}+\frac{1}{2}{\mathrm{k}}^2{\mathrm{v}}_{\mathrm{e}}^2=-\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}$

$\Rightarrow -\frac{\mathrm{GM}}{\mathrm{R}}+\frac{1}{2}{\mathrm{k}}^2.\frac{2\mathrm{GM}}{\mathrm{R}+\mathrm{h}}=-\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}$

$\Rightarrow -\frac{1}{\mathrm{R}}+\frac{{\mathrm{k}}^2}{\mathrm{R}}=-\frac{1}{\mathrm{R}+\mathrm{h}}$

$\Rightarrow \mathrm{h}=\frac{{\mathrm{Rk}}^2}{1-{\mathrm{k}}^2}$