A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height h is ______ S.

Applying energy conservation from (1) to (2)

$\frac{1}{2}m\cdot \left(\frac{2GM}{R_e}\right)-\frac{GMm}{R_e}=\frac{1}{2}m{v}^2-\frac{GMm}{R+r}$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{GMm}{R+r}$

$\Rightarrow v=\sqrt{\frac{2GM}{R+r}}=\frac{dr}{dt}\Rightarrow \sqrt{2GM}{\int }_0^{t}dt={\int }_{R_e}^{R_e+h}\left(\sqrt{R+r}\right)dr$

$\sqrt{2GM}\cdot t=\frac{2}{3}{\left[\right(R+r{)}^{3/2}]}_{R_e}^{R_e+h}$

$t=\frac{2}{3}\sqrt{\frac{R_e^2}{2GM}}[{(1+\frac{h}{R_e})}^{3/2}-1]$

$\frac{GM}{R_e^2}=g,t=\frac{1}{3}\sqrt{\frac{2R_e}{g}}[{(1+\frac{h}{R_e})}^{3/2}-1]$