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 \(....\,S.\)

\(\frac{1}{2} m_{.}\left(\frac{2 G M}{R_{e}}\right)-\frac{G M m}{R_{e}}=\frac{1}{2} m v^{2}-\frac{G M m}{R+r}\)

\(\Rightarrow \frac{1}{2} m v^{2}=\frac{G M m}{R+r}\)

\(\Rightarrow v=\sqrt{\frac{2 G M}{R+r}}=\frac{d r}{d t}\)

\(\Rightarrow \sqrt{2 G M} \int_{0}^{t} d t=\int_{R_{e}}^{R_{e}+h}(\sqrt{R+r}) d r\)

\(\sqrt{2 G M} \cdot t=\frac{2}{3}\left[(R+r)^{3 / 2}\right]_{R_{e}}^{R_{e}+h}\)

\(t=\frac{2}{3} \sqrt{\frac{R_{e}^{3}}{2 G M}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]\)

\(\frac{G M}{R_{e}^{2}}=g\)

\(t=\frac{1}{3} \sqrt{\frac{2 R_{e}}{g}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]\)