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.
Answer: (D) $\frac13\sqrt{\frac{2R_e}{g}}\left[\left(1 + \frac{h}{R_e}\right)^{\frac32} - 1\right]$
Launched at escape speed, the total energy is zero, so at distance $r$ from the centre
$$v = \sqrt{\frac{2GM}{r}} = R_e\sqrt{\frac{2g}{r}}$$
$\dfrac{dr}{dt} = R_e\sqrt{2g}\,r^{-1/2} \Rightarrow \displaystyle\int_{R_e}^{R_e + h} r^{1/2}\,dr = R_e\sqrt{2g}\,t$
$$\frac23\left[(R_e + h)^{3/2} - R_e^{3/2}\right] = R_e\sqrt{2g}\,t$$
$$t = \frac{2}{3}\cdot\frac{R_e^{3/2}}{R_e\sqrt{2g}}\left[\left(1 + \frac{h}{R_e}\right)^{3/2} - 1\right] = \frac13\sqrt{\frac{2R_e}{g}}\left[\left(1 + \frac{h}{R_e}\right)^{3/2} - 1\right]$$
Solution by Sreeraj P, M.Sc Physics