Physics - Gravitation Question with Solution | TestHub

Applying energy conservation from ($\left(1\right) \text{to} \left(2\right)$,
$\frac{1}{2} m·\left(\frac{2\mathrm{GM}}{R_e}\right)-\frac{GMm}{R_e}=\frac{1}{2} m{v}^2-\frac{GMm}{R+r}$

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

$\Rightarrow v=\sqrt{\frac{2GM}{R+r}}=\frac{dr}{dt}$

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

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

$t=\frac{2}{3}\sqrt{\frac{{\mathrm{R}}_{\mathrm{e}}^3}{2\mathrm{GM}}}\left[{\left(1+\frac{h}{R_e}\right)}^{\frac{3}{2}}-1\right]$

$\frac{GM}{R_e^2}=g$

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