Problem source

A comet, which may be regarded as a particle of mass $m$, moving in the sun's gravitational field, at a distance $x$ from the sun, experiences a force $Gm/x^{2}$ (where $G$ is a constant) directly towards the sun. Show that if, at some time, $x=h$ and the comet is travelling directly away from the sun with speed $V$, then $x$ cannot become arbitrarily large unless $V^{2}\geqslant2G/h$. 
A comet is initially motionless at a great distance from the sun. If, at some later time, it is at a distance $h$ from the sun, how long after that will it take to fall into the sun?

Solution source

Consider $E = \frac12 m \dot{x}^2 - \frac{Gm}{x}$, notice that
\begin{align*}
&& \dot{E} &= m \dot{x} \ddot{x} + \frac{Gm}{x^2} \dot{x} \\
&&&= \dot{x} \underbrace{\left (m\ddot{x} + \frac{Gm}{x^2} \right)}_{=0 \text{ by N2}}
\end{align*}

Therefore $E$ is conserved. Therefore if $x \to \infty$ $\frac12 m V^2 - \frac{Gm}{h} = \frac12 m u^2 - 0 \geq 0$ so $V^2 \geqslant 2G/h$

Since $E \approx 0$ we want to solve 
\begin{align*}
&& \dot{x} &= -\sqrt{\frac{2G}{x}} \\
\Rightarrow && -\int_h^0 \sqrt{x} \d x &= \int_0^T \sqrt{2G} \d t \\
\Rightarrow && \frac{2h^{3/2}}{3} &= \sqrt{2G}T \\
\Rightarrow && T &= \frac{\sqrt{2}h^{3/2}}{3\sqrt{G}} = \frac13 \sqrt{\frac{2h^3}{G}}
\end{align*}