= Solution
Write $h=r^2\dot\theta$ for the conserved <specific angular momentum>. Zero <specific orbital energy> in the inverse-square <Newtonian gravitational field> gives
$$
\frac12\left(\dot r^2+\frac{h^2}{r^2}\right)-\frac{GM}r=0.
$$
Define $r_0=h^2/(2GM)$, the <periapsis> radius for a nonradial <parabolic trajectory>. Then
$$
\dot r^2=\frac{2GM(r-r_0)}{r^2}.
$$
On the outgoing branch, integrate from the <periapsis> time $t_0$ and put $s=\sqrt{r-r_0}$:
$$
\begin{aligned}
t-t_0
&=\frac1{\sqrt{2GM}}\int_{r_0}^r\frac{r'\,dr'}{\sqrt{r'-r_0}}\\
&=\frac2{\sqrt{2GM}}\left(\frac{s^3}3+r_0s\right)
=\frac{\sqrt2}{3\sqrt{GM}}(r+2r_0)\sqrt{r-r_0}.
\end{aligned}
$$
The incoming branch has the opposite sign. Squaring gives the <radius-time relation for a parabolic Kepler orbit>, a version of the <Barker equation>:
$$
\boxed{(r-r_0)(r+2r_0)^2=\frac92GM(t-t_0)^2.}
$$
Here $r_0$ and $t_0$ are fixed by <angular momentum> and the time origin at <periapsis>. The relation applies on both branches, with $r\ge r_0$; it is not necessary to solve for the polar angle.
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