= Solution
For a fixed central gravitational source, the radial and transverse components of <Newton's second law> in <polar coordinates> are
$$
\ddot r-r\dot\theta^2=-\frac{GM}{r^2},\qquad
r\ddot\theta+2\dot r\dot\theta=0.
$$
The second equation gives $r^2\dot\theta=h$, the constant specific <angular momentum>. Take $h>0$ by choosing the orientation of $\theta$, and put $u=1/r$. Writing $u'$ and $u''$ for <derivatives> with respect to $\theta$,
$$
\dot r=-hu',\qquad\ddot r=-h^2u^2u'',\qquad
r\dot\theta^2=h^2u^3.
$$
Substituting into the radial equation and dividing by $-h^2u^2$ proves the <Binet equation>
$$
\boxed{u''+u=\frac{GM}{h^2}=k.}
$$
At perihelion the <velocity> is tangential. The <escape velocity> is $v_0=\sqrt{2GM/r_0}$, so $h=r_0v_0$ and $h^2=2GMr_0$. Consequently $k=1/(2r_0)$. The initial conditions are $u(0)=1/r_0$ and $u'(0)=0$. Solving the <linear ordinary differential equation> gives the <parabolic Kepler orbit>
$$
\boxed{u(\theta)=\frac{1+\cos\theta}{2r_0}
=\frac1{r_0}\cos^2(\theta/2),\qquad
r=r_0\sec^2(\theta/2).}
$$
Using <angular momentum> conservation again,
$$
\dot\theta=\frac h{r^2}=\frac h{r_0^2}\cos^4(\theta/2),
\qquad\boxed{\sec^4(\theta/2)\dot\theta=\frac h{r_0^2}.}
$$
On the outgoing branch $r=2r_0$ is first reached at $\theta=\pi/2$. Integrate from release at perihelion, using $w=\tan(\theta/2)$:
$$
t=\frac{r_0^2}{h}\int_0^{\pi/2}\sec^4(\theta/2)d\theta
=\frac{2r_0^2}{h}\int_0^1(1+w^2)dw
=\boxed{\frac{8r_0^2}{3h}.}
$$
More generally the same integral gives $t=(2r_0^2/h)(w+w^3/3)$, the <Barker equation> for this parabolic orbit.
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