Barker equation

For a parabolic trajectory with periapsis radius $r_p$, put $D=\tan (f/2)$, where $f$ is true anomaly. The Barker equation is $t-t_p=\sqrt{2r_p^3/(GM)}(D+D^3/3)$. Since $r=r_p(1+D^2)$, it is equivalent to the radius-time relation for a parabolic Kepler orbit. The sign of $D$ distinguishes approach from recession; $t_p$ is the periapsis time.

On a nonradial zero-energy Kepler orbit, conserved specific angular momentum $h$ gives periapsis $r_p=h^2/(2GM)$ and ${\dot{r}}^2=2GM(r-r_p)/r^2$. Integrating with $s=\sqrt{r-r_p}$ yields $t-t_p=\pm \sqrt{2/(GM)}(r+2r_p)\sqrt{r-r_p}/3$. Squaring gives the displayed version of the Barker equation, valid on both incoming and outgoing branches. The time origin $t_p$ is passage through periapsis, not an arbitrary point on the orbit.

A parabolic Kepler orbit obeys $t=\sqrt{2q^3/(GM)}(D+D^3/3)$ with $D=\tan (f/2)$. At the high-eccentricity angular-speed crossover, $q\ll r_x$, so $n_p{t}_x\simeq (\sqrt{2}/3)(r_x/a_p{)}^{3/2}=2^{7/8}(q/a_p{)}^{3/8}/3$. The small time fraction explains why an eccentric orbit turns through almost $\pi$ of true anomaly while a distant perturber barely moves.