Pericentre-to-crossover flight time
= Pericentre-to-crossover flight time
{title2=$n_pt_x\simeq2^{7/8}(q/a_p)^{3/8}/3$}
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 $n_pt_x\simeq(\sqrt2/3)(r_x/a_p)^{3/2}=2^{7/8}(q/a_p)^{3/8}/3$. The small time fraction explains why an eccentric orbit turns through almost $\pi$ of <true anomaly> while a distant perturber barely moves.