= Zero-energy escape takes infinite time
{title2=$r(t)=[r_0^{3/2}+\tfrac32\sqrt{2\mu}\,t]^{2/3}$}
“Arrives at infinity with zero <speed>” is an asymptotic statement. For outward radial motion at zero <specific orbital energy>,
$$
\dot r=\sqrt{\frac{2\mu}{r}},\qquad t(r)=\frac{2}{3\sqrt{2\mu}}\left(r^{3/2}-r_0^{3/2}\right),\qquad r(t)=\left(r_0^{3/2}+\frac32\sqrt{2\mu}\,t\right)^{2/3}.
$$
Every finite distance takes finite <time>, but $t(r)\to\infty$ as $r\to\infty$. There is no instant of rest at an infinitely distant destination. Differentiating the displayed solution gives $\ddot r=-\mu/r^2$, independently checking <Newton's second law>. In the synthetic example, reaching $r=4\,\mathrm m$ takes exactly $(4-\sqrt2)/3\,\mathrm s\approx0.861929\,\mathrm s$. For nonzero <specific angular momentum>, the <Barker equation> gives the corresponding infinite-<time> limit along a <parabolic trajectory>. Positive-<energy> escape also takes infinite <time> to reach infinity; it retains nonzero asymptotic <speed>.
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