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“Arrives at infinity with zero speed” is an asymptotic statement. For outward radial motion at zero specific orbital energy,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 $(4-\sqrt{2})/3\mathrm{s}\approx 0.861929\mathrm{s}$. 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.

The zero-torque inner boundary condition sets the internal disk torque to zero at a chosen inner radius. It still permits a nonzero accretion rate: the gas carries its own specific angular momentum across the boundary. For a steady Keplerian disk, it gives $\nu \Sigma =\dot{M}[1-\sqrt{r_{\mathrm{in}}/r}]/(3\pi )$. A rapid plunge near the innermost stable circular orbit motivates this approximation, but magnetic coupling or a stellar boundary layer requires examination of the actual stresses.^[ref]