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Use the leading vertically averaged horizontal momentum equations of a thin disk. For an isothermal ideal gas, write $P=\int pdz=c_s^2\Sigma$, with constant isothermal sound speed $c_s$. The disk scale height is of order $H=c_s/{\Omega }_z$; this is exact for local isothermal vertical hydrostatic balance and remains the thickness scale for the estimate. Put $ϵ=H/r\ll 1$. The alpha disk prescription givesAt fixed radii of order $a$ inside the plunge and away from the marginal orbit, $|v_r|,v_{\varphi }$ and $r{\Omega }_z$ are fixed numerical multiples of $(GM/a{)}^{1/2}$. Hence the inflow time is dynamical, while the viscous timescale is longer by $1/(\alpha {ϵ}^2)$.

The Paczyński-Wiita ballistic plunge obeys $v_r{v}_r^{\prime }-v_{\varphi }^2/r=-{\Phi }^{\prime }$ and $v_r{j}_0^{\prime }=0$ exactly. For smoothly varying disk quantities on scale $r$, the pressure correction to radial acceleration has magnitude $c_s^2/r$, and the viscous correction is of order $\nu v/r^2$. Relative to the orbital acceleration scale $r{\Omega }_z^2$, these are respectivelyFinite-height corrections to the mid-plane gravitational force are also $O({ϵ}^2)$ by reflection symmetry. Thus the leading horizontal flow is ${\mathbf{u}}_h=\mathbf{v}+O({ϵ}^{2}v)$, with specific angular momentum differing from $j_0$ by a fraction $O(\alpha {ϵ}^2)$ over a dynamical transit, for $\alpha \lesssim 1$. Vertical structure and the smaller vertical velocity must be supplied by the thin-disk expansion; they are not specified by the planar ballistic expression.

This estimate is not uniform at the marginal orbit. In the ballistic solution, steady mass conservation givesConsequently $d\ln \Sigma /dr=[2(r-a){]}^{-1}+3[2(3a-r){]}^{-1}$. Writing $\delta =3a-r\ll a$, the pressure acceleration scales as $c_s^2/\delta$, while the net ballistic radial force scales as $(GM/a^2)(\delta /a{)}^2$. Their ratio is $O[{ϵ}^2(a/\delta {)}^3]$, evaluated with the characteristic thickness near $3a$. A transonic matching layer of width $\delta /a=O({ϵ}^{2/3})$ is therefore needed. The approximation concerns the bulk plunging flow outside that layer, not a uniform relative approximation arbitrarily close to $3a$.

For the torque comparison, the steady angular-momentum flux is $\mathcal{G}-\dot{M}j=\mathrm{constant}$, where $\mathcal{G}=-2\pi \nu \Sigma r^3{\Omega }^{\prime }$ is the viscous torque in an accretion disk. Inside the plunge, $\Omega =j_0/r^2$, soOutside, the approximately circular disk has $\mathcal{G}=\dot{M}[j(r)-j_0]+O(\alpha {ϵ}^2\dot{M}{j}_0)$, since the inner plunge torque is small. At $4a$,a fixed number independent of disk thickness. At $2a$, radial infall is dynamical and the inner torque is $O(\alpha {ϵ}^2\dot{M}{j}_0)$. Therefore the weak viscous torque in a thin-disk plunging region estimate givesThe numerical prefactor need not be small; the suppression is an asymptotic statement in the thin-disk parameter. Rapid infall and its low surface density explain why a nearly zero inner torque is a useful leading boundary condition for the outer accretion disk.