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The plunging region begins near the innermost stable circular orbit, where nearly circular motion becomes unstable and radial inflow becomes rapid. If the inflow time is shorter than the stress-transport time, the gas approximately conserves its specific angular momentum there.

For an attractive inverse-square force, $r=p/(1+e\cos f)$, where $f$ is true anomaly, $e$ is orbital eccentricity, and $p=h^2/\mu$ is the semilatus rectum. The specific angular momentum relation $\dot{f}=h/r^2$ converts this geometric equation into an angular evolution law.

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.

For inviscid cylindrical rotation without a magnetic field, an axisymmetric radial interchange conserves each parcel's specific angular momentum $l=r^2\Omega$. Its restoring coefficient is ${\kappa }^2=r^{-3}d(l^2)/dr$. Increasing $l^2$ gives centrifugal stability; decreasing $l^2$ gives an exponentially growing interchange mode when radial displacement is allowed by the boundaries. A vanishing derivative is marginal, not a strictly restoring condition.

// Target: astrophysics.bigb

For a slowly evolving circular orbit with tangential deceleration $D>0$, the specific angular momentum is $j=r{V}_c(r)$ and its torque is $\dot{j}=-rD$. Differentiating $j$ gives the displayed drift when $1+q\ne 0$. The positive local decay time $r/(2|\dot{r}|)$ is $(1+q)V_c/(2D)$. In a uniform-density core $q=1$, so it is $V_c/D$; for a flat galaxy rotation curve $q=0$, it is $V_c/(2D)$. Simply setting ${\dot{V}}_c=-D$ neglects the radial contribution to angular momentum change.

The same initial distance can lead to strikingly different orbits. Let $s=v_0/\sqrt{\mu /r_0}\ge 1$ and launch tangentially in one orbital plane. Then periapsis is the launch point, specific angular momentum is $h=r_0{v}_0$, andwhere $f$ is true anomaly. These follow by combining the initial specific orbital energy with the orbital eccentricity relation and the Kepler orbit equation. With $\mu =8{\mathrm{m}}^3/{\mathrm{s}}^2$ and $r_0=2\mathrm{m}$, the initial speeds $2$, $\sqrt{6}$, $2\sqrt{2}$, and $2\sqrt{3}m/s$ give $e=0,1/2,1,2$: a circular orbit, an elliptic orbit with apoapsis distance $6\mathrm{m}$, a parabolic trajectory, and a hyperbolic Kepler orbit. The two open paths have asymptotic speeds zero and $2m/s$ respectively. Their different directions at large distance are part of the Kepler orbit, not an extra scalar escape threshold.

A thin galactic disk is a stellar component whose vertical extent is small compared with its radial size. Small vertical velocity dispersion and substantial specific angular momentum are characteristic. Its survival constrains cumulative galactic disk heating, but its stars need not all have the same formation epoch.

A stationary axisymmetric galactic distribution function depending on the specific orbital energy and axial specific angular momentum is a two-integral distribution. These quantities are conserved in a time-independent axisymmetric Newtonian gravitational potential, so the Collisionless Boltzmann equation holds by the chain rule. Even dependence on $L_z$ removes net azimuthal streaming velocity; an odd part can supply rotation without changing the mass density.

For an instantaneous in-plane change of velocity of magnitude $\gamma v_K$ on a circular orbit, with angle $\theta$ measured from the forward tangential direction, the new specific orbital energy gives $a^{\prime }/a=(1-2\gamma \cos \theta -{\gamma }^2{)}^{-1}$. The specific angular momentum gives $e^{\prime 2}=1-(1-2\gamma \cos \theta -{\gamma }^2)(1+\gamma \cos \theta {)}^2$. The signed semi-major axis becomes negative for an unbound hyperbolic Kepler orbit.

Subtract local specific angular momentum times mass conservation from conservation of angular momentum including wind losses S,T. With outward radial mass flux $\mathcal{F}$ and outward viscous torque in an accretion disk $\mathcal{G}=-2\pi \bar{\nu }\Sigma r^3{\Omega }^{\prime }$, this gives $\mathcal{F}{h}^{\prime }+{\mathcal{G}}^{\prime }=2\pi r(Sh-T)$. For Keplerian rotation and $T=qSh$, substitution in mass conservation yields the displayed equation. A wind with q=1 removes local angular momentum but supplies no additional drift term; q>1 can drive accretion even with zero viscosity.

For a Keplerian disk, let $S$ be the total loss through both faces per unit projected area and $\dot{M}>0$ the inward accretion rate. Annular mass conservation gives ${\dot{M}}^{\prime }=2\pi RS$. A circular orbit has specific orbital energy $-GM/(2R)$, so a marginally escaping wind requires additional energy $GM/(2R)$ per unit lost mass. Equating this sink to a fraction $f$ of the two-face viscous dissipation $(9/4)\nu \Sigma {\Omega }^2$ gives the displayed relation. If the wind carries the local specific angular momentum $h=\sqrt{GMR}$, conservation of angular momentum gives $(\dot{M}h-\mathcal{G}{)}^{\prime }=h{\dot{M}}^{\prime }$, with viscous torque in an accretion disk $\mathcal{G}=3\pi \nu \Sigma h$. Eliminating the torque yields $R^2{\dot{M}}^{\prime \prime }+\frac{3}{2}R{\dot{M}}^{\prime }=\frac{3}{2}f\dot{M}$.

“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.