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Work in the planet's rest frame, neglecting its recoil and tidal forces during the close encounter. A hyperbolic Kepler orbit has positive specific orbital energy $\epsilon =v_{\infty }^2/2$ and specific angular momentum $h=b{v}_{\infty }$, where $b$ is the impact parameter. Its orbital eccentricity isAt the pericentre distance, $r_{\min }=h^2/[\mu (1+e_h)]$. Substituting $h^2={\mu }^2(e_h^2-1)/v_{\infty }^2$ yieldsThe two asymptotes have true anomalies $\pm f_{\infty }$, where $\cos f_{\infty }=-1/e_h$. Their directions of motion give the gravitational scattering angle $\theta =2f_{\infty }-\pi$, so $\sin (\theta /2)=-\cos f_{\infty }=1/e_h$. ThereforeAs a useful independent form, $b^2=r_{\min }^2+2\mu r_{\min }/v_{\infty }^2$, displaying gravitational focusing. Here $\mu =G{M}_p$ to the stated small-comet accuracy; the exact relative-coordinate formula replaces it by $G(M_p+M)$.

Let $r_0$ be the separation of the two centres at release. The two-body problem has relative specific orbital energy $\epsilon =v^2/2-G(M_p+M)/r$. Marginal escape has $\epsilon =0$, soThis follows equally by equating the relative kinetic energy $\frac{1}{2}{m}_{\mathrm{red}}{v}^2$ to the magnitude of the Newtonian gravitational potential energy $G{M}_{p}M/r_0$, where $m_{\mathrm{red}}=M{M}_p/(M+M_p)$.

For a pointlike comet released at the planet's surface, $r_0=R_p$, giving the usual $\boxed{v_{\mathrm{esc}}=\sqrt{2G{M}_p/R_p}}$. If the finite comet touches that surface, its centre is at $r_0=R_p+R$; the same formula uses this larger separation. The assumption $M_p\gg M$ by itself does not imply $R\ll R_p$. Subsequent formulas written in terms of the planet-only escape velocity use the additional small-comet approximation.

This is local escape from an isolated planet. In a stellar Newtonian gravitational field, reaching the edge of the Hill sphere and escaping the stellar system are different energy thresholds.

Let $\chi =v_{\mathrm{esc}}/v_k$, with $v_k=\sqrt{G{M}_{\star }/a_p}$. The Safronov number is $\Theta ={\chi }^2/2$. In the instantaneous planet frame let the asymptotic vectors be ${\mathbf{u}}_{\mathrm{in}}$ and ${\mathbf{u}}_{\mathrm{out}}$; their magnitudes agree. In the stellar frame the corresponding specific orbital energy change during a short encounter isThe quadratic relative-speed terms cancel. A comet with semi-major axis comparable to $a_p$ has stellar binding energy per unit mass of order $v_k^2/2$. Consequently $\chi$ measures whether a single strong encounter can change a substantial fraction of that binding.

Large escape-to-orbital-speed ratios favour ejection; small ratios favour collision or accretion during repeated encounters. For $\chi \gg 1$, a suitably oriented gravitational assist can eject the comet before it strikes the planet. For $\chi \ll 1$, most individual kicks are too weak; many close passages may be needed, providing repeated chances of collision.

The crossover is of order unity, not a sharp theorem about each orbit. The initial binding energy, encounter speed and geometry matter: an already nearly parabolic comet may be ejected by a low-$\chi$ planet, while an unfavourably directed kick can make a comet more tightly bound even when $\chi$ is large.

Write $x=1/a$ for the comet's inverse semi-major axis. Its specific orbital energy is $-G{M}_{\star }x/2$, so escape corresponds to $x=0$. The printed quantityhas dimensions of inverse length. It cannot, literally, be a standard variance-per-time diffusion coefficient. Interpret it as the characteristic root mean square step in $x$ per periapsis passage, as in a discrete comet energy diffusion model.

If successive kicks are unbiased and uncorrelated, after $N$ passages the root mean square displacement is $s_x\sqrt{N}$. Starting at $a_0$ gives $x_0=1/a_0$, hence $N_{\mathrm{ej}}\sim (x_0/s_x{)}^2$. Taking the characteristic interval to be the initial orbital period $P(a_0)$ givesHere $P_p=2\pi \sqrt{a_p^3/(G{M}_{\star })}$. In particular, for $a_0\sim a_p$ this is $P_p(M_{\star }/M_p{)}^2/100$. Without specifying the initial semi-major axis and the time per statistically independent encounter, the PDF cannot determine a unique time.

With the convention $⟨(\Delta x{)}^2⟩=2D_{x}t$, the actual diffusion coefficient would be $D_x=s_x^2/[2P(a_0)]$. The estimate requires $N_{\mathrm{ej}}\gg 1$, orbit crossing, and sufficient phase decorrelation. Near escape the orbital period grows, and resonant or secular correlations invalidate the simple constant-step clock. This is a characteristic diffusion scale, not an exact mean first-passage time: even an unbiased Brownian motion on an unbounded half-line has an infinite mean time to reach its absorbing endpoint.

Put ${\mu }_{\star }=G{M}_{\star }$. Conservation of specific orbital energy and specific angular momentum givesThe last comet expression is the vis-viva equation at the planet's orbit. The planet has only tangential velocity, while the comet's tangential component is $v_t=h/a_p$. Consequently its prograde relative velocity is exactly, before planetary deflection,In the near-parabolic encounter limit $Q\gg a_p$, this reduces toThe displayed approximation also requires $Q\gg a_p$. The assumptions $Q\gg q$ and $q\ll a_p$ alone do not guarantee it: an orbit with $Q=a_p$ encounters the planet near apoapsis and has $v_{\mathrm{rel}}\simeq v_p$, not $\sqrt{3}{v}_p$.

If all coplanar near-parabolic planet-crossing comets are admitted, so that $0\le q\le a_p$, the tangential component ranges from $-\sqrt{2}{v}_p$ to $+\sqrt{2}{v}_p$ when retrograde orbits are included. ThusFor prograde orbits alone the upper limit is $\sqrt{3}{v}_p$; for the stated deep-crossing limit $q\ll a_p$, both prograde and retrograde encounters approach $\sqrt{3}{v}_p$. These are the speeds outside the planet's gravitational well. The surface impact speed is $v_{\mathrm{imp}}=\sqrt{v_{\mathrm{rel}}^2+v_{\mathrm{esc},\mathrm{p}}^2}$, where $v_{\mathrm{esc},\mathrm{p}}$ is the planetary escape velocity; under negligible gravitational focusing the two speeds agree.

In a planetary gravitational assist, the incoming and outgoing relative speeds have the same magnitude. If $\mathbf{u}$ is the planet-frame velocity, the heliocentric specific orbital energy change is consequentlynot an arbitrary addition of $|\Delta \mathbf{u}|$ to the heliocentric speed. The initial binding energy per unit mass is $G{M}_{\star }/(Q+q)\simeq G{M}_{\star }/Q$.

For a weak deflection, $b\gg b_{90}=G{M}_p/v_{\infty }^2$, the hyperbolic scattering formula gives $|\Delta \mathbf{u}|\simeq 2G{M}_p/(b{v}_{\infty })$. Its direction is perpendicular to the incoming relative velocity. At the planet, in the near-parabolic prograde limit,On the favorable side of the planet the energy gain is approximately $2G{M}_p{v}_p\kappa /(b{v}_{\infty })$. Equating this to the binding energy gives the ejection impact parameterOnly one sign of impact parameter gains energy for each radial branch. Noncolliding ejections have $R_p<|b|<b_{\mathrm{ej}}$, giving width $(b_{\mathrm{ej}}-R_p{)}_{+}$, where $x_{+}=\max (x,0)$. For $b_{\mathrm{ej}}\gg R_p$, the planetary ejection of a comet rate is approximately ${\Gamma }_{\mathrm{ej}}={\Sigma }_c{v}_{\infty }{b}_{\mathrm{ej}}$. With ${\mu }_p=M_p/M_{\star }$ this givesThe scaling applies when $b_{90}\ll b_{\mathrm{ej}}\ll r_H$ so that the local two-body encounter and weak-deflection approximations are consistent. The favorable-side coefficient assumes a coplanar orbit; averaging over other encounter geometries changes it. Cumulative diffusion by many subthreshold encounters is not included in this one-encounter estimate.

Assume negligible fragment release speed relative to a circular parent planetesimal. Its initial speed is $v_b=\sqrt{G{M}_{\star }/r_b}$, but radiation pressure reduces the grain's gravitational parameter to $(1-\beta )G{M}_{\star }$. Its specific orbital energy and specific angular momentum areFor $0\le \beta <1/2$, the resulting dust orbit released from a circular parent ring hasRelease is at periapsis: the inherited tangential speed exceeds the new circular speed. Low-$\beta$ grains remain near the dust birth ring; higher bound $\beta$ gives increasingly eccentric orbits extending well outside it. Uniform release longitude produces an approximately axisymmetric dust halo, with long residence times near apoapsis. At $\beta =1/2$ the conservative release orbit is parabolic; above this radiation-pressure blowout threshold it is initially unbound. For $\beta \ge 1$, zero or repulsive effective gravity also prevents a bound ellipse.

The left panel shows one release longitude; rotating it around the ring fills the halo. At longer times Poynting–Robertson drag also carries bound grains inward, so an initially outward-spanning release distribution does not imply an empty inner region in a continuously supplied, collision-free population.

The specific orbital energy released in a black-hole accretion flow reaching an inner radius $r_{\mathrm{in}}$ is of order $GM/r_{\mathrm{in}}$. For a circular Keplerian orbit, the binding energy is more precisely $GM/(2r_{\mathrm{in}})$, with the other half of the potential-energy decrease retained as orbital kinetic energy. Multiplying by the mass accretion rate gives the accretion luminosity scaleFor an inner radius of order the Schwarzschild radius, $r_{\mathrm{in}}\sim 2GM/c^2$, this becomesThis is an energy-release scaling, not unit radiative efficiency of black-hole accretion. The actual inner orbit, stresses and advected energy determine the coefficient and the emitted luminosity; the innermost stable circular orbit of a Schwarzschild black hole lies outside its horizon.

The corotation radius is where the stellar spin matches Keplerian rotation:A parcel forced into corotation at $r_m$ has outward centrifugal acceleration ${\Omega }_m^2{r}_m$. This exceeds the inward gravitational acceleration whenThis is the onset condition for a centrifugal propeller effect. It does not by itself guarantee escape to infinity. For example, a parcel released with only azimuthal speed ${\Omega }_m{r}_m$ has nonnegative specific orbital energy only ifAdditional magnetic work, radial motion or pressure can alter that escape condition.

With $\hat{\mathbf{r}}=(\cos \theta ,\sin \theta )$ and $\hat{\mathbf{\theta }}=(-\sin \theta ,\cos \theta )$, project the relative gas drag to obtainThe specific orbital energy is $\epsilon =v^2/2-\mu /r=-\mu /(2a)$. Differentiation cancels the conservative gravitational work and gives $\dot{\epsilon }=\mathbf{v}\cdot \mathbf{f}$. For the circular orbit,Using $v^2=\mu /a$, the orbit-averaged drag work relation therefore gives the instantaneous leading-order changeThe differential headwind term can change sign during the orbit; the damping term always removes orbital energy.

Let $\mathbf{V}$ be the centre of mass velocity and $\mathbf{v}$ the binary relative velocity. Summing the two linear gas drag forces givesThe last term averages to zero over the internal circular orbit. On the nearly circular stellar orbit, $\mathbf{V}\simeq v_K\hat{\mathbf{\theta }}$ and ${\mathbf{v}}_g=(1-\eta )v_K\hat{\mathbf{\theta }}$, so the averaged tangential acceleration is $-\eta {\kappa }_b{v}_K$.

With specific orbital energy $-G{M}_{\star }/(2a_b)$, the orbit-averaged drag work is $-\eta {\kappa }_b{v}_K^2$, where $v_K^2=G{M}_{\star }/a_b$. Thus gas-driven migration of a binary centre of mass givesThis uses the weak-drag, nearly Keplerian approximation implicit in applying the circular-orbit work relation. Strongly gas-coupled orbits require a coupled radial-azimuthal drift solution.

Put $r_S=2R_g$. On the midplane, circular orbit balance gives $r{\Omega }^2={\partial }_r\Phi =GM/(r-r_S{)}^2$. The specific angular momentum, specific orbital energy and orbital shear parameter of a Paczyński-Wiita circular orbit are thereforeFor a small radial displacement at fixed specific angular momentum, the effective potential stability criterion gives the radial epicyclic frequencyHere ${\kappa }_r$ is the question's epicyclic $\kappa$. Its square is negative for $2R_g<r<6R_g$, so a radial displacement grows instead of undergoing epicyclic motion. The orbit at $r_{\mathrm{in}}=6R_g$ is marginal, and the outer orbits are stable to small radial displacements. At this inner edge,This efficiency belongs to the Paczyński-Wiita potential; it is an approximation to the relativistic black-hole result.

In a steady state, mass conservation makes $\mathcal{F}$ independent of radius. Taking the accretion rate $\dot{M}>0$ for inward flow gives $\mathcal{F}=-\dot{M}$. The conservation of angular momentum equation then makes $-\dot{M}h+\mathcal{G}$ constant. The zero-torque inner boundary condition fixes this constant to $-\dot{M}{h}_{\mathrm{in}}$, soThe viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^3\bar{\nu }\Sigma {\Omega }^{\prime }=2\pi qh\bar{\nu }\Sigma$. Combining these expressions gives the steady accretion disk with arbitrary rotation lawFor $x=r/R_g$, the Paczyński-Wiita circular orbit formulas giveConsequentlyIn particular, $f(6)=0$ and $f\to 1$ far from the black hole, recovering the outer Keplerian accretion disk.

Inside the innermost stable circular orbit, the gas enters the plunging region of a black-hole accretion disk. Its rapid inward motion leaves little time for stresses to exchange angular momentum, motivating the zero-torque inner boundary condition. This is a thin-disc approximation; a strong magnetic stress could change it.

Matter supplied from very large radius has negligible specific orbital energy, while matter crossing the inner edge carries ${\epsilon }_{\mathrm{in}}=-\eta c^2$. With zero inner torque, no energy is supplied by a stress at that edge. If the heat released outside it escapes by radiative transfer, rather than being lost through inward advection, the integrated conservation of energy balance gives

The Euler-Lagrange equations of the particle Lagrangian in a shearing sheet areThe terms coupling $x$ and $y$ are the Coriolis acceleration. The cyclic coordinate $y$ has conserved canonical momentumFor a Newtonian potential of a point mass, $q_0=3/2$ and ${\Omega }_z={\kappa }_r={\Omega }_0$. Set $p_y={\Omega }_0{x}_0/2$ and substitute $\dot{y}=p_y-2{\Omega }_{0}x$ in the radial equation. It becomes $\ddot{x}+{\Omega }_0^2(x-x_0)=0$, a harmonic oscillator equation about the epicyclic guiding center $x_0$. Integration givesThe four real constants in $A,y_0,x_0$ and the two in $B$ account for the six initial position and velocity data.

Expanding the inertial specific angular momentum gives $h=h_0+r_0{p}_y+O(2)$. Thus $p_y$ measures the angular-momentum offset from the reference circular orbit, and $x_0$ specifies the radius of its associated epicyclic guiding center.

The conserved horizontal energy in the rotating frame isIt is the horizontal part of the local Jacobi energy in a shearing sheet, rather than the inertial specific orbital energy. Its positive term is the epicyclic energy; its negative term is the energy of the background shear at guiding-center position $x_0$. Independently, the vertical harmonic oscillator has conserved energy

For the relative position $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$, Newton's law of universal gravitation gives the two-body problem in its reduced one-body form,Taking the dot product with the relative velocity $\mathbf{v}=\dot{\mathbf{r}}$ givesIntegration therefore yields conservation of specific orbital energy:

In the planetocentric hyperbolic Kepler orbit, the speed at infinity is $\Delta v_{pc}$ and the impact parameter is $b$. Conservation of specific angular momentum givesConservation of specific orbital energy at the moon's radius giveswhere $v_m^2=G{M}_p/a_m$. ThusThe moon moves tangentially at $v_m$, soTherefore the correctly dimensioned reading of the displayed result is

The same absorption and re-emission that produce radial radiation pressure also produce Poynting–Robertson drag. This velocity-dependent force removes specific orbital energy and angular momentum, destroys the exact Jacobi constant, and turns the formal equilibrium points into slowly drifting configurations. Stellar-wind drag can provide a similar correction.

Immediately after release, the grain has the comet's position and velocity, but its effective stellar gravitational parameter is ${\mu }_d=(1-\beta )\mu$. Using the comet's specific orbital energy,the grain energy isThereforeRelease with zero relative velocity preserves the specific angular momentum $h^2=\mu a(1-e^2)$. Applying $h^2={\mu }_d{a}_d(1-e_d^2)$ then givesThe paper instead prints $e^2-2e\beta +{\beta }^2$ in the first three terms. That expression is incompatible with both its printed $a_d$ and conservation of $h$, except at $e=1$. The next part is the result obtained from the printed eccentricity, so both consequences are recorded below.

The shell feels only radial gravity and the radial force due to the cosmological constant, so its torque vanishes and its specific angular momentum $L=r^2\dot{\theta }$ is conserved. Multiplyingby $\dot{r}$ and integrating gives the conserved specific orbital energy

For a uniform sphere, assembling concentric shells gives its gravitational potential energyThe cosmological-constant potential per unit mass is $-\Lambda r^2/6$. Since $⟨r^2⟩=3r_{\mathrm{ta}}^2/5$ in a uniform sphere,

The scalar virial theorem weights a potential homogeneous of degree $n$ by $-nW$. Gravity has degree $-1$ and the $\Lambda$ potential degree $2$, so the final state obeysAt turnaround $T=0$, while the virial relation gives $E_f=W_{G,f}/2+2W_{\Lambda ,f}$. Conservation of energy, together with $M=4\pi {\rho }_{\mathrm{ta}}{r}_{\mathrm{ta}}^3/3$, then gives, for $x=r_f/r_{\mathrm{ta}}$ and $\eta =\Lambda /(4\pi G{\rho }_{\mathrm{ta}})$,The root connected continuously to the $\Lambda =0$ solution has $x=1/2-\eta /8+O({\eta }^2)$, equivalentlyto first order in $\eta$. With $\Lambda =0$, virialization occurs at half the turnaround radius. At fixed turnaround state, positive $\Lambda$ makes this equilibrium root slightly smaller because its repulsive quadratic potential enters both energy conservation and the virial relation; sufficiently strong repulsion instead prevents a bound virialized state. Negative $\Lambda$ shifts the root in the opposite direction.

Write $k=\mu (1-\beta )$. Taking the cross product of the equation of motion with $\mathbf{r}$ givesSince $\mathbf{r}\cdot \mathbf{h}=0$ at every time, the orbit lies in the fixed plane perpendicular to the conserved specific angular momentum $\mathbf{h}$. Taking the inner product with $\dot{\mathbf{r}}$ gives the conserved specific orbital energyFor $\beta >1$, the effective inverse-square force is repulsive and the potential term is positive.