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Write the gravitational potential with spherical symmetry as $\Phi (r)$ and let $\mathbf{n}$ be the unit vector normal to a circular orbit, oriented in the direction of its angular velocity. Put $\mathbf{\Omega }=\Omega \mathbf{n}$ and denote the specific angular momentum by $\mathbf{h}=h\mathbf{n}$. Circular force balance and the specific orbital energy giveDifferentiating along the family of circular orbits,A change in orbital orientation contributes $hd\mathbf{n}$ to $d\mathbf{h}$, but $\mathbf{n}\cdot d\mathbf{n}=0$. ConsequentlyThis circular-orbit energy gradient accounts for radial and inclination changes together; spherical symmetry makes the orbital energy independent of the plane.

For either variable-mass particle, $\mathbf{H}=M\mathbf{h}$ and $E=M\epsilon$. Thus $d\mathbf{H}=Md\mathbf{h}+\mathbf{h}dM$, soConservation of mass and conservation of angular momentum impose $d{M}_2=-d{M}_1$, $d{\mathbf{H}}_2=-d{\mathbf{H}}_1$. Adding the two particle contributions provesTo interpret the signs, label the inner orbit 1 and the outer orbit 2. For $f(r)=\epsilon -\Omega h$ the first identity gives $f^{\prime }=-h{\Omega }^{\prime }$. The assumed outward decrease of angular velocity therefore makes $f_1<f_2$, so the mass-exchange part lowers the energy when $d{M}_1>0$: mass is transferred inward. In a coplanar, co-rotating pair, ${\Omega }_1>{\Omega }_2$ and an outward angular momentum transfer has $d{\mathbf{H}}_1=-dH\mathbf{n}$ with $dH>0$. Its contribution is $-({\Omega }_1-{\Omega }_2)dH<0$.

The vector difference also identifies the inclination-reducing exchange. For a fixed small magnitude of admissible angular momentum exchange, its energy contribution is most negative whenby the Cauchy-Schwarz inequality. This gives $d{E}_{\mathrm{angular}}=-s|{\mathbf{\Omega }}_1-{\mathbf{\Omega }}_2{|}^2$. Let $q={\mathbf{n}}_1\cdot {\mathbf{n}}_2=\cos \iota$ and $H_i=|{\mathbf{H}}_i|$, taking fixed particle masses for this exchange. Projecting $d{\mathbf{H}}_i$ perpendicular to its own direction givesThereforeThe downhill exchange reduces the relative orbital inclination. Its inner angular momentum component is ${\mathbf{n}}_1\cdot d{\mathbf{H}}_1=-s({\Omega }_1-{\Omega }_{2}q)<0$; in the aligned limit the outer orbit receives that angular momentum directly. For strongly inclined orbits, part of the exchange instead cancels differently directed angular momenta as the planes align. Together these gradients explain the tendencies toward inward mass transfer, outward angular momentum transport and coplanarity.

Assume an axisymmetric thin disk in the fixed Newtonian potential of a point mass, negligible disk self-gravity, and slow evolution compared with an orbital period. Radial pressure and radial inertia are small compared with gravity and centrifugal acceleration, so $u_{\varphi }=r\Omega (r)$ with $\Omega =(GM/r^3{)}^{1/2}$, independent of height and time. Assume no mass loss or torque through the upper and lower disk faces. Define the surface density of a disk, mass-weighted radial velocity and integrated shear stress byVertical integration of the continuity equation givesThe azimuthal momentum equation, including the geometric term $u_r{u}_{\varphi }/r$, gives conservation of angular momentum in terms of $j=r^2\Omega$:Subtracting $j$ times mass conservation eliminates the time derivative and leaves $\Sigma {\bar{u}}_r{j}^{\prime }=r^{-1}(r^{2}W{)}^{\prime }$.

Define the effective density-weighted viscosity of a disk through the actual stress:If the local viscous stress is $T_{r\varphi }=\rho \nu r{\Omega }^{\prime }$, this reduces to $\bar{\nu }={\Sigma }^{-1}\int \rho \nu dz$. The definition can also describe an effective turbulent transport stress, provided the same shear closure is appropriate. Since ${\Omega }^{\prime }<0$, outward transport corresponds to $W<0$ and positive $\bar{\nu }$.

For Keplerian rotation, $j=\sqrt{GMr}$ and $r^{2}W=-(3/2)\sqrt{GM}{r}^{1/2}\bar{\nu }\Sigma$. ThereforeSubstitution into mass conservation yields the Keplerian viscous diffusion equationThe absence of surface fluxes, fixed rotation law and leading thin-disk approximation are essential: a wind, appreciable radial inertia or evolving gravitational potential would alter this reduction.

In cylindrical coordinates the particle's radial and vertical equations are $\ddot{r}-r{\dot{\varphi }}^2=-{\Phi }_r$ and $\ddot{z}=-{\Phi }_z$. Axisymmetry gives conservation of angular momentum, with specific angular momentum $h=r^2\dot{\varphi }$. Eliminating the cyclic azimuthal motion therefore givesThus $\boxed{{\Phi }_1=\Phi +h^2/(2r^2)}$ governs the meridional motion. Its conserved energy is $\frac{1}{2}({\dot{r}}^2+{\dot{z}}^2)+{\Phi }_1$. The additional term is the centrifugal barrier appropriate to fixed angular momentum.

For a point mass, $\Phi =-GM/\sqrt{r^2+z^2}$. If $h\ne 0$, the vertical derivative vanishes only at $z=0$, and the radial stationary condition there gives $r_0=h^2/(GM)$. This is the sole stationary point with $r>0$, at energy $-GM/(2r_0)$. Its Hessian matrix isConsequently the local contours are closed, approximately circular curves about a strict minimum. Finite negative-energy contours above the minimum remain closed; the centrifugal barrier prevents approach to the axis and the negative energy prevents escape to infinity. The zero contour satisfies $z^2=4r^4/r_0^2-r^2$, while positive-energy contours can be unbounded. The left panel illustrates these features; portions of the outer contours extend outside the plotting window.

Each panel uses its own circular radius as length unit and $GM/r_0$ as potential unit. Linearizing near the minimum givesThe radial and vertical epicyclic frequencies both equal the circular orbital frequency. Circular Kepler orbits are stable to small meridional perturbations of a freely moving particle. With no dissipation the oscillations persist rather than decay. A change in h shifts the circular radius to a nearby orbit; azimuthal phase itself is neutral by rotational symmetry. If $h=0$, there is no stationary point or centrifugal barrier and this circular-orbit stability discussion does not apply.

Set $S=T=0$ and choose $\boxed{\bar{\nu }(r)={\nu }_{0}r}$, with ${\nu }_0>0$ constant. Define $y=\sqrt{r}$ and $g=r^{3/2}\Sigma$. Then $\sqrt{r}\bar{\nu }\Sigma ={\nu }_{0}g$, and ${\partial }_r=(2y{)}^{-1}{\partial }_y$. Directly transforming the Keplerian viscous diffusion equation givesThe heat-equation transform for linear-radius disk viscosity is completed by $x=2y/\sqrt{3{\nu }_0}$:Equivalently one can use $g=\sqrt{r}\bar{\nu }\Sigma$, which differs only by a constant. This is the viscosity law making the transformed diffusivity constant under this square-root-radius reduction. Initial and boundary conditions transform too; the change of variables does not supply unspecified disc-edge data.

The physical diffusion arises from angular momentum transport. Differential rotation creates a viscous shear stress, which transfers angular momentum between neighboring annuli. Gas that loses angular momentum moves inward, while gas that gains it can move outward. Combining conservation of angular momentum with mass conservation turns the stress gradient into a radial mass flux and its divergence into a second radial derivative. Positive kinematic viscosity smooths the appropriately weighted profile; matter need not all move in one direction, and an accreting disc can simultaneously spread outward. Viscosity dissipates orbital energy but redistributes angular momentum internally except for boundary losses.

For the polar decomposition of an invertible real matrix, start with $C=F^{T}F$. For nonzero $y$, $y^{T}Cy=|Fy{|}^2>0$, since $\det F>0$ makes $F$ invertible. The spectral theorem for real symmetric matrices gives an orthonormal basis $n_a$ with $C{n}_a={\lambda }_a^2{n}_a$ and ${\lambda }_a>0$. DefineThen $U$ is a positive-definite symmetric matrix, $U^2=C$, andbecause $\det U=\sqrt{\det (F^{T}F)}=\det F$. Thus $R$ is an orthogonal matrix with positive determinant. Set $V=RU{R}^T$; it too is a positive-definite symmetric matrix, and $VR=RU=F$. Moreover $V^2=F{F}^T$. This provesThe positive square root is unique: any positive symmetric square root of $C$ commutes with $C$, preserves its eigenspaces and has the positive scalar square root on each of them. Hence $U$, then $R$ and $V$, are unique. These are the right stretch tensor, rotation and left stretch tensor of the polar decomposition in continuum mechanics.

A standard spectral strain measure isTake $f\in C^1(0,\infty )$, $f(1)=0$ and $f^{\prime }(1)=1$, with $f$ strictly increasing; the customary stronger condition $f^{\prime }(\lambda )>0$ gives a regular one-to-one parameterization of all positive principal stretches. The zero condition makes a rigid rotation strain free, monotonicity distinguishes extension from contraction and prevents nonunit stretch from being strain free, and the derivative normalization gives the ordinary infinitesimal strain. Indeed, for $F=I+H$,Under a superposed spatial rotation $F\mapsto QF$, $F^{T}F$ and hence $E^f$ remain unchanged, giving material frame indifference. Choosing $f(\lambda )=\lambda -1$ satisfies all these conditions and gives the Biot strain tensor

The meaning of work-conjugate stress and strain here is that a symmetric material tensor $T^f$ satisfiesfor every admissible rate, with both powers measured per reference volume. To keep the nominal-stress index convention explicit, write $M_{iI}=P_{Ii}$, so $M=P^T$ and the power is $M:\dot{F}$. Differentiating $F=RU$ gives $\dot{F}=R(\Omega U+\dot{U})$, where $\Omega =R^T\dot{R}$ is skew. Conservation of angular momentum gives symmetry of $M{F}^T$, or equivalently of $R^{T}MU$. ThereforeThe first term vanishes because a symmetric tensor has zero contraction with a skew tensor. Since ${\dot{E}}^{(1)}=\dot{U}$, the symmetric Biot stress isIf the nominal stress is instead stored spatial-index first as $M$, the same answer is $\mathrm{sym}(R^{T}M)$. The symmetrization is essential: $PR$ need not itself be symmetric.

For an equatorial circular orbit in the Paczyński-Wiita potential, radial force balance givesIts specific angular momentum is $j=R^2\Omega =\sqrt{GM}{R}^{3/2}/(R-R_S)$. Perturbing the circular radius at fixed $j$ gives the radial epicyclic frequency squared ${\kappa }^2=R^{-3}d(j^2)/dR$. ThereforeCircular motion is radially stable for $R>3R_S$, marginal at $3R_S$, and unstable between $R_S$ and $3R_S$. Thus the inner edge of the nearly circular thin accretion disk is $R_{\mathrm{in}}=3R_S$, joining a plunging region of a black-hole accretion disk inside the innermost stable circular orbit. This conclusion concerns the circular thin-disk model, not an assertion that all gas disappears inside this radius.

Take the inward mass accretion rate $\dot{M}$ as positive, and write the positive outward viscous torque in an accretion disk asSteady conservation of angular momentum makes the net inward flux $C=\dot{M}j-\mathcal{G}$ independent of radius. Impose a zero-torque inner boundary condition at $3R_S$, appropriate to an approximately stress-free transition to plunging motion. ThenThis is the steady accretion disk with arbitrary rotation law angular-momentum balance.

The local viscous dissipation integrated through the full disk thickness is $D=\nu \Sigma (R{\Omega }^{\prime }{)}^2=-\mathcal{G}{\Omega }^{\prime }/(2\pi R)$. It is the power per unit planar area summed over both radiating faces. Sincesubstitution gives the two-face dissipation of a Paczyński-Wiita disk:In particular $D$ vanishes at the stress-free inner edge and is nonnegative outside it because $j$ increases there. A one-face radiative flux is $D/2$; the total accretion luminosity is therefore $L={\int }_{3R_S}^{\infty }2\pi RDdR$, with no additional factor of two.

For a direct evaluation put $x=R/R_S$. ThenThe first integral is $7/4$, using $(3x-1)/(x-1{)}^3=3/(x-1{)}^2+2/(x-1{)}^3$. For the second,so it equals $2/[\sqrt{3}(3-1)]=1/\sqrt{3}$. HenceThis is the radiative efficiency of black-hole accretion within this pseudo-Newtonian model.

Independently, the specific orbital energy of a circular particle isThus $E(\infty )=0$ and $E(3R_S)=-GM/(8R_S)=-c^2/16$. The energy released per unit accreted mass is exactly $c^2/16$, confirming the integrated accretion luminosity. The Paczyński-Wiita circular orbit reproduces the Schwarzschild inner radius but gives an approximate binding energy; the efficiency derived here is for the stated potential.

At infinity, the meteorite's incoming asymptote has speed $V$ and perpendicular distance $b$ from the Earth’s centre. The magnitude of its specific angular momentum is the magnitude of $\mathbf{r}\times \mathbf{v}$. On the asymptote, this is $bV$, so conservation of angular momentum gives $\boxed{h=bV}$. Its mechanical energy at infinity is $E=m{V}^2/2$. Comparing with part (ii) givesFor $b>0$ this Kepler orbit is a hyperbola. The smallest $r=1/u$ occurs at periapsis, where $\cos (\theta -{\theta }_0)=1$. ThusThe middle equality follows by substituting $e^2-1=b^2{V}^4/(G^2{M}^2)$. A trajectory entering the Earth satisfies $r_{\min }<R$, equivalentlyTherefore the sufficient collision condition isThis is gravitational focusing: Newtonian gravity increases the collision radius measured at infinity. Equality is the ideal grazing trajectory, which touches the surface; a strict inequality penetrates it. For $b=0$ the incoming motion is radial and collides, with $r_{\min }=0$ as the limiting value of the above expression.

Assume an axisymmetric, geometrically thin disc in a fixed Keplerian rotation field, neglecting the disk's self-gravity. Write $v_R$ positive outwards and letThe ring supplies the local specific angular momentum. Its source per unit area iswhere the Dirac delta function is normalized so that $\int 2\pi RsdR=\dot{M}(t)$. Define the positive outward viscous torque in an accretion disk byThe height-integrated conservation of mass and conservation of angular momentum equations areSubtract $h$ times the mass equation. Since $h$ is time-independent, this givesThe last term is zero as a distribution, because the source is supported where $h=h_0$. Since $h^{\prime }=h/(2R)$, substituting the torque givesInserting this into the mass equation proves the Keplerian disk evolution with a ring source:The viscosity may be a prescribed function or a function of the local disk state; the conservation derivation is the same.

To track signs in the steady problem, define the inward mass fluxThen $0=F_M^{\prime }/(2\pi R)+s$. Away from the source $F_M$ is constant, with $F_M={\dot{M}}_{\mathrm{in}}$ on the inner side and $F_M=-{\dot{M}}_{\mathrm{out}}$ on the outer side, where both loss rates are positive. Integrating through the source gives $F_M(R_0^{+})-F_M(R_0^{-})=-\dot{M}$. Therefore

The outward angular momentum flux is $F_J=\mathcal{G}-F_{M}h$. The steady angular-momentum equation gives $F_J^{\prime }=2\pi Rs{h}_0$. At the two torque-free boundaries its values are $F_J(R_{\mathrm{in}})=-{\dot{M}}_{\mathrm{in}}{h}_{\mathrm{in}}$ and $F_J(R_{\mathrm{out}})={\dot{M}}_{\mathrm{out}}{h}_{\mathrm{out}}$. Integrating across the disk therefore yieldsEliminate ${\dot{M}}_{\mathrm{out}}=\dot{M}-{\dot{M}}_{\mathrm{in}}$ and use $h\propto R^{1/2}$. The two-boundary mass split of a ring-fed disk isThe complementary outward fraction is $(\sqrt{R_0}-\sqrt{R_{\mathrm{in}}})/(\sqrt{R_{\mathrm{out}}}-\sqrt{R_{\mathrm{in}}})$. Both fractions lie between zero and one for a source strictly between the boundaries.

As a local check, integrating the steady mass flux from each zero-torque boundary givesContinuity of this quantity at $R_0$ gives the same split. The split depends on boundary and source angular momenta, rather than on the detailed viscosity law.

Each annulus has a torque proportional to ${\mathbf{J}}_h\times \delta {\mathbf{J}}_d$, so its scalar product with ${\mathbf{J}}_h$ is zero. Summing over annuli preserves this property even when the proportionality coefficient varies with radius. Therefore $\boxed{\mathbf{T}\cdot {\mathbf{J}}_h=0}$ for the total torque due to Lense-Thirring precession.

The torque specified is the torque on the disk. Assuming a closed disk-plus-hole angular-momentum budget with no external torque, its reaction on the black hole has the opposite sign:Thus ${\dot{\mathbf{J}}}_t=0$. The $K_1$ term describes precession, because it is perpendicular to both spins. Since ${\mathbf{J}}_h\times {\mathbf{J}}_d={\mathbf{J}}_h\times {\mathbf{J}}_t$, the hole precesses about the fixed total vector. The $K_2$ term changes the spin directions dissipatively. Writing ${\mathbf{J}}_{d,\perp }$ for the disk component perpendicular to the hole, the vector triple product identity givesA positive $K_2$ removes this component from the disk and transfers it into the hole's changing direction. This is black-hole and disk alignment under internal torque.

Since $\mathbf{T}$ is perpendicular to ${\mathbf{J}}_h$,Hence $\boxed{J_h\text{ is constant}}$, even if the torque coefficients vary in time.

Now differentiate the scalar product with the conserved total vector. The precession term contributes zero, and the vector triple product identity givesAlso ${\mathbf{J}}_h\cdot {\mathbf{J}}_t=J_h^2+{\mathbf{J}}_h\cdot {\mathbf{J}}_d$, so constancy of $J_h$ givesFinally, $J_t^2=J_h^2+J_d^2+2{\mathbf{J}}_h\cdot {\mathbf{J}}_d$ is constant. Differentiating provesThus the disk's spin magnitude decreases while the hole's magnitude is fixed; their vector sum remains fixed as well.

For the alignment conclusion, assume $J_h,J_t>0$ and let $c=\cos \theta ={\mathbf{J}}_h\cdot {\mathbf{J}}_t/(J_h{J}_t)$. Since ${\mathbf{J}}_h\times {\mathbf{J}}_d={\mathbf{J}}_h\times {\mathbf{J}}_t$,For a noncollinear initial state, $|c(0)|<1$, this integrates exactly toConsequently a fixed positive $K_2$, or more generally a positive coefficient of divergent integrated strength, gives $c(t)\to 1$. In this persistent dissipative evolution, $\boxed{{\mathbf{J}}_h\to J_h{\hat{\mathbf{J}}}_t}$: the hole aligns asymptotically with the total angular momentum. The exactly antiparallel configuration $c=-1$ is a stationary, nongeneric exception; when $J_t=0$ there is no total-vector direction to align with.

If the phrase $K_2>0$ allows arbitrary time dependence, positivity alone does not logically force complete eventual alignment. The persistent-torque criterion for asymptotic spin alignment is the needed qualification. For example, take $J_h=J_t=1$, $K_1=0$, $K_2(t)=e^{-t}$ and $c(0)=0$. The same torque equations give $c(t)=\tanh (1-e^{-t})$, with limit $\tanh 1<1$. The final configurations below use the intended persistent-torque regime, including constant positive $K_2$.

Once the hole has aligned, conservation of angular momentum fixes the disk without any additional torque calculation:The disk and hole therefore finish parallel only if $J_t>J_h$. If $J_t<J_h$, they finish antiparallel: this is counteralignment of a black hole and accretion disk. In terms of the initial disk-hole angle ${\theta }_0$, squaring $J_t$ shows that counteralignment occurs whenIf $J_t=J_h$, the final disk angular momentum vanishes and has no spin direction. Thus alignment of the hole with the total vector does not imply alignment of both component spins.

Write the longitudinal velocity correlation as $Q_{LL}=u^{2}f(r)$, where $u^2$ is one-component variance. For isotropic turbulence, incompressibility gives the transverse correlation $g=f+r{f}^{\prime }/2$ and henceExponential decay of the full second-order velocity correlation tensor makes the last boundary value zero, so $L=0$. For invariance of $I$ we additionally need $r^{4}K\to 0$, as part (ii) shows. This follows if the premise of rapid decay of all two-point correlations includes the two-position triple correlation and persists during evolution; exponential second-order decay at one initial time alone does not establish it. Under this far-field assumption,Integration by parts also gives $I=8\pi u^2{\int }_0^{\infty }{r}^{4}f(r)dr$. Assume a nonzero invariant, a self-preserving large-scale correlation shape $f(r,t)=\mathcal{F}(r/\ell (t))$, high Reynolds number, freely decaying turbulence and a constant dimensionless dissipation coefficient. Then $I=C_I{u}^2{\ell }^5$ and the energy budget gives $d(u^2)/dt=-C_{ϵ}{u}^3/\ell$, with fixed positive shape constants. Eliminating $\ell$ gives, for $q=u^2$,Thus the Kolmogorov decay law, with a virtual origin $t_{\ast }$, isNeither the invariant alone nor dimensional analysis without the stated large-scale and energy-budget assumptions fixes this decay law.

Now let $\mathbf{J}={\int }_V\mathbf{x}\times \mathbf{u}dV$. Integrate the supplied identity over both position variables. For a genuinely confined incompressible flow with $\mathbf{u}\cdot \mathbf{n}=0$ on the boundary, each divergence produces a zero surface term, because its flux contains the normal velocity at the corresponding point. Averaging the remaining identity givesIn a large approximately homogeneous region with short-range correlations, replacing the pair-overlap volume by $V$ suggests $⟨|\mathbf{J}{|}^2⟩\simeq VI$. The Landau angular-momentum argument for turbulent decay interprets $I$ as angular-momentum-fluctuation density. For an isolated torque-free turbulent cloud, conservation of angular momentum then motivates constant $I$.

The weaknesses concern precisely the passage to an infinite homogeneous field. A fixed large subvolume is not an isolated cloud: it exchanges momentum and angular momentum by advection and fluid pressure forces. Impermeable walls eliminate the stated velocity-flux terms but need not eliminate viscous or fluid pressure torques. Moreover nonlocal fluid pressure can generate long-range correlations, invalidating the assumed decay, and boundary terms with large lever arms require control before taking a volume limit. Thus the exact integrated identity is sound under its boundary conditions, but the inferred conservation law requires additional hypotheses.

The particle's angular momentum about the force centre is $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$. A central force exerts zero torque about that centre, soBy conservation of angular momentum, $\mathbf{L}$ is a constant vector. If it is nonzero, the identity $\mathbf{r}\cdot \mathbf{L}=0$ places every point of the orbit in the fixed plane through the centre perpendicular to $\mathbf{L}$:If $\mathbf{L}=0$, position and velocity are parallel. Away from the force centre, ${\dot{\mathbf{e}}}_r=(\dot{\mathbf{r}}-\dot{r}{\mathbf{e}}_r)/r=0$, so the motion is radial along a fixed line, which lies in any plane containing that line. Thus the zero-angular-momentum case also gives planar motion. As usual, these assertions apply while the force law supplies a regular orbit; continuation through a singular force centre is a separate issue.

With the Sun fixed, its gravitational force on the planet is a central force. In the fixed orbital plane, the transverse acceleration in polar coordinates therefore vanishes:During $dt$, the position vector sweeps an infinitesimal sector of area $dA=\frac{1}{2}{r}^{2}d\theta$. Hence the areal velocity isIntegrating over any two equal time intervals gives equal swept areas, proving Kepler's second law. Here $L$ is the signed component of angular momentum normal to the orbital plane. The argument uses conservation of angular momentum; an inverse-square dependence is not needed for this particular law, only a central force.

Take $\dot{M}>0$ to mean inward accretion rate, and let $S$ count loss through both faces per unit projected area of the accretion disk. An annulus of width $dR$ loses $2\pi RSdR$ to the wind. In steady state, its incoming mass flux at its outer edge equals the inward flux at its inner edge plus this loss. Hence mass conservation givesIf a loss rate is instead defined per individual face, it must be doubled before substituting for $S$ here.

Use Keplerian rotation ${\Omega }^2=GM/R^3$. The vertically integrated viscous dissipation, summed over both disk faces, isHere $\nu$ is the kinematic viscosity, and $\Sigma$ is the surface density of a disk; for height-dependent viscosity the product means $\int \rho \nu dz$. Material already in a circular orbit has specific orbital energy $v_K^2/2-GM/R=-GM/(2R)$. Giving it zero total binding energy therefore costs an additional $GM/(2R)$ per unit mass. The total launch speed is the escape velocity $\sqrt{2GM/R}$, but one must not charge the wind for the orbital kinetic energy it already possesses. Equating wind power to $f{Q}^{+}$ givesCombining this with mass conservation yields the wind-powered mass loss from a steady accretion disk relation

Define the outward viscous torque in an accretion disk asThe net inward angular momentum flux is $\dot{M}h-\mathcal{G}$. Because each unit mass lost to the wind carries the local specific angular momentum $h$, annular conservation of angular momentum readsIn the wind region, $\mathcal{G}=Rh{\dot{M}}^{\prime }/(3f)$. For constant $f$, differentiation and cancellation of $h$ givesThus the requested Cauchy-Euler differential equation is