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Synchronous radius and its limits. Kepler's third law for a circumplanetary orbit givesHere $G$ is the gravitational constant. A planet-synchronous orbit repeats after one spin period. A planet-stationary orbit additionally requires a circular orbit, zero orbital inclination to the equator and prograde motion. Then an antenna fixed on the ground points continuously at the same satellite; merely matching the orbital period does not give this property.

Write the planetary radius as $R_p=(3M/(4\pi \rho ){)}^{1/3}$. Requiring the semi-major axis to exceed $R_p$ gives $P^2>3\pi /(G\rho )$. The stellar tidal force restricts the circumplanetary orbit to the Hill sphere, of radius $r_H=a(M/(3M_{\star }){)}^{1/3}$. ThusThese are the surface and Hill-radius constraints in the idealized spherical, small-$M/M_{\star }$ model. Long-term prograde stability generally requires a radius appreciably inside the Hill sphere; for nonzero orbital eccentricity, the surface constraint applies to the periapsis, not just to the semi-major axis.

Collision time and the launch population. A phase-mixed isotropic swarm occupies a shell of radial scale $e{a}_s$. Its number density scales as $N_s/(e{a}_s^3)$, its relative speed as $a_s/P$, and its geometric collision cross-section as $r^2$. Consequently the total collision rate scales as $N_s^2{r}^2/(Pe{a}_s^2)$. To give the numerical normalization used here, adopt an effective shell volume $V=4\pi e{a}_s^3$. At a fixed position the velocity directions are uniformly distributed in the tangent plane, so their mean relative speed is $4v_K/\pi =8a_s/P$. For an unordered pair, the collision cross-section is $4\pi r^2$, givingThus the mean interval between collisions is $A/N_s^2$ in this large-population kinetic model. The effective shell width fixes an order-one coefficient: small orbital eccentricity and random planes alone do not specify a unique radial probability density. Phase mixing, negligible gravitational focusing, $r\ll e{a}_s$, and uncorrelated encounters are implicit in this estimate. Exactly identical orbital periods with perfectly fixed phases do not themselves produce a memoryless collision process.

For nearly planet-stationary orbits with $e\ll I\ll 1$, the swarm volume is smaller by a factor of order $I$, while the relative speed is smaller by the same factor, since vertical motion of scale $I{v}_K$ dominates the eccentric motion. The two changes cancel in the rate $n\sigma v$: there is no parametric factor $I$ in the collision time within the same phase-mixed kinetic approximation. Numerical factors and phase correlations can differ. This is not an argument that bringing all satellites into one nearly circular plane makes their phases random.

With $N_s=Rt$, an Inhomogeneous Poisson process has cumulative hazard functionThe survival function of the first collision is $\exp [-H(t)]$. Setting the expected number of collisions to one givesThis is a characteristic first-event population, with probability $1-e^{-1}$ of an earlier event. It is not a median: the median has an additional factor $(\log 2{)}^{1/3}$.

Which population collides next? Immediately after the first disruption let $S=N_{s0}-2$ and $F=2x^{-3}$. The latter follows from mass conservation for equal-density spherical pieces. Using the same unordered-pair counting and geometric collision cross-sections as above, define $b=(1+x{)}^2/2$; thenHere $S^2,F^2$ stand for the usual large-population approximations to $S(S-1),F(F-1)$. The factor two distinguishing identical and different species is essential. The probability that the next collision is fragment–fragment isIt is the most likely type if $S<4/[x(1+x{)}^2]$ and $S<2/x^2$. It has probability greater than one half ifFor $x\ll 1$, a strongly fragment-dominated next event therefore requires $Sx\ll 4$; the fragment–satellite comparison is more restrictive than the satellite–satellite comparison. If $S=0$, the next event must be fragment–fragment, provided fragments remain.

Population equations and the normalization discrepancy. Every satellite–satellite event produces $2x^{-3}$ fragments and destroys two satellites. Every satellite–fragment event produces a net $x^{-3}-1$ fragments and destroys one satellite; every fragment–fragment event destroys two fragments. Consistent collision counting therefore givesFor $x\ll 1$, $b\simeq 1/2$ and $x^{-3}-1\simeq x^{-3}$. The last two terms in $A{\dot{N}}_f$ are then twice those in the printed equation. This is a genuine factor-of-two inconsistency: equal-size fragment pairs have a collision cross-section smaller by $x^2$, so their event rate must be $x^2{N}_f^2/A$ if the satellite event rate is $N_s^2/A$; destroying both fragments necessarily gives the sink $-2x^2{N}_f^2/A$.

If the printed equation is taken as a prescribed approximate rate model instead, its implicit event rates are $N_s^2/A$, $N_f{N}_s/(4A)$ and $x^2{N}_f^2/(2A)$. Under precisely that mixed normalization, its corresponding satellite equation isThe printed source term also neglects the consumed fragment in a satellite–fragment event, a legitimate relative $O(x^3)$ approximation. That approximation does not repair the pair-counting discrepancy.

The ensuing cascade. This is a two-size fragmentation cascade. First use the printed approximate model, dropping satellite–satellite events as requested. Put $S=N_s$, $F=N_f$, $k=4x^2$ and take $x<1/2$. ThenIntegrating this linear differential equation, with $F_0=2x^{-3}$ and $S_0=N_{s0}-2>0$, givesInitially the collisional cascade grows if $S_0>8x^2$, with approximate early exponential growth time $4A{x}^3/S_0$ when the satellite population is nearly fixed. The fragment population reaches its maximum atAfterwards satellites are depleted and fragment–fragment losses dominate. For $S_0>0$ the continuum solution has $S\to 0$, $F\to 0$, withWhen no satellites remain initially, $F(t)=F_0/[1+x^2{F}_{0}t/A]$ directly. Small integer populations eventually invalidate these deterministic differential equations.

The consistently counted model has, to leading order in $x$, exactly the same curve $F(S)$ and peak, but both retained time derivatives are twice as large. Its growth time is $2A{x}^3/S_0$, and $F(t)\sim A/(2x^{2}t)$. Keeping $(1+x{)}^2$ and the consumed fragment replaces $k$ by $4x^2/(1+x{)}^2$ and $x^{-3}$ in the curve by $x^{-3}-1$. This explicitly separates the physical collision bookkeeping from the printed normalization while giving the evolution under both conventions.

Let $a=({\rho }_b-{\rho }_t)/H={\rho }_0{N}^2/g$. The initial density is $\rho (z)={\rho }_t-az$. Conservation of mass in the material homogenized over $-h<z<0$ fixes its mean density to ${\rho }_u={\rho }_t+ah/2$, while the untouched lower layer has interfacial density ${\rho }_i={\rho }_t+ah$. Hence the reduced gravity difference across the interface isThe reference contribution involving ${\rho }_t$ cancels from the difference.

Only the upper layer changes its potential energy. The potential-energy cost of homogenizing a linear stratification issoThe positive sign reflects work against stable stratification.

The rotating boundary supplies energy on scales set by $\Omega$ and $R$. Turbulent drag and dissipation remove energy, and entrainment adds initially stationary fluid that must be accelerated. If upper-layer energy grows, the increasing turbulent loss provides a restoring tendency; after a short adjustment, input and losses can approximately balance while the interface moves on a slower time scale. This gives a physical rationale for a fixed-energy turbulent mixed layer, not a consequence of mass conservation alone. Its characteristic kinetic energy is $K\sim {\rho }_0\pi R^{2}h{u}_u^2/2$. With the stipulated $N$-independent energy closure,so the characteristic speed decreases as $h^{-1/2}$ rather than remaining constant.

The stress does work at a rate proportional to $\pi R^2\tau u_u=c_D{\rho }_0\pi R^2{u}_u^3$. Combine this with the energy increase and absorb fixed drag/geometrical factors into $C>0$:Here the exponents $\alpha ,\beta$ are unrelated to the plume entrainment coefficient in Question 2. The interfacial Richardson number isSet $q=\Omega /N$, $d=\delta /R$ and $r=h/R$. Then $u_u^2/(N^2{h}^2)=d/(2r{\mathrm{Ri}}_i)$ and $r=q(2A{\mathrm{Ri}}_i/d{)}^{1/2}$. Substitution givesThe assumption that $E$ depends on the local Richardson number alone excludes separate dependences on $q,d$. The entrainment exponent from a local interfacial Richardson closure is consequentlyThis comparison treats $q$ and $d$ as independently variable external controls; fitting only a single apparatus would not by itself determine the exponents.

Finally, use the fixed-energy closure again:Integrating from a positive initial mixed-layer depth $h_0$ at time $t_0$ givesWhen the growth term dominates the initial offset, the rotating-disc mixed-layer depth law isTaking the effective time origin to be zero gives the stated scaling. Its validity is limited to the regime of the closures and $h<H$; the singular velocity predicted at $h=0$ is not a model for the initial boundary-layer formation.

Use the correct Gaussian entropy decompositionThe last coefficient is $1/2$, as follows from $-\log {\gamma }_N=N\log (2\pi )/2+|\mathbf{v}{|}^2/2$; the printed hint omits it. Under the allowed differentiability and integrability assumptions, the supplied collision invariants conserve mass and energy. Differentiating therefore giveswhere the extra derivative term $\int {\partial }_{t}F$ vanishes by mass conservation. Thus the Kac entropy production isFor one pair, call the double integral $A_{ij}$. The measure-preserving substitution $(\mathbf{v},\theta )\mapsto (R_{ij,\theta }\mathbf{v},-\theta )$ interchanges $F$ and $F\circ R_{ij,\theta }$, with angles taken modulo $2\pi$. Averaging the original and substituted expressions givesSince $N/(4\pi C_N)=1/[2\pi (N-1)]$, the desired Kac entropy dissipation formula isFor positive values $a,b$, $(a-b)(\log a-\log b)\ge 0$ because the logarithm is increasing. At two zeros use value zero; at one zero and one positive value use the nonnegative extended value $+\infty$. One may first use positive densities and then regularize by $(F+\epsilon {\gamma }_N)/(1+\epsilon )$; rotation invariance of ${\gamma }_N$ preserves the formula and permits the usual limit at zeros under the stated assumptions.

Thus relative entropy is nonincreasing along the evolution. When the dissipation is finite, zero dissipation means pairwise rotation invariance and hence radiality, by the preceding kernel argument. Radial normalized densities other than ${\gamma }_N$ can be stationary with positive relative entropy: vanishing dissipation is not a claim that the unique stationary density is Gaussian.

Effective permeability of a porous medium. Work to leading order in the slender-layer ratio $H/L$. Take the growing lower wedge to have porous permeability $k_1$, with interface $y=Hx/L$, and the upper wedge to have porous permeability $k_2$. The leading pressure is independent of $y$; vertical flow is smaller than horizontal flow by $H/L$. DefineDarcy's law and the fixed two-dimensional volume flux giveHere $q_i$ is Darcy velocity; the parcel speed is $q_i/\varphi$. Integrating the pressure gradient over the length and defining $Q=H{k}_{\mathrm{eff}}\Delta p/(\mu L)$ yieldsThe equal-porous permeability limit is $k_{\mathrm{eff}}=k_1=k_2$. This logarithmic mean comes from parallel layers at each cross-section followed by series addition of their local hydraulic resistances. It is a slender-layer result; a full two-dimensional transmission problem has small end and interface corrections.

Parcel paths and travel times. Pressure equalization does not mean that parcels stay at fixed $y$: mass conservation requires a small vertical flow. Define a streamfunction by $\psi (x,0)=0$ and $\psi (x,H)=Q$. Its leading expression isLet $r=y_0/H$ label a parcel released at the inlet. There $\psi =Qr$. At the inclined interface $\psi /Q=k_1\xi /K(\xi )$, so the parcel crosses from the upper wedge into the lower one atBefore crossing, $y/H=1-(1-r)K/k_2$; afterwards, $y/H=rK/k_1$. These expressions show why integrating the speed along a horizontal line would give the wrong parcel time.

Put $d=k_1-k_2$ and ${\tau }_V=\varphi HL/Q$, the pore-volume throughput time. Integrating $dt=\varphi dx/q_i$ on the two portions of the path givesIts derivative with respect to ${\xi }_c$ is ${\tau }_{V}dK/(k_1{k}_2)$, so it is monotone. The limiting streamline times at the lower and upper boundaries are ${\tau }_V(k_1+k_2)/(2k_1)$ and ${\tau }_V(k_1+k_2)/(2k_2)$ respectively. HenceThe printed expression assumes $k_1\ge k_2$. The absolute value is needed for a nonnegative maximum difference without that ordering. For equal porous permeabilities every parcel has time ${\tau }_V$.

Oil recovery. In the ideal passive-displacement model with $k_1/k_2=10$, the earliest and latest travel times are $0.55{\tau }_V$ and $5.5{\tau }_V$, a spread of $4.95{\tau }_V$. Preferential paths through the high-porous permeability wedge therefore give early water breakthrough while oil on slower paths remains unswept. Continued injection sends much water through paths already swept; complete displacement requires several pore volumes. The flux-weighted residence time in a porous layer remains ${\tau }_V$. Real waterflooding also depends on phase mobilities, relative permeabilities, capillary pressure and mixing, so these numbers illustrate heterogeneity rather than a quantitative two-phase recovery prediction.

Saturation equation. Introduce intrinsic permeability of a porous medium $K$ and porosity $\varphi$. The relative porous permeabilities depend on the wetting fluid saturation. Define the phase mobilities and fractional flowUse the conventional capillary pressure $p_c=p_{nw}-p_w$, generally decreasing as wetting saturation increases. Neglect gravity in this horizontal model. The two Darcy fluxes are $q_w=-K{\lambda }_w{p}_{w,x}$ and $q_n=-K{\lambda }_n(p_{w,x}+p_c^{\prime }(s)s_x)$. Eliminate the common pressure gradient using $q_w+q_n=Q$ to obtainThe wetting-phase mass conservation law is consequentlyFor $p_c^{\prime }\le 0$, $D_c\ge 0$ and the capillary pressure provides nonlinear diffusion. The initially uniform saturation is $s_0$. Pure wetting-fluid injection prescribes $q_w(0,t)=Q$ and $q_n(0,t)=0$; in the zero-capillarity, zero-residual-nonwetting idealization this corresponds to inlet saturation one. With residual nonwetting fluid, replace one by the appropriate maximum accessible saturation.

Shock formation and its speed. Neglect capillarity first. Smooth saturation values travel along characteristic curves at speed $(Q/\varphi )F^{\prime }(s)$. When the trailing values have larger characteristic speeds than the values ahead, the curves intersect, and the Buckley-Leverett equation requires a shock selected as an entropy solution. This is typical for the increasing-convex part of a physical fractional-flow curve; shocks are not inevitable for every possible $F$.

Integrating mass conservation across a shock from upstream $s_s$ to downstream $s_0$ gives the Rankine-Hugoniot conditionIf an upstream rarefaction wave joins the shock smoothly in similarity coordinates, its terminal characteristic speed equals $V_s$. Therefore the saturation at that join obeys the fractional-flow tangent constructionThis is the printed equality. It is a tangent-selection condition, not the dimensional speed by itself; the speed includes $Q/\varphi$. An arbitrary shock between prescribed constant states satisfies the jump condition but need not satisfy this additional tangency relation.

Positive capillary diffusion spreads a jump into a transition layer. For a travelling profile $s(x-V_{s}t)$, integration givesThe endpoint states still satisfy the same Rankine-Hugoniot condition. Thus small capillarity primarily gives the front a finite thickness rather than changing the selected limiting speed; vanishing phase mobility at an endpoint can make the regularization degenerate.

The specified reciprocal fractional flow. On its stated branch, $F^{\prime }=1/s^2$ and $F^{\prime \prime }=-2/s^3<0$. If $F$ extends continuously to the initial saturation and the inlet is at one, the decreasing saturation from behind to ahead makes characteristic speeds increase forwards: there is a rarefaction wave, not a compressive shock. Write $a=Q/\varphi$ and $\xi =x/t$. The self-similar solution isThe fan connects continuously to the initial state and broadens linearly in time. This interpretation requires $F(s_0)=2-1/s_0$. A nonnegative physical fractional flow on the entire interval $[s_0,1]$ also requires $s_0\ge 1/2$.

The PDF specifies the formula only for $s>s_0$ and earlier calls $s_0$ residual. If that means an immobile wetting phase with $F(s_0)=0$, the endpoint is an extra constraint and the continuous-branch answer applies without conflict only at $s_0=1/2$. For $0<s_0<1/2$, the printed branch would give negative fractional flow for $s_0<s<1/2$, so it cannot be the complete physical constitutive law. With an additional admissible extension, for example $F=0$ up to $1/2$ and the given positive branch above it, a compound fan–shock is possible. The tangent condition then givesThe fan $s=\sqrt{a/\xi }$ ends at $\xi =V_s$, followed by a jump to $s_0$. This is a conditional physical extension, not data supplied by the question. For $s_0>1/2$ an immobile residual endpoint is likewise incompatible with a continuous version of the stated branch. These distinctions identify what is, and is not, determined by the printed constitutive assumption.

Let $K$ be intrinsic permeability of a porous medium, $\varphi$ the mobile pore fraction, and let $h$ measure thickness normal to the sloping cap. Define cold and reservoir-temperature properties byThe imposed inequality makes ${\delta }_c>0$, so the cold fluid is still buoyant. Assume $K>0$, $\sin \theta >0$, positive densities and viscosities, and the usual positive thermal-expansion and viscosity coefficients. Away from localized fronts, neglect the along-current thickness gradient. Hydrostatic water pressure and Darcy's law then give the upslope Darcy velocitiesTypically $u_h>u_c$: warming increases buoyancy and decreases dynamic viscosity.

Current depths and the moving moving thermal front in a porous current. For this slope-driven porous gravity current, the injected mass flux per unit well length fixes the cold plateau thickness:The moving thermal front in a porous current moves at $V_T=\Gamma u_c$. Since $u_c$ is explicitly a Darcy velocity, the fluid mass stored per unit horizontal area is $\varphi \rho h$. Apply mass conservation across the moving moving thermal front in a porous current:Writing $\gamma =\varphi \Gamma$ and $r=u_c/u_h$, this gives the distal warm depthFor a retarded moving thermal front in a porous current take $0\le \gamma <1$; otherwise this two-plateau geometry needs reconsideration. The factor $(1-\gamma )/(1-\gamma r)$ is important: the distal mass flux is not generally $Q$, because mass is being stored as the cold region replaces the warm region. Setting the two fluxes equal would silently assume a stationary moving thermal front in a porous current.

If the model absorbs porosity into storage and uses pore velocity as its transport speed, the same formulas use $\gamma =\Gamma$. The explicit Darcy convention in the PDF instead gives $\gamma =\varphi \Gamma$; porosity must be specified or absorbed consistently. In the stationary-front limit $\gamma =0$, the two steady depths reduce to $Q/({\rho }_c{u}_c)$ and $Q/({\rho }_h{u}_h)$. In the Boussinesq approximation, use a common reference density in the mass factors while retaining ${\delta }_c,{\delta }_h$ in the driving force.

Leakage thresholds. A normal thickness $h_i$ generates cap overpressure ${\Pi }_i={\delta }_{i}g{h}_i\cos \theta$. Before leakage modifies the current, the two plateau values areFor the caprock leakage threshold $\Delta p>0$, the candidate critical mass-injection rates are thereforeCold-only leakage occurs for $Q_c^{\ast }\le Q<Q_h^{\ast }$, provided $Q_c^{\ast }<Q_h^{\ast }$. Both plateaus can leak once $Q\ge \max (Q_c^{\ast },Q_h^{\ast })$. In the usual common-density approximation, ${\mu }_c>{\mu }_h$ and $r<1$ ensure $Q_c^{\ast }<Q_h^{\ast }$, giving the expected sequence: no leakage, cold-only leakage, then leakage from both cold and warm regions. Without that approximation the ordering must be checked; the given inequality $\Delta \rho >\alpha \Delta T$ alone does not establish it. These are onset criteria computed on the nonleaking current, not a post-leakage mass budget.

For small slopes $\cos \theta \simeq 1$ and $\tan \theta \simeq \sin \theta \simeq \theta$. If another thickness convention is used, its hydrostatic column and projected flux must be changed consistently; one should not mix a normal thickness with a vertical-pressure formula lacking the cosine.

Cross-current heat conduction. Heat transfer from the warm formation makes temperature vary across the carbon dioxide depth and introduces a warming time controlled by thickness, thermal diffusivity and the surrounding rock's heat capacity. It smooths the sharp thermal adjustment and causes gradual changes in buoyancy, viscosity and the velocity profile. The cold fluid generally warms, becomes more mobile and requires less depth to carry a prescribed mass flux; the enhanced cold-region overpressure and its distinct leakage zone tend to shrink. Warming may occur before a parcel reaches the idealized advective moving thermal front in a porous current, especially for a thin current. A single supplied $\Gamma$ no longer describes all heat transport. Quantitative depths and thresholds then require a coupled temperature equation and thermal boundary data; their exact changes cannot be inferred from the linear property laws alone.

Introduce water dynamic viscosity $\mu$ and the net driving liquid-pressure difference $\Delta P=p_0-p_f>0$. If $p_c>0$ denotes capillary suction relative to an ambient gas pressure $p_a$, then $p_f=p_a-p_c$ and $\Delta P=p_0-p_a+p_c$. If $p_c$ is instead the prescribed liquid front pressure, use $\Delta P=p_0-p_c$. This makes the capillary sign convention explicit.

Hemispherical front and injection flux. A hemispherical surface at radius $r$ has area $2\pi r^2$. Quasi-steady mass conservation makes the total volume flux independent of $r$ in the saturated region, so Darcy's law givesThe changing pore volume is $2\pi \varphi (R^3-R_s^3)/3$, giving $Q=2\pi \varphi R^2\dot{R}$. HenceFor the initially dry idealization $R(0)=R_s$, integration gives the implicit front lawIts right-hand side increases for $R>R_s$, defining one advancing radius and thus the flux implicitly in time. Equivalently, with $r=R/R_s$ and $t_{\ast }=\mu \varphi R_s^2/(k\Delta P)$, $t/t_{\ast }=r^3/3-r^2/2+1/6$. The infinite initial flux is an idealization of zero wetted resistance; inertia, finite inlet geometry and pore-scale effects regularize its earliest stage.

For a thin wetted shell $\ell =R-R_s\ll R_s$, the law reduces toFor $R\gg R_s$, the hemispherical capillary imbibition model instead predictswith $Q/Q_{\infty }=1+R_s/R+O((R_s/R{)}^2)$. Radial spreading makes the hydraulic resistance approach a finite value, while the increasing front area reduces its advance speed.

Comparison with one-dimensional imbibition. Under the same constant-driving-pressure, negligible-gravity assumptions, a planar front at depth $z$ has Darcy flux $q=k\Delta P/(\mu z)$ and storage law $\varphi \dot{z}=q$. ThereforeA fixed-area one-dimensional flux decays as $t^{-1/2}$, whereas the hemispherical total flux tends to a constant and its radius grows as $t^{1/3}$. If purely vertical imbibition includes gravity, upward penetration eventually approaches a capillary-rise height and downward penetration has a different gravity-driven limit. The hemispherical assumption itself neglects that directional gravity effect, so the power-law comparison is within its capillary-dominated regime.

Evaporation-limited steady radius. Take $F_e$ as volumetric water loss per unit front area per unit time. At equilibrium the incoming flux equals $2\pi R^2{F}_e$, givingIf the stated loss is a mass flux, replace $F_e$ by $F_e/{\rho }_w$. Porosity cancels from this equilibrium but affects the time to approach it. The incoming flux decreases with $R$ while the evaporating area increases, so the equilibrium is stable in this model. As $F_e\to 0$, $R_{\ast }\sim [k\Delta P{R}_s/(\mu F_e){]}^{1/2}$; sufficiently large radii can invalidate the negligible-gravity or semi-infinite homogeneous-medium assumptions.

Write the radial velocity as $u(r)$, allowing $u<0$ for accretion flow, and let $c^2=\gamma p/\rho$ be the squared adiabatic sound speed. The steady continuity equation, radial Euler equations for an inviscid fluid and entropy advection equation giveOn a nonzero smooth flow branch, the polytropic equation of state is $p=K{\rho }^{\gamma }$ with constant $K$. Thus the two useful first integrals for spherically symmetric adiabatic flow areHere $\mathcal{J}$ is the signed mass flux and $\mathcal{B}$ is the Bernoulli function. The specific enthalpy is $c^2/(\gamma -1)$; the prescribed Newtonian gravitational potential has no contribution from the gas's own gravity. Eliminating the density derivative using mass conservation gives the differential formThis sonic point equation displays both the singular coefficient and the numerator that must vanish for a smooth transonic branch.

At a finite smooth sonic point $r_{\ast }$, the equation from (a) requiresSet $\delta =\gamma -1$ and $q=2/\beta -1/2$. Substitution into the Bernoulli function givesAn inflow from a warm reservoir at infinity has $\mathcal{B}=c_{\infty }^2/\delta >0$; an outflow reaching infinity has $\mathcal{B}=u_{\infty }^2/2+c_{\infty }^2/\delta \ge 0$. Therefore a nondegenerate transonic spherical flow in a power-law potential requiresFor the usual range $0<\beta <4$, the critical adiabatic exponent for spherical power-law flow is consequentlyThe printed assumption $\beta >0$ also permits $\beta \ge 4$. In that range the boxed inequality holds for every finite $\gamma >1$: there is no finite upper bound. Equivalently, take $f(\beta )=+\infty$ for $\beta \ge 4$. Extending the rational expression beyond $\beta =4$ would give an incorrect restriction.

The strict inequality also follows from local sonic-point slope discriminant analysis, including the potentially cold zero-energy endpoint. Put $X=(r_{\ast }/u_{\ast })u_{\ast }^{\prime }$. Since mass conservation implies $(c^2{)}^{\prime }/c^2=-\delta (u^{\prime }/u+2/r)$, differentiating the sonic point equation at $r_{\ast }$ yieldsIts discriminant is $8[2\beta -(4-\beta )\delta ]$. Moreover, the logarithmic Mach number derivative at the sonic point isA genuine crossing has two distinct branches and a nonzero derivative. At equality the crossing degenerates. In particular, for $\gamma =f(\beta )$ and $\mathcal{B}=0$, mass conservation and the polytropic equation of state make the Mach number constant along the scale-invariant solution; a solution that is sonic there is sonic everywhere, rather than crossing an isolated sonic point.