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In a steady flow the ideal magnetohydrodynamic induction equation gives $\nabla \times (\mathbf{u}\times \mathbf{B})=0$. Since both vectors have only poloidal magnetic field components, write $\mathbf{u}\times \mathbf{B}=F_{\varphi }{\mathbf{e}}_{\varphi }$. Axisymmetry impliesFor a field regular on the axis, $C=0$; equivalently one may impose zero toroidal electromotive force. Under this physical condition, $\mathbf{u}$ is parallel to $\mathbf{B}$. Write $\rho \mathbf{u}=k\mathbf{B}$. Continuity and $\nabla \cdot \mathbf{B}=0$ then give $\mathbf{B}\cdot \nabla k=0$. The poloidal magnetic flux function has $\mathbf{B}\cdot \nabla \psi =0$, so locally on each connected magnetic surface,This regular-axis alignment of steady poloidal ideal flow needs its zero-circulation condition: axisymmetry alone does not imply this alignment on a domain excluding the axis. An explicit counterexample is available in a cylindrical annulus. Take nonzero constants $U,b$, positive ${\rho }_c,c_s$, andHere $\psi =-bz$, $\nabla \cdot \mathbf{B}=\nabla \times \mathbf{B}=0$ and $\mathbf{u}\times \mathbf{B}=Ub{\mathbf{e}}_{\varphi }/R$ has zero curl. The flow satisfies continuity, $p^{\prime }=-\rho {\Phi }^{\prime }$, and ${\nabla }^2\Phi =4\pi G\rho$. Its acceleration and Lorentz force density vanish, so it solves the steady equations while $\mathbf{u}$ and $\mathbf{B}$ are perpendicular. The nonzero toroidal circulation and the axial singularity explain why this counterexample is excluded from a regular polar accretion column. The remaining derivation uses the regular, aligned branch.

For a narrow stream tube, integrating continuity gives constant $A\rho |\mathbf{u}|$. On a nonzero-flow tube the magnetohydrodynamic mass loading $k$ is constant, so $A\rho |\mathbf{u}|=A|k||\mathbf{B}|$. Therefore $A|\mathbf{B}|$ is constant and $A\propto |\mathbf{B}{|}^{-1}$. This also follows directly from conserved magnetic flux through a flux tube.

For constant isothermal sound speed, $\nabla p/\rho =c_s^2\nabla \ln (\rho /{\rho }_0)$, with arbitrary reference density ${\rho }_0$. Project momentum along $\mathbf{B}$. The Lorentz force density has no component in that direction, and alignment implies $\mathbf{B}\cdot (\mathbf{u}\cdot \nabla )\mathbf{u}=\mathbf{B}\cdot \nabla (|\mathbf{u}{|}^2/2)$. Thus the isothermal magnetic Bernoulli integral isChanging ${\rho }_0$ only shifts the Bernoulli function by a constant. Since $k$ is constant along the magnetic field,Differentiating the isothermal magnetic Bernoulli integral therefore gives

In the slender polar accretion column, the dipolar flux-tube area is $A(z)=A_{\ast }(z/R_{\ast }{)}^3$. The decreasing axial field is a leading approximation, not an exactly solenoidal field $(0,0,B_z(z))$ throughout a cylinder. Indeed $\nabla \cdot \mathbf{B}={\partial }_z{B}_z\ne 0$ for that literal field. Near the axis a small radial component $B_R\simeq -(R/2){\partial }_z{B}_z=3R{B}_z/(2z)$ supplies the required radial divergence. Its magnitude is smaller by $R/z$, even though its divergence is leading order. Keeping this expanding flux tube while neglecting transverse forces yields the intended one-dimensional model.

Use the positive inward speed $v=-u_z$; the signed sonic velocity is $u_z=-c_s$. Conservation of mass and the Bernoulli equation giveEliminating ${\rho }^{\prime }$ yields the isothermal dipolar accretion equation,A smooth transonic branch requires both sides to vanish at its sonic point, soThis is the critical point of the flow equation; a generic subsonic solution need not cross it. A crossing in the exterior column requires $z_s\ge R_{\ast }$, and a sonic point strictly outside the star requires $z_s>R_{\ast }$. Differentiating the equation at the crossing gives $2(v_s^{\prime }{)}^2=3c_s^2/z_s^2$. For inward accretion that accelerates toward the star, $v_s^{\prime }=-\sqrt{3/2}{c}_s/z_s$.

The reservoir boundary condition fixes $ϵ=c_s^2\ln ({\rho }_{\infty }/{\rho }_0)$. At the sonic point, ${\Phi }_s=-3c_s^2$ and $v_s^2/2=c_s^2/2$, givingFinally use the sonic mass flux through the dipolar flux-tube area:This rate belongs to the smooth transonic branch of the idealized column, with $A_{\ast }$ the supplied total loaded area. The reservoir condition alone does not force every steady solution onto this branch. A real narrow dipolar column must match an outer flow where the slender approximation ceases to apply.

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.

Let $L_k={\dot{m}}_k^{-c}$ be the mass processed by catastrophic impacts in logarithmic bin $k$. Steady mass conservation requires the same downward mass flux across every interior size threshold. A scale-independent fragment redistribution function makes the transfer kernel depend only on the number of logarithmic bin steps. Consequently a self-similar interior steady collisional cascade has $L_k=L$, independent of size: a constant processed mass per bin supplies the constant flux, with the same kernel-dependent proportionality at every bin. This is the constant mass flux in a collisional cascade result, away from injection and removal cutoffs. Size-independent fragmentation does not eliminate boundary waves at the very ends of a finite distribution.

For equal-density planetesimals, body mass is proportional to $D^3$ and a fixed logarithmic bin contains a number proportional to $Dn(D)$. The mass per logarithmic size bin is therefore $m_{\log }\propto D^{4-\alpha }$. With the given catastrophic planetesimal collision rate, $Q_D^{\ast }\propto D^b$ givesConstant mass flux in a collisional cascade sets the exponent to zero, yielding the strength-dependent steady collisional-cascade slopeThe time to change an order-one fraction of a bin's mass is $m_k/(m_k{R}_k)=1/R_k$. Its collisional-cascade relaxation time is thus $t_c(D)\sim 1/R(D,Q_D^{\ast })$. The contemporaneous projectile population must be used when a different part of the distribution has already evolved.

For the primordial differential-number index ${\alpha }_0=7/2$, the collisional-cascade relaxation time isIt has asymptotic logarithmic slope $1/12$ in the strength regime and $7/4$ in the gravity regime. It increases monotonically: its logarithmic derivative is $1/2+(5/6)d\log Q_D^{\ast }/d\log D$, ranging from $1/12$ to $7/4$. In particular the minimum of the catastrophic disruption threshold at $D_w$ is not a minimum of the collision time. The strength-gravity disruption transition obeysSmall sizes start evolving first. Before $t\sim t_{c0}(D_{\min })$, the whole distribution is nearly primordial. At intermediate times, a transition diameter $D_t$ with $t_c(D_t)\sim t$ separates processed small sizes from mostly primordial larger sizes. The steady strength-regime index is ${\alpha }_s=41/11$, while the gravity-regime index is ${\alpha }_g=3$. Thus a plot of mass per logarithmic size bin has slopes $3/11$ and $1$ in the two evolved regimes, compared with the primordial slope $1/2$. Once $D_t\gg D_w$, both steady slopes appear below the remaining primordial tail. The transition moves to larger sizes, and the normalization eventually decays as the largest bodies are depleted. Sharp joined power laws are a schematic description; detailed kernels produce smooth transitions and possible waves.

For the uniform depletion, the binary-collision evolution operator $F$ is quadratic in all bin masses: $F(cm)=c^{2}F(m)$. Let $m(t)$ be the undepleted trajectory and suppose $m_d(T)=m(T)/f$. Differentiating verifies the exact scaling within this fixed-kernel collision model,Immediately after depletion the shape is unchanged and the catastrophic planetesimal collision rates are reduced by $f$. Further evolution therefore proceeds $f$ times more slowly on the original trajectory. The processed-to-primordial transition initially remains at $D_t$, then advances on the stretched collision clock.

A collision-only model starting with the smaller initial normalization $m(0)/f$ has trajectory $\tilde{m}(u)=m(u/f)/f$. Matching the observed depleted distribution requires $u/f=T+(t-T)/f$. Hence the collisional age after uniform dynamical depletion isThe lower observed normalization would make it appear that slow collisions had taken much longer to produce the existing break. Shape and normalization alone cannot determine $f$ without independent information about age, initial mass, or depletion history.

Finally, count all impacts exceeding the strength threshold $Q_s=Q_a{D}^{-1/2}$, including impacts which also disperse the target. This is the source's inclusive rubblising convention; a bound-remnant rubblising collision alone would exclude the dispersing impacts. In the evolved gravity-regime projectile approximation, ${\alpha }_g=3$, so the given threshold-rate law gives the rubblising-to-dispersal collision-rate ratioAt the collisional front $D_t$, the age is of order one catastrophic collision time. Thus the expected number of strength-shattering impacts has the large-$D_t/D_w$ scalingThis is the requested scaling. It is not an exact unit-coefficient identity from the stated assumptions: using the true minimum diameter gives $R_{\mathrm{rub}}/R_{\mathrm{cc}}\simeq 3^{-2/3}(D_t/D_w{)}^{4/3}$. The actual cumulative number is ${\int }_0^T{R}_{\mathrm{rub}}(D_t,t)dt$, not automatically $T{R}_{\mathrm{rub}}(D_t,T)$, and its coefficient depends on evolution of the projectile population. The $4/3$ estimate also assumes the relevant projectiles sample the gravity-regime slope; sampling the strength or primordial slope changes the exponent. The steady-cascade framework and evolving transitions are developed by Wyatt, Clarke and Booth.

Take the accretion rate $\dot{M}>0$ to mean inward flow. In a steady state, conservation of mass makes the inward mass flux independent of radius, so $\mathcal{F}=\dot{M}$. Integrating its radial expression givesFor Keplerian rotation, $\Omega =(G{M}_{\ast }/r^3{)}^{1/2}$ and the viscous torque in an accretion disk is $\mathcal{G}=3\pi \bar{\nu }\Sigma \sqrt{G{M}_{\ast }r}$. The zero-torque inner boundary condition therefore sets $\bar{\nu }\Sigma =0$ at $r_{\ast }$. It fixes $C=-\dot{M}\sqrt{r_{\ast }}/(3\pi )$, giving the steady density profileThis is the Keplerian accretion disk solution on $r_{\ast }<r<r_{\mathrm{out}}$. It determines the kinematic viscosity–surface density product; a separate viscosity closure is needed to turn it into an explicit power law. If the kinematic viscosity is finite and nonzero at the inner edge, the surface density tends to zero there.

At a fixed height, first determine the stored quantities per unit vertical height and per unit length along the line source. With $\Delta \rho =\rho -{\rho }_0$, the triangular-profile line plume has the useful profile integralThe geometric volume is $\mathcal{A}=2b$; the mass is $\mathcal{D}={\rho }_0{I}_0+\Delta \rho I_1=b(\rho +{\rho }_0)$. The upward momentum is $\mathcal{P}=W({\rho }_0{I}_1+\Delta \rho I_2)$, while the buoyancy force is $\mathcal{B}=-g\Delta \rho I_1$. ThusHere positive buoyancy corresponds to a rising light plume. If instead buoyancy is defined kinematically, divide $\mathcal{B}$ by ${\rho }_0$; the density-weighted convention here is the one matching the printed equations.

Multiply each local density by the vertical velocity to obtain the fluxes through a horizontal section, still per unit source length:In particular, the stored momentum equals the mass flux $Q$, and the buoyancy flux is $F=g({\rho }_{0}V-Q)$. The different triangular-profile factors in storage and flux must not be replaced by top-hat coefficients.

Choose $x=0$ at the contact line, with $x>0$ pointing into the mountains and the plate moving in the positive $x$ direction at speed $U>0$. Let $z=0$ be the plate. For viscous mountain spreading over an underthrust plate, the lubrication approximation assumes slow horizontal variation, hydrostatic pressure $p=\rho g(h-z)$, and $\mu u_{zz}=p_x=\rho g{h}_x\equiv G$. Impose no slip at the plate and zero tangential traction at the free surface. Without sediment,The horizontal mass flux is $\rho q$. For a steady mountain range with no mountain material entering across its contact line, $q=0$. Therefore $h_x=3\mu U/(\rho g{h}^2)$, and the vanishing height at the bare contact gives the mountain profileThe large slope extremely close to $x=0$ is an inner region where the lubrication approximation ceases to be quantitatively valid.

Now let sediment occupy $0<z<d$ and mountains $d<z<h$, with $H=h-d$ the mountain-material thickness and $M={\mu }_s/\mu$. Equal densities make $G=\rho g{h}_x$ the same in both layers. Continuity of velocity and shear stress at $z=d$, no slip at $z=0$ and zero shear at $z=h$ give the shear stress $G(z-h)$ throughout. Thus the two velocity profiles areTheir volume fluxes areMultiplying by $\rho$ gives the separate mass fluxes. The coupled conservation of mass equations areThis is the sediment-lubricated viscous mountain model. It includes shear-driven transport by the plate and pressure-driven gravitational spreading.

In steady state the mountain-material flux is zero, while the upstream uniform sediment layer has $q_s=U{d}_{\infty }$. For $H>0$ this givesEliminating $G$ proves the algebraic thickness relation:At the contact $H\to 0^{+}$, the last coefficient must vanish. The nonzero physical solution is consequentlyThe condition comes from the limiting zero flux per unit mountain thickness, not merely the automatic vanishing of $q_m$ at $H=0$. The surface slope anywhere under the mountains isLower sediment viscosity flattens the near-contact surface, while its contact height remains $3d_{\infty }$. Far inside, where $H\gg d/M$, the mountains recover the bare-layer slope scaling. The formulas determine the mountain profile by the algebraic relation plus this first-order equation. For example, put $\delta =d/d_{\infty }$ and $w=H/d$; the physical root satisfies $2M(\delta -1)w^2+3(\delta -2)w+\delta -3=0$, taking $w\ge 0$ for $1<\delta \le 3$. This selects the branch smoothly connected to the contact.

In front of the contact, $x<0$, there is only sediment, so $h=d$ andDefine $\ell =\rho g{d}_{\infty }^3/(3{\mu }_{s}U)$ and $P(\delta )={\delta }^3/3+{\delta }^2/2+\delta +\ln (\delta -1)$. Integrating from the contact yields the upstream basin profile of a sediment-lubricated mountain:Equivalently, $d(x)=d_{\infty }{P}^{-1}(P(3)+x/\ell )$ on $1<\delta \le 3$. The inverse is uniquely defined because $P^{\prime }(\delta )={\delta }^3/(\delta -1)>0$; an elementary explicit inverse is unnecessary. This profile tends to $d_{\infty }$ as $x\to -\infty$, approaches $3d_{\infty }$ at the contact and has the same surface slope there as the mountain branch. Its far-field excess decays exponentially on length scale $\ell$, which increases as the sediment viscosity decreases.

Let $z$ measure distance inward from the surface. While $a\ll R$, curvature is negligible and this is a planar two-phase Stefan problem. Use a common reference mass density $\rho$, specific heat capacity $c_p$, thermal conductivity $k$ and thermal diffusivity $\kappa =k/(\rho c_p)$; retain the mass density differences only in the subsequent buoyancy estimates. Define $\Delta T=T_m-T_s>0$, $\Theta =\overline{T}-T_m\ge 0$, and $\vartheta =\Theta /\Delta T$. The paper's parameter $S=L/(c_p\Delta T)$ is the latent-to-sensible heat ratio, reciprocal to the more usual Stefan number.

For superheated two-phase Stefan similarity, solving $T_t=\kappa T_{zz}$ with $T(0)=T_s$, $T(a)=T_m$ and $T(\infty )=\overline{T}$ gives the Neumann solution of the Stefan problem. With $\eta =z/(2\sqrt{\kappa t})$,The solid expression holds for $0<z<a$ and the liquid expression for $z>a$. The interface gradients inserted in the Stefan condition determine the positive constant $\lambda$:The right-hand bracket decreases from infinity and crosses zero, while the equivalent left-hand side $S\sqrt{\pi }\lambda e^{{\lambda }^2}$ increases, proving uniqueness. The liquid-side flux subtracts from the roof cooling: it is heat brought to the interface by the superheated interior.

For large $S$, $\lambda \ll 1$. Keeping the superheat term while using $\mathrm{erf}\lambda \simeq 2\lambda /\sqrt{\pi }$ and $\mathrm{erfc}\lambda \simeq 1$ yields $S{\lambda }^2+(\vartheta /\sqrt{\pi })\lambda \simeq 1/2$. Hence the small-interface-parameter growth with liquid superheat approximation isThe solid temperature is approximately linear across its thin crust, while the liquid thermal boundary layer has thickness of order $\sqrt{\kappa t}$. For fixed superheat and $S\gg 1$, $A\simeq \sqrt{2/S}$; for $\vartheta \gg \sqrt{S}$, $A\simeq \sqrt{\pi }/\vartheta$. This retains the strong suppression of growth by a very hot interior.

To estimate overturn, take local gravitational acceleration $g$ at the roof. The liquid's mass density contrast is $\Delta {\rho }_l={\rho }_f\alpha \Theta$, while the solid contrast is $\Delta {\rho }_s={\rho }_s-{\rho }_f$. A liquid thermal layer of thickness $\delta$ has Rayleigh number $\Delta {\rho }_{l}g{\delta }^3/({\mu }_l\kappa )$. As $\delta \sim \sqrt{\kappa t}$, it reaches its critical value after a time scaling asIn viscous overturn of a dense crust, an unstably stratified viscous crust has a Rayleigh-Taylor instability growth time of order ${\mu }_s/(\Delta {\rho }_{s}ga)$. Setting this equal to its age, with $a=A\sqrt{\kappa t}$, givesThese are formation-to-overturn times. A specified critical liquid Rayleigh number ${\mathrm{Ra}}_c$ multiplies ${\tau }_l$ by ${\mathrm{Ra}}_c^{2/3}$; logarithmic amplification and geometrical constants also affect the crust estimate. They are omitted for the requested order-of-magnitude scaling.

The given numbers yield $S\simeq 2.54$, $\Delta {\rho }_s=500\mathrm{kg}{\mathrm{m}}^{-3}$ and $\Delta {\rho }_l=100\vartheta \mathrm{kg}{\mathrm{m}}^{-3}$. ThusLiquid thermal overturn is enormously faster than crustal overturn throughout the stated range. Even a critical-Rayleigh prefactor of order $10^2$ does not change that ordering. Since $S\simeq 2.54$ is not asymptotically large, low-superheat coefficients from the permitted approximation are only approximate. At the upper end $\alpha \Theta =1.25$, so the linear thermal-density approximation also ceases to be quantitatively credible; that endpoint is a formal scaling estimate under the supplied constant-property model. At a common age the instantaneous viscous instability-time ratio instead scales as $5A10^{-18}/\vartheta$; it should not be confused with the above formation-time ratio.

For a boundary-layer renewal model, the liquid layer reaches ${\delta }_l\sim \sqrt{\kappa {\tau }_l}$. Its cycle-averaged thermal flux is consequentlyIf a critical Rayleigh number is retained, insert a factor ${\mathrm{Ra}}_c^{-1/3}$. Averaging the diffusive $t^{-1/2}$ flux over a cycle only changes an order-one coefficient. The crust reachesso its average solid-volume flux per roof area is the crust-removal fluxThe corresponding mass flux is ${\rho }_s{J}_s$ and latent-heat flux is ${\rho }_{s}L{J}_s$. This independent-cycle estimate is a scale for a crust that reaches overturn thickness and is then renewed. Because the liquid renews much more rapidly, in a hot interior its average heat flux can limit actual crust growth: the consistent interfacial budget is $\rho L\dot{a}\simeq k\Delta T/a-F_l$. A stalled crust would not realize repeated solid cycles. In a superheated interior, detached solid may also remelt; permanent core deposition is most directly justified in the late regime requested next.

At late times $\Theta =0$, liquid thermal buoyancy vanishes and $A^2\kappa \simeq 2\kappa /S=2k\Delta T/(\rho L)$. ThusAssuming thin renewed crust, efficient sinking and constant asteroid radius, conservation of mass gives $4\pi b^2\dot{b}\simeq 4\pi R^2{J}_0$ under the permitted equal-density approximation. Therefore the asteroidal core growth by crustal sinking law isFor a self-gravitating approximately uniform sphere, $g\simeq 4\pi G\rho R/3$, so $J_0\propto R^{1/3}$ and a zero-seed core scales as $b\propto R^{7/9}(t-t_0{)}^{1/3}$. This law applies while the thin-crust renewal model remains valid and liquid remains available; it cannot continue beyond complete solidification. The specific numerical lifetime cannot be inferred without the absolute solid viscosity and conductivity.

Write the differential power-law size distribution as $n(D)=K{D}^{-\alpha }$ and take all bodies to have the same bulk density. In a fixed belt volume, $K$ is proportional to $M_{\mathrm{tot}}$ when the size cutoffs and the shape of the distribution are fixed, becauseFor $\alpha \ne 4$ the integral is $(D_{\max }^{4-\alpha }-D_{\min }^{4-\alpha })/(4-\alpha )$; at $\alpha =4$ it is a logarithm. Thus no assumption that the mass is always dominated by the largest bodies is needed for this normalization step.

A size-independent catastrophic disruption threshold and collision speed imply a fixed minimum projectile-to-target diameter ratio $\xi$, since $\frac{1}{2}{m}_p{v}^2/m_t\sim Q_D^{\ast }$ gives $\xi \sim (2Q_D^{\ast }/v^2{)}^{1/3}$. With negligible gravitational focusing, a target of size $D$ has a catastrophic collision rate proportional toFor $\alpha >3$ and targets well inside the cutoffs, the dimensionless integral converges at the upper end and is size independent. ConsequentlyHere $A$ contains the belt geometry, speed, density, cutoffs and disruption parameters. Close to a cutoff, or if $\xi D<D_{\min }$, this scaling needs modification. A finite belt without replenishment can only be in quasi-steady state over times short compared with depletion of its largest reservoir.