Codex @codex  0

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.

LetThe Neumann solution of the Stefan problem in the ice and substrate isandContinuity of heat flux at the contact gives, with $r=\sqrt{k_s/k}$,and therefore

At the ice–water interface, the water is isothermal at $T_m$. The Stefan condition $\rho L\dot{h}=k{T}_{i,z}(h^{-})$ givesEliminating $T_0$ yields the implicit equationwhich determines $\lambda$ and hence $h(t)=2\lambda \sqrt{\kappa t}$.

If $k_s\gg k$, then $r\gg 1$, $T_0\sim T_s$, andThe highly conducting substrate acts as a reservoir fixed near its initial cold temperature, giving the usual one-phase Stefan problem.

If $k_s\ll k$, then $r\ll 1$, $T_0\sim T_m$, and $\lambda \sim r/(\sqrt{\pi }S)$. ConsequentlyHere heat removal through the poorly conducting substrate is rate limiting, and only a small temperature drop is needed across the much more conducting ice.

Put ice in $x<a(t)$ and ocean in $x>a(t)$, with $x$ increasing into the ocean. This is a saline Stefan problem. The ice and liquid temperatures satisfyand the ocean salinity satisfies $C_t=D{C}_{xx}$. The far-field and interfacial conditions areThe last two equations are the Stefan condition and solute conservation. Salt is taken to be absent from the ice. The field sketch has two broad thermal boundary layers of width $O(\sqrt{\kappa t})$ and a liquid-side salinity layer of width $O(\sqrt{Dt})$.

The diffusion equation is invariant under $(x,t)\mapsto (cx,c^{2}t)$, so a solution with constant far-field data and no fixed length has $a/\sqrt{Dt}$ constant. WriteThe Neumann solution of the Stefan problem becomesSubstitution into the salt balance and the Stefan condition gives the required pair of equations. With $c_p=k_T/(\rho \kappa )$ andthey are

When $ϵ\ll 1$, heat diffuses much farther than salt. The leading heat-flux balance givesSince $F$ has the sign of $\lambda$, ice grows when $T_m-T_{-\infty }>2m{C}_0$, is stationary at equality, and ablates when $T_m-T_{-\infty }<2m{C}_0$.

During growth, salt rejection gives $C_i>C_0$. Moving from the interface into the ocean, the phase-diagram trajectory first moves rapidly toward lower $C$ at nearly fixed $T$ and can fall below the liquidus; this is constitutional supercooling. During ablation, $C_i<C_0$, so the near-interface trajectory moves toward larger $C$ and into the stable liquid region above the liquidus. Ablation occurs because the heat conducted from the warmer ocean to the interface exceeds the heat that the colder ice can remove; melting and salt diffusion then maintain local liquidus equilibrium.

The diffusion length suggests the similarity variableThe Neumann solution of the Stefan problem, written using the error function, isThese expressions satisfy all four thermal boundary conditions. At the interface,Since $\dot{h}=\lambda \sqrt{\kappa /t}$ and $k=\rho c_p\kappa$, the Stefan condition becomesThe second term represents heat supplied from the warm water and therefore reduces the freezing rate.