Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 78 (Fluid Dynamics of the Solid Earth)

For a stationary, uniformly permeable skeleton, Darcy law gives $\mathbf{u}=-(\Pi /\mu )\nabla p$. Constant-density conservation of mass gives $\nabla \cdot \mathbf{u}=0$, hence the pressure is harmonic:With uniform gravitational acceleration, the same argument applies to pressure after subtracting its hydrostatic part. The spherical calculation below neglects body-force variation and uses the excess pressure relative to the far field.

Set $ϵ=1-{\rho }_s/\rho >0$ and $\beta =\varphi ϵ$. When a shell freezes, its pore mass decreases at rate $4\pi a^2\varphi (\rho -{\rho }_s)\dot{a}$; the difference must leave as liquid. Since Darcy velocity is volume discharge per total area, not pore-fluid speed, the interfacial and radial balances giveThe positive sign means outward expulsion by the expansion on freezing. Integrating Darcy law with $p(\infty )=0$ givesThere is no liquid Darcy velocity in the ice-filled region.

To state a closed Stefan problem, let $k_{-},k_{+}$ be the effective thermal conductivities of the frozen and unfrozen composites and $C_{-},C_{+}$ their volumetric heat capacities. If the matrix mass density is ${\rho }_m$, the common specific heat gives $C_{-}=c_p[(1-\varphi ){\rho }_m+\varphi {\rho }_s]$ and $C_{+}=c_p[(1-\varphi ){\rho }_m+\varphi \rho ]$. These material coefficients are not numerically specified. With $\mathcal{L}=\varphi {\rho }_{s}L$, the thermal equations areThe exterior equation retains advected sensible heat. A more detailed mechanical-energy model adds the Darcy dissipation source $\mu u^2/\Pi$; its smallness at leading order is checked below. At the origin require ${\partial }_r{T}_{-}(0)=0$, and at infinity require $T_{+}\to T_{\infty }$, $p\to 0$. At the phase boundary impose temperature continuity, phase equilibrium and the Stefan condition:These equations, the radial liquid-mass balance and Darcy law, together with an initial radius and compatible initial temperature fields, specify the moving-boundary model. No curvature correction is imposed. A vanishing initial seed corresponds to the leading growth law below with $a(0)=0$.

Write $\Delta =T_m-T_{\infty }$ and $S=L/(c_p\Delta )$, the latent-to-sensible heat ratio. A conductive interfacial balance gives $a\dot{a}\lesssim k_{+}\Delta /\mathcal{L}$. On length scale $a$, the ratio of thermal diffusion time to growth time is consequently bounded byIt is small for $S\gg 1$ with fixed porosity and comparable constituent densities and thermal properties. Very small porosity requires the sharper condition $C_{\pm }\Delta /(\varphi {\rho }_{s}L)\ll 1$; the bare heat-capacity ratio alone should not be used uniformly as $\varphi \to 0$. The advective-to-conductive ratio isEven if mechanical heating is retained, its size relative to latent-heat release is $p_i\beta /\mathcal{L}=(T_m-T_i)/T_m\le \Delta /T_m$. For fixed water properties this is also small as $S\to \infty$, consistently with the linearized Clausius-Clapeyron relation. Thus unsteadiness, thermal advection and mechanical heating are higher-order effects. The local leading temperature fields satisfy Laplace's equation. The approximation describes radii of order $a$; the distant transient diffusion tail supplies the far-field matching and need not be globally steady.

Regularity at zero and the interface/far-field values giveThe Stefan condition becomes $\mathcal{L}a\dot{a}=k_{+}(T_i-T_{\infty })$. Introduce the dimensionless pressure-feedback factorCombining this heat balance with pressure-dependent phase equilibrium yields the growth and interface values:This is pressure-limited spherical freezing in a porous medium. The interface values are constant during this leading solution, although $p(r,t)$ changes at a fixed exterior radius.

As permeability of a porous medium decreases, $\Lambda$ increases: the interface pressure rises toward the finite value $\Delta /\gamma$, while the interface temperature falls toward $T_{\infty }$. Hydraulic resistance suppresses freezing through pressure-induced freezing-point depression. In the limit $\Pi \to 0$, $a\dot{a}\sim \Pi \Delta /(\gamma \mu \beta )$; a zero-seed radius grows as $\sqrt{\Pi t}$ and tends to zero at fixed time. A finite seed instead hardly changes. At exactly zero permeability an incompressible rigid model cannot expel the expansion volume, so sustained growth stops; deformation or compressibility would require a different model.

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.