Past exam of the physics of the Earth as a Planet course of the University of Cambridge / 2012 / Paper 1 / 1 / c / Solution

For flux-driven convective layer cooling, use $E=C{\theta }^{4/3}$ with the same unspecified coefficient as in (b). The steady heat balance givesThe corresponding boundary-layer thickness isSet $\theta ={\theta }_{\ast }+ϵ$ and expand the heat loss to first order. Since $E^{\prime }({\theta }_{\ast })=(4/3)C{\theta }_{\ast }^{1/3}=4F/(3{\theta }_{\ast })$,This is the near-equilibrium relaxation time. The full equation is nonlinear, so an arbitrary large initial temperature departure need not decay as one exponential. Positive heat loss derivative makes the steady state stable. In particular, $\tau$ scales with $d{\delta }_{\ast }$, rather than $d^2$, because convection shortens the effective heat-transport distance to the boundary layer.