= Solution
The approximation in part v requires the dimensionless stagnant depth
$$
\eta=
\frac{1-3\theta/5}
{\mathcal F\theta^{5/3}}
$$
to remain small. Since $0<\theta\leq1$ makes the numerator order one, this condition is
$$
\boxed{\theta\gg\mathcal F^{-3/5}}.
$$
Using the solution from part v, the equivalent time range is
$$
1+\frac23\mathcal F\tau
\ll\mathcal F^{2/5},
$$
or, at the level of <asymptotic equivalence>,
$$
\boxed{0\leq\tau\ll\mathcal F^{-3/5}}.
$$
The cooling becomes appreciable on the shorter scale $\tau=O(\mathcal F^{-1})$, so these ranges overlap widely when $\mathcal F\gg1$.
The <Rayleigh number> based on the convecting depth is
$$
\operatorname{Ra}
=\frac{g\alpha(T_w-T_m)^2(H-h)^3}{\nu\kappa}.
$$
The definition of $\mathcal F$ implicit in part iii gives
$$
\mathcal F
=C_0H
\left(\frac{g\alpha\Delta T^2}
{\operatorname{Ra}_c\nu\kappa}\right)^{1/3}.
$$
It follows that
$$
\boxed{
\operatorname{Ra}
=\operatorname{Ra}_c
\left(\frac{\mathcal F}{C_0}\right)^3
\theta^2(1-\eta)^3}.
$$
If $1-\eta=O(1)$ and $\theta\gg\mathcal F^{-3/5}$, then $\operatorname{Ra}\gg O(\mathcal F^{9/5})$. The <thermal convection> therefore remains strongly supercritical throughout the range in which the thin stagnant-layer approximation is valid.
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