At depths satisfying $\gamma c\tau\gg1$, the exponential term is negligible. The atmosphere remains approximately isothermal while its intrinsic contribution is also small:
$$
T_{\rm int}^4(b+\tau)
\ll fT_{\rm irr}^4
\left(b+\frac1{\gamma c}\right).
$$
Thus the plateau occupies approximately
$$
\boxed{
\frac1{\gamma c}\ll\tau\ll
f\left(\frac{T_{\rm irr}}{T_{\rm int}}\right)^4
\left(b+\frac1{\gamma c}\right)-b}
$$
and has temperature
$$
\boxed{T_{\rm iso}\simeq T_{\rm irr}
\left[af\left(b+\frac1{\gamma c}\right)\right]^{1/4}}.
$$

For example, take $T_{\rm irr}=2000\,\mathrm K$, $T_{\rm int}=200\,\mathrm K$, $\gamma=1$, $f=1/2$, and the Eddington constants. Then $T_{\rm iso}\simeq1650\,\mathrm K$ and the formal plateau extends from $\tau$ of order unity to several thousand. At still greater depth,
$$
T^4\simeq aT_{\rm int}^4\tau+	ext{constant},
$$
so the radiative solution rises as $T\propto\tau^{1/4}$ until convection replaces it.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 315 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-315/1/b/solution