A useful first estimate places the <radiative-convective boundary> where the intrinsic term becomes comparable to the deep irradiation plateau:
$$
T_{\rm int}^4\tau_{\rm RCB}
\sim fT_{\rm irr}^4
\left(b+\frac1{\gamma c}\right).
$$
Thus
$$
\boxed{\tau_{\rm RCB}\sim
f\left(\frac{T_{\rm irr}}{T_{\rm int}}\right)^4
\left(b+\frac1{\gamma c}\right)},
\qquad
\boxed{P_{\rm RCB}\sim\frac g{\kappa_{\rm IR}}\tau_{\rm RCB}}.
$$
With the assumptions of part (b), $g=10\,\mathrm{m\,s^{-2}}$ and $\kappa_{\rm IR}=10^{-3}\,\mathrm{m^2\,kg^{-1}}$, one finds $\tau_{\rm RCB}\sim6.2\times10^3$ and $P_{\rm RCB}\sim6.2\times10^7\,\mathrm{Pa}\simeq620\,\mathrm{bar}$.

More precisely one equates the local radiative logarithmic gradient to the adiabatic gradient. A constant-opacity grey profile approaches $\nabla_{\rm rad}=1/4$, so a diatomic adiabat with $\nabla_{\rm ad}=2/7$ requires the realistic increase of opacity with depth to become convective. The numerical pressure is therefore an order-of-magnitude estimate, sensitive mainly to $T_{\rm int}$ and deep opacity.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

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