 Solution

Let $D\ll 1$ be the remaining absorber optical depth, and retain the mixture's convective exponent ${\kappa }_a=R_a/c_{pa}$. The outgoing surface contribution is $\sigma T_s^4{e}^{-D}\simeq \sigma T_s^4(1-D)$. To first order, atmospheric emission is the column-mean source times $D$. If $4{\kappa }_a\simeq 1$, the adiabat gives $T(p{)}^4\simeq T_s^{4}p/p_a$, a source linear in mass/optical depth. Therefore its mean is $T_s^4/2$, or equivalently it is evaluated at $p_a/2$:Setting this equal to $L_{⊛}/4$ givesThe zero-opacity limit correctly gives $T_e$, while a small positive opacity raises the surface temperature slightly. The expression is a first-order thin-column result, not a formula to extend to its apparent pole at ${\tau }_{\infty }=2$. More generally, direct thin-column averaging gives $F_{\mathrm{atm}}\simeq D\sigma T_s^4/(1+4{\kappa }_a)$; the half-column value is exact at this order when $4{\kappa }_a=1$. The assumed convective thermal structure remains a modelling condition even when the gas becomes radiatively thin.