= Solution
Adding the two flux solutions gives
$$
\sigma T(\tau)^4=\frac{L_\infty}{2}(1+D-\tau).
$$
The supplied <pressure>–optical-depth law gives $D-\tau=Dp/p_0$. With the exponential <pressure> model $p=p_0e^{-z/H}$,
$$
\boxed{T(z)=T_\infty\left(1+ae^{-z/H}\right)^{1/4},\quad
T_\infty=\left(\frac{L_\infty}{2\sigma}\right)^{1/4},\quad a=\tau_\infty.}
$$
It approaches the finite atmospheric top <temperature> $T_\infty$ and has the lowermost-air value $T(0)=T_\infty(1+D)^{1/4}$. Its lapse rate is
$$
-\frac{dT}{dz}=\frac{T(z)}{4H}\frac{De^{-z/H}}{1+De^{-z/H}}.
$$
The stated approximate sign is appropriate physically: a nonisothermal <ideal gas> does not have an exactly constant hydrostatic <scale height> $R T/g$. Treating $H$ as a representative constant is an additional vertical-coordinate approximation. Algebraically the displayed profile is exact for the prescribed exponential <pressure> and <two-stream approximation>[two-stream] equations.
Back to article page
