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Interpret every logarithm in the empirical profile as base ten with the dimensionless argument $P/(1\mathrm{mbar})$. PutThe upper atmosphere has $T=800\mathrm{K}$ for $x\le 0$, while integration of $dT/d{\log }_{10}P=Ax$ below it givesAssume that the stated $1250\mathrm{K}$ planetary equilibrium temperature is a reasonable brightness temperature at the $1\mathrm{bar}=10^3\mathrm{mbar}$ photosphere. ThenAt $10\mathrm{bar}$, $x=4$, and thereforeThis estimate neglects day-night variation, wavelength-dependent photospheric pressure, and a possible radiative-convective boundary; it treats the retrieved profile as representative of the dayside disk.

Let the stellar radius and temperature be $R_{\ast }$ and $T_{\ast }$. Assume Bond albedo $A_B$, isotropic stellar emission, blackbody planetary emission, and complete redistribution over the tidally locked planet. The absorbed stellar power isAdding the isolated internal luminosity $4\pi R_p^2\sigma T_0^4$ and balancing the total against $4\pi R_p^2\sigma T_{\mathrm{eq}}^4$ gives the planetary equilibrium temperatureIf heat is reradiated uniformly only over the dayside, replace the denominator $4a^2$ by $2a^2$. The latter is often more plausible for inefficient redistribution on a tidally locked bare planet.