<p><a href="https://ui.adsabs.harvard.edu/abs/2010A%26A...520A..27G/abstract" rel="nofollow noreferrer">Guillot et al. 2010</a> give a comparatively simple solution (their Eqn. 27) for a double-gray purely radiative atmosphere. It reproduces the equilibrium temperature in the optically thin limit.</p>
<p><a href="https://ui.adsabs.harvard.edu/abs/2012ApJ...757..104R/abstract" rel="nofollow noreferrer">Robinson &amp; Catling 2012</a> extended the previous calculations to include convection. The expression for the temperature profile as function of optical depth can however not be written down explicitely anymore in that case, it has to be iterated on.</p>
<p>Those are merely solutions that have become popular in recent years in the exoplanet community. Less well known is that <a href="https://ui.adsabs.harvard.edu/abs/1906WisGo.195...41S/abstract" rel="nofollow noreferrer">Schwarzschild 1906</a>, <a href="https://ui.adsabs.harvard.edu/abs/1921MNRAS..81..361M/abstract" rel="nofollow noreferrer">Milne 1921</a> and <a href="https://ui.adsabs.harvard.edu/abs/1969Icar...10..290S/abstract" rel="nofollow noreferrer">Sagan 1969</a> already contributed solutions of the form of
<span class="math-container">$$ T^4(\tau) = T^4_{eq}\left(\frac{1}{2}+\frac{3}{4}\tau  \right) $$</span>
where <span class="math-container">$\tau$</span> is the optical depth of the atmospheric gas, taken as vertical coordinate and the unity factors in the brackets (particularly the 3/4 in front of <span class="math-container">$\tau$</span>) will depend on many details of the derivation, especially boundary conditions and various assumptions about geometry and radiation transport. The variable <span class="math-container">$\tau$</span> increases as we go deeper into the atmosphere.</p>
<p>Note that this solution predicts <span class="math-container">$T=T_{eq}$</span> only at an optical depth of <span class="math-container">$2/3$</span>, i.e. for moderately massive atmospheres. At lower optical depth, the atmospheric temperature will be <span class="math-container">$T&lt;T_{eq}$</span>, even if there is a surface at <span class="math-container">$T=T_{eq}$</span>.</p>
<p>To get a surface temperature, you have to know the infrared optical depth of the surface though <span class="math-container">$\tau_0$</span>, which you have to obtain from the complicated and pressure-dependent opacity functions of the entire gas mixture in question. Then you can evaluate <span class="math-container">$T^4(\tau_0)$</span> for an estimate of the surface (gas) temperature.</p>
<p>For Earth, with <span class="math-container">$\tau_{IR}\approx2$</span> and <span class="math-container">$T_{eq, E}=255K$</span>, this would yield <span class="math-container">$T_{surf,E}=303K$</span> (an ok estimate, considering that the purely radiative model ignores convection), and for Venus with <span class="math-container">$\tau_{IR}\approx 100$</span> (taken from the Sagan paper), and <span class="math-container">$T_{eq,V}=226K$</span> one would get <span class="math-container">$T_{surf,V}=666K$</span> (clearly farther off, but still shows the point that the surface is much hotter than the equilibrium temperature).</p>

CC BY-SA 4.0
2023-11-09T11:43:16.120Z
You wouldn't go to far wrong with an estimate of the height to reach optical depth 1, the adiabatic lapse rate and the equilibrium temperature. This gives a good way to understand the greenhouse effect.
ProfRob
CC BY-SA 4.0
2023-11-09T11:52:03.623Z
@ProfRob: Depends on the surface gravity, which will determine how far away $r_{\tau=1}$ and the radiative-convective boundary are from each other.
AtmosphericPrisonEscape
CC BY-SA 4.0
2023-11-09T13:26:10.323Z
@AtmosphericPrisonEscape lapse rate incorporates gravity.  It's part of the equation.  Maybe that's obvious, but I thought I'd point it out.    That raises some interesting math for a very fast rotating body, where the gravity varies considerably equator to pole . . . but I digress.
userLTK
CC BY-SA 4.0
2023-11-09T14:38:38.663Z
@userLTK Not sure what you are trying to say. The above solution is a purely radiative solution with the optical depth $\tau$ as vertical coordinate. It contains no information about realspace, hence no information about temperature gradients. Note that your usual radiative temperature gradient follows from this via $dT/dx = dT/d\tau d\tau/dx = dT/d\tau \rho \kappa$, where $\rho$ is the density profile, which is how the gravity enters.
AtmosphericPrisonEscape
CC BY-SA 4.0
2023-11-10T15:13:52.633Z
I just meant that on other planets, different gravities, the lapse rate changes.   Which, I think you probably knew.   Maybe my initial comment was unnecessary.  https://image2.slideserve.com/3857728/dry-adiabatic-lapse-rate1-l.jpg
userLTK
CC BY-SA 4.0
2023-11-08T05:19:34.190Z
2023-11-09T09:40:20.020Z
2023-11-09T09:40:20.020Z
AtmosphericPrisonEscape
<p>For example, a thicker atmosphere would probably lead to less temperature variation . . . I assume that much is obvious.   And, greenhouse gases trap heat.</p>
<p>But when posed with a question like &quot;what temperature would Venus be if it had an oxygen-nitrogen atmosphere, same 90 earth atmospheric pressure, but an earth-like atmosphere (minus the water vapor).</p>
<p>Planetary equilibrium temperature is easy to calculate and it's a relatively simple formula.     I know that with an atmosphere it becomes much more complicated and there are questions such as cloud cover, but I'm curious if anything exists that makes an estimate for rocky planet surface temperature adjusted for various atmospheres.</p>
<p>And lapse rate gives a good calculation for how temperature changes with altitude but provides little to nothing on the initial surface temperature.</p>
<p>This question below touches on some of the ideas, but the answer on temperature estimates is a work in progress.</p>
<p><a href="https://astronomy.stackexchange.com/questions/54475/public-data-for-calculating-equilibrium-temperature-of-exoplanets">Public data for calculating equilibrium temperature of exoplanets</a></p>

<planet><planetary-atmosphere><temperature>
Is there a formula for calculating surface temperature of a rocky world based on atmosphere and solar input?
userLTK
# Is there a formula for calculating surface temperature of a rocky world based on atmosphere and solar input?

(Asked by: username_0 on 2023-11-08)

For example, a thicker atmosphere would probably lead to less temperature variation . . . I assume that much is obvious. And, greenhouse gases trap heat.

But when posed with a question like "what temperature would Venus be if it had an oxygen-nitrogen atmosphere, same 90 earth atmospheric pressure, but an earth-like atmosphere (minus the water vapor).

Planetary equilibrium temperature is easy to calculate and it's a relatively simple formula. I know that with an atmosphere it becomes much more complicated and there are questions such as cloud cover, but I'm curious if anything exists that makes an estimate for rocky planet surface temperature adjusted for various atmospheres.

And lapse rate gives a good calculation for how temperature changes with altitude but provides little to nothing on the initial surface temperature.

This question below touches on some of the ideas, but the answer on temperature estimates is a work in progress.

[Public data for calculating equilibrium temperature of exoplanets](https://astronomy.stackexchange.com/questions/54475/public-data-for-calculating-equilibrium-temperature-of-exoplanets)

**Tags:** planet, planetary-atmosphere, temperature

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## 1 answers

### Answer by username_1 on 2023-11-08. Score: 11 | Accepted answer

[Guillot et al. 2010](https://ui.adsabs.harvard.edu/abs/2010A%26A...520A..27G/abstract) give a comparatively simple solution (their Eqn. 27) for a double-gray purely radiative atmosphere. It reproduces the equilibrium temperature in the optically thin limit.

[Robinson & Catling 2012](https://ui.adsabs.harvard.edu/abs/2012ApJ...757..104R/abstract) extended the previous calculations to include convection. The expression for the temperature profile as function of optical depth can however not be written down explicitely anymore in that case, it has to be iterated on.

Those are merely solutions that have become popular in recent years in the exoplanet community. Less well known is that [Schwarzschild 1906](https://ui.adsabs.harvard.edu/abs/1906WisGo.195...41S/abstract), [Milne 1921](https://ui.adsabs.harvard.edu/abs/1921MNRAS..81..361M/abstract) and [Sagan 1969](https://ui.adsabs.harvard.edu/abs/1969Icar...10..290S/abstract) already contributed solutions of the form of
$$ T^4(\tau) = T^4\_{eq}\left(\frac{1}{2}+\frac{3}{4}\tau \right) $$
where $\tau$ is the optical depth of the atmospheric gas, taken as vertical coordinate and the unity factors in the brackets (particularly the 3/4 in front of $\tau$) will depend on many details of the derivation, especially boundary conditions and various assumptions about geometry and radiation transport. The variable $\tau$ increases as we go deeper into the atmosphere.

Note that this solution predicts $T=T\_{eq}$ only at an optical depth of $2/3$, i.e. for moderately massive atmospheres. At lower optical depth, the atmospheric temperature will be $T, even if there is a surface at $T=T\_{eq}$.

To get a surface temperature, you have to know the infrared optical depth of the surface though $\tau\_0$, which you have to obtain from the complicated and pressure-dependent opacity functions of the entire gas mixture in question. Then you can evaluate $T^4(\tau\_0)$ for an estimate of the surface (gas) temperature.

For Earth, with $\tau\_{IR}\approx2$ and $T\_{eq, E}=255K$, this would yield $T\_{surf,E}=303K$ (an ok estimate, considering that the purely radiative model ignores convection), and for Venus with $\tau\_{IR}\approx 100$ (taken from the Sagan paper), and $T\_{eq,V}=226K$ one would get $T\_{surf,V}=666K$ (clearly farther off, but still shows the point that the surface is much hotter than the equilibrium temperature).

> **Comments:**
> **username_2**: You wouldn't go to far wrong with an estimate of the height to reach optical depth 1, the adiabatic lapse rate and the equilibrium temperature. This gives a good way to understand the greenhouse effect.  
> **username_1**: @ProfRob: Depends on the surface gravity, which will determine how far away $r_{\tau=1}$ and the radiative-convective boundary are from each other.  
> **username_0**: @AtmosphericPrisonEscape lapse rate incorporates gravity.  It's part of the equation.  Maybe that's obvious, but I thought I'd point it out.    That raises some interesting math for a very fast rotating body, where the gravity varies considerably equator to pole . . . but I digress.  
> **username_1**: @userLTK Not sure what you are trying to say. The above solution is a purely radiative solution with the optical depth $\tau$ as vertical coordinate. It contains no information about realspace, hence no information about temperature gradients. Note that your usual radiative temperature gradient follows from this via $dT/dx = dT/d\tau d\tau/dx = dT/d\tau \rho \kappa$, where $\rho$ is the density profile, which is how the gravity enters.  
> **username_0**: I just meant that on other planets, different gravities, the lapse rate changes.   Which, I think you probably knew.   Maybe my initial comment was unnecessary.  https://image2.slideserve.com/3857728/dry-adiabatic-lapse-rate1-l.jpg  

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