7
$\begingroup$

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

Improve this question
$\endgroup$

1 Answer 1

Reset to default
11
$\begingroup$

Guillot et al. 2010 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 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, Milne 1921 and Sagan 1969 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 $T<T_{eq}$, 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).

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – ProfRob
    Commented Nov 9, 2023 at 11:43
  • $\begingroup$ @ProfRob: Depends on the surface gravity, which will determine how far away $r_{\tau=1}$ and the radiative-convective boundary are from each other. $\endgroup$
    – AtmosphericPrisonEscape
    Commented Nov 9, 2023 at 11:52
  • $\begingroup$ @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. $\endgroup$
    – userLTK
    Commented Nov 9, 2023 at 13:26
  • $\begingroup$ @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. $\endgroup$
    – AtmosphericPrisonEscape
    Commented Nov 9, 2023 at 14:38
  • $\begingroup$ 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. image2.slideserve.com/3857728/dry-adiabatic-lapse-rate1-l.jpg $\endgroup$
    – userLTK
    Commented Nov 10, 2023 at 15:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .