In that the green house effect, as caused by various atmospheric gases (e.g., H2O, CO2, CH4, etc), acts as an insulating layer which only retards the escape of heat from the planet, it would seem that the above statement is correct. Yet many commentaries on climate change seem to allow for the potential of much greater heating due to "run-away" feedback effects.

Such "run-away" heating scenarios do seem to violate the Stephan-Boltzman Law, and the general laws of thermodynamics, even after allowing for the "back-radiation", that effectively provides the "insulation" that slows the escape of the long-wave (i.e., infrared) radiation. While "greenhouse" gas insulation is the colder body (in daytime), it cannot do a net transfer heat to the plant's surface, which is the hotter body. At night time, however, the greenhouse gas may be the hotter body -- for a while -- and thus transfer some heat back to surface, before morning sun rise. This is how I see it. Is that wrong?

  • $\begingroup$ In theory, positive feedback loops can lead to "runaway" greenhouse effect. There's a big list of them here: en.wikipedia.org/wiki/Climate_change_feedback $\endgroup$
    – f.thorpe
    Commented Jun 8, 2021 at 5:34
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    $\begingroup$ Why would you think that the runaway greenhouse violates any physics, when it has been computed using the laws of physics? The wikipedia page en.wikipedia.org/wiki/Runaway_greenhouse_effect has the formula for the temperature in an optically thick medium $\tau>1$, which can rise above the stellar surface temperature, if $\tau$ is large enough. This doesn't violate anything. Furthermore, please observe how a dense atmosphere, like that of Venus, doesn't know day or nightside on ground-level anymore. $\endgroup$ Commented Jun 8, 2021 at 12:00
  • $\begingroup$ Re Is that wrong? Yes, it is. Every single bit. $\endgroup$ Commented Jun 8, 2021 at 14:14
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    $\begingroup$ A simple-minded black-body estimate for Venus yields a surface temperature of $328\,\mathrm{K}$. The actual surface temperature is $\sim 730\,\mathrm{K}$. $\endgroup$
    – user18801
    Commented Jun 9, 2021 at 8:49

1 Answer 1


We'll take an example of Venus as a poster child for runaway greenhouse effect.

First of all, let's talk of day/night changes. At the surface level there is little to no change as Venus has an albedo of 0.77, i.e. its atmosphere reflects back 77% of light. Compare this to Earth's albedo of 0.3. So in the overall energy balance of Venusian atmosphere the amount of heat coming from the Sun plays a rather small role and most of it will affect only the upper layers of the atmosphere.

And even what light is not immediately reflected back is mostly absorbed by the atmosphere and as a result the day/night temperature changes are practically nonexistent. The lower atmosphere has approximately the same temperature during day or night, at the equator or at the pole.

At the same time, like most of terrestrial planets Venus continues to produce heat primarily as a result of decay of radioactive elements in the mantle. This heat replaces the amount that is lost from the atmosphere into space.

What results is a stable system and really the only thing that can affect it at this stage is the gradual reduction of the nuclear decay heat over time.

EDIT: Having a very long discussion in the comments so I figured I'll update the post to better explain what I meant in the post.

A planet's atmosphere is a thermodynamic system that in the case of Venus appears to be in the state of equilibrium, i.e. heat received in the system equals heat lost to space primarily via IR radiation.

The main parts of planetary energy balance are as follows: incoming solar radiation and geothermal heat as sources of income and loss is primarily radiative.

Geothermal heat also consists of residual heat left over from planet's formation and additional heat generated by the decay of radionuclides. In the case of the Earth geothermal heat is estimated at some 47 TW about half of which is estimated to come from the residual heat and half - from radioactive decay. I couldn't find any reliable sources for Venus, but due to similarities in composition it's reasonable to say that it would exhibit similar energy flux but due to lover volume it's likely to be around 40 TW.

So now the overall simplistic view at the components that result in constant atmospheric temperature of Venus:

  • Albedo of 0.77 limits the amount of solar radiation from an initial flux of $\approx 2601 W/m^2$ to only about $157 W/m^2$ enter link description here
  • Additional minor energy income comes from planet's interior as discussed above.
  • Finally, heat escapes primarily via IR radiation.
  • High density of atmosphere distributes energy more evenly.
  • $\begingroup$ Not sure what you are even trying to say, but beyond that this answer is wrong (same as the question): 1. The high albedo has nothing to do with the high and uniform surface temperature on Venus. 2. The high and uniform surface temperature on Venus has nothing to do with nuclear decay. Rather it is a direct result of the density of those atmospheres. The high density imparts a large heat conductivity onto the gas, hence it is globally uniform, and the high optical depth is responsible for the high temperature. See amongst others ui.adsabs.harvard.edu/abs/2014NatGe...7...12R/abstract . $\endgroup$ Commented Jun 14, 2021 at 12:06
  • $\begingroup$ @AtmosphericPrisonEscape, I'm not sure what your point is. Are you saying that if Venus had a lower albedo, we wouldn't see any daily changes in temperature? Ok, so thermal conductivity now generates heat? $\endgroup$
    – NotAName
    Commented Jun 14, 2021 at 13:28
  • $\begingroup$ My point is, as I said, that your answer is wrong. You are aware, that 'daily changes in temperature' by which you presumably mean 'dayside-nightside temperature differences', only exist up to about the cloud level? And that below that, the Venusian atmosphere has a horizontally uniform temperature? No, conductivity transports heat. $\endgroup$ Commented Jun 14, 2021 at 14:32
  • $\begingroup$ @AtmosphericPrisonEscape, I'm aware of those facts, in fact I said almost exactly that in my answer. $\endgroup$
    – NotAName
    Commented Jun 14, 2021 at 15:05
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    $\begingroup$ @AtmosphericPrisonEscape, think of it as a simple thermodynamic system. So now if energy lost to space were exactly matched by energy received from the Sun, yet we keep dumping around 30 TJ of radiogenic heat into the system every second, the temperature would rise. But since radiative heat losses are proportional to T^4, the system will very soon settle at the balance point where the (radiogenic heat + solar heat) = heat loss to space. All I claimed, that is that in the current state this balance point has already been reached. $\endgroup$
    – NotAName
    Commented Jun 14, 2021 at 23:31

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