<p>I think your ~2 figure for CO2 greenhouse amplification is in error. 
* In the last 200 years it's gone from the high 280s to low 400s.  This is an increase of about 50%  Good thing it's not linear as the previous greenhouse effect is overall about 18K.  If there was no effect, the earth would average about 0 C.
* As CO2 increases it takes more bounces for an infrared photon to escape.  But you get some degree of diminishing returns.  Doubling the CO2 doesn't double the greenhouse effect.</p>

<p>But to your question.</p>

<p>Right now we have about 1 degree of additional greenhouse effect.  Compared to the 18 degrees of existing greenhouse effect this is about 5%.  </p>

<p>I would expect that this is the factor to use.  So the eccentricity amplification should be about 5%  So 5.25K instead of 5.  or observationally 1.05 * 3.8K </p>

CC BY-SA 4.0
2019-05-27T18:36:31.450Z
The factor of 2 was not relating to CO2 concentration, but to the doubling of the temperature effect by means of the vapor feedback... So like 0.5 degrees without feedback (theroretically) and 1 degree with feedback.
oliver
CC BY-SA 4.0
2019-05-27T18:41:52.933Z
I am not sure if you are right. Since percentage increases/feedback processes are described by logarithmic equations, it does not matter how much the current greenhouse effect is, in my opinion. I will try an put this into equations in my question...
oliver
CC BY-SA 4.0
2019-05-27T13:39:57.320Z
2019-05-27T13:39:57.320Z
Sherwood Botsford
<p>I think I understand the mechanism of water vapor feedback in climate change pretty well: a (theoretical) temperature increase due to some isolated factor (e.g. increase in <span class="math-container">$CO_2$</span> concentration) causes a shift in water vapor saturation density and an additional evaporation rate. The consequential shift in actual water vapor concentration causes additional absorption together with isotropic re-emission, slowing down the radiation transport. So the theoretical temperature rise <span class="math-container">$\Delta T_{nofb}$</span> without consideration of feedback is "amplified" by some factor <span class="math-container">$\beta$</span>, giving the real temperature rise, with feedback</p>

<p><span class="math-container">$\Delta T_{wfb} = \beta \Delta T_{nofb}$</span></p>

<p>Of course I know that this argument has to be replaced by solving a differential equation in a more rigorous treatment. There is no "before" and "after", there is only the combined effect.</p>

<p><strong>But what if we consider the "theoretical" temperature change due to the change in solar irradiation from orbital eccentricity? Shouldn't this be amplified by the same water vapor feedback mechanism as well?</strong></p>

<p>Solar irradiation varies over the year between <span class="math-container">$1310$</span> <span class="math-container">$W/m^2$</span> and <span class="math-container">$1420$</span> <span class="math-container">$W/m^2$</span>. According to Stefan-Boltzmann's Law in differential form we would have a heating of the ground by direct absorption of sunlight </p>

<p><span class="math-container">$\frac{dP}{P}=4 \frac{dT}{T}$</span></p>

<p>or</p>

<p><span class="math-container">$\frac{\Delta P}{P}\approx 4 \frac{\Delta T}{T}$</span></p>

<p>So I would expect the temperature change due to yearly changes in solar irradiation (but without the effect of water vapor feedback) to be about</p>

<p><span class="math-container">$\Delta T \approx \frac{T}{4} \frac{\Delta P}{P} = \frac{290 K}{4} \frac{90 W/m^2 }{1366 W/m^2}\approx 4.8 K$</span></p>

<p>The favorite answer to <a href="https://earthscience.stackexchange.com/questions/11913/does-the-elliptical-orbit-of-the-earth-have-effect-on-earths-climate">this question</a> shows a plot that seems to indicate that the order of magnitude is indeed like that, the global line shows a variation of about <span class="math-container">$3.8$</span> <span class="math-container">$K$</span>. I don't know if this was derived from measurements, but I assume that it is somehow substantiated.</p>

<p>But if we now adopt the assumption that these ~5 Kelvin get amplified by the water vapor feedback as well, and take into account that this amplification is said to be in the order of magnitude ~2 in the context of <span class="math-container">$CO_2$</span> entry by mankind, I would expect the amplified yearly temperature change due to the orbital eccentricity of earth to be <span class="math-container">$~10$</span> <span class="math-container">$K$</span>.</p>

<p>Is the linked answer wrong, or is there some flaw in the argument?</p>

<climate-change><water-vapour>
Yearly changes of solar irradiation and atmospheric feedback
oliver
# Yearly changes of solar irradiation and atmospheric feedback

(Asked by: username_0 on 2019-05-27)

I think I understand the mechanism of water vapor feedback in climate change pretty well: a (theoretical) temperature increase due to some isolated factor (e.g. increase in $CO\_2$ concentration) causes a shift in water vapor saturation density and an additional evaporation rate. The consequential shift in actual water vapor concentration causes additional absorption together with isotropic re-emission, slowing down the radiation transport. So the theoretical temperature rise $\Delta T\_{nofb}$ without consideration of feedback is "amplified" by some factor $\beta$, giving the real temperature rise, with feedback

$\Delta T\_{wfb} = \beta \Delta T\_{nofb}$

Of course I know that this argument has to be replaced by solving a differential equation in a more rigorous treatment. There is no "before" and "after", there is only the combined effect.

**But what if we consider the "theoretical" temperature change due to the change in solar irradiation from orbital eccentricity? Shouldn't this be amplified by the same water vapor feedback mechanism as well?**

Solar irradiation varies over the year between $1310$ $W/m^2$ and $1420$ $W/m^2$. According to Stefan-Boltzmann's Law in differential form we would have a heating of the ground by direct absorption of sunlight

$\frac{dP}{P}=4 \frac{dT}{T}$

or

$\frac{\Delta P}{P}\approx 4 \frac{\Delta T}{T}$

So I would expect the temperature change due to yearly changes in solar irradiation (but without the effect of water vapor feedback) to be about

$\Delta T \approx \frac{T}{4} \frac{\Delta P}{P} = \frac{290 K}{4} \frac{90 W/m^2 }{1366 W/m^2}\approx 4.8 K$

The favorite answer to [this question](https://earthscience.stackexchange.com/questions/11913/does-the-elliptical-orbit-of-the-earth-have-effect-on-earths-climate) shows a plot that seems to indicate that the order of magnitude is indeed like that, the global line shows a variation of about $3.8$ $K$. I don't know if this was derived from measurements, but I assume that it is somehow substantiated.

But if we now adopt the assumption that these ~5 Kelvin get amplified by the water vapor feedback as well, and take into account that this amplification is said to be in the order of magnitude ~2 in the context of $CO\_2$ entry by mankind, I would expect the amplified yearly temperature change due to the orbital eccentricity of earth to be $~10$ $K$.

Is the linked answer wrong, or is there some flaw in the argument?

**Tags:** climate-change, water-vapour


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## 1 answers

### Answer by username_1 on 2019-05-27. Score: 1

I think your ~2 figure for CO2 greenhouse amplification is in error.
\* In the last 200 years it's gone from the high 280s to low 400s. This is an increase of about 50% Good thing it's not linear as the previous greenhouse effect is overall about 18K. If there was no effect, the earth would average about 0 C.
\* As CO2 increases it takes more bounces for an infrared photon to escape. But you get some degree of diminishing returns. Doubling the CO2 doesn't double the greenhouse effect.

But to your question.

Right now we have about 1 degree of additional greenhouse effect. Compared to the 18 degrees of existing greenhouse effect this is about 5%.

I would expect that this is the factor to use. So the eccentricity amplification should be about 5% So 5.25K instead of 5. or observationally 1.05 \* 3.8K

> **Comments:**
> **username_0**: The factor of 2 was not relating to CO2 concentration, but to the doubling of the temperature effect by means of the vapor feedback... So like 0.5 degrees without feedback (theroretically) and 1 degree with feedback.  
> **username_0**: I am not sure if you are right. Since percentage increases/feedback processes are described by logarithmic equations, it does not matter how much the current greenhouse effect is, in my opinion. I will try an put this into equations in my question...  

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