Mainstream climate science claims that without an atmosphere, Earth’s surface temperature would be, on average, below freezing. It would be about 33 degrees C colder than it is with an atmosphere. Actually, they claim that without greenhouse gases (GHGs) that would be the case. Someone at NASA even made this ridiculous statement:
Remove carbon dioxide, and the terrestrial greenhouse effect would collapse. Without carbon dioxide, Earth’s surface would be some 33°C (59°F) cooler.— NASA: What is the greenhouse effect
Ha ha! I’m sure they meant to say GHGs, not CO2, but let’s laugh at NASA for employing someone secretly biased like this.
There’s a link on that site to UCAR (Center for Science Education): The Greenhouse Effect. Here we read:
Without the greenhouse effect, Earth’s temperature would be below freezing.— UCAR
Clicking on that link:
Without any heat-trapping greenhouse gases in our atmosphere, Earth would be a frozen ball of ice.
You can learn lots more details about the math at our Calculating Planetary Energy Balance & Temperature web page.— UCAR
Great, we finally got to the meat. I was looking for this math page, but apparently it’s gone now from UCAR and links to an archive.org page. Maybe they are embarrassed? Anyway, here’s how this “iceball” theory works:
The math is mostly fine (Earth is not a perfect Sphere), but the real problem is the choice of 0.31 albedo. Most of the albedo comes from the atmosphere itself! And also this albedo value is only useful in observing Earth from space. ~30% of incoming shortwave solar radiation is indeed reflected from the Earth onto an observing satellite. But this is NOT a metric for figuring out how much solar radiation reaches the surface, which is what is in question. And so this entire calculation is utterly superficial and meaningless.
Some have suggested to use the Moon’s albedo to simulate Earth’s surface temperature without an atmosphere. Moon’s albedo is about 0.12, and the standard calculation method would yield:
$ qalc -t '1361*(1-0.12)/4=5.670367e-8*x^4' x ≈ 269.5674246 # (x = Temperature in Kelvin)
But the moon is the moon, and it’s not like the Earth, at least I don’t think.
Let us look at an Energy Budget diagram to see what the proper parameter is:
So 23 W/m2 is reflected at the surface, and that’s from (23+164)= 187 W/m2 that makes it past the atmosphere. 23/187 = 0.1229…
Ah, so that is indeed very close to the moon. Not a bad assumption.
Then again, 23 out of a total of 340 was reflected at the surface. Maybe the 187 is irrelevant? Can I be sure that the surface would not have rejected the same as the atmosphere? No, really, can I? I don’t know, but I don’t think so. If I’m right, then the surface albedo would be 23/340 = 0.0676.
This gives me more confidence:
The lunar average Bond albedo (at normal solar incidence) A is 0.12 (Vasavada et al. 2012). This is in agreement with the mean value of 0.122 found by Saari & Shorthill (1972). Vasavada et al. (2012) derived a mean albedo of 0.07 for mare and 0.16 for highland surfaces from measurements taken by the Diviner Lunar Radio Experiment. In a NASA summary of the Moon’s bulk parameters19, the Bond albedo is given by 0.11 and the geometric albedo by 0.12— The Moon at Thermal Infrared …
Mares are generally smooth and flat and take up ~16% of moon’s surface. On Earth, our equivalent of lunar mares would be the oceans, and any body of water in general. Water bodies have a small albedo, like 0.06 … similar to lunar mares. But our “mares” take up 71% of our surface. This fact alone makes me confident Earth’s surface albedo is much lower than the moon.
The main problem with the mainstream no-atmosphere formula shown above is that it lacks EMISSIVITY, which is a huge mistake. Here’s the correction:
Now things get interesting.
What is the emissivity of the moon? This paper suggests it is definitely somewhere between 0.92 and 0.97. But this is computed from a narrow set of channels. Such an analysis on Earth also leads to a high result: 0.97-0.98, whereas the actual emissivity is found in my previous article: What is Earth’s Surface Emissivity? : 0.93643.
I would even bet that the moon’s albedo is just one minus its emissivity.
For Earth, let’s go back to the energy budget … 23/340 is 0.067647.
1 minus 0.067647 is 0.932353
0.932353 and 0.93643 are too close to be a coincidence. I think they are saying the same thing:
Screw albedo and emissivity, and just assume Kirchoff’s Law of Radiation for this type of calculation? This implies:
A more accurate S/4 value is taken from my previous article: 339.22
$ qalc -t '339.22=5.670367e-8*x^4' x ≈ 278.1106181
If we take the surface outgoing longwave radiation from CERES: 399.56 (Average, year 2020), we get …
$ qalc -t '399.56=5.670367e-8*x^4' x ≈ 289.7294947
And the difference is about 11.62 degrees C
Now that is much less of a “greenhouse effect” than the fantastical 33 or 34 degrees claimed by mainstream climate scientists!
Without the atmosphere, the average temperature would be nearly 5 degrees C
Have a good day. -Zoe
I don’t know how I missed this chart:
Averaging the emissivity accross 7 channels yields:
(0.965+0.938+0.912+0.766+0.777+0.827+0.787)/7 = ~ 0.853
That’s not exactly the right way to do it, as different channels yield different intensities via Planck’s Law, but I still think emissivity is very close if not exactly (1-albedo).
Just in case I’m wrong, I’ll show the calculation for an albedo of 0.1229 (derived from energy budget):
$ qalc -t '339.22*(1-0.1229)=0.93643*5.670367e-8*x^4' x ≈ 273.5968031
The GH effect would then be 16.13 K.