• # The basic problem

There is a simple formula our understanding of the greenhouse effect (GHE) is based on.

((1-0.3) * 342 / 5.67e-8) ^0.25 = 255K

With:
0.3 = albedo
5-67e-8 = SB (Stefan-Boltzman) constant
0.25 = 4th root, representing the SB law according to which radiation emitted as a function of temperature by the power of 4. And if you start out with radiation given you will have to invert it to get to the temperature.

If we compare this theoretical temperature to the average observed surface temperature of Earth, about 288K, we find a difference of 33K. So that is the GHE. Or that is what it should be. So far so simplified.

The above formula is not quite complete. The term 1-0.3, or 1 minus albedo has a name on it's own, which is absorptivity. Indeed the absorptivity of planet Earth is about 0.7. Yet absorptivity has a little brother called emissivity, which is equally important to calculate the temperature of any surface with given insolation. What we actually need to do is to divide absorptivity by emissivity. So how does the above formula yet yield the right result? Quite simple, it is being assumed emissivity was 1, and dividing any term by 1 will not change the result.

With that knowledge in mind and a given surface temperature of 288K, we can determine the actual emissivity of Earth relative to its surface temperature. We can solve the following equation for x = 0.614.

(((1-0.3)/x) * 342 / 5.67e-8) ^0.25 = 288K

All other things held constant, the surface temperature of Earth is a function of Earths emissivity. We know that with the given temperature emissivity must be about 0.614, or 61.4%. And with an emissivity of 1 the temperature would only be 255K.

That brings up the question what causes emissivity to be so much lower than 1. "Climate science" has a very strict position on this.

Without greenhouse gases, the average temperature of Earth's surface would be about −18 °C 1

The claim is, that without GHGs Earth would be a perfect emitter. All the substantial deviation in emissivity from 1 was ONLY due to GHGs. Of course it as position totally indefensible. There are a number of other factors reducing emissivity apart from GHGs. The surface is NOT a perfect emitter, clouds reduce emissivity and even aerosols. Clouds btw. are not a gas, but either droplets of water or ice crystals, and they are meant to net cool the planet. So they fail any thinkable definition of a GHG.

GHGs have serious competition and the true denialism is about denying this inconvenient fact. The GHE has NOT a magnitude of 33K, and more importantly, GHGs do not provide some 33K of warmth to the surface. The big taboo, the unseen gorilla in the room so to say, is the question how much competition there is, and what role actually remains for GHGs. It is a question that not just poses a threat to the global warming agenda, but potentially might even doom it.

There is a myriad of tricks and excuses to fool around this problem, you would not believe it. It is like a needle to a balloon, and "climate science" does its best to keep the two separated from each other. The scope of this site will be no other than bringing them together..

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• ##### Christos Vournas
at 07.10.2021
1.<b> Earth’s Without-Atmosphere Mean Surface Temperature calculation</b>
Tmean.earth

So = 1.361 W/m² (So is the Solar constant)
S (W/m²) is the planet's solar flux. For Earth S = So
Earth’s albedo: aearth = 0,306

Earth is a smooth rocky planet, Earth’s surface solar irradiation accepting factor Φearth = 0,47
(Accepted by a Smooth Hemisphere with radius r sunlight is S*Φ*π*r²(1-a), where Φ = 0,47)

β = 150 days*gr*oC/rotation*cal – is a Rotating Planet Surface Solar Irradiation Absorbing-Emitting Universal Law constant
N = 1 rotation /per day, is Earth’s axial spin
cp.earth = 1 cal/gr*oC, it is because Earth has a vast ocean. Generally speaking almost the whole Earth’s surface is wet. We can call Earth a Planet Ocean.

σ = 5,67*10⁻⁸ W/m²K⁴, the Stefan-Boltzmann constant

Earth’s Without-Atmosphere Mean Surface Temperature Equation Tmean.earth is:
<b>Tmean.earth= [ Φ (1-a) So (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴</b>

Τmean.earth = [ 0,47(1-0,306)1.361 W/m²(150 days*gr*oC/rotation*cal *1rotations/day*1 cal/gr*oC)¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =
Τmean.earth = [ 0,47(1-0,306)1.361 W/m²(150*1*1)¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =
Τmean.earth = ( 6.854.905.906,50 )¹∕ ⁴ = 287,74 K

Tmean.earth = 287,74 Κ

And we compare it with the
Tsat.mean.earth = 288 K, measured by satellites.
These two temperatures, the calculated one, and the measured by satellites are almost identical.

Conclusions:
The planet mean surface temperature equation
<b>Tmean = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴</b>
produces remarkable results.
The calculated planets temperatures are almost identical with the measured by satellites.
Planet....Tmean...Tsat.mean
Mercury...325,83 K...340 K
Earth.....287,74 K...288 K
Moon......223,35 Κ...220 Κ
Mars......213,21 K...210 K

Te.correct vs Tsat.mean comparison table
Planet....Te......Te.correct....Tmean...Tsat.mean
Mercury....439,6 K...364 K......325,83 K...340 K
Earth......255 K.....210 K......287,74 K...288 K
Moon......270,4 Κ....224 K......223,35 Κ...220 Κ
Mars.....209,91 K....174 K......213,21 K...210 K

The 288 K – 255 K = 33 oC difference does not exist in the real world.
There are only traces of greenhouse gasses.
The Earth’s atmosphere is very thin. There is not any measurable Greenhouse Gasses Warming effect on the Earth’s surface.

There is<b> NO +33°C</b> greenhouse enhancement on the Earth's mean surface temperature.
Both the calculated by equation and the satellite measured Earth's mean surface temperatures are almost identical:

Tmean.earth = 287,74K = 288 K

https://www.cristos-vournas.com
• ##### Fencesitter
at 02.02.2022
Hi GHD, would be interested in your critique of Christos post. Have spent a bit of time reading what I can find on his theory. As usual if you use Google nearly everything is suppressed. Had to use duckduckgo to find any discussion but have not found much.
• ##### GHD
at 08.02.2022
Christos believes the drag factor from fuild dynamics was applicable in thermodynamics. The drag coefficient of a sphere is 0.47, and that of a cube is 1.05 btw. Since celestial bodies are usually of spheric shape, he applies this figure when calculating their should-be temperature. At least so when their surface is reasonably smooth.

Quote:
For Europa Φ=0,47
For Io Φ=1

When there are a lot of craters however, the otherwise perfectly round moon Io should more behave like a cube?! I wish it made more sense. No, the drag factor is not applicable in thermodynamics, nor is Io a cube, nor can you apply some random numbers as you like..
• ##### Christos Vournas
at 20.05.2022
What factor is NOT part of the effective temperature formula that so dramatically affects the actual temperature of the moon?

Why is the actual mean temperature of the moon so much lower than the effective temperature?

NASA lists the effective temperature of the moon at 270.6 kelvin. The mean temperature of the moon at the equator is 220 kelvin.

With no atmospheric effects, why is the surface temperature so much lower than the effective temperature predicts?

What factor is NOT part of the effective temperature formula that so dramatically affects the actual temperature of the moon?

I'll tell you what it is:

It is the Φ -the planet solar irradiation accepting factor. For smooth surface Moon Φ= 0,47.

Te.correct.moon = [ Φ (1-a) So /4σ ]¹∕ ⁴

Te.correct.moon = [ 0,47 (1-0,11) 1.362 W/m² /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =

Te.correct.moon = [ 0,47 (0,89) 1.362 W/m² /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =

Te.correct.moon = [ 2.510.168.871,25 ]¹∕ ⁴ =

Te.correct.moon = 223,83 Κ

This simple example clearly demonstrates the CORRECTNESS of the Φ -the planet solar irradiation accepting factor.

For smooth surface planets, like Moon, Φ= 0,47.

Conclusion:

From now on, for every smooth surface planet and moon, we should take in consideration instead of the planet blackbody effective temperature Te , the corrected VALUES of the planet blackbody effective temperature - the Te.corrected.

Table of results for Te and Te.corrected compared to Tsat and to Rotations/day for smooth surface planets and moons with Φ=0,47

Planet…….. Te.... Te.corrected…..Tsat…Rot/day

Mercury…..440 K…….364 K…...340 K…0,00568

Moon……….270 K…….224 K…...220 K.…0,0339

Earth………255 K…….210 K…...288 K..….1

Mars……….210 K…….174 K…..210 K..…0,9747

Europa…….95,2 K…...78,8 K….102 K...0,2816

Ganymede..107,1 K.....88,6 K…110 K….0,1398

Notice:

The number 0,47 for smooth surface in a parallel fluid flow is taken from the well measured and long ago known Drag Coefficient Data, where Cd =0,47 is for sphere. It is the portion of incident on sphere energy which should be resisted by sphere to remain in balance.
• ##### Christos Vournas
at 21.05.2022
The city of New York area reflectance is much less than the nearby rural area’s. Incident solar energy gets captured in the city’s deep streets’ canyons.
When we say “surface roughness” in relation to the incoming solar energy reflection, it is not on the microscopical level.

The power flux measured by CERES instruments from Earth orbits measures the planet’s diffuse reflection only.
The power flux measured by CERES instruments from Earth orbits completely ignores planet’s specular reflection, because planet’s specular reflection never reaches the CERES instruments’ measuring sensors.

https://www.cristos-vournas.com
• ##### Christos Vournas
at 24.05.2022
https://cdn.simplesite.com/i/2d/39/285978583434475821/i285978589400487356._szw1280h1280_.jpg

The Budget considers the planet's energy balance in <b>Total</b>, and not in average as the Greenhouse warming theory very mistakenly does. The Planet Radiative Energy Budget can be applied to all planets and to all moons.

We have <b>Φ</b> for different planets' and moon's surfaces varying
<b>0,47 ≤ Φ ≤ 1</b>

And we have surface average Albedo <b>"a"</b> for different planets' and moon's varying
<b>0 ≤ a ≤ 1</b>

Notice:
<b>Φ</b> is never less than <b>0,47</b> for planets and moons (<b>spherical shape</b>).
Also, the coefficient <b>Φ</b> is "bounded" in a product with (1 - a) term, forming the <b>Φ(1 - a)</b> product cooperating term.

So the <b>Φ and Albedo</b> are always bounded together.
The <b>Φ(1 - a)</b> term is a coupled physical term.
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