### Everything started with error

Everything started with a logical error.

And here is why.

The planet's old incomplete effective temperature equation:

Te = [ (1-a) S / 4 σ ]¹∕ ⁴

is defined as a planet's equilibrium temperature in the absence of atmosphere.

When calculated, the Earth's Te = 255 K, instead of the satellite measured actual Tmean = 288 K the cause was obvious.

Earth's surface was considered warmer by +Δ33oC because of the planet's Earth atmosphere.

It was error.

### The entire Greenhouse Gases Global Warming theory is an old Climate science

And here is why:

Comparison of results the planet's Te calculated by the Incomplete Equation: Te = [ (1-a) S / 4 σ ]¹∕ ⁴

the planet's Tmean calculated by the Planet's Surface Mean Temperature Equation:

Tmean.planet = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (1)

and the planet's Tsat.mean measured by satellites:

Planet…Te.incompl...Tmean...Tsat.mean

Mercury……439,6 K….325,83 K…….340 K

Earth………..255 K…...287,74 K…….288 K

Moon………..271 Κ…....221,74 Κ…..220 Κ

Mars………209,91 K…..213,42 K…..210 K

Te.incomplete = [ (1-a) S /4σ ]¹∕ ⁴ gives very confusing results:

Planet…Te. incomplete

Mercury……….439,6 K

Earth…………..255 K

Moon…………..271 Κ

Mars…………209,91 K

When inserting the values, this calculation yields a surface temperature of about Te = 255 K. Even though this value is not quite bad for a very simple model, it still deviates significantly from the actual value of Tmean = 288 K.

So the old equation yields Te = 255 K instead of the actual Tmean = 288 K...

But we have not finished yet:

This equation Te.incomplete = [ (1-a) S /4σ ]¹∕ ⁴ also deviates significantly for the Mercury Te = 439,6 K, instead of the actual Tmean = 340 K

and it deviates significantly for the Moon Te = 271 K, instead of the actual Tmean = 220 K.

It is time to abandon the old effective temperature

Te.incomplete = [ (1-a) S /4σ ]¹∕ ⁴ equation, because it gives very confusing results.

The 288 K – 255 K = 33 oC means the by atmosphere +Δ33 oC Earth's surface warming phenomenon, which 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.

The entire Greenhouse Gases Global Warming theory is based on the old effective temperature Te.incomplete = [ (1-a) S /4σ ]¹∕ ⁴ equation the Earth's effective temperature calculation

Te = 255 K which does not exist

When applying the Planet's Surface Mean Temperature Equation

Tmean = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (1)

we have identical to the satellite planet's surface mean temperatures measured results:

Planet........Tmean..Tsat.mean

Mercury…..325,83 K……..340 K

Earth……….287,74 K……..288 K

Moon……….221,74 Κ……..220 Κ

Mars………..213,42 K……..210 K

The Earth's Tmean is calculated Tmean.earth = 287,74 K and it is identical to the measured by satellites the actual value of Tsat.mean.earth = 288 K.

Also almost identical are the calculated by the Equation

Mercury…….325,83 K = 340 K

Moon……….221,74 Κ = 220 Κ

and

Mars………..213,42 K = 210 K

Now everything falls in place and the entire Greenhouse Gases Global Warming theory becomes an old Climate science.

Because Te = 255 K

does not exist.

### 1. Earth’s Without-Atmosphere Surface Mean Temperature calculation

Tmean.earth

So = 1.362 W/m² (So is the Solar constant)

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 INTERACTING-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 Effective Temperature Complete Formula Te.earth is:

Tmean.earth = [ Φ(1-a) So(β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴

Τmean.earth = [ 0,47(1-0,306)1.362 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)1.362 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 complete formula produces remarkable results. The calculated planets’ temperatures are almost identical with the measured by satellites.

Planet…Te.incompl.....Tmean…Tsat.mean

Mercury…..437 K…….346,11 K……..340 K

Earth………255 K……..287,74 K……..288 K

Moon…….271 Κ….…..221,74 Κ……..220 Κ

Mars……209,91 K…...213,42 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.

### Earth / Europa (Jupiter's moon) satellite measured mean temperatures 288 K and 102 K comparison

All the data below are satellites measurements.  All the data below are observations.

Planet….Earth….Europa

Tsat.mean 288 K….102 K

R…...........1 AU…5,2044 AU

1/R²………1…….0,0369

N………....1……1/3,5512 rot./day

a…………..0,30……0,63

(1-a)………0,70……0,37

coeff...0,91469...0,3158

Comparison coefficient calculation

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴

Earth:

Tsat.mean = 288 K

[ (1-a)*(1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ =

= ( 0,70 * 1 * 1 )¹∕ ⁴ = 0,91469

Europa:

Tsat.mean = 102 K

[ (1-a)*(1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ =

= [ 0,37*0,0369*(1/3,5512)¹∕ ⁴ ] ¹∕ ⁴ = 0,3158

Let's compare

Earth coeff. /Europa coeff. =

= 0.91469 /0,3158 = 2,8964

And

Tmean.earth /Tmean.europa =

= 288 K /102 K = 2,8235

Conclusion: Everything is all right. Everything is based on observations.

Notice:

We could successfully compare Earth /Europa ( 288 K /102 K ) satellite measured mean temperatures because both Earth and Europa have two identical major features.

Φearth = 0,47 because Earth has a smooth surface and Φeuropa = 0,47 because Europa also has a smooth surface.

cp.earth = 1 cal/gr*°C, 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.

Europa is an ice-crust planet without atmosphere, Europa’s surface consists of water ice crust, cp.europa = 1cal/gr*°C.

And It is a confirmation that the planet's axial spin (rotations per day) "N" should be considered in the Tmean.planet equation in the fourth root:

Tmean.planet = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴.

### Φ factor explanation

Φ factor explanation

"Φ" is an important factor in the Planet's Surface Mean Temperature Equation:

Tmean.planet = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (K)

It is very important the understanding what is really going on with by planets the solar irradiation absorption.

What we have now is the following:

Jsw.incoming - Jsw.reflected = Jsw.absorbed

Φ = (1-0,53) = 0,47

Φ = 0,47

Φ is the planet's spherical surface solar irradiation accepting factor.

Jsw.reflected = (0,53 + Φ*a) * Jsw.incoming

And

Jsw.absorbed = Φ* (1-a) * Jsw.incoming

Where

(0,53 + Φ*a) + Φ* (1-a) = 0,53 + Φ*a + Φ - Φ*a =

= 0,53 + Φ = 0,53 + 0,47 = 1

The solar irradiation reflection, when integrated over a planet sunlit hemisphere is:

Jsw.reflected = (0,53 + Φ*a) * Jsw.incoming

Jsw.reflected = (0,53 + Φ*a) *S *π r²

For a planet with albedo a = 0 we shall have

Jsw.reflected = (0,53 + Φ*0) *S *π r² =

= Jsw.reflected = 0,53 *S *π r²

The fraction left for hemisphere to absorb is:

Φ = 1 - 0,53 = 0,47

or

Jabs = Φ (1 - a ) S π r²

The factor Φ = 0,47 "translates" the absorption of a disk into the absorption of a hemisphere with the same radius.

When covering a disk with a hemisphere of the same radius the hemisphere's surface area is 2π r².

The incident Solar energy on the hemisphere's area is the same as on disk:

Jdirect = π r² S

The absorbed Solar energy by the hemisphere's area of 2π r² is:

Jabs = 0,47*( 1 - a) π r² S

It happens because a hemisphere of the same radius "r" absorbs only the 0,47 part of the directly incident on the disk of the same radius Solar irradiation.

In spite of hemisphere having twice the area of the disk, it absorbs only the 0,47 part of the directly incident on the disk Solar irradiation.

Jabs = Φ (1 - a ) S π r² ,

where Φ = 0,47 for smooth without atmosphere planets. and Φ = 1 for gaseous planets, as Jupiter, Saturn, Neptune, Uranus, Venus, Titan.

Gaseous planets do not have a surface to reflect radiation. The solar irradiation is captured in the thousands of kilometers gaseous abyss. The gaseous planets have only the albedo "a".

And Φ = 1 for heavy cratered planets, as Calisto and Rhea ( not smooth surface planets, without atmosphere ). The heavy cratered planets have the ability to capture the incoming light in their multiple craters and canyons. The heavy cratered planets have only the albedo "a".

Another thing that I should explain is that planet's albedo actually doesn't represent a primer reflection. It is a kind of a secondary reflection ( a homogenous dispersion of light also out into space ). That light is visible and measurable and is called albedo.

The primer reflection from a spherical hemisphere cannot be seen from some distance from the planet. It can only be seen by an observer being on the planet's surface.

It is the blinding surface reflection right in the observer's eye. That is why the albedo "a" and the factor "Φ" we consider as different values.

Both of them, the albedo "a" and the factor "Φ" cooperate in the Planet Rotating Surface Solar Irradiation Absorbing-Emitting Universal Law:

Φ*S*(1-a) = 4σTmean⁴ /(β*N*cp)¹∕ ⁴

And they are also cooperate in the Planet's Surface Mean Temperature Equation:

Tmean.planet = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ ( K )

Planet Energy Budget:

Solar energy absorbed by a Hemisphere with radius "r" after reflection and dispersion:

Jabs = Φ*πr²S (1-a) ( W )

Total energy emitted to space from a whole planet:

Jemit = A*σΤmean⁴ /(β*N*cp)¹∕ ⁴ ( W )

Φ - is a dimensionless Solar Irradiation accepting factor

(1 - Φ) - is the reflected fraction of the incident on the planet solar flux

S - is the Solar Flux at the top of atmosphere ( W/m² )

Α - is the total planet surface area ( m² )

A = 4πr² (m²), where "r" – is the planet's radius

Tmean - is a Planet's Surface Mean Temperature ( K )

(β*N*cp)¹∕ ⁴ - dimensionless, is a Rotating Planet Surface Solar Irradiation Warming Ability

Jabs = Φ (1- a ) S W/m² sunlit hemisphere

Jabs.earth = 0,47 ( 1 - 0,30 ) So π r² =

= 0,47*0,70 * 1.362* π r² ( W ) =

Jabs.earth = 0,329 So π r² =

= 0,329* 1.362 π r² =

= 448,10 π r² ( W )

What is going on here is that instead of

Jabs.earth = 0,7* 1.362 π r² ( W )

we should consider

Jabs.earth = 0,329* 1.362 π r² ( W ).

Averaged on the entire Earth's surface we obtain:

Jsw.absorbed.average = [ 0,47*(1-a)*1.362 W/m2 ] /4 =

= [ 0,47*0,7*1.362W/m2 ] /4 = 448,098 W/m2 /4 = 112,029 W/m2

Jsw.absorbed.average = 112 W/m2

### The Planet Mercury’s Φ = 0,47 Paradigm Confirmation

We have now chosen the planet Mercury for its very low albedo a=0,068 and for its very slow rotation spin N=1/175 rotations/day.

The planet Mercury is most suitable for the incomplete effective temperature formula definition - a not rotating planet, or very slow rotating. Also it is a planet where albedo (a=0,088) plays small role in planet's energy budget.

These (Tmean, R, N, and albedo) planets' parameters are all satellites measured. These planets' parameters are all observations.

Planet….Mercury….Moon….Mars

Tsat.mean.340 K….220 K…210 K

R…......0,387 AU..1 AU..1,525 AU

1/R²…..6.6769….....1….…0,430

N…1 /175..1 /29,531..0,9747

a......0,068......0,136......0,250

1-a…0,932……0,864…….0,75

Let’s calculate, for comparison reason, the Planet Mercury’s effective temperature with the old incomplete equation:

Te.incomplete.mercury = [ (1-a) So (1/R²) /4σ ]¹∕ ⁴

We have

(1-a) = 0,932

1/R² = 6,6769

So = 1.362 W/m² - it is the Solar constant ( the solar flux on the top of Earth’s atmosphere )

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

Te.incomplete.mercury = [ 0,932* 1.362 W/m² * 6,6769 /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =

Te.incomplete.mercury = ( 37.369.999.608,40 )¹∕ ⁴ = 439,67 K

Te.incomplete.mercury = 439,67 K = 440 K

And we compare it with the

Tsat.mean.mercury = 340 K - the satellite measured Mercury’s mean temperature

Amazing, isn’t it? Why there is such a big difference between the measured Mercury’s mean temperature, Tmean = 340 K, which is the correct, ( I have not any doubt about the preciseness of satellites planets' temperatures measurements ) and the Mercury's Te by the effective temperature incomplete formula calculation Te = 437 K?

Let’s put these two temperatures together:

Te.incomplete.mercury = 440 K

Tsat.mean.mercury = 340 K

Very big difference, 100°C higher!

But why the incomplete effective temperature formula gives such a wrongly higher result?

The answer is simple – it happens because the incomplete formula assumes planet absorbing solar energy as a disk and not as a sphere.

We know now that even a planet with a zero albedo reflects 0,53*S of the incident solar irradiation.

Imagine a completely black planet; imagine a completely invisible planet, a planet with a zero albedo.

This planet will still reflect 53 % of the incident on its surface solar irradiation.

The satellites measurements have confirmed it.

Tsat.mean.mercury = 340 K

Te.incomplete.mercury = 440 K

Very big difference, 100°C higher!

The Planet Mercury’s Φ = 0,47 Paradigm has confirmed it:

Φ = 1 - 0,53 = 0,47

Φ = 0,47

Φ - is the dimensionless planet surface solar irradiation accepting factor

### 4. Mercury’s Surface Mean Temperature calculation

Tmean.mercury

Surface temp..Tmin..Tmean..Tmax

Kelvin...........100.K...340.K...700.K

N = 1/175 rotations/per day,

Planet Mercury completes one rotation around its axis in 175 days.

Mercury average distance from the sun is R=0,387AU.

The solar irradiation on Mercury is

(1/R)² = (1AU/0,387AU)²= 2,584²= 6,6769 times stronger than that on Earth.

Mercury’s albedo is:

amercury = 0,068

Mercury is a rocky planet, Mercury’s surface irradiation accepting factor: Φmercury = 0,47

Cp.mercury = 0,20 cal/gr oC, Mercury’s surface is considered as a basalt rock

β = 150 days*gr*oC/rotation*cal – it is a Rotating Planet Surface Solar Irradiation INTERACTING-Emitting Universal Law constant

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

So = 1.362 W/m² the Solar constant

Mercury’s Surface Mean Temperature Equation is:

Tmean.mercury =[Φ (1-a) So (1/R²) (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴

Planet's Mercury surface mean temperature Tmean.mercury is:

Tmean.mercury = { 0,47(1-0,068) 1.362 W/m²*6,6769*[150* (1/175)*0,20]¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ }¹∕ ⁴ =

= ( 11.271.047.334,17 )¹∕ ⁴ = 325,83 K

Tmean.mercury = 325,83 K

The calculated Mercury’s Surface Mean Temperature Tmean.mercury = 325,83 K is only 4,35 % lower than the measured by satellites

Tsat.mean.mercury = 340 K !

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