The Planet's Mean Surface Temperature Equation: Tmean = [Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ]¹∕ ⁴ (1) 

We have moved  further from the incomplete effective temperature equation

Te = [ (1-a) S / 4 σ ]¹∕ ⁴

(which is in common use right now, but actually it is an incomplete planet Te equation and that is why it gives us very confusing results)

a - is the planet's surface average albedo

S - is the solar flux, W/m²

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

We have discovered the Planet Without-Atmosphere Mean Surface Temperature Equation

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

The Planet Without-Atmosphere Mean Surface Temperature Equation is also based on the radiative equilibrium and on the Stefan-Boltzmann Law.

The Equation is being completed by adding to the incomplete Te equation the new parameters Φ, N, cp and the constant β.

Φ - is the dimensionless Solar Irradiation accepting factor

N - rotations /day, is the planet’s axial spin

cp – cal /gr*oC, is the planet's surface specific heat

β = 150 days*gr*oC/rotation*cal – is the Rotating Planet Surface Solar Irradiation Absorbing-Emitting Universal Law constant.

........................................

The Planet Without-Atmosphere Mean Surface Temperature Equation is also based on the radiative equilibrium and on the Stefan-Boltzmann Law.

But the New Equation doesn't consider planet behaving as a blackbody, and the New Equation doesn't state planet having a uniform surface temperature.

Comparison of results the planet's Te calculated by the Incomplete Equation:

Te = [ (1-a) S / 4 σ ]¹∕ ⁴

the planet's mean surface temperature Tmean calculated by the Planet's Without-Atmosphere Mean Surface Temperature Equation:

Tmean = [ 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

            equation     equation  measured

Mercury    439,6 K    325,83 K      340 K

Earth         255 K        287,74 K      288 K

Moon         270,4 Κ     223,35 Κ     220 Κ

Mars         209,91 K    213,21 K     210 K

To be honest with you, at the beginning, I got by surprise myself with these results.

You see, I was searching for a mathematical approach…

Moon rotates around its axis at a slow rate of 29,5 days.

The day on the Moon is 14,75 earth days long, and the night on the Moon is also 14,75 Earth days long.

Moon is in our immediate neighborhood. So Moon is at the same distance from the sun, as Earth, R=1 AU (astronomical unit).

The year average solar irradiation intensity on the top of atmosphere for Moon and Earth is the same

So = 1361 W/m².

We say "on the top of the atmosphere", it means the solar intensity which reaches a celestial body and then falls on it.

It is all right then, that during these 14,75 earth days long lunar day the Moon's surface gets warmed at much higher temperatures than the Earth.

I’ll try here in few simple sentences explain the very essence of how the planet rotational warming Phenomenon occurs.

Lets consider two identical planets F and S at the same distance from the sun.

Let’s assume the planet F spins on its axis Faster, and the planet S spins on its axis Slower.

Both planets F and S get the same intensity solar flux on their sunlit hemispheres. Consequently both planets receive the same exactly amount of solar radiative energy.

The slower rotating planet’s S sunlit hemisphere surface gets warmed at higher temperatures than the faster rotating planet’s F sunlit hemisphere.

The surfaces emit at σT⁴ intensity – it is the Stefan-Boltzmann emission law.

Thus the planet S emits more intensively from the sunlit side than the planet F.

There is more energy left for the planet F to accumulate then.

That is what makes the faster rotating planet F on the average a warmer planet.

That is how the Planet Surface Rotational Warming Phenomenon occurs.

................................................................................................................

Moon gets baked hard during its 14,75 earth days long lunar day.

And Moon also  emits from its very hot daytime surface hard.

What else the very hot surface does but to emit hard, according to the Stefan-Boltzmann emission Law.

The very hot surface emits in fourth power of its very high absolute temperature.

Jemit ~ T⁴

A warm object in space loses heat via emission. The hotter is the object, the faster it loses heat. 

So there is not much energy left to emit during the 14,75 earth days long lunar night.

And it becomes very cold on the Moon at night.

It is in our Earth's immediate neighborhood happens.

As you know, to maintain a Planet Universal Law Equation one has to study all the planets' behavior. In that way only one may come to general conclusions. That is why I call our Earth as a Planet Earth. After all Earth is a Planet and as a Planet it behaves in accordance to the Universal Laws - as all Planets in the Universe do.

The Planet's Mean Surface Temperature Equation has the wonderful ability the calculated results closely matching to the measured by satellites planets' mean temperatures.

This New Universal Equation can be applied to all the without atmosphere planets and moons in a solar system.

We now have chosen Mercury for its very low albedo a=0,088 and for its very slow rotational spin N=1/175,938 rotations/day.

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 plays little role in energy budget.

These (Tmean, R, N, and albedo) parameters of the planets are all satellite measured. These parameters of the planets 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 /175,938..1 /29,531..0,9747

a.........0,068......0,11......0,250

1-a...…0,932……0,89…….0,75

Let’s calculate, for comparison reason, the 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

And we compare it with the

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

Amazing, isn’t it? Why there is such a big difference between the measured Mercury’s mean surface temperature, Tmean = 340 K, which is the correct, ( I have not any doubt about the preciseness of satellite planets' temperatures measurements ) and the Mercury's Te by the effective temperature incomplete formula calculation Te = 439,67 K?

Let’s put these two temperatures together:

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

Tsat.mean.mercury = 340 K

Very big difference, a 100°C higher! But why the incomplete effective temperature equation gives such a wrongly higher result?

The answer is simple – it happens because the incomplete equation 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 still reflects 53 % of the incident on its surface solar irradiation.

The satellite measurements have confirmed it.

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

Φ = 1 - 0,53 = 0,47

Φ = 0,47

Φ - the dimensionless planet surface solar irradiation accepting factor

Planet reflects the (1-Φ + Φ*a) part of the incident on the planet's surface solar irradiation. Here "a" is the planet's average albedo. 

So we always have:

Jreflected = (1-Φ + Φ*a)S

Jabsorbed = Φ(1-a)S

R = 1 AU, is the Earth's distance from the sun in astronomical units

Earth’s albedo: aearth = 0,306

Earth is a smooth rocky planet, Earth’s surface solar irradiation accepting factor Φearth = 0,47

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

N = 1 rotation /per day, is Earth’s sidereal rotation 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

So = 1.361 W/m² (So is the Solar constant)

Earth’s Without-Atmosphere Mean Surface Temperature Equation Tmean.earth is:

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

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 =

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.

The satellites measured:

Surface temp..Tmin..Tmean..Tmax

Kelvin.............100.K...220.K..390.K

Moon’s albedo: amoon = 0,11

R = 1 AU, is the Moon's distance from the sun in astronomical units

Moon’s sidereal rotation period is 27,32 days. But Moon is Earth’s satellite, so the lunar day is 29,5 days

Moon is a smooth rocky planet, Moon’s surface solar irradiation accepting factor Φmoon = 0,47

cp.moon = 0,19cal/gr*oC, moon’s surface is considered as a dry soil

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

N = 1/29,5 rotations /per day

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

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

Moon’s Mean Surface Temperature Equation Tmean.moon:

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

Tmean.moon = { 0,47 (1-0,11) 1.362 W/m² [150* (1/29,5)*0,19]¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ }¹∕ ⁴ =

= (2.488.581.418,96)¹∕ ⁴ = 223,35 K

Tmean.moon = 223,35 Κ

The newly calculated Moon’s Mean Surface Temperature differs only by 1,54% from that measured by satellites!

Tsat.mean.moon = 220 K, measured by satellites.

The satellites measured:

Surface temp..Tmin..Tmean..Tmax

Kelvin.............130.K...210.K..308.K

R = 1,524 AU, is the Mars' distance from the sun in astronomical units

(1/R²) = (1/1,524²) = 1/2,32

Mars has 2,32 times less solar irradiation intensity than Earth has

Mars’ albedo: amars = 0,25

Mars performs 1 rotation every ( 24,622hours / 24hours/day ) = 1,026 day

N = 1 /1,026 = 0,9747 rotations /day

Mars is a smooth rocky planet, Mars’ surface solar irradiation accepting factor: Φmars = 0,47

cp.mars = 0,18 cal/gr*oC, on Mars’ surface is prevalent the iron oxide

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

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

So = 1.361 W/m² the Solar constant

Mars' Mean Surface Temperature Equation is:

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

Tmean.mars = [ 0,47 (1-0,25) 1.361 W/m²*(1/2,32)*(150*0,9747*0,18)¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =

= ( 2.066.635.457,46 )¹∕ ⁴ = 213,21 K

Tmean.mars = 213,21 K

The calculated Mars’ mean surface temperature

Tmean.mars = 213,21 K is only by 1,53% higher than that measured by satellites

Tsat.mean.mars = 210 K !

The faster (than Moon) Earth's rotation smooths the average heat. The higher (than Moon) Earth's surface specific heat (oceanic waters vs dry regolith), also smooths the average heat. Consequently the daytime Earth's surface temperature (compared to Moon) lessens, and the nighttime Earth's surface temperature (compared to Moon) rises. Earth receives the same amount of solar heat (per unit area) from sun as Moon – for the same albedo. And Earth emits the same amount of solar heat, as the Moon does.

But there is something else very interesting happens.

It is the difference between Earth’s and Moon’s emitting temperatures. At the daytime Earth’s surface is warmed at a much lower temperatures and therefore at the daytime Earth’s surface emits IR radiation at a much lower intensities. So the intensity of Earth’s daytime IR radiation is much lower (than Moon's).

As a result, there is a great amount of energy – compared to Moon – “saved” on Earth during the daytime emission..

This “saved” energy should be emitted by Earth's surface during the nighttime then. At the night-time Earth’s surface is warmer than Moon's and therefore Earth’s surface at night-time is at a higher temperatures.

So the intensity of Earth’s night-time IR radiation is higher.

There is always a balance

The energy in = the energy out

But there is something else very interesting happens.

In order to achieve that balance Earth’s night-time IR emitting intensity should be much higher than the night-time IR emitting intensity of the Moon.

Now we should take notice of the nonlinearity of the Stefan-Boltzmann emission law. Consequently the night-time temperatures on Earth rise higher (compared to Moon) than the daytime temperatures on Earth lessens.

So the average Earth’s surface temperature is warmer (compared to the Moon). Thus Earth’s Tmean.earth = 288 K and Moon’s Tmean.moon = 220 K

The faster rotation and the higher specific heat does not make sun to put more energy in the Earth’s surface. What the faster rotation and the higher specific heat do is to modify the way Earth’s surface emits, the same amount as Moon, of energy (per unit area).

Earth emits IR radiation at lower temperatures during the daytime and at higher temperatures at night-time. Because of the nonlinearity of this process according to the Stefan-Boltzmann emission law, Earth ends up to have on average warmer surface than Moon.

The night-time temperatures on Earth rise higher (compared to Moon) than the day-time temperatures on Earth lessens. Earth receives (for the same albedo and per unit area) the same as Moon amount of solar energy. This energy is “welcomed” on each planet and processed in a unique for each planet way.

To illustrate the above conclusions I’ll try to demonstrate on the Earth-Moon temperatures comparison rough example:

Surface temperatures

………….min……mean……max

          Tmin↑↑→T↑mean ←T↓max

Moon…100 K...220 K …390 K

Δ………..+84 K +68 K….- 60 Κ

Earth…184K↑↑.288 K↑.330 K↓

 So we shall have for the faster rotating Earth, compared to the Moon:

Tmin↑↑→ T↑mean ← T↓max

+84↑↑→ +68↑mean ← -60↓

The faster a planet rotates (n2>n1) the higher is the planet’s average (mean) temperature T↑mean.

To emphasize we should mention that Moon’s max and min temperatures are measured on Moon’s equator, and Earth’s max and min temperatures are not.

Earth’s max and min temperatures are measured on continents, and not on oceanic waters. Otherwise the Δmin would have been even bigger and the Δmax would have been much smaller.

This rough example nevertheless illustrates that for the faster rotating and covered with water (higher cp) Earth compared with Moon the average temperature should be higher.

The planet’s faster rotation and the planet’s higher specific heat “cp” not only smooths, but also processes ( Δmin > Δmax ), the same incoming solar heat, but in a different emission pattern.

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 the entire planet:

Jemit = A*σΤmean⁴ /(β*N*cp)¹∕ ⁴  (W)

a - is the planet's average albedo

Φ - is the dimensionless Solar Irradiation accepting factor

Φ = 0,47 for a smooth spherical surface

(1 - Φ + Φ*a)S - is the reflected fraction of the incident on the planet solar flux

Φ(1 - a)S - is the absorbed fraction of the incident on the planet solar flux

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

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

Tmean - is the Planet's Mean Temperature (K)

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

Φ*S*(1-a) = 4σTmean⁴ /(β*N*cp)¹∕ ⁴

Solving for Tmean we obtain the mean surface temperature equation:

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

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

N rotations/day, is the planet’s sidereal rotation spin 

cp – is the planet's surface specific heat

cp.earth = 1 cal/gr*oC, it is because Earth has a vast ocean. Generally speaking almost the entire Earth’s surface is wet.

We can call Earth a Planet Ocean.

cp = 0,19 cal/gr*oC, for dry soil rocky planets, like Moon and Mercury. Mars has an iron oxide F2O3 surface, cp.mars = 0,18 cal/gr*oC

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

The year-round averaged energy flux at the top of the Earth's atmosphere is Sο = 1.361 W/m². With an albedo a = 0,306 and a factor Φ = 0,47 we have:

Tmean.earth = 287,74 K or 15°C.

This temperature is confirmed by the satellites measured

Tsat.mean.earth = 288 K.

So far we came to the end of this presentation. Its topic was to present the Planet's Without-Atmosphere Mean Surface Temperature Equation:

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

This Equation is based on the incomplete effective temperature equation:

Te = [ (1-a) S / 4 σ ]¹∕ ⁴

And also it is based on the discovered of the Rotating Planet Spherical Surface Solar Irradiation Absorbing-Emitting Universal Law:

Jemit = σΤmean⁴/(β*N*cp)¹∕ ⁴ (W/m²)

The Rotating Planet Surface Solar Irradiation Absorbing-Emitting Universal Law is based on a simple thought.

It is based on the thought, that physical phenomenon which distracts the black body surfaces from the instant emitting the absorbed solar radiative energy back to space, warms the black body surface up.

In our case those distracting physical phenomena are the planet’s sidereal rotation, N rotations/day, and the planet’s surface specific heat, cp cal/gr oC.

Thus we have the measured by satellites Earth’s

Tsat.mean.earth = 288 K

to be the same as the calculated by the mean surface temperature equation

Tmean.earth = 287,74 K.

A planet reflects incoming short wave solar radiation.

A planet's surface has reflecting properties.

1. The planet's albedo "a". It is a surface quality's dependent value.

2. The planet's spherical shape. For a smooth planet the solar irradiation reflection is (0,53 + Φ*a)*Jincoming.

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

Φ = 0,47 for a smooth sphere

Thus

1 - Φ = 0,53

0,53 is for smooth sphere's specular reflection

and "a" albedo is for diffuse reflection.

What we had till now:

Jsw.incoming - Jsw.reflected = Jsw.absorbed

Here

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

And

Jsw.reflected = a* Jsw.incoming

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.

(1 - Φ + Φ*a)S - is the reflected fraction of the incident on the planet solar flux

Φ(1 - a)S - is the absorbed fraction of the incident on the planet solar flux

or

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

Conclusion: The planet's absorbed fraction of the SW incoming radiation in total is:

Jsw.absorbed = 0,47 * (1-a) * Jsw.incoming

Total energy emitted to space from the entire planet:

Jemit = A*σΤmean⁴ /(β*N*cp)¹∕ ⁴ (W)

Α - is the planet's surface (m²)

For Planet Earth

So = 1.361 W/m², the Solar constant

a = 0,306  the Earth's albedo

Jsw.absorbed = 0,47 * (1-a) * 1.361 W/m² =

= 0,47*0,694*1.361 W/m² = 444,26 W/m²

Averaged on the entire Earth's surface we obtain:

Jsw.absorbed.average = [ 0,47*(1-a)*1.361 W/m² ] /4 =

= [ 0,47*0,694*1.361 W/m² ] /4 = 444,26 W/m² /4 =

= 111,07 W/m²

Jsw.absorbed.average = 111 W/m²

A Real Planet's Case

When a planet is considered as a blackbody's properties celestial body we are having very confusing results.

A real planet, not a theoretical one, but a real planet "in flesh and blood" is very much different from a blackbody's properties celestial object.

The Real Planet's Surface Properties:

1. The planet's surface has not an infinitive conductivity. Right the opposite takes place. The planet's surface conductivity is very small, when compared with the solar irradiation intensity and the planet's surface infrared emissivity intensity.

2. The planet's surface has thermal behavior properties. The planet's surface has a specific heat, cp.

3. The incident on the planet solar irradiation does not being distributed instantly and evenly on the entire planet's surface area.

4. Planet does not accept the entire solar irradiation incident in planet's direction. Planet accepts only a small fraction of the incoming solar irradiation. This happens because of the planet's albedo, and because of the planet's smooth and spherical surface reflecting qualities, which we refer to as "the planet's solar irradiation accepting factor Φ".

Planet reflects the (1-Φ + Φ*a) part of the incident on the planet's surface solar irradiation.

And

Planet absorbs only the Φ(1 - a) part of the incident on the planet's surface solar irradiation.

Here "a" is the planet's average albedo and "Φ" is the planet's solar irradiation accepting factor.

For smooth planet without thick atmosphere, Earth included, Φ=0,47

5. Planet's surface has not a constant intensity solar irradiation effect. Planet's surface rotates under the solar flux. This phenomenon is dicisive for the planet's surface infrared emittance distribution.

The real planet's surface infrared radiation emittance distribution intensity is a planet's rotational speed dependent physical phenomenon.

Planet's Surface Radiative Equilibrium

What solar irradiated planet's surface does is to develop the incoming energy to achieve a radiative equilibrium.

It is an equation: energy in = energy out.

The mechanisms planet has to achieve the surface radiative equilibrium:

1. The negative feedbacks.

a). The rising precipitation, due to warming from the excess energy, rises the Earth's cloud cover. This in turn magnifies the planet's albedo. So some less energy reaches the surface.

b). The loss of Arctic oceanic ice cover, due to warming from the excess energy, opens the Arctic oceanic waters. This in turn magnifies the Arctic oceanic surface emissivity (water has higher emissivity compared to the ice, and water has a much higher emissivity compared to the snow covered Arctic oceanic ice fields). So some more energy is emitted to space from the Earth's surface to come closer to the radiative equilibrium (radiative balance).

2. The heat accumulation and the rise of the planet's average temperature.

Some of the solar energy that is not emitted out to space forms the accumulated heat, mostly in the oceanic waters and also in the land masses. The heat accumulation rises the planet's temperature. When the planet's temperature risen planet's infrared radiation energy emissions rise too.

Thus the Planet's Surface remains in Radiative Equilibrium.

A Planet-Without-Atmosphere Effective Temperature Equation, the Te Equation which is based on the radiative equilibrium and on the Stefan-Boltzmann Law, and which is in common use right now:

Te = [ (1-a) S / 4 σ ]¹∕ ⁴

is actually an incomplete Te equation and that is why it gives us very confusing results.

A planet-without-atmosphere effective temperature equation

Te = [ (1-a) S / 4 σ ]¹∕ ⁴

is incomplete because it is based only on two parameters:

1. On the average solar flux S W/m² on the top of a planet’s atmosphere and

2. The planet’s average albedo "a".

We use much more parameters to calculate a planet mean temperature. Planet is a celestial body with more major features when calculating planet mean temperature to consider.

The planet-without-atmosphere mean temperature equation has to include all the planet’s major properties and all the characteristic parameters.

3. The planet's axial spin N rotations/day

4. The thermal property of the surface (the specific heat cp)

5. The planet's surface solar irradiation accepting factor Φ (the spherical surface’s primer geometrical quality). For Mercury, Moon, Earth and Mars without atmosphere Φ = 0,47.

Earth is considered without atmosphere because Earth’s atmosphere is very thin and it does not affect Earth’s Mean Surface Temperature.

Altogether these parameters are combined in the Planet-Without-Atmosphere Mean Surface Temperature Equation:

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

A Planet-Without-Atmosphere Mean Surface Temperature Equation produces very reasonable results:

Tmean.earth = 287,74 K, calculated with the Equation, which is identical with the Tsat.mean.earth = 288 K, measured by satellites.

Tmean.moon = 223,35 K, calculated with the Equation, which is almost the same with the Tsat.mean.moon = 220 K, measured by satellites.

The planet Earth's and the planet Mars' faster rotation creates the necessary level of the "solar irradiation - planet surface" interaction phenomenon...

which results in the day-time much lower surface temperatures and, consequently, in much lower day-time surface infrared radiation emissions

and which results in higher planet surface 24-hours average temperatures.

Φ factor explanation

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

Tmean = [ 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

and

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.

(1 - Φ) 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 also cooperate in the Planet's Mean Surface Temperature Equation:

Tmean = [ 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

a - is the planet's average albedo

(1-Φ + Φ*a )S - 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's surface area ( m² )

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

Tmean - is the Planet's 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 on sunlit hemisphere

Jabs.earth = 0,47 ( 1 - 0,306 ) So π r² =

= 0,47*0,694 * 1.361* π r² ( W ) =

Jabs.earth = 0,326 So π r² =

= 0,326* 1.361 π r² =

= 444,26 π r² ( W )

What is going on here is that instead of

Jabs.earth = 0,694* 1.361 π r² ( W )

we should consider

Jabs.earth = 0,326* 1.361 π r² ( W ).

Averaged on the entire Earth's surface we obtain:

Jsw.absorbed.average = [ 0,47*(1-a)*1.361 W/m² ] /4 =

= [ 0,47*0,694*1.361W/m² ] /4 = 444,26 W/m2 /4 = 111,07 W/m²

Jsw.absorbed.average = 111,07 W/m² 

or 111 W/m²

Mars and Moon 

210 K and 220 K

Let’s compare then:

Moon: Tsat.moon = 220K

Moon’s albedo is amoon = 0,11

What is left to absorb is (1 – amoon) = (1- 0,11) = 0,89

Mars: Tsat.mars = 210 K

Mars’ albedo is amars = 0,25

What is left to absorb is (1 – amars) = (1 – 0,25) = 0,75

Mars /Moon satellite measured temperatures comparison:

Tsat.mars /Tsat.moon = 210 K /220 K = 0,9545

Mars /Moon what is left to absorb (which relates in ¼ powers) comparison, or in other words

the Mars /Moon albedo determined solar irradiation absorption ability:

( 0,75 /0,89 )¹∕ ⁴ = ( 0,8427 )¹∕ ⁴ = 0,9581

Conclusions:

1. Mars /Moon satellite measured temperatures comparison

( 0,9545 )

is almost identical with the Mars /Moon albedo determined solar irradiation absorption ability

( 0,9581 )

2. If Mars and Moon had the same exactly albedo, their satellites measured temperatures would have been exactly the same.

3. Mars and Moon have two major differences which equate each other:

The first major difference is the distance from the sun both Mars and Moon have.

Moon is at R = 1 AU distance from the sun and the solar flux on the top is So = 1.361 W/m²

( it is called the Solar constant ).

Mars is at 1,524 AU distance from the sun and the solar flux on the top is

S = So*(1/R²) = So*(1/1,524²) = So*1/2,32 .

(1/R²) = (1/1,524²) = 1/2,32

Mars has 2,32 times less solar irradiation intensity than Earth and Moon have.

Consequently the solar flux on the Mar’s top is 2,32 times weaker than that on the Moon.

The second major difference is the sidereal rotation period both Mars and Moon have.

Moon performs 1 rotation every 29,531 earth days.

Mars performs 1 rotation every ( 24,622hours / 24hours/day ) = 1,026 day.

Consequently Mars rotates 29,531 /1,026 = 28,783 times faster than Moon does.

So Mars is irradiated 2,32 times weaker, but Mars rotates 28,783 times faster.

And… for the same albedo, Mars and Moon have the same satellite measured mean temperatures.

We shall use now from the Planet's Mean Surface Temperature Equation Tmean = [ Φ (1-a) So (1/R²) (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴

In the Planet's Mean Surface Temperature Equation the planet's axial spin "N" is in fourth root

the (β*N*cp)¹∕ ⁴

The rotation difference's fourth root is

(28,783)¹∕ ⁴ = 2,3162 

And the irradiating /rotating comparison 

2,32 /(28,783)¹∕ ⁴  = 2,32 /2,3162 = 1,001625

It is only 0,1625% difference

When rounded the difference is 0,16 %

It is obvious now, the Mars’ 28,783 times faster rotation equates the Moon’s 2,32 times higher solar irradiation.

That is why the 28,783 times faster rotating Mars has almost the same average satellites measured temperature as the 2,32 times stronger solar irradiated Moon.

Thus we are coming here again to the same conclusion:

The Faster a Planet Rotates, the Higher is the Planet's Average Temperature.

And

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

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

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......0,11......0,250

Mercury:

Tsat.mean = 340 K

[ (1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ = [ 6,6769*(1/175)¹∕ ⁴ ] ¹∕ ⁴ = 1,1640

Moon:

Tsat.mean = 220 K

[ (1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ = [ 1*(1/29,531)¹∕ ⁴ ] ¹∕ ⁴ = 0,8093

Mars:

Tsat.mean = 210 K

[ (1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ = [ 0,430*(0,9747)¹∕ ⁴ ] ¹∕ ⁴ = 0,8090

Let's compare

Mercury [ (1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ / Moon [ (1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ = 1,1640 /0,8093 = 1,4383

Tmean.mercury /Tmean.moon = 340 K /220 K = 1,5454…

They are almost close, because Mercury and Moon have different (1-albedo) values: amoon = 0,11 and amercury = 0,068. For Moon (1-0,11)=0,89 and for Mercury (1-0,68)=0,932

Moon /Mars = 0,8093 /0,8090 = 1,00037 or 0,037 %

If Moon and Mars had the same albedo amoon = 0,11 they would both have the same satellite measured mean temperature Tmean = 220 K. Mars' albedo is amars = 0,25.

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......0,11......0,250

1-a….0,932……0,89…….0,75

coeff..1,2179...............0,7524

Comparison coefficient calculation

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴

Mercury:

Tsat.mean = 340 K

[ (1-a)*(1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ =

= [ 0,932*6,6769*(1/175)¹∕ ⁴ ] ¹∕ ⁴ = 1,1437

Mars:

Tsat.mean = 210 K

[ (1-a)*(1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ =

= [ 0,75*0,430*(0,9747)¹∕ ⁴ ] ¹∕ ⁴ = 0,7524

Let's compare

Mercury coeff. / Mars coeff. =

= 1,1437 /0,7524 = 1,5200

And

Tmean.mercury /Tmean.mars =

= 340 K /210 K = 1,6190 

There is only 6,50 % difference

Conclusion:

Everything is all right. Everything is based on observations.

The comparison coefficient is:

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴

And

The planet's axial spin (rotations per day) "N" should be considered in the Tmean equation in the fourth root:

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

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.

The comparison coefficient is:

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴

And

It is the 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) So (1/R²) (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴

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

These planets' parameters are all observations.

Planet…....Earth.….Moon….Mars

Tsat.mean.288 K….220 K…210 K

R…...............1... AU..1 AU..1,525 AU

1/R²…..........1…........1….…0,430

N…..............1....1 /29,531..0,9747

cp................1.........0,19.......0,18

a..............0,30......0,136......0,250

1-a…........0,70……0,864…….0,75

coeff...........1...................0,72748 

As we can see Earth and Mars have very close (1-a); for Earth 0,70 and for Mars 0,75.

Also Earth and Mars have very close N; for Earth N = 1 rotation /day, and for Mars N = 0,9747 rotation /day.

Earth and Mars both have the same Φ = 0,47 solar irradiation accepting factor.

Thus the comparison coefficient can be limited as follows:

Comparison coefficient calculation

[ (1/R²) (cp)¹∕ ⁴ ]¹∕ ⁴

Earth:

Tsat.mean = 288 K

[ (1/R²)*(cp)¹∕ ⁴ ]¹∕ ⁴ =

= [ 1*(1)¹∕ ⁴ ] ¹∕ ⁴ = 1

Mars:

Tsat.mean = 210 K

[ (1/R²)*(cp)¹∕ ⁴ ]¹∕ ⁴ =

= [ 0,430*(0,18)¹∕ ⁴ ] ¹∕ ⁴ = ( 0,430*0,65136 )¹∕ ⁴ =

= ( 0,2801 )¹∕ ⁴ = 0,72748

Let's compare

Earth coeff. / Mars coeff. =

= 1 /0,72748 = 1,3746

And

Tmean.earth /Tmean.mars =

= 288 K /210 K = 1,3714

Conclusion:

Everything is all right. Everything is based on observations in the comparison coefficient

[ (1/R²) (cp)¹∕ ⁴ ]¹∕ ⁴

And It is the confirmation that the planet's specific heat "cp" should be considered in the Tmean equation in the fourth root:

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

All data are satellites measurements.

R – semi-major axis in AU (Astronomical Units)

a – planet’s average albedo

N – rotations /day – planet’s spin

T1bar – planet atmosphere at 1 bar average temperature in Kelvin

Planet.…Jupiter…..Saturn….Neptune

R………....5,2044…..9,5826.….30,33

1/R²…….0,0369…0,01089…0,001087

a……………0,503…0,342…….0,290

1-a………..0,497…0,658…….0,710

N………….2,4181…2,2727….1,4896

T 1 bar…...165 K….134 K……72 K

Coeff...0,388880... 0,306264…0,170881

Comparison coefficient calculation

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴

Jupiter

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴ =

= [ 0,497*0,0369*(2,4181)¹∕ ⁴ ]¹∕ ⁴ =

= ( 0,497*0,0369*1,2470 ]¹∕ ⁴ = ( 0,0228691 )¹∕ ⁴ =

= 0,388880

Saturn

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴ =

= [ 0,658*0,01089*(2,2727)¹∕ ⁴ ]¹∕ ⁴ =

= ( 0,658*0,01089*1,2278 )¹∕ ⁴ = ( 0,0087980 )¹∕ ⁴ =

= 0,306264

Neptune [ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴ =

= [ 0,710*0,001087*(1,4896)¹∕ ⁴ ]¹∕ ⁴ =

= ( 0,710*0,001087*1,1048 )¹∕ ⁴ = ( 0,000852651 )¹∕ ⁴ =

= 0,170881

Let’s compare

Jupiter coeff. /Saturn coeff. =

= 0,388880 /0,306264 = 1,2698

T1bar.jupiter /T1bar.saturn = 165 /134 = 1,2313

Jupiter coeff. / Neptune coeff. =

= 0,388880 /0,170881 = 2,2757

T1bar.jupiter /T1bar.neptune = 165 /72 = 2,2917

Saturn coeff. /Neptune coeff. =

= 0,306264 /0,170881 = 1,7923

T1bar.saturn /T1bar.neptune = 134 /72 = 1,8611

Notice: The atmospheric composition of Jupiter is closer to the atmospheric composition of Neptune, that is why their mutual comparison matches best.

Conclusion:

Gaseous planets Jupiter, Saturn and Neptune average T1bar temperatures relate the same way as the rocky inner planets Mercury, Moon and Mars average surface temperatures, and also as the Earth with Europa average surface temperatures.

The planets are being separated in groups (Jupiter, Saturn and Neptune - H2), (Mercury, Moon and Mars - regolith), and (Earth with Europa - H2O) only by their similar specific heat.

The comparison coefficient is the same:

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴

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) So (1/R²) (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴.

According to Milankovitch Ice Ages are generally triggered by minima in high-latitude Northern Hemisphere summer insolation, enabling winter snowfall to persist through the year and therefore accumulate to build Northern Hemisphere glacial ice sheets. Similarly, times with especially intense high-latitude Northern Hemisphere summer insolation, determined by orbital changes, are thought to trigger rapid deglaciations, associated climate change and sea level rise. But, at second thought, I concluded that Earth cannot accumulate heat on the continents’ land masses. Earth instead accumulates heat in the oceanic waters.

It is well known that when a planet rotates faster its daytime maximum temperature lessens and the night time minimum temperature rises.

But there is something else very interesting happens.

When a planet rotates faster it is a warmer planet. (it happens because Tmin↑↑ grows higher than T↓max goes down).

The understanding of this phenomenon comes from a deeper knowledge of the Stefan-Boltzmann Law.

It happens so because when rotating faster, the planet's surface has emission temperatures Te the new distribution to achieve.

So that is what happens:

The faster a planet rotates (n2>n1) the higher is the planet’s average (mean) temperature T↑mean.

It happens in accordance to the Stefan-Boltzmann Law.

Let's explain:

Assuming a planet rotates faster and Tmax2 -Tmax1 = -1°C.

Then, according to the Stefan-Boltzmann Law:

Tmin2 -Tmin1 > 1°C

Consequently

Tmean2 > Tmean1.

Assuming a planet rotates faster (n2>n1).

If on the solar irradiated hemisphere we observe the difference in average temperature

Tsolar2-Tsolar1 = -1°C

Then the dark hemisphere average temperature

Tdark2 -Tdark1 >1°C

Consequently the total average

Tmean2 > Tmean1

So we shall have:

Tdark↑↑→ T↑mean ← T↓solar

The faster a planet rotates (n2>n1) the higher is the planet’s average (mean) temperature T↑mean.

A numerical example

Here it is the improved numerical example which proves, the Tmean > Te when the planet rotates fast enough:

As you will see in the numerical example, which I have shown below, when planet rotates faster, on the planet's solar irradiated side the Te.solar temperature subsides from 200 K to 199 K.

On the other hand on the planet's dark side, when planet rotates faster, the Te.dark temperature rises from 100 K to 107,126 K.

So when the solar irradiated side gets on Te.solar cooler by -1 degree °C, the dark side gets on Te.dark warmer by +7,126 degrees °C.

And as a result the planet's total Te temperature gets higher.

It happens so because when rotating faster (n2>n1) the planet's surface has emission temperatures Te the new distribution to achieve.

It happens so, because we have assumed planet emitting as a blackbody with two separate hemispheres. The solar hemisphere emitting some of the absorbed incident solar flux's energy, and the dark hemisphere emitting the rest of the absorbed solar flux's energy.

Also, when the two hemispheres blackbody planet rotating faster the energy in = energy out

balance should be met.

The faster rotation does not change the real planet's energy balance.

Also, the real planet never achieves uniform temperature on both sides, because it receives the solar flux only on the sunlit side.

Because we consider the faster rotating real planet at the same distance from the sun, with the same albedo and Φ factor -

energy in = energy out

balance should be met.

The numerical example:

Assuming a planet with two hemispheres' Te temperatures

Te.solar1 = 200 K, and Te.dark1 = 100 K

Assuming this planet rotates somehow faster (n2 > n1), so assuming the new Te.solar2 average temperature resulting Te.solar2 = 199 K.

What would be the planet's Te.dark2 then?

Jemit.solar1 = σ*(Te.solar1)⁴ ,

σ*(200 K)⁴ = 1.600.000.000*σ for (n1) rot/day

Jemit.solar2 = σ*(Te.solar2)⁴ ,

σ*(199 K)⁴ = 1.568.000.000*σ for (n2) rot/day

Jemit.solar2 - Jemit.solar1 =

= 1.568.000.000*σ - 1.600.000.000*σ =

= - 31.700.000*σ is the difference in the Te solar side emitting intensity when (n2>n1) and 199 K - 200 K = - 1°C

So we have ( - 31.700.000*σ ) less emitting intensity on the solar side (2) when n2>n1.

It should be compensated by the increased emission on the dark side ( + 31.700.000*σ ) for the energy balance equation to get met:

Jemit.dark1 = σ*(Te.dark1)⁴ ,

(100 K)⁴ = 100.000.000

Jemit.dark2 = σ*(Te.dark2)⁴ ,

σ*(Te.dark2)⁴ = (100.000.000 + 31.700.000)*σ = 131.700.000*σ

The dark side higher temperature (2) to compensate the solar side cooler emission (2) by ( - 31.700.000 ) would be

Te.dark2 = (131.700.000)¹∕ ⁴ = 107,126 K

As we see in this numerical example, when the planet rotating faster (n2>n1) the Te temperature on the solar irradiated side subsides from 200 K to 199 K.

On the other hand the Te temperature, when planet rotating faster (n2>n1) on the dark side rises from 100 K to 107,126 K.

So when rotating faster (n2>n1) the solar irradiated planet's side gets on Te cooler by -1 degree °C, the planet's dark side gets on Te warmer by +7,126 degrees °C.

And as a result the planet's total Te temperature gets higher.

It happens so because when rotating faster (n2>n1) the planet's surface has emission temperatures Te the new distribution to achieve.

Consequently, when rotating faster, the planet's mean temperature rises.

Because Te.solar2-Te.solar1 = -1°C

Then the dark hemisphere's Te temperature

Te.dark2 -Te.dark1 = +7,126°C

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

Tmean = [ Te⁴ (β*N*cp)¹∕ ⁴ ]¹∕ ⁴

Tmean.solar2 = [ (Te.solar1 -1°C)⁴ (β*N*cp)¹∕ ⁴ ]¹∕ ⁴

Tmean.dark2 = [ (Te.dark1 +7,126°C)⁴ (β*N*cp)¹∕ ⁴ ]¹∕ ⁴

(Tmean.solar1 + Tmean.dark1) / 2 < (tmean.solar2="" +="" tmean.dark2)="">

Consequently the total average

Tplanet.mean2 > Tplanet.mean1

So we shall have:

when n2>n1

Tmean.dark↑↑→ T↑mean ← T↓mean.solar

The faster a planet rotates (n2>n1) the higher is the planet’s average (mean) temperature T↑mean.

Because Te.dark2↑↑ grows higher (+7,126°C) than the T↓e.solar2 lessens (-1°C).

Thus when a planet rotates faster its mean temperature is higher.

Conclusion:

Earth's faster rotation rate, 1 rotation per day, makes Earth a warmer planet than Moon.

Moon rotates around its axis at a slow rate of 1 rotation in 29,5 days.

Notice:

In the above numerical example we assumed a rotating planet blackbody with two hemispheres' Te.solar and Te.dark.

It was an assumption, because the blackbody by definition has a uniform temperature on its entire surface.

Also we assumed, that the blackbody somehow had accumulated some of the daytime solar energy.

In this numerical example we have a combination of the blackbody and the real planet emitting behavior. And it is also an assumption.

Real planet does not emit according to the exact Stefan-Boltzmann emission law. Real planet emits exactly according to the new Universal law:

Jemit.planet = 4σ Tmean⁴ /(β*N*cp)¹∕ ⁴ .