The Planet's Effective Temperature Complete Formula Te = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴

Plus the introduction to the Reversed Milankovitch Cycle. Click above on the box for more

The Planet's Effective Temperature

Te.complete = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ gives wonderful results Te.earth=288,36 K, Te.moon=221,74 K, Te.mars=213,2 K and 

Te.mercury=346,11 K

identical to the Tmean measured by satellites

Tmercury=340 K

Tearth=288 K

Tmoon=220K

Tmars=210 K

 

Planet...Te.incompl...Te.compl…Tsat.mean

................formula......formula...measured

Mercury…...437 K……..346,11 K……..340 K

 

It is time to abandon the old

Te = [ (1-a) S /4σ ]¹∕ ⁴ incomplete formula.

.

The Planet's Without-Atmosphere Effective Temperature Complete Formula

from the incomplete

Te = [ (1-a) S / 4 σ ]¹∕ ⁴

which is in common use right now, but actually it is an incomplete planet's Te formula and that is why it gives us very confusing results.

to the complete

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

The Planet's Without-Atmosphere Effective Temperature Complete Formula is also based on the radiative equilibrium and on the Stefan-Boltzmann Law.

The Formula is being completed by adding to the incomplete Te formula 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 sidereal rotation spin

cp – 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.

Interesting, very interesting what we see here:

Planet..Tsat.mean..Rotations..Tmin..Tmax

............measured.....per day...................

Mercury..340 K.......1/58,7....100 K...700 K

Earth.......288 K..........1..........................

Moon......220 Κ.......1/29,5...100 K...390 K

Mars......210 K.......0,9747...130 K...308 K

Earth and Moon are at the same distance from the Sun R = 1 AU.

Earth and Mars have almost the same axial spin N = 1rotation /day.

Moon and Mars have almost the same satellites measured average temperatures 220 K and 210 K.

Mercury and Moon have the same minimum temperature 100 K.

Mars' minimum temperature is 130 K, which is much higher than for the closer to the Sun Mercury's and Moon's minimum temperature 100 K.

The planet's effective temperature old Te = [ (1-a) S /4σ ]¹∕ ⁴ incomplete formula gives very confusing results.

And the faster rotating Earth and Mars appear to be relatively warmer planets.

We ended up to the following remarkable results

Comparison of results the planet Te calculated by the Incomplete Formula:

Te = [ (1-a) S / 4 σ ]¹∕ ⁴

the planet Te calculated by the Complete Formula:

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

and the planet Tsat.mean measured by satellites:

Planet  Te.incompl  Te.compl  Tsat.mean

               formula     formula    measured

Mercury    437 K       346,11 K      340 K

Earth         255 K       288,36 K      288 K

Moon         271 Κ        221,74 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…

.

 

   

The Planet's Effective Temperature Complete Formula: Te = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (1)

 

My name is   Christos J. Vournas, M.Sc. mechanical engineer.

I live in Athens Greece.

My e-mail address is:   vournas.christos@yahoo.com

The date is   October 11, 2019

I launched this site to have an opportunity to publish my scientific discoveries on the Climate Change.

I have been studying the Planet Earth’s Climate Change since November 2015;

First I discovered the Reversed Milankovitch Cycle.

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

Φ - the next discovery - is the dimensionless Solar Irradiation accepting factor - very important

The further studies led me to discover the Rotating Planet Spherical Surface Solar Irradiation Absorbing-Emitting Universal Law

and the Planet's Effective Temperature Complete Formula.

.

The first step

At very first look at the data table we distinguish the following:

Planet..Tsat.mean..Rotations..Tmin..Tmax

...........measured...per day......................

Mercury..340 K....1/58,7.....100 K....700 K

Earth.....288 K........1.............................

Moon....220 Κ.....1/29,5.....100 K...390 K

Mars.....210 K....0,9747.....130 K...308 K

The Earth's and Mars' by satellites temperatures measurements, in relation to the incident solar irradiation intensity, appear to be higher,

and it happens because of Earth's and Mars' faster rotation. 

I should say here that I believe in NASA satellites temperatures measurements. None of my discoveries would be possible without NASA satellites very precise planet temperatures measurements.

It is the "magic" of the planet's spin. 

When it is understood, it becomes science.

 

The closest to the sun planet Mercury receives

15,47 times stronger solar irradiation intensity than the planet Mars does.

However on the Mercury's dark side

Tmin.mercury = 100 K,

when on the Mars' dark side

Tmin.mars = 130 K.

These are observations, these are the by satellites planets temperatures measurements.

And they cannot be explained otherwise but by the planet Mars'

56,5 times faster rotation than planet Mercury's spin.

Moon is in our immediate neighborhood

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 = 1362 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.

Let's continue our observations:

Planet.Tsat.mean..Rotations.Tmax..Tmax /2

............measured.....per day....................

Mercury..340 K.......1/58,7.....700 K...350 °C

Earth.......288 K..........1............................

Moon.......220 Κ.......1/29,5....390 K...195 °C

Mars........210 K.......0,9747...308 K...154 °C

As we can see for the slow rotating Mercury and for the slow rotating Moon:

Mercury: 340 K - 350 °C = - Δ 10 °C

Moon: 220 K - 195 °C = Δ 25 °C

On the other hand, for the fast rotating Mars

Tmax.mars /2 = 154 °C

Tsat.mars.mean = 210 K

210 K - 154 °C = Δ 56 °C

It proves that planet Mars, due to its fast spin, appears to be a relatively much warmer planet, than planet Mercury and Moon are.

Mars and Moon satellite measured mean temperatures comparison: 210 K and 220 K

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

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

Moon is at R = 1 AU distance from the sun and the solar flux on the top is So = 1.362 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

2,32 times weaker than that on the Moon.

Mars rotates 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.

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

Thus we are coming here again to the same conclusion:

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

We may conclude that for a faster rotating planet there is the phenomenon of its warmer surface...

And it becomes very cold on the Moon at night

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

And Moon also  emits from its very hot 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.

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A Planet Universal Law Formula

As you know, to maintain a Planet Universal Law Formula 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 Effective Temperature Complete Formula has the wonderful ability to calculate Planets Surface Effective Temperatures (mean temperatures) getting almost the same results as the measured by satellites planets mean temperatures.

This Complete Formula can be applied to all the without atmosphere planets and moons in a solar system.

The more we compare the planets' surface temperatures, the more we understand the planets' surface warming phenomenon.

 

 

Φ - is the dimensionless Solar Irradiation accepting factor - very important.

It is a realizing that a sphere's surface absorbs the incident solar irradiation not as a disk of the same diameter, but accordingly to its spherical shape.

   For a smooth spherical surface Φ = 0,47

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

We have now chosen the planet Mercury for its very low albedo a=0,088 and for its very slow rotation spin N=1/58,646 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 plays small role in 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 /58,646..1 /29,531..0,9747

a......0,088......0,136......0,250

1-a…0,912……0,864…….0,75

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

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

We have

(1-a) = 0,912

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 = [ 0,912* 1.362 W/m² * 6,6769 /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =

Te.incomplete.mercury = ( 36.568.215.492,06 )¹∕ ⁴ = 437,296 K

Te.incomplete.mercury = 437,296 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 = 437 K

Tsat.mean.mercury = 340 K

Very big difference, nearly 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.

The Planet 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

.

1. Earth's Without-Atmosphere Effective Temperature Calculation: Te.earth

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

Earth’s albedo: aearth = 0,30

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

Earth’s Without-Atmosphere Effective Temperature Complete Formula Te.earth is:

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

Τe.earth = [ 0,47(1-0,30)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 =

Τe.earth = [ 0,47(1-0,30)1.362 W/m²(150*1*1)¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =

Τe.earth = ( 6.914.170.222,70 )¹∕ ⁴ =

Te.earth = 288,36 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.

.

2. Moon’s Effective Temperature Calculation: Te.moon

The satellites measured:

Surface temp..Tmin..Tmean..Tmax

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

Moon’s albedo: amoon = 0,136

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 Effective Temperature Complete Formula Te.moon:

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

Te.moon = { 0,47 (1-0,136) 1.362 W/m² [150* (1/29,5)*0,19]¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ }¹∕ ⁴ = 221,74 K

Te.moon = 221,74 Κ

The newly calculated Moon’s Effective Temperature differs only by 0,8% from that measured by satellites!

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

.

3. Mars’ Effective Temperature Calculation: Te.mars

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 = 1.362 W/m² the Solar constant

Mars' Effective Temperature Complete Formula is:

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

Te.mars = [ 0,47 (1-0,25) 1.362W/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

Te.mars = 213,21 K

The calculated Mars’ effective temperature

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

Tsat.mean.mars = 210 K !

.

4. Mercury's Effective Temperature Calculation: Te.mercury

The satellites measured:

Surface temp..Tmin..Tmean..Tmax

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

N = 1/58,646 rotations/per day, Planet Mercury completes one rotation around its axis in 58,646 days.

Mercury's average distance from the sun is R=0,387AU. The solar irradiation on Mercury is (1/R²) = (1AU/0,387AU)²= 2,584²= 6,6769 times stronger than that on Earth.

Mercury’s albedo is: amercury = 0,088

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

Cp.mercury = 0,19 cal/gr*oC, Mercury’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

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

So = 1.362 W/m² the Solar constant

Planet Mercury’s Effective Temperature Complete Formula is:

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

Te.mercury = { 0,47(1-0,088) 1.362 W/m²*6,6769*[150* (1/58,646)*0,19]¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ }¹∕ ⁴ =

Te.mercury = 346,11 K

The calculated Mercury’s effective temperature Te.mercury = 346,11 K is only 1,80% higher than the measured by satellites

Tsat.mean.mercury = 340 K !

.

 

We can confirm now with great confidence

We have collected the results now:

        Te. incompl  Te.compl  Tsat.mean

Mercury   437 K      346,11 K    340 Κ

Earth        255 K      288,36 K    288 K

Moon       271 Κ      221,74Κ     220 Κ

Mars       209,91 K  213,21 K    210 K

the calculated with Planet's Without-Atmosphere Effective Temperature Complete Formula and the measured by satellites are almost the same, very much alike.

It is a situation that happens once in a lifetime in science.

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.

The planet without-atmosphere effective temperature calculating formula has to include all the planet surface basic properties and all the characteristic parameters.

3. The sidereal rotation spin N rotations/day.

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

5. The planet surface solar irradiation accepting factor Φ ( the spherical surface’s primer solar irradiation absorbing property).

Altogether these parameters are combined in the Planet's Without-Atmosphere Effective Temperature Complete Formula:

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

.

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

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 Tmax goes down)

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

It happens so because when rotating faster a planet's surface has new radiative equilibrium temperatures to achieve.

So that is what happens:

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's average temperature

Tdark2 -Tdark1 >1°C

Consequently the total average

Tmean2 > Tmean1

So we shall have:

Tdark↑↑ Tmean Tsolar

And as a result the planet's total average temperature Tmean 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.

.

The Rotating Planet Surface Solar Irradiation Absorbing-Emitting Universal Law

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*σΤe⁴ /(β*N*cp)¹∕ ⁴  (W)

Φ - is the dimensionless Solar Irradiation accepting factor

Φ = 0,47 for a smooth spherical surface

(1 - Φ) - is the reflected 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

Te - is the Planet's Effective Temperature (K)

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

Φ*S*(1-a) = 4σTe⁴ /(β*N*cp)¹∕ ⁴

Solving for Te we obtain the effective temperature:

Te = [ 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 = 1.362 W/m². With an albedo a = 0,3 and a factor Φ = 0,47 we have:

Te.earth = 288,36 K or 15°C.

This temperature is confirmed by the satellites measured

Tsat.mean.earth = 288 K.

.

The Fast Rotating Planet Earth

So far we came to the end of this presentation. Its topic was to present the Planet's Without-Atmosphere Effective Temperature Complete Formula:

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

This Formula is based on the incomplete effective temperature formula:

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 = σΤe⁴/(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

Tmean.earth = 288 K

to be the same as the calculated by the effective temperature complete formula

Te.earth = 288,36 K.

.

Earth is warmer because Earth rotates faster and because Earth’s surface is covered with water

We had to answer these two questions:

1. Why Earth’s atmosphere doesn’t affect the Global Warming?

It is proven now by the Planet's Effective Temperature Complete Formula calculations. There aren’t any atmospheric factors in the Complete Formula. Nevertheless the Complete Formula produces very reasonable results:

Te.earth = 288,36 K,

calculated by the Complete Formula, which is the same as the

Tsat.mean.earth = 288 K,

measured by satellites.

Te.moon = 221,74 K, calculated by the Complete Formula, which is almost identical with the

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

2. What causes the Global Warming then?

The Global Warming is happening due to the orbital forcing.

And… what keeps the Earth warm at Te.earth = 288 K, when the Moon is at Te.moon = 220 K? Why Moon is on average 68 oC colder? It is very cold at night there and it is very hot during the day…

Earth is warmer because Earth rotates faster and because Earth’s surface is covered with water.

Does the Earth’s atmosphere act as a blanket that warms Earth’s surface?

No, it does not.

.

Earth's average SW absorbed solar energy: Jsw.absorbed.average = 112,029 W/m²

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 dependent value.

2. The planet's spherical shape.

For a smooth planet the solar irradiation reflection is 0,53*Jincoming.

 

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

Here 

Jsw.reflected = 0,53 * a * Jsw.incoming

And

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

Where

Φ = (1-0,53) = 0,47 Φ = 0,47

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

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*σΤe⁴ /(β*N*cp)¹∕ ⁴ (W)

Α - is the planet's surface (m²)

For Planet Earth

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

= 0,47*0,7*1.362W/m² = 448,098 W/m²

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 = 112,029 W/m²

What was the Earth's mean temperature 3,5 billion years ago ( 3,5 Bya ) ?

Lets see:

Sun's irradiating intensity was weaker.

It was only 75 % of the present,

S = 0,75*So

And Earth rotated twice as fast then,

1 rotation in 12 hours.

Lets calculate:

Earth’s Without-Atmosphere Effective Temperature calculation

Te.earth.3,5 Bya

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

Earth’s albedo: aearth = 0,30

Φearth = 0,47 

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

N = 2 rotations /per day, was Earth’s sidereal rotation spin at 3,5 Bya

cp.earth = 1 cal/gr*oC

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

Earth’s Without-Atmosphere Effective Temperature Complete Formula Te.earth is:

Te.earth = [ Φ (1-a) 0,75*So (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴

Τe.earth.3,5 Bya = [ 0,47(1-0,30)*0,75*1.362 W/m²(150 days*gr*oC/rotation*cal *2rotations/day*1 cal/gr*oC)¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =

Τe.earth.3,5 Bya = [ 0,47(1-0,30)*0,75*1.362 W/m²(150*2*1)¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =

Τe.earth.3.5 Bya = ( 6.167.274.722,0 )¹∕ ⁴ = 280,23 K

Te.earth.3.5 Bya = 280,23 Κ

or Tmean.earth.3,5 Bya = 7 oC.

Conclusion: 3,5 billion years ago planet Earth had a sustainable for life mean temperature of 7 oC.

From Wikipedia:

" Among the earliest fossil evidence for life is microbial mat fossils found in 3.48 billion-year-old sandstone in Western Australia,[63]".

.

A Blackbody Surface Equilibrium Temperature (A Blackbody Effective Temperature)

A blackbody planet surface is meant as a classical blackbody surface approaching.

Here are the blackbody's properties:

1. Blackbody does not reflect the incident on its surface radiation.

2. Blackbody interracts with the entire incident on the blackbody's surface radiation.

3. Blackbody's emission temperature depends only on the quantity of the incident radiative energy per unit area.

4. Blackbody is considered only as blackbody's surface physical properties. Blackbody is only a surface without "body".

5. Blackbody does not consist from any kind of a matter. Blackbody has not a mass. Thus blackbody has not a specific heat.

Blackbody's cp = 0.

6. Blackbody has surface dimensions. So blackbody has the radiated area and blackbody has the emitting area.

7. The whole blackbody's surface area is the blackbody's emitting area.

8. The blackbody's surface has an infinitive conductivity.

9. All the incident on the blackbody's surface radiative energy is instantly and evenly distributed upon the whole blackbody's surface.

10. The radiative energy incident on the blackbody's surface the same very instant the blackbody's surface emitts this energy away.

The emission temperature the blackbody's surface has according to the Stefan-Boltzmann Law is:

Te = (Total incident W /Total area m² *σ)¹∕ ⁴ K

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

Planet is not a blackbody.

Planet reflects the (1-a)Φ 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

.

A Real Planet's case

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".

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 Temperature

Planet's Surface Radiative Equilibrium Temperature

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 Radiative Equilibrium Temperature being formatted.

Sun's reflection is blinding

I dedicate this work to the great mathematician and astronomer of the 20th century Milutin Milankovitch

The NASA planets surface temperatures measurements are all we have to work with.

And we believe in NASA measurements, because they are very precise and very professionally performed.

And I underline here again, we have to rely only on the NASA measurements. The NASA planets surface temperatures measurements are all we have to work with.

Resume

A Planet-Without-Atmosphere Effective Temperature Calculating Formula, the Te formula 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 formula and that is why it gives us very confusing results.

A planet-without-atmosphere effective temperature calculating formula

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 effective temperature. Planet is a celestial body with more major features when calculating planet effective temperature to consider.

Sunset in the sea

The planet-without-atmosphere effective temperature calculating formula has to include all the planet’s major properties and all the characteristic parameters.

3. The sidereal rotation spin N rotations/day

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

5. The planet 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 Effective Temperature.

Altogether these parameters are combined in a Planet-Without-Atmosphere Effective Temperature Complete Formula:

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

A Planet-Without-Atmosphere Effective Temperature Complete Formula produces very reasonable results:

Te.earth = 288,36 K, calculated with the Complete Formula, which is identical with the Tsat.mean.earth = 288 K, measured by satellites.

Te.moon = 221,74 K, calculated with the Complete Formula, which is almost the same with the Tsat.mean.moon = 220 K, measured by satellites.

Sun's reflection is mirroring

The Planet-Without-Atmosphere Effective Temperature Complete Formula gives us planet effective temperature values very close to the satellite measured planet mean temperatures.

We have collected the results now:

Comparison of results the planet Te calculated by the Incomplete Formula, the planet Te calculated by the Complete Formula, and the planet Tsat.mean measured by satellites:

Planet  Te.incompl. Te.compl. Tsat.mean

                formula    formula    measured

Mercury    437,30 K   346,11 K    340 K

Earth          255 K       288,36 K    288 K

Moon          271 Κ      221,74 Κ    220 Κ

Mars          209,91 K  213,21 K    210 K          

As you can see Te.complete.earth = 288,36 K.

That is why I say in the real world the 288 K - 255 K = Δ 33 oC difference does not exist.

Sun's reflection is when you see sun in the mirror, and it is blinding.

A more detailed analysis follows below

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.

 

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 higher planet surface 24-hours average temperatures.

The planet Earth’s and the planet Mars’ faster rotation is that what creates the necessary interaction for the incident on the planets surface solar energy much more efficient absorption.

 

It is all in the details...

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

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 a planet's surface has a new radiative equilibrium temperatures 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:

Assuming a planet with

Tsolar1 = 200 K, and Tdark1 = 100 K

Assuming this planet rotates faster,

so Tsolar2 = 199 K.

When rotating faster what is the planet's Tdark2 ?

J1emit.solar ~ (T1solar)⁴ ,

(200 K)⁴ = 1.600.000.000

J2emit.solar ~ (T2solar)⁴ ,

(199 K)⁴ = 1.568.000.000

 

J2emit.solar - J1emit.solar =

= 1.568.000.000- 1.600.000.000 =

= - 31.700.000

So we have ( - 31.700.000 ) less emitting on the solar side. It should be compensated by the increased emission on the dark side ( + 31.700.000 ).

On the other hand on the dark side we should have a greater warming than a one degree

( 199 K - 200 K = -1 oC ) cooling we had on the solar irradiated side.

 

J1emit.dark ~ (T1dark)⁴ ,

(100 K)⁴ = 100.000.000

J2emit.dark ~ (T2dark)⁴ ,

(107,126 K)⁴ = 131.698.114

J2emit.dark - J1emit.dark =

= 131.698.114 -100.000.000 =

= 31.698.114

The dark side's higher temperature to compensate the solar side's cooler emission by

( - 31.700.000 ) would be

T2dark = 107,126 K

As we see in this numerical example, when rotating faster maximum temperature on the solar irradiated side subsides from

200 K to 199 K.

On the other hand the minimum temperature on the dark side rises from

100 K to 107,126 K.

So when the solar irradiated side gets on average cooler by 1 degree oC, the dark side gets on average warmer by 7,126 degrees oC.

And as a result the planet total average temperature gets higher.

That is how when a planet rotating faster the radiative equilibrium temperatures are accomplished.

It happens so because when rotating faster a planet's surface has a new radiative equilibrium temperatures to achieve.

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

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.

.

The by a smooth spherical body solar irradiation absorption

 

Φ = 0,47

Φ factor explanation

Φ factor explanation

"Φ" is an important factor in the Planet's Effective Temperature Complete Formula:

Te = [ 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.

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

0,53 [ ( a ) 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) = 4σTe⁴ /(β*N*cp)¹∕ ⁴

And they are also cooperate in the Planet's Effective Temperature Complete Formula:

Te = [ 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*σΤe⁴ /(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 a 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

Te - is a Planet Effective 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 = 112,029 W/m2

Mars and Moon satellite measured mean temperatures comparison: 210 K and 220 K

Mars and Moon 

210 K and 220 K

Let’s compare then:

Moon: Tsat.moon = 220K

Moon’s albedo is amoon = 0,136

What is left to absorb is (1 – amoon) = (1- 0,136) = 0,864

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 /0,864 )¹∕ ⁴ = ( 0,8681 )¹∕ ⁴ = 0,9652

Conclusions:

1. Mars /Moon satellite measured temperatures comparison

( 0,9545 )

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

( 0,9652 )

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 = 1.362 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 Effective Temperature Complete Formula Te = [ Φ (1-a) So (1/R²) (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴

In the Planet's Effective Temperature Complete Formula the 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 rotations per day "N" should be considered in the Te planet complete formula in the fourth root :

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

.

Mercury / Moon / Mars satellite measured mean temperatures 340 K, 220 K and 210 K comparison

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 /58,646..1 /29,531..0,9747

a......0,088......0,136......0,250

 

Mercury:

Tsat.mean = 340 K

[ (1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ = [ 6,6769*(1/58,6460)¹∕ ⁴ ] ¹∕ ⁴ = 1,2463

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 = 1,2463 /0,8093 = 1,53997

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

They are almost identical, because Mercury and Moon have close (1-albedo) values: amoon = 0,136 and amercury = 0,088. For Moon (1-0,136)=0,864 and for Mercury (1-0,88)=0,912

 

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

If Moon and Mars had the same albedo amoon = 0,136 they would both have the same satellite measured mean temperature Tmean = 220 K. Mars' albedo is amars = 0,25.

.

Mercury / Mars satellite measured mean temperatures 340 K and 210 K comparison

 

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 /58,646..1 /29,531..0,9747

a......0,088......0,136......0,250

1-a…0,912……0,864…….0,75

Mercury:

Tsat.mean = 340 K

[ (1-a)*(1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ =

= [ 0,912*6,6769*(1/58,6460)¹∕ ⁴ ] ¹∕ ⁴ = 1,2179

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

= 1,2179 /0,7524 = 1,6187

And

Tmean.mercury /Tmean.mars =

= 340 K /210 K = 1,6190

Conclusion:

Everything is all right. Everything is based on observations.

And

The planet's rotations per day "N" should be considered in the Te complete formula in the fourth root:

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

.

Earth / Europa 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

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

= 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 the confirmation that the planet's rotations per day "N" should be considered in the Te planet complete formula in the fourth root:

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

.

Planet’s Mars Te misfortunate coincidence

We should mention here, that a smooth surface spherical body, as the planet Mars is, doesn’t accept and absorb all the solar radiation falling on the hemisphere.

Only the 0,47*S of the solar energy’s amount is accepted by the hemisphere. The rest 0,53*S is reflected back to space.

That is why Φ= 0,47 what is left for surface to absorb.

Now we have to say about the planet’s albedo "a". The planet’s albedo describes the dispersed on the surface secondary reflection to space fraction of the falling on the hemisphere solar light.

Thus a planet’s surface absorbs only the Φ*(1– a) fraction of the incident on the hemisphere solar energy. 

Comparison of results planet's Mars Tsat.mean.mars measured by satellites, and the planet's Mars Te calculated with Incomplete and Complete formula:

Planet..Tsat.mean…Te.incomplete…Te.complete

Mars ………210 K ……...209,91 K …….213,21 K

We have here planet's Mars mean temperature measured by satellites:

Tsat.mean.mars = 210 K

calculated by incomplete:

Te.mars.incompl = [ (1-a) S /4 σ ]¹∕ ⁴ = 209,91 K

and complete effective temperature formula:

Te.mars.compl = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ]¹∕ ⁴ = 213,21 K

The difference between Incomplete and Complete planet's Effective Temperature calculation formula is:

*[ Φ (β* N*cp)¹∕ ⁴ ]¹∕ ⁴

For planet Mars we have:

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

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

Mars is a 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 Fe2O3

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

(β* N*cp)¹∕ ⁴ = (150*0,9747*0,18)¹∕ ⁴ =

=(26,3169) ¹∕ ⁴ = 2,265 

Φ = 0,47

[ Φ (β* N*cp)¹∕ ⁴ ]¹∕ ⁴ =

=( 0,47* 2,265 )¹∕ ⁴ =

=( 1,06455)¹∕ ⁴ = 1,01576 

So the difference between these two formulas for planet Mars is only 1,576 % !

And it is a coincident.

It is a coincident, but with very important consequences.

Let's explain:

Tsat.mean.mars = 210 K measured by satellites.

is almost equal with Te.mars.incomplete = 209,91 K

and Te.mars.compl = 213,21 K.

When measuring by satellites the

Tsat.mean.mars = 210 K

and calculating with incomplete effective temperature formula

Te.mars.incomplete = 209,91 K

scientist were led to mistaken conclusions.

(They didn’t know about the Planet Effective Temperature Complete Formula yet).

First they concluded that the planet's effective and mean temperatures should normally be equal, which is right.

Secondly they concluded that Earth without atmosphere should have effective temperature, according to the incomplete formula calculation,

Te.earth.incomplete = 255 K.

The measured by satellites Tsat.mean.earth = 288 K.

The difference of Δ 33 oC was, according to scientists, due to the Earth’s atmosphere greenhouse warming effect.

 

Now we have the effective temperature complete formula that gives

Te.mars.compl = 213,21 K.

This result is almost identical to the measured by satellites

Tsat.mean.mars = 210 K.

And the Complete Formula gives very reasonable results for all the other planets without-atmosphere in the solar system.

We know now the Δ 33 oC does not exist.

The Te.earth = 288,36 K = Tsat.mean.earth = 288 K.

Te.earth = 255 K does not exist.

And I dare to assume now, that Complete Formula may be applied to all the planets without atmosphere in the whole Universe.

And I am convinced there is not any measurable greenhouse effect in Earth's atmosphere. Earth's atmosphere is very thin to have a detectable greenhouse effect.

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The Original Milankovitch Cycle

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.

The Reversed Milankovitch Cycle

Milankovitch’s main idea was that the glacial periods are ruled by planet’s movements forcing. At the right we have the Reversed Milankovitch cycle. The minimums in the reversed Milankovitch cycle are the maximums in the original. These two cycles, the original Milankovitch cycle and the reversed differ in time only by a half of a year. According to the reversed Milankovitch cycle there are long and very deep glacial periods and small and very short interglacial. The reversed cycle complies with the paleo geological findings. As we can see in the reversed Milankovitch cycle, we are getting now to the end of a long and a slow warming period. What we are witnessing as a Global Climate Change are the culmination moments at the end of that warming period.

Arctica solar irradiation intensity