### 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. **

**What is the planet's Tdark2 then ?**

**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 higher temperature to compensate the solar side 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.**

.

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

**Mars and Moon satellite measured mean temperatures comparison:**

**210 K and 220 K**

**Let's see what we have here:**

**Planet Tsat.mean**

** measured**

**Mercury 340 K**

**Earth 288 K**

** Moon 220 Κ **

**Mars 210 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 differencies 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.**

**Let’s take out the calculator now and make simple calculations:**

**The rotation difference fourth root**

** (28,783)¹∕ ⁴ = 2,316**

**The irradiation /rotation comparison**

** 2,32 /(28,783)¹∕ ⁴ = 2,32 /2,316 = 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 stronger solar irradiaton.**

**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.**

.

### A 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 entire planet:**

** Jemit = A*σΤmean⁴ /(β*N*cp)¹∕ ⁴ (W) **

**Φ - is a dimensionless Solar Irradiation accepting factor**

** (1-Φ + Φ*a) - 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 surface (m²)**

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

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

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

** Jemit = 4πr²σTmean⁴ /(β*N*cp)¹∕ ⁴ (W) **

**global Jabs = global Jemit**

** Φ*πr²S (1-a) = 4πr²σTmean⁴ /(β*N*cp)¹∕ ⁴**

** Or after eliminating πr² **

**Φ*S*(1-a) = 4σTmean⁴ /(β*N*cp)¹∕ ⁴**

** The planet average**

** Jabs = Jemit per m² planet surface:**

** Jabs = Jemit**

** Φ*S*(1-a) /4 = σTmean⁴ /(β*N*cp)¹∕ ⁴ (W/m²) **

**Solving for Tmean we obtain the planet's surface Mean temperature:**

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

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

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

** cp – is the planet surface specific heat**

** 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. **

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

** 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 = 0,306 and a factor Φ = 0,47 we have:**

** Tmean.earth = 287,74 K or 15°C.**

** This temperature is confirmed by the satellite measured Tsat.mean.earth = 288 K.**

.

### A Planet Without-Atmosphere Surface Mean Temperature Equation

**A Planet Without-Atmosphere Surface Mean Temperature Equation derives from the incomplete Te equation which is based on the radiative equilibrium and on the Stefan-Boltzmann Law.**

** from the incomplete**

** Te = [ (1-a) S / 4 σ ]¹∕ ⁴**

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

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

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

** to the Planet Without-Atmosphere Surface Mean Temperature Equation**

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

** (.......)¹∕ ⁴ is the fourth root**

** S = So(1/R²), where R is the average distance from the sun in AU (astronomical units)**

** S - is the solar flux W/m²**

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

** Planet’s albedo: a**

** Φ - is the dimensionless solar irradiation spherical surface accepting factor**

** Accepted by a Hemisphere with radius r sunlight is S*Φ*π*r²(1-a), where Φ = 0,47 for smooth surface planets, like Earth, Moon, Mercury and Mars…**

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

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

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

** cp – is the planet surface specific heat**

** 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.**

** 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**

** This Universal Formula (1) is the instrument for calculating a Planet-Without-Atmosphere Surface Mean Temperature. **

**The results we get from these calculations are almost identical with those measured by satellites.**

* Planet Te.incompl Tmean Tsat.mean*

* Mercury 439,6 K 325,83 K 340 Κ*

* Earth 255 K 287,74 K 288 K*

* Moon 271 Κ 221,74 Κ 220 Κ*

* Mars 209,91 K 213,21K **210 K*

.

### We can confirm now with great confidence

**So, we can confirm now with great confidence, that a Planet or Moon Without-Atmosphere Surface Mean Temperature Equation, according to the Stefan-Boltzmann Law, is:**

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

**We have collected the results now:**

** Comparison of results**

** the planet's Te calculated by the Incomplete Equation,**

** the planet's surface Tmean calculated by the Equation,**

** and the planet Tsat.mean measured by satellites:**

** Te. incompl Tmean Tsat.mean**

* Mercury 439,6 K 325,83 K 340 Κ*

* Earth 255 K 287,74 K 288 K*

* Moon 271 Κ 221,74Κ 220 Κ*

* Mars 209,91 K 213,21 K 210 K*

** These data, the calculated by a Planet Without-Atmosphere Surface Mean Temperature Equation and the measured by satellites are almost the same, very much alike.**

** They are almost identical, within limits, which makes us conclude that the Planet's Without-Atmosphere Surface Mean Temperature Equation**

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

** can calculate the planets' mean temperatures.**

** It is a situation that happens once in a lifetime in science. Although the evidences existed, were measured and remained isolated information so far.**

** It was not obvious one could combine the evidences in order to calculate the planet’s surface mean temperature.**

** 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 more major parameters for the planet's surface temperature equation.**

** Planet is a celestial body with more major features when calculating planet effective temperature to consider. The planet without-atmosphere effective temperature calculating formula has to include all the planet’s basic 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 solar irradiation absorbing property ).**

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

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

** A Planet Without-Atmosphere Surface Mean 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 = 221,74 K,**

** calculated with the Equation, which is almost the same with the**

** Tsat.mean.moon = 220 K,**

** measured by satellites.**

** A Planet Without-Atmosphere Surface Mean Temperature Equation gives us a planet surface mean temperature values very close to the satellite measured planet mean temperatures.**

** It is a Stefan-Boltzmann Law Triumph! And it is a Milankovitch Cycle coming back! And as for NASA, all these new discoveries were possible only due to NASA satellites planet temperatures precise measurements!**

.

### The Fast Rotating Planet Earth

**So far we came to the end of this presentation.**

** Its topic was to present the Planet Without-Atmosphere Surface Mean Temperature Equation:**

** Tmean.planet = [ 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 = σΤe⁴/(β*N*cp)¹∕ ⁴ (W/m²)**

** Here the (β*N*cp)¹∕ ⁴ is a dimensionless Rotating Planet Surface Solar Irradiation Warming Ability.**

** Φ - is the dimensionless solar irradiation spherical surface accepting factor.**

** Accepted by a Hemisphere with radius r sunlight is S*Φ*π*r²(1-a), where Φ = 0,47 for smooth surface planets, like Earth, Moon, Mercury and Mars…**

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

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

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

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

** 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 surface mean temperature equation**

** Tmean.earth = 287,74 K.**

** These physical phenomena distracting Earth from the instant emitting back to space are the Earth’s rotation around its axis and the Earth’s surface specific heat.**

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

**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*a) *S *π r² = **

**= Jsw.reflected = 0,53 *S *π r²**

** Only the 0,47*So of the solar energy’s amount is accepted by the hemisphere. **

**The rest 0,53*So 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. **

**That is why we have the**

** Φ (1-a) So (1/R²)**

** expression in the surface mean temperature equatio. **

.

### Observations resulting in Equation- the perfect fitting !

**Observations –**

**the planets’ measured temperatures,**

**the planets’ surface specific heat cp,**

**the planets’ sidereal rotation period,**

**the distance from the sun,**

**the measured by space crafts planets’ albedo,**

**the planets’ smooth or heavy cratered surface.**

**The discovery of the “The faster a planet rotates (n2>n1) the higher is the planet’s average temperature:**

**Tmin ↑↑→T↑mean←T↓max,**

**because Tmin grows faster”.**

**The understanding that a planet’s surface does not behave as a blackbody surface and it does not emit as a blackbody.**

**The understanding that:**

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

**And**

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

**All these observations together led to the discovery of the Rotating Planet Spherical Surface Solar Irradiation Absorbing-Emitting Universal Law:**

**Jabs=Φ*S*(1-a)/4=**

**=Jemit=σΤmean⁴/(β*N*cp)¹∕ ⁴ (W/m²)**

**And only then, solving for Tmean we obtain the Planet's Surface Mean Temperature Equation:**

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

**and then we calculate**

**Tmean.earth = 287,74 K,**

**calculated with the Equation, which is identical with the**

**Tsat.mean.earth = 288 K,**

**measured by satellites.**

**Tmean.moon = 221,74 K,**

**calculated with the Equation, which is almost the same with the**

**Tsat.mean.moon = 220 K,**

**measured by satellites.**

**The calculated planet’s temperatures were confirmed by space crafts’ measurements.**

**It is the Rotating Planet Spherical Surface Solar Irradiation Absorbing-Emitting Universal Law:**

**Jabs=Φ*S*(1-a)/4=**

**=Jemit=σΤmean⁴/(β*N*cp)¹∕ ⁴ (W/m²)**

**confirmation.**

**There is a lot of physics here to consider.**

**It is a new Universal Law. That is why it fits in observations.**

**A Universal Law has to fit in observations.**

**A Universal Law has the ability to describe the observations.**

**And has the ability to explain the observations.**

**A new Universal Law becomes then a powerful instrument for the further scientific research.**

.

### The 0,53*S solar irradiation reflected part slides off the sphere as the air flow slides off the aerodynamic car

**Lambert’s cosine law says that the power transmitted to a surface varies with the cosine of the zenith angle.**

**So Lambert’s cosine law says that the power transmitted to a surface, not the power absorbed by the surface.**

** Now, there is also the actual reflection from the sphere**

** 0,53 *S.**

** Well, let’s imagine a sphere with an albedo**

** a =0.**

** A sphere painted black will still reflect**

** 0,53*S**

**(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 **

** What is left (to absorb) to transform in LW outgoing radiation for the planet then is**

** Φ(1 – a)S,**

** where**

** Φ = 1 - 0,53 = 0,47.**

** This 0,53*S reflection part cannot be seen and measured from a spacecraft.**

** It is not diffused on the surface and it is not seen from the perpendicular to the radiant propagation direction.**

** The 0,53*S solar irradiation reflected part slides off the sphere as the air flow slides off the aerodynamic car. **

.

### The surface responses so fast, we can skip the absorption phase and go straight to the emission

It is common to believe that the irradiated surface first absorbs and warms and only then emits according to the Stefan-Boltzmann Law

** No, it is not like this. It does not happen like this.**

Conduction and convection are very slow energy transfer processes.

The radiation is very fast, it “works” at the very instant.

When irradiated the surface responses at the very instant.

It does not absorb first, rise the temperature and then emit.

** The surface responses so fast, we can skip the absorption phase and go straight to the emission.**

Surface emission’s intensity determines surface temperature rise.

The temperature rise creates a higher temperature gradient with the soil or the water.

It is then when the absorption starts.

There is a very small amount energy absorbed and it depends on the surface properties.

.

### Earth's Without-Atmosphere Surface Mean Temperature Equation

**Tmean.earth**

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

Earth’s albedo: aearth = 0,306

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

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

N = 1 rotation /per day, is Earth’s 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

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

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

Τmean.earth = [ 0,47(1-0,30)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 =

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

Τmean.earth = ( 6.854.905.906,50 )¹∕ ⁴ = 287,74 K

** Tmean.earth = 287,74 Κ**

And we compare it with the

** Tsat.mean.earth = 288 K,** measured by satellites.

These two temperatures, the calculated one, and the measured by satellites are almost identical.

Conclusions:

The equation produces remarkable results.

The calculated planets’ temperatures are almost identical with the measured by satellites.

*Planet...Te. incompl.... Tmean..Tsat.mean*

*Mercury…440 K…….325,83 K……..340 K*

*Earth…….255 K……….. 287,74 K……..288 K*

*Moon……271 Κ………..223,39 Κ……..220 Κ*

*Mars….209,91 K……..213,21 K……..210 K*

** The 288 K – 255 K = 33 oC difference does not exist in the real world.**

There are only traces of greenhouse gasses.

The Earth’s atmosphere is very thin. There is not any measurable Greenhouse Gasses Warming effect on the Earth’s surface.

.