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- EARTH effective 210K
- Pluto's Tmean 44 K
- Io's Tmean 110 K
- Jupiter Tmean 165 K
- Earth/Mars 288K/210K
- ALL PLANETS TEMPERAT
- AN IMPORTANT THEOREM
- ANSWERS TO OPPONENTS
- DISCUSSION
- More

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

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 = 1.361 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.

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.

We are ready now to make two very important observations.

1. Moon and Mars Moon's satellite measured Tsat.mean.moon = 220 K Mars' satellite measured Tsat.mean.mars = 210 K

These two observed temperatures on the different planets (Mars and Moon) are very close.

The solar flux on Moon is So = 1.361 W/m². The solar flux on Mars is S.mars = 586,4 W/m².

Thus we observe here that there can be planets with different solar irradiation fluxes, and yet the planets may have (for equal albedo) the same mean surface temperatures.

So we may have:

Many planets with different solar irradiation fluxes, and yet the planets may have (for equal albedo) the same mean surface temperatures.

Conclusion:

Many different solar fluxes (for equal albedo) can create the same mean surface temperatures.

2. Moon and Earth Moon's satellite measured Tsat.mean.moon = 220 K Earth's satellite measured Tsat.mean.earth = 288 K

These two observed temperatures on the different planets (Moon and Earth) are very different.

The solar flux on Moon is So = 1.361 W/m². The solar flux on Earth is So = 1.361 W/m².

Thus we observe here that there can be planets with the same solar irradiation fluxes, and yet the planets may have (for equal albedo) very different mean surface temperatures.

So we may have:

Many planets with the same solar irradiation fluxes, and yet the planets may have (for equal albedo) different mean surface temperatures.

Conclusion:

Many different global temperature distributions (for equal albedo) may balance the same solar flux.

These two very important observations - conclusions lead us to the formulation of the Mean Surface Temperature Equation.

A Planet Without-Atmosphere Mean Surface 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 = 1.361 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 INTERACTING-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 270,4 Κ 223,35 Κ 220 Κ

Mars 209,91 K 213,21K 210 K

Planet Energy Budget:

Solar energy INTERACTING (not reflected) with a Hemisphere surface with radius "r" after reflection (diffuse and specular)

Jnot.reflected = 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 Jnot.reflected = 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

Jnot.reflected = Jemit per m² planet surface:

Jnot.reflected = Jemit

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

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

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 INTERACTING-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 = 1.361 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 satellite measured Tsat.mean.earth = 288 K.

So, we can confirm now with great confidence, that a Planet or Moon Without-Atmosphere Mean Surface 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 270,4 Κ 223,35Κ 220 Κ

Mars 209,91 K 213,21 K 210 K

These data, the calculated by a Planet Without-Atmosphere Mean Surface 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 Mean Surface 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 mean surface temperature equation.

Planet is a celestial body with more major features when calculating planet mean surface temperature to consider. The planet without-atmosphere mean surface 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 Mean SurfaceTemperature 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.

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

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 NOT EMITTED 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 iNTERACTING-Emitting Universal Law:

Jnot.reflected =Φ*S*(1-a)/4=

=Jemit=σΤmean⁴/(β*N*cp)¹∕ ⁴ (W/m²)

And only then, solving for Tmean we obtain the Planet Mean Surface 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 = 223,35 K,

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

Tsat.mean.moon = 220 K,

measured by satellites.

The calculated planets' temperatures were confirmed by space crafts’ measurements.

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

Jnot.reflected=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 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.

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 INTERACTS so fast, surface EMITS and ACCUMULATES at the same instant.

Tmean.earth

So = 1.361 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 INTERACTING-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 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 = ( 6.854.897.370,96 )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……270.4 Κ…...223,35 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.

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