# 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 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 or         Tsat.mean

moon               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 = 0,136

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 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

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*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 surface (m²)

Te - is a Planet Effective 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 = 4πr²σTe⁴ /(β*N*cp)¹∕ ⁴ (W)

global Jabs = global Jemit

Φ*πr²S (1-a) = 4πr²σTe⁴ /(β*N*cp)¹∕ ⁴

Or after eliminating πr²

Φ*S*(1-a) = 4σTe⁴ /(β*N*cp)¹∕ ⁴

The planet average

Jabs = Jemit per m² planet surface:

Jabs = Jemit

Φ*S*(1-a) /4 = σTe⁴ /(β*N*cp)¹∕ ⁴ (W/m²)

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 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 = 0,3 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.

### A Planet Without-Atmosphere Effective Temperature Complete Formula

A Planet Without-Atmosphere Effective Temperature Complete Formula derives from the incomplete Te formula 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 formula and that is why it gives us very confusing results.

A Planet 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.

to the complete

Te = [ 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 Effective Temperature.

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

Planet  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        221,74 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 Effective Temperature Complete Formula, according to the Stefan-Boltzmann Law, is:

Te.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 Te calculated by the Incomplete Formula,

the planet Te calculated by the Complete Formula,

and the planet Tsat.mean measured by satellites:

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

These data, the calculated with a Planet Without-Atmosphere Effective Temperature Complete Formula 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 Without-Atmosphere Effective Temperature Complete Formula

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

can calculate a planet 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 temperature. 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 more major parameters for the planet effective temperature calculating formula.

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 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 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.

A Planet Without-Atmosphere Effective Temperature Complete Formula gives us a planet effective 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 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)

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

Tmean.earth = 288 K to be the same as the calculated by the effective temperature complete formula

Te.earth = 288,36 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.

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 complete effective temperature formula.

### Observations resulting in Formula - 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 discovery of the Φ=0,47

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=σΤe⁴/(β*N*cp)¹∕ ⁴ (W/m²)

And only then, solving for Te we obtain the Planet Effective Temperature Complete Formula:

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

and then we calculate

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.

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=σΤe⁴/(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.

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 Effective Temperature (radiative emission temperature) calculation formula

Te.earth

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

Earth’s albedo: aearth = 0,30

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 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 = 288,36 K

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.

Conclusions:

The complete formula produces remarkable results.

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

Planet..Te. incompl..Te.compll..Tsat.mean

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

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.