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There are three known ways matter gets rid of heat.

At the Top of earth's Atmosphere sun CONSTANTLY "delivers" 1.362 W/m² Short Way (SW) radiative Electro-Magnetic Solar Energy.

What Earth does is on the instant to reflect a portion of the incoming SW to the outer space.

Another portion is at the same instant transformed  from SW into LW (IR) outgoing radiation and instantly emitted also to the outer space.

And a fraction of those W/m² on instant is transformed into heat...

Thus, a very formidable part of the incident solar energy is at the very instant it hits surface immediately gets rid off to space (as a SW reflected and IR emitted EM energy).

The fraction of EM energy transformed to heat is the energy accumulated in the inner layers.

Only after being accumulated in the inner layers heat "belongs" to the planet surface matter.

What happens to the heat next... Heat is a form of energy, the MATTER has to get rid of the heat...

There are three known ways matter gets rid of heat.

1). The IR emission - it is the most effective, especially when matter is warmed at high temperatures.

2). Convection - let's say an oceanic current moving North, it IR emits, and also mixes with colder waters, thus mitigating the ability to IR emission's fast from the water the accumulated heat removal.

3). Conduction - a very slow process, that is why there is the thermocline is present at the equatorial waters.

Eventually all the accumulated heat will be re-radiated. What affects differences in the heat removal delay (storage ?) is the incident radiation /matter interaction process.

Thus we observe the latitudinal differences, the seasonal differences. the decadal and the millennial cycles' differences.

When comparing different planets with a single similar factor, the rest of factors affect the differences.

Moon and Earth are at the same distance from Sun.

Earth's higher rotational spin and Earth's higher the average surface specific heat make Earth surface on average a warmer planet.

Also Earth has atmosphere we live in. It also takes a small part in the heat accumulation - heat removal process.

But Earth's atmosphere is very thin, it does not warm earthen surface by +33C.

(Thus, the not reflected portion of incident SW solar flux when considered in TOTAL - is EQUAL to the planet TOTAL amount of EM IR outgoing emission energy).

The above is obvious now, because the otgoing IR emission is not distributed evenly over the entire planet surface area. 

Also we realize the over the planet surface area distribution of outgoing IR emission is a function of planet spin and the planet surface specific heat.

CONCLUSION: ...because it is not being entirely "captured" by the surface as HEAT.

Below is a very interesting dialog I had with opponent.

While disputing the things become clearer. New approaches and new sentences suddenly emerge in the dialog, like this very important insight:

"Not the entire non-reflected portion should be considered as heat, since it never been conducted in the inner layers."

And

"That is why we should not average flux over entire planet surface by dividing by 4.

Because almost all the incident EM energy gets out the very moment of the incidence on the every spot, no matter how fast planet rotates.

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Opponent:

“It is not possible to get a mean surface temperature of 288 from your calculated surface emission.

When you say 288 mean it would mean that the emission would fluctuate close to 390 Watts/m².

Some areas would have more than this emission, some less, but the averaged emission would have to be around this value.”

Me:

The issue we discuss is very important. What I suggest is to compare Moon mean surface temperature with the Moon effective temperature.

Then please comment what you think about the differences.

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Me:

The “Effective Temperature” removes any effect of spin. It is the temperature a body would reach if input energy were uniformly spread over the entire surface.

It negates the effect of spin. Slow spinning can result in surface temperatures below the effective temperature but not above.

The Solar energy on the sunlit side of the Moon at first increases the lunar surface, but only up to the point the lunar surface emits away all the energy it receives, no more temperature increase (at first, when cool it emits far less than it receives so the energy can be used to increase temperature).

Once the lunar surface is at steady state temperature with the input solar, all the solar energy that could be used to heat the surface is now just radiated away.

If the Moon span faster than lots more energy would be used to increase the surface temperature. But the rate of spin will not exceed the temperature of the effective temperature which is what you want to demonstrate.

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Me:

Opponent

“The Effective Temperature removes any effect of spin. It is the temperature a body would reach if input energy were uniformly spread over the entire surface. It negates the effect of spin.

Slow spinning can result in surface temperatures below the effective temperature but not above.”

Effective temperature is a mathematical abstraction. Every planet has a calculated effective temperature no matter how fast planet rotates.

Why should we consider the planet effective temperature to be the highest mean surface temperature a real planet mean surface temperature can theoretically achieve?

The Stefan-Boltzmann emission law states:

Jemit = σ*T⁴ W/m²

where T is a uniform surface emission temperature.

The amount of energy emitted from a surface with A m² area is:

energy out = A (m²) * σ*T⁴ W/m² = A * σ*T⁴ W

or

energy out = A * Jemit

The Stefan-Boltzmann blackbody surface already has the uniform temperature T, because it is uniformly warmed to that T temperature.

Planet surface cannot be considered as a blackbody uniformly warmed surface.

The incident solar flux cannot be averaged over the planet surface.

Flux is not heat… The fact that planet receives a solar flux, does not mean its energy first warms planet surface and only then the warmed surface emits the same incident amount of solar energy as IR EM emission.

Only a fraction of the not reflected portion is transformed into HEAT and is accumulated in the inner layers.

CONCLUSION:

When realizing it, we come to the conclusion that incident not reflected portion of solar flux cannot be averaged over the entire planet surface... because it is not being entirely "captured" by the surface as HEAT.

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Opponent:

YOU:

“The fact that planet receives a solar flux, does not mean its energy first warms planet surface and only then the warmed surface emits the same incident amount of solar energy as IR EM emission.”

Yes that is basically what happens. The Solar Flux is ABS by surface and converted to heat which increases the molecular vibrations of the the surface.

Some of the energy can be conducted to lower layers of the surface but the greater vibrations of the molecules in the surface layer will emit more IR and will reach a point that they are emitting the same amount of energy in the IR bands and they are receiving in the Visible spectrum from the Sun.

Not sure why you do not accept this as correct.

Do you have good material to demonstrate this is not what takes place?

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Me:

Opponent:

“Not sure why you do not accept this as correct. Do you have good material to demonstrate this is not what takes place?”

I have observed it for many years happening… When I understood its exact meaning I sat and started the research.

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Me:

I think it is almost the same.

What I think is the solar irradiation interacts with matter on the very instant it hits the surface.

The following happen:

1). Partial reflection (specular and diffuse).

2). Partial transformation to other frequencies (IR) and IR emission, as a kind of “instant reflection like process”.

3). A  fraction TRANSFORMED IN HEAT is conducted in the inner layers in form of heat.

Not the entire non-reflected portion should be considered as heat, since it never been conducted in the inner layers.

Opponent:

“Some of the energy can be conducted to lower layers of the surface but the greater vibrations of the molecules in the surface layer will emit more IR and will reach a point that they are emitting the same amount of energy in the IR bands and they are receiving in the Visible spectrum from the Sun.”

It is almost the same, as me. That is why we should not average flux over entire planet surface by dividing by 4.

Because the most part of the incident EM energy gets out the very moment of the incidence on the every spot...

The amount of EM energy which gets out the very moment of the incidence on the every infinitesimal spot depends on the planet rotational spin and on planet surface specific heat.

That is why the amount of accumulated by the innere layers heat is also a function of planet rotational spin and planet surface specific heat.

The faster a planet rotates and the higher is planet surface specific heat, the warmer on average surface the planet is.

The planet simple blackbody Effective Temperature equation Te DEFINITION. There is not any spherical correction factor four (4) in the first place.

Let's continue the efforts to further developing and explaining the New Equation!

Let's have another look to the planet simple blackbody Effective Temperature equation Te DEFINITION:

πr²S*(1-a) = 4πr²σTe⁴ (W)

Let's look close to it:

On the left side there is the SW not reflected portion of the incident solar flux's energy in TOTAL in Watts.

πr²*S*(1-a) (W)

On the right side there is the LW radiative energy a planet radiates over its entire surface area in TOTAL in Watts.

4πr²σTe⁴ (W)

This two amounts of energy (the SW incident and the LW emitted) are equal only energetically-wise, both amounts are expressed in Watts, so we can write the PLANET RADIATIVE ENERGY BALLANCE equation:

energy in = energy out

or

πr²*S*(1-a) = 4πr²σTe⁴ (W)

and, solving for Te we obtain:

Te = [ (1-a) S /4σ ]¹∕ ⁴

Thus, what we do is to solve algebraic equation.

There is not any spherical correction factor four (4) in the first place.

I am not dividing the incident not reflected solar energy by four. The SW incident and the LW emitted are equal amounts of EM energy only energetically-wise.

Thank you, for your very much appreciated help to further developing and explaining the New Equation!

Thank you again for your passionate participation in that, not always easy, task to understand what exactly it is this about Your idiotic equation as you characterized it above...

Yes, I had to explain it more meticulously, thank you, it is a New equation, and it needs us to be focused on the every tiniest detail.

Everything should be thoroughly explained to the very end.

Let's proceed now with the:

“I am not dividing the incident not reflected solar energy by four.”

"Yeah, you are. From your website: ..."

Please, again, let's focus on it:

πr²Φ*S*(1-a) = 4πr²σTmean⁴ /(β*N*cp)¹∕ ⁴ (W)

"See what you did. On the left side, you multiply the solar flux by the area of a circle to get the watts. BTW that is correct because the sun radiates to the surface of a circle.

On the right side, you multiply the S-B equation by the area of a sphere to get the watts. That’s correct as well because a planet radiates over its entire surface area."

Let's look close to it:

On the left side there is the SW not reflected portion of the incident solar flux's energy in TOTAL in Watts.

πr²Φ*S*(1-a) (W)

On the right side there is the LW radiative energy a planet radiates over its entire surface area in TOTAL in Watts.

4πr²σTmean⁴ /(β*N*cp)¹∕ ⁴ (W)

This two amounts of energy (the SW incident and the LW emitted) are equal only energetically-wise, both amounts are expressed in Watts, so we can write the equation:

energy in = energy out

or

πr²Φ*S*(1-a) = 4πr²σTmean⁴ /(β*N*cp)¹∕ ⁴ (W)

and, solving for Tmean we obtain:

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

Thus, what we do is to solve algebraic equation.

There is not any spherical correction factor four (4) here.