The Planet Surface Rotational Warming Phenomenon

The Planet Mean Surface Temperature Equation Tmean = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴

Mercury in color

Φ - dimensionless solar irradiation accepting factor

Φ - factor is proven by the observed quantities

 Φ - factor is proven by the observed quantities

I know what was missing in my syllogisms. There was a need of observed quantities.

I have to offer observed numbers to prove that a planet (a sphere) follows the basic law of the parallel solar rays energy reflection /absorption ratio.

The observed numbers should be very convincing.

We have chosen Mercury for its very low albedo a=0,068 and for its very slow rotational spin N = 1/175,938 rotations/day.

Mercury is most suitable for the blackbody effective temperature equation definition - a not rotating planet, or very slow rotating. Also it is a planet where albedo (a=0,068) plays little role in planet's energy budget.

These (Tmean, R, N, and albedo) parameters of the planets are all satellite measured. These parameters of the planets are all observations.


Tsat.mean.340 K….220 K…210 K

R…......0,387 AU....1 AU...1,525 AU


N…1 /175,938..1 /29,531..0,9747




Let’s calculate the Mercury’s effective temperature with the old blackbody equation:

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

We have

(1-a) = 0,932

1/R² = 6,6769

So = 1.361 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.mercury = [ 0,932* 1.361 W/m² * 6,6769 /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =

Te.mercury = ( 37.369.999.608,40 )¹∕ ⁴ =

439,67 K Te.mercury = 439,67 K = 440 K

And we compare it with the Tsat.mean.mercury = 340 K - the satellite measured Mercury’s mean surface temperature

Let's analyze what we have here.

Because of Mercury having a very low albedo (a = 0,68) the what is left to absorb for Mercury's surface is (1-a) = 0,932 or 93,2 % of the incident solar flux on the Mercury (according to the theory of planets absorbing the incoming parallel solar rays as a cross section disk).

So it was thought that the 93,2 % is absorbed.

Te.mercury = 440 K

Tsat.mean.mercury = 340 K


Let's calculate the Mercury's corrected effective temperature by inserting in the above equation the Φ = 0,47

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

Te.mercury.correct = [ 0,47*0,932* 1.361 W/m² * 6,6769 /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =

Te.mercury.correct = ( 17.536.114.624,63 )¹∕ ⁴ = 363,9 K

Te.mercury.correct = 363,9 K = 364 K


Let’s put these temperatures together:

Te.mercury = 440 K

Te.mercury.correct = 364 K

Tsat.mean.mercury = 340 K


The difference is 440 K - 364 K = 76°C

This difference is due to the absence of the Φ(1-a) coupled term.

And Te.mercury.correct = 364 K is very much closer to Tsat.mean.mercury = 340 K measured by satellite, and it is due to the Φ = 0,47 for the smooth surface planets.

The answer is simple – it happens because the old equation (Te.mercury = 440 K) 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 a

[1 - Φ(1-a)]S portion of the incident on it's surface solar irradiation.

Imagine a completely black smooth surface planet; imagine a completely invisible planet, a planet with a smooth surface and with a zero albedo. This planet still reflects [1 - Φ(1-a)]S portion of the incident on its surface solar irradiation.

The satellite measurements have confirmed it.



The Planet's Surface Mean Temperature Equation produces remarkable results. The calculated planets’ temperatures are almost identical with the measured by satellites.


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.