Mercury in color
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.
Planet….Mercury….Moon….Mars
Tsat.mean.340 K….220 K…210 K
R…......0,387 AU....1 AU...1,525 AU
1/R²…..6.6769….....1….…0,430
N…1 /175,938..1 /29,531..0,9747
a.........0,068.........0,11......0,250
1-a…...0,932….....0,89......0,75
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.
Link:
What factor is NOT part of the effective temperature formula?
What factor is NOT part of the effective temperature formula that so dramatically affects the actual temperature of the moon?
Why is the actual mean temperature of the moon so much lower than the effective temperature? NASA lists the effective temperature of the moon at 270.6 Kelvin. The mean temperature of the moon at the equator is 220 Kelvin.
With no atmospheric effects, why is the surface temperature so much lower than the effective temperature predicts? What factor is NOT part of the effective temperature formula that so dramatically affects the actual temperature of the moon?
I'll tell you what it is:
It is the Φ -the planet solar irradiation accepting factor. For smooth surface Moon Φ= 0,47.
Te.correct.moon = [ Φ (1-a) So /4σ ]¹∕ ⁴
Te.correct.moon = [ 0,47 (1-0,11) 1.362 W/m² /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =
Te.correct.moon = [ 0,47 (0,89) 1.362 W/m² /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =
Te.correct.moon =
[ 2.510.168.871,25 ]¹∕ ⁴ =
Te.correct.moon = 223,83 Κ
This simple example clearly demonstrates the CORRECTNESS of the Φ -the planet solar irradiation accepting factor. For smooth surface planets, like Moon, Φ= 0,47.
Conclusion:
From now on, for every smooth surface planet and moon, we should take in consideration instead of the planet blackbody effective temperature Te , the corrected VALUES of the planet blackbody effective temperature - the Te.corrected.
Table of results for Te and Te.corrected compared to Tsat and to Rotations/day for smooth surface planets and moons with Φ=0,47
Planet…….. Te.... Te.corrected…..Tsat…Rot/day
Mercury…..440 K…….364 K…...340 K…0,00568
Moon……….270 K…….224 K…...220 K.…0,0339
Earth………255 K…….210 K…...288 K..….1
Mars……….210 K…….174 K…..210 K..…0,9747
Europa…….95,2 K…...78,8 K….102 K...0,2816
Ganymede..107,1 K.....88,6 K…110 K….0,1398
Notice:
The number 0,47 for smooth surface in a parallel fluid flow is taken from the well measured and long ago known Drag Coefficient Data, where Cd =0,47 is for sphere. It is the portion of incident on sphere energy which should be resisted by sphere to remain in balance.
Conclusions:
The Planet's Surface Mean Temperature Equation produces remarkable results. The calculated planets’ temperatures are almost identical with the measured by satellites.
Planet..…Te.incomplete…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.
Φ - dimensionless solar irradiation accepting factor
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To formally prove Φ -Factor's correctness in the Ein = Eout formula.
The Energy in:
Ein = (1-a)S W/m²
used in the blackbody planet effective temperature Te is an empirical assertion, which is not based on any theoretical research, not to say, its correctness has not been demonstrated, quite the opposite…
–
The Energy in:
Ein = Φ(1-a)S W/m²
is based on measurements (the Drag Coefficient for smooth spheres in a parallel fluid flow Cd = 0,47), and it is demonstrated to be the correct one.