Planets' term (β*N*cp)¹∕ ⁴ comparison
What I have found is that the real planet does not behave as a blackbody.
1. Blackbody planet the major assumption is that the incident radiation is distributed evenly on the entire blackbody planet surface.
Which corresponds to the conclusion, that rotates or not, the Te would be the same.
And it is a wrong assumption, that the incident radiation is distributed evenly on the entire blackbody planet surface, because it assumes that the blackbody has shape, or the blackbody, which is not evenly irradiated, but only evenly distributes the blackbody itself the incident radiation over the entire surface.
Because the radiation cannot go and get distributed spontaneously on the entire sphere, just because we say so.
What kind of special radiation it is? The sun irradiates the blackbody planet, and on the very instant the radiation is evenly distributed on the entire surface... And is distributed on the dark side hemisphere also - but how?
It is not the blackbody' property whatsoever.
And it is not the solar flux's property to go distributing around the globe... and evenly... Light doesn't travel around the globe.
What a special kind of a blackbody it is?
It is a perfect world they describe, they wish to live in...
2. The real planet the faster it rotates, the warmer its surface is. Also, the real planet's surface never will have a uniform temperature, no matter how fast it rotates.
Because it is a real planet, and not a like blackbody planet.
And because there is not a like blackbody planet.
The real planet's surface never will have a uniform temperature, no matter how fast it rotates. Because the real planet is always solar irradiated only on the daytime hemisphere...
And there is NO physical principle to distribute the solar flux's energy income evenly on the entire planet's surface. The conduction and the convection distributions cannot be compared with the speed of light the planet's daytime hemisphere surface is irradiated.
The daytime hemisphere will always be warmer, because only from the daytime hemisphere planet receives energy.
And there is NO physical principle to distribute the solar flux's energy income evenly on the entire planet's surface.
What we have is a real planet in one hand, and the blackbody classical theory - the Stefan-Boltzmann emission law - in the other hand.
And we start developing these wonderful two hands from the very roots.
Now, to what I have discovered...
Let's compare the (β*N*cp)¹∕ ⁴ term calculated for the planets:
Earth: (β*N*cp)¹∕ ⁴ = 3,5
Mars: (β*N*cp)¹∕ ⁴ = 2,28
Moon: (β*N*cp)¹∕ ⁴ = 0,9661
Mercury: (β*N*cp)¹∕ ⁴ = 0,63436
And because
Te.correct = [ Φ (1 - a) S /4σ ]¹∕ ⁴
and
Tmean = [ Φ (1 - a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴
then
Tmean = [ Te.correct⁴ * (β*N*cp)¹∕ ⁴ ]¹∕ ⁴
So, consequently:
For planets or moons with (β*N*cp)¹∕ ⁴ < 1,="" the="">Te.correct is higher than Tsat
(Mercury 0,63436, Moon 0,9661)
For planets or moons with (β*N*cp)¹∕ ⁴ > 1, the Te.correct is lower than Tsat
(Mars 2,28, Earth 3,5)
And, For planets or moons with (β*N*cp)¹∕ ⁴ = 1,
Te.correct = Tsat
For the planet Mercury's case we have:
Te.correct.mercury = 362 K > Tsat = 340 K
The same phenomenon we observe, but not so much intense for the faster than Mercury rotating Moon.
Te.correct.moon = 224 K >Tsat = 220 K
On the other hand, the much faster rotating Mars, and the much faster rotating Earth (Earth also having a much higher than Mars cp) appear to have much higher Tsat than Te temperatures.
Te.correct.mars = 174 K < tsat="210">
Te.correct.earth = 211 K < tsat="288" k="">
and this happens (for Mars and for Earth) because their planet surfaces emit much less IR energy during the daytime.
.
Table 1. Comparison of Predicted vs. Measured Temperature for All Rock-type Planets
Table 1. Comparison of Predicted vs. Measured Temperature for All Rock-type Planets
Te - planet's effective temperature
Te = [ (1-a) S /4σ ]¹∕ ⁴
Te.correct - the planet's corrected effective temperature
Te.correct = [ Φ (1-a) S /4σ ]¹∕ ⁴
Φ - is the solar irradiation accepting factor
Φ = 0,47 - for smooth surface planets without atmosphere
Φ = 1 - for gases planets and for heavy cratered without atmosphere planets
Te.correct = [ Φ (1-a) S /4σ ]¹∕ ⁴
Tmean = [ Φ (1-a) So (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴
Or
Tmean = Te.correct * [ (β*N*cp)¹∕ ⁴ ]¹∕ ⁴
......................Φ..Te.correct ..[(β*N*cp)¹∕ ⁴]¹∕ ⁴..Tmean ...Tsat
................................°K ......................................°K ........°K
Mercury .....0,47....364,0 ........0,8953............. 325,83 ...340
Earth ..........0,47....211 ...........1,368................287,74 ...288.
Moon ..........0,47....224 ...........0.9978..............223,35 ...220
Mars ...........0,47....174 ............1,227.................213,11 ....210
Io ..................1.......95,16 ........1,169..............111,55 .....110
Europa ........0,47....78,83 ........1,2636.............99,56 .....102
Ganymede...0,47....88,59 ........1,209.............107,14 ....110
Calisto ..........1.....114,66 ........1,1471...........131,52 ....134 ±11
Enceladus .... 1 ......55,97 ........1,3411............75,06 .......75
Tethys ..........1.......66,55 .........1,3145 ...........87,48 .......86 ± 1
Titan .............1.......84,52 .........1,1015 ...........96,03 .......93,7
Pluto .............1.......37 ..............1,1164 ...........41,6 .........44
Charon ..........1......41,90 ..........1,2181 ...........51,04 .......53
Conclusion:
We can calculate planet's mean surface temperature obtaining very close to the satellite measured results.
Tmean = Te.correct * [ (β*N*cp)¹∕ ⁴ ]¹∕ ⁴