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

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It is the Solar Irradiated Planet Surface ROTATIONAL Warming Phenomenon which does the job

Very interesting !

Mars and Moon satellite measured mean surface 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 = 0,11

What is left to absorb is (1 – amoon) = (1- 0,11) = 0,89

 

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 /0,89 )¹∕ ⁴ = ( 0,8427 )¹∕ ⁴ = 0,9581

Conclusions:

1. Mars /Moon satellite measured temperatures comparison( 0,9545 ) is almost identical with the Mars /Moon albedo determined solar irradiation absorption ability ( 0,9581 )

2. If Mars and Moon had the same exactly albedo, their satellite measured mean surface temperatures would have been exactly the same.

And this is very interesting !

Mars rotates N = 0,9747 rotation /day

Moon rotates N= 1 /29,5 rotation day

Mars solar flux S = 586 W/m²

Moon S = 1361 W/m²

 

Mars is at 1,53 AU distance from the sun,

Moon is at 1 AU from the sun.

That is why Mars receives much weaker (586 W/m²) vs (1361 W/m² ) than Moon solar flux.

Nevertheless, Mars’ surface develops almost the same average surface temperature ( 210 K ) as the Moon ( 220 K ).

Both Mars and Moon do not have atmosphere. It is the Solar Irradiated Planet Surface ROTATIONAL Warming Phenomenon which does the job.

Because it is a fact, a scientifically observed happening!

Mars’ surface develops almost the same average surface temperature ( 210 K ) as the Moon ( 220 K ).

Both Mars and Moon do not have atmosphere.

What keeps warm Mars then?

Because it is a fact, a scientifically observed happening!

What else… It is the Solar Irradiated Planet Surface ROTATIONAL Warming Phenomenon which does the job.

The method I use is Planet Temperatures Comparison.

the basic approach formula

(1-a)S /4

gives for Earth =240 W/m²

for Moon

(1-0,11)1.361 /4 = 303 W/m²

for Mars

(1-0,25)586 /4 = 110 W/m²

On Mars: "The resulting mean surface pressure is only 0.6% of that of Earth 101.3 kPa (14.69 psi)." (From Wikipedia)

What I think is Mars' less than 1% of Earth's atmosphere is not capable accumulate almost three times the amount of solar flux hitting Mars' surface.

It is the Surface Rotational Warming which does the work.

Like-wise it happens on Earth too. Only Venus has strong greenhouse effect. Earth, compared to Venus has a very thin atmosphere... 

Planets' Rotational Spins sixteenth root [ (N¹∕ ⁴ ]¹∕ ⁴

Gaseous Planets Jupiter, Saturn, Neptune T1bar mean temperatures 165 K, 134 K, 72 K comparison

All data are satellites measurements.

R – semi-major axis in AU (Astronomical Units)

a – planet’s average albedo

N – rotations /day – planet’s spin

T1bar – planet atmosphere at 1 bar average temperature in Kelvin

Planet.…Jupiter…..Saturn….Neptune

R………....5,2044…..9,5826.….30,33

1/R²…….0,0369…0,01089…0,001087

a……………0,503…0,342…….0,290

1-a………..0,497…0,658…….0,710

N………….2,4181…2,2727….1,4896

T 1 bar…...165 K….134 K……72 K

Coeff...0,388880... 0,306264…0,170881

Comparison coefficient:

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴

Let’s compare

Jupiter coeff. /Saturn coeff. =

= 0,388880 /0,306264 = 1,2698

T1bar.jupiter /T1bar.saturn = 165 /134 = 1,2313

 

Jupiter coeff. / Neptune coeff. =

= 0,388880 /0,170881 = 2,2757

T1bar.jupiter /T1bar.neptune = 165 /72 = 2,2917

 

Saturn coeff. /Neptune coeff. =

= 0,306264 /0,170881 = 1,7923

T1bar.saturn /T1bar.neptune = 134 /72 = 1,8611

Conclusion:

Gaseous planets Jupiter, Saturn and Neptune average T1bar temperatures relate the same way as the rocky inner planets Mercury, Moon and Mars average surface temperatures, and also as the Earth with Europa average surface temperatures. The comparison coefficient is the same:

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴

The planets are being separated in groups (Jupiter, Saturn and Neptune - H2, He), (Mercury, Moon and Mars - regolith), and (Earth with Europa - H2O) only by their similar specific heat.

Notice:

The observed Jupiter and Neptune having the closest coefficient vs planet's temperatures at 1 bar level comparison rates, can be explained by the fact that Jupiter and Neptune have the closest the H2% and He% atmospheric gases content.

Thus, everything else equals, Gaseous Planets' average at 1 bar level Temperatures relate to the

Planets' Rotational Spins sixteenth root [ (N¹∕ ⁴ ]¹∕ ⁴. 

One of the many mistakes in “climate science” is dividing Earth’s incoming solar flux by 4

Clint,

> One of the many mistakes in “climate science” is dividing Earth’s incoming solar flux by 4. That division reduces the temperature of the target. In the case of Earth, the average temperature would be only 255K (-18.1C, -0.5F). By claiming Earth should be that cold, they can then claim CO2 must be supplying the extra warming.

Clint R says: May 9, 2021 at 7:53 PM (in [https://www.drroyspencer.com/2021/05/uah-global-temperature-update-for-april-2021-0-05-deg-c/#comment-688348] )

The highlighting is mine...

Exactly!!! And they do that mistake - dividing the solar flux by 4 - for every planet and moon in the solar system. The solar flux division by 4 is a first-step very primitive theoretical approach for the planet's average temperature very approximate estimation.

By basing the entire climate science on the solar flux's division by 4 the +33oC non existing Earth's atmosphere greenhouse effect was established. And the non existing +33oC number haunts the climate science since 1981!

Also they ignore smooth surface planets reflect not only diffusely, but they also have very strong specular reflection.

Another very important issue they ignore is the existence of the solar irradiated planet surface Rotational Warming Phenomenon.

Tyndall 1861: Even by the aid of a magnifying lens I could not detect the slightest change of position.

Tyndall 1861:

> “Dry air was now admitted into the tube, while the needle of the galvanometer was observed with all possible care. Even by the aid of a magnifying lens I could not detect the slightest change of position. Oxygen, hydrogen, and nitrogen, subjected to the same test gave the same negative result. The temperature of the water was subsequently lowered so as to produce a deflection of 20ºC and 10ºC; but in no case did the admission of air, or any of the above gases into the exhausted tube, produce any sensible change in the position of the needle.”

Tyndall (1861, p. 172)

http://tyndall1861.geologist-1011.mobi/

What we read here is that Tyndall in his 1861 work says dry air:

"Dry air was now admitted into the tube, while the needle of the galvanometer was observed with all possible care. Even by the aid of a magnifying lens I could not detect the slightest change of position."

Tyndall could not detect the slightest change of needle’s position… when measuring dry air IR absorption … Dry air consists of N2 O2 and CO2 (0,04 % or 400 ppm). CO2 is a trace gas in dry air, that is why Tyndall,

“Even by the aid of a magnifying lens I could not detect the slightest change of position. “

The Absolute rightness of the Φ = 0,47 for the smooth surface planets

Now that we have completed the planet (N*cp) product sixteenth root comparisons, which are related with the Planet Surface Rotational Warming Phenomenon...

It is time to demonstrate on the observed data the Absolute rightness of the Φ = 0,47 for smooth surface planets

and Φ = 1 for rough surface planets (heavy cratered).

The data we need for this are available at:

[https://en.wikipedia.org/wiki/Io_%28moon%29]

[https://en.wikipedia.org/wiki/Europa_(moon)]

 

We shall compare the heavy cratered Jupiter's moon Io (Φ = 1) with another Jupiter's moon Europa with the smoothest surface of any known solid object in the Solar System (Φ = 0,47).

Io - Europa (Jupiter's moons) satellite measured mean surface temperatures comparison.

Io

Albedo a = 0,63 ± 0,02

Rough, heavy cratered surface Φ = 1

N = 1 /1,799 rot /day

S = 50,37 W/m²

Tmean = 110 K

Io's surface is composed of sulfur and sulfur dioxide frost

cp.io = 0,145 cal/gr*oC

 

Europa

Albedo a = 0,67 ± 0,03

Europa has the smoothest surface Φ = 0,47

N = 1 /3,5512 rot /day

S = 50,37 W/m²

Tmean = 102 K

Europa's surface consists of water ice crust

cp.europa = 1 cal/gr*oC

 

What we see at first examination of the data is that both, Io and Europa have the same solar flux S = 50,37 W/m² and a very close satellite measured Albedo (Io Albedo a = 0,63 ± 0,02; Europa Albedo a = 0,67 ± 0,03).

The comparison coefficient will be limited then to

[ Φ*(N*cp)¹∕ ⁴ ]¹∕ ⁴

Io coefficient = { 1*[ (1 /1,799)*0,145]¹∕ ⁴ }¹∕ ⁴ =

= [ (0,0806)¹∕ ⁴ ]¹∕ ⁴ = 0,85437

Europa coefficient = { 0,47*[ (1 /1,35512)*1]¹∕ ⁴ }¹∕ ⁴ =

= ( 0,47*0,92684 )¹∕ ⁴ = ( 0,43562 )¹∕ ⁴ = 0,81241

 

Io /Europa coefficients comparison

0,85437 /0,81241 = 1,051647

Tmean.io /Tmean.europa = 110 K /102 K = 1,07843

 

1,07843 /1,051647 = 1,0255

With only 2,55 % difference - it is a wonderful result !!!

 

"With over 400 active volcanoes, Io is the most geologically active object in the Solar System.... Most of Io's surface is composed of extensive plains with a frosty coating of sulfur and sulfur dioxide." [4]

"Europa has the smoothest surface of any known solid object in the Solar System. The apparent youth and smoothness of the surface have led to the hypothesis that a water ocean exists beneath the surface, which could conceivably harbor extraterrestrial life." [5]

What we have witnessed here is that the smoothest celestial body in solar system Europa very much rightfully have been applied the Φ = 0,47 since smooth spheres reflect the incident solar light not only diffusely, but also specularly!

The very satisfactory result is a solid testimony for the Absolute rightness of the Φ = 0,47 for the smooth surface planets.

A model of planet surface with a much prevailing diffuse reflection (without specular reflection)

Here it is a planet Φ = 1 concept. It can be visualized as a sphere covered with vertical blades, which are meridionally oriented... When parallel solar beams hit a surface (no matter how tiny the surface's dimensions are) there always is a specular and a diffuse reflection.

The blades are close enough between them, so to be able not to permit the specular reflected solar energy to escape from planet's surface out to space.

When solar hitting the blades the specular reflection is directed downwards and only the scattered light- the dispersed light -the diffuse reflection is being able to partly get oriented out to space.

When hitting the blades, solar rays are "smutched" over a larger areas compared to the smooth surface planets, and, therefore, there is a larger amounts of solar energy being diffusely dispersed. 

Another model can be considered as a planet surface being very much porous. The dense porous surface is also capable not to permit the specular reflected solar energy to escape to space.

Conclusion

Planets with high Albedo are more likely planets with a very small-to-zero specular reflection portion. For high Albedo planets the solar irradiation accepting factor (at first approach) should be chosen  as Φ = 1.

High diffuse reflection model - A planet surface blades mode.

Φ - factor is proven by the observed quantities

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 planet Mercury are all satellite measured. These parameters of the planet Mercury are all observations.

Planet...Mercury

Tsat.mean.340 K

R.......0,387 AU

1/R²....6.6769

N.....1 /175,938 rotations/day

a........0,068

1-a......0,932

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 Mercury's satellite measured 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.

The calculated Mercury's effective temperature Te.mercury = 440 K is a 100 degrees oC higher than the satellite measured Mercury's mean surface temperature Tsat.mean.mercury = 340 K

Let's now 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

Te.mercury - Te.correct.mercury = 440 K - 364 K = 76°C

This difference is because in the old Te.mercury equation the solar irradiation accepting factor Φ = 0,47 was omitted.

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 presence of the solar irradiation accepting factor Φ = 0,47 in the Mercury's correct Te equation.

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 100 % - 47 % = 53 % of the incident on it's surface solar irradiation.

Imagine a completely black smooth surface planet; imagine a completely invisible but smooth surface planet, a planet with a zero albedo. This planet still reflects 53 % of the incident on its surface solar irradiation.

The satellite measurements have confirmed it.

Link to Wikipedia Mercury (planet) [https://en.wikipedia.org/wiki/Mercury_(planet)]