A Greenhouse Effect?
Posted on: December 19, 2020
THE CLIMATE SCIENCE ANALYSIS OF PLANETARY EQUILIBRIUM TEMPERATURE: {FROM UCAR}
LINK TO SOURCE: https://scied.ucar.edu/planetary-energy-balance-temperature-calculate
The simple case of a rocky planet with no air or water. Sunlight is absorbed by the planet’s surface, heating the ground. Any object with a temperature above absolute zero emits electromagnetic radiation. For planets, that outgoing EM radiation takes the form of infrared. The planet will continue to warm until the outgoing infrared energy exactly balances the incoming energy from the sun. We can calculate the temperature at which thermal equilibrium is reached. Solar energy arriving at the earth’s surface (insolation) is 1,361 watts per square meter. We now multiply insolation by area to compute total incoming energy. The amount of light intercepted by our spherical planet is the cross sectional area of the earth computed as pi*r-squared where r, the radius of the earth, is 6371km or 6,371,000 meters. Thus, he equation for the solar energy intercepted by the earth is:
- E = total energy intercepted
- KS = solar insolation (“solar constant”) = 1,361 watts per square meter
- RE = radius of Earth = 6,371 km = 6,371,000 meters
- Using these values we compute
Using the values described above we find that the earth intercepts about 1.735E18 watts of sunlight.
Since Earth is not a blackbody, some of this energy is reflected away and not absorbed by our planet (albedo). To determine how much energy Earth absorbs from sunlight, we multiply the energy intercepted by (1-albedo) as shown below.
The sunlight Earth absorbs causes the temperature of the planet to rise and that causes the earth to emit heat in the form of infrared (IR) and the temperature to fall to equilibrium where outgoing and incoming energy balance and cancel out. Using the Stefan Boltzmann equation we compute that equilibrium temperature as:
- j* = energy flux
- σ = Stefan-Boltzmann constant = 5.670373 x 10-8 watts / m2 K4 (m = meters, K = kelvins)
- T = absolute temperature (degrees-K)
The Stefan-Boltzmann law tells us how much infrared energy Earth will emit per unit area. We need to multiply this by the total area of Earth’s surface to calculate the total amount of energy emitted by Earth. Since Earth rotates, all of it’s surface is heated by sunlight. Therefore, the whole surface of the spherical planet emits infrared radiation. We can’t use the same shortcut we used for incoming sunlight by treating Earth as equivalent to a disk. Geometry tells us that the surface area of a sphere is four times pi times the radius of the sphere squared. Multiplying the energy emissions per unit area times the surface area of Earth, we derive an expression for Earth’s total infrared energy emissions:
Setting energy coming in = energy going out, we compute the equilibrium temperature
Earth’s overall, average albedo is about 0.31 (or 31%). The value of the Stefan-Boltzmann constant (σ) is 5.6704 x 10-8 watts / m2 K4. Using these numbers and the value for KS we can calculate Earth’s equilibrium temperature as:
The temperature of the earth measured by satellite is 287.15K. Therefore the difference, 287.15 – 253.7 = 33.45K of heat cannot be explained in terms of this heat balance. The usual explanation of this imbalance is that the atmosphere absorbs outgoing heat and re-radiates IR in all directions, a phenomenon described as the greenhouse effect.
But it’s not really that simple because there is an unexplained anomaly here called the Faint Young Sun Paradox (FYSP). You can google that phrase for information on the FYSP. The bottom line is that it’s a paradox and that’s a fancy way of saying we don’t know. Below is a Youtube video discussion of this issue.
THE CHRISTOS VOURNAS BLOG LINK: https://www.cristos-vournas.com/
The Christos Vournas blog presents an estimation of the equilibrium surface temperature of the earth without an atmosphere. His finding is that the equilibrium temperature is 287.74K equal to the measured earth temperature from space and that therefore there is no role for the atmosphere in the temperature equation. A significant difference in the two theoretical planetary models is this sentence in the Vournas blog: “Earth is warmer because Earth rotates faster and because Earth’s surface is covered with water“.
These two variables are missing in the standard climate science model of equilibrium planetary temperature where planets are smooth rocky balls and where the spin velocity is not relevant to temperature. The contradiction here is that in the study of the Faint Young Sun Paradox described in a related post : LINK: https://tambonthongchai.com/2020/12/21/the-faint-young-sun-paradox-geological-co2/ , both rotational speed and oceans are considered in the mean earth temperature computation.
.
.
.
26 Responses to "A Greenhouse Effect?"
December 20, 2020 at 7:20 am
Understood it better in the second read. Good analysis. Fully agree. Thanks again. Trust all is well. Merry Christmas.
December 21, 2020 at 2:32 am
Thank you very much, Cha-am Jamal, for your introduction of
The Planet Without-Atmosphere Mean Surface Temperature Equation
(the equilibrium surface temperature of the earth without an atmosphere).
“The Christos Vournas blog presents an estimation of the equilibrium surface temperature of the earth without an atmosphere. His finding is that the equilibrium temperature is 287.74K equal to the measured earth temperature from space and that therefore there is no role for the atmosphere in the temperature equation. A significant difference in the two theoretical planetary models is this sentence in the Vournas blog: “Earth is warmer because Earth rotates faster and because Earth’s surface is covered with water“. These two variables are missing in the standard climate science model of equilibrium planetary temperature where planets are smooth rocky balls and where the spin velocity is not relevant to temperature.”
We have moved further from the incomplete effective temperature equation
(which is in common use right now, but actually it is an incomplete planet Te equation and that is why it gives us very confusing results)
Te = [ (1-a) S / 4 σ ]¹∕ ⁴
a – is the planet’s surface average albedo
S – is the solar flux, W/m²
σ = 5,67*10⁻⁸ W/m²K⁴, the Stefan-Boltzmann constant
The Planet Without-Atmosphere Mean Surface Temperature Equation
Tmean = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (1)
The Planet Without-Atmosphere Mean Surface Temperature Equation is also based on the radiative equilibrium and on the Stefan-Boltzmann Law.
The Equation is being completed by adding to the incomplete Te equation the new parameters Φ, N, cp and the constant β.
Φ – is the dimensionless Solar Irradiation accepting factor
N – rotations /day, is the planet’s axial spin
cp – cal /gr*oC, is the planet’s surface specific heat
β = 150 days*gr*oC/rotation*cal – is the Rotating Planet Surface Solar Irradiation Absorbing-Emitting Universal Law constant.
We ended up to the following remarkable results.
Comparison of results the planet’s Te calculated by the Incomplete Equation:
Te = [ (1-a) S / 4 σ ]¹∕ ⁴
the planet’s mean surface temperature Tmean calculated by the Planet’s Without-Atmosphere Mean Surface Temperature Equation:
Tmean = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (1)
and the planet’s Tsat.mean measured by satellites:
Planet..Te.incompl….Tmean…Tsat.mean
…………. equation…..equation…measured
Mercury… 439,6 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
To be honest with you, at the beginning, I got by surprise myself with these results.
You see, I was searching for a mathematical approach…
http://www.cristos-vournas.com
December 21, 2020 at 6:01 am
Φ factor explanation
There is need to focus on the Φ factor explanation.
Φ comes from the realization that a sphere reflects differently than a flat surface perpendicular to the Solar rays.
Φ – is the dimensionless Solar Irradiation accepting factor
“Φ” is an important factor in the Planet’s Surface Mean Temperature Equation:
Tmean.planet = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (K)
It is very important the understanding what is really going on with by planets the solar irradiation reflection.
There is the specular reflection and there is the diffuse reflection.
The planet’s surface albedo “a” accounts for the planet’s surface diffuse reflection.
So till now we didn’t take in account the planet’s surface specular reflection.
A smooth sphere, as some planets are, have the invisible from the space and so far not detected and not measured the specular reflection.
The sphere’s specular reflection cannot be seen from the distance, but it can be seen by an observer situated on the sphere’s surface.
Thus, when we admire the late afternoon sunsets on the sea we are blinded from the brightness of the sea surface glare. It is the surface specular reflection what we see then.
When we integrate the specular reflection from the parallel solar rays hitting the disk of radius “r” and the cross-section “π r²” over the sunlit hemisphere of radius “r”, the result is 0,53π r² S.
Thus the
0,53π r²*S is the specular reflected fraction of the incident on the smooth planet’s spherical surface solar flux.
Φ = 1 – 0,53 = 0,47
Φ = 0,47
Thus the
0,47π r²S – is The What Is Left Fraction of the incident solar flux for the planet’s smooth spherical surface to absorb because of the spherical surface’s specular reflection.
What we have now is the following:
Jsw.incoming – Jsw.reflected = Jsw.absorbed
Jsw.reflected = (0,53 + Φ*a) *Jsw.incoming
And
Jsw.absorbed = Φ*(1-a) *Jsw.incoming
Where
(0,53 + Φ*a) + Φ* (1-a) = 0,53 + Φ*a + Φ – Φ*a =
= 0,53 + Φ = 0,53 + 0,47 = 1
The solar irradiation reflection (the specular plus the diffuse) over the planet’s sunlit hemisphere is:
Jsw.reflected = (0,53 + Φ*a) * Jsw.incoming
Jsw.reflected = (0,53 + Φ*a) *S *π r²
For a planet with albedo a = 0 (completely black surface planet) we would have
Jsw.reflected = (0,53 + Φ*0) *S *π r² =
= Jsw.reflected = 0,53 *S *π r²
For a planet without any outgoing specular reflection
we would have Φ =1
And
Jsw.reflected = a *S *π r² =
In general:
The fraction left for hemisphere to absorb is:
Jabs = Φ (1 – a ) S π r²
The factor Φ = 0,47 “translates” the absorption of a disk into the absorption of a smooth hemisphere with the same radius.
When covering a disk with a hemisphere of the same radius the hemisphere’s surface area is 2π r².
The incident Solar energy on the hemisphere’s area is the same as on the disk:
Jdirect = π r² S
But the absorbed Solar energy by the hemisphere’s area of 2π r² is:
Jabs = 0,47*( 1 – a) π r² S
It happens because a smooth hemisphere of the same radius “r” absorbs only the 0,47 part of the directly incident on the disk of the same radius Solar irradiation.
In spite of hemisphere having twice the area of the disk, it absorbs only the 0,47 part of the directly incident on the disk Solar irradiation.
Jabs = Φ (1 – a ) S π r² ,
where Φ = 0,47 for smooth without atmosphere planets.
and Φ = 1 for gaseous planets, as Jupiter, Saturn, Neptune, Uranus, Venus, Titan.
Gaseous planets do not have a surface to reflect radiation. The solar irradiation is captured in the thousands of kilometers gaseous abyss. The gaseous planets have only the albedo “a”.
And Φ = 1 for heavy cratered planets, as Calisto and Rhea ( not smooth surface planets, without atmosphere ).
The heavy cratered planets have the ability to capture the incoming light in their multiple craters and canyons. The heavy cratered planets have only the albedo “a”.
Another thing that I should explain is that planet’s albedo actually doesn’t represent a primer reflection. It is a kind of a secondary reflection ( a homogenous dispersion of light also out into space ). That light is visible and measurable and is called albedo.
The primer reflection from a spherical hemisphere cannot be seen from some distance from the planet. It can only be seen by an observer being on the planet’s surface. It is the blinding surface reflection right in the observer’s eye.
That is why the albedo “a” and the factor “Φ” we consider as different values.
Both of them, the albedo “a” and the factor “Φ” cooperate in the Planet Rotating Surface Solar Irradiation Absorbing-Emitting Universal Law:
Φ*(1-a)*Sπ r² = 4π r²*σTmean⁴ /(β*N*cp)¹∕ ⁴
And they are also cooperate in the Planet’s Surface Mean Temperature Equation:
Tmean.planet = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ ( K )
Planet Energy Budget:
Solar energy absorbed by a Hemisphere with radius “r” after reflection and diffusion:
Jabs = Φ*πr²S (1-a) ( W )
Total energy emitted to space from a whole planet:
Jemit = A*σΤmean⁴ /(β*N*cp)¹∕ ⁴ ( W )
Φ – is a dimensionless Solar Irradiation accepting factor
(1 – Φ) – is the reflected fraction of the incident on the planet solar flux
S – is a Solar Flux at the top of atmosphere ( W/m² )
Α – is the total planet surface area ( m² )
A = 4πr² (m²), where “r” – is the planet’s radius
Tmean – is the Planet’s Surface Mean Temperature ( K )
(β*N*cp)¹∕ ⁴ – dimensionless, is a Rotating Planet Surface Solar Irradiation Warming Ability
Thus
energy in = energy out
Φ*(1-a)*Sπ r² = 4π r²*σTmean⁴ /(β*N*cp)¹∕ ⁴
For Earth’s surface we would have:
Jabs.earth = 0,47 ( 1 – 0,306 ) So π r² =
= 0,47*0,694 * 1.361* π r² ( W ) =
Jabs.earth = 0,326 So π r² =
= 0,326* 1.361 π r² =
= 444,26 π r² ( W )
Jabs.earth = 444,26 π r² ( W )
I am not averaging Jabs.earth = 444,26 π r² ( W ) over the entire Earth’s surface, because it is a wrong approach.
It is a misleading mistake to average the “absorbed” incident solar flux’s fraction over the entire earth’s surface.
Also I have put the word “absorbed” in brackets, because it is not exactly absorbed, but instantly emitted back to space as an IR radiation.
http://www.cristos-vournas.com
December 21, 2020 at 11:55 am
See also
https://wp.me/pTN8Y-5Nw
December 21, 2020 at 9:38 pm
Please tell me, Jamal, should I continue explaining the theory supporting the equation in this page?
Would you like to put me your questions, which I will answer then?
Or is it better for me to proceed with explaining everything I consider important?
I am with this equation since February 2019.
There is not any settled method explaining these findings. The method is developing via dialogs.
Christos
December 21, 2020 at 9:51 pm
Ευχαριστούμε για την προσφορά. Αν έχω απορίες, θα επικοινωνήσω μαζί σας.
December 21, 2020 at 9:56 pm
Thank you. I will contact you if i have questions about your work.
December 24, 2020 at 12:06 am
Ben Wouters:
“A tidally locked Moon like planet at our distance of the sun will have a hot and a cold side, average temperature ~150K. Rotation will increase the average temperature, but never to Te (effective temperature) since the temp. difference between equator and poles will remain.”
The planets blackbody equilibrium temperatures (Te planets effective temperatures) cannot be considered as planets without atmosphere average (mean) surface temperatures.
It is a very mistaken concept. No planet has a uniform surface temperature.
And it is not proven by the observations.
For Mars Te.mars = 210 K
and Tsat.mean.mars = 210 K
these two temperatures – the theoretical mathematical abstraction’s value of Te.mars to be equal to the satellite measured Tsat.mean.mars is a coincident.
These planets Te and Tsat temperatures equality is never observed again in the entire (measured) solar system.
Also it should be underlined that Tsat.mean.mars = 210 K is not planet’s Mars uniform surface temperature.
From Wikipedia, the free encyclopedia
Mars
Surface.. temp.. min……mean…..max
Kelvin………….. 130 K….210 K… 308 K
Instead of accepting this fundamental observation as an undeniable fact, we are comforting ourselves trying to explain the observed differences between the every planet calculated Te and the satellite measured Tsat.mean.planet.
Thus for Earth we have the greenhouse warming effect theories.
For some planets (Jupiter, Saturn, Neptune) we have the huge inner sources of heat theories.
For some other cases we have the tidal warming theories (Jupiter’s and Saturn’s satellites).
For every planet-case we are looking for an excuse-explanation to keep the mistaken concept about the Te = Tmean equality for planets without atmosphere.
But the truth is, there is not any measurable greenhouse warming effect on the Earth’s surface.
The Earth’s atmosphere is very thin and it is very transparent both ways – in and out.
And as for carbon dioxide – there are only traces of CO2 in Earth’s atmosphere. Only traces…
December 24, 2020 at 2:31 am
Christos Vournas December 24, 2020 at 12:06 am
The planets blackbody equilibrium temperatures (Te planets effective temperatures) cannot be considered as planets without atmosphere average (mean) surface temperatures.
Agree, it is nonsense. Just look at our Moon:
Te ~270K, actual avg. temp. ~197K (Diviner project)
Afaik Te originates in astrophysics, and was used as a first approximation of the avg. temp. of a newly discovered object based on its albedo and TSI from its local star.
But the truth is, there is not any measurable greenhouse warming effect on the Earth’s surface.
Agree, but without atmosphere Earth would be colder. The atmosphere does reduce the energy loss to space.
At ~288K Earth would lose ~400 W/m^2 without atmosphere (and need ~400 W/m^2 energy input to maintain that temp.) Atmosphere reduces this to ~240 W/m^2.
Lacis ea 2010 claim that
The Sun is the source of energy that heats Earth. Besides direct solar heating of the ground, here is also indirect longwave (LW) warming arising from the thermal radiation that is emitted by the ground, then absorbed locally within the atmosphere, from which it is re-emitted in both upward and downward directions, further heating the ground and maintaining the temperature gradient in the atmosphere.
If the atmosphere “further heats” the ground then it must also have heated our ~4km deep oceans which is obvious nonsense since the deep ocean temperature is ~78K above the avg. Lunar surface temp.
Agree that CO2 is good as plant food, not for heating our oceans 😉
December 24, 2020 at 3:24 am
If the atmosphere “further heats” the ground then it must also have heated our ~4km deep oceans which is obvious nonsense since the deep ocean temperature is ~78K above the avg. Lunar surface temp.
should read
If the atmosphere “further heats” the ground then it must also have heated our ~4km deep oceans since their temperature is ~78K above the avg. Lunar surface temp. Which is obvious nonsense.
December 24, 2020 at 6:48 am
Ben Wouters December 24, 2020 at 2:31 am
“but without atmosphere Earth would be colder. The atmosphere does reduce the energy loss to space.
At ~288K Earth would lose ~400 W/m^2 without atmosphere (and need ~400 W/m^2 energy input to maintain that temp.) Atmosphere reduces this to ~240 W/m^2.”
” At ~288K Earth would lose ~400 W/m^2.”
Yes, it is correct if ~288K was Earth’s uniform equilibrium temperature. But ~288K is Earth’s average surface temperature.
Earth’s surface doesn’t emit IR radiation out to space at that temperature (~288K).
Every infinitesimal spot and at every infinitesimal space of time emits at different intensity (W/m^2).
And, consequently, every infinitesimal spot and at every infinitesimal space of time emits at different temperature Tχ (K).
When integrated over the entire planet’s sphere area and over the 24 hours long day-night solar irradiation cycle
the total planet’s emission energy will be
Jemit.planet = Jemit1 + Jemit2 + …+ Jemitχ
χ → ∞
or, maybe better like this
Jemit.planet = { [ dA1*σΤ1⁴ /(β*N*cp)¹∕ ⁴ + dA2*σΤ2⁴ /(β*N*cp)¹∕ ⁴ + …+ dAχ*σΤχ⁴ /(β*N*cp)¹∕ ⁴ ]dt1 + [ dA*1σΤ1⁴ /(β*N*cp)¹∕ ⁴ + dA2*σΤ2⁴ /(β*N*cp)¹∕ ⁴ + …+ dAχ*σΤχ⁴ /(β*N*cp)¹∕ ⁴ ]dtψ +… + etc } / Σ dtψ
dAχ → 0, Σ dAχ = 4πr² (m²)
dtψ → 0, Σ dtψ = 24 hours
Jemit.planet = 4πr²σΤmean⁴ /(β*N*cp)¹∕ ⁴ 
Earth’s surface doesn’t emit
Jemit = 4πr²σ288K⁴
But Earth’s surface emits
Jemit = 4πr²σ288⁴ /(β*N*cp)¹∕ ⁴
We are not justified to average Jemit = 4πr²σ288⁴ /(β*N*cp)¹∕ ⁴
over the entire surface and to write
Jemit = σ288⁴ (W/m^2)
because surface doesn’t emit on average temperature.
Thus, the result is very much different.
Jemit.earth = 4πr²σ288⁴ /(β*N*cp)¹∕ ⁴
Planet Energy Budget:
Jabs = Jemit
πr²Φ*S*(1-a) = 4πr²σTmean⁴ /(β*N*cp)¹∕ ⁴
Solving for Tmean we obtain the Planet Mean Surface Temperature Equation:
Tmean.planet = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (K)
Let’s calculate Tmean.earth
So = 1.361 W/m² (So is the Solar constant)
S (W/m²) is the planet’s solar flux. For Earth S = So
Earth’s albedo: aearth = 0,306
Earth is a smooth rocky planet, Earth’s surface solar irradiation accepting factor Φearth = 0,47
(Accepted by a Smooth Hemisphere with radius r sunlight is S*Φ*π*r²(1-a), where Φ = 0,47)
β = 150 days*gr*oC/rotation*cal – is a Rotating Planet Surface Solar Irradiation Absorbing-Emitting Universal Law constant
N = 1 rotation /per day, is Earth’s axial spin
cp.earth = 1 cal/gr*oC, it is because Earth has a vast ocean. Generally speaking almost the whole Earth’s surface is wet. We can call Earth a Planet Ocean.
σ = 5,67*10⁻⁸ W/m²K⁴, the Stefan-Boltzmann constant
Earth’s Without-Atmosphere Mean Surface Temperature Equation Tmean.earth is:
Tmean.earth= [ Φ (1-a) So (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴
Τmean.earth = [ 0,47(1-0,306)1.361 W/m²(150 days*gr*oC/rotation*cal *1rotations/day*1 cal/gr*oC)¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =
Τmean.earth = [ 0,47(1-0,306)1.361 W/m²(150*1*1)¹∕ ⁴ /4*5,67*10⁻⁸ W/m²K⁴ ]¹∕ ⁴ =
Τmean.earth = ( 6.854.905.906,50 )¹∕ ⁴ = 287,74 K
Tmean.earth = 287,74 Κ
When we apply the Tmean.planet equation
Tmean.planet = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (K)
by substituting the terms
(Φ, S, N, cp) for each different planet, then we calculate the planets’ mean surface temperatures, and they match very closely with the planets’ mean surface temperatures measured by satellites.
Planet………..Te…………Tmean….Tsat.mean
Mercury….439,6 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
🙂
December 24, 2020 at 9:57 pm
Correction:
I think the below should be divided also by “χ”
and instead of
Jemit.planet = { [ dA1*σΤ1⁴ /(β*N*cp)¹∕ ⁴ + dA2*σΤ2⁴ /(β*N*cp)¹∕ ⁴ + …+ dAχ*σΤχ⁴ /(β*N*cp)¹∕ ⁴ ]dt1 + [ dA*1σΤ1⁴ /(β*N*cp)¹∕ ⁴ + dA2*σΤ2⁴ /(β*N*cp)¹∕ ⁴ + …+ dAχ*σΤχ⁴ /(β*N*cp)¹∕ ⁴ ]dtψ +… + etc } / Σ dtψ
it will more likely look:
Jemit.planet = { [ dA1*σΤ1⁴ /(β*N*cp)¹∕ ⁴ + dA2*σΤ2⁴ /(β*N*cp)¹∕ ⁴ + …+ dAχ*σΤχ⁴ /(β*N*cp)¹∕ ⁴ ]dt1 + [ dA*1σΤ1⁴ /(β*N*cp)¹∕ ⁴ + dA2*σΤ2⁴ /(β*N*cp)¹∕ ⁴ + …+ dAχ*σΤχ⁴ /(β*N*cp)¹∕ ⁴ ]dtψ +… + etc } / χ*Σ dtψ
dAχ → 0, Σ dAχ = 4πr² (m²)
dtψ → 0, Σ dtψ = 24 hours
χ → ∞
Any mathematical suggestion and any help will be welcomed and very much appreciated. 🙂
January 2, 2021 at 6:02 am
Any mathematical suggestion and any help will be welcomed and very much appreciated.
Little time, but have a look at this text:
Click to access utc_blog_reply_part1-1.pdf
N&Z give a imo good calculation of the avg surface temp. of a non rotating moon like planet.
I do NOT agree with their ATE effect.
January 2, 2021 at 11:16 pm
Simple calculation for a tidally locked moon like planet at our distance of the sun is to just spread incoming solar over half the sphere (divide by 2 iso 4), apply albedo and calculate the radiative balance temperature.
1364 W/m^2/2*0,89 = ~607 W/m^2
SB => ~321 K
For the dark side use 3K (Cosmic Background Radiation).
Avg. surface temp. 324/2= 162K Slightly higher than the more correct N&Z calculation, but much simpler 😉
January 3, 2021 at 4:34 am
The planet mean surface equation is very simple:
Tmean.planet = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (K)
What we were left to do was substituting the terms for every planet or moon without atmosphere (or with a thin atmosphere) in the solar system – the equation “works” for all of them.
January 5, 2021 at 3:57 am
The Planet Surface Rotational Warming Phenomenon
I’ll try here in few simple sentences explain the very essence of how the planet rotational warming Phenomenon occurs.
Lets consider two identical planets F and S at the same distance from the sun.
Let’s assume the planet F spins on its axis Faster, and the planet S spins on its axis Slower.
Both planets F and S get the same intensity solar flux on their sunlit hemispheres. Consequently both planets receive the same exactly amount of solar radiative energy.
The slower rotating planet’s S sunlit hemisphere surface gets warmed at higher temperatures than the faster rotating planet’s F sunlit hemisphere.
The surfaces emit at σT⁴ intensity – it is the Stefan-Boltzmann emission law.
Thus the planet S emits more intensively from the sunlit side than the planet F.
So there is more energy left for the planet F to accumulate then. That is what makes the faster rotating planet F on the average a warmer planet. That is how the Planet Surface Rotational Warming Phenomenon occurs.
January 5, 2021 at 7:27 pm
Similarly we can evaluate the planet surface specific heat Cp importance.
The Planet Surface Specific Heat Cp Warming Phenomenon
I’ll try here in few simple sentences explain the very essence of how the planet surface specific heat warming Phenomenon occurs.
Lets consider two identical planets H and L at the same distance from the sun.
Let’s assume the planet H has a Higher average surface specific heat, and the planet L has a Lower average surface specific heat.
Both planets H and L get the same intensity solar flux on their sunlit hemispheres. Consequently both planets receive the same exactly amount of SOLAR RADIATIVE ENERGY.
The Lower average surface specific heat planet L sunlit hemisphere surface gets warmed at higher temperatures than the Higher average surface specific heat planet H sunlit hemisphere.
The surfaces emit at σT⁴ intensity – it is the Stefan-Boltzmann emission law.
Thus the planet L emits more intensively from the sunlit side than the planet H.
So there is more energy left for the planet H to accumulate then. That is what makes the Higher average surface specific heat planet H on the average a warmer planet. That is how the Planet Surface Specific Heat Cp Warming Phenomenon occurs.
January 5, 2021 at 9:42 pm
Let’s proceed the syllogism.
N – is the planet’s rotational spin
cp – is the planet’s average surface specific heat
N*cp is the product of planet’s N and cp
Now, let’s have two identical planets, but with different rotational spin N1 and N2, and with different average surface specific heat cp1 and cp2.
Which planet has the highest mean surface temperature Tmean ?
Of course, since every planet has its own unique rotational spin (diurnal cycle) and every planet has its own unique average surface specific heat… we should compare for the two planets
N*cp – the product of N and cp.
Consequently, the planet with the highest N*cp product should be the planet with the highest mean surface temperature Tmean.
Example:
Earth’s N.earth = 1 rot /day
Moon’s N.moon = 1 /29,5 rot /day
Earth’s cp.earth = 1 cal /gr.oC (watery planet)
Moon’s cp.moon = 0,19 cal /gr.oC (regolith)
(N.earth)*(cp.earth) = 1*1 = 1 rot.cal /day.gr.oC
(N.moon)*(cp.moon) = (1 /29,5)*0,19 = 0,00644 rot.cal /day.gr.oC
Let’s compare the products:
(N.earth)*(cp.earth) / [(N.moon)*(cp.moon)] = 1 /0,00644 = 155,3
What we see here is that the Earth’s N*cp product is 155,3 times higher than the Moon’s N*cp product.
And the satellite measured mean surface temperatures are
Tmean.earth = 287,16 K
https://en.wikipedia.org/wiki/Earth
Tmean.moon = 220 K
https://simple.wikipedia.org/wiki/Moon
Thongchai Thailand 
December 20, 2020 at 2:30 am
A tidally locked Moon like planet at our distance of the sun will have a hot and a cold side, average temperature ~150K. Rotation will increase the average temperature, but never to Te (effective temperature) since the temp. difference between equator and poles will remain.
Adding geothermally heated oceans will completely change the dynamics.
Sun only increases the surface temperature a bit.
Regarding fluxes the solar one dominates completely, but temperature wise geothermal dominates since the deep oceans are at ~275K and solely heated by geothermal.
The FYSP disappears when you realize that young Earth was covered with hot oceans. They were hot simply because the crust was very thin or even non-existent. Perhaps these oceans were (close to) boiling.
FYS was only able to increase the temperature of the Mixed Surface Layer a little. Currently the sun increases the temperature of the MSL from ~275K (deep ocean temperature) to ~288K.
https://tallbloke.wordpress.com/2014/03/03/ben-wouters-influence-of-geothermal-heat-on-past-and-present-climate/
December 20, 2020 at 4:15 am
Thank you