Guest Post by Reed Coray
On Dec. 6, 2011 12:12 am Lord Monckton posted a comment on a thread entitled Monckton on sensitivity training at Durban that appeared on this blog on Dec. 5, 2011. In that comment he wrote:
“First, it is not difficult to calculate that the Earth’s characteristic-emission temperature is 255 K. That is the temperature that would obtain at the surface in the absence of any greenhouse gases in the atmosphere. Since today’s surface temperature is 288 K, the presence as opposed to absence of all the greenhouse gases causes a warming of 33 K”.
Since I’m not sure what the definition of the “Earth’s characteristic-emission temperature” is, I can’t disagree with his claim that its value is 255 K. However, I can and do disagree with his claim that 255 K is “the temperature that would obtain at the surface in the absence of any greenhouse gases in the atmosphere”.
When computing the Earth’s surface temperature difference in “the presence as opposed to the absence of all greenhouse gases”, (i) two temperatures (A and B) must be measured/estimated and (ii) the difference in those temperatures computed. The first temperature, A, is the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases. The second temperature, B, is the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only–i.e., an atmosphere that contains non-greenhouse gases but is devoid of greenhouse gases.
For temperature A almost everyone uses a “measured average” of temperatures over the surface of the Earth. Although issues may exist regarding the algorithm used to compute a “measured average” Earth surface temperature, for the purposes of this discussion I’ll ignore all such issues and accept the value of 288 K as the value of temperature A (the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases).
Thus, we are left with coming up with a way to measure/estimate temperature B (the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only). We can’t directly measure B because we can’t remove greenhouse gases from the Earth’s atmosphere. This means we must use an algorithm (a model) to estimate B. I believe the algorithm most commonly used to compute the 255 K temperature estimate of B does NOT correspond to a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only”. As will be evident by my description (see below) of the commonly used algorithm, if anything that algorithm is more representative of a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases” than it is representative of a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only.”
If I am correct, then the use of 255 K in the computation of the Earth surface temperature difference with and without greenhouse gases is invalid.
Although there are many algorithms that can potentially lead to a 255 K temperature estimate of B, I now present the algorithm that I believe is most commonly used, and discuss why that algorithm does NOT represent “the temperature of the Earth’s surface in the presence of an atmosphere that is devoid of greenhouse gases”. I believe the algorithm described below represents the fundamental equation of radiative transfer for the Earth/Sun system assuming (a) an Earth absorption albedo of 0.3, and (b) an Earth emissivity of 1.
(1) The “effective temperature” of the Sun [i.e., the temperature of a sun-size spherical blackbody for which the radiated electromagnetic power (a) is representative of the total solar radiated power, and (b) has a power spectral density similar to the solar power spectral density] is approximately 5,778 K.
(2) For a spherical blackbody of radius 6.96×10^8 meters (the approximate radius of the sun) at a uniform surface temperature of 5,778 K, (a) the total radiated power is approximately 3.85×10^26 Watts, and (b) the radiated power density at a distance of 1.5×10^11 meters from the center of the blackbody (the approximate distance between the center of the Sun and the center of the Earth) is approximately 1,367 Watts per square meter.
(3) If the center of a sphere of radius 6.44×10^6 meters (the approximate radius of the Earth) is placed at a distance of 1.5×10^11 meters from the center of the Sun, to a good approximation the “effective absorbing area” of that sphere for blackbody radiation from the Sun is 1.3×10^14 square meters; and hence the solar power incident on the effective absorbing area of the sphere of radius 6.44×10^6 meters is approximately 1.78×10^17 Watts (1.3×10^14 square meters x 1,367 Watts per square meter).
(4) If the sphere of radius 6.44×10^6 meters absorbs electromagnetic energy with an “effective absorption albedo” of 0.3, then the solar power absorbed by the sphere is 1.25×10^17 Watts [1.78×10^17 Watts x (1 – 0.3)].
(5) A spherical blackbody (i.e., a spherical body whose surface radiates like a surface having an emissivity of 1) of radius 6.44×10^6 meters and at a temperature 254.87 K (hereafter rounded to 255 K) will radiate energy at the approximate rate of 1.25×10^17 Watts.
(6) If independent of the direction of energy incident on a sphere, the surface temperature of the sphere at any instant in time is everywhere the same, then the sphere possesses the property of perfect-thermal-conduction. Thus, for (a) an inert (no internal thermal energy source) perfect-thermal-conduction spherical body of radius 6.44×10^6 meters and uniform surface temperature 255 K whose center is placed at a distance of 1.5×10^11 meters from the center of an active (internal thermal energy source) spherical blackbody of radius 6.96×10^8 meters and uniform surface temperature 5,778 K, and (b) the inert perfect-thermal-conduction spherical body (i) absorbs electromagnetic energy with an effective absorption albedo of 0.3, and (ii) radiates electromagnetic energy with an emissivity of 1 then the perfect-thermal-conduction inert spherical body at temperature 255 K will be in radiation rate equilibrium with the active spherical blackbody at temperature 5,778 K. If the phrase “inert perfect-thermal-conduction spherical body of radius 6.44×10^6 meters” is replaced with the word “Earth,” and the phrase ” active spherical blackbody of radius 6.96×10^8 meters and uniform surface temperature 5,778 K” is replaced with the word “Sun”, it can be concluded that: If (a) an “Earth” at temperature 255 K is placed at a distance of 1.5×10^11 meters from the “Sun” and (b) the “Earth” (i) absorbs electromagnetic energy with an effective absorption albedo of 0.3, and (ii) radiates energy with an emissivity of 1, then the “Earth” will be in radiation rate equilibrium with the “Sun.” For the above conditions, the temperature of the “Earth” in radiation rate equilibrium with the “Sun” will be 255 K.
This completes the algorithm that I believe is commonly used to arrive at an “Earth’s characteristic-emission temperature” of 255 K, and hence is used to compute the 33 K temperature difference.
Even ignoring the facts that (1) it is incorrect to use the “average surface temperature” when computing radiation energy loss from a surface, and (2) in the presence of an atmosphere, (a) the blackbody radiation formula may not apply, and (b) blackbody radiation from the surface of the Earth is not the only mechanism for Earth energy loss to space (the atmosphere even without greenhouse gases will be heated by conduction from the Earth surface and both conduction and convection will cause that thermal energy to be distributed throughout the atmosphere, and the heated atmosphere will also radiate energy to space), the problem with using the 255 K temperature computed above to determine the difference between the Earth’s temperature with and without greenhouse gases is that the effective Earth absorption albedo of 0.3 used to generate the 255 K temperature is in part (mainly?) due to clouds in the atmosphere, and atmospheric clouds are created from water vapor, which is a greenhouse gas.
Thus an effective absorption albedo of 0.3 is based on the presence of a greenhouse gas–water vapor. It is illogical to compute a difference between two temperatures both of whose values are based on the presence of greenhouse gases and then claim that temperature difference represents the temperature difference with and without greenhouse gases. Without water vapor, there won’t be any clouds as we know them. Without clouds, the effective absorption albedo of the Earth will likely not be 0.3, and hence without the greenhouse gas water vapor, the Earth’s surface temperature in the absence of greenhouse gases is likely to be something other than 255 K. Thus, the 255 K “Earth characteristic-emission temperature” as computed using the algorithm above is NOT relevant to a discussion of the Earth surface temperature difference for an atmosphere that does and an atmosphere that does not contain greenhouse gases. Only if 0.3 is the effective absorption albedo of the Earth in the presence of an atmosphere devoid of all greenhouse gases is it fair to claim the presence of greenhouse gases increases the temperature of the Earth by 33 K.
Because clouds reflect a significant amount of incoming solar power, without water vapor I believe the effective absorption albedo of the Earth will be less than 0.3. If true, then more of the Sun’s energy will be absorbed by an Earth whose atmosphere is devoid of greenhouse gases than by an Earth whose atmosphere contains clouds formed from the greenhouse gas water vapor. This implies a higher Earth surface temperature in the absence of water vapor than the “Earth’s characteristic-emission temperature of 255 K”.
For an effective absorption albedo of 0, the temperature of the Earth in radiation rate equilibrium with the Sun will be approximately 278.64 K (hereafter rounded to 279 K). If this value is used as the Earth temperature in the presence of an atmosphere devoid of greenhouse gases, then it can be argued that the presence of greenhouse gases introduces a warming of approximately 9 K (288 Kelvin minus 279 K).
In summation, using the simplified arguments that I believe are also used to arrive at the 33 K temperature difference (i.e., assumed perfect-thermal-conduction Earth, blackbody Earth emission, greybody Earth absorption with an effective absorption albedo between 0 and 0.3, and ignoring atmospheric radiation to space for an Earth atmosphere devoid of greenhouse gases), I conclude the presence of greenhouse gases in the Earth’s atmosphere increases the Earth’s temperature by somewhere between 9 K and 33 K. Thus, I believe the claim that the presence of atmospheric greenhouse gases increases the temperature of the Earth by 33 K is based on an argument that has little relevance to the Earth’s temperature in the presence of an atmosphere devoid of greenhouse gases; and hence at best is misleading and at worst incorrect.
Note: Upon first publication – the guest author Reed Coray was accidentally and unintentionally omitted.
Actually, Dave , I should have said ‘solar influence’ rather than ‘solar activity’ which would have brought in astronomical changes such as the Milankovitch cycles.
I think we are all guessing at vague somethings at this distance in time. The snowball Earth idea is a concept rather than a fact.
The continents were in very different positions back then so that there are many bits of continental land mass bearing signs of glaciation that are now near the equator.
Luther Wu says:
December 26, 2011 at 10:41 am
R. Gates says:
December 26, 2011 at 10:27 am
I guess then that the moon, which is also a spinning sphere, would not have a boiling side and freezing side? But it does.
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What?
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A shocking revelation isn’t it. At the moon’s equator on the sunlit side the temperature is around 120C whereas on the opposite dark side, the temperature is about -230C. Amazing what an ocean and greenhouse atmosphere can do!
Sorry, not one of my best efforts, left out the all important link:
http://www.vukcevic.talktalk.net/TS.htm
John Brookes says [December 26, 2011 at 4:18 am]: “Anyway, well done for writing this, but make a bit more effort to think things through.”
John, I have given more thought to the problem. I don’t claim to know either (a) how to calculate or (b) what the temperature of the Earth’s surface would be in the absence of greenhouse gases. I believe such a computation is extremely complex and is might be meaningless except as an exercise. What I argue is that the 33 K temperature difference arrived at by (a) measuring an average Earth surface temperature and (b) the temperature of the Earth based on a measured absorptivity of 0.7 (albedo of 0.3, which is in large part due to clouds) and an emissivity of 1 is not relevant to a computation of the Earth with and without greenhouse gases. I, like others who have commented on my post, believe (and I admit I may be wrong) CAGW proponents want the temperature difference associated with greenhouse gases to be as large as possible because to do so increases the importance of greenhouse gases.
Richard M says [December 26, 2011 at 4:37 am]: “I’ve mentioned before that I thought this computation was missing from the literature. I believe you’ve made a great start. However, there is more to the question than just albedo.”
I agree, there is much more to the problem. I’ll outline the problem I would like to see someone attack. Assume you have a spherical solid (non-liquid) body that exists in the vacuum of space isolated from all other objects. Assume the spherical body has a radius, “R”, and uniformly-distributed mass equal to the mass of the Earth–this assumption will approximate the gravitational effects of the Earth at the spherical body’s surface. Assume the spherical body is NOT rotating–I know this does not represent the Earth, but I want to create spherical symmetry which in part requires the removal of all non-radial forces (primarily coriolis). Assume the surface of the sphere radiates like a graybody (blackbody with emissivity greater than 0 and less than 1) with the same emissivity, “e”, over the entire surface. Assume at the center of the sphere a source of energy exists (e.g., radioactive decay) at a constant but known energy rate “W” Watts. For these conditions, I believe it is relatively simple to compute the spherical body’s surface temperature, “T” when radiation-rate equilibrium is reached–i.e., when the power radiated to space from the spherical body’s surface is equal to the power generated internal to the sphere. Specifically, I argue that by symmetry the surface temperature of the spherical body will be everywhere the same and equal to
T = [W/(sigma*e*A)]^0.25
where “sigma” is the Stefan-Boltzmann constant, and “A = 4*pi*R^2” is the surface area of the sphere.
Okay, so much for the easy problem. Now, to the spherical body characterized above (a) uniformly over the surface of the sphere add oxygen and nitrogen where the masses of the added oxygen and added nitrogen are equal, respectively, to the mass of the oxygen and nitrogen in the Earth’s atmosphere, and (b) in addition to the atmosphere of (a), add elements/compounds of greenhouse gases uniformly over the sphere’s surface. I don’t care what the greenhouse gases are, but it would be nice to consider only adding CO2 as well as adding both CO2 and water vapor. If the latter (both CO2 and water vapor), let the masses of the added gases be equal to the “average” masses of CO2 and water vapor in the Earth’s atmosphere. Then for both (a) and (b) compute in radiation-rate equilibrium (1) the surface temperature of the sphere and (2) the temperature as a function of altitude above the sphere’s surface.
davidmhoffer says [December 26, 2011 at 5:01 am]: “Further, I don’t think the 255K number is supposed to represent the surface temperature of the earth without greenhouse gases. It is supposed to represent the temperature of the earth without the greenhouse EFFECT. If that is the case, then the logic presented to arrive at 255K seems quite reasonable.”
Lord Monckton specifically used the words “greenhouse gases” when discussing the 33 K temperature difference. I believe most people when discussing the source of the 33 K temperature difference either directly or indirectly also treat the source to primarily be “greenhouse gases.” If as you say “the 255 K number is supposed to represent the temperature of the earth without the greenhouse EFFECT“, then why mention greenhouse “GASES” at all? If you remove “greenhouse gases” from the “greenhouse effect”, any mention of “greenhouse gases” in a discussion of the 33 K temperature difference becomes irrelevant.
R. Gates says:
December 26, 2011 at 10:27 am
I guess then that the moon, which is also a spinning sphere, would not have a boiling side and freezing side?
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The moon… spinning huh, like Venus you must mean, right? R. Gates , that sure is a cheap shot to Lars P. Or, I guess using your logic, that even with a greenhouse gas-less Earth spinning at 1 RPM, it would have the sun lit side boiling hot and the dark side freezing cold. Why don’t you stop quoting from your AGW manual every now and then and just think of what you are saying. Any intelligent person could understand Lars point.
Once again, only Gerlich & Tscheuschner have explained where the 255 K comes from : a wrong calculation (that more scientists use in their models, which is ridiculus) meaning nothing of a non rotating earth without any atmosphere using Stefan-Boltzmann law the other around. The only reason we are not burning or freezing to death on earth is because of our 1 bar atmosphere temperates the climate.They also explain why calculating the real temperature of a rotation earth year along (still without an atmosphere though) is impossible, even with the most powerful computers.
Short of slight variations in mineral content affecting actual albedo, using the moon’s temperature on the sunlit and then the dark side should give you a pretty good idea of what the “average” temperature would be on earth. The daytime side of earth would be approximately 120C and the nighttime side would fall to a frosty -230C (but of course no frost would form as there would be no water vapor). The moons albedo is about .12, and you could make a reasonable assumption that the earth’s would be somewhere in the region without atmosphere, oceans, or clouds. But of course the earth has a different mineral mix than the moon, perhaps making a bit darker, so a slightly lower albedo is probably in order. But as the above extreme dirurnal temperature extremes the earth would experience indicate– the notion of an average temperature in a world devoid of ocean and atmosphere becomes meaningless.
The Earth does warm itself with heat from the mantle.
The oil man’s rule of thumb is that the ambient temperature goes up about 1 degree C for ever 1000 ft deeper you go down, so obviously the surface is cooler than the ground below it.
It’s trivial to calculate the temperature of the Earth without the conjecture of the “greenhouse effect” using Stefan-Boltzmann Law and the equation of state of gases in the atmosphere.
Simply calculate the temperature using the Sun’s flux, locate the altitude of the temperature playing with the albedo, then use the equation of state of the atmosphere to reach the surface.
The temperature will be to high since the calculation assumes a static atmosphere, i.e., no rotation, no slipping at the boundary and no wind.
The Stefan-Boltzmann Law can not be applied to gases since they do not have a continuous frequency distribution.
In short, your calculation is nonsense since you ignored gravity and what you refer to as the “greenhouse” effect is nothing more than the heat capacity of gases.
Dave Springer says:December 26, 2011 at 9:56 am
‘On must explain why temperature probes buried a meter or more deep in lunar regolith at 45 degrees latitude record a year-round constant temperature of -23C while the same probe buried a meter deep at the latitude on the earth records a year-round temperature of 11C.
Something is making the earth ~33C warmer than the moon’
Yes, correct. It’s the combination of GHG warming and lapse rate heating. The IPCC fails to take into account the lapse rate heating.
Next problem of the Kindergarten-Model:
The “surface temperature” of 288K which is measured in a Stevenson screen in ABSENCE of radiation and in 2m height is a “convective” air temperature.
Thus one can not compare an air temperature of 2 m height in absence of radiation with a radiation temperature in 0 m height.
Thanks cal for your response. However, the conduction of air itself is not what I was referring to. I was only referring to conduction from the surface to the air right above it. After that we would still have convection and the Coriolis effect to move the heat around. And, don’t forget time. Even a small effect can grow large over millions of years (unless something else counters it). I think the response by Alan D McIntire covers more closely what I am wondering about.
In addition, I believe the point has been made that O2 and N2 still radiate a small amount. Hence, we would still have some cooling at higher altitudes and a lapse rate.
I believe Myrrh attempted to answer my question by asserting a temperature of 69C. Ok fine. How is that determined? I’d really like to understand in more detail how such a system would behave. I really think we’d learn a lot if we could understand this GHG-less Earth.
Maybe the effect would be minimal and the answer is 255.x or maybe it would be large as Myrrh stated. I still don’t feel comfortable with the various responses.
Away from tectonic plate boundaries, the geothermal gradient is 22.1°C per km of depth (1°F per 70 feet of depth) in most of the world.
R. Gates says:
December 26, 2011 at 11:00 am
A shocking revelation isn’t it.
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What’s shocking is that you would try to sidetrack this thread with nonsense.
Only if you don’t count the temps in deep geological time, which were much longer at much higher temps, ~24°C. The “Hot House” state is more common than the “Ice Box”.
It would be advisable for the “carbon fanatics” to start walking naked and eating stones. Cotton is made by plants from CO2, Water and Sunlight,and making sugar, floor, food,etc too, in a process you ignore called photosynthesis, …and you know what? plants breath the CO2 you exhale, so you will have to stop breathing at all, …but rejoice! so you will cooperate with the world sustainability. Let Gaia take care of you once and for all!
The concept of temperature is only possible with an atmosphere.
This is a very good discussion but it’s clear that the very complexity of the human generated models from all sides is dwarfed by natures complexity.
Albedo changes with latitude and time of year when you consider the incident angle of incoming solar radiation, however outgoing radition must vary on a different schedule. I know the models are very complex and they have big computers, but nature would Demand many many thousands of variables which are full of discontinuous functions.
Good stuff and I would agree with the view that sensitivity is over estimated.
Bill
Too much eggnog. Far too many half-truths I’ll just comment on one:
Dave Springer says: “…The ocean has an albedo of effectively 0 and it covers 70% of the surface.”
Although these statements are independently true, juxtaposing them creates a false impression. The zenith angle of the sun varies across the globe, and reflectance (albedo) varies with zenith angle. To put it in simple terms for the noggists, have you ever stood on a cliff above the ocean and seen the sun reflecting off the water? It’s often too bright to look at, corresponding to an albedo close to 1.00. Albedo of the ocean is a function of zenith angle, clouds, temperature, waves, wind, ice, tide, flotsam, foam, and plankton (!). The range of ocean albedo overlaps that of ice.
Dave Springer also says: “.No one knows exactly what the earth’s average albedo really is. Climate models use values which differ by as much as 0.07 from one to another.”
Correct. You can see why variability in ocean albedo makes computation of average albedo problematic. It’s constantly changing. Worse yet, the ocean absorptivity-emissivity ratio is not 1.00, just another example of how the complexity of the earth will forestall development of anything close to a valid climate model within our lifetimes.
Why make the analysis so needlessly long and complex. Surely its much simpler to say, the solar energy in space near the earth is 1367 watts/sqM. The earth presents a disc to the sun while the actual surface is that of a sphere. The ratio of areas is pi *R^2/4*pi*R^2 or 1:4 hence the energy averaged over the surface of the Earth is 1367/4 = 342 watts/sqM. Now the Earth has an albedo of 0.3 which means 30% of this energy is reflected back out to space so the Earth only absorbs 243 watts/sqM. For a black body to emit 243 watts/sqM it must have a temperature of 255K. However without any green hosue gases in the atmosphere there would be no water vapour, hence no clouds and clouds are responsible for most of Earth’s albedo. If without clouds the albedo dropped to say 0 Earth woudl be receiving 342 watts/sqM and a black body radiating 342 watts/sqM would be at a temperature of 279K. In fact the Earth would still have an albedo greater than 0 but so the temperature would be somewhat less than 279K but non the less your argument is very sound. However it does not go far enough!!!!
What also needs to be recognised is that the green house warming of water vapour is logarithmic like all green house gases while the cloud cooling is close to linear. At very low concentrations of water vapour green hosue warming dominates greatly and the presence of water vapour warms the Earth, however as water vapour levels rise the INCREMENTAL warming impact of water vapour falls but as total concentration rises the impact of cloud cooling becomes far more significant and eventually dominates INCREMENTALLY over the warming impact. Water vapour content is determined by temperature (exponential relationship) so put another way, at very low temperatures water vapour warms the Earth but as the temperature rises the warming is increasingly offset by cloud cooling and eventually cloud cooling dominates so at high temperatures water vapour cooling the Earth,
In short, water vapour is a powerful stabilising influence that sets an operating point for Earth’s temperature and maintains it by strong negative feedback.
How strong; consider water vapour content of the atmosphere doubles for each 10C rise in temperature. Doubling water vapour at a first approximation doubles cloud mass (either denser clouds or more of them both of which increase albedo) and we knwo that clouds at present contribute about 80 watts/sqM of cooling. So potentially a 10C rise would incrementally double that for an incremental 80 watts/sqM of cooling. The green house warming due to water vapour is generally taken to be about 80-100 watts/sqM but this is the result of about 10 doublings so the incremental warming per doubling is about 8-10 watts/sqM. Thus the net impact of doubling water vapour is +10 – 80 or 70 watts/sqM of cooling. Or about 7 watts.sqM per degree C. Compare that with the claimed impact of doubling CO2 of 3 watts/sqM. A rise of 3/7 of a degree C would cancel it out and yes a net impact of 0.4-0.5C per doubling of CO2 sounds about right.
Clearly the anaysls above is very crude and approximate and could not be relied on for a good estimate but non the less it gives an idea of just how strong the negative feedeback from water vapour is.
jorgekafkazar said @ur momisugly December 26, 2011 at 1:00 pm
“You can see why variability in ocean albedo makes computation of average albedo problematic. It’s constantly changing. Worse yet, the ocean absorptivity-emissivity ratio is not 1.00, just another example of how the complexity of the earth will forestall development of anything close to a valid climate model within our lifetimes.”
The modelers admit to needing computers at least a couple of orders of magnitude greater computational power. Certainly not going to happen in my lifetime. One wonders if they will require orders of magnitude more energy to run…
In any complex system, with many variables, the complexity overwhelms simple analysis. Our minds try to extract simple cause & effect, but the system isn’t amenable to such analysis. While a living cell is even more complex, it is a suitable analogy, In the cell we have a complex mix of feedback, adaption and adjustment. While we can examine components of the cell and understand them, we cannot grasp the total complexity of what happens. We CAN describe the overall function of a cell, and we can come up with a simplified model. But we cannot remove any major part without destroying the whole. The complexity of climate involves chemical cycles and the biosphere, not just simple physics and climatology.
We can’t just remove all the green house gases and see what would happen. It makes as much sense as a cell sans mitochondria.
Both life and climate appeared to pre-modern man as evidence of the hand of God.
Maybe there was some truth in that!
Don’t forget that the core of the Earth is very hot – millions of degrees acording to the Goreacle….
kiwistonewall says:
December 26, 2011 at 1:24 pm
We can’t just remove all the green house gases and see what would happen. It makes as much sense as a cell sans mitochondria.
That is true if you wanted to do something with that model, but it doesn’t mean you can’t learn something interesting. If we want to understand the temperature impact of the gases called GHGs then it would be nice to understand what might happen if they were removed. Of course we have to be careful, but we might just learn a few things we didn’t know before. If the answer is 69C as Myrrh indicated that should set off alarm bells all over the place since the basic assumption of climate science is 15C.
Oops, sorry … I meant -18C for the last temperature.
David Springer at 9:56 stated
On must explain why temperature probes buried a meter or more deep in lunar regolith at 45 degrees latitude record a year-round constant temperature of -23C while the same probe buried a meter deep at the latitude on the earth records a year-round temperature of 11C.
The reason is much simpler than any claims of lapse rate or green house warming. The moon in effect rotates once every 28 days relative to the sun (28 times slower than Earth), it has no atmosphere to distribute heat and no water with high thermal mass to store it. Thus the surface has a very short time constant and tracks solar iput. However the relationship between energy input and temperature is a fourth power law which is extremely non linear. Thus at lunar noon the solar input is the full 1367 watts/sqM which yields a temperature of 121C yet during the long lunar night with no energy input the temeprature falls to -233C or just 40K. Burying a probe 3 M down under the surface does not remove this non linearity because the 3M only serves to average out the surface temperature fluctuation not the non linear relationship between energy input and surface temperature. When you average a T^4 relationsip over a large range of T the average will come out too low.