Guest Post by Willis Eschenbach
I’ve been considering the effect that temperature swings have on the average temperature of a planet. It comes up regarding the question of why the moon is so much colder than you’d expect. The albedo (reflectivity) of the moon is less than that of the Earth. You can see the difference in albedo in Figure 1. There are lots of parts of the Earth that are white from clouds, snow, and ice. But the moon is mostly gray. As a result, the Earth’s albedo is about 0.30, while the Moon’s albedo is only about 0.11. So the moon should be absorbing more energy than the Earth. And as a result, the surface of the moon should be just below the freezing temperature of water. But it’s not, it’s much colder.
Figure 1. Lunar surface temperature observations from the Apollo 15 mission. Red and yellow-green short horizontal bars on the left show the theoretical (red) and actual (yellow-green) lunar average temperatures. The violet and blue horizontal bars on the right show the theoretical Stefan-Boltzmann temperature of the Earth with no atmosphere (violet), and an approximation of how much such an Earth’s temperature would be lowered by a ± 50°C swing caused by the rotation of the Earth (light blue). Sunset temperature fluctuations omitted for clarity. DATA SOURCE
Like the Earth, averaged over its whole surface the moon receives about 342 watts per square metre (W/m2) of solar energy. We’re the same average distance from the sun, after all. The Earth reflects 30% of that back into space (albedo of 0.30), leaving about 240 W/m2. The moon, with a lower albedo, reflects less and absorbs more energy, about 304 W/m2.
And since the moon is in thermal equilibrium, it must radiate the same amount it receives from the sun, ~ 304 W/m2.
There is something called the “Stefan Boltzmann equation” (which I’ll call the “S-B equation” or simply “S-B”) that relates temperature (in kelvins) to thermal radiation (in watts per square metre). It says that radiation is proportional to the fourth power of the temperature.
Given that the moon must be radiating about 304 W/m2 of energy to space to balance the incoming energy, the corresponding blackbody lunar temperature given by the S-B equation is about half a degree Celsius. It is shown in Figure 1 by the short horizontal red line. This shows that theoretically the moon should be just below freezing.
But the measured actual average temperature of the lunar surface shown in Figure 1 is minus 77°C, way below freezing, as shown by the short horizontal yellow-green line …
So what’s going on? Does this mean that the S-B equation is incorrect, or that it doesn’t apply to the moon?
The key to the puzzle is that the average temperature doesn’t matter. It only matters that the average radiation is 304 W/m2. That is the absolute requirement set by thermodynamics—the average radiation emitted by the moon must equal the radiation the moon receives from the sun, 304 W/m2.
But the radiation is proportional to the fourth power of temperature. This means when the temperature is high, there is a whole lot more radiation, but when it is low, the reduction in radiation is not as great. As a result, if there are temperature swings, they always make the surface radiate more energy. As a result of radiating more energy, the surface temperature cools. So in an equilibrium situation like the moon, where the amount of emitted radiation is fixed, temperature swings always lower the average surface temperature.
For confirmation, in Figure 1 above, if we first convert the moment-by-moment lunar surface temperatures to the corresponding amounts of radiation and then average them, the average is 313 W/m2. This is only trivially different from the 304 W/m2 we got from the first-principles calculation involving the incoming sunlight and the lunar albedo. And while this precise an agreement is somewhat coincidental (given that our data is from one single lunar location), it certainly explains the large difference between simplistic theory and actual observations.
So there is no contradiction at all between the lunar temperature and the S-B calculation. The average temperature is lowered by the swings, while the average radiation stays the same. The actual lunar temperature pattern is one of the many possible temperature variations that could give the same average radiation, 304 W/m2.
Now, here’s an oddity. The low average lunar temperature is a consequence of the size of the temperature swings. The bigger the temperature swings, the lower the average temperature. If the moon rotated faster, the swings would be smaller, and the average temperature would be warmer. If there were no swings in temperature at all and the lunar surface were somehow evenly warmed all over, the moon would be just barely below freezing. In fact, anything that reduces the variations in temperature would raise the average temperature of the moon.
One thing that could reduce the swings would be if the moon had an atmosphere, even if that atmosphere had no greenhouse gases (“GHGs”) and was perfectly transparent to infrared. In general, one effect of even a perfectly transparent atmosphere is that it transports energy from where it is warm to where it is cold. Of course, this reduces the temperature swings and differences. And that in turn would slightly warm the moon.
A second way that even a perfectly transparent GHG-free atmosphere would warm the moon is that the atmosphere adds thermal mass to the system. Because the atmosphere needs to be heated and cooled as well as the surface, this will also reduce the temperature swings, and again will slightly warm the surface in consequence. It’s not a lot of thermal mass, however, and only the lowest part has a significant diurnal temperature fluctuation. Finally, the specific heat of the atmosphere is only about a quarter that of the water. As a result of this combination of factors, this is a fairly minor effect.
Now, I want to stop here and make a very important point. These last two phenomena mean that the moon with a perfectly transparent GHG-free atmosphere would be warmer than the moon without such an atmosphere. But a transparent atmosphere could never raise the moon’s temperature above the S-B blackbody temperature of half a degree Celsius.
The proof of this is trivially simple, and is done by contradiction. Suppose a perfectly transparent atmosphere could raise the average temperature of the moon above the blackbody temperature, which is the temperature at which it emits 304 W/m2.
But the lunar surface is the only thing that can emit energy in the system, because the atmosphere is transparent and has no GHGs. So if the surface were warmer than the S-B theoretical temperature, the surface would be emitting more than 304 W/m2 to space, while only absorbing 304 W/m2, and that would make it into a perpetual motion machine. Q.E.D.
So while a perfectly transparent atmosphere with no GHGs can reduce the amount of cooling that results from temperature swings, it cannot do more than reduce the cooling. There is a physical limit to how much it can warm the planet. At a maximum, if all the temperature swings were perfectly evened out, we can only get back to S-B temperature, not above it. This means that for example, a transparent atmosphere could not be responsible for the Earth’s current temperature, because the Earth’s temperature is well above the S-B theoretical temperature of ~ -18°C.
Having gotten that far, I wanted to consider what the temperature swings of the Earth might be like without an atmosphere. Basic calculations show that with the current albedo, the Earth with no atmosphere would be at a blackbody temperature of 240 W/m2 ≈ -18°C. But how much would the rotation cool the planet?
Unfortunately, the moon rotates so slowly that it is not a good analogue to the Earth. There is one bit of lunar information we can use, however. This is how fast the moon cools after dark. In that case the moon and the Earth without atmosphere would be roughly equivalent, both simply radiating to outer space. At lunar sunset, the moon’s surface temperature shown in Figure 1 is about -60°C. Over the next 30 hours, it drops steadily at a rate of about 4°C per hour. At that point the temperature is about -180°C. From there it only cools slightly for the next two weeks, because the radiation is so low. For example, at its coolest the lunar surface is at about -191°C, and at that point it is radiating a whopping two and a half watts per square metre … and as a result the radiative cooling is very, very slow.
So … for a back of the envelope calculation, we might estimate that the Earth would cool at about the lunar rate of 4°C per hour for 12 hours. During that time, it would drop by about 50°C (90°F). During the day, it might warm about the same above the average. So, we might figure that the temperature swings on the Earth without an atmosphere might be on the order of ± 50°C. (As we would expect, actual temperature swings on Earth are much smaller, with a maximum of about ± 20-25 °C, usually in the desert regions.)
How much would this ±50° swing with no atmosphere cool the planet?
Thanks to a bit of nice math from Dr. Robert Brown (here), we know that if dT is the size of the swing in temperature above and below the average, and T is the temperature of the center of the swing, the radiation varies by 1 + 6 * (dT/T)^2. With some more math (see the appendix), this would indicate that if the amount of solar energy hitting the planet is 240 W/m2 (≈ -18°C) and the swings were ± 50°C, the average temperature would be – 33°C. Some of the warming from that chilly temperature is from the atmosphere itself, and some is from the greenhouse effect.
This in turn indicates another curiosity. I’ve always assumed that the warming from the GHGs was due solely to the direct warming effects of the radiation. But a characteristic of the greenhouse radiation (downwelling longwave radiation, also called DLR) is that it is there both day and night, and from equator to poles. Oh, there are certainly differences in radiation from different locations and times. But overall, one of the big effects of the greenhouse radiation is that it greatly reduces the temperature swings because it provides extra energy in the times and places where the solar energy is not present or is greatly reduced.
This means that the greenhouse effect warms the earth in two ways—directly, and also indirectly by reducing the temperature swings. That’s news to me, and it reminds me that the best thing about studying the climate is that there is always more for me to learn.
Finally, as the planetary system warms, each additional degree of warming comes at a greater and greater cost in terms of the energy needed to warm the planet that one degree.
Part of this effect is because the cooling radiation is rising as the fourth power of the temperature. Part of the effect is because Murphy never sleeps, so that just like with your car engine, parasitic losses (losses of sensible and latent heat from the surface) go up faster than the increase in driving energy. And lastly, there are a number of homeostatic mechanisms in the natural climate system that work together to keep the earth from overheating.
These thermostatic mechanisms include, among others,
• the daily timing and number of tropical thunderstorms.
• the fact that clouds warm the Earth in the winter and cool it in the summer.
• the El Niño/La Niña ocean energy release mechanism.
These work together with other such mechanisms to maintain the whole system stable to within about half a degree per century. This is a variation in temperature of less than 0.2%. Note that doesn’t mean less than two percent. The global average temperature has changed less than two tenths of a percent in a century, an amazing stability for such an incredibly complex system ruled by something as ethereal as clouds and water vapor … I can only ascribe that temperature stability to the existence of such multiple, overlapping, redundant thermostatic mechanisms.
As a result, while the greenhouse effect has done the heavy lifting to get the planet up to its current temperature, at the present equilibrium condition the effect of variations in forcing is counterbalanced by changes in albedo and cloud composition and energy throughput, with very little resulting change in temperature.
Best to all, full moon tonight, crisp and crystalline, I’m going outside for some moon-viewing.
O beautiful full moon! Circling the pond all night even to the end Matsuo Basho, 1644-1694
w.
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
G’morn from sunny Brisbane Willis
You replied
I don’t think we’re quite on the same wavelength, which is my fault for not articulating my thoughts well enough. With your permission I’ll try an analogy, I like analogies.
I’ve designed a new cook top element. It consists of an inner ring R1 and a larger outer ring R2. Total diametre is 6 inches. Each ring has a seperate on/off temperature dial.
I switch on R1 and turn the dial to a point where R1 heats up to 60c. I leave R2 dead.
What is the AVERAGE temperature of the cook top element? (R1+R2)/2=30c
Now I take a pot with a base diametre of 6 inches, fill it with tap water and place it on the element.
Will I be able to heat that water to a temperature above 30c? I believe I can.
Even though heat is coming from R1 only (at 60c), conduction and convection within the pot of water will ensure the temperature of the water will be well above THE AVERAGE element temperature of 30c.
If I understand your point, you are saying this (what is now) warmer water will also conduct to the cold R2 heating it up until equilibrium, hence the average temperature of the WHOLE element goes above 30c thereby the average temperature of the water can never be above that of the element.
This maybe true…HOWEVER
In the case of a gaseous atmosphere, once the parcel of atmosphere near the cold surface (at the pole) cools down, it can no longer convect/advect like the water in the pot does. Warm parcel of gas above, cool parcel of gas below, equals temperature inversion.
Therefore, so long as some portion of the atmosphere aloft is blocked from contacting the ground, it can not conduct away its heat nor will it radiate away its heat due to it being a non-GHG gas.
So, by the time we get to daybreak on day 2, we have an atmosphere that is now warmer (as an average) than it was the day before. The whole process begins again with the very warm equator warming more of the atmosphere etc etc.
When will this process stall? that’s a discussion all of its own, but it won’t stall before the AVERAGE temperature of the WHOLE atmosphere is higher than the AVERAGE temperature of the WHOLE surface.
A few of dot points that keep sticking in my mind when I think about the above
* The surface is 2 dimensional but the atmosphere is 3 dimensional
* During the warming phase, convection and overturning warms the whole atmosphere (effectively overcoming the 3rd dimesion)
* During the cooling phase, denser cooler parcel of atmosphere near the surface prevents the parcel above from contacting the surface. The cooling phase cannot overcome the 3rd dimension.
* temperature inversions on Earth are most common at the poles and at coastal upwelling zones (from wiki “In the polar regions during winter, inversions are nearly always present over land.”)
http://en.wikipedia.org/wiki/Inversion_%28meteorology%29
* Conductive warming of the atmosphere is ALWAYS quicker/more efficient, than conductive cooling. This leads to residual warmth left in the atmosphere at the end of a day/night cycle.
p.s. At the wiki link I provided above it also says this..
I though that was interesting.
best regards
Dewitt writes “Then he’s wrong. The temperature determines the density. ”
…and density defines temperature (in a planetary/gravity situation) is what they’re saying in my brief readings of this. I certainly dont proclaim to be an expert and I’m not defending this argument but I will defend against those who put up strawman arguments.
sky says:
January 10, 2012 at 4:20 pm
” ALL substances with a temperature above Kelvin zero–not just solid bodies and GHGs–radiate energy whose flux density is related to its mass density and specific heat”
Sky, this is simply not correct. If you had said that solid bodies radiate energy related to their temperature and emissivity then this would have been correct. Mass and specific heat have nothing to do with it.
The pressure-depth relates to the actual depth via the integral of densities above the point of measurement. That requires knowing the density profile at the time of measurement. But if the dive takes several days (and the probe drifts), the previous densities no longer apply.
I’m surprised that inertial sensors weren’t used as an adjunct measurement system; to provide a second point of reference and to measure lateral translation, sideways drift due to current as different depths. The inertial sensors detects acceleration and orientation (6 degrees of freedom). They could be re-zeroed when surfaced.
Previous comment should have made it clear that I am writing about Argo probes that are ostensibly used to determine the amount and of energy and its convective flow within the major thermal mass of the climate system; the oceans.
But I got ahead of myself in commenting…
Willis, further my article @ur momisugly January 10, 3:03 pm, and your footnoted comment thereon; Thanks for your support; it is invaluable. I repeat my précis on your article, and seek your help:
If we consider Earth without an atmosphere or vegetation etc, it is rather speculative as to what the surface conditions might be compared with the moon. It would be affected by how long in this condition, increased gravitational attraction to uninvited planetoids and stuff, what volcanism, continental masses versus what we currently know as ocean crust, tectonics, lack of sedimentation and precipitative erosion, thermal/albedo qualities of the regolith and, and…
Putting all that aside, if we consider the hotspot under the sun to be plus/minus 20 degrees in longitude, that would give a rotation period of ~ 2.7 hours based on the current day length. Intuitively, I feel that it would take less than 2 hours for the hotspot to reach equilibrium T. Does this seem to be reasonable conjecture to you?
TimTheToolMan says:
January 10, 2012 at 3:59 pm
Tim, you can’t explain Nikolov’s hypothesis. Neither can anyone else. Nikolov might be able to, but he has bowed out.
So you can’t say that “Nickolov at the very least has made the point that …”, because Nikolov hasn’t shown anything at all. He has not shown that density determines temperature, that’s fifty steps beyond what he has shown.
And yet you and others still claim that somehow he is right, despite the fact that no one can explain to me how his hypothesis is supposed to work!
Man. Talk about the Emperor and no clothes. You sit there and freely admit that you don’t know what Nikolov is saying, but you still believe him. Not only that, you advise me to “think about his meaning”, even though you can’t explain his meaning. Why the hell should I think about his meaning, when no one (including Nikolov) will say what he means?
I swear, sometimes I think I’ve fallen down the rabbit hole, where I’m supposed to believe six impossible things before breakfast …
w.
Tim, one further thought. You say that Nikolov made the point that “atmosphere density determines its temperature”. In that regard, NASA refers to the “equation of state“. In one of its variants, that says
p = R ρ T
where p is pressure, R is the gas constant, ρ (rho) is density, and T is temperature. Solving for T gives us
T = p / (R ρ)
This means that density does not uniquely determine temperature as you say. You need to know pressure as well.
So if you still claim that Nikolov showed that only density is needed to determine temperature, you’ll have to explain how that works.
w.
Willis writes “Man. Talk about the Emperor and no clothes. You sit there and freely admit that you don’t know what Nikolov is saying, but you still believe him. ”
Whereas I specifically said “I certainly dont proclaim to be an expert and I’m not defending this argument but I will defend against those who put up strawman arguments.”
So no I dont believe him because I’m not defending his argument. But Frankly to make a strawman argument about pressure (not) doing work and therefore Nikolov having nothing useful to say diverts any useful discussion towards something you know you didn’t understand in the first place.
You appear to agree that hypothetically adding non-GHG atmosphere to the moon will raise the average temperature of the moon. Ira doesn’t believe this. Whether you follow through with that thought or not is up to you. I expect less than 1% of WUWT readers read my posts but far more will read your thoughts…
Willis writes “This means that density does not uniquely determine temperature as you say. You need to know pressure as well.”
Where did I say uniquely?
ferd berple says:
January 9, 2012 at 12:01 am
First, ignore all wave numbers larger than 2,500 cm-1 and smaller than 50 cm-1, then look at the line strengths.
H2O and CO2 have line strengths greater than 1e-19
N2 has a line strength around 1e-28
O2 has two groups of lines, one around 1e-28 and another at 1e-24.
These are very big differences. In addition, those plots assume zero pressure. When typical pressures are considered, the individual lines merge into bands, which explains how CO2 and H2O bands overlap even though individual lines do not.
BTW how do you get CO2 is 1e-23? I don’t see it in the plots.
Willis Eschenbach says:
January 9, 2012 at 10:13 am
Baa Humbug says:
January 9, 2012 at 2:19 am
Easy, the atmosphere will be the same temperature as the maximum daytime temperature. Only the lower few meters would conduct heat back to the surface. Above that, the atmosphere would be at a constant (isothermal) temperature. The only reason the Earth’s atmosphere cools with increasing height is because Greenhouse gases radiate energy. The troposphere is cooled by water vapor. The stratosphere and mesosphere are cooled by CO2.
Robert Brown says:
January 10, 2012 at 9:19 am
—————————-
Thank you for your educational comments. I have a question if you are willing. My questions at this point are primarily directed at the effects of water vapor and clouds on SWR entering the ocean and on the earths energy budget relative to the “residence time” of changes in SWR versed changes in LWR. You spoke of water vapor (clear sky) in this comment and its well known GHE. My question is related and will contain an assertion or two which are also questions.
As I understand it, about 98% of TSI energy lies between about 250 nm in the UV and 4.0 microns; with the remaining as 1% left over at each end. Spectral graphs often have superimposed on them the actual ground level (air Mass once) spectrum; that shows the amounts of that energy taken out by primarily O2, O3, and H2O, in the case of H2O which absorbs in the visible and near IR perhaps 20% of the total solar energy is capture by water VAPOR (clear sky) clouds are an additional loss over and above that. So water vapor alone prevents 20% of TSI from reaching the surface. It is understandable how this is thought to heat the atmosphere, however, is it preventing that energy from impacting a part of the earth/ocean/atmosphere system which has an even longer residence time then the GHG atmosphere, thus lowerering earths energy budget, despite increasing energy to the atmosphere? This is stated within two assertions as follows…
1. At its most basic only two things can effect the heat content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system.
2. It therefore follows that any effect which increases the residence time of LW energy in the atmosphere, but reduces the input of SW and LWIR energy entering the oceans, (such as clear sky water vapor and clouds) causes a net reduction in the earth’s energy balance, proportioned to the energy change involved, relative to the residence time of the radiations involved. (To paraphrase an old maxium, one unit of energy absorbed into the oceans is worth two in the atmosphere)
Thanks for your thoughts.
Bob Fernley-Jones says:
January 10, 2012 at 7:05 pm
Willis, further my article @ur momisugly January 10, 3:03 pm, and your footnoted comment thereon; Thanks for your support; it is invaluable. I repeat my précis on your article, and seek your help:
Thanks, Bob. The application of S-B is not “nonsense”. It gives us a theoretical upper limit to the temperature of the planet. We know that if the planet’s average temperature exceeds that value, it is emitting more than it is radiating. This is useful information, because it helps us identify impossible claims.
Indeed it would, which is why I described my calculations as “back of the envelope”. I wanted to at least get some sense of the size of the swings. The size I arrived at (±50°C) seems like a reasonable first cut, given the modern values of about half of that.
You are talking about a main difference between the earth and the moon, which is that the moon rotates slowly enough that there is little thermal lag in the system. I just took a years worth of data for the nearest weather station to me (Santa Rosa, CA). Peak insolation, of course, is at noon. Peak temperature is at about 2:45 PM. So that’s the lag between the peaks here. Not sure if that helps.
All the best,
w.
TimTheToolMan says:
January 10, 2012 at 8:30 pm
You said that “atmosphere density determines its temperature”. But density doesn’t determine temperature, so your statement is untrue. The word “determines” contains the claim of exclusivity, because if other things were required, then density couldn’t determine the temperature. You’d need other information.
w.
Robert Clemenzi says:
January 10, 2012 at 11:36 pm
Ah, there’s the problem. You misunderstand the reason the atmosphere cools with increasing height. The reason that the atmosphere has a dry adiabatic lapse rate (cools with increasing altitude) has nothing at all to do with greenhouse gases. It is a consequence of the distribution of energy within the air parcel, a distribution created by gravity. In fact, the size of the lapse rate is given by
Dry Adiabatic Lapse Rate = g / Cp
where g is the force of gravity, and Cp is the specific heat of air at constant pressure. For the Earth’s atmosphere, it is a decrease of 9.8°C per kilometre of elevation. Note that the equation says nothing about GHGs, as they are immaterial in calculating the dry adiabatic lapse rate.
w.
Tim, you started out the Nikolov discussion by saying the following:
Now, for you to tell me that I don’t understand Nikolov, that must mean you understand him, no? Otherwise, how could you claim I don’t understand him?
You go on to say what you think Nikolov is saying, something about pressure and temperature.
Now, you say the following:
Indeed, you are defending his argument, you are sounding very much like you believe him about how “pressure defines the temperature” whatever that means, and you are telling me I don’t understand Nikolov’s argument in the bargain.
Tim, I don’t understand Nikolov’s theory. And by your own admission, you don’t understand Nikolov’s theory. In fact, nobody I can find understands Nikolov’s theory. I asked Nikolov to explain it in clear simple terms. He disappeared. I asked if anyone can explain the core of it in a few clear sentences. No one has been able to do so.
I don’t know whether or not Nikolov has anything useful to say because I can’t understand his theory, and neither can you, and neither can anyone else I can find. I find it absolutely ludicrous that so many people are defending something that they don’t understand … humans are indeed strange.
w.
David says:
January 10, 2012 at 11:40 pm
I’m not Robert Brown, David, and I hope he answers you. My problem with what you say is that I have no idea what you are calling the “residence time” of energy. It would be extremely useful if you could answer the following two questions:
How do you measure the residence time of energy, and in what units is it measured?
All the best,
w.
Willis Eschenbach says:
January 11, 2012 at 1:02 am
The cooling of 9.8°C per kilometre of elevation is simply the dry adiabatic lapse rate and I agree that this has nothing to do with greenhouse gases. However, the actual troposphere lapse rate is 6.5 °C per kilometre of elevation. It is water vapor that cools the atmosphere and creates this difference.
To be perfectly clear, without greenhouse gases to cool the atmosphere, the atmosphere would be a single, very hot, temperature and there would be no convection to cool the surface.
Sorry, I was not clear. Without greenhouse gases, the actual troposphere lapse rate would be zero because the atmosphere would be isothermal. The greenhouse gases cool the atmosphere producing a lapse rate of 6.5 °C per kilometre of elevation.
Robert Clemenzi,
Absent greenhouse gases, the atmosphere would not be isothermal. The surface temperature at the poles would still be colder than the surface temperature at the equator. That means the pressure will decrease less rapidly with altitude at the equator than at the poles. That causes what’s called a pressure gradient force which in turn causes air circulation. Any air circulation will move heat around and force the lapse rate towards the adiabatic rate. The free energy for the work needed to do this comes from the temperature difference between the poles and the equator. The circulation will increase the temperature at the poles and decrease the temperature at the equator by turbulent convection from the moving air. So even without greenhouse gases in the atmosphere, you would still have something like the Hadley cells and probably jet streams in the Northern and Southern hemispheres as well.
@Willis
Do you have no shame? That was SO out of context. It was preceded by saying the ocean primarily cools by evaporation. I have given the facts many times which anyone may verify in the literature. Ocean cools approximately 70% via evaporation, 25% via radiation, and 5% via conduction. Given that’s true, and it IS true, then IF the ocean doesn’t cool by longwave radiation then neither is it heated by it.
You have continually failed to dispute this, Willis. Calling it lunacy doesn’t make it that. It just makes you look weak in that you lack a substantive rebuttal.
Dear Mr. Eschenbach,
At the risk of making myself rediculous, I want to ask you something.
After a closer view to this discussion I came to the following consideration:
The main difference between Moon and Earth is that the the Earth is covered with water (7/10). So I ask myself, how is the behavior of water regarding radiation.
If my understanding of physics is right, water is able to transmit absorbed energy upwards (mostly radiation if a transparent atmosphere is assumed) as well as downwards (by convection, other thermodynamical processes and radiation) by nearly the same amounts. If so, you need the energy twice to get a radiative equilibrium at TOA, because only half of the stored energy can be radiated upwards. Calculating the required S-B-temperature of water considereing an albedo of 0.1 for surface only (no clouds!) we get approx. 320K for a fully transparent atmosphere. Assuming that Ramanathan (1997) was right and we have a net cooling of approx. 50W/m² by clouds, we get about 305K.
Summarize this with the equilibrium for solids of 255K in their parts (about 7/10 of water and 3/10 of land) we get an equilibrium that is very close to the current temperatures, approx. 286K.
Now my question: is this too easy thinking or is it really that simple?
Willis writes “I find it absolutely ludicrous that so many people are defending something that they don’t understand … humans are indeed strange.”
Well thats your perception I suppose. I’m not defending his argument, I’m merely stating its not the strawman argument you’ve put forward (as did Ira) and given my own brief understanding of what his argument is very roughly about. I dont pretend to understand the detail and yet you put words into my mouth about exclusivity.
Anyway, it seem to me that its inescapable that the amount of atmosphere has a bearing on the temperature of a planet but its clearly not a simplistic argument that you’ve been putting forward. There is time of rotation, heat capacity as well as inversion layers vs convection and all sorts of possibilities beyond PV=nRT.
So whilst Nickolov may not even be correct about his version of the detail (whatever that is) the idea may still have merit.
Robert Brown says:
January 10, 2012 at 10:01 am
After that it became ellipsoid to the degree that it isn’t quite a completely rigid body.
“Don’t mind me, I just teach this stuff. Or if you prefer, can you provide evidence to back your description?”
Don’t mind me either. I’m just an engineer who designs mechanical systems around these principles. Teaching is a hobby.
The tides slow down the rotation they don’t lock it into any specific configuration. This loss in angular momentum is of course translated into heat. If it weren’t for mass distribution asymmetry what would determine which half of the moon would come to rest facing toward us? And if there is no preference for any particular face how can a particular face be the chosen one? In other words why isn’t it the other side of the moon that’s facing us?
One might also want to think about Mercury which is tidally locked into 1.5 revolutions per orbit and how that’s possible.
Per your request (and an answer to my question about Mercury’s wonky tidal lock):
Planetary Siences
Imke De Pater, Jack Jonathan Lissauer
http://books.google.com/books?id=RaJdy3_VINQC&pg=PA34&lpg=PA34&ots=oozdNGPUJ0&dq=tidal+lock+mass+distribution+asymmetry
Willis Eschenbach says:
January 11, 2012 at 1:02 am
You are shedding more heat than light (pun intended) on this gravitational heating canard. In your missive above you quote the formula for dry adiabatic lapse rate. Dry adiabatic lapse rate is for a RISING parcel of air i.e. a parcel of air in vertical motion. Nicholov and Huffman are describing static parcels of air. In a static parcel the decrease in temperature is simply a matter of moving farther away from the source of heat which in dry air the heat source is the surface until you reach thermopause where highly energetic regions of solar spectrum begin to directly heat the air in a big way. At the edge of the atmosphere the temperature is in the thousands of degrees!