The Moon is a Cold Mistress

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.

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jae
January 9, 2012 7:56 pm

Willis:
OMG: YOU SAID THIS:
What I and others have been consistently calling the “S-B temperature” in this thread is the temperature the planet would have if it were an airless blackbody. It is the temperature calculated from the S-B equation for the amount of solar energy hitting the planet.””
WOW. Please explain how this relates to the real Planet Earth, sir! I am very confused about this statement. We do NOT have an “airless body!” WTF ARE YOU SAYING, MAN?

Rosco
January 9, 2012 8:16 pm

Surely the “day” side of the Moon is radiating way more than 304 W/sq m – up to almost equal to the solar constant whilst the “dark” side is radiating way less – say even as low as a few watts/sq m if NASA is correct for the minimum temperatures.
These would be the absolute maximum and minimum and there would be variation over the whole sphere.

January 9, 2012 8:25 pm

It is the same thing coldlynx said, with the same problem. You are claiming that a surface can warm an atmosphere to a temperature warmer than the surface. To do that, you’d have to have heat flowing from the cooler object (the surface) to the warmer object (the atmosphere).

I’m sorry I didn’t make myself clear.
I am claiming the equator can warm the atmosphere to a level higher than the higher lattitudes hence the atmosphere will have a higher average temperature than the average temperature of the surface.
The warmest “object” is the equator. Lets forget about averages for a moment.
It is the equator which warms the atmosphere (the most) via conduction. And because the atmosphere is a gas, it rises and spreads when warmed via conduction. However, the reverse, i.e. cooling via conduction, cannot happen as quickly, or as efficiently if you will, due to temperature inversion.
I do not claim the atmosphere will ever get warmer than the surface at the equator at noon, but the average temperature of the (whole) atmosphere will be warmer than the average temperature of the (whole) surface.
Why would the atmosphere…say 100mtrs altitude at the poles… be warmer than the surface at the poles? Because temperature inversion will not allow the atmosphere to cool down to the level of the surface.
The energy comes from the very warm equator via conduction. Is distributed globally via thermals and convection and conduction but it cannot cool as efficiently without radiation, hence the build up.
regards

Rosco
January 9, 2012 8:28 pm

I simply must disagree with this “But a transparent atmosphere could never raise the moon’s temperature above the S-B blackbody temperature of half a degree Celsius.”
Clearly the Moon gets much hotter than half a degree celsius so I find the reference to blackbody temperatures misleading. The radiation from the Sun at ~1367 W/sq m hits half of the sphere of the moon thus the blackbody temperature of this part is ~ 381 K whilst illuminated. As the Moon spins the radiation levels drop very low on the dark side – I’ll say again I see no reason to even consider an average – it is basically meaningless and a tool of the AGW to use to justify their theories.
This clearly demonstrates that probably the most significant feature of the Earth with respect to habitability is the period of rotation followed by the amazing properties of water followed by a freely convecting atmosphere.

Tim Folkerts
January 9, 2012 9:06 pm

Willis,
Here is an interesting thought experiment. Set a large rock out on a sunny day on the moon (or lifeless earth-like object). Once it reaches ~ 100 C, place the rock into a very well insulated container. By reducing convection and conduction and radiation, you could in principle keep that rock very close to 100 C all thru the night. By repeating this process, you could in principle heat an arbitrarily large amount to rock up toward the MAXIMUM daily temperature. As long as the RADIATING SURFACE is at the S-B effective temperature, other parts could be warmed up to a level approaching the MAX temperature with clever insulation.
In a similar way, in principle it could be possible to heat the IR-transparent atmosphere up to a temperature approaching the MAXIMUM surface temperature.
* heat some air by putting it into thermal contact with the hottest part of the ground
* move the hottest air so it is out of thermal contact with the ground
* repeat.
So I can sort of see the argument some are making. It could be possible to get the whole bottom layer of the atmosphere to ~ the maximum surface temperature. (I do think the lapse rate would keep the whole atmosphere from getting this warm
**********************************************
In practice I don’t think this would work without some sort of “compartments” in the atmosphere. The actual atmosphere would warm and rise in the “hot afternoon” areas (but cool as it rose). This would tend to set up a circulation with hot air rising, spreading out at high altitudes, and heading toward the night side, where the air would fall, warming the cool side. The cool air from the night side would “get sucked back” toward the hot side to replace the rising warm air. Only if you could “trap” the hot air in some compartment above the ground (out of thermal contact with the ground as well) (akin to putting the rocks in the well-insulated containers) would this even start to work effectively (at least in my opinion).

Dave
January 9, 2012 9:10 pm

” So in an equilibrium situation like the moon, where the amount of emitted radiation is fixed, temperature swings always lower the average surface temperature.”
Hi Willis,
Seems like a mistake here before you’ve even got started. The moon has thermal mass, so incoming radiation and outgoing radiation only balance in the longer term. Temperature variation is a short-term effect, You can only assume the balance when working in the long-term, so to do so whilst working in the short-term invalidates the analysis.
Regards
Dave

Rosco
January 9, 2012 9:16 pm

I hate to bring up argument four from Radiating the Ocean but I cannot agree with the argument :-
“In addition there are losses of sensible heat (~ 30 w/m2) and evaporative losses (~ 70 w/m2). That’s a total loss of 390 + 30 + 70 = 490 w/m2.
But the average solar input to the surface is only about 170 watts/square metre.
So if the DLR isn’t heating the ocean, with heat gains of only the solar 170 w/m2 and losses of 390 w/m2 … then why isn’t the ocean an ice-cube?”
As I see it the answer is simple – during the day with the Sun vertically overhead the insolation is probably 4 times 170 W/sq m – 1367 W/sq m solar constant (Trenberth) X 51 % heats the Earth’s surface (IPCC). The physics I was taught said the insolation could be calculated by multplying half the solar constant by the latitude of the point of interest at the equinox. So for where I live, 27 S, the maximum insolation in those circumstances would be in excess of 600 W/sq m.
Clearly this would fry us all – although we can get fairly warm here – if it weren’t for the the energy soaked up by evaporating water, a convecting atmosphere and the relatively short day length – ie 12 hours max baking followed by 12 hours cooling.
There seems to be supported by the solar panels I once foolishly bought. They are rated at 175 W per 1.3 sq m. This reduces to ~135 W/sq m which is a reasonable 25 odd % efficiency.
Accordingly I believe the role of the Sun has been downplayed to support AGW theory. Also I refuse to believe 99% of the atmosphere comprised of Nitrogen and Oxygen does not become heated and as such radiate IR and therefore rendering the amount of IR from CO2 negligible – negligible to my way of thinking is less than 1 % and CO2 is way less than that.

Rosco
January 9, 2012 9:31 pm

I think that the oceans are responsible for most of the radiation at night, they warm the air above the surface which begins to convect as the land has lost the “stored heat”. Thus a “land breeze” sets in drawing the cooler air to the ocean to be warmed and humidified and this warmed air helps slow heat loss and provides the radiative effect observed at night – explains why the seaside is always warmer than inland at night and why Palm Trees can grow in Scotland – I’ve seen ’em.
I still think trace gases with low specific heat properties cannot create a large radiative effect and the amount of energy is small compared to latent heat of water – just my opinion but AGW theory seems to be the CO2 tail wagging the water dog.
It’s all a cycle as someone once said.

DeWitt Payne
January 9, 2012 9:56 pm

Hi Willis,
Good article.
The surface temperature of some part of the moon can exceed 120 C (1368 W/m2 for a black body) if the emissivity of the surface for long wavelength IR is both less than 1 and less than the absorptivity for solar radiation. There are a number of substances known to have this property (see the table here: http://www.redrok.com/concept.htm ). Freshly galvanized metal plate, for example, has an absorptivity for solar radiation of 0.65 and an emissivity for thermal IR of 0.13. If we put a sheet of galvanized metal at the lunar equator, it would absorb 0.65 * 1368 = 889.2 W/m2. But to radiate that much energy, it would have to have a temperature of (889.2/(5.67E-7 * 0.13))^0.25 = 1805 K or 1532 C. Of course, that’s more than 1000 C above the melting point of zinc, but you see the point. There’s even a company, Almeco-TiNOX Solar ( http://www.almeco-tinox.com/en/products/solar_absorber/coating_technology ) that makes a coating that they claim has a solar absorptivity of 0.95 and a thermal emissivity of less than 0.04. Now that would get really hot!

Warren
January 9, 2012 10:09 pm

Rosco
The “Palm Trees” you see in Scotland are Cordyline australis, they grow in a wide range of climates in New Zealand, from snow level to sea level, this has been discussed previously in WUWT
I have photographs of them from my time in Scotland and Ireland, along with various Hebe spec. that have been grown from stock from New Zealand.
They are not a true “Palm tree”

CRISP
January 9, 2012 10:12 pm

“But a characteristic of the greenhouse radiation (downwelling longwave radiation, also called DLR)……”
The 2nd Law of Thermodynamics: You cannot tranfer heat from a cold body (upper atmosphere) to a hotter body (the earth surface) without doing work. This applies to absolute transfer, not to nett transfer, despite the bullshit being spouted by the alarmists. This is an iron-clad law. There is no getting around it. Any so-called radiative ‘down-welling’ is totally irrelevant to warming. The Greenhouse Theory is based on this nonsensical idea.
You must distinguish between Greenhouse Theory and atmospheric insulation. Two very different concepts.
They have had 100 years to prove the Greehouse Theory – and have utterly failed. Forget it guys. It is bogus. Move on.

January 9, 2012 10:42 pm

Willis
I disagree and here is why.
1. Even in that perfectly transparent atmosphere there is an atmosphere.
2. The temperature of the surface is about 60-90C during the day.
3. Atmospheric molecules would still physically impact the surface of the Earth.
4. The molecules would pick up that thermal energy and pass it on to other molecules.
5. Convection would heat the atmosphere
6. Due to the frequency of collision of molecules it would take time for the atmosphere to mechanically radiate its warmth back to the ground with the ground colder at night.
And thus without an IR at all you have been abel to average the temperature. To what?
The molecules are absorbing energy from a body radiating at 70-90 c and the molecules are incapable of emission in the IR so how do the get rid of their now excess mechanical energy?

izen
January 9, 2012 10:49 pm

@- CRISP says: January 9, 2012 at 10:12 pm
“The 2nd Law of Thermodynamics: You cannot tranfer heat from a cold body (upper atmosphere) to a hotter body (the earth surface) without doing work. This applies to absolute transfer, not to nett transfer, despite the bullshit being spouted by the alarmists. This is an iron-clad law.”
So does The 2nd Law of Thermodynamics: mean that you cannot tranfer heat from a cold body (a coat) to a hotter body (me) without doing work. ?
No point in wearing a coat on a cold morning then…..

coldlynx
January 9, 2012 10:57 pm

Willis You write
“… the surface can never warm the atmosphere to be warmer than the surface is. No matter what the difference in efficiency is, the surface can only warm the atmosphere up to the temperature of the surface. Beyond that it can’t go, you can’t make heat flow uphill. ”
Right. But that is not what I say. (Or try to…, sorry for my english). I say the average atmosphere temperature will be warmer than average surface temperature, and below maximum surface temperature. The “daily” temperature swing will pump up the temperature in the atmosphere to above average surface temperature and below maximum surface temperature.
Just because of how heat is transported. From warm to cold. And warm gases have less density than cold gases.

R. Gates
January 9, 2012 11:25 pm

For those who’d like to watch an actual NASA lecture given to potential future Astronauts headed to the moon, with lots of great detail about the thermal, geological, radiation, environment, I highly recommend:
http://www.lpi.usra.edu/lunar/moon101/mendell/

ferd berple
January 9, 2012 11:33 pm

Isn’t dividing solar output by 4 a form of averaging that will tend to distort the black-body calculations and result in too high a calculated temperature? Isn’t the radiation perpendicular to the sun 1368 W/m2, but as you approach the poles and terminator this drops off COS. Don’t you need to raise this to the 4th power and then average? Similarly, isn’t using a constant albedo also a form of averaging. Surely the moon’s surface is not homogenized. Both these factors would appear to increase the calculated BB temperature above what could be expected without averaging.

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