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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Dan in California
January 9, 2012 11:47 pm

I enjoy your posts, Willis.
Two points. First, Luna’s surface warms rapidly after dawn partially because the thermal conductivity of the regolith is very low. The effective surface depth is thin because it’s not rock or compacted; it has good vacuum insulation between the particles. I haven’t a clue how to quantify this, or whether it would significantly change the thought experiment results.
Second, I was around when the exterior coatings were chosen for the Space Station. That’s another example of a body at the same distance from the sun. The coatings were chosen to have absorption and emission characteristics to keep the structure at about 20C without active thermal control. Turns out that’s not terribly difficult.

ferd berple
January 10, 2012 12:07 am

Rosco says:
January 9, 2012 at 9:16 pm
Also I refuse to believe 99% of the atmosphere comprised of Nitrogen and Oxygen does not become heated and as such radiate IR
It does seem remarkable that the emissivity of N2/O2 would be 0.0000. That would seem imply that N2/O2 would never cool in space. Hardly seems possible.

J. Radefahrt (Ger)
January 10, 2012 1:00 am

You may call me stupid but I wonder, if one wants to know the equilibrium of the atmosphereless Earth, why does this one not recognize that there were no clouds, therefore no cloud albedo effect and so only 10% albedo – the pure albedo of the surface? Clouds are, as I understand it, an effect bound on an atmosphere.
The next thing I wonder is, why does one not consider that a comparison with a grey/black body is only feasible for 3/10 of the Earth surface because of the Oceans? The behavior of absorption and emission by water is seriously different to a solid surface, because short wave radiation is mostly absorbed below the area where the emission of longwave radiation (upwards) is possible, what means that not all heat can be radiated upwards what finally reduces the emission coefficient significantly (radiation is emitted in both directions up and down within the water).
In result this would give a much higher equilibrium as the stated 255K, possibly higher as the current average (what would make sense regarding to Ramanathan who stated a net cooling by clouds by about 50 W/m²).

metamars
January 10, 2012 1:24 am

“Hardly any scientists believe that the earth’s core and mantle is producing any energy, other than a modest amount of fission. ”
Sorry, wikipedia says the opposite. http://en.wikipedia.org/wiki/Geothermal_gradient
“The Earth’s internal heat comes from a combination of residual heat from planetary accretion (about 20%) and heat produced through radioactive decay (80%).[2] ”
Lord Kelvin had estimated the age of the earth at ~40 million years, by calculating how long it would have taken to cool off to present temperature (and unaware of nuclear fission).
The same wikipedia article says, “An estimated 45 to 90 percent of the heat escaping from the Earth originates from radioactive decay of elements within the mantle.[8]”. Given such a degree of imprecision, and given that I doubt the composition of the earth, including nuclear species, is anything like certain, I still hope LaViolette’s ideas are pursued.

David
January 10, 2012 1:40 am

How much of the GHE on earth is actually due to the oceans where the residence time of energy is far far longer then any GHG?
Willis answers,,,”None, as I understand the greenhouse effect.”
——————————————————————————————————–
Willis, please understand that of course I do not literally mean GHE, when I refer to the oceans; except in the context of THERMAL CAPACITY. At its most basic only two things can effect the energy 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. (You may henceforth refer to this as David’s law. (-;) There is a fairly exact correlation between residence time of energy and thermal capacity. As the ocean thermal capacity is thosands of times that of the atmosphere, it appears logical that it is a more effective GHLiquid, then any GHG; your thoughts in this regard are welcome aand requested.

David
January 10, 2012 1:46 am

Also, although the average albedo of earth is higher then the moon’s, is it higher at laditudes where TSI is stongest?
Willis responds…”The earth albedo varies by location, by time of day, and by time of year, so it’s hard to answer your question. Where the TSI is the strongest (tropics) the albedo is part of the dynamic system keeping the earth from overheating, This means albedo also varies by temperature. It is higher where the temperature is highest, which in turn is where the TSI is greatest, so the answer to your question is generally yes, at least in the tropics.” W. ———————————————————-
At first glance this does not appear logical to me. In general the oceans are a blackbody, absorbing whatever radiation reaches the surface with little reflectivity,(especially in the tropics) The NH has a great deal of landmass north of the tropics, as well as year round snow and ice in the arctic, antarctic as well as tremendous winter albedo (land mass snow cover) beyond year round ice. Additionally the incident angle of sunligh creates ever greater reflectance as one moves further from the tropics. A further factor is the poles appear to have a great deal of consistent.cloud cover as I look at the global map on the right side of WUWT home page. For these reasons I would have to see actual meauserment to accept your assertion here, as I suspect that the tropics. especially the southern tropics have the lowest albedo as well as the greatest TSI, especially in January when the earth is thee million miles closer to the sun and TSI is close to 100 W/m2 greater then in July. Your thought here are appeciated as well.

David
January 10, 2012 2:07 am

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.”
It therefore follows that any effect which increases the residence time of LW energy in the atmosphere, (water vapor for instance) but reduces the input of SW energy entering the oceans, 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.
So a further question for anyone wishing to do more then a thought experiment. Sunlight radiating on the Earth when it’s about 3,000,000 miles closer to the sun in January, is about 7% more intense than in July. (TOA of plus 100 W/m2 ) Despite the increased insolation the atmospheres average temperature is about 4 degrees cooler in January. Because the Northern Hemisphere has more land, which heats easier then water, most people state that the Earth’s average temperature is about 4 degrees F higher in July than January, when in fact they should be stating that the ATMOSPHERE is 4 degrees higher in July. In January this extra SW energy is being pumped into the oceans where the “residence time” within the Earth’s ocean land and atmosphere is the longest. There is not only 7% more intense radiation, there is more ocean to receive this radiation. So, the atmosphere in January loses energy two ways. One to the oceans until it is re radiated as LWIR, and two, to space due to greater NH albedo resulting in greater total earth albedo during the NH winter months. Now the question. Is the earth gaing or losing energy in January, and please quantify your answer?

January 10, 2012 4:26 am

Firstly, when anyone quotes the above mentioned assumption that backradiation is similar day or night they might do well to read about Prof. Nahle’s experiment* and his night time measurements of downwelling radiation which ranged from 61.93 W/m^2 down to 46.45 W/m^2 seven hours later. These are far less than the values shown on any energy diagram.
Nahle explains how his instrument views a verticle cone which has a 5km diameter at the top, where it seems most of the measured radiation is coming from. Now his instrument cannot tell what height the radiation is actually coming from, so I suspect it assumes an average within its range which goes up to 30,000 metres. But if most radiation is coming from that 5km diameter disc, that’s a lot of square metres to divide by. How on Earth (or in the atmosphere) can we expect any high degree of accuracy in such measurements?
Secondly, on the subject of Moon temperatures, the use of the S-B equation should be averaged over the full day/night cycle, which is 26.3 Earth days on the Moon. This gives far lower values than the proverbial -18 deg.C figure.
Remembering that the incident solar radiation is of course zero at night, it might appear that the temperature on the Moon would also be zero (0 deg.K) at night.
But neither the Earth’s outer surface nor the Moon’s act like strict blackbodies. The reason is that a perfect blackbody has no conductive interface with its surrounds. The Earth and even the Moon do. These surrounds are not just the atmosphere in the case of the Earth, but also the deep ocean waters and the underground mass of crust, mantle and core. The Sun’s radiation causes thermal energy to flow into these regions during the day, and that energy then takes a similar time to flow back out at night, these rates depending upon conductivity or convection in the oceans. Of course some of this energy (especially in the oceans) may take months or years to fully exit again.
The common error in all the blackbody calculations relating to surface temperatures is that this conducted energy, and the rates of conduction and rates of revolution of the planet or moon, all seem to be ignored when they should not be. Blackbody theory is also incorrectly applied to gases and even individual molecules in the atmosphere. Whenever it is applied, there should always be a compensating calculation which not only deducts energy transferred by means other than radiation, but which also deducts a compensating calculation based on the temperature of the surrounding region into which the radiation is directed. How then can we calculate positive radiation from a cool layer of the atmosphere to a warmer layer below, let alone to a warmer surface? Nahle suggests that the only radiation he is measuring is actually coming from warm globules of air radiating to cooler regions below them, as is possible in the stratosphere where there is temperature inversion.
* http://climate-change-theory.com/RadiationAbsorption.html

DeWitt Payne
Reply to  Doug Cotton
January 10, 2012 7:33 am

Doug Cotton:
“Firstly, when anyone quotes the above mentioned assumption that backradiation is similar day or night they might do well to read about Prof. Nahle’s experiment* and his night time measurements of downwelling radiation which ranged from 61.93 W/m^2 down to 46.45 W/m^2 seven hours later”
Nahle’s experiment is bogus, as usual. For one thing, his instrumentation is inadequate for the task. His IR thermometer only goes down to -60C or 117 W/m2, so I don’t see how gets values below the lower limit of the device. See my reply to Jack Frost at Science of Doom: http://scienceofdoom.com/2010/07/31/the-amazing-case-of-back-radiation-part-three/#comment-15176 for more detail. You might have to wait a while, though. It was long with a lot of links so it’s stuck in moderation.

January 10, 2012 4:54 am

DeWitt:
Your sheet of metal does not have to radiate the amount you calculate for the simple reason that thermal energy will be conducted from its underneath side into the surface, as it is hotter than the surface.
You, like others, ignore the conduction of thermal energy from the thin outer layer of the surface deep into the underground regions where it builds up during the day, and then exits out of the surface at night. On Earth much of that exiting of energy from the surface to the atmosphere is by means other than radiation anyway. This is why the Earth’s surface cools at a slower rate than the atmosphere at night, the atmosphere being colder all the time of course. The crust itself is doing most of the insulating at night, not the atmosphere, and there is certainly no warming effect by any backradiation at night, nor even any resultant slowing of the rate of cooling of the surface.
There is no atmospheric greenhouse effect.

DeWitt Payne
Reply to  Doug Cotton
January 10, 2012 7:20 am

It does if I insulate the other side. Actually, the thermal conductivity of the Lunar surface is quite low so even without extra insulation, there would be little heat lost from the back side.

Tim Folkerts
January 10, 2012 5:46 am

CRISP expounds: “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. ”
I hear this unsupported assertion quite often. I tell you what, Crisp, the more fundamental version of the 2nd law is “the entropy of a thermally isolated system can only increase”. When you can calculate that the entropy change for the atmosphere/earth system & show that it increases when the cool atmosphere radiates back a fraction of the energy it receives, then I will put some confidence in your conclusions.

Spector
January 10, 2012 5:53 am

RE: CRISP: (January 9, 2012 at 10:12 pm)
(i)”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.”
Not quite. You cannot move against an opposing force without doing work. No force opposes the motion of a photon emitted in the cold upper atmosphere before it is absorbed, either in the atmosphere or on the ground. Back radiation is real. The only thing that thermodynamics says is that there must be *more* energy in the up-welling radiation than in the down-welling radiation if the upper air is cooler than the surface.
The primary problem with the carbon-dioxide greenhouse alarm is an assumed broad-spectrum effect for a narrow-band process. When the amount of CO2 is doubled, the effect of almost all of the added CO2 is hidden behind the mask created by the CO2 that was already in the atmosphere. It would be like adding to the length but not to the blocking width of a tree in the middle of a stream.

Tim Folkerts
January 10, 2012 5:56 am

ferd berple says: “It does seem remarkable that the emissivity of N2/O2 would be 0.0000. That would seem imply that N2/O2 would never cool in space. Hardly seems possible.”
Then it is a good thing that science is not limited by your imagination. Or my imagination. Or Willis’s imagination.
The simple fact is that N2 has a MUCH lower emissivity than CO2, and consequently N2 would cool MUCH slower than CO2 out in space. Just like a polished piece of aluminum would cool MUCH slower than the same piece of Al painted black. (Actually any color paint would do, since the IR properties are mostly independent of the visible properties).

January 10, 2012 5:57 am

Hi Willis, do you perhaps know if anyone ever measured the change in humidity over the years?
After looking at the daily average readings from about 20 weather stations all over the world I am finding a change of about -0.02%RH per annum.
http://www.letterdash.com/HenryP/henrys-pool-table-on-global-warming
So, if this estimate is not far from correct, then the average global humidity is now about o.75% RH lower than it was 37 years ago.
If I am not mistaken (at 15 degrees C) that translates again to a loss of about 0.1% in absolute humidity.
You see how that compares with the increase in of CO2? (0.01% increase over the last 50 years)

Tom in recently clold Florida.
January 10, 2012 6:10 am

Alexander Feht says:
January 9, 2012 at 11:43 am
“P.S. Couple of people here propose a weird argument about “dead man under a blanket.” Be it known to them that plants are protected from freezing by blankets, though, last time I checked, plants had no internal sources of heat (unless you burn them)”
However, plants and the surrounding ground retain heat from the day and preventing that from escaping is what will keep the plants under the blankets warm enough to prevent damage. .

January 10, 2012 6:20 am

Interesting post. Incidentally I know the moon orbits and keeps one face to the earth. But interestingly the idea that it has an axial rotation as opposed to just an orbit is widespread, even to a report on the BBC a few days ago. Maybe I should not be surprised by the BBC, but otherwise it’s a common mistake even in scientific discussion.
The term is “gravitationally locked”. 4 billion (or so) years ago the moon orbited much closer to the earth than it is today — its orbit is receding by a few centimeters a year — and the tidal force exerted on the Earth (proportional to 1/r^3 where r is the radius of the orbit) was many times stronger. Tides in the Earth’s early oceans would have been spectacular things, rising five or ten meters nearly everywhere. Tidal heating was much larger as well, and the gravitational coupling turned the moon’s initial axial rotation, whatever it was, into heat.
Eventually the moon “froze” into an orbit such that its period of revolution around the Earth and period of rotation around its axis are the same, deformed by the tide so that it is very slightly football shaped, with the long axis pointing towards the Earth (the Earth is similarly a flattened sphere due to its rotation with a running tidal bulge that faces/opposes the moon).
I don’t think this is a common mistake in scientific discussion, though, at least not discussion among physicists or astronomers. However, I occasionally teach astronomy, and in the process learned a disturbing thing. Some bright (but cynical) person did a survey of Harvard students out on the quad one fine day. One of the survey questions asked the students to draw the relative position of the Earth, Moon and Sun when the Moon is new, first quarter (half full), full, last quarter (half full).
Only one student in four, if I recall correctly, could do so. A sad commentary, really. Readers are invited to try it for themselves, now, without looking. Bear in mind that you’ve been looking at the Moon as it proceeded through its phases for your entire life, and you have in addition the powerful guide of your reason (your answer must make sense and “work”).
Astronomy is really an amazing thing to study — it’s simple enough that anybody can understand it (even without a lot of calculus or physics in their background), it includes things like the Stephen-Boltzmann law and Wien’s Law (both used, along with I = P/(4\pi R^2), as key steps in determining how far away everything is from the Earth), and amazingly good backyard telescopes are cheap and readily available as never before. Amateurs can photograph things like the Orion nebula or the planets as easily as professionals did 100 years ago, with telescopes that are vastly surpass the ones they used to make awesome discoveries way back then.
rgb

January 10, 2012 6:38 am

@- HenryP says:
January 10, 2012 at 5:57 am
“Hi Willis, do you perhaps know if anyone ever measured the change in humidity over the years?”
The Radiative Signature of Upper Tropospheric Moistening
http://www.sciencemag.org/content/310/5749/841
Identification of human-induced changes in atmospheric moisture content
http://www.pnas.org/content/104/39/15248.abstract
Data from the satellite-based Special Sensor Microwave Imager (SSM/I) show that the total atmospheric moisture content over oceans has increased by 0.41 kg/m2 per decade since 1988.

RobB
January 10, 2012 6:57 am

“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?”
Has anyone explains what actually happens to an atmosphere that can’t absorb/emit radiation (IR). Is this even realistic. What happens if you replace the zero absorption/emission atmosphere with a slightly more realistic pure N2 atmosphere??

January 10, 2012 6:58 am

I have strong bias towards not changing the Moon [or Mars].
Good, because it wouldn’t work. The problem is this. The molecules of a gas at some temperature at equilibrium have what is called a Maxwell-Boltzmann distribution of velocities. That is, some molecules are moving fast, some slow, most in between, at any given temperature. The MB distribution is a peaked curve with a long tail at higher speeds, and it is different (scaled) for particles of different mass because of equipartition, where 1/2 m v-average-squared = 1/2 kT as a constraint (if you like) on the distribution so that more massive molecules move more slowly, on average at the same temperature are lighter ones
A planet has something called an “escape speed” — a speed for which something thrown away from the planet’s surface will never fall back, assuming no drag forces. It is easily computed, for the Earth it is 11.2 km/second. For the moon it is 2.4 km/second, as it is smaller and has weaker gravity.
The Moon’s escape speed is sufficiently low that any of the lighter molecular mass gases have a relatively short lifetime on the hot side. Too large a fraction of the MB distribution consists of molecules up above escape speed, and as those molecules fly off into space the remaining ones re-thermalize and constantly replace them. It is actually a cooling mechanism, as the lost molecules carry away energy from the high tail of the MB distribution (a similar mechanism is used as an ultracooling refrigerator in certain physics experiments).
Perhaps you could give the Moon an atmosphere of Krypton, or Xenon (no, not the warrior princess:-), if you had enough.
Incidentally, this is why the atmosphere of Mars is mostly CO_2, and why the Earth’s atmosphere has very little hydrogen or helium or methane in it. The molecular weight of CO_2 is greater than that of O_2 (by the weight of one carbon atom) or N_2 (by the weight of one oxygen atom). It takes longer for CO_2 to outgas from a planet via the MB tails because the whole distribution is shifted down so a lot smaller percentage of the molecules are above the escape velocity cutoff.
The Earth is constantly losing atmosphere in exactly this way, lightest molecules first. To some extent, the atmosphere lost is replaced by outgassing from the Earth’s crust and e.g. biological activity, but once it is gone it is gone. I do not know precisely how atmospheric density is thought to have varied over the last 4 billion years (aside from a few things about the very early atmosphere in the pre-oxygen era) but I’m guessing that on average it is decreasing and will continue to decrease gradually over geological time. This could even be an explanation for why the Earth has been gradually slipping into long term cooling and glaciation, although I’m not certain it is a necessary explanation given periods of geological glaciation hundreds of millions or billions of years ago.
I am most skeptical about “heating” of the atmosphere due to pressure and PV=NkT — that is nonsense, frankly. However, Willi’s observation above that an atmosphere acts like “thermal ballast” and helps reduce the hot side and cold side temperature differential, which de facto increases the mean temperature of the planet closer to that predicted by SB for a uniform ball seems to me to be dead on the money. The more uniform the temperature of a planet, the warmer the planet on average. The denser the atmosphere, the better the greenhouse gas trapping, the more uniform the temperature, where the latter may well be less important than the former — I’d have to do some estimates of the heat gain/loss of a volume of atmosphere (plus the top whatever centimeters of the surface) to get a feel for it.
Note well that while PV=NkT may not be the mechanism at all, it is closely related to the heat capacity of a gas, and the heat capacity of a gas may be an important component of the mechanism. It would be very interesting to see if this assumption can actually quantitatively reproduce the NZ curve for the planetary temperatures as a function of atmospheric density. I intuitively think that it might.
rgb

Joules Verne
January 10, 2012 7:26 am

Robert Brown says:
January 10, 2012 at 6:20 am
Standing on north pole facing the sun.
New moon is at your 12 o’clock.
Quarter moon is 1:30 and 10:30
Half moon is 3:00 and 9:00
Three quarter moon at 4:30 and 7:30
Full moon at 6:00
That a bunch of Harvard students would get that wrong is no surprise. Formal education is a racket that enriches the employees of the institution at the expense of others including students, parents, and taxpayers. The system became increasingly worse as independent low-cost access to learning materials became more widespread. The internet greatly accelerated the decline into a real sham.
That question you posed, by the way, is typical of what appears on the math portion of the Scholastic Aptitude Test which was dumbed down considerably in 1995. I aced the math portion of the SAT back in 1978 before it was dumbed down. So maybe I’m just a bigoted autodidact with an IQ exceeded only by my cynicism but it’s rare that I’m wrong so that’s not a very likely scenario.
Tidal locking (the more commonly used term than gravitational locking) is a little misleading as it has nothing to do with tides on the earth. It begins with a mass distribution asymmetry in a smaller rigid body. The earth’s moon is large enough to form a gravitationally shaped sphere but the mass distribution won’t be perfect even if the spheroid shape is perfect. The heavier side of the smaller body will eventually come to constantly face the larger body.

Joules Verne
January 10, 2012 7:36 am

CRISP expounds: “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. ”
That is true but it should be qualified by saying there is no net transfer of energy from the colder to the warmer.
What actually happens is that the rate of net energy transfer from the warmer object (the earth) to the colder object (the cosmos) is slowed down. So it is quite correct to say that the cooler cannot heat the warmer but the rate at which energy moves from warmer to cooler can be throttled by intermediaries like London Fog coats and London fog banks. The principle is pretty much the same for both. It’s passive insulation in both cases. If you believe that insulation such as clothing works to limit the loss rate of body heat then you should also believe that insulation such as water vapor works to limit the loss rate of the earth’s heat. Word.

January 10, 2012 7:39 am

(If you add an atmosphere, then the situation could change a bit, and the inner surface could indeed be warmer than the outer shell by an amount related to the lapse rate.)
How’s that again? Read your own argument. If the system is at equilibrium it will all be at the same temperature, including the atmosphere, including gravity, under any and all circumstances. The only circumstance whereby the inner (idealized) sphere can be warmer than the outer one is if it starts out warmer and one hasn’t waited long enough for the system to equilibrate.
To put it microscopically, the average energy per molecule per degree of freedom will eventually be 1/2 kT, period. As you said, heat will always flow from warmer areas to cooler ones until equilibrium is reached. So very good reply, except for the parentheses…;-)
rgb

January 10, 2012 7:42 am

http://www.tech-know.eu/uploads/Greenhouse_Effect_on_the_Moon.pdf
To quote NASA’s analysis,
During lunar day, the lunar regolith absorbs the radiation from the sun and transports it inward
and is stored in a layer approximately 50cm thick. As the moon passes into night, the radiation
from the sun quickly approaches zero (there is still a bit of radiation from the earth) and, in
contrast with a precipitous drop in temperature if it was a simple black body, the regolith then
proceeds to transport the stored heat back onto the surface, thus warming it up significantly over
the black body approximation.
The above is a link to a article several years old regarding this subject. The quote is from the article.

January 10, 2012 7:51 am

Willis, thanks for a very interesting, thought-provoking article!
Also, very good and polite answers to comments, many not so thoughtful or polite.

Joules Verne
January 10, 2012 7:52 am

clivehbest says:
January 10, 2012 at 1:40 am
“Alan – that’s right. Albedo reduces for lower incident radiation . As the Earth’s ocean surfaces get heated, the Earth starts to “swet” – producing clouds which reduces the Albedo.”
The earth sweats more as it gets warmer.
That’s the simplest and most insightful analogy I’ve seen in quite some time. A wonderful way to convey the concept of negative feedback in the climate system. Thank you.

Joules Verne
January 10, 2012 8:18 am

Tom_R says:
January 9, 2012 at 8:38 am
>> son of mulder says:
January 9, 2012 at 3:42 am
but on a clear night the air will cool quicker than the surface. <<
"The surface will cool quicker than the air, which is the reason we can have frost form at air temperatures above freezing. Your point about a difference between air temps and surface temps is still valid."
Good point but it depends on the surface. I have a high emissivity surface (white mineral roofing) on a flat (1:14 slope) roof with an unobstructed view of the night sky. Beneath it is R-30 insulation. The shapes that appear (rafters, underlying heat sources, etc) in the dew and frost patterns are fascinating and I see it every morning from above. I've observed frost on it when the air temperature as recorded by sheltered a min/max thermometer 10 feet above it never fell below 40F during the night. I've never really seen, that I can recall, frost forming on natural surfaces when air temperature is above freezing.

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