Lunar atmosphere?

[Brotoro]

13 hours ago, Snark said:

It's worth noting that the exact thermodynamics-- for example, how much of the solar radiation goes to heating atmosphere, and what the equilibrium temperature is-- depends a lot on various messy specifics that haven't been at all specified in the problem statement.  Climate modeling is hard.

For example:

• What's the content of greenhouse gases in the gas mix?  Makes a huge difference to the equilibrium temperature.  Just ask Venus. 
• How much convection "mixing" happens between surface atmosphere and upper atmosphere?  This is really important, because the only gas that can escape is the upper atmosphere, whereas the solar heating's going to happen mainly down at the surface.  And I have no idea what the convection model would look like for the Moon, given the extreme differences from Earth even if it had a similar atmosphere (for example, the very slow day/night cycles, the overall smaller scale of the globe, the much lower gravity, etc.)  Would it be a "dead" atmosphere that hardly moves, and quickly stratifies into stagnant layers?  Or would it be a continuous raging hurricane that keeps everything well-mixed?  I haven't an inkling.
• Is there any water involved?  Water has a huge effect on atmospheric thermodynamics.  Very large specific heat in the liquid phase makes it a big heat reservoir, if there's a lot of it.  Large heat of vaporization and heat of fusion means lots of heat transfer happens when it changes phase.  In its vapor state, it's a powerful greenhouse gas.  As ice (or clouds), it's highly reflective.  And so forth.

 

 

Yes, but the details of the situation (how much of the incoming solar radiation is reflected directly, how much is absorbed by the surface and atmosphere) does not change the fact that most of the incoming radiation is re-emitted as infrared (be it by the surface of by the atmosphere), so that outgoing energy is not available to lift the atmosphere out of the gravity well. Only the difference between what goes in and what goes out as radiation (the vast majority) is available to remove the atmosphere. So your calculation assuming all the energy goes into removing the atmosphere (in 500 million seconds) is wildly off.

Edited October 4, 2017 by Brotoro

[Snark]

5 minutes ago, Brotoro said:

Yes, but the details of the situation (how much of the incoming solar radiation is reflected directly, how much is absorbed by the surface and atmosphere) does not change the fact that most of the incoming radiation is re-emitted as infrared

Actually, yes it does.  That's my point.  It matters enormously.  If the surface is getting warmed up, and there's a lot of convection mixing in the atmosphere, and/or there's a fair amount of greenhouse gas involved, then you could have a situation in which most of the heat goes into warming the atmosphere rather than re-emitting.

On the other hand, if the atmosphere is dead and stagnant and stratified, and there's little or no greenhouse gas present, you could have a situation in which most of it does get emitted as IR, as you say.

Or if you are completely covered in ice and/or clouds, your albedo may be high enough that most of the incoming energy gets reflected away before it has a chance to be converted to heat in the first place.

That's my whole point:  the devil is hugely in the details, here.

[Brotoro]

No. It will only keep putting lots of energy into the atmosphere until you reach approximate equilibrium temperature. Then the surface and atmosphere will be radiating the infrared energy to space about as fast as the planet is receiving it from the Sun. If you add greenhouse gasses (in addition to the oxygen/nitrogen specified in the original question), that will only mean the equilibrium temperature will end up being higher before the outgoing IR balances the incoming solar radiation.

[Green Baron]

2 hours ago, Brotoro said:

Yes, but the details of the situation (how much of the incoming solar radiation is reflected directly, how much is absorbed by the surface and atmosphere) does not change the fact that most of the incoming radiation is re-emitted as infrared (be it by the surface of by the atmosphere), so that outgoing energy is not available to lift the atmosphere out of the gravity well. Only the difference between what goes in and what goes out as radiation (the vast majority) is available to remove the atmosphere. So your calculation assuming all the energy goes into removing the atmosphere (in 500 seconds) is wildly off.

I couldn't find a spectrum of moonlight, sure the must be one somewhere, and if it's only of the sodium "atmosphere".

The moon emits IR because there is no atmosphere. Earths atmosphere actually absorbs much of the infrared radiation and so warms up (a bit).

Also the moon reflects shorter wavelengths of the sun's spectrum better than longer ones one should think, so infrared emission is rather less.

Also i can't imagine that overall the moon emits less than it receives :-)

I don't think it is wildly off to assume that most of the incoming energy is converted into heat and / or absorbed by an atmosphere, thus helping it escape. There are no plants that convert sunlight, the energy has no other sink.

 

Furthermore one should not underestimate the effect of a missing magnetosphere. Ionized particles from the upper atmosphere will quickly been blown away.

2 hours ago, Brotoro said:

No. It will only keep putting lots of energy into the atmosphere until you reach approximate equilibrium temperature. Then the surface and atmosphere will be radiating the infrared energy to space about as fast as the planet is receiving it from the Sun. If you add greenhouse gasses (in addition to the oxygen/nitrogen specified in the original question), that will only mean the equilibrium temperature will end up being higher before the outgoing IR balances the incoming solar radiation.

That i don't understand. Putting lots of energy into an atmosphere does heat it. The atmosphere of the moon would not be an adiabatic system if there were no hefty inversion layers, an equilib. temp would be difficult to apply because there would not be a lasting equilibrium. It warms, partly ionizes in the hard radiation, it looses mass through mechanical escape and magnetic forces/particles, pressure sinks, that cools a bit but surely not enough to keep radiation from heating /ionizing enough particles to allow them to escape.

The surface doesn't directly radiate IR to space; only if there is no atmosphere. It exchanges heat with the contacting lower atmosphere, which leads to high differences between space (~100K or so ?) and surface (400K in daylight). That's bubbling :-)

But of course, i may be wrong.

 

Edit: i always assume that the atmosphere in question resembles earth's. A Venus like CO2 soup behaves totally different, but that's not the point here, right ?

Edited October 3, 2017 by Green Baron

[Brotoro]

Yes, energy is put into the atmosphere and heats it up (by radiation coming in and by the surface radiating outward, and by convection). THEN that warmed air re-radiates the energy as infrared...some making it out to space, and some warming the surface more than it would be without the atmosphere. The temperature of the atmosphere and surface will increase until the outgoing infrared (from surface and atmosphere...mostly from atmosphere) balances the incoming solar radiation. After the equilibrium is reached, not a lot more energy is going to be captured to heat up the atmosphere (it's already warmed up to the equilibrium state).

The Earth absorbs about 235 Watts per square meter of energy from the Sun: 168 W/m^2 by the surface, 67 W/m^2 by the atmosphere. Some light energy is reflected directly back into space by the clouds, air, and surface). The warked-up surface radiates The surface re-radiates only 40 W/m^2 directly from the surface into space...most of the re-radiated IR from the surface gets trapped by the atmosphere. The warmed up atmosphere radiates about 195 W/m^2 into space, and the surface radiates 40 W/m^2 directly to space. So you get almost the same amount going out as coming in (some small amount of energy comes out of the interior of the Earth, and some energy is lost from the atmosphere by escaping gas molecules, but these are small amounts compared to the equilibrium radiative input and output.

On the hypothetical Moon with a thick atmosphere, the same thing would be going on. Most of the energy input by solar radiation is going to be re-radiated by the atmosphere and surface as infrared (with less coming from internal heat and more being lost as kinetic energy of escaping molecules compared to Earth). But you are not going to have anywhere close to all of the incoming solar energy to use in removing the atmosphere (in 500 million seconds)...most of the energy will get re-radiated by the warmed-up atmosphere as infrared.

(The numbers in paragraph 2 come from a diagram in the Wikipedia article on the greenhouse effect.)

 

Edited October 4, 2017 by Brotoro

[DerekL1963]

7 hours ago, Brotoro said:

But you are not going to have anywhere close to all of the incoming solar energy to use in removing the atmosphere (in 500 seconds)

Nobody said 500 seconds to remove the atmosphere - the figure calculated was 500 million seconds (16 years) as a theoretical minimum.  You're picking a fight based on words you're putting into someone else's mouth.

[Brotoro]

1 hour ago, DerekL1963 said:

Nobody said 500 seconds to remove the atmosphere - the figure calculated was 500 million seconds (16 years) as a theoretical minimum.  You're picking a fight based on words you're putting into someone else's mouth.

Ah...You are correct in that I misquoted 500 seconds. But my disagreement comes from his calculation where he uses all of the incoming energy to remove the atmosphere...when most of the incoming energy is going to get re-radiated back to space as infrared radiation. Only a small fraction will be usable for removing the atmosphere.

(earlier misquotes now fixed)

Edited October 4, 2017 by Brotoro

[Green Baron]

I think the assumption that there exists an equlibrium for an earth like atmosphere on the moon is wrong or is reached when the atmosphere is gone, but i have no data to support that except that even the Mars with a weak magnetic field and higher gravity and volcanism couldn't hold one and even the earth looses a very small part of it's blanket.

And i realize that many others in similar discussions share these thoughts, but that's of course no argument ;-)

 

An eq. would mean that different forces' and processes' effects cancel each other out, but once the atmopshere has magically appeared on the moon there is only the way into space, supported by:

the sun's spectrum coming in,

lack of a magnetic field,

high differences between day and night,

the lack of a stabilizing biosphere (that btw. is part of the absorption on earth in "good times" and may set the energy free again in "bad times", but that leads far away).

ionization and carry-away by solar winds / particles

the lack of inversion layers (the differences between day&night would "burn" any away every morning)

 

Otoh there are no processes to renew lost atmosphere, like volcanism, carbon cycle, etc., buffering in oceans, biosphere, blabla :-) So, simply put and without discussing how much stays here or there: Energy comes in. It heats. Some of the heated gases escape. No new ones are generated. No eq. until most of the gases are elsewhere.

 

The basic question is how long does it take until the atmosphere is reduced to a negligible amount. And there we can part our little group in the "shorties" and the "longies". I am a shorty :-)

 

Edited October 4, 2017 by Green Baron

[MatterBeam]

4 hours ago, Brotoro said:

Ah...You are correct in that I misquoted 500 seconds. But my disagreement comes from his calculation where he uses all of the incoming energy to remove the atmosphere...when most of the incoming energy is going to get re-radiated back to space as infrared radiation. Only a small fraction will be usable for removing the atmosphere.

(earlier misquotes now fixed)

I believe this fraction will depend on the thickness of the atmosphere, its composition and the albedo of the surface. This will determine how much energy is absorbed by the atmosphere following this general equation.

• Energy absorbed: Atmosphere thickness * Atmosphere Absorption * ( 1 + Albedo) * Solar Power

The thickness will be dependant on gravity. For example, on Earth, atmospheric density drops by about 63.2% every 7.64km. This is its scale height - density decreased by a factor 1/e every 7.64km. Convection, solar heating and different elements complicate this slightly. On the moon, gravity is 6 times lower, so the scale height will be 45.8km. For the same surface pressure, the atmosphere will be 6 times taller and/or 6 times denser.  Absorption is a factor of the atmospheric composition. Empirical studies have tell us 77% of the solar intensity reaches the ground. A lunar atmosphere will have to be 6 times denser to achieve 1 atm on the ground. Albedo is between 0 and 1. Average lunar albedo is 0.12, so 12% of the incident radiation is reflected back through our hypothetical atmosphere. 

Edited October 4, 2017 by MatterBeam

[DerekL1963]

 

Ah...You are correct in that I misquoted 500 seconds. But my disagreement comes from his calculation where he uses all of the incoming energy to remove the atmosphere...

He specified that said calculation was a naive assumption, which to me means that it's not meant to be taken (as you seem to be doing) as a model of reality.  He's said so many times - you even quoted and replied to a message where points out the difficulties of creating said model.

You're arguing with something nobody claimed as fact.  He keeps telling you that you're correct (it's a complex situation), and you keep telling him he's wrong.

[Snark]

 

No. It will only keep putting lots of energy into the atmosphere until you reach approximate equilibrium temperature. Then the surface and atmosphere will be radiating the infrared energy to space about as fast as the planet is receiving it from the Sun. If you add greenhouse gasses (in addition to the oxygen/nitrogen specified in the original question), that will only mean the equilibrium temperature will end up being higher before the outgoing IR balances the incoming solar radiation.

Well, first of all, without greenhouse gases, the atmosphere doesn't radiate IR, by definition.

But aside from that:  I agree with your statement above... for a physically closed system.  Which the Earth is (for all practical purposes), and which the Moon would not be.

The Earth is effectively a closed system (for matter, though not for energy).  Atmospheric gases can't escape, in thermodynamically significant quantities, because Earth's escape velocity is so much higher than the equilibrium temperatures involved.  For example, an unusually high-energy molecule in the upper atmosphere might get launched "into space", thus leaving the remainder of the atmosphere that much cooler... except that it's a vanishingly tiny probability that the molecule would have enough energy to escape.  So it would fall back to earth, hitting the atmosphere with the same energy with which it left it, and that little bit of energy that it carried away will get put right back in.  So therefore, the only significant way for the Earth to lose heat to space is via radiation.

With the Moon, however, that would not be the case.  Its escape velocity is much lower than the Earth's, and there are a thermodynamically significant number of molecules that can escape.  And when they do so, they carry away energy with them (because it's only the most-energetic ones that escape).  To all intents and purposes, it would undergo what amounts to evaporative cooling.  Not literally, of course-- there's no phase change involved here-- but the point is that the Moon can shed energy not just via radiation, but also via shedding energetic molecules.

Example scenario #1, with no significant greenhouse gases:

The atmosphere can neither absorb nor emit IR; it's completely IR-transparent.  The only place where absorption and emission happens is at the surface.

I don't know what kind of weather pattern the Moon would end up forming, but here's a scenario that seems plausible to me:  It has a really really slow rotation rate, which means the day side will be a lot warmer than the night side.  So I could picture the entire globe becoming one big toroidal convection cell:  Warm air rises over the noon equator, then spreads out laterally, flowing towards the dark side, where it converges and sinks over the midnight equator down to the surface, where it spreads out and heads towards the daytime side.  Would this actually happen?  I dunno, but it seems plausible so let's go with it.

The surface on the daytime side receives heat from the sun and gets hot.  It can lose heat again via radiation, of course, and will lose some that way.  But it's also got a continuous cold wind blowing along the surface from the nighttime side, which it transfers a lot of heat to via direct contact.  That warm air now rises and ends up in the upper atmosphere-- not losing heat other than via adiabatic expansion, since it doesn't have greenhouse gas and therefore can't radiate IR.  So now you have this warmed-up air in the upper atmosphere, where it can lose some of its heat from escaping molecules.  Repeat the cycle.

Since the continuous flow of cold air at the surface keeps the surface from getting super hot, it's not going to radiate a lot of IR-- it transfers the heat mostly by conduction to the air rather than by radiation.

In a scenario like that, I could see the lion's share of solar energy going to help molecules escape.

Sun shining on Earth is like sun shining on a metal object in a vacuum:  the metal gets hotter until it reaches thermal equilibrium by radiating as much as it receives.

Sun shining on the Moon may be more like sun shining on a rain puddle:  it doesn't make the water get super hot and start radiating, it just makes it evaporate faster.

Example scenario #2:  Atmosphere with plenty of greenhouse gases

I expect it would still tend to create a toroidal convection cell as described above.  The difference is that the atmosphere is now capable of radiating heat, yes.

But... now it's hard for the surface to rid itself of heat via radiation (because the atmosphere gets in the way).  And it's also hard for the lower atmosphere to rid itself of heat (because the atmosphere above it also gets in the way).  Any IR radiation that happens will be in the upper reaches of the atmosphere... which is also where high energy molecules have the ability to physically escape.  There will be some heat loss there through radiation, and also some heat loss through "evaporative cooling".  Which one is bigger may depend on the mix of atmospheric gases and various other factors.  I wouldn't be surprised if the percentage of heat loss to radiation might be higher in this case than in the no-greenhouse-gases case... but how much higher still seems unclear.  Certainly it's not obvious to me that IR would win by a landslide.

[Brotoro]

 

Certainly it's not obvious to me that IR would win by a landslide.

The velocity of the gas molecules scales as the square root of the temperature (T^1/2). The rate of radiation loss from an opaque gas or solid scales asT⁴. That's a T⁸ difference. So, as the temperature of the atmosphere (and Moon) rises, the rate of IR radiation out to space, be it from either the lunar surface or from the opaque layers of the atmosphere (depending on which scenario above you want to discuss), would vastly outstrip the energy loss by escaping molecules. The energy will take the easy way out (IR radiation), so you are only going to have a small fraction of the incoming energy to cook off the air molecules.

Edited October 5, 2017 by Brotoro

[Snark]

 

The velocity of the gas molecules scales as the square root of the temperature (T^1/2). The rate of radiation loss from an opaque gas or solid scales asT⁴. That's a T⁸ difference.

Yep.

 

So, as the temperature of the atmosphere (and Moon) rises, the rate of IR radiation out to space, be it from either the lunar surface or from the opaque layers of the atmosphere (depending on which scenario above you want to discuss), would vastly outstrip the energy loss by escaping molecules.

Yes, and at lower temperatures it would be the other way around.  That's an equation that says colder temperatures favor one, and higher temperatures favor the other-- without any implication at all about where the cross-over point would be, unless we've got some numbers to play with.  Depending on what the equilibrium temperature is, and what the emissivity and molecular mass of the atmosphere is, IR or evaporation might predominate.  Your point being?

 

The energy will take the easy way out

Okay.

 

(IR radiation)

Or not.  Got some numbers?

[Brotoro]

Look at the time scales. Even you say that it would take 16 years to lift all of the air out of there with 100% maximum efficiency. At that rate, escaping molecules simply can't carry away energy very quickly over short timescales. At the rate sunlight is pouring energy into the system, the temperature will rise very rapidly in a much shorter period of time (the Moon currently goes from  -150°C to over 100°C in a matter of days, even when radiating IR unimpeded to space). The temperature in the proposed atmosphere would rise rapidly to the point where IR emission dominates -- there is no other way for the energy to escape in the short term.

[p1t1o]

Presented without comment:

https://gizmodo.com/that-time-when-volcanic-eruptions-created-a-temporary-a-1819215649

 

[Brotoro]

2 hours ago, p1t1o said:

Presented without comment:

https://gizmodo.com/that-time-when-volcanic-eruptions-created-a-temporary-a-1819215649

 

They have an obvious error in that article where it says "The peak thickness for this lunar atmosphere, the researchers claim, would have occurred about 3.5 million years ago, and it managed to persist around the Moon for about 70 million years before it all trickled out into space." ...but that should read '3.5 BILLION years ago' (judging from the original paper).

[p1t1o]

37 minutes ago, Brotoro said:

They have an obvious error in that article where it says "The peak thickness for this lunar atmosphere, the researchers claim, would have occurred about 3.5 million years ago, and it managed to persist around the Moon for about 70 million years before it all trickled out into space." ...but that should read '3.5 BILLION years ago' (judging from the original paper).

Oh gawd yeah. The writing at gizmodo is the worst (hence "without comment"), but they push alot of articles about tons of stuff

[Ten Key]

Here's a link to the actual paper.

http://www.sciencedirect.com/science/article/pii/S0012821X17304971?via%3Dihub

[Green Baron]

2 hours ago, Ten Key said:

Here's a link to the actual paper.

http://www.sciencedirect.com/science/article/pii/S0012821X17304971?via%3Dihub

Hmmm.

So i read it and i must say i find it meagre. It is based almost completely on assumption, and a lot of questions remain open or came up when reading.

Where do the gases come from ? A moon formed from debris of an impact can as well be expected to have mostly "outgassed" during formation by collision. The sun was colder then than it is today, is that included ?

Maybe we should gather forces and write a paper :-) ?