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Venus Atmosphere Puzzle, one man's struggle with atmospheric physics
qraal
post Jun 5 2006, 12:15 PM
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Hi All

This might seem like a really dumb question, but what's the mass of the Cytherean atmosphere per unit area?

At first pass I thought it was easy - same as for an isothermal atmosphere, Po/g, where Po is surface pressure and g is surface gravity. Simple. Except Venus doesn't come close to approximating an isothermal atmosphere. From a graph in Mark Bullock's PhD thesis (Hi Mark if you're visiting) I pulled the figures for Po and To as 92 bar and 735 K, while the left-side of the temperature curve was 250 K at 0.1 bar and 63 km. At about 210 K the temperature drop with altitude stops, then slowly rises into the Cytherean stratosphere.

Ok. My atmospheric physics is pretty limited - I 'modelled' that lapse rate pressure curve as a power law:

P/Po = (T/To)^n

and likewise for density, d/do = (T/To)^n.

Temperature, T, as a function of altitude, Z, I computed as T(Z) = To*(1-Z/(n.Zo)).

Zo = (k.T/m.g), where k is Boltzmann's constant and m is the molecular mass of the atmosphere.

These equations I then integrated between 210 K and 0.033 bar, 70 km, and 735 K and 92 bar, zero altitude.

The resulting equation is m = (n/(n+1))*(do.Zo)*(1 - (T/To))^(n+1) - a bit of simple algebra and the Gas equation shows that do.Zo = Po/g.

Thus the mass is lower than for a simple isothermal atmosphere by roughly (n/(n+1)). In this case n = 6.33, higher than the dry adiabat for CO2 which gives n = 4.45.

Now an adiabatic or polytropic atmosphere is an idealisation, but it seems odd to me that whenever Venus' atmospheric mass is discussed people always use the higher isothermal value. Have I missed something important in the physics, or is Venus's atmospheric mass just 86.4% of the usually quoted value?
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remcook
post Jun 5 2006, 02:15 PM
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QUOTE
do.Zo = Po/g


This is the equation for hydrostatic equilibrium:

dp = rho * g * dz

For isothermal atmosphere, this gives an exponential density profile:

rho = rho0 * exp(-z/H) - H is scale height.

Integrating this will give you what you want I think. If you know temperature and pressure vs. height (from probes for instance), you can determine the density from the ideal gas law at different altitudes. Integrating that should be more accurate.
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qraal
post Jun 5 2006, 11:45 PM
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Hi rem

That's exactly what I did and I still get the non-intuitive answer that a polytropic atmosphere with the same surface pressure masses less than an isothermal atmosphere.

I just want to know if that's correct. Earth's atmospheric mass is quoted as being 10,333 kg/m^2, but if you integrate the standard atmosphere it's only ~ 9,561 kg.

Adam

QUOTE (remcook @ Jun 6 2006, 02:15 AM) *
This is the equation for hydrostatic equilibrium:

dp = rho * g * dz

For isothermal atmosphere, this gives an exponential density profile:

rho = rho0 * exp(-z/H) - H is scale height.

Integrating this will give you what you want I think. If you know temperature and pressure vs. height (from probes for instance), you can determine the density from the ideal gas law at different altitudes. Integrating that should be more accurate.
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ngunn
post Jun 6 2006, 11:37 AM
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In an atmosphere in which temperature decreases rapidly with height for the first few scale heights the upper parts will not be as high above the surface as they would be in the isothermal case. This means they will experience greater g and hence contribute more to the pressure at the surface. More pressure for the same mass = less mass for the same pressure.

I like to imagine two simplistic pictures - One is the thin atmosphere approximation where all the atmospheric mass is assumed to be just above the surface. Two is the isothermal case. Venus should be somewhere in between.
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qraal
post Jun 7 2006, 12:14 PM
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Hi ngunn

Thanks for the reply. You know I wondered if that was the case, but it seems incorrect.

Firstly in a true adiabatic atmosphere the atmosphere has a sharp boundary at an altitude of n.Zo i.e. (Cp/Cv)/(Cp/Cv-1) scale heights up. As the scale height for both isothermal and adiabatic atmospheres is the same for the same surface temperature an increase in altitude of just 4.5 scale heights doesn't seem enough to change the gravity very much. For a dry adiabat of a CO2 atmosphere with To=735 K that's an altitude of just 70.5 km. That's a gee-change of just 2%.

Secondly, the isothermal atmosphere modified to take into account the change of gravity with radius has an infinite mass per unit area - the average density approaches a finite value which summed to an infinite altitude gives an infinite mass.

Adam

QUOTE (ngunn @ Jun 6 2006, 11:37 PM) *
In an atmosphere in which temperature decreases rapidly with height for the first few scale heights the upper parts will not be as high above the surface as they would be in the isothermal case. This means they will experience greater g and hence contribute more to the pressure at the surface. More pressure for the same mass = less mass for the same pressure.

I like to imagine two simplistic pictures - One is the thin atmosphere approximation where all the atmospheric mass is assumed to be just above the surface. Two is the isothermal case. Venus should be somewhere in between.
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ngunn
post Jun 7 2006, 12:46 PM
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Well you've done the detailed calculations, not me, but I'm surprised by your statement that an isothermal atmosphere would have an infinite mass per unit area. This would be true only for an infinitesimally small planet (!) assuming such an object could have an atmosphere. Are you starting your integrals from r=0 or h=0 ?
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Phil Stooke
post Jun 7 2006, 01:50 PM
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There's some great Venus atmosphere stuff (and some on Mars) from a presentation here:

http://www.mrc.uidaho.edu/entryws/full/pro...e_detailed.html

Check out the zipped powerpoint presentation I-4.1 on the Venera missions.

Phil


--------------------
... because the Solar System ain't gonna map itself.
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Guest_DonPMitchell_*
post Jun 7 2006, 08:33 PM
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Guests






QUOTE (Phil Stooke @ Jun 7 2006, 06:50 AM) *
There's some great Venus atmosphere stuff (and some on Mars) from a presentation here:

http://www.mrc.uidaho.edu/entryws/full/pro...e_detailed.html

Check out the zipped powerpoint presentation I-4.1 on the Venera missions.

Phil


Kerzhanovich was in charge of the doppler shift studies in the Venera missions, and now works at JPL. He was famous for his arguments with mission planners, to land the Venera capsules further off center, so he could get better horizontal wind measurements. Astronomers thought the atmosphere was superrotating, from UV cloud studies, but there was a long history of people misjudging the rotation of Venus, and not complete certainty that the UV features were some kind of atmospheric wave phenomenon. So Kerzhanovich's dopper measurements were the first real proof of superrotating zonal winds on Venus.

[attachment=6124:attachment]

I see he liked my annotated version of the Venera-7 doppler data (grin). But the best analysis of that data was made by John Ainsorth at Goddard SFC. The Soviets really didn't want to talk about the parachut failures too much, but Ainsorth did some pretty detailed simulations and paints a fairly hair-raising picture of the probe's descent. After complete parachute failure, the probe hit the surface of Venus at 38 miles per hour, bouncing on impact and then coming to rest at a 50-degree tilt which severely lowered its radio signal to Earth. Russian engineers later used the doppler data and the fact that Venera-7 was not destroyed on impact, to calculate properties of the surface soil.

One man's parachute failure is another man's penetrometer!
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qraal
post Jun 9 2006, 12:43 PM
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Hi ngunn

I've read it before, but replicated the calculations. Basically the 'energy height' is finite for a realistic spherical planet so a finite density for the atmosphere is achieved all the way to infinity (the usual limits for the computation of atmospheric mass are from z =0, at R=surface radius, out to z= infinity.) The usual integral, assuming constant gravity with altitude is thus in an infinite potential well.

In the real universe the mean free path becomes large enough so that mixing is no longer an issue and all the atmospheric components fractionate due to their different masses. And then along comes the solar wind, Van Allen Belts etc. But in abstract you get an infinite mass.

Adam

BTW the average density at infinity is very low d = do*e^-[(11.18/0.5)^2] = 9 x 10^-218 kg/m^3, one molecule every 8.6 x 10^47 light years, but integrated to infinity it's still infinite. Calculus taken to its absurd limits.

QUOTE (ngunn @ Jun 8 2006, 12:46 AM) *
Well you've done the detailed calculations, not me, but I'm surprised by your statement that an isothermal atmosphere would have an infinite mass per unit area. This would be true only for an infinitesimally small planet (!) assuming such an object could have an atmosphere. Are you starting your integrals from r=0 or h=0 ?
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ngunn
post Jun 9 2006, 12:57 PM
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Very interesting, and I follow the reasoning you give but I still find the result surprising.

Consider the following thought experiment:
1/ Take one large solid globe, in vacuo.
2/ Release in it's vicinity finite amount of gas.
3/ Wait for the gas to form an atmosphere around the globe.
4/ Arrange for the atmosphere to be at a uniform temperature throughout.

Question: At what point does the mass of this finite amount of gas jump to infinity?
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The Messenger
post Jun 9 2006, 01:57 PM
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QUOTE (qraal @ Jun 5 2006, 05:45 PM) *
That's exactly what I did and I still get the non-intuitive answer that a polytropic atmosphere with the same surface pressure masses less than an isothermal atmosphere.

cool.gif
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ngunn
post Jun 10 2006, 10:39 AM
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Hi graal. I think that the messenger's messageless message is trying to say we have gone round in a circle . . . however I still need to understand things in my slow, hand-waving way.

First 2 questions. Are you now satisfied that you have found the reason why your integral doesn't go on to infinity, and are you now getting an answer for the atmospheric mass of Venus that is in line with what you expected intuitively?

For me there had to be a common sense reason why the integral terminates, and it had to be independent of external factors. A finite atmosphere must be able to restrain it's own urge to become infinite without being disciplined by the Van Allen police! So I did another thought experiment:

This time I provided my gravitating globe with an extremely thin atmosphere. In fact I gave it just one single molecule of gas. The molecule hops about receiving thermal kicks from the surface at each bounce. I quickly realised that this atmosphere could not be isothermal, because the molecule slows down as it rises and accelerates again as it falls. At the very top of the highest bounces (which must be vertical) the molecule stops altogether, thus momentarily reaching zero Kelvin. The atmosphere therefore has zero volume - and zero mass - from here on out.

So it is the fact that molecules are of finite size that causes the atmosphere to terminate. If the molecules were infinitely many and infinitesimally small we would never reach the point where the mean free paths become long and thermal collisions give way to ballistic trajectories. The atmosphere would indeed go on for ever whilst still having a finite total mass, a contradiction, therefore molecules are of finite size.

I don't know if this is of any use but thanks anyway for making me think.
Regards
Nigel
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qraal
post Jun 10 2006, 12:34 PM
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Hi Nigel

Ok. Let's try again. Everything you've said about the discreteness of the atmosphere is true and as I've already noted it's what really obtains - eventually the gases stop running into each other and get separated out by gravity. I know that, though your eloquent restatement was better than my clumsy exposition.

What I was discussing was an idealisation, the hydrostatic model of an isothermal atmosphere that's in a gravity field dropping off via the inverse square law. Assume the gas is infinitely divisible and you get an infinite mass at infinity - infinity BTW is the usual limit when integrating the equation (rho) = (rho)o*e^-([Z/Zo), which as you can see only achieves a zero value at Z/Zo = infinity. That's just the nature of the maths and I know its limitations. When molecules only collide every few metres or so things change drastically from the case of micrometre path lengths.

My quandary concerns the case of the adiabatic atmosphere - a whole different ballgame, to coin a phrase. An adiabatic atmosphere in its idealisation has a finite altitude at which temperature, pressure and density fall to zero. That's because of the temperature-altitude relation - i.e. T/To = [(n.Zo - Z)/(n.Zo)] so T/To = 0 at Z = n.Zo, and because (rho) = (rho)o*(T/To)^n it means it drops to zero too.

So the mass integral has finite limits rather than the unphysical "infinity". Integrated, the equation gives:
m = (n/(n+1))*(n.Zo) and n.Zo = Po/g. So as I've already noted the mass is lower than the isothermal case for the same surface pressure.

Why is this so? And is it correct? The maths says "yes" and it means that the atmospheric mass on Venus is less than the usually quoted figures. So 88.8 bar of CO2 is actually 76.8 'bar' in weight. And 865 tons/sq.m in mass not the usual 1,000.

Ok. So that's the state of play. Why is it so? My first thought is that because expansion cooling lowers the pressure as you climb in height the hydrostatic balance can be maintained without relying on just the weight of the gas column itself. The temperature gradient acts like an extra potential. Am I even close?

Adam

QUOTE (ngunn @ Jun 10 2006, 10:39 PM) *
Hi graal. I think that the messenger's messageless message is trying to say we have gone round in a circle . . . however I still need to understand things in my slow, hand-waving way.

First 2 questions. Are you now satisfied that you have found the reason why your integral doesn't go on to infinity, and are you now getting an answer for the atmospheric mass of Venus that is in line with what you expected intuitively?

For me there had to be a common sense reason why the integral terminates, and it had to be independent of external factors. A finite atmosphere must be able to restrain it's own urge to become infinite without being disciplined by the Van Allen police! So I did another thought experiment:

This time I provided my gravitating globe with an extremely thin atmosphere. In fact I gave it just one single molecule of gas. The molecule hops about receiving thermal kicks from the surface at each bounce. I quickly realised that this atmosphere could not be isothermal, because the molecule slows down as it rises and accelerates again as it falls. At the very top of the highest bounces (which must be vertical) the molecule stops altogether, thus momentarily reaching zero Kelvin. The atmosphere therefore has zero volume - and zero mass - from here on out.

So it is the fact that molecules are of finite size that causes the atmosphere to terminate. If the molecules were infinitely many and infinitesimally small we would never reach the point where the mean free paths become long and thermal collisions give way to ballistic trajectories. The atmosphere would indeed go on for ever whilst still having a finite total mass, a contradiction, therefore molecules are of finite size.

I don't know if this is of any use but thanks anyway for making me think.
Regards
Nigel
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ngunn
post Jun 12 2006, 11:23 AM
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Fine. I agree the adiabatic model is probably a better approximation for the real Venus atmosphere than the isothermal (or quasi-isothermal with a ballistic 'cap'). I would therefore prefer your lower estimate of the mass to one calculated on the quasi-isothermal model. The only thing I don't quite understand is this quote from your post #3:- 'I still get the non-intuitive answer that a polytropic (adiabatic?) atmosphere with the same surface pressure masses less than an isothermal atmosphere.' To me this result is exactly what one WOULD expect intuitively. The atmosphere is cooler at the top, therefore closer to the surface and experiencing higher g, therefore less mass is required to account for the observed surface pressure.

My extended ramble into the isothermal model was just because I was intrigued by the infinite integral and its implications, especially the fact that the particulate nature of gases can be 'deduced' in this way, which I had not realised before.
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The Messenger
post Jun 12 2006, 08:56 PM
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QUOTE (ngunn @ Jun 12 2006, 05:23 AM) *
Fine. I agree the adiabatic model is probably a better approximation for the real Venus atmosphere than the isothermal (or quasi-isothermal with a ballistic 'cap'). I would therefore prefer your lower estimate of the mass to one calculated on the quasi-isothermal model. The only thing I don't quite understand is this quote from your post #3:- 'I still get the non-intuitive answer that a polytropic (adiabatic?) atmosphere with the same surface pressure masses less than an isothermal atmosphere.' To me this result is exactly what one WOULD expect intuitively. The atmosphere is cooler at the top, therefore closer to the surface and experiencing higher g, therefore less mass is required to account for the observed surface pressure.

Confusing, but I find myself agreeing with Gaal. The fact that the temperature is decreasing with altitude means the upper atmosphere is more dense at that altitude than it would be under isothermal conditions, and each inversion in temperature would lead to more compacting of the molecules relative to isothermal conditions. At the altitude after which there are no further inversions, the expansion is the same as under the isothermal condition, except that there is a denser layer under the inversion than in the isothermal case.

The only exception would be if the last inversion is very close to the surface, and the temperature increase after this inversion is much much greater than the gradient between the surface and the inversion.

It is worth noting that in the current model of the Titan atmosphere, they use many inversion layers in order to support the density distribution found by Huygens - (Titan's atmosphere is very thick in general, relative to the earth's, but also has a more exaggerated vertical scale due to the lower mass of the moon.) We will know more about Titan after the limb and bistatic radar measurements are evaluated...The Cassini altimeter data should help, too.
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