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qraal
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?
remcook
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.
qraal
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.
ngunn
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.
qraal
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.
ngunn
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 ?
Phil Stooke
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
DonPMitchell
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.

Click to view 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!
qraal
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 ?
ngunn
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?
The Messenger
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
ngunn
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
qraal
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
ngunn
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.
The Messenger
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.
ngunn
Hi Messenger. You found my post confusing??!!

I don't think graal and I are in disagreement at all, just thinking through the same thing in different ways, which I for one have found illuminating.

The Titan case is very different from Venus because after decreasing to the tropopause the temperature rises strongly but erratically with height. No simple model will do in this case, neither isothermal nor adiabatic. One would have to integrate by the computer equivalent of counting squares under the graph. To make things still more complicated I wouldn't be surprised if the temperature profile varies quite a bit with latitude, season, cryovolcanism and precipitation processes, maybe also interaction with Saturn's magnetosphere. But we have wandered quite a long way from Venus now and should really be in 'Titan's atmosphere and weather'.
remcook
just a note: Titan's temperature does vary significant with latitude, but not with longitude. at the north, the lower stratosphere gets very cold (at least 20 degrees colder than the equator) and the higher stratosphere gets very warm.
qraal
Hi All

Well I decided to do a numerical experiment on a model Venus atmosphere and sum up 100 metre high blocks of atmosphere, computing Cp, p, rho, T, m, and energy with each step. I looked up the behaviour of CO2's heat capacity, Cp, over the temperature range I was interested in and discovered it varied dramatically - from 0.735 KJ/kg.K at 200 K to 1.148 KJ/kg.K at 750 K. Quite unlike the behaviour of air for which Cp is constant over the terrestrial temperature range. I could fit the behaviour of Cp with temperature well enough with a quadratic over the range 275 K - 750 K, and a linear fit between 275-200 K.

I baulked at an analytical solution, so I chucked the equations of state into an Excel spread-sheet and summed it up, from 735 K and 92 bar to 0.17 bar at 198 K. What I found was the mass is exactly Po/g, which was quite a "surprise" considering my first doubts. The total atmospheric heat energy is 0.706 teraJoules, which would radiate away to space in about 138 years at a constant 231 K effective temperature - if you could shade Venus. Of course at 304 K the lower reaches would start condensing as liquid, adding extra latent heat.

My final task is to work out the average temperature.

Adam
qraal
Hi again

And the average temperature is 630 K.

Hmmm... that was an interesting exercise. One of these days I will have to do it with smaller increments in a C program and calculate a new value for gravity at each step. But I can't imagine it will be much different in final outcome.

Of course the really thorough approach is to do a General Circulation Model of the whole planet... whew!

My curiosity was first sparked by Paul Birch's paper "Terraforming Venus" quickly, which used a figure of 630 K for Venus' average temperature - how did he arrive at that figure? He also used 0.84 KJ/kg.K as Venus' heat capacity - the average I got is more like 1.075 KJ/kg.K. His figures were fine for a rough estimate, but the problem had me intrigued, so I did my first naive analytical solution and quickly got very puzzled by the discrepancy between our results.

Now it seems the numerical approach has verified his estimate, more or less.

My rough model has a lot of room for improvement, but it's kind of fun in a nerdy way ;-)

Adam
ngunn
QUOTE (qraal @ Jun 23 2006, 02:11 PM) *
Hmmm... that was an interesting exercise. One of these days I will have to do it with smaller increments in a C program and calculate a new value for gravity at each step. But I can't imagine it will be much different in final outcome.


Yes, someone's still watching this space! Nice to have your updates.

From the quote above I was wondering what you did do about g in your latest approximation? You're bound to get a mass per unit area of Po/g if you use the same g throughout, regardless of how the other parameters vary. The only thing the temperature and pressure do is control the mean height of an air molecule, hence the mean g it experiences and therefore its mean weight. So if you allowed for a decrease of g with height you should have got a higher mass . . .
The Messenger
[quote name='qraal' date='Jun 23 2006, 06:42 AM' post='59569']
Hi All

Well I decided to do a numerical experiment on a model Venus atmosphere and sum up 100 metre high blocks of atmosphere, computing Cp, p, rho, T, m, and energy with each step. I looked up the behaviour of CO2's heat capacity, Cp, over the temperature range I was interested in and discovered it varied dramatically - from 0.735 KJ/kg.K at 200 K to 1.148 KJ/kg.K at 750 K.
[/quote]
That is good news - now we know that as the average Earth temperature increases, the additional CO2 will act as a heat sink and the glaciers will come back unsure.gif

Seriously, this was good analytical work, eliminating a gross inconsistency and demonstrating that there is almost order in the universe...

[/quote]
qraal
Hi Messenger & ngunn

Hey thanks for the nice comments and useful gedankenexperiments. More update as follows...

I've 'discovered' that the US NIST has the proper fit equation for Cp of CO2 online - I was close as it's almost a quadratic. I just did a rough fit from a table of values, but I was pretty close.

Now I need to work out a simplified equation for 'g' for a future C version of the model. Of course I could just use the good odd inverse square law, but why make it easy on me and hard on the number cruncher? Gravity varies almost linearly with altitude in the range we're discussing. The less maths functions I need to call from the C library the better.

Also I've managed to find a discussion of cloud formation with data on Venus's sulphuric acid clouds, so I'll be able to add that little refinement to the higher altitude computations.

Also I did an Earth version of the model based on the International Standard Atmosphere's lapse rates and altitudes, all the way to the Stratopause's ceiling at 51 km. The match is pretty good and the mass of the air column is exactly as expected, Po/g. The ISA uses 'geopotential altitude' so I didn't need to compute 'g'.

Now all I need is to figure out the heat transfer equations and I can model the atmosphere temporally too. I've got a copy of Mark Bullock's thesis and he covers that in some detail, probably more detail than my poor laptop can number crunch in a meaningful time.

Adam
RNeuhaus
Venus' Double Vortex Confirmed in New Animation

A huge "double-eye" atmospheric vortex has been confirmed to exist at the South Pole of the planet Venus.

http://www.space.com/scienceastronomy/0606...nus_vortex.html



There are two vortex in the South Pole. That is odd!! ohmy.gif

Scientists think the vortexes are created by a combination of a natural cycling of hot air in the planet's atmosphere and high velocity, westward-blowing winds that take only four days circle the planet. It is still unclear, however, why there are two vortexes at each pole.

Rodolfo
DonPMitchell
At last, some pictures from VEX. These are fascinating.

Click to view attachment Click to view attachment

Here are the best previous images of the double vortex, by Pioneer Venus, and the Venera-15 IR spectrometer.
qraal
Thanks Don

Man that's so bizarre.

Venus is a surprising planet - not 'dead' at all, except in a restricted sense, and I'm still hanging out for sulphur munching bugs in the clouds.

There's a surprising amount of energy locked up in that 4 day rotation.

Adam

QUOTE (DonPMitchell @ Jun 28 2006, 01:33 PM) *
At last, some pictures from VEX. These are fascinating.

(snipped pix)

Here are the best previous images of the double vortex, by Pioneer Venus, and the Venera-15 IR spectrometer.
qraal
Hi All

Updated the gravity - now it's a linear approximation good out to about 200 km. Considering the quintic equations I used working out Cp(T) for N2 and CO2 the computational improvement is slight, but the mental exercise was worthwhile. Also I had made a basic error working out the enthalpy of the gas mix for each cell, so I've corrected that too. The total energy, using 0 K as the reference point, is about 588 GJ/sq.m.

Having done all that I've been trying to educate myself on cloud condensation physics and getting a headache in the process. Does anyone here know if the cloud layers have been measured in their vertical extent and optical depth? I'm pretty sure I've read some such data, but a good reference on the topic would be handy. The article I have is more theoretical than factual and gave me a few hints on how to modify the model, but now that I'm tweaking it I'm dissatisfied with a 'fudge factor' and want some real understanding.

Chilling Venus's atmosphere out, as a precursor to terraforming, was the main problem I wanted to investigate when I began this exercise, but as I've progressed the fun has come from comparing it against the data. Some very odd things would happen on a darkened Venus - CO2 snow would probably fall (and sublime on the way down) before CO2 rain did, for example. I've been imagining a scenario in which hydrogen was crashed into Venus to burn the CO2 into carbon and water, via the Bosch reaction, then wondering what condensing water would do to a surface made of carbon-deprived oxides of calcium and the like. Would a mad upheaval of the regolith ensue, rapidly forming carbonates? Kind of like 'cooking' limestone to make lime, but in reverse.

Adam
edstrick
The soviet venus descent probes measured light levels and then (later missions) visible and near-IR spectra. They also took cloud samples and got composition data. They may also have carried "nephelometers" as well. Quite a bit of data.

The american Pioneer venus probes carried nephelometers on the small probes and more sophisticated cloud particle size spectrometer and infrared radiometer on the large probe. There's a rather vast amount of information for a rather small number of atmospheric descents.

The most global data on clouds is from radio occultation temperature/pressure profiles down to some 35 km altitude, and occultation data from Magellan provides cloud absorption profiles.

The starting place for all current studying of Venus is the Univ. Arizona Press book "VENUS" from the early 90's.
qraal
Hi ed

Thanks for the heads up on that reference, which I'll have to track down.

Adam
edstrick
The Univ of Arizona has had a method of producing "Weighty Tomes" for some decades.

They announce a conference on a subject, recently held was "Protostars and Planets V". With the aid of experts in the field, they pick a set of topics and the authors who are most expert in those topics. "You have just been volunteered to write a summary article on what we know about the vertical cloud structure of Venus' atmosphere, and we have volunteered the Russian Expert, Academician Boris, to be your co-author on this paper."

It's a sort of "you can't refuse" honor.

Other reseachers are invited (open invitation) to present papers at the meeting. The review papers <first versions ready at the time of the meeting> and selected other papers are published in a thousand or so page volume with a title like "Venus" or "Protostars and Planets V" at what used to be and probably still is a remarkably low price for such an academic "tome".

These books become landmarks and mileposts in the field. For studying pre-Magellan Venus studies, the "Venus" volume is 1,000% essential. You START there and work your way through relevant articles and the relevant papers referernced in those articles.
DonPMitchell
The U of Arizona books are essential. There are actually two Venus books, and you want them both (Venus and Venus II). Also check out their Mars book.

There have been many descent probes on Venus. The first really detailed information came from Venera-9 and 10 in 1975, which had nephelometers, spetrometers, and various other instruments. The three-layer cloud structure was discovered then, as well as the first photos of the surface.

Venera-11, 12 and the Pioneer probes arrived in 1978. The Venera probes contained more sophisticated spectrometers than the 1975 versions, and also mass spectrometers, gas chromatography and x-ray fluorescence spectrometers. The latter provided the first real data on the composition of the cloud material, including the discovery of Iron Chloride as a component. The PV large probe had a particle-size device which provided unique data. There has been much debate about the PV results, which some believe show a distinct large particle size that may be crystals -- the so-called mode 3 controversy.

Venera descent probes relayed data through the high-gain connection of the main spacecraft, so they were able to send almost 100 times as much data as the Pioneer probes, including large numbers of mass spectra and optical spectra as they descended.

The last word has been from the Vega descent probes. They were particularly geared to study the clouds and try to answer questions raised by the Pioneer particle-size spectrometer. Two different particle analyzers were on the Vega probes, as well as a more sophisticated gas chromatography experiment, which discovered the profile of chlorine, sulphur and phosphorus abundance as a function of altitude. The clouds are a lot more complex than just sulphuric acid droplets, particularly the lower layers.

The big names in Venusian atmospheric chemistry are Vladimir Krasnopolsky, the late Vasily Moroz and Larry Esposito. They have a joint paper in Venus II, and Krasnopolsky has a good book on the chemistry of Mars and Venus.
qraal
Hi Don

Hey thanks for the insights - Venus is truly weird. Would be dull if it was easy.

Your webpages on the Russian Venus efforts are amazing too.

What do you think the 'crystals' in the atmosphere might be?

Adam

QUOTE (DonPMitchell @ Jul 3 2006, 05:56 AM) *
The U of Arizona books are essential. There are actually two Venus books, and you want them both (Venus and Venus II). Also check out their Mars book.

There have been many descent probes on Venus. The first really detailed information came from Venera-9 and 10 in 1975, which had nephelometers, spetrometers, and various other instruments. The three-layer cloud structure was discovered then, as well as the first photos of the surface.

Venera-11, 12 and the Pioneer probes arrived in 1978. The Venera probes contained more sophisticated spectrometers than the 1975 versions, and also mass spectrometers, gas chromatography and x-ray fluorescence spectrometers. The latter provided the first real data on the composition of the cloud material, including the discovery of Iron Chloride as a component. The PV large probe had a particle-size device which provided unique data. There has been much debate about the PV results, which some believe show a distinct large particle size that may be crystals -- the so-called mode 3 controversy.

Venera descent probes relayed data through the high-gain connection of the main spacecraft, so they were able to send almost 100 times as much data as the Pioneer probes, including large numbers of mass spectra and optical spectra as they descended.

The last word has been from the Vega descent probes. They were particularly geared to study the clouds and try to answer questions raised by the Pioneer particle-size spectrometer. Two different particle analyzers were on the Vega probes, as well as a more sophisticated gas chromatography experiment, which discovered the profile of chlorine, sulphur and phosphorus abundance as a function of altitude. The clouds are a lot more complex than just sulphuric acid droplets, particularly the lower layers.

The big names in Venusian atmospheric chemistry are Vladimir Krasnopolsky, the late Vasily Moroz and Larry Esposito. They have a joint paper in Venus II, and Krasnopolsky has a good book on the chemistry of Mars and Venus.
DonPMitchell
That is a mystery. Many believe there are no crystals, just larger sized droplets. Sulphur, Chlorine, Phosphorus and Iron have all been identified as important elements in the cloud material. There are many theories about how these elements might combine.
qraal
Hi Don

The exobiological theory is the most exciting, but we know so little about the chemistry of such vapours on such a scale. Maybe some process is making odd carbon allotropes?

Adam

QUOTE (DonPMitchell @ Jul 3 2006, 01:01 PM) *
That is a mystery. Many believe there are no crystals, just larger sized droplets. Sulphur, Chlorine, Phosphorus and Iron have all been identified as important elements in the cloud material. There are many theories about how these elements might combine.
edstrick
The Pioneer Large Probe Cloud Particle Size Spectrometer took images of the shadows of particles as they flowed through the field-of-view / focal plane of the instrument. These data were processed to count particles and sort them into size-bins, producing particle size "spectra" or more accurately, "size-distribution histograms".

As I vaguely recall, the largest, better resolved particles were analyzed for departures from circuarity or "equi-dimensional" form.. some "length-vs-width" number. In some parts of the clouds, small numbers of particles considerably larger than the approx 1 micrometer sulphuric acid droplets were supposedly detected and these particles were flagged as not round.

There was an extended arguement over the validity of the data, and it (as far as I used to know -- my info's obsolete), was never resolved as there wasn't enough data to go any further. No other missions took comparable data. One Mission, One Instrumet, One descent profile..... "Tiz a Puzzlement".
DonPMitchell
Vega-1 and Vega-2 performed similar experiments. The LSA particle-size spectrometer and the ISAV-A optical aerosol analyzer were installed on the last two spacecrafts to land on Venus. The ISAV-A investigators concluded that the large particles were approximately spherical with a refractive index of 1.4 +/- 0.1.

The Vega instruments also measured sub-micron particles in considerable abundance, which were too small for the PV instrument to see.

The Venera-9 and 10 nephelometers measured scattering at several angles, but that data was never as conclusive as the aerosol particle sensors on PV and Vega.
JRehling
QUOTE (RNeuhaus @ Jun 27 2006, 03:21 PM) *
Venus' Double Vortex Confirmed in New Animation
Scientists think the vortexes are created by a combination of a natural cycling of hot air in the planet's atmosphere and high velocity, westward-blowing winds that take only four days circle the planet. It is still unclear, however, why there are two vortexes at each pole.


I didn't know about the double vortex -- it is odd.

I wonder if it has something to do with a vortex forming on the solar side then rotating into shade in time for a new vortex to form when the first one has moved into night. As for why that would happen in discrete pairs per rotation is bound to be a tricky question, but cyclical patterns do arise in turbulent systems -- the devil is at a level of detail my [non]study of fluid dynamics has never gotten to.
qraal
Thanks Don & Ed

More data to cram into my brain.

I'd really like to get my hands on the Venus International Reference Atmosphere, but VIRA isn't available even second hand on all the usual web-shops. I have found journal articles which give some values from VIRA, but they're kind of empty without the underlying rationale - the scale heights vary so much between levels I can't see any pattern except for large poly-nomials. Why? Tighter fit to the data I guess, but the underlying processes seem obscure.

Will have to try and dredge up those Venus books.

Adam
DonPMitchell
Planetary circulation is fascinating. Rotating patterns of vortices are not unusual.

Click to view attachment

This old Tiros-9 photo of the Earth shows a number of large cyclonic and anti-cyclonic weather patterns extending cross the north and south temperate zones. You get a sort of standing-wave pattern like that, with about six of these giant vortices moving around the Earth in the north and south. Called Rossby Waves or planetary waves.

Around the equator, where the change in angular momentum is less, you just have simple hadley-cell circulation, tilted at an angle by coriolis force.

On Jupiter you see many bands of this stuff. A trade-wind band around teh equator like Earth, then some Rossby waves in the temperate bands, and then nearer the poles, the powerful coriolic forces lead to chaotic circulation.
MichaelT
QUOTE (qraal @ Jun 5 2006, 12:15 PM) *
Hi All

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

If I get your question right you don't have to integrate, because you know the surface pressure and the acceleration of gravity of Venus. That makes it much simpler:

pressure = force / area
force = mass * acceleration
=> pressure = mass * acceleration / area

This translates to the following formula:

p = M * a / A

=> M / A = p / a

All you have to know is the (equatorial) acceleration of gravity of Venus, which is a = 8.87 m/sē, and the surface pressure p = 9.3 MPa.

Therefore:
M / A = 9,300,000 Pa / 8.87 m/sē
M / A = 1,048,478 kg/mē

The Venusian atmosphere has a mass of more than 1000 tonnes per square meter. It's almost exactly 1/100th of that for Earth (p = 101300 Pa; a = 9.81 m/sē; M/A = 10,326 kg/mē). For Mars it is just 217 kg/mē.

Michael
edstrick
I'd forgotten the instruments on the Vega probes, remembering the collected particle composition measurements and the like. Ultimately, Venus atmosphere science needs a long-lived balloon system that can make repeated up and down "bobbing" oscillations from under the sub-cloud haze layers to at least the bottom of the upper haze layer.
Bob Shaw
QUOTE (DonPMitchell @ Jul 4 2006, 02:26 AM) *
Planetary circulation is fascinating. Rotating patterns of vortices are not unusual.

This old Tiros-9 photo of the Earth shows a number of large cyclonic and anti-cyclonic weather patterns extending cross the north and south temperate zones. You get a sort of standing-wave pattern like that, with about six of these giant vortices moving around the Earth in the north and south. Called Rossby Waves or planetary waves.

Around the equator, where the change in angular momentum is less, you just have simple hadley-cell circulation, tilted at an angle by coriolis force.

On Jupiter you see many bands of this stuff. A trade-wind band around teh equator like Earth, then some Rossby waves in the temperate bands, and then nearer the poles, the powerful coriolic forces lead to chaotic circulation.



Don:

Remember the early National Geographic paintings of our world as seen from space, back in the mid 1950s? They tended to show Earth with belts, just like Jupiter!

Bob Shaw
ngunn
QUOTE (MichaelT @ Jul 4 2006, 10:25 AM) *
If I get your question right you don't have to integrate, because you know the surface pressure and the acceleration of gravity of Venus. That makes it much simpler:

pressure = force / area
force = mass * acceleration
=> pressure = mass * acceleration / area

This translates to the following formula:

p = M * a / A

=> M / A = p / a

All you have to know is the (equatorial) acceleration of gravity of Venus, which is a = 8.87 m/sē, and the surface pressure p = 9.3 MPa.

Therefore:
M / A = 9,300,000 Pa / 8.87 m/sē
M / A = 1,048,478 kg/mē


Graal and I have been round this a couple of times since his 5th June post which you reply to here.
The expression 'mass times acceleration due to gravity over surface area' is fine, but to calculate the total weight of the atmosphere you have to take account of the variation of gravity with height. How much of the atmosphere is at which height depends in turn on how the integrals of pressure, temperature etc shape up. By measuring the surface pressure we in fact measure the weight of the atmosphere, not it's mass directly. To infer its mass we need to refer to the same model of the atmospheric parameters as we would if reasonong in the opposite direction, from mass to weight, and hence pressure. This is what graal has been doing (I think!)
MichaelT
QUOTE (ngunn @ Jul 4 2006, 03:51 PM) *
measuring the surface pressure we in fact measure the weight of the atmosphere, not it's mass directly. To infer its mass we need to refer to the same model of the atmospheric parameters as we would if reasonong in the opposite direction, from mass to weight, and hence pressure. This is what graal has been doing (I think!)

Oh, I see smile.gif
Yes, that is what I forgot. Gravity acceleration is not constant with height and the change is not negligible... So the formula that I gave would just be the lower mass limit wouldn't it?

That's certainly a difficult problem.

Michael
The Messenger
QUOTE (MichaelT @ Jul 4 2006, 10:40 AM) *
Oh, I see smile.gif
Yes, that is what I forgot. Gravity acceleration is not constant with height and the change is not negligible... So the formula that I gave would just be the lower mass limit wouldn't it?

One would think so, but if I have followed Gaal correctly, the quoted mass of the Venus Atmosphere is less than this 'simple' limit. There are a number of possibilities that could lead to this unintuitive result - including the non-linear thermal capacity CO2.
ngunn
Which is what I still don't understand (or believe). In this limiting 'skinny atmosphere' aproximation every air molecule is assumed to be near the surface experiencing the maximum possible g. The atmosphere cannot weigh more in any other configuration than it does in this one and the thermal and other properties of the gas become irrelevant in this approximation. There is one, and only one, way that the air molecules could exert a reaction force on the surface of the planet greater than their own weight and that is if they were being fired off ballistically faster than the escape velocity and leaving the planet altogether.
qraal
Hi ngunn

For the first 200 km of altitude Venus's gravity is 8.87 - 0.0028z (m/s^2) with z in km.

Not bad really, but it adds a bit of mass to the air column overall.

Thing is I am not yet convinced by the hydrostatic argument, but it seems sound on numerical integration.

Adam

and it's 'Qraal' - a bit like 'kraal' and 'kroll'.


QUOTE (ngunn @ Jul 5 2006, 09:19 PM) *
Which is what I still don't understand (or believe). In this limiting 'skinny atmosphere' aproximation every air molecule is assumed to be near the surface experiencing the maximum possible g. The atmosphere cannot weigh more in any other configuration than it does in this one and the thermal and other properties of the gas become irrelevant in this approximation. There is one, and only one, way that the air molecules could exert a reaction force on the surface of the planet greater than their own weight and that is if they were being fired off ballistically faster than the escape velocity and leaving the planet altogether.
ngunn
Ah! qraal with a Q - my apologies. It's the underlining that does it. I keep getting Messenqer wronq as well.
qraal
Hi ngunn & MichaelT

As you might've guessed I have performed a numerical integration using a variable 'g' and the results are similar. Atmospheric mass only differs slightly between the two (<1%).

My real puzzle is why integration of the density equation, r = ro(T/To)^(Cp/(Cp-Cv)), didn't give the mass as Po/g. Constant lapse rate, as used in the International Earth Atmosphere, still gives Po/g after numerical integration - but I must've made a stupid assumption somewhere when I derived an analytical solution that differed so strongly. Over the temperature range in question, for Earth, Cp/Cv is practically constant, and only begins to differ from 1.4 for temperatures over 350 K.

That being said the Venus model is doing ok, but I've discovered the learning curve on modelling atmospheric absorption of solar radiation is quite steep. The (z,T) curve for Venus rapidly approaches a dT/dz of ~0 at a certain altitude, I guess due to increased heat gain from radiation. Now I could throw in an empirical fit, just like VIRA, but I'd like to read more of my references and come to some understanding of what's going on. I'm guessing that the atmosphere is becoming stable against convection forming an isothermal 'lid' just like the temperate zone tropopause on Earth.

If so there should be some literature around on just how that works, at least for Earth, and by extrapolation for Venus too.

Adam Crowl aka qraal
DonPMitchell
You're looking at Mark Bullock's thesis I assume. The only problem there is he is doing a 1-dimensional climate simulation, just the radiation balance calculation. That's the simplest of models, and doesn't take into account the important and complex effect of heat transport by convection. As soon as you try to take that into account, you must begin simulating general circulation, and the calculation becomes a lot more complex and expensive.

There are probably also latent heat issues due to phase changes and chemical changes, which are not fully understood yet on Venus. In Earth simulations, the phase changes in water are important, but I don't know if that's true for Venus. The role of water vapor in Earth climate modelling is still one of the biggest unknown factors -- water vapor is a greenhouse gas, but when it increases so does cloud formation, which then increases the reflection of heat.
MichaelT
QUOTE (qraal @ Jul 5 2006, 10:32 PM) *
If so there should be some literature around on just how that works, at least for Earth, and by extrapolation for Venus too.

Adam Crowl aka qraal

On Earth the absorption of UV radiation by the ozone layer is mainly responsible for the increasing temperature in the stratosphere, and, thus for the existence of the tropopause (dT/dz = 0). So there should be plenty of literature around on how that radiation absorption works on our planet.

Michael
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