Venus Atmosphere Puzzle, one man's struggle with atmospheric physics |
Venus Atmosphere Puzzle, one man's struggle with atmospheric physics |
Jun 5 2006, 12:15 PM
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#1
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Junior Member Group: Members Posts: 57 Joined: 13-February 06 From: Brisbane, Australia Member No.: 679 |
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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Jun 10 2006, 10:39 AM
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#2
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Senior Member Group: Members Posts: 3516 Joined: 4-November 05 From: North Wales Member No.: 542 |
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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Jun 10 2006, 12:34 PM
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#3
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Junior Member Group: Members Posts: 57 Joined: 13-February 06 From: Brisbane, Australia Member No.: 679 |
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 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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