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 23 2006, 01:11 PM
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#2
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Junior Member Group: Members Posts: 57 Joined: 13-February 06 From: Brisbane, Australia Member No.: 679 |
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 |
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Jun 23 2006, 01:51 PM
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#3
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Senior Member Group: Members Posts: 3516 Joined: 4-November 05 From: North Wales Member No.: 542 |
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 . . . |
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