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


Junior Member Group: Members Posts: 57 Joined: 13February 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 leftside 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*(1Z/(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? 


Jun 7 2006, 12:46 PM
Post
#2


Senior Member Group: Members Posts: 3423 Joined: 4November 05 From: North Wales Member No.: 542 
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 ?



Jun 9 2006, 12:43 PM
Post
#3


Junior Member Group: Members Posts: 57 Joined: 13February 06 From: Brisbane, Australia Member No.: 679 
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. 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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