Venus Atmosphere Puzzle, one man's struggle with atmospheric physics Options

[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?

[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.