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?

[MichaelT]

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/s², and the surface pressure p = 9.3 MPa.

Therefore:
M / A = 9,300,000 Pa / 8.87 m/s²
M / A = 1,048,478 kg/m²

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/s²; M/A = 10,326 kg/m²). For Mars it is just 217 kg/m².

Michael

[ngunn]

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/s², and the surface pressure p = 9.3 MPa.

Therefore:
M / A = 9,300,000 Pa / 8.87 m/s²
M / A = 1,048,478 kg/m²

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]

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
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