This is something I would pass on to a student that partied too much before an exam so he(she) could earn a portion of the grade point average they just lower back and somewhat get back into my good graces. I used to get assigned all kinds of interesting problems that had nothing to do with any course I was taking. Actually, I spent most of my educational life attempting to make up for having more fun than allowed at the wrong time and often the wrong place.

So, even though I don't care to, I am going to pretend I care and attempt to set up the problem. Using basic Wikipedia stuff for Earth.

Surface area of the Earth: 510,072,000 km2 or 5.1e14 m2

Velocity, equatorial rotational: 465 m/s

Mass of Atmosphere: 5.15e18 kg

Composition: 80% N2 20% O2 obviously not exact.

Gravity: 10m/s2

Surface pressure 1bar

Thermal properties: N2 http://www.engineeringtoolbox.com/air-properties-d_156.html Specific heat: 1040 J/kgK Thermal Conductivity: 0.024W/mC Specific Volume: 0.872 kg/m3 Gas Constant: 297 J/kgK

Note: The thermal conductivity of N2 is not zero. Heat will be transferred from the surface to the nitrogen in the atmosphere. Nitrogen has a specific heat 1040J/kgK. Nitrogen is not a very effective absorber or emitter of infrared radiation. Basic take away, it can gain heat, but not as easily lose it to space. The specific volume is 0.872 kg/m3, Gas Constant 297J/kgK Temperature: 254.5K Pressure: 1 Bar But first!

5.15e18kg divided by 5.1e14m2 yields 1000kg/m2. The density of air at 1 bar and -50C (223K) is 1.5kg/m3, so 667 m3 for 1000kg/m2 or 667 meter high atmosphere with no significant heat transfer. Yes, that is air not just N2.

With an average surface temperature of -18C, P1V1/T1=P2V2/T2 yields V2=P1V1T2/P2T1, p1=P2, V2=V1*(254.5/223)=1.14V1 1.14*667=760 meters

So if the radiant loss is not greater than the conductive gain, there would be an atmosphere 760 meters high.

Oxygen absorbs strongly in the UV spectrum. There would be virtually no change in the O2/UV relationship in a no GHG Earth. That energy would be lower due to the lower altitude. That energy would cause upper level convection tending to decrease density at the top of the No GHG atmosphere. That is the creation of the tropopause.

So the Earth would have an atmosphere at least 760 meters high, with a tropopause. Its temperature would be roughly 254.5K at the surface with an unknown lapse rate at this point. This is all it takes, but I may dig out the numbers on the weak emissision of long wave and the O2 absorption of UV to better estimate the altitude.

You say: 5.15e18kg divided by 5.1e14m2 yields 1000kg/m2

ReplyDeleteYou left out a zero. So your atmosphere calculation should be 10 times higher.

Except that the pressure drops as you gain height, and so the atmosphere is less dense. So the height goes as high as you like.

The limit shows up when it gets so thin you need to stop thinking of it as a continuous fluid and pay attention to individual molecules.

You are right. Should be ~7600 meters. I never did get back to this post to double check, thanks.

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