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Re: [Phys-l] Atmospheric pressure calculus deduction

Sorry about that - I misinterpreted what you meant when you said "if you already know the weight of the air column, you don't assume hydrostatic balance."

I thought you meant that the pressure equals the weight per area whether there is hydrostatic balance or not. I figured hydostatic balance was the same as "mechanical equilibrium", which you indicated was a necessary assumption.

[I think we agree that the integration is only necessary to show that the P=wt/A can be derived from hydrostatic balance or mechanical equilibrium or whatever we want to call it]


From: on behalf of John Denker
Sent: Mon 6/23/2008 8:35 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Atmospheric pressure calculus deduction

On 06/23/2008 03:42 PM, Robert Cohen wrote:
Can you expand on why one needn't assume hydrostatic balance? I can
easily imagine an atmosphere that is not in hydrostatic balance
(e.g., constant pressure with height) in which the surface pressure
would not equal the weight per area.

Well, I can imagine such things too, and that could be a problem,
but in such cases integrating g rho dz doesn't solve the problem.
In particular, suppose their is a huge wind blowing over an
airfoil-shaped hill. In that case it's not even entirely obvious
what you mean by pressure --(static pressure? dynamic pressure?)--
and even once you decide that, knowing the weight of the air
column and/or g rho dz isn't going to tell you the pressure.

In any case, the point remains that the integral of g rho dz *is*
the weight of the air column, so if you knew the weight at the
beginning you wouldn't need to do the integral. IMHO that's
a pretty robust result. (This assumes g is constant, which is
good to very high accuracy in the troposphere and stratosphere.)

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