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How would an internet climatologist calculate the pressure as a function of height in an isothermal atmosphere?
The only reasonable way I can see this relatively easy and fundamental climatology-related problem being done is by setting up a differential equation and solving it.
I hope I do not get responses saying "I would consult peer reviewed literature" or "It should be determined by consensus.
Some here you deny the predictive nature of science might think that the only way to do it is to make actual measurements!
"With a barometer"
You would need to take an infinite number of measurements. Furthermore, if the situation referred to a hypotheticakl atmosphere in thge future, for example the atmosphere if the Earth heated up 2 degrees C, you could not determine it by direct measurement. Furthermore, my specification of isothermal makes the problem one relating to an atmosphere held under slightly unnatural conditions--therefore you could not just do a measurement on a sysstem that was not expensively prepared.
Perhaps you were partly joking. But I thgink you were partly serious. You people here really do not understand that science's power is from its predictive nature. It was Kepler and Newton, not Ptolrmy, who made the major advances regarding planetary motion. Oh, and Newton did it using a differential equation, and neither were peer reviewed or argued they were proven correct by consensus or by being voted Best Answer by a group of people with poor science skills.
Pegminer and Portland Joe know what they are doing.
Troubling to give kudos to Portland Joe, being that he is a Denier, but he beat out so far all the Non-Deniers other than Pegminer who also knew how to do the problem.
4 Answers
- pegminerLv 78 years agoFavorite Answer
Well, the isothermal atmosphere is very nice, because its differential equation is easily integrable through separation of variables...it's too bad that the actual atmosphere is not isothermal. The exponential drop-off with height is a useful first approximation, though.
However, to answer your question, if I wanted to know the pressure at a particular height, I could start with the pressure at the surface, then subtract off the amount of pressure that comes from a thin layer near the surface, where I can take the density to be constant. Then when I have the pressure at the top of that layer, I could calculate the density there and do the procedure all over again, stepping my way upward. If I make the layers thin enough I could achieve any degree of accuracy I desire.
- Anonymous5 years ago
Pressure is hard to be control, but if you have the determination and have the impeccable attitudes towards pressure, I'm sure everyone are able to function well under pressure.
- Portland-JoeLv 68 years ago
P = Pressure
P(0) = 1 Kg/cm^2 = pressure at 0 centimeters above the surface
A = altitude
dP/dA = (change in P)/(change in A) = -D = -Density in Kg/cm^3
D(0) = 0.029 Kg/22400 cm^3 at 0C = 1.3 e -6 Kg/ml
D = P * (1.3 e -6 Kg/ml)
dP/dA = -P * (1.3 e -6 Kg/ml)
dP/P = -(1.3 e -6 Kg/ml)dA
ln(P) = -(1.3 e -6 kg/cm^3)A +C
P = exp[-(1.3 e-6 Kg/ml)A](exp(C))
where A is in centimeters and C is the constant of integration
C = 0
so that P(0) = 1
otherwise: C = ln(P(0))
Such equations are nice, but can be over sold. Be careful about believing a computer model's output for poorly understood phenomena using an unverified model. http://www.drroyspencer.com/2013/06/epic-fail-73-c...
