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Simple question - barometric formula derivation

  • Thread starter diffusion
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  • #1
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I haven't had extensive experience in solving DE's just yet; I'm curious as to where the P(0) term comes from between the 4th and 5th expressions in the following derivation:

http://en.wikipedia.org/wiki/Barometric_formula#Derivation

My attempt follows:

[itex]\frac{dP}{P} = - \frac{M g\,dz}{kT}[/itex]

[itex]\int\frac{1}{P}dP = - \int\frac{M g}{kT}dz[/itex]

[itex]P = e^{-\int\frac{M g}{kT}dz}[/itex]

[itex]P = e^{-Mgz/kT}[/itex]

Clearly I'm missing something obvious, but I don't know what exactly.
 

Answers and Replies

  • #2
430
2
Limits of the integration.
 

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