- #1

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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.