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Simple question - barometric formula derivation

  1. Mar 8, 2010 #1
    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:


    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.
  2. jcsd
  3. Mar 9, 2010 #2
    Limits of the integration.
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