pressure in earth's atmosphere as a function of height above sea level

Question

While I was studying the measurements of pressure at earth's atmosphere,I found the barometric formula which is more complex equation ($P'=Pe^{-mgh/kT}$) than what I used so far ($p=h\rho g$).

So I want to know how this complex formula build up? I could reach at the point of $${dP \over dh}=-{mgP \over kT}$$ From this how can I obtain that equation. Please give me a Mathematical explanation.

Answer 1

This type of problem is known as a differential equation. In this particular case the solution can be guessed, since you have that the derivative of the function $P$ is just a constant times $P$. The only function satisfying this condition is the exponent, since: $${d(A e^{a x}) \over dx} = a A e^x$$ Thus, if $${dP\over dh} = \left( \frac {-mg}{kT}\right) P$$ Then $P$ must be: $$P(h) = A\exp\left(\frac {-mg}{kT} \cdot h\right)$$ Where $A$ is a constant. In particular, if the pressure $P_0$ at $h=0$ is known, then by substitution, We have: $$P(h) = P_0\exp\left(\frac {-mg}{kT} \cdot h\right)$$

Answer 2

If

$\frac{dP}{dh} = (-\frac{mgP}{kT})$,

then

$\frac{1}{P} \frac{dP}{dh} = (-\frac{mg}{kT})$,

or

$\frac{d(ln P)}{dh} = (-\frac{mg}{kT})$.

Integrating with respect to $h$ over the interval $[h_0, h]$ yields

$ln(P(h)) - ln(P(h_0)) = (-\frac{mg}{kT})(h - h_0)$,

or

$ln(\frac{P(h)}{P(h_0)}) = (-\frac{mg}{kT})(h - h_0)$,

or

$\frac{P(h)}{P(h_0)} = exp(-\frac{mg}{kT}(h - h_0))$,

or

$P(h) = P(h_0)exp(-\frac{mg}{kT}(h - h_0))$.

If we now take $h_0 = 0$, and set $P = P(0)$, $P' = P(h)$, the desired formula is had.