# Why $e$ in the formula for air density?

I am reading a book that says that the density of air is approximately $D = 1.25 e^{(-0.0001h)}$, where h is the height in meters. Why is Euler's number $e$ used here? Was a differential equation used in deriving this formula?

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Also, it can be thought of as the Boltzmann distribution in presence of gravitational field, in which case the exponential is related to the definition of entropy. – auxsvr Jun 25 '14 at 20:40

It's actually a surprisingly straightforward differential equation. If you assume that the acceleration due to gravity $g$ doesn't change with altitude (a good approximation if the atmosphere is thin compared to the radius of the earth), Bernoulli's relation tells you the change in the pressure $P$ with height $h$: $$\frac{dP}{dh} = -\rho g$$ Meanwhile the pressure and the density are also related by the ideal gas law $$PV = NRT$$ or $$P = \rho \frac{RT}{M}$$ where $M$ is the mass of one mole of the gas. If you're willing to neglect the changes in temperature $T$ and mean molar mass $M$, you can differentiate with respect to height and find \begin{align} \frac{dP}{dh} = \frac{d\rho}{dh} \frac{RT}M &= -\rho g \\ \frac{d\rho}{dh} &= -\rho \frac{gM}{RT} = -\frac{\rho}{h_0} \end{align} This is the classic differential equation for an exponential.

If I use nice round numbers $R=8\,\mathrm{\frac{J}{mol\cdot K}}$, $T=300\,\mathrm K$, $M=30\,\mathrm{g/mol}$, $g=10\,\mathrm{m/s^2}$, I get a scale height of 8000 meters, different from your textbook's approximation of $10^4$ meters by about 20%.

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Euler's constant appears naturally in phenomena where the spatial gradient of a quantity (or rate of change with time) is proportional to the quantity itself: $$\frac{\mathrm{d}X}{\mathrm{d}x} = X/x_0$$ ($x_0$ determines the strength of the proportionality, and keeps units straight.)

The solution of this differential equation is $$X=X_o e^{x/x_0}$$ $X_0$ sets the "vertical" scale: it's the value of $X$ at $x=0$. Air density turns out to behave this way. If you want more detail, you should amend the question, or start a new question!

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You can rearrange the terms to have any constant as the base of the exponent:

$D = 1.25 e^{(-0.0001h)}$

$= 1.25 (e^{0.0001})^{-h}$

$= 1.25 (2^{\frac{0.0001}{ln 2}})^{-h}$

$\approx 1.25 (2^{0.00014})^{-h}$

$= 1.25 \times 2^{(-0.00014h)}$

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This is of course true of any power or log term. Changing the base is pretty arbitrary. For example if you converted it to "base Pi" it might look like there was some amazing relationship with Pi in the equation when the reality is that it was completely arbitrary. – Brandon Enright Jun 25 '14 at 20:29