Variation of air density with height

[brotherbobby]

TL;DR Summary

Is there a way to determine how air density varies with height? We all know that it falls, but what is the relation. I am looking for a function like $\rho_{\text{air}}(x)$ where $x$ is the height from the earth's surface.

Using the ideal gas equation $PV = nRT\Rightarrow PV = \frac{m}{M} RT$ where $m,M$ are the mass and molecular weights of the gas respectively.

This yields $\frac{m}{V} = \frac{PM}{RT} = \rho$, the density of the gas at a point with pressure $P$.

If only we can obtain the variation of pressure with height from the Earth's surface : $P(x) = ?$, we could use it to find $\rho(x)$.

Does anyone know?

 

[A.T.]

Is there a way to determine how air density varies with height?

https://en.wikipedia.org/wiki/Density_of_air#Variation_with_altitude

 

[brotherbobby]

https://en.wikipedia.org/wiki/Density_of_air#Variation_with_altitude

Thank you and sorry to bother. Actually I found the wikipedia page a bit confusing. It's better explained on here : Barometric formula .

 

[vanhees71]

Caveat: That's for an isothermal atmosphere!

 

[brotherbobby]

Caveat: That's for an isothermal atmosphere!

Yes, I noticed, thank you. This means that the variation of density of air with height, $\rho (x)$ is dependent both on pressure and temperature of air with height. Is there a resource besides wikipedia on how to derive the temperature dependence? Wikipedia only gives the results.

 

[vanhees71]

For the derivation of the isothermal barometric formula assuming an ideal gas you have two posibilities.

(1) Thermodynamics

The phase-space distribution function is the Maxwell-Boltzmann function
$$f(\vec{x},\vec{p})=\exp\{[-\beta (H(\vec{x},\vec{p})-\mu \}.$$
The Hamiltonian reads
$$H=\frac{\vec{p}^2}{2m} + m g z.$$
Integrating over $\vec{p}$ gives the density. We need it not with the absolute norm but just relative to $z=0$ and thus get
$$\rho(\vec{x})=\rho_0 \exp(-\beta m g z)$$
with $\beta=1/(k_{\text{T}} T)$ and $m$ is the mass of the gas molecules.

(2) Hydrostatic equation

$$\vec{\nabla} P=\rho \vec{g}.$$
For an isothermal change of state you have
$$P=\rho k_{\text{B}} T/m$$
and thus the equation gets
$$\vec{\nabla} \rho = -\frac{\rho}{k_{\text{B}} T} m g \vec{e}_z$$
and thus through integration
$$\rho=\rho_{0} \exp \left (-\frac{m g z}{k_{\text{B}} t} \right).$$
Of course the isothermal atmosphere is only a good approximation for not too large changes of height.

For a realistic barometric formula you need an empirical equation of state. E.g., the pressure as function of height. For details see

https://en.wikipedia.org/wiki/Barometric_formula

 

[Frabjous]

There are also tables
NACA Report 1235 Standard Atmosphere
U.S. Standard Atmosphere, 1962
U.S. Standard Atmosphere Supplements, 1966
https://apps.dtic.mil/dtic/tr/fulltext/u2/a008098.pdf

There is probably more recent stuff.

 

[rcgldr]

From what I recall, the atmosphere is split into 3 or 4 regions, each region with it's own curve, for ballistic missile and space craft.

 

[DrStupid]

This means that the variation of density of air with height, $\rho (x)$ is dependent both on pressure and temperature of air with height. Is there a resource besides wikipedia on how to derive the temperature dependence?

The derivation is similar to the derivation for the isothermal case as demonstrated by @vanhees71 in #6. It also starts with the hydrostatic equation

$p' = - \rho \cdot g$

but assuming isententropic processes instead of isothermal changes:

$p \cdot V^\kappa = const.$

Together with the density

$\rho = \frac{m}{V} = \frac{{M \cdot n}}{V}$

and the ideal gas equation

$p \cdot V = n \cdot R \cdot T$

this results in the differential equation

$p' = - \frac{{M \cdot g \cdot p_0^{\frac{{\kappa - 1}}{\kappa }} }}{{R \cdot T_0 }}p^{\frac{1}{\kappa }}$

with the solution

$p = p_0 \cdot \left[ {1 - \left( {1 - \frac{1}{\kappa }} \right) \cdot \frac{{M \cdot g}}{{R \cdot T_0 }} \cdot z} \right]^{\frac{\kappa }{{\kappa - 1}}}$

and

$T = T_0 - \left( {1 - \frac{1}{\kappa }} \right) \cdot \frac{{M \cdot g}}{R} \cdot z$

But that only works within the troposphere because it is convection dominated. It fails in the stratosphere (which is radiation dominated).