Scale height

 ( Atmospheric Dynamics ) 

In atmospheric, earth science, and planetary sciences, a scale height, usually denoted by the capital letter H, is a distance (vertical or radial) over which a physical quantity decreases by a factor of e (the base of natural logarithms, approximately 2.718).

H = \frac{k_\text{B}T}{mg},
     

H = \frac{RT}{Mg},
     

k_B = Boltzmann constant =

R = molar gas constant = 8.31446 J⋅K⁻¹⋅mol⁻¹

T = mean atmospheric temperature in = 250 K for Earth

m = mean mass of a molecule

M = mean molar mass of atmospheric particles = 0.029 kg/mol for Earth

g = acceleration due to gravity at the current location

The pressure (force per unit area) at a given altitude is a result of the weight of the overlying atmosphere. If at a height of z the atmosphere has density ρ and pressure P, then moving upwards an infinitesimally small height dz will decrease the pressure by amount dP, equal to the weight of a layer of atmosphere of thickness  dz.

Thus: $$\backslash frac\{dP\}\{dz\}\; =\; -g\backslash rho,$$ where g is the acceleration due to gravity. For small dz it is possible to assume g to be constant; the minus sign indicates that as the height increases the pressure decreases. Therefore, using the equation of state for an ideal gas of mean molecular mass M at temperature T, the density can be expressed as $$\backslash rho\; =\; \backslash frac\{MP\}\{RT\}.$$

Combining these equations gives $$\backslash frac\{dP\}\{P\}\; =\; \backslash frac\{-dz\},$$ which can then be incorporated with the equation for H given above to give $$\backslash frac\{dP\}\{P\}\; =\; -\; \backslash frac\{dz\}\{H\},$$ which will not change unless the temperature does. Integrating the above and assuming P₀ is the pressure at height z = 0 (pressure at sea level), the pressure at height z can be written as $$P\; =\; P\_0\backslash exp\backslash left(-\backslash frac\{z\}\{H\}\backslash right).$$ This translates as the pressure decreasing exponentially with height.

In Earth's atmosphere, the pressure at sea level P₀ averages about , the mean molecular mass of dry air is , and hence m = × = . As a function of temperature, the scale height of Earth's atmosphere is therefore H/ T = k_B/ mg = / ( × ) = . This yields the following scale heights for representative air temperatures:

T = 290 K, H = 8500 m,

T = 273 K, H = 8000 m,

T = 260 K, H = 7610 m,

T = 210 K, H = 6000 m.

These figures should be compared with the temperature and density of Earth's atmosphere plotted at NRLMSISE-00, which shows the air density dropping from 1200 g/m³ at sea level to 0.125 g/m³ at 70 km, a factor of 9600, indicating an average scale height of 70 / ln(9600) = 7.64 km, consistent with the indicated average air temperature over that range of close to 260 K.

Note:

• Density is related to pressure by the ideal gas laws. Therefore, density will also decrease exponentially with height from a sea-level value of ρ₀ roughly equal to .
• At an altitude over 100 km, the atmosphere is no longer well-mixed, and each chemical species has its own scale height.
• Here temperature and gravitational acceleration were assumed to be constant, but both may vary over large distances.

\frac{dP}{dz} = -\frac{GM_*\rho z}{(r^2 + z^2)^{3/2}},
     

G = Newtonian constant of gravitation ≈

r = the radial cylindrical coordinate for the distance from the center of the star or centrally condensed object

z = the height/altitude cylindrical coordinate for the distance from the disk midplane (or center of the star)

M_* = the mass of the star/centrally condensed object

P = the pressure of the gas in the disk

$\backslash rho$ = the gas mass density in the disk

In the thin disk approximation, $z\; \backslash ll\; r$, and the hydrostatic equilibrium, the equation is $$\backslash frac\{dP\}\{dz\}\; \backslash approx\; -\backslash frac\{GM\_*\backslash rho\; z\}\{r^3\}.$$

To determine the gas pressure, one can use the ideal gas law: $$P\; =\; \backslash frac\{\backslash rho\; k\_\backslash text\{B\}\; T\}\{\backslash bar\{m\}\}$$ with

T = the gas temperature in the disk, where the temperature is a function of r, but independent of z

$\backslash bar\{m\}$ = the mean molecular mass of the gas

Using the ideal gas law and the hydrostatic equilibrium equation, gives $$\backslash frac\{d\backslash rho\}\{dz\}\; \backslash approx\; -\backslash frac\{GM\_*\; \backslash bar\{m\}\backslash rho\; z\}\{k\_\backslash text\{B\}\; Tr^3\},$$ which has the solution $$\backslash rho\; =\; \backslash rho\_0\; \backslash exp\backslash left(-\backslash left(\backslash frac\{z\}\{h\_\backslash text\{D\}\}\backslash right)^2\; \backslash right),$$ where $\backslash rho\_0$ is the gas mass density at the midplane of the disk at a distance r from the center of the star, and $h\_\backslash text\{D\}$ is the disk scale height with $$$$

h_\text{D} = \sqrt{\frac{2k_\text{B}Tr^3}{GM_* \bar{m}}} \approx
0.0306 \sqrt{\frac{(T/100~\text{K})(r/1~\text{au})^3}{(M_* / M_\odot)(\bar{m}/2~\text{Da})}}~\text{au},
     

As an illustrative approximation, if we ignore the radial variation in the temperature $T$, we see that $h\_\backslash text\{D\}\; \backslash propto\; r^\{3/2\}$ and that the disk increases in altitude as one moves radially away from the central object.

Due to the assumption that the gas temperature T in the disk is independent of z, $h\_\backslash text\{D\}$ is sometimes known

\rho(r, z) = \rho_0(r) \exp\left(-\left(\frac{z}{h_\text{D}}\right)^2\right) -
 \rho_\text{cut}(r) \left[1 - \exp\left(-\left(\frac{z}{h_\text{D}}\right)^2\right)\right],
     

\rho_\text{cut}(r) = (\mu_0 \sigma_\text{D} r)^2 \frac{B_z^2}{\mu_0}
 \left(\frac{\Omega_*}{\Omega_\text{K}} - 1\right)^2,
     

H_\text{B} = h_\text{D} \sqrt{\ln\left(1 + \rho_0/\rho_\text{cut}\right)},
     

h_\text{B} = h_\text{D} \sqrt{\ln\left(1 + \frac{1 - 1/e}{1/e + \rho_\text{cut}/\rho_0} \right)}.