Scale height

The earth atmosphere's scale height is about 8.5km,as can be confirmed from this diagram of air pressurepby altitudeh:At an altitude of 0, 8.5, and 17 km, the pressure is about 1000, 370, and 140hPa,respectively.

Inatmospheric,earth,andplanetarysciences, ascale height,usually denoted by the capital letterH,is a distance (verticalorradial) over which aphysical quantitydecreases by a factor ofe(the base ofnatural logarithms,approximately 2.718).

Scale height used in a simple atmospheric pressure model

For planetary atmospheres, scale height is the increase in altitude for which theatmospheric pressuredecreases by a factor ofe.The scale height remains constant for a particular temperature. It can be calculated by

$$H={\frac {k_{\text{B}}T}{mg}}$$ or equivalently $$H={\frac {RT}{Mg}}$$ where:

• k_B=Boltzmann constant=1.381×10⁻²³J⋅K⁻¹‍
• R=gas constant
• T= mean atmospherictemperatureinkelvins= 250 Kfor Earth
• m= mean mass of a molecule (units kg)
• M= mean mass of one mol of atmospheric particles = 0.029 kg/mol for Earth
• g=accelerationdue togravityat the current location (m/s²)

The pressure (force per unit area) at a given altitude is a result of the weight of the overlying atmosphere. If at a height ofzthe atmosphere hasdensityρand pressureP,then moving upwards an infinitesimally small heightdzwill decrease the pressure by amountdP,equal to the weight of a layer of atmosphere of thicknessdz.

Thus: $${\frac {dP}{dz}}=-g\rho$$ wheregis the acceleration due to gravity. For smalldzit is possible to assumegto be constant; the minus sign indicates that as the height increases the pressure decreases. Therefore, using theequation of statefor anideal gasof mean molecular massMat temperatureT,the density can be expressed as $$\rho ={\frac {MP}{RT}}$$

Combining these equations gives $${\frac {dP}{P}}={\frac {-dz}{\frac {k_{\text{B}}T}{mg}}}$$ which can then be incorporated with the equation forHgiven above to give: $${\frac {dP}{P}}=-{\frac {dz}{H}}$$ which will not change unless the temperature does. Integrating the above and assumingP₀is the pressure at heightz= 0 (pressure atsea level) the pressure at heightzcan be written as: $$P=P_{0}\exp \left(-{\frac {z}{H}}\right)$$

This translates as the pressuredecreasing exponentiallywith height.

InEarth's atmosphere,the pressure at sea levelP₀averages about1.01×10⁵Pa,the mean molecular mass of dry air is 28.964uand hence m = 28.964 ×1.660×10⁻²⁷=4.808×10⁻²⁶kg.As a function of temperature, the scale height of Earth's atmosphere is thereforeH/T=k/mg= (1.38/(4.808×9.81))×10³=29.26 m/K.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 atNRLMSISE-00,which shows the air density dropping from 1200 g/m³at sea level to 0.5³= 0.125 g/m³at 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 theideal gaslaws. Therefore, density will also decrease exponentially with height from a sea level value ofρ₀roughly equal to 1.2 kg m⁻³
• At heights over 100 km, an atmosphere may no longer be well mixed. Then 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.

Planetary examples

Approximate atmospheric scale heights for selected Solar System bodies:

• Venus:15.9 km
• Earth:8.5 km
• Mars:11.1 km
• Jupiter:27 km
• Saturn:59.5 km
  • Titan:21 km
• Uranus:27.7 km
• Neptune:19.1–20.3 km
• Pluto:~50 km

Scale height for a thin disk

A schematic depiction of the force balance in a gas disk around a central object, e.g., a star.

For a disk of gas around a condensed central object, such as, for example, a protostar, one can derive a disk scale height which is somewhat analogous to the planetary scale height. We start with a disc of gas that has a mass which is small relative to the central object. We assume that the disc is in hydrostatic equilibrium with thezcomponent of gravity from the star, where the gravity component is pointing to the midplane of the disk:

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

where:

• G=Gravitational constant≈6.674×10⁻¹¹m³⋅kg⁻¹⋅s⁻²‍
• r= the radialcylindrical coordinatefor the distance from the center of the star or centrally condensed object
• z= the height/altitudecylindrical coordinatefor 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
• $\rho$= the gas mass density in the disk

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

To determine the gas pressure, one can use theideal gas law: $$P={\frac {\rho k_{\text{B}}T}{\bar {m}}}$$ with:

• T= the gas temperature in the disk, where the temperature is a function ofr,but independent ofz
• ${\bar {m}}$= the mean molecular mass of the gas

Using theideal gas lawand the hydrostatic equilibrium equation, gives: $${\frac {d\rho }{dz}}\approx -{\frac {GM_{*}{\bar {m}}\rho z}{kTr^{3}}}$$ which has the solution $$\rho =\rho _{0}\exp \left(-\left({\frac {z}{h_{D}}}\right)^{2}\right)$$ where$\rho _{0}$is the gas mass density at the midplane of the disk at a distancerfrom the center of the star and$h_{D}$is the disk scale height with

$$h_{D}={\sqrt {\frac {2kTr^{3}}{GM_{*}{\bar {m}}}}}\approx 0.0306{\sqrt {\frac {\left(T/100\ {\text{K}}\right)\left(r/1{\text{ au}}\right)^{3}}{\left(M_{*}/M_{\odot }\right)\left({\bar {m}}/2{\text{ amu}}\right)}}}\ {\text{ au}}$$ with$M_{\odot }$thesolar mass,${\text{au}}$theastronomical unitand${\text{amu}}$theatomic mass unit.

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

Due to the assumption that the gas temperature in the disk,T,is independent ofz,$h_{D}$is sometimes known as the isothermal disk scale height.

Disk scale height in a magnetic field

Amagnetic fieldin a thin gas disk around a central object can change the scale height of the disk.For example, if a non-perfectly conducting disk is rotating through a poloidal magnetic field (i.e., the initial magnetic field is perpendicular to the plane of the disk), then a toroidal (i.e., parallel to the disk plane) magnetic field will be produced within the disk, which willpinchand compress the disk. In this case, the gas density of the disk is:

$$\rho (r,z)=\rho _{0}(r)\exp \left(-\left({\frac {z}{h_{D}}}\right)^{2}\right)-\rho _{\text{cut}}(r)\left[1-\exp \left(-\left({\frac {z}{h_{D}}}\right)^{2}\right)\right]$$ where thecut-offdensity$\rho _{\text{cut}}$has the form $$\rho _{\rm {cut}}(r)=(\mu _{0}\sigma _{D}r)^{2}\left({\frac {B_{z}^{2}}{\mu _{0}}}\right)\left({\frac {\Omega _{*}}{\Omega _{K}}}-1\right)^{2}$$ where

• $\mu _{0}$is thepermeability of free space
• $\sigma _{D}$is theelectrical conductivityof the disk
• $B_{z}$is the magnetic flux density of the poloidal field in the$z$direction
• $\Omega _{*}$is the rotationalangular velocityof the central object (if the poloidal magnetic field is independent of the central object then$\Omega _{*}$can be set to zero)
• $\Omega _{K}$is thekeplerianangular velocityof the disk at a distance$r$from the central object.

These formulae give the maximum height,$H_{B}$,of the magnetized disk as

$$H_{B}=h_{D}{\sqrt {\ln \left(1+\rho _{0}/\rho _{\rm {cut}}\right)}},$$ while the e-folding magnetic scale height,$h_{B}$,is $$h_{B}=h_{D}{\sqrt {\ln \left(1+{\frac {1-1/e}{1/e+\rho _{\text{cut}}/\rho _{0}}}\right)}}\.$$

See also

• Time constant