The Hydrostatic Equations

]> The Hydrostatic Equations

PDAS Home > Contents > Difficult Problems > Hydrostatic Equations

Public Domain Aeronautical Software (PDAS)

Geopotential and Geometric Altitude

If the mathematical expressions do not display properly on your browser, you may view the page in PDF. I have a page to help you get started with MathML.

The standard atmosphere is defined in terms of geopotential altitude. The idea behind this concept is that a small change in geopotential altitude will make the same change in gravitational potential energy as the geometric altitude at sea level. Mathematically, this is expressed as g dZ = G dH where H stands for geopotential altitude and Z stands for geometric altitude, g is the acceleration of gravity and G is the value of g at sea level. The value of g varies with altitude and is shown in elementary physics texts to vary as

\frac{g}{G} = {(\frac{E}{Z + E})}^{2}

(1)

where E is the radius of the earth. So,

dH = \frac{g}{G} dZ = {(\frac{E}{Z + E})}^{2} dZ

(2)

and integrating yields

\int_{0}^{H} dH = \int_{0}^{Z} {(\frac{E}{Z + E})}^{2} dZ

(3)

H = \frac{E Z}{E + Z}

(4)

Z = \frac{E H}{E - H}

(5)

While Z and H are virtually identical at low altitudes, you can calculate that Z = 86 km corresponds to H=84.852 km. (Use 6356 km for the radius of the earth). At this altitude, g is 0.9735 times the value at sea level. If you don't like the definition of H as a differential, you can regard H=EZ / (E+Z) as the definition of H and then derive dH/dZ=g/G.

The Perfect Gas Law

The equation of state of a perfect gas is

ρ = \frac{M P}{R T}

(6)

where P is the atmospheric pressure, R is the universal gas constant, rho is the density, T is the absolute temperature and M is the mean molecular weight of air. M is assumed constant (=28.9644) up to 86 km where dissociation and diffusive separation become significant. R is 8.31432 joules K^-1mol^-1.

The Hydrostatic Equations

The fundamental equation is

dP = - ρ gdZ = - ρ G dH

(7)

and using the perfect gas law gives

dP = - \frac{M P}{R T} G dH

(8)

This equation leads directly to the calculation of pressure in the standard atmosphere. Within an atmospheric layer, the temperature T is a linear function of the geopotential altitude H.

T = T_{b} + L (H - H_{b})

(9)

where $L$ is the constant gradient of temperature and $T_{b}$ and $H_{b}$ are the temperature and geopotential altitude at the base of the layer. The hydrostatic equation then becomes

dP = - \frac{G M}{R} \frac{P}{(T_{b} + L (H - H_{b}))} dH

(10)

and the pressure at any value of H within this layer is found by integration of this equation

\int_{P_{b}}^{P} \frac{dP}{P} = - \int_{H_{b}}^{H} \frac{G M}{R (T_{b} + L (H - H_{b}))} dH

(11)

The right hand integral takes different forms, depending upon whether L is zero or not. When L=0, the integral is

\ln (\frac{P}{P_{b}}) = - \frac{G M}{T_{b} R} (H - H_{b})

(12)

and when L is not zero, the integral is

\ln (\frac{P}{P_{b}}) = - \frac{G M}{R L} \ln (\frac{T_{b} + L (H - H_{b})}{T_{b}})

(13)

Writing these equations in exponential form, when L=0, then

\frac{P}{P_{b}} = \exp (- \frac{G M (H - H_{b})}{{RT}_{b}})

(14)

and when L is not zero,

\frac{P}{P_{b}} = {(\frac{T_{b} + L (H - H_{b})}{T_{b}})}^{- \frac{G M}{R L}}

(15)

You can see now why geopotential altitude is used for the definition of the standard atmosphere. If Z were used, then g would appear in the equations instead of G and the variation of g with altitude would have to be included in the integration, making a rather complicated equation.

PDAS Home > Contents > Difficult Problems > Hydrostatic Equations

Public Domain Aeronautical Software (PDAS)