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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

$$\frac{g}{G}={\left(\frac{E}{Z+E}\right)}^{2}$$ | (1) |

where E is the radius of the earth. So,

$$\mathrm{dH}=\frac{g}{G}\mathrm{dZ}={\left(\frac{E}{Z+E}\right)}^{2}\mathrm{dZ}$$ | (2) |

and integrating yields

$${\int}_{0}^{H}\mathrm{dH}={\int}_{0}^{Z}{\left(\frac{E}{Z+E}\right)}^{2}\mathrm{dZ}$$ | (3) |

$$H=\frac{EZ}{E+Z}$$ | (4) |

$$Z=\frac{EH}{E-H}$$ | (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 equation of state of a perfect gas is

$$\rho =\frac{MP}{RT}$$ | (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^{-1}mol^{-1}.

The fundamental equation is

$$\mathrm{dP}=-\rho \mathrm{gdZ}=-\rho G\mathrm{dH}$$ | (7) |

and using the perfect gas law gives

$$\mathrm{dP}=-\frac{MP}{RT}G\mathrm{dH}$$ | (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

$$\mathrm{dP}=-\frac{GM}{R}\frac{P}{({T}_{b}+L(H-{H}_{b}))}\mathrm{dH}$$ | (10) |

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

$${\int}_{{P}_{b}}^{P}\frac{\mathrm{dP}}{P}=-{\int}_{{H}_{b}}^{H}\frac{GM}{R({T}_{b}+L(H-{H}_{b}))}\mathrm{dH}$$ | (11) |

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

$$\mathrm{ln}\left(\frac{P}{{P}_{b}}\right)=-\frac{GM}{{T}_{b}R}(H-{H}_{b})$$ | (12) |

and when L is not zero, the integral is

$$\mathrm{ln}\left(\frac{P}{{P}_{b}}\right)=-\frac{GM}{RL}\mathrm{ln}\left(\frac{{T}_{b}+L(H-{H}_{b})}{{T}_{b}}\right)$$ | (13) |

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

$$\frac{P}{{P}_{b}}=\mathrm{exp}(-\frac{GM(H-{H}_{b})}{{\mathrm{RT}}_{b}})$$ | (14) |

and when L is not zero,

$$\frac{P}{{P}_{b}}={\left(\frac{{T}_{b}+L(H-{H}_{b})}{{T}_{b}}\right)}^{-\frac{GM}{RL}}$$ | (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.