# Deriving Equations for Atmospheric Pressure and Density

The levels of air pressure and air density in the atmosphere depend on the altitude, the amount and composition of the atmosphere and the temperature profile. Atmospheric pressure, density and temperature are related via the ideal gas law.

We can't measure the amount of atmosphere directly, but we can measure a rough temperature profile with balloons. The overall dependence of atmospheric pressure with altitude can be derived by observing the pressures acting on a small volume. Using calculus we can then derive the pressure curve for the whole atmosphere. Using the temperature profile and the ideal gas law we can also derive the density gradient from the pressure gradient.

1. determining the pressure on a small volume element
2. creating an emirical simplified model for the temperature profile of the atmosphere
3. deriving the atmospheric pressure profile as a function of altitude and temperature profile using calculus
4. deriving the density profile from the pressure and temperature profile using the ideal gas law
(1)
where'
 $P$ ' =' 'air pressure $\rho$ ' =' 'air density $T$ ' =' 'absolute temperature $R_\mathrm{S}$ ' =' 'Spezific Gas Constant; dry air = 287.058 J/(kg·K)
(2)
where'
 $\mathrm{d} P$ ' =' 'air pressure differenctial $g$ ' =' 'gravitational acceleration $\rho$ ' =' 'air density $\mathrm{d} h$ ' =' 'height differential

Using the ideal gas law (1) we can replace the density in (2) by the pressure to get the following differential equation:

 (3)

## Solving Nummerically

The differential equation (3) can be solved for the pressure gradient nummerically. In this case g and T can be arbitrary functions or measurements. Nummerically we solve the following equation iteratively:

 (4) with and

We begin with some start conditions, which could be measurements, for P0(h0), g0(h0), T0(h0). After each iteration we get the new pressure for the next height increment and we may have different g and T values. There are more complicated ways for nummerical integrations, that yield more accuracy.

## Analytical Solutions

If we make the simplifying assumptions that g is constant for the height range we want to solve for and T is constant or a linear function of h, we can solve the differential equation (5) analytically.

 (5)

We can bring the pressure P from the right side to the left side, so we have all pressure terms on the left, which simplifies solving the differential equation:

(6)

Note: Pressure and temperature are commonly dependent on altitude. The temperature has to be measured empirically. It can also be constant over some altitude range.

In the International Standard Atmosphere model the empirically measured temperature is divided into some ranges and approximated either as constant (isotherm) or linearly dependent on altitude.

## Pressure Isotherm

Lets first derive the pressure equation for the simpler case of constant temperature.

We can solve (6) by integrating both sides. The integration 1/P dP = ln(P) is applied to the left side, the integration of the right side is straight forward A · ∫ dh = A·h:

 (7) (8)

Applying the limits:

 (9)

Pref is constant. Lets bring it to the right hand side:

 (10)

To get rid of the logarithms we raise both sides to the power of e, because eln(X) = X:

 (11)

Applying eln(C)−X = eln(C) · eX we get:

 (12)

So we get finally the formula for pressure:

(13)
where'
 $P(h)$ ' =' 'pressure at altitude h above sea level $P_\mathrm{ref}$ ' =' 'pressure at the reference altitude href above sea level $g$ ' =' '9.80665 m/s2 = gravitational acceleration $R_\mathrm{S}$ ' =' '287.058 J/(kg·K) = Spezific Gas Constant for dry air $T$ ' =' 'constant temperature in the range from href to h $h$ ' =' 'altitude above sea level $h_\mathrm{ref}$ ' =' 'reference altitude above sea level

This formula makes the simplification that the gravitational acceleration g is constant in the altitude range href ... h.

## Pressure with Linear Temperature Gradient

We start again with the differential equation derived at (6), repeated here:

 (14)

But this time temperature T is not constant, but a linear function of altitude h, so that T(href) = Tref:

 (15)

Inserting (15) into (14) we get the new differential equation:

 (16)

We can solve (16) by integrating both sides. The general integral ∫ (1/X) dX = ln(X) is applied to both sides:

 (17) (18)

Putting in the limits yields:

 (19)

Applying ln(A) − ln(B) = ln(A/B) on the right hand side and bringing the constant ln(Pref) to the right hand side yields:

 (20) with

To get rid of the logarithms we raise both sides to the power of e:

 (21)

Applying eln(P)−X = eln(P) · eX = P · eX we get:

 (22)

Applying eβ·ln(X) = eln(X)·(−β) = (eln(X))β = Xβ we finally get our formula for pressure in the case of a linear temperature gradient:

(23)
with
where'
 $P$ ' =' 'pressure at altitude h above sea level $P_\mathrm{ref}$ ' =' 'pressure at reference altitude href above sea level $T_\mathrm{ref}$ ' =' 'temperature at reference altitude href $a$ ' =' 'temperature gradient (negative lapse rate) $h$ ' =' 'altitude above sea level $h_\mathrm{ref}$ ' =' 'reference altitude above sea level $g$ ' =' '9.80665 m/s2 = mean gravitational acceleration at sea level $R_\mathrm{S}$ ' =' '287.058 J/(kg·K) = Spezific Gas Constant for dry air