Air pressure decreases with altitude because there is less air above you at higher elevations.
The atmosphere is held close to Earth by gravity, so every layer of air supports the weight of the air above it.
This calculator models that idea using hydrostatic equilibrium and the ideal gas law.
1. Hydrostatic equilibrium
Consider a thin horizontal layer of air. The pressure decreases upward because the layer must support the weight of air above it.
The hydrostatic equation is
\[
\begin{aligned}
\frac{dP}{dh} &= -\rho g.
\end{aligned}
\]
Here \(P\) is pressure, \(h\) is altitude, \(\rho\) is air density, and \(g\) is gravitational field strength.
2. Ideal gas law for air density
For an ideal gas, density can be written as
\[
\begin{aligned}
\rho &= \frac{MP}{RT}.
\end{aligned}
\]
In this formula, \(M\) is the molar mass of air, \(R\) is the gas constant, and \(T\) is absolute temperature.
Substituting this into the hydrostatic equation gives
\[
\begin{aligned}
\frac{dP}{dh}
&= -\frac{MPg}{RT}.
\end{aligned}
\]
To solve this equation, we need a temperature model.
3. Isothermal atmosphere
In an isothermal atmosphere, temperature is constant with altitude:
\[
\begin{aligned}
T(h) &= T.
\end{aligned}
\]
Then the pressure solution is exponential:
\[
\begin{aligned}
P(h)
&= P_0 e^{-\frac{Mgh}{RT}}.
\end{aligned}
\]
The pressure ratio is therefore
\[
\begin{aligned}
\frac{P(h)}{P_0}
&= e^{-\frac{Mgh}{RT}}.
\end{aligned}
\]
This model is mathematically simple, but it is not realistic over large altitude ranges because real atmospheric temperature changes with height.
4. Scale height
The isothermal formula can also be written using a scale height \(H\):
\[
\begin{aligned}
H &= \frac{RT}{Mg}.
\end{aligned}
\]
Then
\[
\begin{aligned}
P(h) &= P_0e^{-h/H}.
\end{aligned}
\]
A larger scale height means pressure decreases more slowly with altitude.
5. Linear lapse-rate atmosphere
In the lower atmosphere, temperature often decreases approximately linearly with altitude.
This is modeled as
\[
\begin{aligned}
T(h) &= T_0 - Lh.
\end{aligned}
\]
Here \(T_0\) is the reference temperature at sea level and \(L\) is the lapse rate.
Substituting this temperature profile into the hydrostatic equation gives
\[
\begin{aligned}
P(h)
&= P_0\left(1-\frac{Lh}{T_0}\right)^{\frac{Mg}{RL}}.
\end{aligned}
\]
This is commonly used as a simple troposphere estimate when \(T(h)\) remains positive.
6. Pressure ratio
The pressure ratio tells what fraction of the reference pressure remains at altitude:
\[
\begin{aligned}
\text{pressure ratio}
&= \frac{P(h)}{P_0}.
\end{aligned}
\]
A ratio of \(0.5\) means the pressure is half of the reference pressure.
A ratio of \(0.25\) means the pressure is one quarter of the reference pressure.
7. Air density at altitude
After pressure and temperature are known, density follows from the ideal gas relation:
\[
\begin{aligned}
\rho(h)
&= \frac{MP(h)}{RT(h)}.
\end{aligned}
\]
Density matters for breathing, aircraft lift, drag, sound propagation, and weather behavior.
8. Pressure gradient
The local pressure gradient is
\[
\begin{aligned}
\frac{dP}{dh}
&= -\rho g.
\end{aligned}
\]
Since density decreases with altitude, the pressure gradient also becomes smaller in magnitude at higher altitude.
9. Half-pressure altitude
In the isothermal model, the altitude where pressure drops to half of \(P_0\) is found from
\[
\begin{aligned}
\frac{1}{2}
&= e^{-h/H}.
\end{aligned}
\]
Solving gives
\[
\begin{aligned}
h_{1/2}
&= H\ln 2.
\end{aligned}
\]
In the lapse-rate model, the same idea can be solved using the power-law pressure expression.
10. Worked example: altitude 5000 m
Use a standard lower-atmosphere estimate:
\[
\begin{aligned}
P_0 &= 101325\ \mathrm{Pa},\\
T_0 &= 288.15\ \mathrm{K},\\
L &= 0.0065\ \mathrm{K/m},\\
M &= 0.0289644\ \mathrm{kg/mol},\\
g &= 9.80665\ \mathrm{m/s^2},\\
R &= 8.314462618\ \mathrm{J/(mol\,K)},\\
h &= 5000\ \mathrm{m}.
\end{aligned}
\]
First compute the temperature at altitude:
\[
\begin{aligned}
T(h)
&= T_0 - Lh\\
&= 288.15 - (0.0065)(5000)\\
&= 255.65\ \mathrm{K}.
\end{aligned}
\]
Then use the lapse-rate pressure formula:
\[
\begin{aligned}
P(h)
&= P_0\left(1-\frac{Lh}{T_0}\right)^{\frac{Mg}{RL}}\\
&= 101325
\left(1-\frac{(0.0065)(5000)}{288.15}\right)^
{\frac{(0.0289644)(9.80665)}{(8.314462618)(0.0065)}}\\
&\approx 54020\ \mathrm{Pa}.
\end{aligned}
\]
So at \(5000\ \mathrm{m}\), the pressure is about \(54.0\ \mathrm{kPa}\), or roughly \(0.533\) atmospheres.
11. Formula summary
| Goal |
Formula |
Use |
| Hydrostatic equilibrium |
\(\frac{dP}{dh}=-\rho g\) |
Relates pressure decrease to air weight |
| Air density |
\(\rho=\frac{MP}{RT}\) |
Connects pressure, temperature, and density |
| Isothermal pressure |
\(P=P_0e^{-Mgh/(RT)}\) |
Constant-temperature model |
| Scale height |
\(H=\frac{RT}{Mg}\) |
Altitude scale for exponential pressure decrease |
| Lapse-rate temperature |
\(T(h)=T_0-Lh\) |
Lower-atmosphere temperature approximation |
| Lapse-rate pressure |
\(P=P_0(1-Lh/T_0)^{Mg/(RL)}\) |
Troposphere-style pressure estimate |
| Pressure ratio |
\(P(h)/P_0\) |
Fraction of reference pressure remaining |
12. Assumptions and limitations
- The air is treated as an ideal gas.
- The molar mass of air is treated as constant.
- Gravity is treated as constant over the altitude range.
- The isothermal model assumes constant temperature.
- The lapse-rate model assumes a linear temperature decrease.
- The simple lapse-rate model is mainly intended for lower-atmosphere estimates.
- For high altitudes, a multi-layer standard atmosphere is more realistic.
- Local weather, humidity, and temperature inversions can make real pressure different from the model.
Key idea: air pressure decreases with altitude because the weight of the air column above you becomes smaller.
