In fact, the atmosphere consists of compressible gases, so that the mass density \(\rho(z)\) of the air decreases continuously with the altitude \(z\). However, the weight of the gas layers must be supported by the pressure gradient \(\frac{dP}{dz}\) in the atmosphere. The balance of compressive forces and gravitation is expressed by the **hydrostatic equation**:
\[ \frac{dP(z)}{dz} = -\rho(z) g \]
with the gravitational acceleration \(g \approx 9.81 \,\text{m}/\text{s}^2\).

This equation can be derived as follows: We consider a cuboid volume element of the atmosphere with the base area \(A \) parallel to the Earth's surface and the height \(\Delta z \) (see the diagramm on the right). The mass of this volume is \(\Delta m \approx \rho(z) A \Delta z\), so that a weight \(\Delta F_g = -\Delta m g \) acts on the entire volume. In addition, the pressures \(P(z + \frac{1}{2} \Delta z)\) and \(P(z - \frac{1}{2}\Delta z)\) act respectively on the top and bottom of the cuboid. In the static case, the sum of all forces on the volume element must be zero. In the limit \(\Delta z \to 0\), the pressure difference changes into the derivative \(\frac{dP}{dz}\), so that the difference equation becomes a differential equation. \[ \begin{align*} \sum F = 0 &= - \Delta m g - P(z + \tfrac{1}{2} \Delta z)A + P(z + \tfrac{1}{2} \Delta z)A \\ &= -\left(\rho(z) g - \frac{P(z + \tfrac{1}{2} \Delta z)- P(z + \tfrac{1}{2} \Delta z)}{\Delta z} \right) A \Delta z \\ &\stackrel{\Delta z \to 0}{\to} -\left(\rho(z) g - \frac{d P(z)}{dz} \right) dV \end{align*} \] Since the expression in parentheses must be zero it follows \[ \boxed{\dfrac{d P(z)}{dz} = - \rho(z) g } \]

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