Atmospheric physics 3: Hydrostatic equation

In fact, the atmosphere consists of compressible gases, so that the mass density ρ(z)\rho(z) of the air decreases continuously with the altitude zz. However, the weight of the gas layers must be supported by the pressure gradient dPdz\frac{dP}{dz} in the atmosphere. The balance of compressive forces and gravitation is expressed by the hydrostatic equation: dP(z)dz=ρ(z)g \frac{dP(z)}{dz} = -\rho(z) g with the gravitational acceleration g9.81m/s2g \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 AA parallel to the Earth's surface and the height Δz\Delta z (see the diagramm on the right). The mass of this volume is Δmρ(z)AΔz\Delta m \approx \rho(z) A \Delta z, so that a weight ΔFg=Δmg\Delta F_g = -\Delta m g acts on the entire volume. In addition, the pressures P(z+12Δz)P(z + \frac{1}{2} \Delta z) and P(z12Δz)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 Δz0\Delta z \to 0, the pressure difference changes into the derivative dPdz\frac{dP}{dz}, so that the difference equation becomes a differential equation. F=0=ΔmgP(z+12Δz)A+P(z+12Δz)A=(ρ(z)gP(z+12Δz)P(z+12Δz)Δz)AΔzΔz0(ρ(z)gdP(z)dz)dV \begin{aligned} \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{aligned} Since the expression in parentheses must be zero it follows dP(z)dz=ρ(z)g \boxed{\dfrac{d P(z)}{dz} = - \rho(z) g }

Note by Markus Michelmann
3 years, 5 months ago

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