Hydrostatic equilibrium is one of the most important fundamental principles in atmospheric physics and astrophysics. It defines the properties of stable gaseous systems confined within a gravitational field, and can be used to estimate the properties of Earth's atmosphere, Jovian planets, stars, and even clusters of galaxies. In its simplest form, it states that the inward force of gravity on an infinitesimal parcel of gas is balanced by the outward force of pressure by the gas underneath it.

The simplest physical case of an isothermal (uniform temperature), ideal gas is the most physically enlightening.

Take a cube of gas, with sides dx=dy=dz and density ρ, bounded top and bottom by gas pressures P1 and P0 at a height z in a (nearly) constant gravitational field.

        /         /|
       /         / |
      /         /  |dz  gravity (g)
     |---------|   |       |
     |         |   |       V
     |         |  /
     |         | / dy


The sides of the cube have area dA = dxdy. We assume that gravity is only acting in the downwards direction, and there are no lateral accelerations, so we ignore motions of the cube in the X and Y directions. For the cube to be static, all vertical forces on the cube must sum to zero. The net downwards force is the force of gravity on the cube plus the downwards pressure from gas above the cube, or

Fdown = - Mcube g - P1 dA (eq. 1)

Note the minus signs, indicating that the force is acting in the negative (downwards) direction along the z-axis. The net upwards force is just the pressure of the gas below the cube, P0 dA. The sum of the forces is then

Fdown + Fup = -Mcube g + (P0 - P1) dA = 0 (eq. 2)

Since Mcube is just ρ dx dy dz = ρ dz dA, the equation reduces to

-ρg dz = P1 - P0 = dP (eq. 3)

Now, since we're talking about an ideal gas, P = ρ k T / μ, where k is Boltzmann's constant, T is the temperature, and μ is the mean molecular weight. We can substitute this into equation (3) and get

P (-μg)/(kT) dz = dP (eq. 4)

Integration of this equation yields

ln(P) + const = -(μg)/(kT) z (eq. 5)

where ln is the natural logarithm, resulting from the integration of dP/P. This is then usually written in exponential form, as

P = P(z0) exp(-μgz/(kT)) (eq. 6)

where P(z0) is the pressure at the (arbitrary) point z0 (on Earth, this would be the equivalent of sea level). The quantity (kT/μg) is usually rewritten as H, the scale height, leading to the simple equation

P = P(z0) exp (-z/H) (eq. 7)

On Earth, the value of the scale height is about 8 km, meaning that the air pressure drops by a factor of e (about 36 percent) when you increase your altitude by one scale height.

Obviously, there are some caveats to using the above formulation. First, the local acceleration due to gravity, g, is not a constant, so to be precise, it should be replaced with Newton's Law of Gravitation:

g = GM/R2 (eq. 8)

Second, the gas you deal with will be neither isothermal nor ideal, so they introduce a further difficulty into the equation. However, as written, equation (7) provides a good, first-order approximation to the behavior of gaseous atmospheres. If you happen to be an astronomy or atmospheric physics grad student, it will almost certainly appear on an exam at some point so this derivation is worth knowing.

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