An Introduction to Hydrostatic Equilibrium

We can use a model of a parcel of air at some distance r above earth's surface to determine different characteristics of the atmosphere including how density changes at different r values, and how gravitational acceleration changes at different r values. We can assume that the earth's atmosphere is comprised of an ideal gas which allows us to use the ideal gas law: $P=n \times k_B \times T$ n is the number density of particles (in $cm^3$) $k= 1.4 \times 10^{-16} erg K^{-1}$ which is the Bolzmann constant. 

a) Think of a small cylindrical parcel of gas, with the axis running vertically in the Earth's atmosphere. The parcel sits a distance r from the Earth's center, and the parcel's size is defined by a height $\delta r$ is much less than r and a circular cross sectional area A. The parcel will feel pressure pushing up from gas below $P_{up} = P(r)$ and down from above $P_{down} = P(r+ \delta r)$ where pressure is expressed as a function of r. Draw a picture to describe this scenario (Figure 1).

Figure 1:

What other force will the parcel feel, assuming it has a density $\rho (r)$ and the Earth has a mass $M_o$

 The parcel will feel the force from pressure up $P_{up} \times A$ , the force from pressure down $P_{down} \times A$ , and the force from gravity down $F_g$.

c) If the parcel is not moving, give a mathematical expression relating the various forces, remember that force is a vector and pressure is a force per unit area.

If the parcel is not moving, the net force will be 0.

$P_{up} \times A - P_{down} \times A - F_g = 0$

d)Give an expression for the gravitational acceleration, g, at a distance r above the Earth's center in terms of the physical variables of this situation

$F_g = P_{up} \times A - P_{down} \times A$

$F_g = P(r) \times A - P(r+ \delta r) \times A$

$F_g = mg$

$m= \rho \times V$, $V = \delta r \times A$

$\rho \times \delta r \times A \times g = P(r) \times A - P(r+ \delta r) \times A$

Areas cancel out

$\rho \times \delta r \times g = P(r) -  P(r+ \delta r)$ 

$g= \frac {P(r) -  P(r+ \delta r)}{\rho \times \delta r}$

$g(r)= \frac{GM_o}{r^2}$

e) Show that $\frac{dP(r)}{dr} = -g \rho (r)$

This is the equation of hydrostatic equilibrium

In part d we found: $\rho (r) \times \delta r \times g = P(r) -  P(r+ \delta r)$

$P(r) -  P(r+ \delta r) = \delta P$

$\delta P= \rho (r) \times \delta r \times g$

$\frac{\delta P}{\delta r}= \rho (r) \times g$

f)Derive an expression describing how the density of the Earth's atmosphere varies with height, $\rho (r)$

In part e we found $\frac{\delta P}{\delta r}= \rho (r) \times g$

The ideal gas law says $P=n \times k_B \times T$ which can be rewritten as $P=\frac{\rho (r)}{m_{average}} \times k_B \times T$ since n= \frac{N}/{V} and \rho (r) = \frac{m}{V }= \frac{N \times m_{average}}{V}$

Solving for $\rho (r)$, $\rho (r) = \frac{P \times m_{average}}{k_B \times T}$

$ln(\rho (r)) = \frac{-mg}{k_B T} \times r + C$

$\rho (r) = Ce ^{\frac{-mg}{k_B T} \times r }$

$\rho (r) = \rho _o e^{\frac{-m_{average}g}{k+B T}\times r}$

$\rho(0) = \rho _o)$

g) Show that the height H over which the density or pressure falls off by a factor of 1/e is given by $H= \frac{kT}{mg}$

m is the mean mass of a gas particle

We can take the proportion of two different densities $\rho(r)$ and $\rho (r+H)$

$\rho (r) = \rho _o e^{\frac{-m_{average}g}{k+B T}\times r}$

$\rho (r+H) = \rho _o e^{\frac{-m_{average}g}{k+B T}\times (r+H)}$

$\rho (r+H) = \rho _o e^{\frac{-m_{average}g}{k+B T}\times r} \times \rho _o e^{\frac{-m_{average}g}{k+B T}\times H}$ 

$\rho (r+H) = \rho (r) \times \rho _o e^{\frac{-m_{average}g}{k+B T}\times (H)}$

$\frac{\rho (r+H)}{\rho (r)}= e^{\frac{-mg}{k_B T}H}= \frac{1}{e}$ = \frac{-mg}{k_B T} \times H =-1$

Therefore $H= \frac{k_B T}{mg}$

h) What is the Earth's scale height, $H_o$ ?

Therefore $H= \frac{k_B T}{mg}$

The mass of a proton is $m_p$ = $1.7 \times 10^{-24} g$ and the Earth's atmosphere is mostly nitrogen $N_2$ where atomic nitrogen has 7 protons, 7 neutrons. 

Since protons and neutrons have approximately the same mass, we can approximate the mass of a N atom to be 14$m_p$ and since earth's atmosphere is mostly $N_2$, $N_2$ weighs about $28m_p$

$k_B= 1.4 \times 10^{-16} erg K^{-1}$

Temperature is about 288K (Earth Fact Sheet (nasa.gov))

g=$9.8 m/s^2$

$H= \frac{1.4 \times 10^{-16}erg K^{-1} \times 288 K}{4.76 \times 10^{-23}g \times 980 cm/s^2}$

$H=860,000 cm = 8,600 m = 8.6km$