Worksheet 6: Hydrostatic Equilibrium

A Hydrostatic Haiku:
Gravity Pulls Down
Down, Inexorably Down
P briefly impedes

Consider the Earth's atmosphere by assuming the constituent particles comprise an ideal gas,
such that P=nk$_B$T, where n is the number density of particles (with units cm $^{-3}$), $k_B=1.4\times 10^{- 16}$ erg K $^{-1}$ is the Boltzmann constant. We'll use this ideal gas law in just a bit, but first:

a) Think of a small, cylindrical parcel of gas with the axis tinning vertically into 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 $\Delta r<<r$ 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 ($P_{down}=P(r+\Delta r)$).



Make a drawing of this, and discuss the situation and the various physical parameters with your group.  

b) What other force will the parcel feel, assuming it has a density, $\rho$, and the Earth has a mass, $M_\oplus$?

Like any other object with mass on Earth, the parcel will feel the force of gravity pulling it down.  

$F_g=-\frac{GM_{cyl}M_\oplus}{r^2}$

$M_{cyl}=\rho A\Delta r$

$F_g=-\frac{G\rho A\Delta rM_\oplus}{r^2}$  Eq. 1

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

To find this expression, I added up all of the forces together.  The parcel isn't moving, so the net force on it is 0.  

$F_{tot}=F_g+(P_{up}+P_{down})A$

$0=-\frac{GM_\oplus \rho A\Delta r}{r^2}+(P(r)-P(r+\Delta r))A$

$\frac{GM_\oplus \rho \Delta r}{r^2}=-P(r+\Delta r)+P(r)$  Eq. 2

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 the situation.

If nothing else from introductory mechanics stuck, the fact that F=ma did.  Seriously, I might even be able to recite that in my sleep.  Eq. 1 defines the force of gravity.  If $\rho A\Delta r$ defines the parcel's mass (density times volume), then the other components of Eq. 1 must define the acceleration, or g.

                                                        $g=\frac{GM_\oplus}{r^2}$

e) Show that $\frac{dP(r)}{dr}=-gp$.  This is the equation of hydrostatic equilibrium.  What does it "say" in English?  Does it make sense?

First, we'll take Eq. 2 and substitute in g.   

$g\rho \Delta r=-P(r+\Delta r)+P(r)$

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

As annoying as it was to calculate in high school calculus, 

$\frac{dP(r)}{dr}=\frac{P(r+\Delta r)-P(r)}{ \Delta r}$

Therefore, 

$\frac{dP(r)}{dr}=-gp$   Eq. 3

This says that pressure as a function of altitude decreases as density increases.  

f) Now go back to the idea gas law described above.  Derive an expression describing how the density of the Earth's atmosphere varies with height, $\rho (r)$.  Recall that $\frac{dx}{x}=d\ln x$.

If n is number density and m is the mass of an individual particle, $\rho = nm$, which means we can rewrite the ideal gas law as 

$P(r)=\rho(r)\frac{kT}{m}$

Solving Eq. 3 for $\rho$, we can rewrite an equation for $\rho$ in terms of $\rho(r)$ as

$\rho=-\frac{d\rho}{dr}\frac{kT}{gm}$    Eq. 4

If you solve Eq. 4 for dr and integrate, you get the following equation:

$r=-\frac{kT}{gm}\ln\rho+c$

This equation is really easy to solve for $\rho(r)$:

$\rho(r)= e^{-\frac{rmg}{kT}-c}=ce^{-\frac{rmg}{kT}}$

When doing problems like this in class, professors always tell us to remember to consider the initial conditions.  When r=0, the density should equal the initial density, so $c=\rho_o$.

g) Show that the height, H, over which the density falls off by a factor of 1/e is given by

$H=\frac{k_BT}{mg}$

where m is the mean mass of a gas particle. This is the "scale height." First, check the units.

Then do the math. Then make sure it makes physical sense, e.g. what do you think should

happen when you increase m? Finally, pat yourselves on the back for solving a fi rst-order

di fferential equation and finding a key physical result!

This problem is asking us to find an r such that $\frac{\rho(r)}{\rho_o}=e^{-1}$ is true.  We can rewrite this as 

$e^{-\frac{rmg}{kT}}=e^{-1}$

If we take the natural log of both sides of this equation, we find 

$\frac{rmg}{kT}=1$

When we solve this for r, we find that 

$H=r=\frac{kT}{mg}$   Eq. 5

h) What is the Earth's scale height, $H_\oplus$?  The mass of a proton is $1.7\times 10^{-24}$ g, and the Earth's atmosphere is mostly nitrogen, $N_2$, where the atomic nitrogen has 7 protons and 7 neutrons.

This can be found by substituting the values provided into Eq. 5.  

$H=\frac{1.4\times 10^{-16}}{(7)(1.7\times 10^{-24})(9.8\times 10^2)}$

$=1\times 10^4$