Worksheet 7: All of it.

An Intro to Hydrostatic Equilibrium

The problem: Consider the Earth’s atmosphere by assuming the constituent particles comprise an ideal gas, such that P = nkT, where n is the number density of particles (with units cm^3 ), k = 1.4x10^16 erg K^1 is the Boltzmann constant. We’ll use this ideal gas law in just a bit, but first

Part 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 ∆r + r and a circular cross-sectional area A (it’s okay to use r here, because it is an intrinsic property of the atmosphere). The parcel will feel pressure pushing up from gas below (Pup = P(r)) and down from above (Pdown = P(r + ∆r)).

Part B: What other force will the parcel feel, assuming it has a density ρ(r) and the Earth has a mass M?

The parcel will also feel gravitational force, which we will denote as \(\vec{F_g}\).

Part 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.

If the parcel is not moving, we can conclude: \[\vec{F_{up}} + \vec{F_{down}} + \vec{F_g} = 0\] And because pressure is force per unit area, we have \[A\vec{P_{up}} +A \vec{P_{down}} + \vec{F_g} = 0\]

Part D: Give an expression for the gravitational acceleration, g, at at a distance r above the Earth’s center in terms of the physical variables of this situation.

We know that the equation for gravity is: \[F_g = - \frac{GM_{\oplus}M_{gas}}{r^2}\] Using F=ma or in this case, F=mg, allows us to solve for g. \[M_{gas}g= - \frac{GM_{\oplus}M_{gas}}{r^2}\] \[g(r)= - \frac{GM_{\oplus}}{r^2}\] 

Part E: Show that \[\frac{dP}{dr} = - \rho g \] is the equation of hydrostatic equilibrium.

We can rewrite F to give: \[F_g= V \rho g\] where \[V = \Delta r A\] so \[F_g=\Delta r A \rho g\] We can also rewrite the equation found in part C to give \[F_{up} + F_{down}= -F_g\] Plugging in pressures and our new form of F allows us to solve:\[AP(r) - AP( \Delta r + r ) = - \Delta r A \rho g \]\[P(r) - P( \Delta r + r ) = - \Delta r  \rho g \] \[P(r) - P( \Delta r + r ) = - \Delta r  \rho g \]  \[\frac{P(r) - P( \Delta r + r ) }{\Delta r} = - \rho g \] \[ \lim_{r\to\infty} \frac{P(r) - P( \Delta r + r ) }{\Delta r} = - \rho g \] \[\frac{dP}{dr} = - \rho g \] 

Part F: Now go back to the ideal gas law described above. Derive an expression describing how the density of the Earth’s atmosphere varies with height, ρ(r)? (HINT: It may be useful to recall that dx/x = d ln x.)

Because \[n = \frac{particles}{cm^3}\] \[\rho = \frac{ \bar{m} particles}{cm^2}\] \[ \rho = \bar{m} n\] \[n= \bar{m} \rho\] Therefore using P=nkT gives: \[P = kT \frac{\rho}{ \bar{m}}\] Plugging this back into \[\frac{dP}{dr} = - \rho g \] gives \[ \frac{d (kT \frac{ \rho}{ \bar{m} })}{dr} = - \rho g \] \[\frac{kT}{ \bar{m}} \frac{d ( \rho)}{\rho} = - g dr \] Which we can integrate to find the answer \[\frac{kT}{ \bar{m}} \int \frac{d ( \rho)}{\rho} = \int - g dr \]  \[\frac{kT}{ \bar{m}} ln(\rho) = - g r + C \] \[ \rho (r) = C \cdot e^{\frac{-gr \bar{m} }{kT}} \]

Part G: Show that the height, H, over which the density falls off by an factor of 1/e is given by \[H= \frac{kT}{g \bar{m}}\] where m is the mean (average) 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 first-order differential equation and finding a key physical result!

First check units: \[H_{units} = cm \] \[H_{units} = \frac{ \frac{gcm^2}{ks^2} \cdot k}{ g \cdot \frac{cm}{s^2}} = cm \]

Now we can solve. The problem is asking for us to solve for the scenario below:

In the equation found in F, when H=0 \[ \rho (r) = C \cdot e^{0}  = C\] so the equation can be written as \[ \rho (r) = \rho_{\circ}  \cdot e^{\frac{-gr \bar{m} }{kT}} \] For this problem we need \[ \frac{\rho (r) }{\rho_{\circ}} = e^{-1} \] So \[ -1 = \frac{-gH \bar{m}}{kT} \] Solving for H gives the answer \[H= \frac{kT}{g \bar{m}}\]

This makes physical sense because if the mean mass of a gas particle gets heavier, the scale height should be lower. 

Part H: What is the Earth’s scale height, \(H_{\oplus}\)? The mass of a proton is 1.7 x 10^-24 g, and the Earth’s atmosphere is mostly molecular nitrogen, N2, where atomic nitrogen has 7 protons, 7 neutrons.  

Assuming an outside temperature of 300 kelvin and that \(N_2\) has 28 photons and neutrons of the same mass  \[\bar{m} =  28 \times 1.7 \times 10^{-24} = 4.8 \times 10^{-23} \]  Then solve \[H_{\oplus}= \frac{kT}{g \bar{m}}\] \[H_{\oplus}= \frac{1.4 \times 10^{-16} \times 300}{1000 \times 4.8 \times 10^{-23} \times 14 } = 8.8 \times 10^5 cm \]

Acknowledgements: I worked with Barra on this problem.