Worksheet 7, Problem 1: Hydrostatic Equilibrium

1. Consider the Earth’s atmosphere by assuming the constituent particles comprise an ideal gas, such that \(P = nk_BT\), where n is the number density of particles (with units \(cm^{-3}\)), \(k = 1.4 \times 10^{-16} erg \, K^{-1}\) 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 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 \(\delta r << r\) 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 (\(P_{up} =  P(r)\)) 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.

We can represent the small cylindrical parcel of gas with our drawing below. The cylinder has a height of dr, is positioned a distance r away from the earth's center, and has a cross sectional area, A. As we consider the forces acting on the cylinder, there are two primary forces at play. The first is the pressure downwards, \(P_{down} = P( r + \delta  r)\), and the second is the force upwards, \(P_{up} =  P(r)\).

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

Beyond the upward and downward pressure effecting the cylinder, the cylinder is acted upon by the gravitational pull of earth. Before calculating this, we can use what we know about the volume of the cylinder and its density to calculate it's mass, m. The mass is calculated as follows: \[m = \rho(r) \times A \times dr\]Given that the gas has a mass of m, and the earth has a density of M, the cylinder will feel the following force: \[ F_g = G \frac{mM}{r^2}\] We can also write this by substituting in the density and volume of the cylinder for the mass, m. \[ F_g = G \frac{\rho(r) A dr M}{r^2}\] Adding this to our sketch, our drawing now shows all three forces acting on the cylinder.

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.

Since the cylinder is stationary in this situation, we can assume that all forces cancel out. This means that the upwards force of pressure (pressure x area) will be equal to the downward force of pressure (pressure x area) and gravity. \[P_{up} \times A = P_{down} \times A + F_{g}\]
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.

To derive this expression, we can begin by fundamentally realizing that gravity is a force. Since it is a force, we know that it can be expressed by \(F = m \times a\). Setting this force equal to the gravitational force we derived in part B, we have: \[ F_g = ma = mg =  G \frac{mM}{r^2}\] Then by dividing both sides by the mass of the cylinder, we arrive at: \[g =  G \frac{M}{r^2}\] e) Show that \[\frac{dP(r)}{dr} = -g \rho (r)\] This is the equation of hydrostatic equilibrium.

Beginning with our equation for forces being applied to our gas we have: \[P_{up} \times A = P{down} \times A + F_{g}\] \[P(r) \times A = P(r + dr) \times A + A dr \rho (r) g\] Since the area is common to each term, we can divide this out: \[P(r) = P(r + dr) + dr \rho (r) g\] Then, by rearranging out terms, we can put the equation in the form of the (negative) derivative definition: \[\frac{P(r) - P(r + dr)}{dr}= \rho (r) g\] This then simplifies to: \[- \frac{dP(r)}{dr} = \rho(r) g\] or \[ \frac{dP(r)}{dr} = - \rho(r) g\] 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, \(\rho (r)\)? (HINT: It may be useful to recall that \(\frac{dx}{x} =  d ln (x)\)

Beginning with the ideal gas law declared above, we have: \[P(r) = nk_BT\] As we try to relate density to this equation, we can recall that the density and number density of particles relates with the following equation: \[ \rho(r) = \bar{m}n\] Where (\ \={m} \) represents the average molecular weight of particles. Plugging this into the equation, we now have: \[P(r) = \frac{\rho(r)}{\bar{m}}k_BT\] Solving for density we get:  \[\rho(r) = \frac{P(r) \bar{m}}{k_BT}\] Applying the hydrostatic equilibrium equation of \(\frac{dP(r)}{dr} = -g \rho (r)\) we can substitute for \(\rho(r)\) and integrate on both sides: \[\int_{P(0)}^{P(r)} \frac{dP(r)}{dr}dP(r) = \int_{0}^{r} -g \frac{P(r) \bar{m}}{k_BT}dr\] Using the hint provided in the question that \(\frac{dx}{x} =  d ln (x)\) we can easily solve this integral. \[\int_{P(0)}^{P(r)} \frac{dP(r)}{dr}dP(r) = \int_{0}^{r} -g \frac{P(r) \bar{m}}{k_BT}dr\] \[ln (\frac{P(r)}{P(0)}) = \frac{-g r \bar{m}}{k_BT}\] \[P(r) = e^{\frac{-g r \bar{m}}{k_BT}} P(0)\] Finally, returning to the ideal gas law, we can substitute in \(P(r) = \frac{\rho(r)}{\bar{m}}k_BT\) \[\frac{\rho(r)}{\bar{m}}k_BT = e^{\frac{-g r \bar{m}}{k_BT}} P(0)\] \[ \rho (r) = \frac{\bar{m}}{k_B T} e^{\frac{-g r \bar{m}}{k_BT}} P(0)\]
g) Show that the height, H, over which the density falls off by an factor of 1/e is given by \[H =\frac{kT}{mg}\] 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!

In order to solve for the scale height (which represents the height that the density that changes by a factor of 1/e), we need to take a close look at our solution to part f, which defines how density changes as a function of r. Something that stands out from this is that the only term that varies is: \[e^{\frac{-g r \bar{m}}{k_BT}}\] To find the scale height, we simply need to determine when this term changes by a factor of 1/e. Fortunately, this occurs when \( (\frac{-g r \bar{m}}{k_BT}) = 1\). Solving for r, we get: \[H = r = \frac{k_BT}{g r \bar{m}}\] To confirm our answer, we can plug in the units for each expression. \[cm = \frac{gcm^2/s^2K^{-1}K}{gcm/s^2} = cm\]
h) What is the Earth’s scale height, H? The mass of a proton is \(1.7 \times 10^{24} g\), and the Earth’s atmosphere is mostly molecular nitrogen, \(N_2\), where atomic nitrogen has 7 protons, 7 neutrons.

First, we will start with the assumption that the Earth's atmosphere is primarily composed of nitrogen. Now, we want to calculate the average molecular mass of this nitrogen atmosphere by multiplying the number of protons and neutrons in \(N_2\) by the mass of a proton or neutron. \[ \frac{14 \, neutrons + 14 \, protons }{N_2 \, molecule}\times \frac{1.7 \times 10^{-24}g}{proton/neutron} = 5\times 10^{-23}g\] Now substituting this into our scale height at T = 273K, we get:
\[H = \frac{(1.4 \times 10^{-16} erg \, K^{-1}) \times (273K)}{(5 \times 10^{-23}g) \times (9.8 \times 10^2 cm/s^2)} = 8.03 \times 10^5 cm\]

Acknowledgements: Johnathan Budd & Willie Pirc