Worksheet 7, Problem 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 = 1.4x10^-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 ∆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 + ∆r)). 


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


Parcel of Gas

 

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

   Since the parcel is close to Earth, it will experience the gravitational force of the Earth which is given by \(F_{g}\:=\:\frac{GM_{E}M_{parcel}}{r^{2}}\). Since the density, \(\rho (r)\), is given by \(\frac{mass}{volume}\), this equation can be rewritten as \(F_{g}\:=\:\frac{GM_{E}\rho (r)\:\times\:A\:\times\:\Delta r}{r^{2}}\)
   
   
   (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. 

   When the parcel is not moving, the forces acting upon are in equilibrium. The forces acting on it are the downward pressure, the upward pressure, and the force of gravity. Since pressure is force per area, we must multiply by the surface area of the base in order to get the total force from pressure acting on the parcel. Thus, the equilibrium equation is: 
   \[F_{up}\:=\:F_{down}\]
   \[P_{up}A\:=\:P_{down}A\:+\:F_{g}\]
   \[P(r)A\:=\:P(r+\Delta r)A\:+\:F_{g}\]
   
   
   (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. 

   We know that the equation for gravitational force is given by:
   \[F_{G}\:=\:-\frac{GM_{E}M_{parcel}}{r^{2}}\]
   We also know from Newton's laws of motion that \(F\:=\:m\:\times\:a\). Thus,
   \[F_{G}\:=\:m_{parcel}\:\times\:a\]
   In this case a is equal to the gravitational acceleration g. Thus, cancel out the mass of the parcel from the force for gravitational acceleration and we are left with \[g\:=\:\frac{GM_{E}}{r^{2}}\]
   
   
   (e)  Show that

   

   

   dP(r)dr=−gρ(r)

   This is the equation of hydrostatic equilibrium. 

   We know that the gravitational force acting on the parcel is equal to the mass of the parcel times the gravitational acceleration. If we rewrite the mass of the parcel as density times volume, we have \[F_{G}\:=\:\rho (r)\:\times\:A\:\times\:\Delta (r)\]. 
   Dividing by A gives us the per unit gravitational force and cancels out the A from our equation.
   Next we know from our force equilibrium equation that \[-F_{G}\:=\:P(r+\Delta r)\:-\:P(r)\]
   From the definition of the derivative, we know that 
   \[\frac{dP}{dr}\:=\:\frac{P(r+\Delta r)\:-\:P(r)}{\Delta r}\]
   Substituting the negative of gravitational force for the numerator of the fraction on the right hand side of this equation, we get \[\frac{dP}{dr}\:=\:\frac{\Delta r \rho (r) g}{\Delta r}\]
2. Canceling the \(\Delta r\)s leaves us with what we were trying to prove:
3. dP(r)dr=−gρ(r)
   (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.) 

4.  
5. We know that the ideal gas law is as follows:
6. \[P\:=\:nk_{b}T\]
7. \[n\:=\:\frac{\rho (r)}{m}\]
8. m in this case is the average mass of the particles.
9. We can then plug this value of P into the differential equation we were given in the last problem which produces:
10. \[\frac{d(\rho (r) kT)}{dr}\:=\:-g\rho (r)\]
11. Rearranging this equation into an easily integrable form gives us:
12. \[\frac{d\rho (r)}{\rho (r)}\:=\:\frac{-mg}{kT}dr\]
13. We can then integrate both sides of the equation to get:
14. \[ln(\rho (r))\:=\:\frac{-mg}{kT}r\]
15. We then solve this equation for \(\rho (r)\) and find that
16. \[\rho (r)\:=\: e^{\frac{-mgr}{kT}}+k\]
17. where k is some integration constant.

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


H=kTm¯g

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! 


We know that the scale height is the distance r over which the density changes by a factor of \(\frac{1}{e}\).  We can set this equal to a proportion using the equation we derived in the last part to give us:

\[\frac{\rho (r)\:\times\:e^{\frac{-mgH}{kT}}}{\rho (r)}\:=\:\frac{1}{e}\]
 Notice how H has replaced r in this new equation.
Solving for H shows that H does in fact equal \(\frac{kT}{mg}\)

(h)  What is the Earth’s scale height, HC? 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.

We can plug in numbers to the equation that we produced in the previous problem to solve this problem. 

\[g\:=\:9.8\:\times\:10^{2} \frac {cm}{s^{2}}\]

The average particle mass of nitrogen is roughly 1.7 * 10^-24 g 

The rough estimate of the atmospheric temp (T) is 250K

Plugging all that in we come up with the Earth's scale height which is about \(7.8\:\times\:10^{5} cm\)