Earth's atmosphere: how good is our model from class?

In class, we worked out how the density would scale with radius in the Earth's atmosphere using 2 assumptions: 1) taking the gravitational acceleration to be constant and equal to \(g=\frac{GM_{\oplus}}{R_{\oplus}^2}\) and 2) taking the atmosphere to be isothermal (all at the same temperature T). Here, I'll discuss what happens when you relax those assumptions, one at a time, and then go on to discuss a more realistic model of Earth's atmosphere. Let's first relax assumption 1) but keep assumption 2). Set up force balance on a parcel of mass with mass m, area A, and thickness \(\Delta r\) at a radius r above the center of the Earth: $$F_g+A(-P(r+\Delta r)+P(r))=0.$$ Now before we had used \(F_g=mg\) for the force of gravity, but now let's account for the fact that the mass is at some radius \(r>R_{\oplus}\). In other words, \(mg\) is the gravitational force at the Earth's surface; but we want the force at some distance \(h=r-R_{\oplus}\) above the surface. We have $$F_g=-\frac{GM_{\oplus}m}{r^2}.$$ We can then write $$-\frac{GM_{\oplus}m}{r^2}=A(P(r+\Delta r)-P(r )).$$ Now note that for a parcel with density \(\rho( r)\), the mass is \(m=A\Delta r\rho(r )\). Note that we have let the parcel's density be a function of \(r\) here because the density is changing as you go up in the atmosphere. We don't yet know \(what\) function of \(r\) \(\rho\) is, but we'll find out! Rewriting our previous result by replacing the mass and then dividing both sides by area \(A\) and \(\Delta r\), we get $$-\frac{GM_{\oplus}\rho(r )}{r^2}=\frac{dP}{dr}.$$ Note that, just as we did in class, I've rewritten \( (P(r+\Delta r)-P( r))/\Delta r \) as \(dP/dr \). We now use the ideal gas law, which is a good approximation for our atmosphere, provided we use the right average particle mass from the atmosphere's measured composition, to replace \( P(r )=\rho( r) k_B T/\bar{m} \), where we remember assumption 2) is still in force so \(T\) is not a function of \(r\). \(\bar{m}\) is the average mass per particle, which we used to write the number density in the ideal gas law as a mass density over a mass. We now have the differential equation $$ -\frac{GM_{\oplus}\rho(r )} {r^2}=\frac{d\rho}{dr}\frac{k_B T}{\bar{m}}.$$ We can rearrange this to read $$-\frac{GM_{\oplus}\bar{m}}{k_B T r^2}dr=\frac{d\rho}{\rho}.$$ We can now integrate both sides of this equation to find \(\rho (r )\), but before we do so, let's look at the physical meaning. The left-hand side (lhs) compares the gravitational energy due to a fractional increas in radius \(\Delta r/r)\), with the thermal energy, \(k_B T\). To make that clearer, remember, gravitational PE is $$PE=-\frac{GM_{\oplus}m}{r},$$ so we can rewrite $$-\frac{GM_{\oplus}m}{r^2}dr=-\frac{GM_{\oplus}}{r}\frac{dr}{r}=PE\times\frac{dr}{r}.$$ So this is saying, if we go up by a fractional radius \(\dr/r\), then the potential energy will change by PE times that fraction. How much the PE changes relative to the thermal energy determines the fractional change in density, \(d\rho/\rho\); that's what the right-hand side is telling us. Now let's solve: $$-\int_{R_{\oplus}}^r \frac{GM_{\oplus}\bar{m}}{k_B T r^2}dr=\int_{R_{\oplus}}^r \frac{d\rho}{\rho}.$$ We find $$\frac{GM_{\oplus}\bar{m}}{k_B T}\frac{1}{r}\bigg|_{R_{\oplus}}^r=\ln \rho \bigg|_{R_{\oplus}}^r,$$ which can be rearranged and simplifed to give $$\rho ( r)=\rho(R_{\oplus})\exp\bigg[\frac{GM_{\oplus}\bar{m}}{k_B T}\big[\frac{1}{r}-\frac{1}{R_{\oplus}}\big]\bigg].$$ To make sense of this, let's write \(r\) in terms of \( R_{\oplus}\) and height \(h\) above the Earth's surface, as $$r=R_{\oplus}+h=R_{\oplus}\left(1+\frac{h}{R_{\oplus}}\right).$$ We can then Taylor expand \(r^{-1}\) about \(h=0\), meaning \(h/R_{\oplus}\ll 1\). Doing so we have $$r^{-1}\approx \frac{1}{R_{\oplus}}\left(1-\frac{h}{R_{\oplus}}+\left(\frac{h}{R_{\oplus}}\right)^2+\cdots\right).$$ Let's see what this does in our expression for \(\rho\), specifically in the term $$\big[\frac{1}{r}-\frac{1}{R_{\oplus}}\big].$$ We can see the 1 in the expansion will be canceled out by the \(\frac{1}{R_{\oplus}}\) above, and at lowest order in \(h\) we recover the result from class for constant gravitational force, \(\rho \propto e^{-h}\). We have $$\rho (h )=\exp\left[\frac{PE}{E_{th}}\left[-\frac{h}{R_{\oplus}}+\left(\frac{h}{R_{\oplus}}\right)^2\right]\right].$$ Note that I've rewritten this in terms of the ratio of potential energy to thermal energy, for the physical reasons discussed earlier (before we integrated). You can see that accounting for the variation of the force of gravity with \(r\) slightly increases the density, because the term in \( \left(h/R_{\oplus}^2\right) \) is positive. Note that it's also small, because typical heights would be much less than the Earth's radius. Using that idea, we can even Taylor expand the whole exponential, this time by factoring out \(\frac{h}{R_{\oplus}}\) to have $$\exp\left[-\frac{PE}{E_{th}}\frac{h}{R_{\oplus}}\left[1-\frac{h}{R_{\oplus}}\right]\right]=\exp\left[-\frac{PE}{E_{th}}\frac{h}{R_{\oplus}}\right] \times \exp\left[-\frac{PE}{E_{th}}\frac{h}{R_{\oplus}}\frac{h}{R_{\oplus}}\right].$$ Approximating the second exponential, we have . . .