Week 5: WS9 #2

2. Let's apply the equation of hydrostatic equilibrium $$\frac{dP}{dr}=-\rho(r)g(r)$$ to some basic physical problems. 
a) Imagine a tank/lake/bathtub/ocean on Earth filled with water. Use the equation of hydrostatic equilibrium to prove that the pressure at a depth h is given by $$P(h)=\rho gh+P_{atm}$$ where g is the gravitational field of the Earth and P_atm is the atmospheric pressure. 

$\frac{dP}{dr}=-\rho g$
$\int^{P(h)}_{P_{atm}}dP=\int^0_h-\rho gdr$
Integrating this expression gives us $P(h)=\rho gh+P_{atm}$.

b) Noting that on Earth the atmospheric pressure is $P_{atm}\approx10^6dyne/cm^2$ and that water density is 1g/cc, how many times atmospheric pressure is the pressure at the bottom of the Marianas trench (11km deep)? 

$P_{marianas}=\rho gh=(1g/cm^3)(980cm/s^2)(1.1\times10^6cm)\approx10^9dyne/cm^2$
$\frac{P_{marianas}}{P_{atm}}=\frac{10^9}{10^6}=10^3$
The pressure in the Marianas trench is about 1000 times that on the Earth's surface.

c) Now let us explore what is called the isothermal atmosphere model. In this model, the atmosphere of the Earth is at temperature T that is independent of height. Use the equation of hydrostatic equilibrium and the ideal gas law, P = nkT, where n is the number density of particles and $k=1.4\times10^{-16}erg/K$ is the Boltzmann constant to derive an expression describing how the density of the Earth's atmosphere varies with height $\rho(r)$. You may assume the typical mass of particles in the air to have mass m and the density at the Earth's surface is $\rho_0$.

$n=\frac{P}{kT}$
Multiplying both sides by the particle's mass gives us $mn=\rho=\frac{mP}{kT}$.
Differentiating both sides gives us $d\rho=\frac{mdP}{kT}=\frac{m}{kT}(-\rho(r)g(r)dr$.
We can approximate g(r) as constant, so the integral becomes $\int^{\rho(r)}_{\rho_0}\frac{d\rho}{\rho(r)}=\frac{-m}{kT}\int^{r+r_\oplus}_{r_\oplus}gdr$.
Solving and simplifying the integral yields $\rho(r)=\rho_0e^{\frac{-mgr}{kT}}$.

d) The height, H, over which the density falls off by a factor of 1/e, $$H=\frac{kT}{mg}$$ where m is the mean mass of a gas particle, is called the scale height. When astronomers talk about the "thickness" of an atmosphere, we typically mean the scale height of the atmosphere. While formally the solution is defined even when r goes to infinity, atmospheres with greater scale height are "puffier" than those with smaller scale height. What should happen when you increase/decrease m, T, and g? 

Looking at the units of H, we can tell that it's actually a height: $[H]=\frac{[dyne\times cm/K][K]}{[g][cm/s^2}=cm$, since $dyne=\frac{erg}{cm}=\frac{g\times cm}{s^2}$.
This equation indicates that at higher temperatures, the atmosphere should be "puffier." This makes sense, since at higher temperatures the atmospheric gas will expand. With greater values of m (average particle mass) or g, the atmosphere will have a lower scale height. This makes sense, because heavier particles or stronger gravity will result in gas that stays closer to the planet.

e) What is the Earth's scale height, $H_\oplus$? The mass of a proton is $1.7\times10^{-24}g$, and the Earth's atmosphere is mostly molecular nitrogen, N₂. Take the average temperature for Earth's atmosphere to be ~250K. 

$H_\oplus=\frac{(1.4\times10^{-16}erg/K)(250K)}{28\times1.7\times10^{-24}g)(980cm/s^2)}=7.5\times10^5cm=7.5km$