(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 $\Delta r \ll r$ 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 the pressure pushing up from the gas $(P_{up} = P(r))$ and down from below $(P_{down} = P(r + \Delta r))$.
Make a drawing of this, and discuss the situation and the various physical parameters with your group.
In this situation, a parcel of air is selected in the Earth's atmosphere. Now, it is important to think of this parcel of air as a stationary parcel of air, even though technically, air molecules are constantly moving in and out of it. However, they are moving in and out of this small subsection of the atmosphere at an equal rate, so the total number of molecules, and the forces acting on the parcel do not change. In this parcel of air, which is distance $r$ from the center of the Earth, has a height of $\Delta r$. Since air is made up of matter, and all matter has mass, this parcel of air is acted upon by the force of gravity. Also, since the gases in the atmosphere are made up of gas particles, the gases exert pressure from both above and below the parcel of air. The pressure from gasses below the the parcel put an upward force on the parcel $P_{up}$, and this pressure is a function of the distance from the Earth to the bottom of the parcel, $r$, where $P_{up} = P(r)$. The pressure from the gases above the parcel put a downward force on the parcel $P_{down}$, and this pressure is a function of the distance from the Earth to the bottom of the parcel, plus an additional distance to the height of the parcel $P_{down} = P(r + \Delta r)$, since the air pressure at the top of the parcel may not be the same as the pressure from the bottom.
(b) What other force will the parcel feel, assuming it has a density $\rho(r)$ and the Earth has a mass $M_{\bigoplus}$?
As mentioned before, the parcel will feel not only the upward force from the pressure of gasses below, and the downward force from pressure of the gases above the parcel, but the parcel will also feel the downward force of gravity towards the center of the Earth. This is because the parcel of air is made up of gas particles, and any particulate matter has mass, which is subject to the force of gravity. To show the effect of the force of gravity that the air parcel will experience, we can use Newton's law of Universal Gravitation:
\begin{align}
F_g = \frac{GM_{\bigoplus}m_{parcel}}{r^2}
\end{align}
where $M_{\bigoplus}$ is the mass of the Earth, $m_{parcel}$ is the mass of the parcel of air, and $r$ is the distance between the center of the Earth and the parcel of air.
We can figure out the mass of the parcel of air using the definition of density, $\rho = \frac{m}{v}$, where $\rho$ is the density, $m$ is mass, and $v$ is volume. We can rearrange this equation to show that $m = \rho v$. To make this specific for the mass of the parcel of air, $m_{parcel} = \rho (r) \cdot A \Delta r$, where the volume of the cylindrical parcel of air is simply the area of the circle, $A$ of the cylinder, multiplied by the height of the cylinder, $\Delta r$. Plugging this into the equation for the force of gravity, we get:
\begin{align}
F_{g} = (\rho (r) \cdot A \Delta r) \frac{GM_{\bigoplus}}{r^2} = (\rho(r) \cdot \Delta r)g
\end{align}
The other forces on the parcel include the upward force $F_{up}$ caused by the pressure from the gases below the parcel of air, and the downward force $F_{down}$ caused by the pressure from the gases above the parcel of air.
Since pressure $P$ is defined as $P = \frac{F}{A}$, where $F$ is a force, and $A$ is the area over which the force is acting on, we can rearrange this to get force in terms of pressure and area, in the form of $F = P \times A$. Since we know both the density and the area of the parcel of gas, we can solve for $F_{up}$ and $F_{down}$:
\begin{align}
F_{up} &= P_{up} \cdot A = P(r) \cdot A\\
F_{down} &= P_{down} \cdot A = P(r + \Delta r) \cdot A
\end{align}
(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 parcel is not moving, we can think of th parcel of air as a stationary free-body diagram. As seen from the previous part, the upward force that is acting on a parcel $F_{up}$, and the downward forces that are acting on the parcel are $F_{down}$ and the force of gravity $F_{g}$. Since forces are vectors, and the parcel of air is not moving, we can represent the summation of forces as follows:
\begin{align}
\sum F = ma &= 0\\
F_{up} - F_{down} - F_g &= 0\\
F_{up} &= F_{down} + F_g
\end{align}
Since we know from part (b) what $F_{up}$, $F_{down}$, and $F_g$ are, we can substitute them into the summation of forces, and get the following relationship:
\begin{align}
P(r) \cdot A &= P(r + \Delta r) \cdot A + (\rho (r) \cdot A\Delta r)g\\
P(r) &= P(r + \Delta r) + (\rho (r) \Delta r)g
\end{align}
(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.
Using the equation we got in part (c), we can rearrange the equation to solve for $g$ as follows to show how gravitational acceleration $g$ at distance $r$ above the Earth:
\begin{align}
g = \frac{P(r) - P(r + \Delta r)}{\rho (r) \Delta r}
\end{align}
(e) Show that
\begin{align}
\frac{dP(r)}{dr} = -g \rho(r)
\end{align}
This is the equation of hydrostatic equilibrium.
In order to derive the equation for hydrostatic equilibrium, we can use the equation we got for part (d):
\begin{align}
g = \frac{P(r) - P(r + \Delta r)}{\rho (r) \Delta r}
\end{align}
It seems that the equation for hydrostatic equilibrium relates the derivative of the pressure at a distance $r$ to the density of air multiplied by gravitational acceleration. Since the equation from part (d) looks very similar to the definition of a derivative, without the pesky $\rho (r)$ in the denominator, we can move it to the other side of the equation, and we get:
\begin{align}
g\rho (r) = \frac{P(r) - P(r + \Delta r)}{\Delta r}
\end{align}
Whoa... this is starting to really look like the formal definition of the derivative. Let's manipulate it to get the hydrostatic equilibrium equation:
\begin{align}
g\rho (r) &= \frac{P(r) - P(r + \Delta r)}{\Delta r}\\
g\rho (r) &= -\frac{P(r + \Delta r) - P(r)}{\Delta r}
\end{align}
Now we're definitely at the definition of a derivative. Let's see what happens when $\Delta r$ gets arbitrarily small and approaches 0, indicating that instead of dealing with a column of air, we're dealing with the individual particles:
\begin{align}
\lim_{r \to 0} g\rho (r) &= \lim_{r \to 0} -\frac{P(r + \Delta r) - P(r)}{\Delta r}\\
g\rho (r) &= -\frac{dP(r)}{dr}\\
-g\rho (r) &= \frac{dP(r)}{dr}
\end{align}
And so we have derived the equation for hydrostatic equilibrium!!
(f) Now go back to the ideal gas law described above. Derive an expression describing the density of the Earth's atmosphere varies with height, $\rho(r)$? (HINT: It may be useful to recall that $\frac{dx}{x} = d \ln x$.)
So the ideal gas law we defined above was $P = nk_BT$, where $P$ is pressure, $n$ is the number of particulate matter, $k_B$ is the Boltzmann constant $1.4 \times 10^{-16}$ $erg$ $K^{-1}$, and $T$ is temperature in Kelvin. We are looking for an expression that describes the density of the Earth's atmosphere with respect to height, or $\rho (r)$. How can we find $\rho (r)$?
Well for starters, we can use the equation for hydrostatic equilibrium, and solve for $\rho (r)$ to get:
\begin{align}
\rho (r) &= -\frac{dP(r)}{gdr}
\end{align}
Okay, so what exactly is $P(r)$ in this equation? It is the pressure of air at a particular distance from Earth, $r$. We can use the ideal gas law to figure out the pressure of air at any $r$. The ideal gas law requires the total number of particles $n$ in the system. When we are dealing with pressure at a particular density $r$, we can get the number of particles by using the air density at that particular $r$, a.k.a. $\rho (r)$ and dividing it by the average mass of the particles in the air, represented by $\overline{m}$. Therefore, we can represent $n = \frac{\rho (r)}{\overline{m}}$. Plugging this into the equation for the ideal gas law to solve for pressure at a particular height, $P(r)$, we get:
\begin{align}
P(r) = nk_BT = \frac{\rho (r) k_B T}{\overline{m}}
\end{align}
Okay, so we now have $P(r)$. Since we are looking for $\rho (r)$, we can plug this back into the equation for $\rho (r)$ we solved for earlier, and simplify:
\begin{align}
\rho (r) &= -\frac{dP(r)}{gdr}\\
\rho (r) &= -\frac{d\left(\frac{\rho (r) k_B T}{\overline{m}}\right)}{gdr}\\
\rho (r) &= -\frac{k_BT}{\overline{m}g} \frac{d\rho (r)}{dr}
\end{align}
Now, we have $\rho (r)$ on both sides of the equation. In order to isolate $\rho (r)$, we have to solve the differential equation, which will yield the result:
\rho (r) = \rho_0 e^{-\frac{\overline{m}gr}{k_B T}}
\end{align}
(g) Show that the height, $H$, over which the density falls off by a factor of $\frac{1}{e}$ is given by:
\begin{align}
H = \frac{kT}{\overline{m}g}
\end{align}
where $\overline{m}$ 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 $\overline{m}$?
Okay, so we want to see if the density between two points falls by a factor of $\frac{1}{e}$ can be represented by $H$, which is defined as the height $H = \frac{kT}{\overline{m}g}$. In order to see if this is true, let's first check the units:
\begin{align}
H = \frac{kT}{\overline{m}g} = \frac{\frac{erg}{K} \cdot K}{g \cdot \frac{cm}{s^2}} = \frac{\frac{\frac{g \cdot cm^2}{s^2}}{K} \cdot K}{g \cdot \frac{cm}{s^2}} = \frac{\frac{g \cdot cm^2}{s^2}}{\frac{g \cdot cm}{s^2}} = \frac{g \cdot cm^2 \cdot s^2}{g \cdot cm \cdot s^2} = cm
\end{align}
Since centimeters is a unit of height, the units check out!
Now let's look to see how the fall in density corresponds to a factor of $\frac{1}{e}$. We can compare the ratios of the density at two different heights, where $r_1 = r_0$ and $r_2 = r_0 - H$:
\begin{align}
\frac{1}{e} &= \frac{\rho (r_1)}{\rho (r_2)}\\
\frac{1}{e} &= \frac{\rho_0 e^{-\frac{\overline{m}g}{k_BT}(r_1)}}{\rho_0 e^{-\frac{\overline{m}g}{k_BT}(r_2)}}\\
\frac{1}{e} &= \frac{e^{-\frac{\overline{m}g}{k_BT}(r_0)}}{e^{-\frac{\overline{m}g}{k_BT}(r_0 - H)}}\\
\frac{1}{e} &= \frac{e^{-\frac{\overline{m}g}{k_BT}(0)}}{e^{-\frac{\overline{m}g}{k_BT}(- H)}}\\
\frac{1}{e} &= \frac{1}{e^{-\frac{\overline{m}g}{k_BT}(H)}}\\
e &= \rho_0 e^{\frac{\overline{m}g}{k_BT}(H)}\\
1 &= \frac{\overline{m}g}{k_BT}(H)\\
H &= \frac{k_BT}{\overline{m}g}
\end{align}
(h) What is the Earth's scale height, $H_{\bigoplus}$? 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 and 7 neutrons.
The scale height of the Earth, $H_{\bigoplus}$ can be calculated by simply plugging in values into the generic scale height formula derived in the previous part. We just need to figure out the average mass of the gas particles in Earth's atmosphere. Since the Earth is made up of mostly Nitrogen gas, $N_2$, two atoms of Nitrogen make up one gas particle. Since a Nitrogen atom has 7 protons and 7 neutrons, which both weigh approximately the same at about $1.7 \times 10^{-24}:$ g, the mass of the average gas particle is as follows:
\begin{align}
\overline{m} = N_{2_{mass}} = 2 \cdot N_{mass} = 2 \cdot (\#Protons + \#Neutrons) \cdot particle_{mass} = 2 \cdot 14 \cdot (1.7 \times 10^{-24}) = 4.76 \times 10^{-23} g
\end{align}
Okay, so now that we know the mass of the average gas particle in the atmosphere, we can directly solve for the scale height of the Earth. It is important to remember that the average temperature of Earth is 15 degrees Celsius, or 288 Kelvin:
\begin{align}
H_{\bigoplus} &= \frac{(1.4 \times 10^{-16} \text{ erg } K^{-1})(288 K)}{(4.76 \times 10^{-23} g)(9.8 \frac{m}{s^2})} \approx 9 km
\end{align}
Very well done Zahra! This is pretty much an ideal writeup. Every step is shown, your work is very easy to follow, and you've done a nice job with the LaTeX formatting. Keep up the good work!
ReplyDelete