Worksheet 9, Problem 2- Star Formation

Forming Stars - Giant molecular clouds occasionally collapse under their own gravity (their own "weight") to form stars. This collapse is temporarily held at bay by the internal gas pressure of the cloud, which can be approximated as an ideal gas such that $P = nkT$, where $n$ is the number density ($\text{cm}^-3$) of gas particles within a cloud of mass $M$ comprising particles of mass $\overline{m}$ (mostly hydrogen molecules $H_2$), and $k$ is the Boltzmann constant, $k = 1.4 \times 10^{-16} \text{ erg K}^{-1}$.

(a) What is the total thermal energy, $K$, of all the gas particles in a molecular cloud of total mass $M$? (HINT: a particle moving in the $i^{th}$ direction has $E_{thermal} = \frac{1}{2}mv_i^2 = \frac{1}{2}kT$. This fact is a consequence of a useful result called the Equipartition Theorem.)

Okay, so we want to know the total thermal energy, $K$ of the all the gas particles in the molecular cloud. Well let's look at just one gas particle first. Since the gas particle exists in a three-dimentional space, the thermal energy of the single gas particle is three times that of the thermal energy in any given direction, and can be expressed as:

\begin{align}
E_{thermal_{particle}} = \frac{3}{2}mv_i^2 = \frac{3}{2}kT
\end{align}

Okay, so we now have the thermal energy of a single gas particle. To find the thermal energy of all the gas particles in a molecular clouds, we need to know just how many gas particles there are. Since we know the mass of the entire molecular cloud to be $M$, and the mass of a single gas particle to be $\overline{m}$, we can get the total number of gas particles by the expression $\frac{M}{\overline{m}}$. Therefore, we can multiply the thermal energy of a single particle by the number of particles $\frac{M}{\overline{m}}$ to get the thermal energy of the entire molecular cloud, as such:

\begin{align}
E_{thermal_{cloud}} = \frac{3}{2} \frac{ktM}{\overline{m}}
\end{align}

(b) What is the total gravitational binding energy of the cloud of mass $M$?

The total gravitational binding energy of a cloud is simply the gravitational potential energy, $U$ of all the particles in a system, which we derived in Worksheet 8, Problem 2, to be:

\begin{align}
U = -\frac{GM^2}{R}
\end{align}

(c) Relate the total thermal energy to the binding energy using the Virial Theorem, recalling that you used something similar to kinetic energy to get the thermal energy earlier.

We can relate the total thermal energy $E_{thermal}$ of a system to the total binding energy $U$ using the Virial Theorem. The Virial Theorem states:

\begin{align}
K = -\frac{1}{2}U
\end{align}

Since the thermal energy is a result of the kinetic energy, and the binding energy is a result of the gravitational potential energy, we can substitute the kinetic energy and the potential energy of the Virial Theorem with the values we got for the thermal energy and gravitational binding energy and simplify to get the relation between the two:

\begin{align}
\frac{3}{2} \frac{ktM}{\overline{m}} &= -\frac{1}{2}(-\frac{GM^2}{R})\\
\frac{3kT}{\overline{m}} &= \frac{GM}{R}\\
3kT &= \frac{GM \overline{m}}{R}
\end{align}

(d) If the cloud is stable, then the Virial Theorem will hold. What happens when the gravitational binding energy is greater than the thermal (kinetic) energy of the cloud? Assume a cloud of constant density $\rho$.

In order to understand this question, let's think about the two components of the Virial Theorem. When a cloud is stable, the Virial Theorem holds true, such that $K = -\frac{1}{2}U$. In a stable system where the Virial Theorem holds true, the kinetic energy of the entire system balances with the total gravitational potential energy of the system. Kinetic energy $K$ has the tendency to make things spread apart, since the more kinetic energy there is, the more particles move around and disperse radially outward in a diffusion-like pattern. Gravitational potential energy $U$ has the tendency to bring particles come together. Under the Universal Law of Gravitation, any two objects with mass will attract one another. Therefore, in a stable molecular cloud, the individual gas particles are in a tug-of-war between the outward expansion caused by thermal (kinetic) energy, and the inward pull of gravitational potential energy that wants to bring all the particles together.

When the gravitational potential energy (binding energy) is greater than the thermal (kinetic) energy, the tug-of-war is over in favor of gravity. As a result, the inward pull by gravitational potential energy overcomes the outward expansion by kinetic energy, and the gas particles have a net movement inwards. This is when a molecular cloud "collapses" under the weight of its own gravity.

(e) What is the critical mass, $M_J$, beyond which the cloud collapses? This is known as the "Jeans Mass."

In the case of a cloud collapsing under the  weight of its own gravity, we know that the gravitational potential energy of the system is greater than the kinetic energy of the system. We can use the equation from part (c) to model this scenario, except instead of having a stable system where the kinetic energy is equal to the gravitational potential energy, in this case, the gravitational potential energy is greater than the kinetic energy:

\begin{align}
3kT &< \frac{GM \overline{m}}{R}
\end{align}

Now, based on this equation, the gravitational potential energy is only determined by the radius $R$ of the molecular cloud, since the mass of the molecular cloud $M$, and the mass of the individual gas particles $\overline{m}$ are constant along with the gravitational constant.

So now that we know that the gravitational potential energy is determined only by the radius of the molecular cloud. But if the molecular cloud changes its radius because the gas particles are coming closer together since the gravitational potential energy is greater than the kinetic (thermal) energy, then the density of particles in the cloud $\rho$, changes! Therefore, we can relate density and radius using the definition of density as mass over unit volume: $\rho = \frac{mass}{volume}$.

\begin{align}
\rho = \frac{M}{\frac{4}{3} \pi R^3}
\end{align}

And solving for radius, we get:

\begin{align}
R = \left(\frac{3M}{4\rho \pi}\right)^{\frac{1}{3}}
\end{align}

Since we are looking for the critical mass, $M_J$, we can rearrange the equation from part (c), and substitute radius $R$, to solve for the mass:

\begin{align}
3kT &= \frac{GM_J \overline{m}}{R}\\
M_J &= \frac{3kTR}{G\overline{m}}\\
M_J &= \frac{3kT}{G\overline{m}} \cdot \left(\frac{3M}{4\rho \pi}\right)^{\frac{1}{3}}\\
M_J^3 &= \left(\frac{3kT}{G\overline{m}}\right)^3 \left(\frac{3M}{4\rho \pi}\right)\\
M_J^2 &= \left(\frac{3kT}{G\overline{m}}\right)^3 \left(\frac{3}{4\rho \pi}\right)\\
M_J &= \left(\frac{3kT}{G\overline{m}}\right)^{\frac{3}{2}} \left(\frac{3}{4\rho \pi}\right)^{\frac{1}{2}}
\end{align}

(f) What is the critical radius, $R_J$, that the cloud can have before it collapses? This is known as the "Jeans Length."

To find the critical radius, we simply need to solve for the radius $R_J$ from the equation we derived in part (c), and substitute using the density equation.

So, first, let's rearrange the equation from part (c) to solve for $R_j$:

\begin{align}
3kT &= \frac{GM \overline{m}}{R_J}\\
R_J &= \frac{GM \overline{m}}{3kT}
\end{align}

Then, let's rearrange the density equation to solve for $M$ so we can substitute it into the previous equation:

\begin{align}
\rho &= \frac{M}{\frac{4}{3} \pi R_J^3}\\
M &= \frac{4}{3} \rho \pi R^3
\end{align}

Now we can substitute $M$ for the first equation, and solve for the critical radius, $R_J$:

\begin{align}
R_J &= \frac{GM \overline{m}}{3kT}\\
R_J &= \frac{G \left(\frac{4}{3} \rho \pi R_J^3\right) \overline{m}}{3kT}\\
3kTR_J &= G \left(\frac{4}{3} \rho \pi R_J^3\right) \overline{m}\\
3kT &= G \left(\frac{4}{3} \rho \pi R_J^2\right) \overline{m}\\
R^2 &= \frac{3kT}{\frac{4}{3} G \rho \pi \overline{m}}\\
\sqrt{R^2} &= \sqrt{\frac{9kT}{4 G \rho \pi \overline{m}}}\\
R &= \frac{3}{2} \sqrt{\frac{kT}{G \rho \overline{m}}}
\end{align}

Worksheet 8, Problem 4- The Virial Theorem: The M80 Star Cluster

Question: The cluster M80 has an angular diameter about 10 arcminutes and resides about $10^4$ parsecs from the Sun. The average speed $<v> \approx 10$ $ km$ $ s^{-1}$. Approximate how much mass, in solar masses $M_{\bigodot}$, the cluster contains. 

Let's draw this scenario to see if we can get a better understanding of this situation:

In this scenario, we know the distance, $a$ from the Sun to M80 to be $10^4$. We also know that M80 has an angular diameter of 10 arcminutes, therefore, the angle $\theta = 10'$.

Since we are trying to solve for the mass of the M80 cluster, we can use the equation we derived in the previous blog post that relates mass of the cluster $M$ to the average velocity of the individual stars $<v>$ and the radius of the star cluster $R$:

\begin{align}
M = \frac{<v>^2R}{G}
\end{align}

Because we need to know the radius of the star cluster, we can refer back to the diagram and see that instead of using the angular diameter, $\theta = 10'$, we can use $\alpha = 5'$ to get the angular radius.

Using the angular radius, we have a trigonometric problem:

Since we are solving for the radius $R$, we can use trigonometry:

\begin{align}
\tan(\alpha) = \frac{R}{a}
\end{align}

Knowing that the angular radius of the star cluster, $\alpha = 5'$ is a very small angle, we can approximate using the small-angle approximation $tan(\alpha) = \alpha$ to solve for the radius of M80 $R$:

\begin{align}
\alpha &= \frac{R}{a}\\
R &= \alpha \times a\\
R &= 5' \times 10^4 parsecs\\
R &= 5' \cdot \frac{2.9 \times 10^{-4} rads}{1'} \times 10^4 parsecs \cdot \frac{3.09 \times 10^{18} cm}{1 parsec}\\
R &= 4.48 \times 10^{19} cm
\end{align}

Okay, so we now have the radius of the M80 cluster. We also know the average velocity of the individual stars in M80 cluster, so we can directly solve for the mass of the star cluster using the equation we derived earlier:

\begin{align}
M &= \frac{<v>^2 R}{G}\\
M &= \frac{(10 \frac{km}{s} \cdot \frac{10^5 cm}{1 km})^2 (4.48 \times 10^{19} cm)}{6.67 \times 10^{-8} \frac{cm^3}{g \cdot s^2}}\\
M &= \frac{(1 \times 10^12 \frac{cm^2}{s^2}) (4.48 \times 10^{19} cm)}{6.67 \times 10^{-8} \frac{cm^3}{g \cdot s^2}}\\
M &= \frac{4.48 \times 10^{31} \frac{cm^3}{s^2}}{6.67 \times 10^{-8} \frac{cm^3}{g \cdot s^2}}\\
M &= 6.72 \times 10^{38} \text{ g}
\end{align}

So the mass of the supercluster comes out to be $6.72 \times 10^{38}$ grams. However, we want this answer in terms of solar mass. Since the solar mass, $M_{\bigodot}$ of the Sun is approximately $M_{\bigodot} \approx 2 \times 10^{30}$.

Therefore,the mass of the cluster M80 is $\frac{M}{M_{\bigodot}} \approx 3.38 \times 10^5$. In other words, $M \approx 3.38 \times 10^5 M_{\bigodot}$.

Worksheet 8, Problem 3 - The Virial Theorem: All About that Mass

Question: If the average speed of a star in a cluster of thousands of stars is $<v>$, give an expression for the total mass of the cluster in terms of $<v>$, the cluster radius, $R$, and the relevant physics constants.

Okay, so we are not given a lot of information here. Let's extrapolate from the information given to us to see if there is more things we can find out about this scenario. We know we are dealing with a star cluster, so let's draw one below:

A star cluster typically has thousands of stars, each with a mass, $m$. To account for the mass of the entire star cluster, we need to sum the masses of each individual star, $m_i$, such that the mass of the entire cluster is represented by: $M = \sum_0^n m_i$, where $n$ is the number of stars in the star cluster, $m_i$ is the mass of any given individual star in the star cluster, and $M$ is the total mass of the entire star cluster.

Now that we can represent the mass of the entire cluster as $M$, let's consider the energy present in the star cluster. Everything in the universe has gravitational potential energy, and everything not at absolute zero has kinetic energy. Let's try to quantify the kinetic and potential energy of this star cluster.

The kinetic energy $K$ of the star cluster is given by: $K = \frac{1}{2}M<v>^2$, where $M$ is the mass of the star cluster, and $<v>$ is the average speed of an individual star in the cluster.

The potential energy $U$ of the star cluster is given by $U = -\frac{GM^2}{R}$, where $M$ is the mass of the star cluster, $R$ is the radius of the star cluster, and $G$ is the universal gravitational constant.

Knowing the average velocity of the individual stars in the star cluster $<v>$, and the radius of the star cluster $R$, we can use the Virial Theorem, which relates the total kinetic and potential energies of a stable system like the star cluster system, to solve for the mass of the entire cluster:

\begin{align}
K &= -\frac{1}{2}U  \qquad \leftarrow \text{Virial Theorem}\\
\frac{1}{2}M<v>^2 &= -\frac{1}{2}(-\frac{GM^2}{R})\\
<v>^2 &= \frac{GM}{R}\\
M &= \frac{<v>^2 R}{G}
\end{align}

We have derived the equation to solve for the total mass $M$ of a star cluster given it's radius $R$ and the velocity of the average star in the star cluster $<v>$ using the Virial Theorem!

Worksheet 9, Problem 1- Star Formation [I'm All About that Space]

Source: https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiYS5Ml9AzLuW7B64Or5gFWJdTQZhraug5UV6ZcSrUPVxN3_lonHO1hI6bi4T9Lu1qZS_cETidYr2xJeKtMlqu-_SO6BErGhQ0oVIolJpsIBva6JXWZ6boI3kaGgbraI2mAoZURsm5eSg8/
s640/The_Most_Dangerous_Place_in_the_Universe__3_3__medium%25282%2529.gif

Question: The spacial scale of star formation: The size of a modest star forming molecular cloud, like the Taurus region, is about 30 pc. The size of a typical star is, to an order of magnitude, the size of the Sun.

(a) If you let the size of your body represent the size of the star forming complex, how big would the forming star be? Can you come up with an analogy that would help the layperson understand this difference in scale? For example, if the cloud is the size of a human, then the star is the size of what?

Okay, so if we're trying to get an understanding of the scale of the molecular cloud in comparison to the central star, let's create an analogy to the human body. Since the height of the average human male is $175$ cm, the diameter of a hypothetical spherical human would also be $D_{human} = 175$ cm.

Let's compare the proportion of the human body to some unknown part of the human body, $D_{x}$, such that the ratio is the same as the ratio between the diameter of the central star in a molecular cloud, and the Taurus region of the cloud. The proportionality can be set up as follows:

\begin{align}
\frac{D_{human}}{D_{x}} &= \frac{D_{taurus}}{D_{star}}\\
\end{align}

We can rearrange the previous equation to solve for $D_x$. We also know the diameter of a hypothetical human body to be about $175$ cm. The problem tells us that the average star in the middle of a molecular cloud is about the same size as the Sun, which has a diameter of $1.39 \times 10^{11}$ cm. We are also given the size of the molecular cloud, which is $30$ pc. Knowing all this information, we can plug in the values directly and solve for the proportional diameter of an unknown part of the human body, $D_x$:

\begin{align}
D_x &= \frac{D_{human} \cdot D_{star}}{D_{taurus}}\\
D_x &= \frac{(175 \text{ cm}) \cdot (1.39 \times 10^{11} \text{ cm})}{30 \text{ pc} \times \frac{3.09 \times 10^{18} \text{ cm}}{1 \text{ pc}}}\\
D_x &= 2.62 \times 10^{-7} \text{ cm}
\end{align}

As you can see, if an average human, which represents the star forming molecular cloud, has a diameter of $175$ cm, then the star that in the molecular cloud will be of the size $2.62 \times 10^{-7}$ cm, or $2.62 \times 10^{-9}$ meters, which is about as big as a strand of DNA.

Other things that exist at this scale are:

Scale of the Universe 2: http://htwins.net/scale2/

(b) Within the Taurus complex there is roughly $3 \times 10^4 M_{\bigodot}$ of gas. To order of magnitude, what is the average density of the region? What is the average density of a typical star (use the Sun as a model)? How many order of magnitude difference is this? Consider the difference between lead ($\rho_{lead} = 11.34$ $g$ $cm^{-3}$) and air ($\rho_{air} = 0.0013$ $g$ $cm^{-3}$ ). This is four orders of magnitude, which is a huge difference!

Okay, so in order to compare the densities of the Taurus complex and a Sun-like star (the Sun has a mass of $1.99 \times 10^{33}$ g), let's try to find the density of the Taurus complex first. We know that the density $\rho$ is mass per unit volume, which can be represented by $\rho = \frac{mass}{volume}$. We can approximate the Taurus-complex as a spherical body with a radius $15$ pc to get it's volume.

\begin{align}
\rho_{taurus} = \frac{M_{taurus}}{V_{taurus}} = \frac{M_{taurus}}{\frac{4}{3} \pi r_{taurus}^3} = \frac{3 \times 10^4 M_{\bigodot}}{\frac{4}{3} \pi (15 \text{ pc} \cdot \frac{3.09 \times 10^{18} \text{ cm}}{1 \text{ pc}})^3} = \frac{3 \times 10^4 ( 1.99 \times 10^{33} \text{ g})}{\frac{4}{3} \pi (15 \text{ pc} \cdot \frac{3.09 \times 10^{18} \text{ cm}}{1 \text{ pc}})^3} = 1.4 \times 10^{-22} \frac{g}{cm^3}
\end{align}

Now that we have the density of the Taurus-complex, let's solve for the mass of a typical star, similar to the Sun. The mass of the Sun is $1.99 \times 10^{33}$ g, and the radius of the Sun is $6.97 \times 10^{10}$ cm. We can solve for the density of the Sun using the formula $\rho = \frac{mass}{volume}$ and plug the values for mass and radius directly:

\begin{align}
\rho_{\star} = \frac{M_{\star}}{V_{\star}} = \frac{M_{\star}}{\frac{4}{3} \pi r_{\star}^3} = \frac{1.99 \times 10^{33} \text{ g}}{\frac{4}{3} \pi (6.97 \times 10^{10} \text{ cm})^3} = 1.4 \frac{g}{cm^3}
\end{align}

As you can see, the density of a star is 22 orders of magnitude larger than that of a Taurus-complex. To put that in perspective, think air and lead differ in densities by an order of 4 magnitudes!

Worksheet 7, Problem 1: An Intro to Hydrostatic Equilibrium

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 $\Delta r \ll r$ 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: 

\begin{align}
 \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}