Day Lab Final Analysis

After collecting data and making calculations on the angular size of the sun, the rotational speed of the sun, and the rotational period of the sun, we are now ready to determine the distance to the sun!

The first step in completing this is two calculate the solar radius:
\[V_{radial} = \frac{2 \pi R}{P}\] \[R = \frac{V_{radial} \times P}{2 \pi} \] Let's consider our knowns: \[ P = 27.57 \, days = 2.38 \times 10^6 \, seconds\] \[V_{radial} = 1.36 \, km/s = 1.36 \times 10^5 \, cm/s\] \[ \theta = 0.56 \, degrees  = .0098 \, radians\] Now, solving for R, we get: \[ R = \frac{1.36 \times 10^5 \, cm/s \times 2.38 \times 10^6 \, seconds}{2 \pi} = 5.15 \times 10^{10} \, cm \]

Next, can use the small angle approximation to determine the distance to the sun. Below is a diagram of what we are measuring:

Using basic trigonometry, we know that the above relationship holds: \[sin( \theta ) = \frac{2R}{AU}\] Additionally, since theta is such a small angle, we can use the following approximation: \[sin( \theta ) \approx \theta \] We now have the following relationship" \[ \theta = \frac{2R}{AU} \] Now, solving for AU, we get: \[ AU = \frac{2R}{ \theta} = \frac{2 \times 5.15 \times 10^{10} \, cm}{.0098 \, radians} = 1.05 \times 10^{13}  \, cm \pm 1 \times 10^{12} \, cm\]
Compared to the actual measurement of the AU, which is \(1.495 \times 10^{13} \, cm \) we had a percentage error of 30%. In the lab, there were several potential sources of error that could help account for this. In the section where we determined the rotational speed of the sun, we had to round several values, and make some eyeball approximations on the plots. Additionally, the Telluric offset may not have been completely accurate, which would completely effect the doppler effect measurement. Another source of error could likely come from our tracers on the sunspots. To mark the spots, we used a market with a slightly broad tip, which would potentially give inaccurate locations.

Day Lab Part 3: Rotational Period of the Sun

The motion of the sun in the sky tells us something about the rotation of the Earth, and the Earth's orbit around the Sun. An important motion of the sun itself is its rotation about its own axis. This can be measured by using sunspots as tracers of that rotation, as Galileo first did in 1612. 

By capturing images of the sun over time, we can use distinct sun spots to follow the rate of its rotation, or its rotational period. To do this, we used images of the sun, and laid a transparency over each image, marking distinct sunspots on a coordinate system. Below is an image of the sun with the overlaid transparency.

After marking the spots, we then removed the transparency, and collected the data on how many degrees each spot moved.

Finally, we used the data and generated average rotational rates for each spot.

To arrive at our final number, we averaged the three values from each sun spot to get an average rotation of 27.57 earth days per rotation. 

Day Lab Part 2: Rotational Speed of the Sun

The doppler shift allows us to measure radial velocity. We will make several spectral measures of the NaD lines (at 5889 and 5896 Angstroms) from the East limb of the Sun and then the West limb of the Sun. Then with the CCD (and attached microscope lens) we take careful measurements of the Doppler shifts of these two lines relative to that of a Telluric absorption line, which arised from H20 in the Earth's atmosphere. Because we do not know the orientation of the solar equator, we will take 8 measurements around the edge in pairs (left, right; top, bottom; top, right; bottom, left; top, left; bottom right).

To begin, we used several mirrors to focus the light into the spectrophotometer. This device breaks up the light into its component wavelengths, which allows us to see each of their corresponding intensities using a CCD. Below is the layout of the experiment setup:

To generate a sort of coordinate system of the sun, we took measurements at 8 different locations. Once we collected the measurements for each location, we entered the data into an excel spreadsheet for analysis. Below is the Normalized plot of spectral lines:

Normalized Plot of Spectral Lines

We can notice from the plot that there are two distinct spikes in the lines. These depict the sodium absorption lines, where we expect to see the doppler shift. Since we wanted to find the lines that were on opposite sides of the sun, we looked at the minimum values, and chose the lines with the greatest separation. From our plots, we found these to be the top and bottom data sets.

We then plotted these values, and fit a curves to each dataset. These are shown below:

Rather than simply taking the difference, however, we also needed to take into account any shift that should not be attributed to the doppler effect. To do this, we also considered the Telluric absorption lines that should remain the same in both graphs. Below is the plot of the Telluric absorption lines. 

We found the telluric shift to have an offset of 0.9288 pixels, which we factored into the calculation for final rotational velocity. Overall, we found a shift of approximately 3 pixels per dataset, which resulted in a final rotational velocity of: 1.36 km/s

Day Lab Part 1: Measuring the Angular Size of the Sun

The sun appears to move all the way around the Earth (360 degrees) in 24 hours. If we measure how long (minutes and seconds) the sun takes to move its own diameter along the sundial in our lab, we can measure its angular diameter, in degrees. 

To conduct this portion of the lab, we focused the sun's light with a lens onto a large white sheet of paper. This acted as our sundial. Below is a diagram of the setup: 

As a group, we then marked the two extremes of the sun, with a corresponding time. We then allowed the sun to pass one full diameter, and again noted the time. After conducting these measurements several times, we were then able to determine the suns angular diameter using some simple geometry. The average time we measured for one diameter to pass was: 2 min 15 seconds, or 135 seconds. 

From this, we were then able to calculate the angular diameter of the sun, \( \theta\). \[ \frac{24 \, hours}{360 \, degrees} = \frac{diameter \, time}{ \theta }\] \[ \frac{24 \, hours}{360 \, degrees} = \frac{135 \, seconds}{ \theta }\] 
\[ \theta = \frac{135 \, seconds \times 360 \, degrees}{ 24 \, hours \times \, 60 \, minutes \times 60 \, seconds} = 0.56 \, degrees \]

Dark Matter in the Coma Cluster

http://gregg.physics.ucdavis.edu/gregg/coma/coma.html

The Coma Cluster represents the nearest developing cluster of galaxies to earth, at a distance of approximately 100 Mpc. Although it was originally considered to be in virial equilibrium, the Coma Cluster is actually highly dynamic. Astronomers have identified more than 1,000 galaxies within the Coma Cluster! (http://gregg.physics.ucdavis.edu/gregg/coma/coma.html)

From Zwicky's publication, we are given several bits of information about the cluster. This is outlined below:

• Differences in speed of 1500 to 2000 km/sec
• \(R_{cluster} = 1 \, million \, light \, years = 10^{24} \, cm \)
• Contains 800 individual Nebulae
• Each nebulae has a mass of \(M_{nebulae} = 10^9 \, solar \, masses\)

Although Zwicky calculated out the average velocities of the stars, in his paper, we are going to reproduce his results using our expert method!

First, we'll sketch out the system: 

In our sketch, each of the nebulae is represented by a star in motion. To get started, we'll begin by assuming the system is in a state of equilibrium, and by applying the virial theorem. \[ K = - \frac{1}{2}U\] Now let's expand the expressions for both K and U: \[K = \frac{1}{2}M \bar{v}^2 \] \[U = - \frac{GM^2}{R}\] Inserting these into the virial theorem equation, we get: \[ \frac{1}{2}M \bar{v}^2 = \frac{1}{2} \frac{GM^2}{R} \] Solving for the average velocity, we get: \[ \bar{v} = \sqrt{\frac{GM}{R}} \] Now, we can apply the values that Zwicky has provided for the Coma Cluster to calculate this average velocity! \[ \bar{v} = \sqrt{\frac{(6.67 \times 10^{-8}) \times (800 \times 10^9) \times (2 \times 10^{33}) }{10^{24}}} \approx 3 \times 10^6 \, \frac{cm}{sec} \approx 30 \, \frac{km}{sec} \]

On the surface, this calculation is a bit concerning. According to Zwicky, observations have revealed differences in speed between 1500 to 2000 km/sec. Even if we adjusted our assumption that this system is in equilibrium, the change would only change by a relatively small factor. Our calculation, however, is off my several orders of magnitude.

Quite remarkably, this disparity is where Zwicky made an amazing discovery. All of our calculations are based solely on visible mass, but clearly, there is something causing the velocities to be much greater than expected. Zwicky credits this difference to some additional mass that is not visible. This appropriately named matter has been dubbed "Dark Matter."

Another hypothesis tested by Zwicky from these calculations is how the apparent velocities of the Coma Cluster would be attributed to the Einstein redshift. This is given by the relative change in wavelength shown below (considering just visible mass): \[ \frac{ \Delta \lambda } { \lambda} \sim - \frac{e_p}{c^2} \sim 3.5 \times 10^{-8} \] This translates to a speed of ~ 10 m/s. Again, for the immense different in velocity to be accounted for, there would need to be an immense amount of dark matter for the system to hold!

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 (\(cm^3\) ) of gas particles within a cloud of mass M comprising particles of mass \(\bar{m}\) (mostly hydrogen molecules, \(H_2\)), and k is the Boltzmann constant, \(k = 1.4 \times 10^{16} erg \, K^{-1}\) .

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

To calculate the total thermal energy, K, we must consider the movement of particles in each direction - x, y, and z. to account for this, we can multiply the velocities of each particle by the 3 axis: \[Single \, particle \, thermal \, energy  = 3 \times (\frac{1}{2} m{v_i}^2)  = 3 \times \frac{1}{2} kT\] To then find the total thermal energy of all particles, we will sum over all particles in the system: \[ K = \sum_{n=0}^{\infty} \frac{3}{2} kT \]
(b) What is the total gravitational binding energy of the cloud of mass M?

To determine the total gravitational binding energy of the cloud, we can begin by assuming that the cloud is generally in the shape of a sphere with a radius of R. Now, we can represent the binding energy as the gravitational potential energy of the complex. This is represented as: \[ E_{binding} = G \frac{M^2}{R} \]
(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.

According to the Virial Theorem: \( K = - \frac{1}{2}U \)

This represents a state of equilibrium, in which the potential energy of the system equals the gravitational binding energy. As we set these equal, we want to first assure that we are considering these values for the entire system, and not just a single particle. Recall that our total thermal energy was represented by:  \[ K = \sum_{n=0}^{\infty} \frac{3}{2} kT \] We can also represent this sum by: \[ K = \frac{M}{\bar{m}} \frac{3}{2} kT \] Now, plugging this into the Virial Theorem we achieve: \[ \frac{M}{\bar{m}}\frac{3}{2}kT = -G\frac{M^2}{2R}\] \[ \frac{M}{\bar{m}}3kT = -G\frac{M^2}{R}\] (d) If the cloud is stable, then the Viriral 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 \( \rho \).

If the gravitational binding energy is greater than the thermal (kinetic) energy, then the cloud will begin to collapse and begin the process of star formation. As this occurs, particles pull closer together, gradually reducing the radius of the cloud. This produces the following inequality:
\[ \frac{M}{\bar{m}}3kT<G\frac{M^2}{R}\]
(e) What is the critical mass, \( M_J \) , beyond which the cloud collapses? This is known as the “Jeans Mass.”

In our equation from part (d), we can substitute \(M_J\) for M to get the following: \[ \frac{M_J}{\bar{m}}3kT<G\frac{{M_J}^2}{R}\] To determine the critical mass (Jeans Mass), we can solve for \(M_J\): \[ M_J = \frac{3R}{\bar{m} G} kT \]Additionally, we can represent the mass in terms of density using the following relationship: \[ Volume = \frac{M}{ \rho } = \frac{4}{3} \pi R^3 \] \[ M = \rho V = \frac{4}{3} \pi \rho R^3\] Setting this mass expression equal to our critical mass, \(M_J\), we get the following: \[ \frac{4}{3} \pi \rho R^3 =  \frac{3R}{\bar{m} G} kT \] Then, solving for R, we get: \[ R^2 = \frac{9kT}{4 \pi \rho \bar{m} G} \] \[ R = \frac{3}{2} \sqrt{\frac{kT}{\pi \rho \bar{m} G}} \] Now, we can represent \(M_J\) in terms of density instead of R, by plugging in our radius equation: \[ M_J = \frac{9   \sqrt{\frac{kT}{ \pi \rho \bar{m} G}}   }{2 \bar{m} G} kT  = \frac{9}{2} \sqrt{ \frac{k^3 T^3}{ \pi \rho \bar{m}^3 G^3}}\] For this limit, if mass were to be any greater than the Jeans Mass, the cloud would collapse.

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

From part (e), we have already solved \(R_J\) to be: \[ R_J = \frac{3}{2} \sqrt{\frac{kT}{ \pi \rho \bar{m} G}} \] This limit entails that a cloud with a radius greater than Jeans Length would collapse.

Acknowledgements: Johnathan Budd & Willie Pirc

Worksheet 9, Problem 1: Star Scale

The spatial 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 stars be? Can you come up with an analogy that would help a layperson understand this difference in scale? For example, if the cloud is the size of a human, then a star is the size of what? 

In an attempt to understand the extraordinary size of a star forming cloud, we aim to consider what size a newly formed star represents within its parent cloud, with respect to the size of a human being. For this problem, we will scale the star forming cloud to the size of a person. Let's examine just what size a star would be now!

The first step for approaching this problem involves drawing an appropriate figure:

From this, we know the size of an average person (not Shaq!) and how to calculate the size of a star forming complex and the size of our sun. The final piece that we will need is the size of our unknown green object. To quantify their relationship, we can establish the following ratios: \[ \frac{V_{cloud}}{V_{sun}} = \frac{V_{person}}{V_{unknown}} \] For ease of calculation, we'll consider each object as a sphere, with the volume of a sphere as: \[ V = \frac{4}{3} \pi r^3 \] Applying this to our ratio, we reach: \[ \frac{ \frac{4}{3} \pi r_{cloud}^3}{\frac{4}{3} \pi r_{sun}^3} = \frac{ \frac{4}{3} \pi r_{person}^3}{ \frac{4}{3} \pi r_{unknown}^3} \] Simplifying, we reach:  \[ \frac{r_{cloud}}{r_{sun}} = \frac{r_{person}}{r_{unknown}} \] Next, we solve for our unknown r: \[ r_{unknown} =  \frac{r_{person} \times r_{sun}}{r_{cloud}}\] With our equation set, we can now begin to plug in what we know: \[ r_{cloud} = 30 \, pc \times \frac{3 \times 10^{18} \, cm}{1 \, pc} = 9 \times 10^{19} \, cm \] \[ r_{sun} = 7 \times 10^{10} \, cm\] \[ r_{person} \approx 100 \, cm \] Solving for the unknown r, we get: \[ r_{unknown} =  \frac{(100 \, cm) \times (7 \times 10^{10} \, cm)}{9 \times 10^{19} \, cm} = 1 \times 10^7 \, cm \] \[ r_{unknown} = 1 \, nm\] We can now update our figure to show the appropriate values:

This radius corresponds approximately to the size of an amino acid, one of the simplest building blocks of life.

(b) Within the Taurus complex there is roughly \(3 \times 10^4 M_{sun}\) 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 orders 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!

First, let's calculate the density of the Taurus complex. We are given that the total mass is \(3 \times 10^4 M_{sun}\). To find the density, we can divide this by the overall (spherical estimate) volume of the complex. \[ \rho = \frac{Total \, Mass}{Volume} = \frac{(3 \times 10^4) \times ( 2 \times 10^{33} \, g)}{\frac{4}{3} \pi (9 \times 10^{19} \, cm)^3} \] \[ \rho_{complex} = 2 \times 10^{-23} \approx 10^{-22} \, \frac{g}{{cm}^3}\] Next, we can calculate the density of a star, using the sun as a model: \[ \rho = \frac{Sun \, Mass}{Volume} = \frac{2 \times 10^{33}\, g}{\frac{4}{3} \pi (7 \times 10^{10} \, cm)^3} \] \[ \rho_{sun} = 1.39 \approx 1 \, \frac{g}{{cm}^3} \] In comparing these two values, we see that there is a difference of 22 orders of magnitude. Putting this in the context of air and lead differing by only 4 orders of magnitude, it isn't surprising that as a star forms, it pulls in the massive amounts of dust and gas with its gravitational pull!

Acknowledgements: Johnathan Budd & Willie Pirc

Worksheet 8, Problem 3: Star Cluster

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 physical constants.

Let's start again by drawing out our system:

Let's start by recalling the virial theorem: \( K = - \frac{1}{2} U \). Now, let's write out the full expressions for both Kinetic Energy, K, and Potential Energy, U. \[ K = \frac{1}{2} M v^2 \] \[ U = -G \frac{M^2}{R} \] In these expressions, M is equal to the total mass of all stars in the system, and v is the average velocity of the stars. Now, relating our two energies using the virial theorem, we get: \[ \frac{1}{2} M v^2 = \frac{1}{2} G \frac{M^2}{R} \] Solving for M, we get: \[ M = \frac{Rv^2}{G} \]

Acknowledgements: Johnathan Budd & Willie Pirc Team EE

Worksheet 8, Problem 2: Going Virial

Consider a spherical distribution of particles, each with a mass \(m_i\) and a total (collective) mass \(\sum_{i}^{N} m_i = M\), and a total (collective) radius R. Convince yourself that the total potential energy, U, is approximately \[U \approx - \frac{GM^2}{R} \]
To begin this problem, let's start by drawing our system:

Now, let's recall Kepler's Law: \[ \frac{p^2}{R^3} = {4 \pi^2}{GM} \] Rearranging terms, we achieve:  \[ \frac{GM^2}{R} = {4 \pi^2 R^2 M}{p^2} \] Next, we'll want to consider what assumptions we can make about this system. First, we know that the volume of a sphere is represented by: \(V = \frac{4}{3} \pi r^3\). From this, we can draw that for a sphere, the majority of the volume is achieved at the greatest r values (near the surface). For this derivation, we'll assume that the majority of the particles are also at an average distance of R. Based on this, we can approximate the average period of the particles to be: \[average \, period = \bar{p} = \frac{2 \pi R}{\bar{v}} \] Next, we can use this to solve for average velocity, \( \bar{v} \). \[ \bar{v} = \frac{2 \pi R}{\bar{p}} \] To relate this average velocity of a particle to energy, we can now consider the kinetic energy, K, for a particle: \[ K = \frac{1}{2} m \bar{v}^2 \] Plugging in our average velocity of all particles, we get the total kinetic energy: \[ K = \frac{1}{2} M (\frac{2 \pi R}{\bar{p}})^2  = \frac{2M \pi^2 R^2}{\bar{p}^2} \] Next, we can apply the virial theorem. We recall this as: \[K = \frac{1}{2}U\] \[or\] \[U = -2K\] Applying this to our K equation from above, we achieve: \[U =   \frac{4M \pi^2 R^2}{\bar{p}^2} \] Recalling our rearranged form of Kepler's Law, we notice that this expression is equal to: \[ - \frac{GM^2}{R} \] And thus, our assertion is confirmed! The total potential energy, U, is approximately \[U \approx - \frac{GM^2}{R} \]

Acknowledgements: Johnathan Budd & Willie Pirc Team EE

Worksheet 7, Problem 1: Hydrostatic Equilibrium

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 \(cm^{-3}\)), \(k = 1.4 \times 10^{-16} erg \, K^{-1}\) 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 \(\delta r << r\) 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 (\(P_{down} = P( r + \delta  r)\)). Make a drawing of this, and discuss the situation and the various physical parameters with your group.

We can represent the small cylindrical parcel of gas with our drawing below. The cylinder has a height of dr, is positioned a distance r away from the earth's center, and has a cross sectional area, A. As we consider the forces acting on the cylinder, there are two primary forces at play. The first is the pressure downwards, \(P_{down} = P( r + \delta  r)\), and the second is the force upwards, \(P_{up} =  P(r)\).

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

Beyond the upward and downward pressure effecting the cylinder, the cylinder is acted upon by the gravitational pull of earth. Before calculating this, we can use what we know about the volume of the cylinder and its density to calculate it's mass, m. The mass is calculated as follows: \[m = \rho(r) \times A \times dr\]Given that the gas has a mass of m, and the earth has a density of M, the cylinder will feel the following force: \[ F_g = G \frac{mM}{r^2}\] We can also write this by substituting in the density and volume of the cylinder for the mass, m. \[ F_g = G \frac{\rho(r) A dr M}{r^2}\] Adding this to our sketch, our drawing now shows all three forces acting on the cylinder.

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 cylinder is stationary in this situation, we can assume that all forces cancel out. This means that the upwards force of pressure (pressure x area) will be equal to the downward force of pressure (pressure x area) and gravity. \[P_{up} \times A = P_{down} \times A + F_{g}\]
d) Give an expression for the gravitational acceleration, g, at at a distance r above the Earth’s center in terms of the physical variables of this situation.

To derive this expression, we can begin by fundamentally realizing that gravity is a force. Since it is a force, we know that it can be expressed by \(F = m \times a\). Setting this force equal to the gravitational force we derived in part B, we have: \[ F_g = ma = mg =  G \frac{mM}{r^2}\] Then by dividing both sides by the mass of the cylinder, we arrive at: \[g =  G \frac{M}{r^2}\] e) Show that \[\frac{dP(r)}{dr} = -g \rho (r)\] This is the equation of hydrostatic equilibrium.

Beginning with our equation for forces being applied to our gas we have: \[P_{up} \times A = P{down} \times A + F_{g}\] \[P(r) \times A = P(r + dr) \times A + A dr \rho (r) g\] Since the area is common to each term, we can divide this out: \[P(r) = P(r + dr) + dr \rho (r) g\] Then, by rearranging out terms, we can put the equation in the form of the (negative) derivative definition: \[\frac{P(r) - P(r + dr)}{dr}= \rho (r) g\] This then simplifies to: \[- \frac{dP(r)}{dr} = \rho(r) g\] or \[ \frac{dP(r)}{dr} = - \rho(r) g\] 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, \(\rho (r)\)? (HINT: It may be useful to recall that \(\frac{dx}{x} =  d ln (x)\)

Beginning with the ideal gas law declared above, we have: \[P(r) = nk_BT\] As we try to relate density to this equation, we can recall that the density and number density of particles relates with the following equation: \[ \rho(r) = \bar{m}n\] Where (\ \={m} \) represents the average molecular weight of particles. Plugging this into the equation, we now have: \[P(r) = \frac{\rho(r)}{\bar{m}}k_BT\] Solving for density we get:  \[\rho(r) = \frac{P(r) \bar{m}}{k_BT}\] Applying the hydrostatic equilibrium equation of \(\frac{dP(r)}{dr} = -g \rho (r)\) we can substitute for \(\rho(r)\) and integrate on both sides: \[\int_{P(0)}^{P(r)} \frac{dP(r)}{dr}dP(r) = \int_{0}^{r} -g \frac{P(r) \bar{m}}{k_BT}dr\] Using the hint provided in the question that \(\frac{dx}{x} =  d ln (x)\) we can easily solve this integral. \[\int_{P(0)}^{P(r)} \frac{dP(r)}{dr}dP(r) = \int_{0}^{r} -g \frac{P(r) \bar{m}}{k_BT}dr\] \[ln (\frac{P(r)}{P(0)}) = \frac{-g r \bar{m}}{k_BT}\] \[P(r) = e^{\frac{-g r \bar{m}}{k_BT}} P(0)\] Finally, returning to the ideal gas law, we can substitute in \(P(r) = \frac{\rho(r)}{\bar{m}}k_BT\) \[\frac{\rho(r)}{\bar{m}}k_BT = e^{\frac{-g r \bar{m}}{k_BT}} P(0)\] \[ \rho (r) = \frac{\bar{m}}{k_B T} e^{\frac{-g r \bar{m}}{k_BT}} P(0)\]
g) Show that the height, H, over which the density falls off by an factor of 1/e is given by \[H =\frac{kT}{mg}\] 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!

In order to solve for the scale height (which represents the height that the density that changes by a factor of 1/e), we need to take a close look at our solution to part f, which defines how density changes as a function of r. Something that stands out from this is that the only term that varies is: \[e^{\frac{-g r \bar{m}}{k_BT}}\] To find the scale height, we simply need to determine when this term changes by a factor of 1/e. Fortunately, this occurs when \( (\frac{-g r \bar{m}}{k_BT}) = 1\). Solving for r, we get: \[H = r = \frac{k_BT}{g r \bar{m}}\] To confirm our answer, we can plug in the units for each expression. \[cm = \frac{gcm^2/s^2K^{-1}K}{gcm/s^2} = cm\]
h) What is the Earth’s scale height, H? 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, 7 neutrons.

First, we will start with the assumption that the Earth's atmosphere is primarily composed of nitrogen. Now, we want to calculate the average molecular mass of this nitrogen atmosphere by multiplying the number of protons and neutrons in \(N_2\) by the mass of a proton or neutron. \[ \frac{14 \, neutrons + 14 \, protons }{N_2 \, molecule}\times \frac{1.7 \times 10^{-24}g}{proton/neutron} = 5\times 10^{-23}g\] Now substituting this into our scale height at T = 273K, we get:
\[H = \frac{(1.4 \times 10^{-16} erg \, K^{-1}) \times (273K)}{(5 \times 10^{-23}g) \times (9.8 \times 10^2 cm/s^2)} = 8.03 \times 10^5 cm\]

Acknowledgements: Johnathan Budd & Willie Pirc

An Ominous Eclipse

As the world forges on burning fossil fuels and expending non-renewable resources, considerable efforts are being made to promote clean, renewable energy harvesting. Of these, wind and solar have been two forces leading the charge. As shown below, in 2009, renewable energy sources accounted for ~8% of the United States energy production.

http://geology.com/articles/renewable-energy-trends/

Although solar energy hasn't quite reached mainstream status in the United States, Europe has taken on a much fuller adoption in recent years, accounting for 60% of the World's solar use! From the solar farms being built throughout the countryside of the UK (shown below) to photocells being installed on rooftops, solar energy has been gaining steam throughout the EU. 

http://www.aecom.com/Where+We+Are/Europe/Energy/_projectsList/Development+of+solar+farms,+UK

While this may all represent a strong step moving forward in the realm of renewable energy, the European energy grid will face a serious test in the coming month. On March 20th, the Earth will witness it's largest solar eclipse since 1999. While this may on the surface appear to be a relatively minor disruption, it actually poses some major issues. The biggest issue is the timing. While solar cells obviously don't function at night, that also corresponds the lowest energy consumption period of the day. With the solar eclipse occurring in the morning of March 20th, it will coincide with peak energy usage. Energy grid workers are estimating the eclipse to result in a loss of 35,000 MW of solar energy, which is expected to place a major strain on the grid. 

As Gizmodo illustrates the issue, the power grid acts as a river of sorts, that flows at a constant rate. However, when the solar generators stop working, the energy supply must be drawn from a different source, which puts a major strain on the rest of the grid. Additionally, this creates a ripple or wave throughout the grid where the void disrupts the system. 

This solar reliance issue brings to light a large void in technology - large scale battery storage. Currently, there is no truly effective way of storing solar energy, so it must be produced and consumed in the same process. This limitation accounts for much of the hesitation to adopt solar energy on a large scale. Until this sort of technology can be further developed, the ominous solar eclipse may continue to attack the grid and damage the system that Europe has frivolously to build. Perhaps, however, the effects will be negligible. Only time will tell, and we will surely find out come March 20th. Stay posted....

http://gizmodo.com/now-we-actually-have-a-real-reason-to-dread-solar-eclip-1687768305