Bonus Blog: Astronomy and Saving the World from Astroids!

http://www.daniel-irimia.com/2012/12/04/video-simulation-apocalypse-caused-by-asteroids-impact-equals-detonation-of-100-million-tons-of-tnt/

On February 15, 2013, a seventeen meter long astroid came roaring through the air above Chelyabinsk, Russia. Although the astroid mostly dissipated in the air, the event broke windows is reported to have injured more than a thousand people. Despite the damage caused by this relatively small meteor, experts are aware of the consequences that could have occurred. We were particularly lucky that the angle of impact was large enough that the astroid didn't significantly impact the earth.

As a result of this event, however, more awareness has been turned to developing mechanisms to protect our planet from future astroid collisions, and quite importantly to protect our species from the fate of the dinosaurs. In this blog post, we'll examine some of the protocols and techniques being explored to potentially save our planet!

The Chelyabinsk astroid posed an interesting issue to astronomers, since it came in a unique size that was large enough to be dangerous, but small enough that it was undetected until breaching our atmosphere. Scientists set the probability of these types of astroids approaching earth between ten and a hundred years.

A congressional project dubbed Spacedguard mandated that NASA aim to find 90% of all near-Earth astroids greater than 1 kilometer by 2005. This project was severely underfunded, and produced meager results.

ATLAS (Astroid Terrestrial-impact Last Alert System Project)

http://www.huffingtonpost.com/2013/02/21/asteroid-early-warning-system-atlas-space-rocks_n_2728101.html

Beginning this year, a new project is being implemented called ATLAS. This is designed to provide at least a week's notice for astroids around 45 meters, and three week's notice for those around 140 meters. To improve on these types of space surveillance, government agencies are looking more and more towards launching satellite telescopes to that can constantly monitor for meteors. This overcomes the significant limitation of earth telescopes that can only work effectively at night.

Some initial techniques proposed to protect against smaller astroids in the future include laser beams and nuclear missiles, which at this point are still under developed for useful applications in this area. Some more novel approaches, however, are also being examined for development quite a ways into the future. One method is known as a gravity tractor. This relies on early detection of a potential astroid collision. If detected early enough, a small spacecraft could be sent to the astroid, and gradually alter its path. Over several months, this method could push the astroid off it's colliding path with earth, and save the day!

Mirror Bees

http://www.nytimes.com/2007/12/09/magazine/09_5_asteroid.html?_r=0

An even more far off approach known as 'Mirror Bees' would utilize many small spacecrafts, each with focusing mirrors to direct the sun's light at one point on an astroid. This would be used to heat the small area to the point of vaporization, thereby creating a propulsive jet. Again, applying this over a period of time, this would be used to nudge the astroid off course, thus redirecting it away from its trajectory towards earth.

While the track record of neglecting funding, and far into the future approaches being proposed don't exactly instill confidence, it is important and exciting first step that approaches are even being thought of and considered. Particularly with the rapid growth and expansion of technology, it will be exciting to see the next stages of this developing field!

Resources: 
http://www.space.com/13524-deflecting-killer-asteroids-earth-impact-methods.html
http://www.wired.com/2013/02/asteroid-watching-mitigation/

Eclipsing Binary Lab

http://www.space.com/22509-binary-stars.html

Purpose

The objective of this lab was to detect the presence of an eclipsing binary pair of stars, as well as to characterize information regarding the stars' masses and radii. Beyond applying a light curve characterization technique to this eclipsing binary, there is great significance in the method used in this lab. This same light curve technique can be used to find exoplanets throughout the universe, thereby enabling the search for habitable planets.

The data being collected for this eclipsing binary pair adds to a relatively sparse database that allows us to better understand evolutionary models of the universe. Therefore, by collecting new, accurate data, this lab provides valuable information to the greater scientific community.

Black Image Portion: http://en.wikipedia.org/wiki/Variable_star#/media/File:Light_curve_of_binary_star_Kepler-16.jpg

For this experiment, we used the Clay Telescope on the roof of the Harvard Science Center to to track the movement of the low mass eclipsing binary, NSVS01031772. Incorporated with the Clay Telescope is a CCD camera that allows for capturing light data on the targeted stars. As the ecplising binary pair of stars pass in front of and behind one another, the light curve generated by the CCD images create a distinct shape. As the smaller, dimmer star passes in front of the larger, brighter star, the amount of light reaching the CCD is actually reduced, thus causing a primary dip in the light curve. Then, during a second transit, as the larger star passes in front of the smaller star, a secondary dip in the light curve occurs. This dip is smaller, because the full brightness of the larger star is still seen, rather than being partially blocked (as it is in the other primary transit).

Methods and Theory

To help determine the mass, period, velocity, and separation of the system, we used two primary datasets: Light Curve Data & Radial Velocity Data. Additionally, we leveraged several physical properties and equations to ultimately resolve the mass, radii, and separation of our two star system.

The first dataset for the light curve data was collected during the experiment using the Clay Telescope. A sample light curve for a transiting object is shown below:

From this plot, we can see that there are five primary characterized regions: Baseline 1 (T1), Ingress (T2), Transit (T3), Egress (T4), and Baseline 2 (T5). Each of these regions provide valuable information for making calculations on the system. Additionally, we can characterize an important vertical property of the plot, known as the depth of the transit, \( \delta \). This depth of transit can be used to calculate the radii of the two stars in the system. Over the course of the transit, several mathematical equations can be used to express the physical state of the system. First, to derive an expression for the the radial relationships of the two stars, we can consider what the two depths of the light curve transits tell us. While we might expect the relationship to simply be \( \delta = \frac{ {R_2}^2 } { {R_1}^2} \) as we found in an earlier worksheet, we need to account for the fact that both stars are luminous, and we actually need to utilize the difference in transit depths to create a more accurate expression of: \[\frac{ {R_2}^2 } { {R_1}^2}  = \frac{ \delta_1 } { 1 - \delta_2 } \] Now, we can create expressions for R1 and R2 using the transit times from the light curves. Since the two stars are traveling past each other, their relative velocity during the transit is the sum of their individual radial velocities, \( V_1 + V_2 \). Additionally, for the full transit to be completed, the eclipsing star must travel a distance of \( 2R_1 \) across the eclipsed star, and then an additional \( 2R_2 \) for completing the ingress and egress periods. Since we know the total time of the transit from our light curve data, and we can capture the radial velocities of each star from the spectral analysis data, we can then solve for these two radii: \[ R_1 + R_2 = \frac{ ( V_1 + V_2 ) \times t_{transit} } { 2 } \] To solve for the two different radii values individually, we can use our depth equation from above: \[\frac{ {R_2}^2 } { {R_1}^2}  = \frac{ \delta_1 } { 1 - \delta_2 } \] \[ R_2 =  \sqrt{ \frac{ \delta_1 } { 1 - \delta_2 }} R_1 \] Applying this to our above equation, we get:  \[ R_1 + \sqrt{ \frac{ \delta_1 } { 1 - \delta_2 }} R_1 = \frac{ ( V_1 + V_2 ) \times t_{transit} } { 2 } \] \[ R_1 = \frac{ ( V_1 + V_2 ) \times t_{transit} } { 2 (1 +  \sqrt{ \frac{ \delta_1 } { 1 - \delta_2 }}) } \] Applying the same approach to \( R_2 \), we get: \[ R_2 = \frac{ ( V_1 + V_2 ) \times t_{transit} } { 2 (1 +  \sqrt{ \frac{ 1 - \delta_2 } {  \delta_1 }}) } \]

The next physical quantity that we need to utilize is the center of mass of the system. This is characterized by: \[ X_{com} = \frac{ \sum m_i x_i}{m_i} \] By defining \(X_{com}\) to be zero, we will have the following relationship for our system. \[ M_1 a_1 = M_2 a_2 \] \[ \frac{M_1}{M_2} = \frac{a_2}{ a_1} \] Next, we can use the definition of period as: \[ P = \frac{ 2 \pi a }{V} \]  Then, solving for a, we get: \[ a_1 = \frac{V_1 P }{2 \pi } \] \[ a_2 = \frac{V_2 P }{2 \pi } \] Plugging these values into our mass equations from above, we get a simple expression, since the constants \( 2 \pi \, and \, P \) cancel out from each term: \[ \frac{M_1}{M_2} = \frac{V_2}{ V_1} \] This expression tells us that the larger of the two stars will be moving slower. This allows us to use the radial velocity curves obtained by Lopez and Morales and determine which radial velocity is attributed to which star.

The radial velocity was obtained through spectral analysis, and the data is shown below:

http://www.fas.harvard.edu/~astrolab/rvcurve.pdf

From this plot, we can extract some valuable data about the system. However, to first better understand what is going on with this radial velocity plot, we will overlay an illustration of what the two stars are doing during each orbital phase below.

To interpret this graph, it is important to note that the positive radial velocities correspond to a star moving away from us, while a negative radial velocity corresponds to a star moving towards us. Additionally, we can see from this plot that the maximum radial velocities occur when the stars are side-by-side, with one moving towards us, and one moving away. This allows us to use the doppler effect, where the wavelength of the light will either be increased or decreased depending on the direction that the relevant star is moving. Since we know that the slower velocity is attributed to the more massive star, we can extract this radial velocity data as follows: \[ V_1 = 143.85 \, km/sec \] \[ V_2 = 156.06 \, km/sec \] From this, we can now determine the mass ratio of the stars with the expression: \[ M_1 = \frac{V_2}{V_1} M_2 \] However, to determine an actual value for each mass, we have some more work to do. To achieve this, we can now apply Kepler's law to begin solving our system of equations: \[ P^2 = \frac{ 4 \pi^2 (a_1 + a_2)^3}{G(M_1 + M_2)} \] Solving for mass, we get: \[ M_1 + M_2 = \frac{ 4 \pi^2 (a_1 + a_2)^3}{G P^2} \] Substituting in the expressions for a from above into Kepler's equation, we get: \[ M_1 + M_2 = \frac{ 4 \pi^2 (\frac{P V_1}{2 \pi} + \frac{P V_2}{2 \pi})^3}{G P^2} = \frac{ P (V_1 + V_2)^3 }{2 \pi G}  \] Now, since \[ M1 = \frac{V_2}{V_1} M_2 \] We can write a complete expression as: \[ M_1 + \frac{V_1}{V_2} M_1 = \frac{ P (V_1 + V_2)^3 }{2 \pi G} \] \[ M_1 = \frac{ P(V_1 +V_2)^3}{ 2 \pi G ( 1 + \frac{ V_1}{V_2} ) } \] \[ M_2 = \frac{ P(V_1 +V_2)^3}{ 2 \pi G ( 1 + \frac{ V_2}{V_1} ) } \] 

With these expressions, we now have all of the equations needed to solve for each of our quantities.

Separation:
\[ a_1 = \frac{V_1 P }{2 \pi } \] \[ a_2 = \frac{V_2 P }{2 \pi } \]

Radii:
\[ R_1 = \frac{ ( V_1 + V_2 ) \times t_{transit} } { 2 (1 +  \sqrt{ \frac{ \delta_1 } { 1 - \delta_2 }}) } \] \[ R_2 = \frac{ ( V_1 + V_2 ) \times t_{transit} } { 2 (1 +  \sqrt{ \frac{ 1 - \delta_2 } {  \delta_1 }}) } \]

Mass:
\[ M_1 = \frac{ P(V_1 +V_2)^3}{ 2 \pi G ( 1 + \frac{ V_1}{V_2} ) } \] \[ M_2 = \frac{ P(V_1 +V_2)^3}{ 2 \pi G ( 1 + \frac{ V_2}{V_1} ) } \]
Observations

The light curves generated for this lab were created through data collection via the Harvard Clay Telescopes and an Apogee Alta U47 CCD to collect luminosity measurements.

Overall, there were a total of six datasets collected between March 24th and April 12th on nights with mostly clear visibility. We observed using the R-band filter with the telescope directed at our target of NSVS01031772 at RA, DEC = 13:45:35 + 79:23:48. To ensure proper images were taken, we adjusted the exposure length of the CCD to around 30,000, thereby avoiding saturation (~65,000). To then track our target stars during the observation, we selected guide stars using the control software.

Finally, we set the system to automatically take exposures for the duration of the transit and beyond to allow for a baseline to be established before and after on our light curve.

To account for any blemishes or noise due to the CCD or Telescope, we also took sky flats that were used to normalize our data from each night of observation.

To clean and reduce the data, averages were taken over multiple data points to remove random variance and smooth the final light curve.

Analysis

To analyze the collected data, we used MaximDL to generate our light curve. First, to do this, we delineated which objects in our images were reference stars, and which objects were our targets. The reference stars were used to provide a 'reference' to compare the varying luminosity of our eclipsing binary pair.

http://www.fas.harvard.edu/~astrolab/object_field.png

With these objects identified, we then created light curves for each of the six observation nights. This allowed us to not only compare observations and generate a more reliable light curve, it also allowed us to understand the period of the system. To determine the period, we adjusted the different observation days by an offset until they were aligned. This occurred with a period of 8 hours, 50 minutes, and 8 seconds (8.84 hours). The aligned light curves are shown below:

From this curve we could then establish the depth and overall time of both the primary and secondary transit. To calculate the depth of the transits, we subtracted the lowest point of each dip on the curve from the baseline level. With a baseline level of ~1.35 and minimums of 0.68 and 0.75 respectively, we found that \( \delta_1 = 0.67 \) \( \delta_2 = 0.6 \) However, we then normalized the data by dividing by the baseline, giving us normalized values of \[ \delta_1 = 0.496 \] \[ \delta_2 = 0.444 \] Next, we used the light curve to determine the transit time of each dip. For the primary transit, this produced a time of about 1.3 hours, while the secondary transit was slightly shorter at around 1.2 hours.

With this data processed, we are now able to calculate final values for each of our desired characteristics of the system.

Results

From the Methods section above, we can recall equations we will use to calculate each respective value:

Separation:
\[ a_1 = \frac{V_1 P }{2 \pi } \] \[ a_1 = \frac{ (143.85 \, km/s) (8.84 \, hours)(3600 \, s/hr )}{2 \pi } = 7.286 \times 10^{10} \, km \sim 1.047 R \odot \]

\[ a_2 = \frac{V_2 P }{2 \pi } \]\[ a_2 = \frac{ (156.06 \, km/s) (8.84 \, hours)(3600 \, s/hr )}{2 \pi } = 7.9 \times 10^{10} \, km \sim 1.14 R \odot \]

\[ a_1 + a_2 = 1=2.187 R \odot \]

Radii:
\[ R_1 = \frac{ ( V_1 + V_2 ) \times t_{transit} } { 2 (1 +  \sqrt{ \frac{ \delta_1 } { 1 - \delta_2 }}) } \]

\[ R_1 = \frac{ ( 143.85 \, km/s + 156.06 \, km/s) \times 1.3 \, hours \times 3600 \, s/hr } { 2 (1 +  \sqrt{ \frac{ 0.496 } { 1 - 0.444 }}) } = 3.61 \times 10^5 km \sim 0.52 R \odot \]

\[ R_2 = \frac{ ( V_1 + V_2 ) \times t_{transit} } { 2 (1 +  \sqrt{ \frac{ 1 - \delta_2 } {  \delta_1 }}) } \]

\[ R_2 = \frac{ ( 143.85 \, km/s + 156.06 \, km/s) \times 1.3 \, hours \times 3600 \, s/hr } { 2 (1 +  \sqrt{ \frac{ 1 - 0.444 } {  0.496 }}) } = 3.54 \times 10^5 \, km \sim 0.508 R \odot \]

Mass:
\[ M_1 = \frac{ P(V_1 +V_2)^3}{ 2 \pi G ( 1 + \frac{ V_1}{V_2} ) } \]
\[ M_1 = \frac{ (8.84 \, hours)( 3600 s/hr)(143.85 \, km/s + 156.06 \, km/s)^3}{ 2 \pi G ( 1 + \frac{ 143.85 \, km/s}{156.06 \, km/s} ) } = 1.08 \times 10^{33} \, g \sim 0.53 M \odot \]

\[ M_2 = \frac{ P(V_1 +V_2)^3}{ 2 \pi G ( 1 + \frac{ V_2}{V_1} ) } \]
\[ M_2 = \frac{ (8.84 \, hours)( 3600 s/hr)(143.85 \, km/s + 156.06 \, km/s)^3}{ 2 \pi G ( 1 + \frac{ 156.06 \, km/s}{ 143.85 \, km/s} ) } = 9.77 \times 10^{32} \, g \sim 0.48 M \odot \]

Conclusion:

These values match reasonably well with the results obtained by Lopez-Morales, with a slight degree of error. Given that observations were taken in the city environment of Cambridge, it's impressive that our results matched so closely with the previous results!

Astronomy and Consumer Products

While the study of Astronomy has produced many great findings and insights about the nature of the universe, it has created several amazing by-products along the way. In fact, NASA research and development has produced some of the most innovative and valuable pieces of technology used right here on Earth. Here we'll examine some of the amazing by-products of NASA!

Solar Panels

http://science.howstuffworks.com/innovation/nasa-inventions/nasa-improve-solar-energy.htm

Given the lack of power supplies and electricity in space, NASA had to tackle the problem of powering their satellites and space crafts while in orbit. Since the most readily available resource in space is the light from the sun, NASA overcame the challenge by harvesting solar power. Since most satellites sit beyond the range of earth's atmosphere, they never need to worry about a cloudy day! Stemming from NASA's work with solar panels, the technology has expanded to use on homes across the globe!

Memory Foam

http://www.memoryfoam.com/mattress/history-of-memory-foam/

In the late 1960's, NASA worked to develop improved seat cushions to allow astronauts to withstand the extreme G-forces of space travel. This development led to the invention of Memory foam, which has expanded to applications in mattresses and chairs by companies like Tempur-Pedic. 

Smoke Detectors

http://list25.com/25-coolest-nasa-discoveries-that-changed-your-life/2/

In the first United States space station, Skylab, NASA needed to create some mechanism for detecting toxic gases and fires on board. With this, they developed the first smoke detector, which has now become a required device in homes universally!

Cordless Power Tools

http://list25.com/25-coolest-nasa-discoveries-that-changed-your-life/3/

Again stemming from the electrical outlet limitations aboard a spacecraft, and the need to make repairs on the fly, NASA developed the worlds first cordless power tools. The expansion of power tools back on earth has allowed for significant advancements in carpentry and home improvement.

Light Emitting Diodes (LED)

http://list25.com/25-coolest-nasa-discoveries-that-changed-your-life/5/

To provide space shuttles with high-intensity, power efficient light, NASA developed light emitting diodes. These have been applied almost universally in flashlights, and in extensive automotive applications. 

Worksheet 14.2: Tides

Draw a circle representing the Earth (mass M), with 8 equally-spaced point masses, m, placed around the circumference. Also draw the Moon with mass \(M_{moon} \) to the side of the Earth. In the following, do each item pictorially, with vectors showing the relative strengths of various forces at each point. Don’t worry about the exact geometry, trig and algebra. I just want you to think about and draw force vectors qualitatively, at least initially.

(a) What is the gravitational force due to the Moon, \( F_{moon,cen} \) on a point at the center of the Earth? Recall that vectors have both a magnitude (arrow length) and a direction (arrow head).

The gravitational force due to the Moon, \( \vec{F_{moon,cen}} \) is given by: \[ F_{moon,cen} = \frac{ G M_{moon} M_{earth} } {r^2} \] This force vector will be in the direction of the earth towards the moon.

(b) What is the force vector on each point mass, \( F_{moon} \), due to the Moon? Draw these vectors at each point.

For each point mass of mass m, the force vector will be: \[ F_{moon} = \frac{ G M_{moon}m } {a^2} \] In this case, a is equal to the distance from each point mass to the moon.

(c) What is the force difference, ∆F , between each point and Earth’s center? This is the tidal force.

The force difference is simply the vector subtraction of the force at Earth's center and the force at each point mass. This would be given by the following expression: \[ \Delta \vec{F} = \vec{F_{moon}} - \vec{F_{earth, center}} \] This produces different force differences for the point masses around the earth, which accounts for different tides in different areas around the globe.

(d) What will this do to the ocean located at each point?

This will result in the oceans moving according to the tidal forces. On the far side of the earth, away from the moon, the tidal force will create a low tide. At the point closest to the moon, the tidal forces will create a high tide. And at the points on the top and bottom of the earth, the tidal forces will create a slack tide (in-between).

(e) How many tides are experienced each day at a given location located along the Moon’s orbital plane?

During the earth's rotation, each given location experiences 2 high and 2 low tides a day. During these period, the high tides will occur when the location is facing the moon, while the low tides will occur when the location is pointing away from the moon.

(f) Okay, now we will use some math. For the two points located at the nearest and farthest points from the Moon, which are separated by a distance ∆r compared to the Earth-Moon distance r, show that the force difference is given by \[ \Delta F = \frac{2GmM_{moon}}{r^3} \Delta r \] To begin, we can compare the forces at each of these two points: \[ F_{near} = \frac{GmM_{moon}}{r^2} \] \[ F_{far} = \frac{GmM_{moon}}{(r + \Delta r)} \] We can express this difference as: \[ \Delta F = F(r + \Delta r) - F(r) \] This is useful because we can divide both sides by \( \Delta r\) to get the derivative definition expression. This then allows us to calculate \( \Delta F \) by taking the derivative of F(r) and then multiplying by \( \Delta r \) in the end. \[ \Delta F = \frac{d}{dr} ( \frac{GmM_{moon}}{r^2} ) \Delta r \] \[ \Delta F = \frac{2GmM_{moon}}{r^3} \Delta r \]

(g) Compare the magnitude of the tidal force \( \Delta F_{moon} \) caused by the Moon to \( \Delta F_{sun} \) caused by the Sun. Which is stronger and by how much? What happens when the Moon and the Sun are on the same side of the Earth?

The tidal force generated by the sun is expressed by: \[ \Delta F_{sun} = \frac{2GmM_{sun}}{AU^3} \Delta r \] From this, we can see that the primary differences between this expression, and our expression for \( \Delta F_{moon} \) in part (f) is the sun distance and the sun mass. Since \( \Delta r \) is the same value in both expression, we can set those terms equal to each other, and generate a relationship for the sun and moon forces. \[ \Delta F_{sun} = \frac{r^3}{AU^3} \frac{M_{sun}}{M_{moon}} \Delta F_{moon} \] Inserting known quantities for r, AU, \( M_{sun} \), and  \( M_{moon} \), we get the following relationship: \[ \Delta F_{sun} \approx 0.4 \, \Delta F_{moon} \] or \[ \frac{ \Delta F_{moon} } { \Delta F_{sun} } \approx 2.5 \] This tells us that the tidal force caused by the sun is approximately 2.5 times weaker than the tidal force caused by the moon. This makes sense, since we generally don't consider the sun as a factor in our oceans' tides.

(h) How does the magnitude of ∆F caused by the Moon compare to the tidal force caused by Jupiter during its closest approach to the Earth (r ~ 4 AU)?

We can again take the same approach that we followed in part (g): \[ \Delta F_{jupiter} = \frac{2GmM_{jupiter} }{(4AU)^3} \] Now, setting the \( \Delta r \) terms equal, we can solve for \( \Delta F_{jupiter} \) in terms of \( \Delta F_{moon} \): \[ \Delta F_{jupiter} = \frac{r^3}{(4AU)^3} \frac{M_{jupiter}}{M_{moon}} \Delta F_{moon} \] Now, plugging in our known quantities of r, AU, \( M_{jupiter} \), and \( M_{moon} \), we can get our relationship between \( \Delta F_{jupiter} \) and \( \Delta F_{moon} \). \[ \Delta F_{jupiter} \approx 1 \times 10^{-5} \Delta F_{moon} \] \[ \frac{ \Delta F_{moon}}{ \Delta F_{jupiter} } \approx 1 \times 10^5 \]

Acknowledgements: Team EE

Worksheet 14.1, Problem 2: White Dwarf

A white dwarf can be considered a gravitationally bound system of massive particles.

https://prezi.com/jgjtnwr7mgz4/life-cycle-of-a-star-white-dwarf/

(a) Express the kinetic energy of a particle of mass m in terms of its momentum p instead of the usual notation using its speed v.

To begin, we can consider the standard expression of kinetic energy: \[ K = \frac{1}{2} m v^2 \] To convert this to being in terms of momentum, we can use the following expression for momentum: \[ p = mv \] Solving this for v, we get: \[ v = \frac{p}{m} \] Now, we can plug this value into v for our kinetic energy expression: \[ K = \frac{p^2}{2m} \] (b) What is the relationship between the total kinetic energy of the electrons that are supplying the pressure in a white dwarf, and the total gravitational energy of the WD?

To relate the total kinetic energy with potential energy, we can apply the virial theorem: \[ K = - \frac{1}{2} U \] From part (a), we know that \[ K = \frac{p^2}{2m} \]. Since we want to know the total kinetic energy for all electrons, we also want to include the term, n, for number of electrons, giving us the following expression: \[ K_{total} = n \times \frac{p^2}{2m} \] Next, we can apply the equation for potential energy: \[ U = - \frac{GM^2}{R} \] Substituting these into the virial theorem, we get: \[ n \times \frac{p^2}{2m} = \frac{GM^2}{R} \]
(c) According to the Heisenberg uncertainty Principle, one cannot know both the momentum and position of an election such that ∆p∆x > h/4π . Use this to express the relationship between the kinetic energy of electrons and their number density ne (Hint: what is the relationship between an object’s kinetic energy and its momentum? From here, assume p « ∆p and then use the Uncertainty Principle to relate momentum to the volume occupied by an electron assuming Volume ~ (∆x)^3 .)

To begin, we can derive an expression for number density. Since the problem has given that the volume occupied by an electron is \( V = { \Delta x }^3 \), we can assume the number density to be approximately : \[ n_e = \frac{ 1}{ { \Delta x }^3 } \] From part (a), we related Kinetic energy to momentum and mass. For this problem, we need to solve for momentum, so we can rearrange our expression from part (a) to do so: \[ K = \frac{p^2}{2m} \] \[ p = \sqrt{2 \times m \times K} \] Since we now have terms that involve \( \Delta p \) and \( \Delta x \), we can start plugging into the uncertainty principle expression. This expression is: \[ \Delta p \Delta x > \frac{h}{4 \pi } \] We can begin by removing constants from this expression, which are \( 4 \pi \) and h. This gives us the following expression: \[ \Delta p \Delta x > 1 \] \[ \Delta p > \frac{ 1}{ \Delta x} \] Now, from our number density expression, we can achieve the following relationship: \[ \Delta p > { n_e }^{ \frac{1}{3} } \] \[{ \Delta p}^3 > n_e \] We can then insert our expression for momentum in terms of Kinetic energy: \[ { \sqrt{ 2 \times m \times K } }^3 > n_e \] \[ ( 2 \times m \times K )^ { \frac{3}{2} } > n_e \] Again, we can remove the constants from this expression, giving us: \[ K^ { \frac{3}{2}} > n_e \] \[ K > {n_e}^{ \frac{2}{3}} \] \[ K \sim {n_e}^{ \frac{2}{3}} \]
(d) Substitute back into your Virial energy statement. What is the relationship between ne and the mass M and radius R of a WD?

As we've seen, the Virial Theorem states that: \[ K \sim \frac{1}{2} U \] Earlier in part (b), we expanded this expression to: \[ n_e \times K = \frac{GM^2}{R} \] We can now apply our expression for K from part (c): \[ n \times {n_e}^{ \frac{2}{3} } = \frac{ G M^2}{R} \] We can reduce this by canceling out constants: \[ n \times {n_e}^{ \frac{2}{3} } \sim \frac{ M^2}{R} \] To get the mass of a white dwarf, we can use the number density of protons, and multiply it by the mass of a proton and the overall volume. Additionally, since the number of protons and electrons should be equal, we can use the following expression: \[ n \sim M \] \[{n_e}^{ \frac{2}{3} } \sim \frac{ M}{R} \] \[ {n_e} \sim ({ \frac{M}{R}})^ { \frac{3}{2}} \]
(e) Now, aggressively yet carefully drop constants, and relate the mass and radius of a WD.

Given that the number density is equal to the number of electrons divided by \( R^3 \) we have the following expression: \[ n_e \sim \frac{n}{R^3} \] Additionally, from part (d), we showed that \( n \sim M \), which gives: \[ n_e \sim \frac{M}{R^3} \] Inserting this into our expression from part (d), we get: \[ \frac{M}{R^3} \sim ( \frac{M}{R} )^ {\frac{3}{2} } \] \[ M \sim \frac{1}{R^3} \]
(f) What would happen to the radius of a white dwarf if you add mass to it?

Our expression in part (e) shows that ass mass is added to a White Dwarf the radius will, in fact, decrease! This makes sense, since the density will increase, and the gravitational pull will draw the volume of the WD inwards.

Acknowledgements: Team EE

Systemic Summary: Instant Internet

On Greg Laughlin's blog systemic, he discusses the current state of internet latency, and ways that this could potentially be reduced, to allow for near spontaneous data transmission all across the globe.

In order for data transmission to appear spontaneous, it simply needs to beat the human's response-rate processing of ~ \( \frac{1}{30} \) second (30 msec). If data were to travel at the speed of light around the world, the trip from New York to Tokyo would take approximately 1 msec, or \( \frac{1}{30} \) the required amount of time for apparently spontaneous data transmission. Below is a representation of the path this data would have to follow:

http://oklo.org/

Currently, the time that it takes to travel this path is 80 msec along the NTT PC-1 cable. This is 1.44x slower than the speed of light. Although this is the fastest possible transmission speed of the current system, this isn't the standard time required for real-world data to be transmitted. In a study done by Ankit Singla, over 20 million measurements were taken on live data, and the latency was measured. This revealed that actual transmission was in fact 40x slower than the speed of light.

One of the primary causes of this latency is due to a small portion of Internet users that are consuming the vast majority of the Internet band-width. One proposed system to improve latency and narrow in on the apparently spontaneous Internet, is setting up a parallel infrastructure that is limited to standard online users.

A few prototypical networks of this sort have been established between Chicago and New York, aiming to eliminate latency. In testing, these networks can transmit within 2% of the speed of light minimum.

http://oklo.org/

These parallel infrastructure networks follow a Steiner Tree Configuration, which allows for minimizing distances across a sphere. A sample of a Steiner configuration applied to the continental United States is shown below:

http://oklo.org/

By minimizing these transmission distances, and allowing for parallel pathways, major latency improvements can be made. Additionally, incorporating fiber optic pathways could also allow for improvements in global data transmission latency. Initial testing has been conducted by Google, with their project, Google Fiber, to expand availability of fiber optic internet data distribution.

http://www.digitaltrends.com/computing/google-working-10-gigabit-internet-speeds/

Although laying out fiber optic cables over long distances is a challenge, the results are staggering, with download speeds of 1 Gb per second. This is over 100x faster than current Internet providers speeds. Particularly as Internet dependency, and data transmission requirements grow, this will be a fascinating area and problem to tackle.

A 39-day Trip to Mars!

The current six month travel time to Mars is a problem that the Ad Astra Rocket Company is working to solve. Standard chemical rockets provide a very large initial propulsion, but exhaust their fuel during the initial thrust from the earth's surface. The rocket then travels through space reliant on the initial thrust provided by the chemical thrusters.

VASIMR Ion Rocket (http://www.newscientist.com/article/dn17476-ion-engine-could-one-day-power-39day-trips-to-mars.html?full=true#.VS8kLJMbDmY)

The new VASIMR (Variable Specific Impulse Magnetoplasma Rocket) is taking an entirely new approach. Rather than using a large initial thrust, this engine will provide continuous thrust for the duration of the journey. In lieu of using chemical fuel, the Ion engine accelerates ions through an electric field to propel the rocket through space. Although this push is considerably less than the chemical engine's thrust, it can be sustained for a much much longer time. After only a short period of continuous thrust, the rocket will meet and exceed the speeds generated by previous single thrust engines.

http://www.huffingtonpost.com/2015/04/06/vasimr-rocket-mars_n_7009118.html

One of the greatest challenges with these plasma rockets is generating enough power to use them over a sustained period of time on a trip. To address this, engineers are working to build on-board nuclear reactors that can provide the sustained energy needed. Currently, in testing that has been done with non-nuclear power supplies, the engine can only be sustained for approximately one minute. NASA, however, has partnered with Ad Astra Rocket Company to aid in funding and research to develop a prototype that lasts a minimum of 100 hours, over the next three years.

Worksheet 13.2 Problem 1: Exoplanets!

Draw a planet passing in front of its star, with the star on the left and much larger than the planet on the right, with the observer far to the right of the planet. The planet’s semi-major axis is a.

(a) Show that the probability that a planet transits its star is \( \frac{R_*}{a} \), assuming \(R_P << R_* << a \). What types of planets are most likely to transit their stars?

In order to witness an exoplanet transit across a star, the path of the planet's orbit around the star must cross within a direct line of sight from our location on earth. For this reason, we are only able to see a limited number of exoplanets in space. In our drawing above, the necessary path is shown for a planet to be seen by an observer to the right.

To determine the probability that a planet transits a star, let's begin to quantify the system above. From the drawing, we can determine that a transit can only be observed between the two edges of the star. This region creates a solid angle described by: \[ SA = \int_0^{2 \pi} \int_{\frac{-R_*}{a}}^{\frac{R_*}{a}} sin \theta d \theta d \phi \] Since the radius of the star is so much smaller than the distance a, we can use the small angle approximation to create the bounds on the second integral of the above relationship. Evaluating this integral, we produce a solid angle of: \[SA = \frac{4 \pi R_*}{a} \] Given that the total solid angle of a sphere is \( 4 \pi \), then the probability of the visible solid angle of a transit being in line of sight is given by: \[ Probability = \frac{4 \pi R_*}{a} \times \frac{1}{4 \pi} = \frac{R_*}{a} \]
(b) If 1% of Sun-like stars in the Galaxy have a Jupiter-sized planet in a 3-day orbit, what fraction of Sun-like stars have a transiting planet? How many stars would I need to monitor for transits if I want to detect 10 transiting planets?

Unknowns: a, \( R_*\)

To begin, we will use Kepler's law to determine a value for a, the distance between the sun-like star and their jupiter-like planet. Similar to our work in Worksheet 13.1, Problem 1, we can achieve this by setting up a proportion with Earth's distance to the sun, which is in units of AU's. \[ \frac{ (1 year) \times ( 4 \pi^2 ) \times (AU^3) }{ GM} = \frac{ (\frac{3}{365} \, years) \times ( 4 \pi^2 ) \times (a^3) }{ GM} \] Solving for a, we get: \[ a = ( {\frac{3}{365}})^{ \frac{2}{3}}AU \]  The next term that we need to define is \(R_*\). Fortunately, this has a fairly straightforward value of ~ \( \frac{1}{200}AU \). Knowing this value, we can now calculate the percent of the above declared planets that fall within our observable probability: \[ \frac{ \frac{1}{200}AU }{ ( {\frac{3}{365}})^{ \frac{2}{3}}AU } \times 1 \, percent = 5 \times 10^{-3} \] This provides the fraction of transiting planets. To determine the number of stars that would be necessary to monitor to detect 10 transiting planets, we would divide 10 by this fraction, as shown below: \[ Number \, of \, stars \, = \frac{10}{5 \times 10^{-3}} \approx 2000 \, stars \]

Acknowledgements: Team EE

Worksheet 13.1 Problem 1: A Subtle Wobble

(a) We often say that planets orbit stars. But planets and their stars actually orbit their mutual center of mass, which in general is given by \( X_{com} = \frac{ \sum{m_ix_i}}{\sum{m_i}} \) . Set up the problem by drawing an x-axis with the star at \( x = -a_* \) with mass \(M_*\), and the planet at \( x = a_p \, and \, m_p \). Also, set \( x_{com} = 0 \). How do \( a_p \) and \( a_* \) depend on the masses of the star and planet?

To demonstrate the actual orbiting behavior of stars and planets, we began by sketching out the system below:

In this problem, we examine the orbital relationship of a star (left), a planet (right), and their center of mass (Xcom). This position, declared as \(X_{com} \) is derived from the following: \[ X_{com} = \frac{\sum{m_i x_i}}{\sum{m_i}} \] For this system described above, this center of mass would be: \[  X_{com} = \frac{M_* a_* + m_p a_p }{ M_* + m_p} \] Since we're setting our center of mass to the origin of our system, we can set this equation to zero, which allows us to relate our distances and masses. \[ M_* a_* = m_p a_p \] \[ \frac{a_*}{a_p} = \frac{m_p}{M_*} \]

(b) In a two-body orbital system the variable a is the mean semimajor axis, or the sum of the planet’s and star’s distances away from their mutual center of mass: \( a = a_p + a_* \). Label this on your diagram. Now derive the relationship between the total mass \(M_* + m_p \approx M_* \), orbital period P and the mean semimajor axis a, starting with the Virial Theorem for a two-body orbit (assume circular orbits from here on).

First, we recall the virial theorem: \[K= - \frac{1}{2}U \] To use this, we can insert the value of K for the total mass of the system: \[\frac{1}{2}M_* v^2 = - \frac{1}{2} U \] Furthermore, we can insert the gravitational potential energy in the place of U: \[\frac{1}{2}M_* v^2 = - \frac{1}{2} ( \frac{-G {M_*}^2}{a}) \] \[ v^2 = \frac{GM}{a} \] Additionally, since v is equal to angular velocity, we can further expand this to: \[ (\frac{2 \pi a}{P})^2 = \frac{GM}{a} \] Finally, solving for P, we get: \[ P^2 = \frac{4 \pi^2 a^3}{GM} \] Should this look familiar, that's because this is none other than Kepler's Third Law.

(c) By how much is the Sun displaced from the Solar System’s center of mass (a.k.a. the Solar System “barycenter”) as a result of Jupiter’s orbit? Express this displacement in a useful unit such as Solar radii.

Knowns:
\( M_{sun} \approx 1000 M_{Jup}\)
\( P_{Jup} \approx 12 \, years \)
\( R_{sun} = 7 \times 10^{10} \, cm \)

Knowing that Jupiter's period is 12 years, we can determine the distance between jupiter and the sun. Since we are trying to maintain simple units, such as the AU, we can actually calculate the jupiter-sun distance but setting up a proportion to the earth-sun distance: \[ \frac{{a_{jupiter-sun}}^3}{12^2 \, years} = \frac{{a_{earth-sun}}^3}{1^2} \] \[a_{jupiter-sun} = 12^{\frac{2}{3}} a_{earth-sun} = 12^{\frac{2}{3}} AU \] We can now use our equation from part a that relates distances and masses: \[a_p = a_* \frac{M_*}{m_p} \] The total distance, a, then equals: \[ a = a_* + a_* \frac{M_*}{m_p} \] Since we are approximating the sun to be 1000x heavier than jupiter, we can solve for \(a_* \): \[ a_* = \frac{ a } { 1+1000}\] \[ a_* = \frac{ 12^{\frac{2}{3}}AU} { 1+1000}\] \[ 1 \, AU = 214 \, R_{sun} \] \[ a_* = \frac{ 10^{3}R_{sun}} { 1+1000}\]

Acknowledgements: Team EE

Worksheet 12, Problem 2: Habitable Zones

The Earth resides in a “Goldilocks Zone” or habitable zone (HZ) around the Sun. At our semimajor axis we receive just enough Sunlight to prevent the planet from freezing over and not too much to boil off our oceans. Not too cold, not too hot. Just right. In this problem we’ll calculate how the temperature of a planet, Tp, depends on the properties of the central star and the orbital properties of the planet.

(a) Draw the Sun on the left, and a planet on the right, separated by a distance a. The planet has a radius Rp and temperature Tp. The star has a radius R‹ and a luminosity L‹ and a temperature Teff.

(b) Due to energy conservation, the amount of energy received per unit time by the planet is equal to the energy emitted isotropically under the assumption that it is a blackbody. How much energy per time does the planet receive from the star? How much energy per time does the Earth radiate as a blackbody?

Flux is given by the following formula: \[ F = \frac{L_{star}}{4 \pi a^2} \] To uncover the amount of energy that gets absorbed by the planet, we can multiply the flux by the area of the planet, as shown below: \[ E = \frac{L_{star}}{4 \pi a^2} \times 4 \pi R_{planet}^2 \] The energy that is radiated isotropically outward by the planet is then: \[E_{planet} = 4 \pi {R_{p}}^2 \sigma {T_{p}}^4 \]
(c) Set these two quantities equal to each other and solve for TP .

Setting these equal, we get the following equation: \[\frac{L_{star}}{4 \pi a^2} \times 4 \pi R_{planet}^2 = 4 \pi {R_{p}}^2 \sigma {T_{p}}^4\] Then, solving for \(T_p\), we get: \[ T_p = {\frac{L_{star}}{4 \pi a^2 \sigma}}^{\frac{1}{4}}\]
(d) How does the temperature change if the planet were much larger or much smaller?

If the planet were much larger or much smaller, the temperature would not change, since the amount of energy absorbed and radiated are both directly proportional to the size of the planet.

(e) Not all of the energy incident on the planet will be absorbed. Some fraction, A, will be reflected back out into space. How does this affect the amount of energy received per unit time, and thus how does this affect Tp?

The amount of energy received per unit time will be decreased by the fraction A, since it is simply being lost to space. The temperature, \(T_p\) will also be reduced, but not by as much as the flux. This is because the temperature is proportional to the \( \frac{1}{4} \) power of flux.

To derive this exact expression, we can use our equations from part (b) and (c). For both the energy absorbed and the resulting \( T_p \), we can multiply the expression by (1 - A), which gives us the correct fraction for what is absorbed. These expression now become: \[ E = (1-A) \frac{L_{star}}{4 \pi a^2} \times 4 \pi R_{planet}^2 \] \[ T_p = (1 - A){\frac{L_{star}}{4 \pi a^2 \sigma}}^{\frac{1}{4}}\]

Acknowledgements: Team EE

Worksheet 12, Problem 5: Leveraging Proportions

5. Assuming the core temperature, \(T_c\), of a Sun-like star is pretty much constant (nuclear fusion is a threshold process with a steep temperature dependence on the reaction rate), what are the following relationships?

(a) Mass-radius (i.e. assume R „ Mα and find α).

To begin, we can consider the equation of state, which is shown below: \[P = \frac{ \rho k T_c}{ \bar{m}} \] Since we are primarily concerned with ratios in this problem, we can ignore the constants above. This gives us the following relationship \[ P \sim \rho \] Since we know that density is simply mass over volume, we can expand this to: \[ P \sim \rho \sim \frac{M}{R^3}\] Next, we can consider the relationship of pressure within a star to mass and density. This relationship is presented as follows: \[P \sim \frac{M \rho}{R} \sim \frac{M^2}{R^4} \] Setting our Pressure relation proportions equal to each other, we get: \[ \frac{M}{R^3} \sim \frac{M^2}{R^4} \] Reducing this, we get our final Mass-Radius relationship: \[M \sim R\]

(b) Mass-luminosity (L  Mα) for massive stars M ą 1 Md, assuming the opacity (cross-section per unit mass) is independent of temperature κ “ const.

Utilizing the equation of state and pressure relationship established in part A, we know that \[P \sim \frac{M^2}{R^4} \sim \rho T\] Additionally, we know that \[\rho \sim \frac{M}{R^3}\] By combining these, we achieve the relationship that \[T \sim \frac{M}{R} \] Next, we can use the Luminosity proportion of \[L \sim \frac{T^4R^4}{M}\] Inserting \(T \sim \frac{M}{R} \), we get \[L \sim \frac{M^4}{M} \sim M^3\] This gives us our ultimate luminosity-mass equation of \[ L \sim M^3 \]

(c) Mass-luminosity for low-mass stars M <=  1 Md, assuming the opacity (cross-section per unit mass) scales as k ~ ρT^3.5 . This is the so-called Kramer’s Law opacity.

Including constant values in our luminosity proportion, we get the following: \[L \sim \frac{{T_c}^4R}{k \rho}\] Plugging in Kramer's law for opacity, and assuming T as a constant, we get: \[L \sim \frac{{T_c}^4R}{k \rho} \sim \frac{R}{ {\rho}^2}\] Since \(\rho \sim \frac{M}{R^3} \) and \( M \sim R \), our L relationship can be re-written as \[ L \sim \frac{M}{ {\rho}^2}\] Additionally, density can be re-written as \[ \rho \sim \frac{1}{M^2}\] Plugging this density value into our L proportion, we get: \[ L \sim \frac{M}{ \frac{1}{M^4}} \sim M^5\] Therefore, for low-mass stars, the Mass-Luminosity Relationship is \[ L \sim M^5\]

(d) Luminosity-effective temperature (L „ T α eff) for the two mass regimes above. This locus of points in the T-L plane is the so-called Hertzsprung-Russell (H–R) diagram. Sketch this as log L on the y-axis, and log Teff running backwards on the x-axis. It runs backwards because this diagram used to be luminosity vs. B-V color, and astronomers don’t like to change anything. Include numbers on each axis over a range of two orders of magnitude in stellar mass (0.1 ă M ă 10 Md). For your blog post, look up a sample H-R diagram showing real data using Google Images. How does the slope of the observed H-R diagram compare to yours?


We can begin with the Luminosity formula: \[L = 4 \pi R^2 \sigma T^4 \] \[ L \sim R^2 {T_{eff}}^4 \] Applying our proportions of \(L \sim M^3 \) and \( M \sim R \), we get a new relationship of \[ L \sim M^2 {T_{eff}}^4 \sim L^{0.5}{T_{eff}}^4\] Isolating L, we get: \[L^{0.5} \sim {T_{eff}}^4\] \[L \sim {T_{eff}}^8 \]

On a log scale, this produces a slope of 8. Since this would be inverted using the astronomers' unorthodox graphing practice, it would actually show a slope of -8. Comparing this to the H-R plot below, we can see that the relationship holds reasonably well.

http://casswww.ucsd.edu/archive/public/tutorial/HR.html

Acknowledgements: Team EE

Free Form Post: WeighTrack 2.0

Beyond the realm of astrophysics and AY16, I live in the world of Electrical Engineering and Computer Science. For my capstone Senior Thesis, I've worked for the past year on developing a dynamic liquid inventory tracker, known as WeighTrack 2.0. WeighTrack 2.0 utilizes an intricate load cell network and RFID technology to weigh and identify bottles embedded with RFID tags. By using these weight measurements and tag identification, the volumetric content of a bottle can be tracked over time, and monitored in real-time on a personalized web-page. Additionally, the shelf provides users with warnings about recalls, expirations, approaching expirations, and any other important information regarding the bottles.

The major applications for this technology are within chemical labs, restaurants and bars, and home refrigerators. 

In a lab, this allows users to dynamically monitor their chemical usage, as well as keep track of any bottles approaching expiration. This data is also valuable for chemical manufacturers to keep track of how their chemicals are consumed, and for targeted advertising. 

In bars, WeighTrack 2.0 allows managers and owners to track every liquor drink poured over the course of a night. This would allow for tracking a bartender's performance and actions during their shift. 

Integrated into refrigerators, this would allow individuals to monitor any expired or recalled items, as well as track consumption patterns of food. These patterns could then be utilized to set up automatic ordering with Amazon Fresh or Stop and Shop Peapod.

WeighTrack 2.0

Worksheet 11, Problem 2: Composition Change

11. 2 Problem: 3

Derive the third equation for mass conservation (dM/dr) starting with the integral equation that relates M(r) to r and \( \rho(r) \).

In this problem, we study how the mass of a star changes as the radius changes. The composition of star change drastically from their dense core to the more perfuse outer layers.

To begin, we can consider the mass of a star in terms of radius: \[ M(r) = \frac{4}{3} \pi r^3 \rho (r) \]
From this, we can establish an integral relationship by imagining infinitesimally small concentric shells for the sun with radii dr : \[ M(r) = \int_0^r { 4 \pi r^2 \rho (r) dr } \] Taking the derivative \( \frac{dM(r) }{dr}\) , we can resolve the way that mass changes over radius of a star. \[ \frac{dM(r) }{dr} = 4 \pi r^2 \rho (r) \]
Acknowledgements: Team EE

Worksheet 11, Problem 1

1. Consider a photon that has just been created via a nuclear reaction in the center of the Sun. The photon now starts a long and arduous journey to the Earth to be enjoyed by Ay16 students studying on a nice Spring day.

Photons that are generated in the core of a star follow a random pathway to the surface, colliding with each other along the way. Below, we will examine in-depth some of the properties of a photons journey to the surface of the sun.

The figure above shows a representative path traveled by a photon to the edge of a star.

(a) The photon does not travel freely from the Sun’s center to the surface. Instead it random walks, one collision at a time. Each step of the random walk traverses an average distance l, also known as the mean free path. On average, how many steps does the photon take to travel a distance ∆r?

The first step to calculating the number of steps taken by a photon to a star's surface is setting up an appropriate summation. To begin, we can consider the length of each step to be l, and the overall distance to be ∆r. We can then consider each step taken as a vector, \( \vec{r_i} \), and the distance D, as the sum of these vectors, \( \vec{D} = \sum{ \vec{r_i}} \). It's important to consider, however, that this displacement will converge to zero for a large number of random-walk journeys. For this reason, we will \( \vec{D^2} \) and take the square root of the result to be \( \delta r = ( \vec{D^2)}^{\frac{1}{2}} \).

This process is equivalent to calculating the dot product of each vector step, \( \vec{r_i} \) with each other. Below is the result of this dot product multiplication: \[ \vec{D^2} = n \vec{r_0^2} + \vec{r^2} cos \theta \] Considering that the photon's random walk covers all directions and values of theta, the cosine term will favor zero, leaving us with the following relationship: \[ \vec{D^2} = n \vec{r_0^2} \] Since we are treating each step as distance of length l, we can replace \( \vec{r_0} \) with l. Additionally, since D represents the overall distance traveled, we can replace this with ∆r in the equation above. This then gives us the following relationship: \[ { \delta r}^2 = n l^2 \]  By simply rearranging this, we can determine that on average, the number of steps taken by the photon to travel distance ∆r is: \[ n = \frac{ { \delta r}^2}{l^2} \]
(b) What is the photon’s average velocity over the total displacement after many steps? Call this ~vdiff, the diffusion velocity.

Calculating the diffusion velocity is a fairly straightforward process, since we know both the distance traveled by the photon, and the speed that photons (light) travel at. Knowing that our total distance is ∆r, we just need to calculate the total time to resolve the final velocity. Each step n takes \( \frac{l}{c} \) time to complete. So, \[ t_{step} = \frac{l}{c} \] To then find the total time, we multiply this by the total number of steps taken, n. \[t_{total} = n \frac{l}{c} \] From part A, we also know that \[ \delta r = l \sqrt{n} \] Dividing this total distance by the total time gives us our differential velocity. \[ \vec{v_{diff}} = \frac{ \delta r}{t_{total}} = \delta r \frac{c}{nl} = l \sqrt{n} \frac{c}{nl} =  \frac{c}{ \sqrt{n}} \]
(c) The “mean free path” l is the characteristic (i.e. average) distance between collisions. Consider a photon moving through a cloud of electrons with a number density n. Each electron presents an effective cross-section σ. Give an analytic expression for “mean free path” relating these parameters

Determining the "mean free path" l can be achieved through dimensional analysis. Our desired unit for this is length (cm). Let's begin with our units for number density: \( \frac{1}{cm^3} \). Furthermore, our units for cross sectional area, σ, are \( cm^2 \). By multiplying these terms together, we would get units of \( \frac{1}{cm} \). To arrive at our desired units of cm, we simply need to invert this term, to get the "mean free path." This value is expressed below: \[ l = \frac{l}{n \sigma} \]
(d) The mean free path, l, can also be related to the mass density of absorbers ρ, and the “absorption coefficient” κ (cross-sectional area of absorbers per unit mass). How is κ related to σ? Express vdiff in terms of κ and ρ using dimensional analysis.

Again, considering dimensional analysis, we can derive this expression for mean free path in terms of mass density and absorption coefficient. The desired unit of l is again cm. The units of mass density are \( \frac{g}{cm^3} \). The units of the absorption coefficient are \( \frac{cm^2}{g} \). By multiplying these terms, we again would get units of \( \frac{1}{cm} \). To arrive at our desired units, we would again need to invert this term. This final expression is shown below: \[ l = \frac{1}{ \rho \kappa} \] To see how \( \kappa \) is related to \( \sigma \), we can equate our two expressions for l: \[ \frac{1}{n \sigma} = \frac{1}{ \rho \kappa} \] Then solving for \( \sigma \), we get: \[ \sigma = \frac{ \rho \kappa }{n} \] Again applying dimensional analysis, we know that mass density has units \( \frac{g}{cm^3} \), and number density has units \( \frac{1}{cm^3} \). Seeing that these are divided in our \( \sigma \) expression, we know that these terms will cancel to give us just mass and \( \kappa \), as shown below: \[ \sigma = mass \times \kappa \] The l expression from this step can be used to also update our \( \vec{v_{diff}} \) from part b. This provides us with a new expression of \[ \vec{v_{diff}} =\delta r \frac{c}{nl} = \frac{c}{ \delta r \rho \kappa} \]
(e) What is the diffusion timescale for a photon moving from the center of the sun to the surface? The cross section for electron scattering is σT = 7 x 10^25 cm^2 and you can assume pure hydrogen for the Sun’s interior. Be careful about the mass of material through which the photon travels, not just the things it scatters off of. Assume a constant density, ρ, set equal to the mean Solar density (N.B.: the subscript T is for Thomson. The scattering of photons by free (i.e. ionized) electrons where both the K.E. of the electron and λ of the photon remain constant—i.e. an elastic collision—is called Thomson scattering, and is a low-energy process appropriate if the electrons aren’t moving too fast, which is the case in the Sun.)

To begin, let's consider our knowns:
Mass of hydrogen atom: \( 2 \times 10^{24} g \)
Solar mass density: \( \rho = 1.5 \frac{g}{cm^3} \)
Solar radius: \( 7 \times 10^{10} cm \)

Since we are trying to solve for time, which equals distance traveled divided by the velocity traveled, as shown below: \[ t = \frac{d}{v} \] The net distance, d, traveled will be equal to ∆r. Furthermore, we can apply our equation for \( \vec{v_{diff}} \) to give us a complete expression of: \[ t = \frac{ \delta r}{ \frac{c}{ \delta r \rho \kappa}} = \frac{ \delta r^2 \rho \kappa}{c} \] Additionally, we can substitute \( \kappa = \frac{ \sigma }{ m_{hydrogen}} \). This give us a final expression of \[ t = \frac{ \delta r \rho \sigma }{ m_{hydrogen} c } = 9 \times 10^{10} \, s = 3000 years \]

Acknowledgements: Team EE

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, \( \theta\). \[ \frac{24 \, hours}{360 \, degrees} = \frac{diameter \, time}{ \theta }\] \[ \frac{24 \, hours}{360 \, degrees} = \frac{135 \, seconds}{ \theta }\] 
\[ \theta = \frac{135 \, seconds \times 360 \, degrees}{ 24 \, hours \times \, 60 \, minutes \times 60 \, seconds} = 0.56 \, degrees \]