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.