Eclipsing Binary Lab

Introduction/ Purpose:

This blog post discusses and analyzes the data of an eclipsing low-mass binary star system, NSVS01031772, observed by the Astronomy 16 class at Harvard University using the optical Clay Telescope. The goal of the lab is to find the masses, radii, and semi-major axes of the stars of the double-lined spectroscopic binary using data from a radial velocity plot and a light curve.

A binary star is a system of two stars which are gravitationally bound and are orbiting around their common center of mass. When binary stars are very far away, they cannot be resolved and are seen as a single point, but because they are orbiting each other, we see a decrease in brightness as they cross over one another. This is because the unresolved point that we see has the brightness of both stars, but when one is crossing the other, it blocks the light from the other star, as demonstrated in the animation below.

Figure 1: The graph shows the dip in brightness as one of the stars transits the other.
Image Credit:
http://38.media.tumblr.com/fa98981e91a5ea5d13953d3a8086c2ac/tumblr_n0s10t9MS21rnq3cto1_500.gif

For this lab, we observed the transits of the low-mass binary stars and analyzed the depths of the dips in their brightness to find information about the masses, radii, and separation between the two stars. Because these low-mass binary stars have low surface brightness, they are hard to detect, though there are many of these stars. Due to this fact, there is limited data on stars less massive than \(0.6M_\odot\). Finding information on the mass-radius relationship is important, however, since it provides constraints for evolutionary models of main sequence stars. It can also be used to infer similar information on planets orbiting these stars.

Methods:

López-Morales et al. performed an investigation on the binary star, NSVS01031772. In their paper, they included a radial velocity (RV) plot of the two stars in the system, which they found using Doppler spectroscopy. We use this plot to determine important information about the binary system. The RV plot shows the radial velocity of each star at a given time, where negative radial velocity means that the star is moving towards the observer, and positive radial velocity means that the star is moving away from the observer.

Figure 2: The radial velocity (RV) plot of NSVS01031772.
Image Credit: López-Morales et al., 2006

The RV plot can be used to find the maximum radial velocities of each star, which is given by the peak radial velocities of the stars subtracted by the radial velocity of the center of mass. The center of mass has a radial velocity to due relative motion between us and the binary system and it is shown in the plot by the horizontal dotted line at a radial velocity of \(19.0\pm 1.0\) km/s. With the maximum radial velocities, one can determine the mass ratio of the system. The star with a larger maximum radial velocity is less massive than the star with smaller maximum radial velocity. We will call the star represented by a solid line in the RV plot Star 1, and the star represented by the dotted line Star 2. Star 1 has a peak radial velocity of about -120 km/s and Star 2 has a peak radial velocity of about 180 km/s. Subtracting the radial velocity of the center of mass, this gives us maximum radial velocities of around -140 km/s for Star 1 and 160 km/s for Star 2. These are poor approximations, however, since they are measured by eye along the axis of the plot. López-Morales et al. provide data with significantly better accuracy using the Southeastern Association for Research in Astronomy (SARA) data set, getting maximum radial velocities of \(143.85\pm 0.37\) km/s for Star 1 and \(156.06\pm 0.88\) km/s for Star 2. From this, we can conclude that Star 1 is more massive than Star 2.

Solving for the mass ratio, we can begin with a relation derived from the center of mass equation in a previous problem:

\(a_1M_1=a_2M_2\)

\(\dfrac{M_1}{M_2}=\dfrac{a_2}{a_1}\)

Where \(a_1\) is the distance from Star 1 to the center of mass of the system, \(a_2\) is the distance from Star 2 to the center of mass, \(M_1\) is the mass of Star 1, and \(M_2\) is the mass of Star 2. Then, from the equation for the orbital period \(P=\frac{2\pi a}{v}\), we get the relation:

\(\dfrac{a_1}{a_2}=\dfrac{v_1}{v_2}\)

So we find a mass ratio of:

\(\dfrac{M_1}{M_2}=\dfrac{v_2}{v_1}=\dfrac{156.06\text{ km/s}}{143.85\text{ km/s}}\approx 1.08\)

In addition to finding the maximum radial velocities and the mass ratio, we can also use the RV plot to describe the relative motion of the stars at different points on the plot. The image below depicts what the orbit of the two stars would look like based on the RV plot. Star 1 is shown in orange, Star 2 is shown in yellow, and their center of mass is located at the red pluses. 

Figure 3: The orbits of the two stars relative to each other. Star 1 is shown in orange and is the more massive than Star 2, shown in the yellow.

In the first scenario, the stars are at their peak radial velocities, which means that all of their motion is directed towards and away from the observer. In the second scenario, where there is no radial velocity, the stars' motion is directed to the left and right of the observer, with no components of motion along the direction of the observer. Finally, the third scenario shows a point along the orbit where some of the motion is directed towards and away from the observer, and some is directed to the left and right, here the radial velocity is smaller than the peak radial velocity.

Another important plot of data is the light curve, which shows the flux of electromagnetic radiation of an astronomical object over time. They are generally in a particular band or frequency interval, and can sometimes be periodic, as in the case of eclipsing binary stars. Using the Clay telescope to collect optical data of the NSVS01031772 binary star system, we were able to plot the light intensity of the eclipsing binaries over time. At the points where the stars transit one another, there would be a dip in the light intensity. The frequency of these dips in intensity can tell us the period of the stars' orbit. Furthermore, the depth of the dip, its duration, and time of ingress and egress can be determined from the light curve and can be used to find the radii, masses, and semi-major axes of the stars. 

To find the masses of the stars, we need to use Kepler's Third Law, the center of mass equation, the equation for period, and the Newtonian improvement to the Keplerian semi-major axis:

\(P^2=\dfrac{4\pi ^2a^3}{G(M_1+M_2)}\)

\(a_1M_1=a_2M_2\)

\(P=\dfrac{2\pi a_1}{v_1}=\dfrac{2\pi a_2}{v_2}\)

\(a=a_1+a_2\)

We begin with Kepler's Third Law, substituting the Newtonian improvement to the semi-major axis for a, and substituting \(M_2=\frac{a_1}{a_2}M_1\):

\(P^2=\dfrac{4\pi ^2(a_1+a_2)^3}{G(M_1+\frac{a_1}{a_2}M_1)}\)

Then, solving the equations of the periods for the semi-major axes and using the relation \(\frac{a_1}{a_2}=\frac{v_1}{v_2}\):

\(P^2=\dfrac{4\pi ^2\left(\frac{P(a_1+a_2)}{2\pi}\right)^3}{G(M_1+\frac{a_1}{a_2}M_1)}\)

\(\Longrightarrow M_1=\dfrac{v_2P(v_1+v_2)^2}{2\pi G}\)

Repeating a similar process for the second mass, we get:

\(\Longrightarrow M_2=\dfrac{v_1P(v_1+v_2)^2}{2\pi G}\)

Now, to solve for the radii of the stars, we need to use the transit depth equation, which was derived in a previous problem, and the transit time expression, which we must derive. To do this, we must consider the Stefan-Boltzmann law, \(L\propto R^2T^4\). From here, we can see that if the temperatures are relatively similar, a star with a larger radius will have a bigger luminosity. Thus, we can conclude that the primary transit (the transit with the larger depth) occurs when the star with smaller radius passes in front of the star with larger radius, blocking a higher proportion of light; the secondary transit (with a smaller depth) would then occur when the star with a larger radius passes in front of the star with a smaller radius. 

Figure 4: The transit of the smaller star across the bigger star.

From Figure 4, we can see that in the frame of the bigger star, the smaller star has a velocity of \(v_1+v_2\) and must travel a distance of \(2R_1+2R_2\). The transit time, \(t_{trans}\), is given by:

\(t_{trans}=\dfrac{2(R_1+R_2)}{v_1+v_2}\)

We also know the transit depth equation from a previous result:

\(\delta =\left( \dfrac{R_2}{R_1}\right)^2\)

This equation, however, was derived under the assumption that the transiting object was opaque. In this case, the transiting star is emitting its own light. We can correct for this by subtracting the luminosity of the transiting star from the light curve. Figure 5 gives and example of what this would look like. 

Figure 5: The dotted line represents the light curve with luminosity of the smaller star subtracted. \(\delta_1\) is the depth of the primary transit, and \(\delta_2\) is the depth of the secondary transit.

Having subtracted the depth of the secondary transit from the light curve, our transit depth equation for the primary transit now becomes:

\(\left( \dfrac{R_2}{R_1}\right)^2=\dfrac{\delta_1}{1-\delta_2}\)

Solving for \(R_2\) and plugging into the transit time equation, we get:

\(R_1+\sqrt{\dfrac{\delta_1}{1-\delta_2}}R_1=\frac{1}{2}(v_1+v_2)t_{trans}\)

\(\Longrightarrow R_1=\frac{1}{2}(v_1+v_2)t_{trans}\left(\dfrac{1}{1+\sqrt{\frac{\delta_1}{1-\delta_2}}}\right)\)

And for \(R_2\), we get:

\(\Longrightarrow R_2=\frac{1}{2}(v_1+v_2)t_{trans}\left(\dfrac{1}{1+\sqrt{\frac{1-\delta_2}{\delta_1}}}\right)\)

Observations:

The optical observations of the binary NSVS01031772 were done using the Clay Telescope, which is a 0.4 m DFM engineered telescope, located on the roof of the Harvard Science Center. The Clay Telescope has a DFM filter wheel with Bessel filters and an Apogee Alta U47 imaging CCD with a 13' x 13' field of view. The CCD contains \(1024\times 1024\) pixels, which means that each pixel has an angular diameter of approximately 0.76'. Furthermore, the telescope is controlled using a Telescope Control System and also can be controlled using The Sky software, which provides a map of the sky with coordinates to many astronomical objects preprogrammed. This allows one to easily locate and slew the telescope to a particular object. Also, the telescope and the dome have auto-tracking features which follow a target while compensating for the rotation of the Earth. A guide star can also be used for more accurate tracking.

Figure 6: The Clay Telescope
Image Credit:
http://isites.harvard.edu/fs/docs/icb.topic207662.files/Clay_Telescope_3.jpg

The observing process was broken up into groups that gathered data separately on different days beginning from March 24th until April 12th. I was observing on the night of April 11th, 2015. The weather was fairly clear and mostly cloudless, with a temperature of \(50^\circ\) F and a relative humidity of 32%.

We observed the binary NSVS01031772 at a Right Ascension of 13:45:35, and a Declination of +79:23:48. The images were taken as 60 second exposures in the R-band filter since the stars were M-type dwarfs with low enough temperatures that they were emitting significantly in the red wavelengths.

Possible error in our data could result from the slight cloud coverage, light pollution from Cambridge and Boston area, and some difficulties with the MaxIm DL software which caused stoppages in data collection on some of the nights. 

Analysis:

The reductions on the raw data were done using MaxIm DL. MaxIm DL was set to automatically take dark frames and biases. Dark frames are exposures taken with the aperture lens closed in order to account for dark current from the CCD. Dark current is a small electric current running through the CCD that creates noise in images. Bias frames are images that are obtained without any exposure time and contain noise from the internal electronics. Additionally, flat frames were taken shortly after sunset, which are images of the uniformly lit sky. To account for any non-uniformities, the telescope is slewed a little and other flat frames are taken. All of these flat frames are then averaged and they are used to account for blemishes such as dust or smudges on the lens and any varying sensitivity of different regions of the CCD.

Using MaxIm DL, the raw data was calibrated with the dark frames, biases, and flats. Then, photometric analysis was done on the images, which is a process that measures the electromagnetic flux, or the photon count, on selected points on the images.

Figure 7: This is the field of view of the observations. "Obj1" refers to the object we are observing, binary system NSVS01031772, and there are four reference stars selected for photometric purposes. When taking images, we had to make sure all stars were centered in the image.
Image Credit: http://www.fas.harvard.edu/~astrolab/object_field.png 

Four reference stars were selected and their absolute magnitudes were set to zero since we are only concerned with variations in magnitude and not the magnitude itself. The green annuli centered around the reference stars and the object subtract off any light that comes from other sources. Then, the data gathered by MaxIm DL for each image was transferred to an Excel file and plotted as a light curve. The following image shows the resulting light curves from six different nights of observing.

Figure 8: The light curve for the binary system, NSVS01031772, with data taken from six different nights of observing. The curves from the various nights of observations were aligned on the plot by using an orbital period of 8.84 fractional hours. The dip in the curve on the left is the primary transit, which occurs when the smaller star passes in front of the bigger star. The dip on the left is the secondary transit, which occurs when the larger star passes in front of the smaller star.

On the Excel spreadsheet, the period had to be found adjusted until the curves for the six different nights of observation were aligned. The resulting value would be the orbital period of the binary system, which we found to be 8.84 fractional hours, confirmed by López-Morales et al. This value for the period has a negligible uncertainty, and was treated as a fixed constant by López-Morales et al. paper. We can also see by visual inspection of the light curve that the baseline (the flux of both stars before transit) is about 1.30. The depth of the primary transit is about \(0.60\pm 0.03\) (found by subtracting 1.3-0.7), and the depth of the secondary transit is about \(0.51\pm 0.02\) (found by subtracting 1.3-0.79). 

Figure 9: The top panel is a magnified view of the primary transit and the bottom panel is magnified view of the secondary transit. The horizontal dotted lines are the estimated errors for the determined values of the transit depths. The estimated values for the depths are \(0.60\pm 0.03\) for the primary transit, and \(0.51\pm 0.02\) for the secondary transit.

It is also important to note that the flux is normalized to 1.3. In our calculations, we need the light curve to be normalized to 1. In order to do this, we must divide the values of the graph by 1.3. This changes our transit depths to \(0.46\pm 0.02\) for the primary transit, and \(0.39\pm 0.02\) for the secondary transit. We can also see from the curves in Figure 9 that the transit times are approximately \(1.30\pm 0.20\) hours.

Results:

With our knowledge of the light curve, we can again find the motion of the two stars relative to each other over time. 

Figure 10: The relative motion of the stars at different points in time along the light curve. The direction of the orbit is shown only for the sake of consistency, however, since the direction of the orbit is unable to be directly concluded from the information provided by the light curve.

Based on our knowledge of primary and secondary transits, we know that the smaller star will be directly in front of the larger star at the bottom of the primary transit, and it will be directly behind the larger star at the bottom of the secondary transit. On the baseline, neither of the stars are covering each other, so we get the luminosity of both stars. The difference between this method and the radial velocity plot method is that in this case, we do not know the direction of the orbit from the information provided by the light curve.

We can now use the expressions derived in the methods section to calculate the masses, radii, and separation of the two stars. We know that \(v_1\approx 143.85\pm 0.37\) km/s, \(v_2\approx 156.06\pm 0.88\) km/s, and \(P=8.84\) hours. Plugging these values into our mass equations:

\(M_1=\dfrac{v_2P(v_1+v_2)^2}{2\pi G}\)

\(M_1=\dfrac{(156.06\text{ km/s})(8.84\text{ hr}\times 3600\text{ s hr}^{-1})(143.85\text{ km/s}+156.06\text{ km/s})^2}{2\pi (6.67\times 10^{-8}\text{ cm}^3\text{ g}^{-1}\text{ s}^{-1})}\)

\(M_1\approx 1.07\times 10^{33}\) g

And similarly for \(M_2\):

\(M_2=\dfrac{v_1P(v_1+v_2)^2}{2\pi G}\)

\(M_2=\dfrac{(143.85\text{ km/s})(8.84\text{ hr}\times 3600\text{ s hr}^{-1})(143.85\text{ km/s}+156.06\text{ km/s})^2}{2\pi (6.67\times 10^{-8}\text{ cm}^3\text{ g}^{-1}\text{ s}^{-1})}\)

\(M_2\approx 9.83\times 10^{32}\) g

Now, to solve for the radii, we know that the transit time is \(t_{trans}=1.3\pm 0.20\) hrs, and the depths are \(\delta_1 =0.46\pm 0.02\) and \(\delta_2 =0.39\pm 0.02\). Plugging these values into our expressions, we get:

\(R_1=\frac{1}{2}(v_1+v_2)t_{trans}\left(\dfrac{1}{1+\sqrt{\frac{\delta_1}{1-\delta_2}}}\right)\)

\(R_1=\frac{1}{2}(143.85\text{ km/s}+156.06\text{ km/s})(1.3\text{ hr}\times 3600\text{ s hr}^{-1})\left(\dfrac{1}{1+\sqrt{\frac{0.46}{1-0.39}}}\right)\)

\(R_1\approx 3.76\times 10^{10}\text{ cm}\approx 0.54 R_\odot\)

Similarly for \(R_2\), we get:

\(R_2=\frac{1}{2}(v_1+v_2)t_{trans}\left(\dfrac{1}{1+\sqrt{\frac{1-\delta_2}{\delta_1}}}\right)\)

\(R_2=\frac{1}{2}(143.85\text{ km/s}+156.06\text{ km/s})(1.3\text{ hr}\times 3600\text{ s hr}^{-1})\left(\dfrac{1}{1+\sqrt{\frac{1-0.39}{0.46}}}\right)\)

\(R_2\approx 3.26\times 10^{10}\text{ cm}\approx 0.47R_\odot\)

Finally, to solve for the separation between the two stars, we only need to use the Newtonian improvement to the Keplerian semi-major axis and the equations for period:

\(P=\dfrac{2\pi a_1}{v_1}=\dfrac{2\pi a_2}{v_2}\)

\(a_1=\dfrac{P v_1}{2\pi}\)   and   \(a_2=\dfrac{Pv_2}{2\pi}\)

So from here we get:

\(\Longrightarrow a=\dfrac{P(v_1+v_2)}{2\pi}\)

Plugging in values, we get:

\(a=\dfrac{(8.84\text{ hr}\times 3600\text{ s hr}^{-1})(143.85\text{ km/s}+156.06\text{ km/s})}{2\pi}\)

\(a\approx 1.52\times 10^{11}\text{ cm}\approx 2.19R_\odot\)

Conclusion:

In this lab, we observed the eclipse of the binary NSVS01031772 between March 24th and April 12th using the Harvard Clay telescope. From our optical observations, we were able to create a light curve. Drawing information from the light curve, we found the period of the stars' orbit, their radii, masses, and separation. When comparing our calculations to those done in the López-Morales et al. paper, our calculations are fairly accurate. The table below compares our results to those of López-Morales et al.

The uncertainties of our measurements were done by eye. These uncertainties can be propagated through our calculations to find the uncertainties of our results. However, due to the fact that the uncertainties were done in a heuristic manner, it is unnecessary to do so for the purposes of this experiment. 

Possible sources of error might come from noise due to light pollution from Boston area. Also, on some of the observing nights, groups had problems with MaxIm DL, which crashed and lost a few data points, though the effects of this on accuracy are probably negligible. Furthermore, in our calculations, we assumed perfectly circular orbits and zero inclination of the orbit. It is unlikely that these assumptions would hold perfectly true, though the extent to which they might affect our results is unknown. Also, some of the values used in the calculations were visually determined, which allows room for human error. Despite these possible sources of error, our results agreed to a large extent with the results of the paper by López-Morales et al. Thus, we can suppose that our results will have small uncertainties. 



I would like to thank Charles Law, Andy Mayo, Daniel Chen, and the 3 TFs for their help in completing this lab and in performing the necessary analysis of the data.

Citations:
http://en.wikipedia.org/wiki/Binary_star
http://en.wikipedia.org/wiki/Light_curve
http://www.fas.harvard.edu/~astrolab/NSVS01031772.pdf
http://www.fas.harvard.edu/~astrolab/claytelescope.html
http://en.wikipedia.org/wiki/Bias_frame
http://www.fas.harvard.edu/~astrolab/EB_Ay16.html

Astrobites, Daily Paper Summaries: The Moon's Formation

Image Credit:
http://www.thesilverink.com/wp-content/uploads/2015/04/Earth_and_moon.png

Our Moon has a special place in our Solar System. Yes, it is our Moon, but more interestingly, it is the largest known moon compared to its host planet. Some astrobiologists believe that the large size of the moon and the resulting tides were what allowed life to form several billion years ago. Sadly, the moon is receding from the Earth by a few centimeters every year. 

A question we are still pondering is how the Moon got there to begin with. The currently accepted theory is the giant impact theory, which claims that around 4.5 billion years ago, a Mars-sized protoplanet, Theia, collided with the Earth creating a disk of molten rock, gas, and debris which combined to become the Moon. This theory works well in explaining the similar orientations of the spins of the Earth and Moon and the lack of iron on the Moon. Simulations of this theory have been able to accurately recreate the Earth-Moon system. 

An artists rendition of the giant impact theory.
Image Credit:
http://upload.wikimedia.org/wikipedia/commons/4/4a/Artist's_concept_of_collision_at_HD_172555.jpg

This theory also used to be supported by the similarities in composition between the Earth and Moon. We have found that the Earth and Moon have the same quantities of certain isotopes while our observations of other planets and their moons show that there are very different proportions of these isotopes. This would imply that the Earth and Moon have a common origin, and that perhaps the Moon formed from material from the Earth. This could have been a result of the collision with Theia. However, simulations of the giant impact theory actually show that most of the material that forms into the Moon comes from Theia, and not Earth. If this were true, that would mean that Theia would have to have formed in a part of the planetary disk that has similar composition as Earth in order to create a Moon with similar composition as the Earth. 

Two recent papers discuss the likelihood that Theia could have been isotopically similar to Earth, coming to different conclusions. Some of the reasons these two studies could have differed include their assumptions on the orbital characteristics big planets. It isn't necessarily true that these planets would have orbited in the same way back then as they do now. Both studies found that different configurations of the planets Jupiter and Saturn would have affected the regions in the planetary disk where developing planets would acquire their material, thus changing the outcomes in the formations of the Earth and Theia. Also, the distribution of the isotope, oxygen-17, is unknown, though both studies used the same assumption of a linear distribution. Perhaps a significant cause to the difference in the conclusions of the two papers' is the amount that they allowed the Earth to contribute to the formation of the Moon. The study that concluded that the similar composition of Theia was likely also allowed a larger percentage of the Moon to be made from the Earth. Finally, the study which concluded in the unlikeliness of the similar composition of Theia only considered models that were analogous to the Earth and Theia, while the other study considered all possible collisions. 

Both studies require more research and analysis to come to a strong conclusions, though they offer interesting questions to consider. You can find more information on the research papers for each study. The paper concluding that such conditions for the formation of the moon are unlikely can be found here, and the paper concluding that such conditions are plausible can be found here.

Citations:

http://astrobites.org/2015/04/08/was-our-moons-formation-likely-or-lucky/

Free Form: Astronomy and the Mayans

Between 250 and 500 CE, the Mayans began documenting the skies. Many of their calculations and reports of astronomical objects' positions are found in the Dresden Codex, which is a book of 78 pages of astronomical tables. The image below shows some of the pages of the Dresden Codex.

The Mayan Dresden Codex
Image Credit: http://www.latinamericanstudies.org/maya/codex-dresden.jpg

To the Mayans, the stars were signs from the gods that showed their will and desires, and they were considered to be telling of human life on Earth. Various astronomical objects represented different Gods. The Sun was representative of the Sun God, Kinich Ahau, who was believed to shine during the day, and then turn to a jaguar to go to the underworld at night. The Mayans kept close watch of the Sun and had very accurate knowledge for predictions of the Equinoxes, Solstices, and Solar eclipses. 

Mayan Sun God, Kinich Ahau
Image Credit: https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiBDs2EH9PxW16Tb58LkglJCirFoxSV0TRNaE155uraugwKy-hLE9RqxFRF4mXwkdBOk3Rwo4ZU6RSG2gNRf_dV-CumMWfb1CVZQEikJE21zh4M9tyKTd0UFdZnD3o5musro0Xs_BJ4eBM/s1600-h/2484925157_3978c19c0d_o.jpg

The most important planet to the Mayans was Venus. This planet's position in the sky was used to determine when war would occur, and when sacrifices of captured warriors and leaders would take place. The Mayans were able to calculate a year on Venus to be 584 days long (with respect to the Earth, since they believed that the Earth was the center of the Universe), compared to the current value: 583.92. 

Mayan Temple
Image Credit:
http://www.redorbit.com/media/uploads/2012/11/science-110912-001-617x416.jpg

Many of the temples dedicated towards different gods were aligned to objects such as the Sun or Venus, so that on a certain date and time of the year, the Sun would shine directly onto the temple stairs, as in the example of Chichén Itzá. Additionally, Mayan observatories were specially placed so that they would be illuminated on certain dates. The Xochicalco observatory is underground with a hole in the ceiling. The Sun shines directly over this observatory on two dates of the year, projecting an image of the Sun on the floor.

The Mayan calendar was a system of interlocking cycles.
Image Credit: https://explorable.com/mayan-astronomy

The Mayan civilization is perhaps best known for their calendars. They had two calendar systems, one which was for ceremonies, called the Tzolk'in, which was based on a 260-day cycle, and the other calendar was called the Haab, which was based on the Solar year. This calendar was fairly accurate and with the Solar year calculated to be 365 days. It was separated into 18 months of 20 days and an additional five day long month. You probably remember just a few years ago on December 21, 2012, the world was supposed to end because the Mayan calendar came to an end (it obviously didn't, but that is besides the point). Observing the skies and creating a calendar helped the Mayans keep track of the seasons, which provided them with dates for harvest and helped in other agricultural areas. As we can see, astronomy is a field which has been present for over a millennium and has been incorporated into mythology and utilized as a indicator of time. 

Citations:

http://latinamericanhistory.about.com/od/ancientlatinamerica/p/Ancient-Maya-Astronomy.htm

http://www.historymuseum.ca/cmc/exhibitions/civil/maya/mmc07eng.shtml

https://explorable.com/mayan-astronomy

http://www.authenticmaya.com/maya_astronomy.htm

http://ircamera.as.arizona.edu/NatSci102/NatSci102/text/extmayaastronomy.htm

http://www.starteachastronomy.com/mayan.html

Worksheet 14.2, Problem 1: Tides

Draw a circle representing the Earth (mass \(M_\oplus\)), 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.

The Earth has mass \(M_\oplus\) and the Moon has mass \(M_{moon}\). The red dots along the circumference of the
Earth represent point-masses, m.

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

The Earth experiences a gravitational force from the Moon. The red vector
represents the force of the Moon on a point at the center of the Earth.

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

Each red vector arrow represents the force of the Moon on a given point on the Earth. Different points on Earth will experience different forces from the Moon due to their varying distances away from the Moon.

c) What is the force difference, \(\Delta \vec{F}\), between each point and Earth’s center? This is the tidal force.

The black vectors represent the tidal forces, which are the differences between the force vectors at each point along the circumference of the Earth and the force vector at the center of the Earth.

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

The light blue around the Earth depicts the effect of the tidal forces on the ocean.

Since the point on Earth closest to the Moon experiences a stronger gravitational pull from the Moon than the point at the center of the Earth, it accelerates more than the center of the Earth, and thus causes the ocean on this side to 'stretch' towards the Moon. Similarly, the point on Earth furthest from the Moon has a weaker gravitational pull from the Moon than the center of the Earth, causing it to accelerate less than the center. Thus here, the two points stretch apart from each other, giving the appearance that the ocean is stretching away from the Earth. These are what we know as the tides.

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

There are two tides on the Earth at any given moment of the day: on the side facing the moon and the side opposite the moon. Since the Earth completes one full rotation per day, any given location will experience a tide twice a day.

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 \(\Delta r\) compared to the Earth-Moon distance r, show that the force difference is given by 

\(\Delta F=\dfrac{2GmM_{moon}}{r^3}\Delta r\)

(Hint: recall that 

\(\displaystyle\lim_{\Delta x \to 0} \dfrac{f(x+\Delta x)-f(x)}{\Delta x}=\dfrac{d}{dx}f(x)\approx \dfrac{\Delta f(x)}{\Delta x}\)

And \(\Delta r\ll r\).)

The distance between the Moon and closest point on the Earth is r, and the diameter of the Earth (the distance between the closest and furthest points on the Earth) is \(\Delta r\). Let's call the force on the point on the Earth closest to the Moon \(F_1\) and the force on the point furthest from the moon \(F_2\). The difference in the gravitational forces on these two points is equal to:

\(\Delta F=F_1-F_2=\dfrac{GmM_{moon}}{\left(r-\frac{\Delta r}{2}\right)^2}-\dfrac{GmM_{moon}}{\left(r+\frac{\Delta r}{2}\right)^2}\)

           \(=GmM_{moon}\left[\dfrac{1}{\left(r-\frac{\Delta r}{2}\right)^2}-\dfrac{1}{\left(r+\frac{\Delta r}{2}\right)^2}\right]\)

           \(=GmM_{moon}\left[\dfrac{1}{r^2\left(1-\frac{\Delta r}{2r}\right)^2}-\dfrac{1}{r^2\left(1+\frac{\Delta r}{2r}\right)^2}\right]\)

We can Taylor expand this using the Taylor expansions \(\frac{1}{(1-\epsilon )^2}\approx 1+2\epsilon\)   and   \(\frac{1}{(1+\epsilon )^2}\approx 1-2\epsilon\).

 \(\Delta F=GmM_{moon}\left[\dfrac{1}{r^2}\left(1+\frac{\Delta r}{r}\right)-\dfrac{1}{r^2}\left(1-\frac{\Delta r}{r}\right)\right]\)

We simplify this to get the final answer:

\(\Delta F=\dfrac{2GmM_{moon}}{r^3}\Delta r\)

g) Compare the magnitude of the tidal force \(\Delta F_{moon}\) caused by the Moon to \(\Delta F_\odot\) 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?

We can use the expression we found from part (f) to compare the tidal forces from the Moon and from the Sun:

\(\dfrac{\Delta F_{moon}}{\Delta F_\odot}=\dfrac{\frac{2GmM_{moon}}{r_{moon}^3}\Delta r}{\frac{2GmM_\odot}{r_\odot^3}\Delta r}=\dfrac{M_{moon}}{M_\odot}\left(\dfrac{r_\odot}{r_{moon}}\right)^3\)

We know that \(M_{moon}=7.35\times 10^{25}\) g, \(M_\odot =1.99\times 10^{33}\) g, \(r_{moon}=3.63\times 10^8\) m, and \(r_\odot =1.50\times 10^{11}\) m. Plugging these values in, we get:

\(\dfrac{\Delta F_{moon}}{\Delta F_\odot}=\dfrac{7.35\times 10^{25}\text{ g}}{1.99\times 10^{33}\text{ g}}\left(\dfrac{1.50\times 10^{11}\text{ m}}{3.36\times 10^8\text{ m}}\right)^3\approx 2.6\)

The magnitude of the tidal force caused by the Moon is about 3 times stronger than the tidal force caused by the Sun. While the Sun is significantly more massive, it is much further away compared to the Moon, which is why it has a smaller tidal force.

Since both the Moon and the Sun cause a tidal force on the Earth, both forces are interacting with the Earth at the same time in varying ways. When the Sun and the Moon are on the same side of the Earth, their tidal forces will add to create larger tides on Earth. These tides are called Spring Tides. Conversely, when the Sun and the Moon are on opposite sides of the Earth, the tidal forces will subtract from each other to make smaller tides on Earth, called Neap Tides.

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

Again, we can use the expression from part (f) to create a ratio:

\(\dfrac{\Delta F_{moon}}{\Delta F_J}=\dfrac{M_{moon}}{M_J}\left(\dfrac{r_J}{r_{moon}}\right)^3\)

The mass of Jupiter is \(M_J=1.90\times 10^{30}\) g and its distance from the Earth on its closest approach is \(r_J\approx 4\text{ AU}\approx 7.50\times 10^{11}\) m. Plugging these in, we get:

\(\dfrac{\Delta F_{moon}}{\Delta F_J}=\dfrac{7.35\times 10^{25}\text{ g}}{1.90 \times 10^{30}\text{ g}}\left(\dfrac{7.50\times 10^{11}\text{ m}}{3.36\times 10^8\text{ m}}\right)^3\approx 3.4\times 10^5\)

We have found that the tidal force from the Moon is about 400,000 times stronger than the tidal force from Jupiter. So the tidal force from Jupiter can be considered negligible, and we can see that the Moon is the most dominant cause for tides on the Earth.

I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their collaboration in solving this problem.

Worksheet 14.1, Problem 2: The Mass-Radius Relation for White Dwarfs

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

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 \(v\).

We know from the general form of the equation for kinetic energy that:

\(K=\frac{1}{2} mv^2\)

We also know that momentum is given as:

\(p=mv\)

Rewriting kinetic energy to be in terms of momentum instead of speed, we get the equation:

\(K=\dfrac{p^2}{2m}\)

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?

Since this is a gravitationally bound system, we know that it is virtualized, which allows us to use Viral theorem.

\(K=-\frac{1}{2}U\)

Where \(U\) is the total gravitational energy. We can write an expression for the total kinetic energy and the total gravitational energy. In part (a), we already have an equation for the kinetic energy of a single particle of mass m. If we want to find the total kinetic energy of the electrons in the white dwarf, we need to multiply this by the number of particles. So we will get a total kinetic energy of:

\(K=\dfrac{p^2}{2m_e}N\)

Where N is the number of particles. For the total gravitational energy, we have already derived in a previous problem that the total gravitational potential energy for a sphere is given as:

\(U=-\dfrac{3GM^2}{5R}\)

Putting these expressions for total kinetic energy and potential energy into the Virial equation, we get:

\(\dfrac{p^2}{m_e}N=\dfrac{3GM^2}{5R}\)

c) According to the Heisenberg uncertainty Principle, one cannot know both the momentum and position of an election such that \(\Delta p\Delta x >\frac{h}{4\pi}\). Use this to express the relationship between the kinetic energy of electrons and their number density \(n_e\). (Hint: what is the relationship between an object’s kinetic energy and its momentum? From here, assume \(p\approx \Delta p\) and then use the Uncertainty Principle to relate momentum to the volume occupied by an electron assuming Volume\(\sim (\Delta x)^3\).)

We begin with the expression for total kinetic energy (for \(N_e\) electrons) in part (b):

\(K=\dfrac{p^2}{2m_e}N_e\)

If we make the approximation that \(p\approx \Delta p\) for the electron, then this equation becomes:

\(K=\dfrac{\Delta p^2}{2m_e}N_e\)

Furthermore, to find an expression for the number of electrons, \(N_e\), we can assume that the number of protons, \(N_p\) in the white dwarf is equal to the number of electrons, \(N_e\). Since the mass of protons is much larger than the mass of electrons, we can make the approximation that \(N_p\approx N_e\approx \frac{M}{m_p}\), where M is the total mass of the white dwarf. Putting this into our equation, we get:

\(K=\dfrac{\Delta p^2}{2m_e}(\dfrac{M}{m_p})\)

From here, we can introduce the Heisenberg uncertainty principle. If we solve for \(\Delta p\), we find \(\Delta p\sim \frac{1}{\Delta x}\). Also, we know that number density is equal to number per volume. If we approximate the volume occupied by a single electron to be \(\Delta x^3\), then we find that \(n_e=\frac{1}{\Delta x^3}\), or \(\Delta x=\frac{1}{n_e^{1/3}}\). So this means that we have the relation \(\Delta p\sim n_e^{1/3}\). We can rewrite our expression for kinetic energy to be:

\(K=\dfrac{n_e^{2/3}}{2m_e}(\dfrac{M}{m_p})\)

Dropping constants to find a proportionality relationship between kinetic energy and number density, we get an answer of:

\(K\sim Mn_e^{2/3}\)

d) Substitute back into your Virial energy statement. What is the relationship between \(n_e\) and the mass M and radius R of a WD?

From Virial theorem, which says that the total kinetic energy is equal to half of the total potential energy, we can plug in the relation from part (c) to get:

\(Mn_e^{2/3}\sim \dfrac{3GM^2}{5R}\)

Which gives us a relationship between \(n_e\), M, and R of:

\(n_e^{2/3}\sim \dfrac{M}{R}\)

e) Now, aggressively yet carefully drop constants, and relate the mass and radius of a WD.

From part (d) we found a relationship between the number density, mass, and radius of a white dwarf. We also know, however, that the number density of electrons in the white dwarf is the number per volume:

\(n_e\sim \dfrac{N_e}{V}\)

In part (c), we made the claim that \(N_e\approx \frac{M}{m_p}\), and the volume of the white dwarf would be \(V=\frac{4}{3}\pi R^3\). So we can rewrite number density in terms of the mass and the radius of the white dwarf:

\(n_e\sim \dfrac{\frac{M}{m_p}}{\frac{4}{3}\pi R^3}\)

Simplifying this into a scaling relation:

\(n_e\sim \dfrac{M}{R^3}\)

Now plugging this into our relationship from part (d), we get:

\(\left(\dfrac{M}{R^3}\right)^{2/3}\sim \dfrac{M}{R}\)

Now, we can solve for a scaling relationship between the mass and the radius:

\(M\sim \dfrac{1}{R^3}\)

f) What would happen to the radius of a white dwarf if you add mass to it?

Interestingly, we found from part (e) that the mass is inversely proportional to the radius cubed, which means that a white dwarf with a larger mass will have a smaller radius. This makes sense because the mechanism which is keeping the white dwarf from collapsing on its own weight is the electron degeneracy pressure. As we increase mass, the gravitational force becomes stronger at a faster rate than the force due to pressure from electrons, so the electrons can't hold up the white dwarf as much, causing a decrease in size.


I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their help in solving this problem.

Worksheet 13.2, Problem 2: Exoplanet Transits

In this problem, a different method for detecting exoplanets is introduced. In this method, an exoplanet can be found by observing the amount of light we receive from a star over time. If a planet is crossing over the star, it blocks some of the light from reaching us, and we can detect this small change. The image below shows a graph of what this would look like in observations.

Image Credit:
https://github.com/OSCAAR/OSCAAR/wiki/Introduction-to-Differential-Photometry

Draw a star projected on the sky, with a dark planet passing in front of the star along the star’s equator.

A planet of radius \(R_p\) transits a star of radius \(R_\star\).

a) How does the depth of the transit depend on the stellar and planetary physical properties? What is the depth of a Jupiter-sized planet transiting a Sun-like star?

The transit depth describes the extent to which a planet blocks the light from a star. We can already intuitively say that bigger planets will have transits of greater depth and planets transiting smaller stars will have a greater transit depth since they cover a larger proportion of the star's area. More precisely, the transit depth is the ratio of the projected area of the planet to that of the star:

\(d=\dfrac{\pi R_p^2}{\pi R_\star^2}\)

Where \(d\) is depth, \(R_p\) is the radius of the planet, and \(R_\star\) is the radius of the star. So the depth of the transit depends on the square of the ratio of the sizes of the planet and the star:

\(d=\left(\dfrac{R_p}{R_\star}\right)^2\)

For a Jupiter-sized planet transiting a Sun-like star, we know that the radius of Jupiter is approximately a tenth of the radius of the Sun, or that ten Jupiter-radii make one Solar radius. Using this fact, we find the depth of the transit to be:

\(d=\left(\dfrac{R_J}{10 R_J}\right)^2=\dfrac{1}{100}\)

So we have a depth of about 0.01 for a Jupiter-sized planet transiting the Sun, which implies that Jupiter would block 1% of the light from the Sun as it is transiting.

b) In terms of the physical properties of the planetary system, what is the transit duration, defined as the time for the planet’s center to pass from one limb of the star to the other?

A planet of radius \(R_p\) transits a star of radius \(R_\star\) with a transit time of \(t_t\).

To find the transit time, \(t_t\), we can use the equation \(d=rt\) where the distance traveled is the diameter of the star and the rate is the velocity of the planet's orbit. Knowing the orbital period of the planet, \(P\), and the planet's distance from the star, which we can call \(a_p\), we can find the  circumference of the orbit, and therefore the velocity of the planet:

\(v=\dfrac{2\pi a_p}{P}\)

Now, using the relation \(t=\frac{d}{v}\), we get a transit time of:

\(t_t=\dfrac{2R_\star}{\frac{2\pi a_p}{P}}=\dfrac{R_\star P}{\pi a_p}\)

So the transit time depends on the radius of the star, the distance of the planet away from the star, and the period of the planet's orbit. 

c) What is the duration of “ingress” and “egress” in terms of the physical parameters of the planetary system?

The ingress and egress are the motion of the planet crossing over the edge of the star (where ingress describes the planet as it moves into the sun, and egress describes the planet as it moves out of the sun). So the duration of the ingress and egress would be the time it takes for the planet to travel the distance of its own diameter. Again, using the equation \(d=rt\), we find that:

\(t_{in}=t_{eg}=\dfrac{2R_p}{\frac{2\pi a_p}{P}}=\dfrac{R_p P}{\pi a_p}\)

Citations:
http://kepler.nasa.gov/Science/about/characteristicsOfTransits/

I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their help in solving this problem.