Eclipsing Binary Lab

Artist's depiction of a binary system

Binary stars are stellar systems in which two stars orbit their common center of mass. When observed by the unaided eye from a far enough distance away, these dual-star systems appear to be a single point of light and thus can easily be mistaken for individual stars like our own Sun. Interestingly enough, these types of systems are not a rarity in the Universe. It is thought that nearly half of the stars that are visible from Earth are part of a multiple star system. Notable examples of binary systems include our closest stellar neighbor, Alpha Centauri, and the twin suns, Tatoo I and Tatoo II, that are orbited by Tatooine, homeworld of Anakin and Luke Skywalker.

Tatoo I and Tatoo II

While these binary systems are not as easily observed from Earth as they are from the surface of Tatooine, the students of Astronomy 16 set out to observe the NSVS01031772 system using the Clay Telescope on top of the Science Center. This particular binary system is a low-mass system, which makes it particularly interesting to observational astronomers because few low-mass systems have been discovered because, while they are predicted to be extremely common, their low brightness makes them difficult to detect. The properties and observations of this system were first laid out in a paper published in 2006 by López-Morales, et al. entitled "NSVS01031772: A New \(0.50\:+\:0.54\: M_{\odot}\) Detached Eclipsing Binary. The purpose of our lab was to gather information about the binary NSVS01031772 system by observing its eclipse and then subsequently calculate the stellar radii, masses, and separation between the two stars using our data. Since our data, being collected in Cambridge by novice astronomers was expected to be inferior to the data collected by the researchers, we could use their scientific paper to check our results.

The key source of our information comes from the fact that we observed an eclipsing binary system. An eclipse in a binary system happens when one of the stars passes in front of the other, thus blocking some of the flux that would have been detected by our telescope on Earth. When we observe a star using a sophisticated telescope, the telescope can produce a "light curve" that shows how much flux it receives from the observed object over time. In the case of an eclipsing binary, this flux is initially constant. This baseline flux occurs when the full receivable flux from both stars is received by the telescope on Earth. However, since the stars are in orbit around their mutual center of mass, one will eventually pass in front of the other and eclipse it. When this eclipse takes place, the light curve dips down as the flux decreases due to one star blocking the emitted flux of the other. When the smaller star in the system passes in front of the larger star, the dip is greater than when the larger star completely blocks the smaller star.

Methods

In order to calculate the radii, masses, and separation of the two stars, we can use many of the equations that we developed in class, especially in the exoplanet worksheet. Here are the equations that we can use when the system's center of mass is defined as zero.

\[\frac{-M_{1}a_{1}\:+\:M_{2}a_{2}}{M_{1}\:+\:M_{2}}\:=\:0\]
\[-M_{1}a_{1}\:+\:M_{2}a_{2}\:=\:0\]
\[\frac{M_{1}}{M_{2}}\:=\:\frac{a_{2}}{a_{1}}\]
This gives us the relationship between mass and the semi-major axis of the two stars.
To find the semi-major axises, we use the relationship between period and velocity:
\[P\:=\:\frac{2\pi a}{V}\:\rightarrow\:V\:=\:\frac{2\pi a}{P}\:\rightarrow\:a\:=\:\frac{VP}{2\pi}\]
\[a_{1}\:=\:\frac{V_{1}P}{2\pi}\]
\[a_{2}\:=\:\frac{V_{2}P}{2\pi}\]

Then using Kepler's Third Law, we can solve for the masses of the stars.
\[P^{2}\:=\:\frac{4\pi^{2} a^{3}}{G(M_{1}\:+\:M_{2})}\]
\[(M_{1}\:+\:M_{2})\:=\:\frac{4\pi^{2} a^{3}}{GP^{2}}\]
We also know that:
\[\frac{M_{1}}{M_{2}}\:=\:\frac{a_{2}}{a_{1}}\]
Thus:
\[M_{1}\:=\:\frac{M_{2}a_{2}}{a_{1}}\]
\[M_{2}(1\:+\:\frac{a_{2}}{a_{1}})\:=\:\frac{4\pi^{2} a^{3}}{GP^{2}}\]
This means that:
\[M_{1}\:=\:\frac{4\pi^{2} a^{3}}{GP^{2}(1\:+\:\frac{a_{2}}{a_{1}})}\:\times\:\frac{a_{2}}{a_{1}}\]
\[M_{2}\:=\:\frac{4\pi^{2} a^{3}}{GP^{2}(1\:+\:\frac{a_{2}}{a_{1}})}\]

Finally, in order to calculate the radius of the two stars, we need to use the method that we employed in our work with exoplanets. That is, we had to calculate the observed change in flux when one object passes in front of another and then use that to calculate the radius of each object. We know that luminosity is equal to Flux times the surface area of the star. This can be written as:
\[L\:=\:F\:\times\:A\]
\[F\:=\frac{L}{A}\]
We can describe the depth of our transit as the fractional difference in flux that occurs when the primary transit takes place, that is when the smaller star passes in front of the :
\[\frac{F_{initial}-F_{transit}}{F_{initial}}\:=\:\frac{(F_{1}\:+\:F_{2})-((F_{1}-F_{2})\:+\:F_{2})}{(F_{1}\:+\:F_{2})}\]
\[\frac{F_{2}}{F_{1}\:+\:F_{2}}\]
We know from class that this can be simplified to:
\[Depth\:=\:\frac{R_{2}^{2}}{R_{1}^{2}\:+\:R_{2}^{2}}\]
We also know that the duration of the primary transit is the time it takes for the smaller star to completely cross the diameter of the larger star, thus we know that:
\[t\:=\:\frac{2\pi R_{1}}{V_{2}}\]
\[R_{1}\:=\:\frac{V_{2}t}{2\pi}\]
Knowing the radius of the larger star, we can then substitute into the equation for the transit depth to solve for the radius of the second star.

Data and Observations
The Astronomy 16 class collected the data for this lab using the Clay Telescope, located on top of Harvard's Science Center. In total, we took 1004 images of our system. We used the red filter so we could best offset the urban conditions that detract from the clarity of our data collection. The data that we collected about the flux of the system was then normalized and folded so that all of the data from different nights could be recorded on the same graph. This graph, our light curve, is shown here:

As you can see from our light curve, the depth of the transit was about 0.7 and the duration of the transit was about 1.5 hours. These two values will be useful in our subsequent calculations.

Using the Radial Velocity Plot from the López-Morales, et al. paper, we can determine the radial velocity of the objects which will also be useful in our calculations.

The squares refer to the value of \(V_{2}\) and the circles refer to the value of \(V_{1}\).
From this we see that the max value for \(V_{2}\) is about \(1.8\:\times\:10^{2}\:km/s\) and the max value for \(V_{1}\) is about \(1.4\:\times\:10^{2} km/s\).

Analysis and Results
Before we begin to calculate, let's take a look at the values that we know:

From the light curve we know that:
Depth of Transit = 0.7
Time of transit = 1.5hrs = \(5.4\:\times\:10^{3} s\)

From the scientific paper we know:
\[V_{1}\:=\:1.4\:\times\:10^{7}cm/s\]
\[V_{2}\:=\:1.8\:\times\:10^{7}cm/s\]
\[P\:=\:8.83\:hrs\:=\:3.2\:\times\:10^{4}s\]


We can then use that data to find out the separation between the two stars.
\[a_{1}\:=\:\frac{V_{1}P}{2\pi}\:=\:\frac{1.4\:\times\:10^{7}(3.2\:\times\:10^{4}}{2\pi}\]
\[a_{1}\:=\:7\:\times\:10^{10}\:cm\]
\[a_{2}\:=\:\frac{V_{2}P}{2\pi}\:=\:\frac{1.8\:\times\:10^{7}(3.2\:\times\:10^{4}}{2\pi}\]
\[a_{2}\:=\:9.2\:\times\:10^{10}\:cm\]
\[a\:=\:a_{1}\:+\:a_{2}\:=\:1.62\:\times\:10^{11}\:cm\]

Now that we know \(a_{1}\) and \(a_{2}\), we can calculate the masses of the two stars:
\[M_{2}\:=\:\frac{4\pi^{2} a^{3}}{GP^{2}}\]
\[M_{2}\:=\:\frac{4\pi^{2} (1.62\:\times\:10^{11})^{3}}{6.67\:\times\:10^{-8}(3.2\:\times\:10^{4})^{2}}\]
\[M_{2}\:\approx\:1\:\times\:10^{33}g\:=\:.537\:M_{\odot}\]
\[M_{1}\:=\:M_{2}\:\times\:\frac{a_{2}}{a_{1}}\:=\:0.7 M_{\odot}\]

Finally, we just need to calculate the radii of the two stars. We know that:
\[R_{1}\:=\:\frac{V_{2}t}{2\pi}\]
\[R_{1}\:=\:\frac{1.8\:\times\:10^{7}\:\times\:5400s}{6.28}\]
\[R_{1}\:=\:1.55\:\times\:10^{10} cm\:=\:.22\:R_{\odot}\]
Now, to be able to find the radius of the second star, we need to use the equation for the depth of the transit, which we know to be about 0.7.
\[Depth\:=\:\frac{R_{2}^{2}}{(R_{1}^{2}\:+\:R_{2}^{2})}\:=\:0.7\]
\[R_{2}^{2}\:=\:2.33 R_{1}^{2}\]
\[R_{2}\:=\:1.53R_{1}\:=\:.34R_{\odot}\]

So we can see that our results do not perfectly align with the results set out in the paper in which,

\[M_{1}\:=\:0.54M_{\odot}\]
\[M_{2}\:=\:0.5M_{\odot}\]
\[R_{1}\:=\:0.53R_{\odot}\]
\[R_{2}\:=\:0.51R_{\odot}\]

The source of this error stems from the following:
1) The assumptions about the properties of the stars that we used to make our calculations
2) The relatively poor quality of our data collected using the Clay Telescope
3) My own human estimates of the quantities given by the graphs

Regardless of any errors that we may have made during the course of the observing and calculation process, it is nonetheless impressive and extremely exciting that we, a group of first-time astronomers, were able to observe this awesome natural phenomena taking place out there in the Universe. This was a very eye-opening experience for me, as I'm sure it was for my classmates as well, and it makes me excited about astronomy. I'm very grateful for the opportunity to take this class and be part of this lab. Who knows what we'll look at next!

Astronomy and Everyday Life

Astronomers are able to do some pretty interesting things with their modern instruments. They can look billions of light-years away, glimpsing the primordial Universe, detect far-off planets and even determine their atmospheric composition. They've used their telescopes to take detailed images of the Universe and have shared them with the world.

Pillars of Creation, Eagle Nebula, Taken by Hubble Space Telescope

However, in order to accomplish these feats, billions of dollars had to be expended to build and operate their expensive instruments. Considering no one alive today will actually visit these far-off regions of space, the average person might want to ask: Why are we spending so much money to get some cool desktop wallpapers for our computers? Ignoring the long-term usefulness of astronomy and its philosophically noble aim of understanding man's place in the Universe, what direct impact has astronomy had on everyday life? Through some serendipitous observations, astronomers have been able to create technologies with uses outside of the realm of astronomy. Here are a couple of those technologies.

Charge-Coupled Devices (CCDs)
CCDs are sensitive light detectors that were invented by researchers at Bell Labs working on memory storage. They were trying  As anyone who has used the Clay Telescope will know, CCDs are important to astronomers because they allow them to take clear images and data of the objects they are observing. Though not invented by them, astronomers recognized the potential for CCDs to be used for telescope imaging and adapted the technology to suit their needs. This innovation paved the way for digital imaging as we know it and today millions of people carry a CCD in their mobile phones.

WiFi
WiFi is one of the most important technologies in our lives today and its availability can make or break a budding coffee shop. It allows us to connect to the Internet and other devices from many locations without physically connecting our computer. WiFi was invented by John O'Sullivan, an Australian astronomer who used techniques from radio astronomy. Dr. O'Sullivan has even said that "Curiously, it was a failed experiment to detect exploding mini black holes the size of an atomic particle that led to this work."

These inventions are just some of many invaluable technological developments that have come into everyday use because of astronomy. While technology transfer itself hardly justify the economic cost of astronomy research (that justification comes from emphasis on the long-term and understanding the Universe), the importance and impact of astronomy is evident all around us, whether you look up at the stars or down at the lens of your digital camera to take a CCD enabled selfie.

Worksheet 14.2, Problem 1

Draw a circle representing the Earth (mass M_o), with 8 equally-spaced point masses, m, placed around the circumference. Also draw the Moon with mass M_c 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_c,cen, on a point at the center of the Earth? Recall that vectors have both a magnitude (arrow length) and a direction (arrow head). 


 

(b)  What is the force vector on each point mass, F_c, due to the Moon? Draw these vectors at each point. 


 

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


 In order to get the force difference, we subtract the vector of the central force from the vector of the force at each point and are left with the following (where green arrows represent the net force).

 

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


At each point, the ocean will rise slightly in the direction of the net force (green arrows from the previous problem). As you can see, this will cause high tides on both the near and far sides of the Earth and low tides in between.

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


Each given location located along the moon's orbital plane experiences 2 low tides and 2 high tides. This is because the Earth's rotation causes the moon to "cross over" a given location on the Earth two times. The first time it is near or overhead, the location will experience a high tide. As it moves away, the location will experience a low tide. However, once the moon "reaches" the opposite side of the Earth, the location will once again experience a high tide and will then experience a low tide when the moon moves 90 degrees away in the Earth's rotation.

(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:

ΔF=2GmMmoonr3Δr

(HINT: recall that

limx→∞f(x+Δx)−f(x)Δx=ddxf(x)≈Δf(x)Δx

and Δr<<r.

We know that:

\[\frac{d}{dx}f(x)\:\approx\:\frac{\Delta f(x)}{\Delta x}\]

We can then rewrite this with variables that pertain to our specific situation:

\[\frac{d}{dr}F(r)\:\approx\:\frac{\Delta F(r)}{\Delta r}\]

\[\Delta F\:=\:\frac{dF(r)}{dr}\Delta r\]

We know that F(r) is equal to:

\[F(r)\:=\:-\frac{GM_{moon}m}{r^{2}}\]

Taking the derivative and substituting into the equation for \(\Delta F\) we see that:

\[\Delta F\:=\:\frac{2GM_{moon}m}{r^{3}}\Delta r\]

(g)  Compare the magnitude of the tidal force ∆F_c caused by the Moon to ∆F_o 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? 


Using the equation above, we know that:

\[\Delta F_{moon}\:=\:\frac{2GM_{moon}m}{r_{moon}^{3}}\Delta r\]

\[\Delta F_{Sun}\:=\:\frac{2GM_{sun}m}{r_{sun}^{3}}\Delta r\]

Thus, we can see that:

\[\frac{\Delta F_{moon}}{\Delta F_{sun}}\:=\:\frac{M_{moon} r_{sun}^{3}}{M_{sun} r_{moon}^{3}}\]

If we plug actual values into this equation, we see that the ratio is about 2.35. This means that the tidal force from the moon is twice as strong as the tidal force from the Sun. Even though the Sun is much more massive than the moon, it is also much farther away so it doesn't have a larger tidal force.

(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)? 


From the previous part of the problem, we can infer that:

\[\frac{\Delta F_{moon}}{\Delta F_{Jupiter}}\:=\:\frac{M_{moon} r_{Jupiter}^{3}}{M_{Jupiter} r_{moon}^{3}}\]

We can then plug our known values for the radii and mass of both the Moon and Jupiter, and can see that this ratio is equal to about \(1.4\:\times\:10^{5}\). This implies that the tidal force from the moon is about 140,000 times stronger than the tidal force from Jupiter even when Jupiter is on its closest approach.

Worksheet 14.1: Problem 2: White Dwarfs

A white dwarf 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. 


Kinetic Energy is equal to \(\frac{1}{2}mv^{2}\).
Momentum, \(p\), is equal to \(mv\).
We can then rewrite Kinetic Energy as:
\[K\:=\:\frac{p^{2}}{2m}\]

(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? 


White Dwarfs can be assumed to be perfect blackbodies in this problem, and thus we can use the Viral Theorem to relate kinetic and potential energy. The Virial Theorem tells us that:
\[K\:=\:-\frac{1}{2}U\]

(c)  According to the Heisenberg uncertainty Principle, one cannot know both the momentum and the 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 n_e (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.) 


 

From the Heisenberg uncertainty Principle, we can assume the following expression:

\[p\Delta x\:=\:\frac{h}{4\pi}\]

We can then rearrange that to solve for momentum and see that:

\[p\:=\:\frac{h}{4\pi \Delta x}\]

Now that we have another equation that involves momentum, we can set it equal to the first relationship that we obtained in part a:

\[K\:=\:\frac{p^{2}}{2m}\]

\[p\:=\:\sqrt{2Km}\]

We can now set our two expressions equal to each other:

\[\frac{h}{4\pi \Delta x}\:=\:\sqrt{2Km}\]

Solving for K, we find that:

\[K\:=\:\frac{h^{2}}{32 \pi^{2} \Delta x^{2} m}\].

Now, we have to express this relationship in terms of the number density in the region of \(\Delta x\_, which seems tricky at first because we don't have an explicit term for the number density given to us. Number Density is equal to:

\[n_{e}\:=\:\frac{Number\:of\:Electrons}{Volume}\]

We know that volume scales like so:

\[Volume\:\sim\:\Delta x^{3}\].

We can then substitute this into our expression for the number density and get:

\[n_{e}\:=\:\frac{Number\:of\:Electrons}{\Delta x^{3}}\]

\[n_{e}\:=\:\frac{1}{\Delta x^{3}}\]

 Next, we can rearrange our expression for K:

\[K\:=\:\frac{h^{2}}{32 \pi^{2} \Delta x^{2} m}\]

\[\Delta x^{2}\:\sim\:n_{e}^{-\frac{2}{3}}\]

\[K\:=\:\frac{h^{2}n_{e}^{\frac{2}{3}}}{32 \pi^{2}m}\]

Since we used terms pertaining to a single electron, this is the expression for the Kinetic Energy of one electron. In order to find the energy of the entire White Dwarf, we will need to know the number of electrons. The number of protons and electrons is about equal in the star, so the number of electrons is as follows:

\[N_{e}\:=\:N_{p}\:=\:\frac{M}{m_{p}}\]

We can multiply the kinetic energy of one electron by this expression in order to get the total kinetic energy of the white dwarf. Thus:

\[K_{tot}\:=\:\frac{h^{2}n_{e}^{\frac{2}{3}}M}{32 \pi^{2}m_{e}m_{p}}\]

 

(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? 


The Virial Theorem is as follows:

\[K\:=\:-\frac{1}{2}U\]

We know from the previous part that the total kinetic energy of the star is equal to:

\[K_{tot}\:=\:\frac{h^{2}n_{e}^{\frac{2}{3}}M}{32 \pi^{2}m_{e}m_{p}}\]

We can then set these equal to each other and see that:

 \[\frac{h^{2}n_{e}^{\frac{2}{3}}M}{32 \pi^{2}m_{e}m_{p}}\:=\:\frac{GMm_{e}}{2R}\]

Next, we can solve for number density and be left with:

 \[n_{e}\:=\:(\frac{16\pi^{2}Mm_{e}m_{p}}{Rh^{2}})^{\frac{3}{2}}\]

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


Starting with the expression for number density we got in the last part, we can start aggressively dropping constant to get:

\[n_{e}\:=\:(\frac{16\pi^{2}Mm_{e}m_{p}}{Rh^{2}})^{\frac{3}{2}}\]

\[n_{e}\:\sim\:(\frac{M}{R})^{\frac{3}{2}}\]

Then, if we assume that the number density of electrons is equal to the number density of protons, we can arrive at the following relationship:

\[n_{e}\:=\:\frac{Mass}{Volume}\:\sim\:\frac{M}{R^{3}}\]

Thus:

\[\frac{M}{R^{3}}\:\sim\:(\frac{M}{R})^{\frac{3}{2}}\]

Finally, by rearranging, we see that:

\[M\:\sim\:\frac{1}{R^{3}}\]

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


Strangely enough, the above relationship implies that the radius would decrease if you were to add more mass to it. This is contrary to what we have seen in larger, main sequence stars.

Controversy Over a New Telescope in Hawaii

Mauna Kea, a dormant volcano located on the northern side of the island of Hawaii, is a spectacular sight to behold. Rising 13,796 feet above the surface of the Pacific Ocean, Mauna Kea's snow-capped peak starkly contrasts the tropical climate found at its base. Measured from its base on the ocean floor, Mauna Kea is in fact the tallest mountain on Earth, standing over 30,000 feet tall.

The mountain is prized by astronomers because of it is perhaps the best spot on Earth for astronomical observing. It currently hosts 13 operating telescopes that combined are 60 times more powerful than the Hubble Space Telescope. Clearly, this observation post is very important for the study of astronomy.

Since this is such a great spot to put a telescope, there are plans to continue constructing the Thirty Meter Telescope, a 1.4 billion dollar project that, when completed in 2024, will astronomers to see 13 billion light-years away and observe the formation of galaxies at the beginning of the Universe.

A rendered model of the TMT

However, astronomers have been met by large, viral protests by native Hawaiians who feel that this massive construction project will further impinge on their sacred mountaintop.

Protestors on the mountaintop, at the University of Hawaii, and throughout the world have voiced their opposition to the construction of the TMT. Their protests have questioned scientists' right to appropriate their land to further studies, even though those studies benefit all of mankind. In the meantime, while the protests continue, it is uncertain if the project will be completed on schedule if at all.

Astrobites Daily Paper Summary

Are Extrasolar Worlds More Likely to Be Water-rich?
Link: http://astrobites.org/2015/03/06/are-extrasolar-worlds-more-likely-to-be-water-rich/

Water is all around us. From the vast oceans covering the planet to the cells of our body, we are completely surrounded by this abundant molecule. Life as we know it needs water to survive and thus it is very important to us. When considering the existence of extraterrestrial life, scientists, based on our current observations and understanding, believe that water is most likely necessary for the development and propagation of life, and thus planets that have water are more likely to harbor life.

Though we are surrounded by so much water here on Earth, we never seem to think about where it came from. It most likely was deposited over time by water-rich asteroids that impacted with the Earth after its formation. Since the Earth formed closer to the Sun than the asteroids, less water remained in the final planetary formation. The asteroids formed in a region with a lower temperature and therefore had more water incorporated into their final form.

When considering the amount of water that exists in extrasolar systems, we have to consider the conditions under which the solar systems formed. Short-lived radioactive isotopes caused much of the heating of the solar system and thus determined the loss of the amount of volatile substances such as water. However, the amount and concentration of these isotopes is thought to vary across systems and thus some systems may have more volatiles like water.

We can also see that the amount of water varies depending on the distance away from the star because of the difference in temperature.

Worksheet 13.2: Problem 2: Transiting Exoplanets

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

 

 

(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 depth of the transit depends on how much the flux changes when the exoplanet crosses in front of the star. The amount of energy that is blocked depends upon both the visible area of the planet and of the star. Since luminosity is equal to the flux times the area of the star, we know that the flux received by an observer during the transit is equal to:

\[F_{t}\:=\:\frac{L_{*}}{A_{*}-A_{p}}\]

In order to find the depth of the transit, we have to compare the flux received during the transit to the flux normally received:

\[\frac{F_{t}}{F_{*}}\:=\:\frac{\frac{L_{*}}{4 \pi (R_{*}^{2}-R_{p}^{2})}}{\frac{L_{*}}{4 \pi R_{*}^{2}}}\]

\[\frac{F_{t}}{F_{*}}\:=\:(1-\frac{R_{p}^{2}}{R_{*}^{2}})\]

The depth of the transit is represented by the term \(\frac{R_{p}^{2}}{R_{*}^{2}}\).

 Since Jupiter's radius is roughly one-tenth that of the Sun, this term is equal to \(10^{-2}\).

(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? 


We know that time is equal to \(\frac{Distance}{Velocity}\). The distance that the planet's center travels is twice the radius of the star \((2R_{*}\).  We also know that the general equation for velcoity is \(V_{p}\:=\:\frac{2\pi a}{P}\). Using Kepler's Law, we can rewrite that as:

 

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


These terms describe the amount of time required for a planet to eclipse its star. Both terms repreesnt  the same amount of time. This is equal to the amount of time required to travel two radii because the distance the planet travels from the beginning to the end of the eclipse is two of it radii.We can then substitute and arrive at an expression for this time in terms of the physical parameters.