Comet Shoemaker-Levy 9 and the Jovian Impact

Shoemaker Levy - 9: The comet that slammed into Jupiter
Source: http://www2.jpl.nasa.gov/sl9/gif/sl9_home.gif

The Shoemaker-Levy comet is one of the most well-known comets in astronomy, recognized by everyone from ameteur backyard space enthusiast to the veterans at NASA. This comet showed us for the first time the immense power a comet can have if it slams into the planet.

No... Shoemaker-Levy 9 was not aiming for Earth. It's target was bigger. Much bigger. Shoemaker-Levy 9 was captured into Jupiter's orbit, and in July 1994, slammed into the gas giant. This was the first time humans had the chance to witness the destructive power of space debris left over from the early days of the formation of the solar system collide catastrophically with a planet. With nearly every ground and space-based telescope pointed towards Jupiter, we saw the comet, which had broken into 21 pieces, collide on the far side of the planet.

So, how did we know exactly when and where this comet was going to strike, in order to have enough preparation to set up so many ground based telescopes, and even the Hubble Space Telescope, to capture the impact? Well, the comet was discovered in March 1993 by David Levy, and Eugene and Carolyn Shoemaker. When the comet's orbit was mapped, it became apparent that it was, in fact, not orbiting the Sun, but rather orbiting Jupiter. Projecting the orbital calculations, it was clear that the comet would collide into Jupiter in July 1994.



Okay, so a comet is colliding into Jupiter. What is so special about that? The fact that this was the first time humans were able to witness a comet collide into another planet was a once-in-a-lifetime opportunity. It provided scientists with a look at the impact zones of where the comet his Jupiter, and see the massive scarring that it caused. It was rare to see black spots on the surface of Jupiter where the comet hit, especially since Jupiter is made up of mostly gasses. Therefore, the collision had profound impact in the study of the Jovian atmosphere, and helped scientist determine some of its chemical composition and wind speeds.


It was unfortunate that the spacecraft Galileo was still en-route to Jupiter, and was unable to capture the event up-close. However, Galileo was able to observe how the comet collision disrupted Jupiter's rings, and later caused the entire ring to tilt by 2 km. A flyby of New Horizons en-route to Pluto also showed that the ring has not fully recovered, and detected disturbances in the ring.

Watching the destructive power that a comet can have on a gas giant like Jupiter really paved the way for NASA's Near Earth Objects (NEO) program to look for and track the paths of all space rocks that could be potentially be dangerous for life on Earth. It also showed that because of Jupiter's immense gravity, it has been attracting space rocks into its orbit and shielding the inner planets from space debris since the early days of the solar system. In fact, had it not been for Jupiter attracting comets and asteroids en-route towards the inner planets, collisions might have been much more frequent on Earth, and extinction-level events such as those that killed off the dinosaurs may have been more common, to the point that complex life might never have gotten a foot-hold.

Worksheet 6, Problem 4: Binary Star System

Question: You observe two gravitationally bound stars (a binary pair). One is blue and one is yellow. The yellow star is six times brighter than the blue star. Qualitatively compare their temperature and radii, i.e. which is hotter, which is smaller? Next, quantitatively compare their radii (to 1 significant figure). 

A binary star system
Source: http://www.vikdhillon.staff.shef.ac.uk/seminars/lives_of_binary_stars/binary_movie.gif

A binary star system is one where two stars orbit around a central gravitational mass. In this scenario, we have a bright yellow star and a relatively dim blue star that are orbiting each other. Just by this information, we already can tell some important characteristics about these two stars.

Electromagnetic Spectrum: Blue light has shorter wavelength than yellow light
Source: http://www.astronomersgroup.org/images/EMspectrum.jpg

First of all, the different colors of the stars can tell us that the blue star is hotter than the yellow star. This is because on the electromagnetic spectrum, blue wavelengths of light are shorter than yellow wavelengths of light. Since all electromagnetic waves travel at the speed of light, you can relate wavelength and frequency with $c = \nu \lambda$, where $\nu$ is the frequency and $\lambda$ is the wavelength. As wavelength increases, frequency decreases, and vice versa, when wavelength decreases, frequency increases. But what does wavelength and frequency have to do with temperature of the star? Well, the higher the frequency of the star, the more energy it has, as shown by the Planck-Einstein relation, $E = h\nu$. Since a blue star has a shorter wavelength, and therefore higher frequency than a yellow star, it has more energy than a yellow star. Since temperature is simply a measure of the kinetic (and sometimes potential!) energy, the more energy there is, the hotter something is. Therefore, since the blue star has more energy, it has a higher temperature than the yellow star, and is therefore hotter.

In terms of size, since the luminosities of the stars are different, with the yellow star being much brighter than the blue star, you can look at the equation for luminosity for a blackbody (a star is a good blackbody), $L = 4 \pi R^2 \sigma T^4$, where $R$ is the radius of the star, $T$ is the temperature, and $\sigma$ is the Stefan-Boltzmann constant. Removing the constants, it is apparent that the luminosity of a star is proportional to its radius and temperature, $L \propto R^2T^4$. Since we have established that the blue star is much hotter, the fact that it is much dimmer than the yellow star would mean that the yellow star has to be much larger to compensate for lack of temperature in order to be brighter than the blue star.

In order to show that the yellow star is larger than the blue star, we can quantitatively calculate just how much larger the yellow star is. We again use the luminosity equation, $L = 4 \pi R^2 \sigma T^4$:

• The yellow star's luminosity is $L_{Y}$ with a radius $R_{Y}$ and temperature $T_{Y}$
• The blue star's luminosity is $L_{B}$ with a radius $R_{B}$ and temperature $T_{B}$

Since we know that the yellow star is 6 times as bright as the blue star, we can relate the two as follows: 

\begin{align}
 L_{Y} &= 6L_{B} \\
 4\pi R_{Y}^2 \sigma T_{Y}^4 &= 6 \cdot 4\pi R_{B}^2 \sigma T_{B}^4 \\
R_{Y}^2 T_{Y}^4 &= 6 R_{B}^2 T_{B}^4 \\
\frac{R_{Y}^2}{R_{B}^2} &= \frac{6T_{B}^4}{T_{Y}^4}\\
\sqrt{\frac{R_{Y}^2}{R_{B}^2}} &= \sqrt{\frac{6T_{B}^4}{T_{Y}^4}}\\
\frac{R_{Y}}{R_{B}} &= \sqrt{6}\frac{T_{B}^2}{T_{Y}^2}
\end{align}

Hmm... we seem to be stuck here. We don't know the actual temperatures of the two stars. However, we don't need to know their actual temperatures, since we can use the Wien Displacement Law, $\lambda_{max} = \frac{b}{T}$, where $\lambda_{max}$ is the peak of the wavelength, $b$ is the Wien's displacement constant, $0.3$ cm K, and $T$ is the temperature of the blackbody. We know the wavelengths of both blue light (450 nm) and yellow light (580 nm), so we can solve for the temperature using the Wien Displacement Law. By rearranging the Wien's displacement law, we get $T = \frac{b}{\lambda_{max}}$. We can substitute this into our relation of the blue and yellow stars:

\begin{align}
\frac{R_{Y}}{R_{B}} &= \sqrt{6}\left(\frac{T_{B}}{T_{Y}}\right)^2\\
\frac{R_{Y}}{R_{B}} &= \sqrt{6}\left(\frac{\frac{b}{\lambda_{B_{max}}}}{\frac{b}{\lambda_{Y_{max}}}}\right)^2\\
\frac{R_{Y}}{R_{B}} &= \sqrt{6} \left( \frac{\lambda_{Y_{max}}}{\lambda_{B_{max}}}\right)^2\\
\frac{R_{Y}}{R_{B}} &= \sqrt{6} \left( \frac{580 nm}{450 nm}\right)^2\\
\frac{R_{Y}}{R_{B}} &\approx 4
\end{align}

Based on the ratio of the radius of the yellow star to the blue star, the yellow star has a radius that is approximately 4 times as big as that of the blue star!

Worksheet 5, Problem 2: Blackbodies in Space

Blackbodies are nice because they're such simple objects. Their outward appearance is entirely determined by their temperature. If there were cows in space, astronomers would imagine them to be spherical blackbodies (and seriously, it wouldn't be a bad approximation). In this exercise we'll take advantage of the relative simplicity of blackbodies to derive some useful expressions that you'll use during this term, and throughout the your astronomy career.

Questions:

In astronomy, it is often useful to deal with something called the "bolometric flux," or the energy per area per time, independent of frequency. Integrate the blackbody flux $F_{\nu}(T)$ over all frequencies to obtain the bolometric flux emitted from a blackbody, F(T). You can do this by substituting the variable $u = \frac{h{\nu}}{kT}$. This will allow you to split things into a temperature-dependent term, and a term comprising an integral over all frequencies. However, rather than solving for the integral, just set the integral and all constants equal to a new, single constant $\sigma$, which is also known as the Stefan-Boltzmann constant. If you're really into calculus, go ahead and show that $\sigma \approx 5.7 \times 10^{-5} erg s^{-1} cm^{-2} K^{-4}$. Otherwise, commit this number to memory.

In order to start off this derivation, let's look at the equation for the intensity of radiation emitted from a blackbody, which is given by the following equation:

\begin{align}
I_{\nu}(T) = \frac{2{\nu}^2}{c^2} \cdot \frac{h{\nu}}{e^{\frac{h{\nu}}{kT}} - 1}
\end{align}

Now that is a complicated equation! Let's think about what it means. This equation shows the energy per time, per area, per frequency, per solid angle. However, we are looking for the "bolometric flux", which is energy per area, per time, independent of frequency. So the equation we have for intensity, $I_{\nu}(T)$ involves a "per frequency" and "per solid angle" component. In order to get rid of these, let's first think about the "solid angle". You can think of a solid angle as a small patch of area on the surface of a sphere. Since a blackbody emits radiation isotropically from every point, you can integrate $I_{\nu}(T)$ over all the solid angles to get the "blackbody flux", $F_{\nu}(T)$. Since we are dealing with a spherical body, you have to integrate over the $\theta$ dimension, as well as the $\phi$ dimension, and take into account the fact that the "solid angle" gets smaller the higher the latitude, which is represented by $\sin(\theta)$. Also, since we are dealing with isotropic radiation, we have to consider only the radiation that is facing us, which can be determined by the $\cos(\theta)$, the component of the radiation that is actually affecting us. Therefore, in order to get the flux of the blackbody:

\begin{align}
F_{\nu}(T) &= \int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{2}} I_{\nu}(T) \cos \theta \sin \theta d\theta d\phi\\
F_{\nu}(T) &= \pi I_{\nu}(T)
\end{align}

Now, in order to get rid of the "per frequency" part of the the flux equation $F_{\nu}(T)$, and get the "bolometric flux" $F(T)$, we can integrate over all the frequencies possible. We can use the variable $u = \frac{h{\nu}}{kT}$ to help with the integration.

\begin{align}
F(T) &= \int_{0}^{\infty} F_{\nu}(T) d{\nu}\\
&= \int_{0}^{\infty} \pi I_{\nu}(T) d{\nu}\\
&= \int_{0}^{\infty} \pi \cdot \frac{2{\nu}^2}{c^2} \cdot \frac{h{\nu}}{e^{\frac{h{\nu}}{kT}} - 1} d{\nu}\\
&= \frac{2 \pi k^4 T^4}{c^2 h^2} \int_{0}^{\infty} \frac{u^3}{e^u - 1} du\\
&= \frac{2 \pi^5 k^4 T^4}{15 c^2 h^3}\\
F(T) &= \sigma T
\end{align}

The value of $\sigma = 5.7 \times 10^{-5} erg s^{-1} cm^{-2} K^{-4}$ is the coefficient for $T$, also known as the Stefan-Boltzmann constant.

The Wien Displacement Law: Convert the units of the blackbody intensity from $B_{\nu}{T}$ to $B_{\lambda}(T)$. IMPORTANT: Remember that the amount of energy in a frequency interval $d{\nu}$ has to be exactly equal to the amount of energy in the corresponding wavelength interval $d \lambda$.

Okay, so we have the intensity of a blackbody given by $I_{\nu}(T)$, which can also be represented by $B_{\nu}(T)$. However, $B_{\nu}(T)$ represents the intensity per a given interval of frequency, $\nu$. However, we are looking for the intensity of a blackbody per a given wavelength $\lambda$, represented by $B_{\lambda}(T)$.

Since we are told that the amount of energy in a given frequency interval $d_{\nu}$ is equal to the amount of energy in the corresponding wavelength interval $d \lambda$, we can say that $B_{\nu}(T) d{\nu} = B_{\lambda}(T) d \lambda$. Solving for $B_{\lambda}(T)$, we get:

\begin{align}
B_{\lambda}(T) = B_{\nu}(T) \frac{d \nu}{d \lambda}
\end{align}

Also, we know that frequency and wavelength can be related by the equation, $c = \nu \lambda$. Solving for $\nu$ and taking it's derivative with respect to $d \lambda$, we get:

\begin{align}
\nu &= \frac{c}{\lambda}\\
\frac{d\nu}{d\lambda} &= -\frac{c}{{\lambda}^2}
\end{align}

Substituting for both $\frac{d\nu}{d\lambda}$ and $B_{\nu}(T)$, you get:

\begin{align}
B_{\lambda}(T) &= B_{\nu}(T) \frac{d \nu}{d \lambda}\\
&= - \frac{2{\nu}^2 }{c^2} \frac{h\nu}{e^{\frac{h\nu}{kT}} - 1} \cdot \frac{c}{{\lambda}^2}\\
&= -\frac{2h{\nu}^3} {c {\lambda}^2 \left( e^{\frac{h\nu}{kT}} - 1 \right) }\\
&= -\frac{2h{\nu}^2} {{\lambda}^3 \left( e^{\frac{h\nu}{kT}} - 1 \right) }\\
&= -\frac{2hc^2} {{\lambda}^5 \left( e^{\frac{h\nu}{kT}} - 1 \right) }
\end{align}

Therefore,  $B_{\lambda}(T) = -\frac{2hc^2} {{\lambda}^5 \left( e^{\frac{h\nu}{kT}} - 1 \right) }$

Derive an expression for the wavelength $\lambda_{\text{max}}$ corresponding to the peak of intensity distribution at a given temperature T. (HINT: How do you find the maximum of a function? Once you do this, again, substitute $u = \frac{hv}{kT}$). The expression you end up with will be transentental, but you can solve it easily to first order, which is good enough for this exercise.

In order to find the $\lambda_{max}$ corresponding to the peak intensity distribution at a given temperature T, we can find the maximum of the blackbody intensity function with respect to the wavelength $\lambda$. In order to find the maximum, we can set the derivative of $B_{\lambda}(T)$ with respect to wavelength $\lambda$ to 0. It is worth noting that you can substitute $u = \frac{h\nu}{kT}$. You can also use the relationship between frequency and wavelength, $\nu = \frac{c}{\lambda}$ and substitute it into $u = \frac{h\nu}{kT}$ to get $\lambda = \frac{hc}{ukT}$. This will help in the derivation to find the maximum:

\begin{align}
\frac{dB_{\lambda}(T)}{d\lambda} = 0\\
\frac{d}{d\lambda}\left( -\frac{2hc^2} {{\lambda}^5 \left( e^{\frac{h{\nu}}{kT}} - 1 \right) }\right) = 0\\
\frac{d}{du} \left( \frac{2hc^2}{\left(\frac{hc}{ukT}\right)^5 (e^u - 1)} \right) = 0\\
\frac{d}{du} \left( \frac{2u^5k^5T^5}{h^4c^3\left(e^u - 1\right)} \right) = 0
\end{align}

Here, we can drop the constants, so that we're left with a derivation only in terms of $u$:

\begin{align}
\frac{d}{du} \frac{u^5}{e^u - 1} = 0\\
\frac{5u^4(e^u -1) - u^5(e^u)}{(e^u - 1)^2} = 0\\
5(e^u - 1) - ue^u = 0\\
5e^u - 5 - ue^u = 0\\
(5-u)e^u = 5
\end{align}

Now, in order to get the maximum value, we need to solve for $u$. We can use the Taylor approximation of $e^u = 1 + u$ to approximate the value of $u$:

\begin{align}
(5-u)(u + 1) \approx 5\\
4u - u^2 + 5 \approx 5\\
u^2 - 4u \approx 0\\
u - 4 \approx 0\\
u \approx 4.
\end{align}

To get the maximum wavelength $\lambda_{max}$, we can substituting the value of $u$ into the equation $\lambda = \frac{hc}{ukT}$, to get :

\begin{align}
\lambda_{max} = \frac{hc}{4kT}
\end{align}

The Rayleigh-Jeans Tail: Next, let's consider photon energies that are much smaller than the thermal energy. Use a first-order Taylor expansion on the term $e^{\frac{h{\nu}}{kT}}$ to derive a simplified form of $B_{\nu}(T)$ in this low-energy regime. (HINT: The Taylor expansion of $e^x \approx 1 + x$).

The term $e^{\frac{h\nu}{kT}}$ can be simplified using the first order Taylor approximation such that:

\begin{align}
e^{\frac{h\nu}{kT}} \approx 1 + \frac{h\nu}{kT}
\end{align}

We can then simplify the equation for radiation emitted by a blackbody, $B_{\nu}(T)$ as such:

\begin{align}
B_{\nu}(T) &= \frac{2{\nu}^2}{c^2} \cdot \frac{h{\nu}}{e^{\frac{h{\nu}}{kT}} - 1}\\
&\approx   \frac{2{\nu}^2}{c^2} \cdot \frac{h{\nu}}{\frac{h\nu}{kT}}\\
&\approx   \frac{2{\nu}^2kT}{c^2}
\end{align}

This simplified version of intensity $B_{\nu}(T) \approx \frac{2{\nu}^2kT}{c^2}$ is an approximation for energies that are much smaller than thermal radiation, often known as the Rayleigh-Jeans Tail.

Write an expression for the total power output of a blackbody with a radius R, starting with the expression for $F_{\nu}$. This total energy output per unit time is also known as the bolometric luminosity, $L$.

The total energy output per time, also known as bolometric luminosity. We know from part 1 that the bolometric flux of a blackbody is given in terms of energy per area per time. Therefore, in order to get the bolometric luminosity, we need to figure out the surface area over which the blackbody is radiating energy from and multiply it by the bolometric flux to get rid of the "per area" component. Since we approximate all blackbodies to a spherical shapes, the surface area over which the blackbody is radiating is given by $4\pi R^2$.

Therefore, the bolometric luminosity, $L$ is given by:
\begin{align}
L &= \text{Surface Area} \cdot F(T)\\
L &= 4 \pi R^2 \sigma T^4
\end{align}

A Free for All: The Cassini - Huygen's Mission

Source: NASA.gov

The picture on the left is one of the most profound images in all of human history. Take a guess as to what it is a picture of?

...

...

...

It seems to have roundish rocks in a very orange-ish air.

...

Since this is an astronomy class, this is probably not a picture of the Earth. So could it be Mars?

...

But those rocks in the picture don't look like they're from Mars, because they are so round. They look like something from a dried lake on Earth, but the atmosphere looks so much like the pictures taken by the rovers on Mars.

...

In fact, this is not a planet. It is a satellite of the beautiful Saturn. This is Titan, Saturn's largest moon.

Source: NASA.gov

It was a long-held belief in the scientific community that in order to find life and other interesting phenomenon in the solar system, you needed to visit the planets. And since we knew that all the planets beyond Mars were gas giants, there was absolutely no point in looking for signs of life on those planets. However, venturing into the solar system through the Voyager program, it was revealed that the moons of the gas giants were far more robust than the icy worlds we had always imagined them to be. The gravitational pull of their monstrous parent planets, especially Jupiter and Saturn, created tectonic activity, thermal vents, active vol
canoes, and underground oceans within these satellites. Suddenly, there was an uptake interest to explore these moons, one of which was Titan.

Titan had always been a mystery, because looking at it through a telescope, or even through a probe, would only allow you to see is a mysterious yellow-orange haze. However, in 1997, the joint effort of NASA, the European Space Agency, and the Italian Space Agency sent the Cassini-Huygens probe to explore Saturn. The Cassini-Huygens reached Saturn in June 30, 2004, and studied the Saturnian system, including its many moons. One of Cassini's main missions was to see variations in the haze of Titan, and attempt to peer through the clouds through the Visual and Infrared Mapping Spectrometer (VIMS) on board Cassini, which yielded a stitched image seen on the left.

One of the most interesting finds by Cassini-Huygens was that beneath the haze of Titan, there existed an entire lakes and possibly oceans of pure hydrocarbons, particularly liquid methane. Titan seemed to show varying atmospheric conditions, including

Worksheet 4, Problem 2 - Resolving the Telescopes

Question: CCAT is a 25-meter telescope that will detect light with wavelength up to 850 microns. How does the angular resolution of this huge telescope compare to the angular resolution of the much smaller MMT 6.5-meter telescope observing the infrared J-band? 

In this problem, we are comparing two telescopes to see how powerful they are. Just by pure numbers, we are given that the CCAT is 25-meters in diameter, versus the MMT, which is only 6.5 meters in diameter. Now, if you haven't seen my previous post about a telescope's angular resolution, please read it! Like seriously!

But of course, that is a very long post, so if you don't want to read it, here's the short version:

TL;DR - The larger the diameter of the telescope, the more angular resolution it has, which basically means the larger the telescope, the more powerful it is.

The other thing this question talks about is the J-band. In the simplest terms, the J-band is light at 1.25 micrometers ($1.25 \cdot 10^{-4}$ centimeters) in wavelength.

Okay, so the formula for angular resolution, which we derived in the previous post (seriously, read it!) of a telescope is:

\begin{align}
\theta_{\text{min}} = \frac{\lambda}{D}
\end{align}

where $\theta_{\text{min}}$ is the minimum angle that the telescope can resolve (how small/far away things a telescope can see), $\lambda$ is the wavelength of light passing through, and $D$ is the diameter of the telescope.


Okay, so for the CCAT, we know that the telescope is 25 meters (2500 centimeters) in diameter, and can detect wavelength up to 850 microns ($8.5 \cdot 10^{-2}$ centimeters). To calculate it's angular resolution when looking at something 850 microns in size, you get:

\begin{align}
\theta_{\text{min}_{CCAT}} = \frac{\lambda}{D} = \frac{8.5 \cdot 10^{-2} \text{ cm}}{2500 \text{ cm}} = 3.4 \cdot 10^{-5} \text{ radians}
\end{align}

That is pretty good resolution for seeing something that small!

However, since the problem states that the telescope is observing the infrared J-band, which is at 1.25 micrometers ($1.25 \cdot 10^{-4}$ centimeters) in wavelength, the angular resolution of the CCAT when observing the J-band is:

\begin{align}
\theta_{\text{min}_{CCAT}} = \frac{\lambda}{D} = \frac{1.25 \cdot 10^{-4} \text{ cm}}{2500 \text{ cm}} = 5.0 \cdot 10^{-8} \text{ radians}
\end{align}

As for the MMT, which is 6.5-meters (650 centimeter), can gather light at the J-band. It's angular resolution is as follows:

\begin{align}
\theta_{\text{min}_{MMT}} = \frac{\lambda}{D} = \frac{1.25 \cdot 10^{-4} \text{ cm}}{650 \text{ cm}} = 1.9 \cdot 10^{-7} \text{ radians}
\end{align}

The ratio of minimum angular resolution of the CCT to the MMT when observing the J-Band is:

\begin{align}
\frac{\theta_{\text{min}_{CCAT}} }{\theta_{\text{min}_{MMT}}} = \frac{5.0 \cdot 10^{-8} \text{ radians}}{1.9 \cdot 10^{-7} \text{ radians}} = 0.26.
\end{align}

Now, having a higher resolution means having a smaller $\theta_{\text{min}}$, so when observing wavelengths at 1.25 microns, the CCAT is only slightly better at 26% better.

When looking at the ratio of the CCAT to the MMT with respect to the CCAT observing at 850 microns, vs MMT's 1.25 microns, things get more interesting:

\begin{align}
\frac{\theta_{\text{min}_{CCAT}} }{\theta_{\text{min}_{MMT}}} = \frac{3.4 \cdot 10^{-5} \text{ radians}}{1.9 \cdot 10^{-7} \text{ radians}} \approx 180
\end{align}

This shows that the CCAT discerning light at 850 microns would be 180 times worse than the MMT looking at 1.25 microns.

So I guess, bigger isn't always better!

Worksheet 4, Problem 1: The Double Slit - Light as a Wave

Question: To understand the basics of astronomical instrumentation, I find it useful to go back to the classic Young's double slit experiment. Draw a double slit setup as a 1-D diagram. Draw it big, use straight lines, and label features clearly. The two slits are separated by a distance $D$, and each slit is $w$ wide, where $w \ll D$ such that its transmission function is basically a delta function. There is a phosphorescent screen placed a distance $L$ away from the slits, where $L \gg D$. We'll be thinking of light as plane-parallel waves incident on the slit-plane, with the propagation direction perpendicular to the slit plane. Further, the light is monochromatic with a wavelength $\lambda$. 

1. Convince yourself that the brightness pattern of light on the screen is a cosine function. (HINT: Think about the conditions for constructive and destructive interference of the light waves emerging from each slit).
2. Now imagine a second set of slits placed just inwards of the first set. How does the second set of slits modify the brightness pattern on the screen?
3. Imagine a continuous set of slit pairs with ever decreasing separation. What is the resulting brightness pattern?
4. Notice that this continuous set of slits forms a "top hat" transmission function. What is the Fourier transform of a top hat, and how does this compare to your sum from the previous step? 
5. For the top hat function's FT, what is the relationship between the distance between the first nulls and the width of the top hat (HINT: it involves the wavelength of light and the width of the aperture)? Express your results as a proportionality in terms of only the wavelength of light $\lambda$ and the diameter of the top hat $D$. 
6. Take a step back and think about what I am trying to teach you with this activity, and how it relates to a telescope's primary mirror?

WHEW.... that is a long problem, but lets approach it one step at a time to see just what is the significance of a telescope's primary mirror. To start off, just as the problem suggests, lets draw a diagram of the double slit setup. 

Figure 1: In this diagram, the pink dots show the constructive interference of the light waves, and the blue dots show the destructive interference of the light waves. The constructive and destructive interferences are linear, and project onto the phosphorescent screen. 

Now the first part of the problem asks to:

(A) Convince yourself that the brightness pattern of light on the screen is a cosine function. 

If you look at Figure 1, you can see that the constructive (pink) and destructive (blue) interference of light. Constructive interference of light is when the crests, or the troughs of two wave are aligned with one another so as to increase the amplitude of the wave. Destructive interference is when the crest of one wave, and the trough of another wave are aligned with one another so as to decrease the amplitude of a wave. In Figure 1, all the constructive interferences (pink) are linearly aligned as to project a constructive wave pattern on the phosphorescent screen. In a similar manner, all the deconstructive interferences (blue) are also linearly aligned, between the line of constructive interferences, to project a deconstructive wave pattern between the constructive wave pattern on the phosphorescent screen. If you create a wave that is the sum of the constructive and destructive interference projected on the phosphorescent screen, you will see a wave is created that reflects the changed amplitudes. Since the pattern on the phosphorescent screen is a result of constructive and deconstructive interference which results in a new wave, and since all waves can be represented as a cosine function, the pattern of light on the screen is a cosine function. 

Okay, so that was one way to show, intuitively, by constructive and deconstructive interference, you get a wave, which is represented by a cosine function. However, is there a more rigorous way to prove the relationship between the constructive and deconstructive interference that is actually happening? Yup! Let's start on a blank slate, and look at the distances light has to travel, to get a more mathematical understanding of why the pattern of light is a cosine function. 

Figure 2: The light coming through the two slits converges at a point on the phosphorescent screen. 

From Figure 2, it is apparent that light can converge at any point on the phosphorescent screen (this would only be one of the many places light can converge, but we are using one point for the sake of simplicity). Since light must travel different distances to reach the same point because of the distance $D$ between the two slits, the change in distance, $\Delta l$ represents the difference in distance. Since the point light on the phosphorescent screen is at an angle $\theta$ from the incident light, we can use the distance between the slits $D$ and the difference in distance between the light coming from the two slits, $\Delta l$ to show their relationship to each other using trigonometry.

Figure 3: The relationship between the angle of incident light $\theta$, the distance between the two slits $D$, and the difference in distance the light from the two slits needs to travel to meet at any given point $l$.  

 Based on the relationship between distance in light shown in Figure 3, and the constructive, and destructive interference pattern shown in Figure 1, we can establish the following two relationships: 

For constructive interference: 

\begin{align}
 \Delta l = D \cdot \sin \theta = n \lambda\\
\end{align}

For destructive interference:

\begin{align}
 \Delta l = D \cdot \sin \theta = \left( \frac{1}{2} + n \right) \lambda\\
\end{align}

This relationship shows that for any crest $n$, such that $n \in \mathbb{Z}$, constructive interference happens when the distance between the length the light has to travel to converge at the phosphorescent screen is an integer multiple of the wavelength of the monochromatic light. For deconstructive interference, the distance between the individual light waves must be a exactly one-half of a wavelength shifted from an integer multiple of the wavelength of monochromatic light. Because the relationship is related through the $\sin$ function, by making $n$ represent the crest, you are essentially shifting the function by $\frac{1}{2} \lambda$, thereby showing that the relationship of light interference in the light pattern on the phosphorescent screen is actually a cosine function!

But wait... that was a lot of work to show that the light pattern on the screen was a cosine function. In fact, there is an easier way to show the same relationship using Fourier transformations! Fo-ri-what-now?!?!

Fourier transformations are linear transformations that allow us to represent other functions as a sum of sines and cosines. Here is a cheat sheet to show how some common functions look like after a Fourier transformation (you'll thank me for this later): 

Figure 4: Fourier transformation cheat sheet!

Okay, so what does this cheat sheet have to do with proving that the light pattern on the phosphorescent screen creates a cosine function? Well, let's redraw the diagram to look at the which of the common functions provided best represents our double slit experiment:

Figure 5: The Even Pair Function

If you look at the double slit as allowing two beams of light that diffract as they enter the slit, you can see how it resembles the Even Pair function in the Fourier Transformations cheat sheet. And if you look at the Fourier transformation of the Even Pair function, you can see that it turns into a cosine function!

WOWW!!! That was a lot of work to do the first part of the problem. But now that we know how to use Fourier transformation, we'll be able to see other cool connections in how light behaves as a wave.

So as for the second question:

(B) Now imagine a second set of slits placed just inwards of the first set. How does the second set of slits modify the brightness pattern on the screen? 

Just like the first problem, let's think about it conceptually before we let the Fourier transformation just give us the answer. Let's look at the diagram for this scenario:

Figure 6: There are two sets of slits. The outer set of slits is separated by distance $D_0$, and the inner set of slits is separated by distance $D_1$. 

Okay, so we know that we have an outer set of slits separated by distance $D_0$, and an inner set of slits separated by distance $D_1$. Since each set of slits can create its own diffraction pattern, with each slit providing a source of light, there would be more interference, both constructive and destructive. Since the distance $D_0$ is larger than distance $D_1$, the distance between the crests and troughs would be larger, but it would still be a cosine function. Therefore, you are essentially adding 2 cosine functions. Since both of them will have their converging crests be at the point of intersection on the phosphorescent screen, there will be both stronger constructive and destructive interference. Therefore, the brighter spots would be even more brighter, and the darker spots would be even darker.

A visual representation of the constructive and destructive interference is as follows:

Figure 7: The green cosine function represents the outer slits with distance $D_0$, and the pink cosine function represents the inner slits with distance $D_1$. The yellow cosine function represents the resulting cosine function as a linear combination of the pink and green functions following constructive and destructive interference. 

To show this from a Fourier transformation perspective, if you look at Figure 6, you have 2 Even Pair functions instead of 1. As figure 6 shows, the pink set of slits makes one Even pair, and the green pair of slits makes another Even Pair. Since each Even Pair Fourier transforms into a cosine function, you have 2 cosine functions that are linearly added together.

Okay, so now we know that we can use Fourier Transformations to add functions together to represent even more complicated functions. That is going to be a very useful strategy for the next concept:

(C) Imagine a continuous set of slit pairs with ever decreasing separation. What is the resulting brightness pattern?

In this case, we are just dealing with an infinite number of slit pairs. Since we will be adding many different cosine functions together, we will have even stronger constructive interference in the center, and very strong destructive interference as you progress outwards from the center. Since only the center will continually have more and more constructive interference, thereby making it much brighter, while the outskirts away from the center will be a mixture of crests and troughs all over the place, the deconstructive interference will make the outer bands much dimmer. Therefore, this resembles the sinc function:

Figure 8: Sinc Function

(D) Notice that the continuous set of slits forms a "top hat" transmission function. What is the Fourier transform of a top hat, and how does it compare to your sum from the previous step? 

So the claim is that the continuous slits form a "top hat". Let's look at a diagram to show that this is indeed the case:

Figure 9: Continuous slits create a top hat (look at it sideways)!!

According to the cheat sheet in Figure 4, a Top Hat function Fourier transforms into the Sinc function! That is exactly what we got in our previous step!!!

(E) For the top hat function's FT, what is the relationship between the distance between the first nulls and the width of the top hat (HINT: it involves the wavelength of light and the width of the aperture)? Express your results as a proportionality in terms of only the wavelength of light $\lambda$ and the diameter of the top hat $D$. 

Since the top hat is made up of infinitely many slits, we can break each slit down to be the size of the wavelength, $\lambda$. Therefore, the width, $W$ can be represented by $W = \frac{\text{distance}}{\text{wavelength}} = \frac{D}{\lambda}$. The Fourier transformation tells us that the distance to the first null is $\frac{1}{W}$, so the relationship between the first nulls and the width of the Top Hat is $\frac{1}{W} = \frac{1}{\frac{D}{\lambda}} = \frac{\lambda}{D}$.

(F) Take a step back and think about what I am trying to teach you with this activity, and how it relates to a telescope's primary mirror? 

Okay, since we've done a lot of math, lets think as to what is the use of knowing how light diffraction with relation to the distances it travels teaches us about telescopes? You can think of a telescope's entire primary mirror as a light-collecting tool, or a large single slit. As a result, as the telescope sits and gethers light, it measures the diffraction pattern the light makes, which ideally makes a sinc function. 

As we analyzed the relationship between the diameter of the top hat and the wavelength, you can think of the diameter as the diameter of the primary mirror in a telescope. As a result, the relationship in the previous part tells us that the larger the diameter is in a telescope, the more resolution it has.

The relationship to the angular resolution can be provided by:

\begin{align}
\theta \propto \frac{\lambda}{D}
\end{align}

which shows that the larger the diameter of the mirror, the more detail the from the diffraction bands of the wavelength can be gathered.

Wow... this problem really opened up my eyes as to how telescopes actually work!!

LST - Losing Some Time (3.02)

Question: The Local Sidereal Time (LST) is the right ascension that is at the meridian right now. LST = 0:00 is at noon on the Vernal Equinox (the time when the Sun is on the meridian March 20th, for 2013 and 2014)

1. What is the LST at midnight on the Vernal Equinox?
2. What is the LST 24 hours later (after midnight in part 'a')?
3. What is the LST right now (to the nearest hour)?
4. What will the LST be tonight at midnight (to the nearest hour)?
5. What LST will it be at Sunset on your birthday?

What is the LST at midnight on the Vernal Equinox? 

Okay, so the definition of LST provided to us is, "the right ascension at the meridian right now". That sounds like a complicated definition, so let's get familiar with some terminology first. We have always thought of the day as being 24 hours long. However, that is not always true. We have two definitions of the word "day"! There is a solar day, and then there is the sidereal day. Wat!

Solar day vs. sidereal day

• A solar day is the 24 hour day we are used to in everyday life, measured from noon on one day to noon the next day. 
• A sidereal day is the time it takes the Earth to make one full rotation around the with respect to the stars. The Earth takes about 23 hours and 56 minutes to actually make a full rotation. 

The LST is a clock that tells the time according to a sidereal day. As a result, every 24 hours, the sidereal time is 4 minutes more than it was the previous day.

It is useful to think that since you gain 4 minutes every in LST for every 24 hours, you gain 1 minute in LST time every 6 hours. 

On the Vernal Equinox at noon, LST is 0:00. At midnight on the Vernal Equinox, 12 hours have passed, so LST would have increased by 2 minutes (since 2 six-hour time chunks have passed). Therefore, LST would show 12:02 to account for the addition 2 minutes added. 

What is the LST 24 hours later (after midnight from the previous part)?

Since 24 hours have passed since midnight on the Vernal Equinox, LST gains an additional 4 minutes to account for the loss of 4 minutes in sidereal time. Also, since midnight on the Vernal Equinox also contributed to 2 minutes to LST, the total minutes that have been added to LST are 2 + 4 = 6 minutes. Therefore, the LST would be 12:06. 

Another way to look at it is that 12 hours passed from the Vernal Equinox at 0:00 to get to midnight. Then after another 24 hours, we are up to 36 hours from the Vernal Equinox at 0:00. Since we gain 1 minute for every 6 hours in LST, $\frac{36}{6} = 6$ minutes get added to LST. Therefore, it is 12:06.

What is the LST right now (to the nearest hour)?

As of right now, it is 18:00 on February 5, 2015. Since the last Vernal Equinox was on March 20, 2014, 322 days and 6 hours have been passed. To calculate LST, we first need to calculate the right ascension:

\begin{equation}
322.25 \text{ days} \times \frac{24 \text{ hours}}{1 \text{ day}} \times \frac{4 \text{ minutes}}{24 \text{ hours}} \times \frac{1 \text{ hour}}{60 \text{ minutes}} \approx 21.483 \text{ hours} \approx \text{21 hours 29 minutes}
\end{equation}


The right ascension is 21 hours and 29 minutes, which means that the LST would be 2 hours and 31 minutes behind solar time. Therefore, LST would be 3:29, or approximately 3:00.

What will the LST be tonight at midnight (to the nearest hour)?

By midnight tonight, 6 more hours would have passed from the LST time 3:29, so LST would be 9:29, or approximately 9:00. 

What LST will it be at sunset on your birthday?

My birthday is on December 6 (woohoo!!), with a projected sunset time of 16:12. Since the Vernal Equinox for 2015 (which is on March 20th) would have passed by the time my birthday comes, exactly 261 days, 4 hours, and 12 minutes would have passed. Doing the calculations: 

\begin{equation}  261.174 \text{ days} \times \frac{24 \text{ hours}}{1 \text{ day}} \times \frac{4 \text{ minutes}}{24 \text{ hours}} \times \frac{1 \text{ hour}}{60 \text{ minutes}} \approx 17.41 \text{ hours} \approx \text{17 hours 25 minutes}   \end{equation} 

Since the right ascension is 17 hours 25 minutes, LST would be 5 hours and 25 minutes ahead of solar time. Therefore, since solar time would be 4:12, and another 5 hours 25 minutes are added, LST would be 9:37.

When You're Bored on a Desert Island (2.1.04)

Question: Imagine you are on a desert island (without wireless) and for some reason you need Kepler's Third Law of motion. Using dimensional analysis, what is the form of this equation, which relates period, total mass, and separation of a two-body gravitational orbit? 

In order to derive Kepler's Third Law, you need to have a two-body gravitational orbit. Thankfully, since you are stuck in the middle of the desert without wireless (seriously?) you have all the time in the world to think about which heavenly body you can use. brightest object we can use, of course, is the Sun. Since the Earth revolves around the Sun, we have our two-body gravitational orbit. 

Now, here's what we know about the Sun-Earth system: 

• The Earth revolves around the Sun
• The Sun's gravitational force keeps the Earth in orbit
• The Earth's orbit is mostly circular 
• The period, $T$ of one Earth orbit around the Sun is 365 days
• The distance, $a$ between the Earth and the Sun is 1 AU
• The total mass of the Earth can be represented by $m_{Earth}$
• The total mass of the Sun can be represented by $M_{Sun}$
• The total mass of the system, $M_{Total}$

Okay, so we listed some hard numbers, such as the distance between the Earth and the Sun, as well as the period, $T$ of 365 days. However, since we are trying to come up with Kepler's laws with just algebraic physics and dimensional analysis, we'll ignore the numbers. 

Instead, let's look at some of the physics concepts we can use. We see that the Earth revolves around the Sun due to the gravitational force, and revolves around the Sun in mostly a circular fashion. The law of gravity is defined as follows:
\begin{align*}
F_G = \frac{GM_1 M_2}{r^2}
\end{align*}
Therefore we can use Newton's Second Law of motion to derive Kepler's Third Law: 

\begin{align*} F_{net} = ma_c\\
 \frac{GM_1 M_2}{r^2} = ma_c
\end{align*}

Since the Earth revolves around the Sun in an almost-circular orbit, we have to use centripetal acceleration, where $a_c = \frac{v^2}{r}$. We can also use the definition of velocity, where $v = \frac{distance}{time}$ to show the velocity of the Earth when it makes a full orbit around the Sun in a period, $T$. Therefore, velocity can be represented as $v = \frac{distance}{time} = \frac{2 \pi a}{T}$.
 \begin{align*}
 \frac{GM_1 M_2}{r^2} &= ma_c\\
\frac{G \cdot m_{Earth} \cdot M_{Sun}}{a^2} &= m_{Earth}\frac{v^2}{a}\\
\frac{GM_{Sun}}{a} &= v^2\\
\frac{GM_{Sun}}{a} &= \left(\frac{2 \pi a}{T}\right)^2\\
\frac{GM_{Sun}}{a} &= \frac{4 \pi^2 a^2}{T^2}
\end{align*}
Since we are looking for the relationship between the period, $T$ and the distance between two bodies, $a$, and the total mass, $M_total$, we can rearrange the previous equation to show this relationship.
\begin{align*}
T^2 &= \frac{4 \pi^2 a^3}{GM_{Sun}}
\end{align*}
Although the total mass of the system, $M_{total} = M_{Sun} + M_{Earth}$, since the $M_{Sun} \gg M_{Earth}$, we can assume that $M_{total} = M_{Sun}$, which leads to the equation:
\begin{align*}
T^2 &= \frac{4 \pi^2 a^3}{GM_{Total}}
\end{align*}
The equation above is actually Kepler's Third Law!!!

However,we are looking for the relation between period, distance, and total mass. Since $\frac{4 \pi^2}{G}$ is a constant, the relation between period, $T$, distance between an two bodies, $a$, and the total mass, $M_{Total}$ as follows:
\begin{align*}
T^2 &\propto \frac{a^3}{M_{Total}}
\end{align*}
To go a little further, in any given two-body gravitational system, the total mass, $M_{total}$ will also be a constant, so Kepler's Third Law actually shows the proportionality that:
\begin{align*}
T^2 &\propto a^3
\end{align*}
This is the profound relation in Kepler's third law, because it shows the relation between the distance between two bodies, and the orbital period of a smaller body revolving around a much larger body. 

Speed of Sound in a Gas (2.1.03)

As this fighter jet reaches the speed of sound, it leaves behind a
trail of water vapor. This is because of the drop in air pressure!

Question: The speed of sound in a gas $c_s$ is related to the pressure $P$ and density $\rho$ of the gas. Use dimensional analysis to figure out the form of this relationship.

Ahhhh... dimensional analysis! It is one of the most useful tools we have to make sure that any problem we are solving actually makes sense. Because if the units don't make sense, then you messed up somewhere.

Dimensional analysis can also be used to figure out the relationships between various physical properties. In this problem, we are given that the speed of sound is related to the pressure $P$ and density $\rho$ of a gas. But how do pressure and density relate to each other, and how can we make sure that they are accurately representing the relationship to the speed of sound in a gas?

Lets start by defining pressure $P$, density $\rho$, and the speed of sound, $c_s$ in terms of their physical units. To keep things consistent, let's use $d$ for distance (in centimeters), $t$ for time (in seconds), and $m$ for mass (in grams):

• $P = \frac{Force}{Area} = \frac{Newtons}{d^2} = \frac{\frac{m \cdot d}{t^2}}{d^2}$
• $\rho = \frac{Mass}{Volume} = \frac{m}{d^3}$
• $c_s = velocity = \frac{d}{t}$

Based on this, we can see that the distance unity, $d$ needs to end up in the numerator, and the time unit, $t$ needs to be in the denominator in order for the units to make sense relative to the speed of sound. Therefore, since only $P$ has a time unit, $t$ in should be in the numerator. Since $\rho$ has a larger magnitude of distance, $d^3$ in it's denominator than either of the distance units in $P$, it will contribute more by being the denominator of the ratio between $P$ and $\rho$ so that the distance unit is actually in the numerator when simplified. Let's check to see if this reasoning makes sense mathematically using dimensional analysis:  

$\begin{align*}
c_s & \propto \frac{P}{\rho}\\
\frac{d}{t} & \propto  \frac{\frac{\frac{m \cdot d}{t^2}}{d^2}}{\frac{m}{d^3}}\\
\frac{d}{t} & \propto \frac{\frac{m \cdot d}{d^2 \cdot t^2}}{\frac{m}{d^3}}\\
\frac{d}{t} & \propto \frac{m \cdot d^4}{m \cdot d^2 \cdot t^2}\\
\frac{d}{t} & \propto \frac{d^2}{t^2}\\

\therefore c_s & \propto \sqrt{\frac{P}{\rho}}
 \end{align*}$

As you can see, the the speed of sound in a gas is proportional to the square root of the ratio of $P$ to $\rho$.

From the Stars to Your Eyes (2.1.02)

Question: What is the power output of a star that is 100 light years away that you can barely see from a dark site at night? Assume the star emits most of its energy at the peak of the eye's sensitivity. 

Okay, so how do we approach a problem that seems to give us so little information? Do we even have enough information to solve this? One of the goals of Astronomy is to be able to infer the various properties of the things that are involved in a problem.



To start thinking about this problem, lets figure out all the information given to us. The information that is directly given to us is:

• The distance, $r$, between the star and our eyes is about 100 light years away, which is about $10 \times 10^{20}$ centimeters
• Most of the energy emitted by the star is given at the peak of the eye's sensitivity

Okay, so not much information here. But let's unpack the second piece of information given to us, about the "peak of the eye's sensitivity". What are certain thing we know about the human eye:

• The eye can only see visible light of the electromagnetic spectrum, which is at about 500 nm, or $5 \times 10^{-5} cm$
• The eye can process about 10 frames per second (10 Hz)
• The eye can "see" something when it receives 10 photons of light
• The surface area of the eye is about $2$ $cm^2$

Now that we have a little bit more information to work with, lets think about the physical constraints we are working with, and some of the physical constants and equations that can help us come up with a solution. Oftentimes, a picture can help see how the situation looks like, so I created the following diagram:

• The star emits its photons radially outwards, meaning that at any distance, the concentration of photons can be measured by the surface area of the sphere the light reaches. The surface area that the light reaches at any distance can be calculated by $SA_{light}= 4\pi r^2$
• The speed of light is $c = 3 \times 10^{10} cm/s$ 
• The amount of energy in a photon is $E = h\nu$, for frequency $\nu$ and Planck's constant, $h = 6.6 \times 10^{-27} erg/s$
• Frequency, $\nu = \frac{velocity}{\lambda}$, where $\lambda$ is the wavelength of the photon
• Finally, since we're solving for power, the equation for power is: $Power=\frac{Energy}{Time}$ 

Whew! That was a lot of information we just gathered that is relevant to the question. So let's go about solving for the power output of the star.

In order to solve for the power output of the star, we would need to know how much energy the star is emitting, and the amount of time it is emitting at. Unfortunately, based on the information we gathered, we don't know much about the star other than the fact that it is 100 light years away. However, we have a lot of information about the eyes. We can calculate how much power the eye is absorbing when it "sees" a star. We also know how big both the eye is, as well as how big the sphere of light is that reaches the eye. Therefore, we can create a proportionality that relates the power output of the star to the power absorbed by the eye using the surface area of both the eye and the sphere of emitted light.

$
\begin{align*}
\frac{SA_{Light}}{SA_{Eye}} &= \frac{Power_{Star}}{Power_{Eye}}\\
Power_{Star} &= \frac{Power_{Eye} \times SA_{Light}}{SA_{Eye}}\\
&= \frac{\frac{\text{Energy of 10 Photons}}{\text{Time}} \times SA_{Light}}{SA_{Eye}}\\
&= \frac{\frac{10 \cdot h\nu}{\text{frame rate of eye}} \times SA_{Light}}{SA_{Eye}}\\
&= \frac{\frac{\frac{10 \cdot h \cdot velocity}{\lambda}}{\text{frame rate of eye}} \times SA_{Light}}{SA_{Eye}}\\
&= \frac{\frac{\frac{10 \cdot hc}{\lambda}}{\text{frame rate of eye}} \times SA_{Light}}{SA_{Eye}}\\
&= \frac{\frac{\frac{10 \cdot hc}{\lambda}}{\text{frame rate of eye}} \times 4 \pi \times \text{radius of light sphere}^2}{SA_{Eye}}\\
&= \frac{\frac{\frac{10 \cdot hc}{\lambda}}{\text{frame rate of eye}} \times 4 \pi r^2}{SA_{Eye}}\\
&= \frac{\frac{\frac{10 \cdot (6.6 \times 10^{-27} erg \cdot s)(3 \times 10^{10} cm/s)}{5 \times 10^{-5} cm}}{\frac{1}{10} s} \times 4 \pi (10 \times 10^{20} cm)^2}{2 cm^2}\\
Power_{Star} &\approx 2 \times 10^{33} erg/s
\end{align*}
$

As you can see, the power output of the star is $2 \times 10^{33}$ ergs per second.

From Sailor Moon to Astronomer

Imagine a 5 year old girl curiously watching the anime, Sailor Moon. For those who are unfamiliar with the show, Sailor Moon follows the life of a girl named Usagi, who is bestowed magical powers from the Moon, and is descendent from the Moon Kingdom. Her royal guard, the Sailor Senshi, each come from their own respective planets.

From left to right: Sailor Star Healer, Sailor Star Maker, Sailor Star Fighter, Sailor Mercury, Sailor Venus, Sailor Jupiter, Sailor Mars, Sailor Chibi Chibi, Sailor Moon, Sailor Chibi Moon, Sailor Neptune, Sailor Uranus, Sailor Pluto, Sailor Saturn

Because of my interest in Sailor Moon, I read upon the character bios of all the characters. Since each character represented a planet, I further delved into the histories of the planets, which led my curiosity to read about the characteristics of each planet. This was my first exposure to astronomy. 

Reading anything and everything I could find about the planets, I became very interested in astronomy as a subject from a very young age. I sought new materials to read whenever I came across them, including textbooks, videos, and documentaries. Documentaries about astronomy really piqued my interest outside of the solar system, and into the vast cosmos beyond. 

As a result, 7-year old me would look into things like the Big Bang, how the Universe would die, quasars, supermassive black holes, the birth and death of stars, and how the Earth was formed. Everything about the cosmos fascinated me, to the point that I wanted to be an astrobiologist. But honestly, the thing that captivated my childhood brain was all the pretty pictures of stars and planets. The Universe is like shiny glitter everywhere, which is what 7-year old girls like. 

One of the highlights I remember was when I was in 5th grade, the New Horizons probe was launched on a mission to Pluto. I was so excited about it, because I had seen a documentary about it a couple of weeks earlier. I remember I would check the countdown clock every day to see how much time was left until New Horizons would reach Pluto in 2015, down to the minutes and seconds. Now that it is 2015, I am ecstatic that the almost-ten-year wait is over.