Other Binary Systems

In the previous post we explored the idea of a binary star system, in which two stars orbit one another.  This behavior, however, is not exclusive to stars.

Planets:
It is very possible to have 2 planets orbit one another.  A very close-to-home example of this is our very own dwarf planet Pluto and its moon Charon.  These tow bodies are actually fairly close in mass is 11.6 percent of Pluto's.  For reference, the mass of our moon (Luna) is only 1.2% of Earth's mass.  Pluto and Charon, being so close in mass ratio, actually orbit each other around a point that is outside of Pluto known as a barycenter.  One could argue that these two are a binary system of rocks.

Pluto and Charon relative to a fixed barycenter

Black Holes:
Now it gets cool.  Black holes, like stars, can orbit one another, creating wicked gravity phenomena.

Supermassive Black Holes:

Now it gets scary.  Supermassive black holes can have sizes of several orders of magnitude higher than regular black holes.  Binary systems of them are often the result of galactic collisions, and such systems emit some of the most powerful gravity waves in the universe.

An artist's rendering of a black hole system after a galactic collision

Quasars:
Now it gets terrifying.  Quasars are supermassive black holes that actively consume matter and emit insane amounts of light and other radiation.  Scientists have managed to find several binary systems and even a triple system of quasars.

Artist's rendering of a single quasar

Information and pictures from:
http://en.wikipedia.org/wiki/Charon_(moon)
http://en.wikipedia.org/wiki/Moon
http://en.wikipedia.org/wiki/Binary_black_hole
http://en.wikipedia.org/wiki/Quasar

Worksheet 6, Problem 4: Binary Star Comparisons

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). 

Let's think about this system quantitatively first.  We know that hotter objects tend to be bluer in color, since they emit lower frequency of light, like a blue hot coal verses a red hot coal.  So, this means that the blue star is hotter.
Next let's consider size.  We know that the yellow star is 6 times brighter than the blue star.  Intuitively, this means that it should be larger, since it emits light that is less energetic than the blue star.  It needs to emit more of the less energetic light, so it must be larger to do so.

So we have this rough picture:

In order to compare their radii, we first need to quantitatively compare their temperatures.
In a previous blog post we derived the formula for the Wein's Displacement Law which expresses a star's peak wavelength of emitted light as a function of its temperature and fundamental constants.

Let's do the blue star first, knowing that a typical blue wavelength of light is about 475nm.
\[\lambda =2.898\times 10^{-3}m \times T\]
\[T_b=\frac{2.898\times 10^{-3}m}{\lambda}=\frac{2.898\times 10^{-3}m}{475nm}=\boxed{6.1\times 10^3 K}\]


This process can be repeated for the yellow star of peak wavelength 570nm.

\[\lambda =2.898\times 10^{-3}m \times T\]
\[T_y=\frac{2.898\times 10^{-3}m}{\lambda}=\frac{2.898\times 10^{-3}m}{570nm}=\boxed{5.1\times 10^3 K}\]

Next we can apply the Luminosity formula to each star to calculate their relative radii.
Since the Luminosities are relative, let's let \(L_y=6L_b\)
For the blue star:
\[L_b=4\pi R^2 \sigma T_b^4\]
\[R=\left(\frac{L_b}{4\pi \sigma T_b^4}\right) ^{\frac{1}{2}}=\left(\frac{L_b}{4\pi (5.67\times 10^{-8}Wm^{-2}K^{-4}) (6.1\times 10^3K)^4}\right) ^{\frac{1}{2}}=\boxed{\sqrt{L_b}\times 3\times10^{-5}m/W}\]
This may looks somewhat confusing, but because we were never given a numerical value for \(L_b\) we must leave the radius in terms of \(L_b\).
Let's repeat this for the yellow star.
\[L_y=4\pi (R)^2 \sigma T_y^4\]
\[R=\left(\frac{L_b}{4\pi \sigma T_b^4}\right) ^{\frac{1}{2}}=\left(\frac{6L_b}{4\pi (5.67\times 10^{-8}Wm^{-2}K^{-4}) (5.1\times 10^3K)^4}\right) ^{\frac{1}{2}}=\boxed{\sqrt{L_b}\times 1\times10^{-4}m/W}\]

When we calculate the ratio of these values we find that the radius of the yellow star is 3.3 times the radius of the blue star, so we can confidently declare that while our qualitative reasoning is correct, our picture is not to scale.

Worksheet 5, Problem 2: Blackbody Math

(a)  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 Fν(T) over all frequencies to obtain the bolometric flux emitted from a blackbody, F(T). You can do this using by substituting the variable u = hν/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 all constants equal to a new, single constant called σ, which is also known as the Stefan-Boltzmann constant. If you’re really into calculus, go ahead and show that σ « 5.7 ˆ 10 ́5 erg s ́1 cm ́2 K ́4. Otherwise, commit this number to memory.
(b)  The Wien Displacement Law: Convert the units of the blackbody intensity from Bν(T) to Bλ(T) IMPORTANT: Remember that the amount of energy in a frequency interval dν has to be exactly equal to the amount of energy in the corresponding wavelength interval dλ.
(c)  Derive an expression for the wavelength λmax corresponding to the peak of the intensity distribution at a given temperature T. (HINT: How do you find the maximum of a function?) Once you do this, again substitute u = hν/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.
(d) The Rayleigh-Jeans Tail: Next, let’s consider photon energies that are much smaller than hν
the thermal energy. Use a first-order Taylor expansion on the term ekT to derive a simplified form of Bν (T) in this low-energy regime. (HINT: The Taylor expansion of ex « 1 ` x.)
(e) Write an expression for the total power output of a blackbody with a radius R, starting with the expression for Fν. This total energy output per unit time is also known as the bolometric luminosity, L. 

Let's start with a brief discussion of blackbodies.  These are theoretical (yet highly useful) objects that absorb all light that is incident upon them with out reflection it.  They also give off radiation based on their temperature.  Most things can be be approximated as black bodies, even humans and stars.  

(a)  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 Fν(T) over all frequencies to obtain the bolometric flux emitted from a blackbody, F(T). You can do this using by substituting the variable u ” hν{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 all constants equal to a new, single constant called σ, which is also known as the Stefan-Boltzmann constant. If you’re really into calculus, go ahead and show that σ « 5.7 ˆ 10 ́5 erg s ́1 cm ́2 K ́4. Otherwise, commit this number to memory.

We know that frequency specific flux has the formula:

\[B_{\nu}(T)=\pi\frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\]

We want to integrate this over all frequencies to find \(F_{\nu}\) so:

\[F_{\nu}(T)=\int^\infty_0B_{\nu}(T)d\nu=\int^\infty_0\pi\frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}d\nu\]

Let's make a u-substitution of \(u=\frac{h\nu}{kT}\) and \(du=\frac{h}{kT}d\nu\)

\[\int^\infty_0\pi\frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}d\nu=\frac{2\pi k^4T^4}{c^2h^3}\int^\infty_0\frac{u^3}{e^u-1}du\]
Now comes the fun part: We can set everything except temperature to equal a constant \(\sigma\), so:
\[\boxed{F(T)=\sigma T^4}\]

(b)  The Wien Displacement Law: Convert the units of the blackbody intensity from Bν(T) to Bλ(T) IMPORTANT: Remember that the amount of energy in a frequency interval dν has to be exactly equal to the amount of energy in the corresponding wavelength interval dλ.

So we are back to specific intensity.

\[B_{\nu}(T)=\frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\]

We know that \(\nu=\frac{c}{\lambda}\), but there is a catch, we need to make sure that the energy in a \(d\lambda\) equals the energy in a \(d\nu\).  Thus \(B_{\lambda}(T)d\lambda=B_{\nu}(T)d\nu\) or \(B_{\lambda}(T)=B_{\nu}(T)\frac{d\nu}{d\lambda}\)
This gives us \(B_{\lambda}(T)=B_{\nu}(T)\frac{c}{\lambda^2}\)
Finishing the substitution of \(\frac{c}{\lambda}=\nu\) we get:
\[\boxed{B_{\lambda}(T) = \frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}}\]

c)  Derive an expression for the wavelength λmax corresponding to the peak of the intensity distribution at a given temperature T. (HINT: How do you find the maximum of a function?) Once you do this, again substitute u = hν/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.

To find a maximum value of a function, we look for where it's first derivative is zero. So:

\[\frac{d}{d\lambda}B_{\lambda}(T) =\frac{d}{d\lambda}\frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}\]
\[B'_{\lambda}(T)=\frac{-10hc^2}{\lambda^6(e^{\frac{hc}{\lambda kT}}-1)}+\frac{2h^2 c^3 e^{\frac{hc}{\lambda kT}}}{\lambda^7 kT(e^{\frac{hc}{\lambda kT}}-1)^2}=0\]
To simplify, we can remove anything that we can factor from the two expressions since it equals zero simplifying the equation to:
\[B'_{\lambda}(T)=-5+\frac{h c e^{\frac{hc}{\lambda kT}}}{\lambda kT(e^{\frac{hc}{\lambda kT}}-1)}=0\]
That looks slightly less scary.  Now lets do another u-substitution where \(u=\frac{hc}{\lambda kT}\) giving us:
\[5=\frac{u e^u}{e^u-1}\]
Taylor approximations tell us that \(e^u \approx 1+u\) so:
\[5=\frac{u+u^2}{u}=1+u\]
\[u=4=\frac{hc}{\lambda kT}\]
So at the end of it all, the maximum wavelength is is given by:
\[\boxed{\lambda_{max}(T)=\frac{hc}{4kT}}\]

(d) 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 ekT to derive a simplified form of Bν (T) in this low-energy regime. (HINT: The Taylor expansion of ex « 1 ` x.)

So, we start with:
\[B_{\nu}(T)=\frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\]
If \(e^x \approx 1+x, e^{\frac{h/nu}{kT}} \approx 1+\frac{h/nu}{kT}\)
Plugging this in, we get:
\[B_{\nu}(T) \approx \frac{2\nu^2}{c^2}\frac{h\nu}{1+\frac{h\nu}{kT}-1}\]

\[B_{\nu}(T) \approx \frac{2\nu^2}{c^2}\frac{h\nu}{\frac{h\nu}{kT}}\]

\[\boxed{B_{\nu}(T) \approx \frac{2\nu^2kT}{c^2}}\]

(e) Write an expression for the total power output of a blackbody with a radius R, starting with the expression for Fν. This total energy output per unit time is also known as the bolometric luminosity, L.

Earlier in this problem we already established that if one integrates \(F_{\nu}(T)d\nu\) then the rustle is \(F(T)=\sigma T^4\).
With this as a starting point, this is the flux through every area of the black body.  The surface area of a sphere is \(4\pi r^2\), so the total luminosity is:
\[\boxed{L=4\pi r^2 \sigma T^4}\]

This problem was done with collaboration with G. Grell.

Worksheet 4, Problem 2: Comparing Telescopes

CCAT is a 25-meter telescope that will detect light with wavelengths 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 in the infrared J-band? 

Telescope resolution is dependent upon the size of the primary mirror, and the wavelength of light that it observes.  

\(\theta=?\)
\(\lambda=850um,1.25um\)

\(d=25m,6.5m\)

\[\theta=1.220\frac{\lambda}{d}\]

This equation gives the resolution of a telescope (in terms of the smallest resolvable angle) in which \(\lambda\) refers to the wavelength of light and D refers to the aperture diameter.  The factor of 1.220 is a constant to assume a circular aperture.  

First let's solve for the CCAT.

\[\theta=1.220\frac{\lambda}{d}=1.220\frac{850um}{25m}=1.220\frac{8.5\times 10^{-4}m}{25m}=\boxed{4.1\times 10^{-5} radians}\]

Next lets solve for the MMT.  The MMT observes the J-Band of radiation, an infrared band centered at 1.25um.  

\[\theta=1.220\frac{\lambda}{d}=1.220\frac{1.25um}{6.5m}=1.220\frac{1.25\times 10^{-6}m}{6.5m}=\boxed{2.3\times 10^{-7} radians}\]

What does this mean? This means that the MMT can see 177x more clearly than the CCAT, despite it's smaller size.  

J-Band: http://en.wikipedia.org/wiki/J_band

Formula: http://en.wikipedia.org/wiki/Angular_resolution

Worksheet 4, Problem 1: Fundamental Telescope Optics

a) Convince yourself that the brightness pattern of light on the screen is a cosine function. 
b) Now imagine a second set of slits placed just inward of the first set. How does the second set of slits modify the brightness pattern on the screen?
c) Imagine a continuous set of slit pairs with ever decreasing separation. What is the resulting brightness pattern?
d) 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?
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? Express your result as a proportionality in terms of only the wavelenght of light λ and the diameter of the top hat D.
f) Take a step back and think about what I’m trying to teach you with this activity, and how it relates to a telescope primary mirror. 

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

Firstly, let's consider what causes constructive or destructive interference.  Constructive interference will occur when two light rays overlap while in phase.  To be infuse, the parks and troughs of each wave must line up such that the peaks of the two waves strike the surface at the same time.  In a similar condition, destructive interference occurs when the two waves are out of phase, or the peaks of wave 1 align with the troughs of wave 2, effectively canceling out both waves.

Double slit interference occurs because when the light rays from each slit strike the screen together, they have traveled different distances, so they may no longer be in phase with one another.  

Let's quantify this. For constructive interference:
\(n\lambda=dsin\theta\)  
The makes sense because the distance \(dsin\theta\) is approximately the difference in distances that the two light rays travelled, so it must be equal to a full number "n" of wavelengths \(\lambda\).

Brightness is at a maximum when the interference is completely constructive, or when \(sin\theta=0\)  This occurs when \(sin\theta\) equals zero, which occurs at \(\theta=0,\pi\).  We want brightness to be at a maximum at these situations, so a Cosine function makes sense, since \(cos\theta=1\) when \(\theta=0\).


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

This will cause even more diffraction and interference effects, effectively causing each maxima (bright spot) to be made of several smaller maxima, as each of the 4 light rays falls in and out of sync with its neighbors.

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

This will cause the central maxima to be turned into an ever growing series of smaller and smaller maxima that get closer and closer to together, forming what is known as a top hat function.  

d) 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?

The Fourier Transform of a top hat is the \(sinc\) or \(\frac{sinx}{x}\) function.  This makes sense because it resembles the spectra of a single slit (which all of the now infinitesimally tiny silts add up to).  The tiny maxima will form a somewhat even light band in the waveform of a top hat or sinc function. 

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?  Express your result as a proportionality in terms of only the wavelenght of light λ and the diameter of the top hat D.

Since greater wavelengths of light exhibit greater bending or diffraction, a smaller slit induces greater diffraction, and the top hat diameter is approximately the width of the aperture, the relationship should be characterized by:
\[D\propto d_{null}\lambda\]

f) Take a step back and think about what I’m trying to teach you with this activity, and how it relates to a telescope primary mirror. 

A telescope's primary mirror is on e the primary determining factors of the telescope's resolution.  In this case, a larger primary mirror is similar to having a larger aperture, more distinct maxima and minima will be produced, thereby creating a more detailed top hat wave with a greater resolution and more detail.

Solved in colaberation with G. Grell.
Diagram from http://web.utk.edu/~cnattras/Phys250Fall2012/modules/module%201/diffraction_and_interference.htm 

Comet Lovejoy

For the past few months, Comet Lovejoy (C/2014 Q2) has been visible in the night sky to binocular bearing observers or those with good eyesight and a dark night.  Recently discovered on August 17th 2014 by Terry Lovejoy, this comet has just made its approach near the sun as part of its ~8000 year orbital period.  I recently volunteered to be a part of the Harvard Observing Project (HOP) and spent some time in the Harvard Clay Telescope on top of the Science Center with Graduate Students John Lewis and Jane Huang.  Despite the usual Bostonian light pollution (and some spotlights from a Chris Pratt event), we managed to capture this image:

This image conveys an interesting phenomenon, comets actually glow green.  This green glow is due to the icy composition of the comet interacting with the solar wind.  The solar wind is a mess of charged particles that are given off by and orbit the sun at high speeds.  When these ions impact upon the the oxygen in the ice of the comet, the oxygen is briefly ionized and emits its characteristic green spectral lines.

A careful observer will note that the background stars are blurred.  This is because the telescope is tracking the comet, which moves relative to the celestial sphere.  With a 10-30 second exposure on the image, the slight movement of the comet again the background blurs the stars.

Sources:

Harvard Observing Project

http://en.wikipedia.org/wiki/C/2014_Q2_(Lovejoy)

Worksheet 3, Problem 2: LST Events

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).
a) What is the LST at midnight on the Vernal Equinox?
b) What is the LST 24 hours later (after midnight in part ’a’)?
c) What is the LST right now (to the nearest hour)?
d) What will the LST be tonight at midnight (to the nearest hour)? 
e) What LST will it be at Sunset on your birthday? 

First let's discuss the difference between LST and normal time.  LST takes into account the revolution of the Earth around the sun, and has only 23h 56 min per day.  Thus, the current LST is the current time, plus 4 minutes per solar day since the vernal equinox.  I will use military time to avoid having to specify AM/PM.

a) What is the LST at midnight on the Vernal Equinox?
This is fairly easy.  Midnight on the vernal equinox is exactly 12 hours of solar time after the noon of the equinox. Thus:

\[LST=\text{solar time} - 12h + (\text{Days since Vernal equinox} \times 4\text{ minutes/day})\]
\[LST=\text{24:00} -12h + (0.5days \times 4\text{ minutes/day})=\boxed{12:02}\]

b) What is the LST 24 hours later (after midnight in part 'a’)?
This is also straightforward.  Intuitively, another day has passed, so LST is ahead another 4 minutes.
\[LST=\text{solar time} - 12h + (\text{Days since Vernal equinox} \times 4\text{ minutes/day})\]

\[LST=\text{24:00} -12h +(1.5days \times 4\text{ minutes/day})=\boxed{12:06}\]

c) What is the LST right now (to the nearest hour)?
Here is where it gets a bit more complicated because we need to work to find the number of days that have transpired since the equinox.  It's not overly difficult, just time consuming.  Conveniently, the current time is 12:00 on February 9th 2015 in Cambridge, MA, so we don't need to account for fractions of days.  According to Internet sources, it has been 326 days since the vernal equinox.

\[LST=\text{solar time} - 12h + (\text{Days since Vernal equinox} \times 4\text{ minutes/day})\]

\[LST=\text{12:00} -12h + (326days \times 4\text{ minutes/day})\approx \boxed{22:00}\]

d) What will the LST be tonight at midnight (to the nearest hour)? 
This is easy.  We already know the time at noon, so we just need to add another two minute offset like in problem a).
\[LST=\text{solar time} - 12h + (\text{Days since Vernal equinox} \times 4\text{ minutes/day})\]

\[LST=\text{24:00} -12h + (326.5days \times 4\text{ minutes/day})\approx \boxed{10:00}\]

e) What LST will it be at Sunset on your birthday? 

My birthday is July 3rd.  Internet sources claim that the next vernal equinox is on Friday, March 20th.  My birthday is 105 days after the equinox.  Sunset on my birthday will be at 19:25 in Cambridge (ignoring Daylight Savings Time which is only an annoyance on stellar scales).

\[LST=\text{solar time} - 12h + (\text{Days since Vernal equinox} \times 4\text{ minutes/day})\]

\[LST=\text{19:25} -12h + (105.3days \times 4\text{ minutes/day})= \boxed{14:25}\]

Worksheet 3, Problem 1: Martian Solar Day

What is the difference between sidereal and solar day on Mars if Mars has the same rotation period and orbits at 1.5 AU?

\(a_{Mars}=1.5AU\)
\(a_{Earth}=1AU\)
\(p_{Mars}=?\)
\(p_{Earth}=366.25 sidereal\text{ } days\)
\(day_{Mars, Solar}=?\)
\(day_{Earth, Solar}=24h\)
\(day_{Earth, Sidereal}=23h,56'\)
\(day_{Mars, Sidereal}=23h,56'\)

\(\theta=?\)

First let's review the background concepts.  A sidereal day is the time that a planet takes to rotate \(360^{\circ}\) relative to fixed distant stars.  A solar day is how long a planet takes to have the sun appear at the zenith at a given longitude.  Due the the orbit of the planet around the sun, a solar day will result in more than \(360^{\circ}\) of rotation and is thus longer than a sidereal day.  The difference between these two times will be dependent upon the angle through which the planet moves each day.

To begin the calculation, we should find the Mars' Period.
Kepler's Third Law of Planetary motion states:
\[p_{planet}^2\propto a_{planet}^3\]
We can use Earth as a reference point to find the period of Mars.
\[\frac{p_{Mars}^2}{a_{Mars}^3}=\frac{p_{Earth}^2}{a_{Earth}^3}\]
\[p_{Mars}=\left(\frac{p_{Earth}^2}{a_{Earth}^3}\times a_{Mars}^3\right)^{\frac{1}{2}}\]
\[p_{Mars}=\left(\frac{(366.25sidereal\text{ }days)^2}{(1AU)^3}\times (1.5AU)^3\right)^{\frac{1}{2}}\]
\[\boxed{p_{Mars}=673sidereal\text{ }days}\]

Next to find \(\theta\) we divide the degrees in a circle by the time it takes Mars to travel around the circle.
\[\theta =\frac{360^{\circ}}{p_{Mars}}=\frac{360^{\circ}}{673sidereal\text{ }days}=\boxed{.54^{\circ}/sidereal\text{ }day}\]
Converting from a degrees system to an hours system gives us:
\[.54^{\circ}/sidereal\text{ }day\times \frac{1440min}{360^{\circ}}=\boxed{2.1\frac{minutes}{day}}\]
This gives us the difference between a Martian Sidereal Day and a Martian Solar Day.  Our calculations show that a Martian Solar Day is 2.1 minutes longer than a Martian Sidereal Day.
This makes sense, because Marts moves through a lesser angle of orbit per day than Earth, so its time difference should also be less.

Calculated in collaboration with G. Grell, J. Herrmann, and H. Sigurslid.

Worksheet 2.1, Problem 6: A Fly's Push-Up

How much energy does a fly exert if it does a single push up in cgs units?
(Assuming that a fly is a .25cm x .25cm x .25cm cube of water)

\(\rho=1g/cm^3\)
\(L=0.25cm\)
\(\Delta H=6.25\times 10^{-2}cm\)
\(g=9.8 \times 10^2 cm/sec^2\)
\(m=?\)
\(V=?\)
\(E=?\)










First, lets calculate more about the fly.  We know his dimensions, so we can calculate his mass using density.
\[m=\rho V=\rho L^3=1g/cm^3 \times (0.25cm)^3=\boxed{1.6\times 10^{-2}g}\]

Let's assume that our fly decides to be really epic and do a pushup using all four of his limbs such that he lifts his entire body off of the ground instead of just pivoting on his rear legs.  Let's also assume that a cubic fly's legs are about \(\frac{1}{4}\) of its height, giving it a change in height of \(\Delta H=0.0625cm\).

We now have all of the components necessary to solve for energy exerted in a pushup.

\[E=mg\Delta H=\]
\[E=(1.6\times 10^{-2}g) \times (9.8 \times 10^2 cm/sec^2) \times (6.25\times 10^{-2}cm) \approx \boxed{1.0\text{ }erg}\]


That's not a lot of energy, but for a little fly, that's not too bad.

Worksheet 2.1, Problem 1: Seeing a Lightbulb

The eye must receive 10 photons in order to send a signal to the brain that says, “Yep, I see that.” If you are standing in an enormous, completely dark cave and just barely discern a light bulb at a distance of 1 kilometer, what is the power output of the bulb?

D=1km

P=?
λ=532nm
f=50hz
h=6.6×10^-27erg×s
c=3
A_eye= 1cm²
To begin solving this problem we must first understand what is being asked.  A lightbulb is a distance d away from an eye.  The eye must receive at least 10 photons before it refreshes to see the object.  Thus, we are solving for the power output of the lightbulb such that the area occupied by the eye will receive the energy equivalent of 10 photons per refresh cycle.

First we should find how much energy 10 photons is.  
\[E=hf=\frac{hc}{\lambda}\]
We know that an average visible photon of green light has a wavelength of 532nm,so:
\[E=\frac{hc}{\lambda}=\frac{6.6\times 10^{-27} erg\cdot s\times 3 \times 10^{10} cm/s}{532nm}=\]
\[=\frac{6.6\times 10^{-27} erg\cdot s\times 3 \times 10^{10} cm/s}{5.32\times 10^{-5}}=\]
\[3.7\times 10^{-12}erg\]
So, for ten photons this is:
\[\boxed{3.7\times 10^{-11}erg}\]
The eye needs to receive this many photons in one refresh cycle.  So:
\[P_{eye}=Energy/cycle \times f=3.7\times 10^{-11}erg \times 50hz=\]
\[\boxed{P_{eye}=1.9\times 10^{-9}erg/sec}\]
Next, let's calculate the power of the bulb.  Since the lightbulb emits photons in all directions, if we approximate the bulb as a point source, once can imagine a large sphere of radius D whose surface has the energy of the lightbulb spread evenly across it.  Due to this even distribution of energy, the ratio of the surface areas of he eye and sphere will equal the ratio of the powers of the eye and sphere. Using this, relation, we can solve for \(P_{sphere}\).
\[\frac{A_{sphere}}{A_{eye}}=\frac{P_{sphere}}{P_{eye}}\]
\[P_{sphere}=\frac{A_{sphere}}{A_{eye}} \times P_{eye}\]
\[P_{sphere}=\frac{4 \pi D^2}{1cm^2} \times 1.9 \times 10^{-9} erg/sec\]
\[P_{sphere}=\frac{4 \pi (1 \times 10^5 cm)^2}{1cm^2} \times 1.9 \times 10^{-9} erg/sec\]
\[\boxed{P_{sphere}=2.4 \times 10^{2} erg/sec}\]

Solved with collaboration from B. Beak, G. Grell, and F. Young.

Introduction

I am Anthony Taylor, Harvard Class of 2018. I grew up in Endicott, NY (Greater Binghamton). I attended Maine Endwell Public Schools and graduated from Maine Endwell Senior High School in 2014.  I plan to concentrate in both Astrophysics and Physics.  I anticipate attending graduate school and working for either academia or the private sector in a research position.

From a young age I have been interested in astronomy, outer space, and space exploration.  I believe that with the continued improvements in technology, astronomy is a field of study that has great potential in advancing scientific knowledge in a variety of fields across the board form theoretical physics to biology.  I would like to contribute to and to be a part of this process.

On a lighter note, my personal hobbies include distance running, which I partake in most mornings, computers, technology, and engineering.