Blog Post 23, Free Form: Astronomy and Politics: Interplanetary Law

Who owns space? For that matter, who owns the moon, or the orbit of a satellite? While astronomy is generally a very pure science, it has allowed mankind to leave our planet's surface and go to new unclaimed territory. The real question here is, if you go to the moon and build a moon-house there, is that your property? Can you own part of the moon? If you happen to blow up your moon-neighbor's house, under what law code is this illegal? This may seem very futuristic, but the nations of Earth actually put forth legal proceeding on these very question (okay fine, not these exact questions, but close enough).

On October 10th, 1967, the United Nations enacted the "Outer Space Treaty," also known as the "Treaty on Principles Governing the Activities of States in the Exploration and Use of Outer Space, including the Moon and Other Celestial Bodies."  That's quite a mouthful, so we will call it the OST (an initialism that I invented about 15 seconds ago).

The OST was originally put forth as a military treaty which bans the placement of usage of weapons of mass destruction in space, in orbit around Earth, or on the moon or other celestial bodies.  So basically, no satellite nukes.  Interestingly, it does not address conventional weapons, so you can put as many missiles in orbit as you want, but they can't be nuclear.  

The next part of the OST bans individual or national ownership of the any celestial body or resource, leaving them as the "common heritage of mankind."  Unfortunately, this means that you can't own a lunar golf course.  What a shame.  This doesn't mean that all satellites and equipment are public property, because all space equipment still belongs to the state that launched it.  This all means that the owner of the object is liable for any damage caused by said object.  This means that in the hit film Gravity, Russia has to pay for all of the equipment the its defunct satellite broke.  

The OST also addresses individual activities in space.  If a non-governmental party wishes to perform actions in space, they need to work with the jurisdiction of a signing nation of the OST.  For instance, SpaceX works under the United States' jurisdiction.  

IN 1979, a second treaty called the "Moon Treaty" (MT (also my own initialism)) was created.  This treaty tightened restrictions on space resource allocation, military functions, and generally attempted to sum up the OST and a few other treaties that hd been put in place in the years between the OST and MT.  Unfortunately, No space capable nations signed it.  This treaty has only been ratified by 16 nations, including Australia, Chile, Mexico, Pakistan and Turkey, none of which have space programs.  The US has not ratified it, neither has any space capable country.  Thus, it has no real power at all.

So, let's answer our initial questions:

Who owns space? All Humans

Who owns the moon, or the orbit of a satellite? Mankind owns the moon, the launching nation owns the satellite.

If you go to the moon and build a moon-house there, is that your property? The house is your property, but the land you placed it on is not.

Can you own part of the moon? No

If you happen to blow up your moon-neighbor's house, under what law code is this illegal? This is illegal under the OST as it is a "harmful interference with activities in the peaceful exploration and use of outer space."

So, there's a bit of space law for you.  One cool closing point: the OST claims that outer space is "common heritage of mankind."  I guess we just declared ownership of all alien worlds.  No wonder that always try to wipe us out in movies.

Sources:
https://en.wikipedia.org/wiki/Outer_Space_Treaty
https://en.wikipedia.org/wiki/Space_law
https://en.wikipedia.org/wiki/Moon_Treaty

Blog Post 22, WS 7.1, Problem 4: Type Ia Supernova Energy Conservation

Calculate the total energy output, in ergs of the explosion assuming that the white dwarf's mass is converted to energy via fusion of carbon into nickel. Note that the process of carbon fusion is not entirely efficient, and only about 0.1% of this mass will be radiated away as electromagnetic radiation.

How does this compare to the total binding energy, in ergs, of the original white draw? Does the white dwarf explode completely, or is some mass left over in the form of a highly concentrated remnant?

Let's begin.
First let's find the energy of the explosion.  It's time for a classic equation:
\[E=Mc^2\]
We are told that only 0.1% of the mass is converted to energy, so:
\[E=0.001Mc^2\]
Let's plug and chug.
\[E=0.001(1.4M_{\odot})(3\times 10^{10}cm/s)^2\]

\[E=0.001(2.8\times 10^{33}g)(3\times 10^{10}cm/s)^2\]
\[\boxed{E=2.52\times 10^{51} ergs}\]

Now lets find the potential energy of the original system.
\[U=\frac{3GM^2}{5R}\]
\[U=\frac{3(6.7\times 10^{-8}cm^3 g^{-1}s^{-2})(2.8\times 10^{33}g)^2}{5(12\times 10^8 cm)}\]

\[\boxed{U=2.6\times 10^{50} ergs}\]

Thanks to approximations and estimations of mass consumption etc, these values are not equal, however, in the grand scheme of things, they are the same.  This implies that Type Ia SNe leave nothing behind and the white draw is completely blown apart!

I worked on the problem with B. Brzycki, G. Grell, and N. James.

Blog Post 21, WS 7.1, Problem 1: White Dwarf Pressure

White Dwarfs are supported internally again the force of gravity by "electron degeneracy" pressure.  The maximum mass that can be supported by this exotic form of pressure is 1.4 M(sun) (also known as the Chandrasekhar Mass.)  The radius of our white dwarf is approximately twice the radius of the Earth, or \(12\times 10^8\) cm.
Given this mass M, and radius R, derive an algebraic expression for the eternal pressure of a white dwarf with these properties.  Start with Virial Theorem, and assume that the white dwarf is an ideal gas with uniformly distributed mass.

As the question prompted, lets start with Virial Theorem.
\[K=-\frac{1}{2}U\]
Next we should apply the kinetic energy of a particle in 3D space:
\[K_p=\frac{3}{2}kT\]
Where k is the Boltzmann Constant and T is the temperature.
I few assume that we have N particles, the the total kinetic energy is:
\[K=\frac{3}{2}NkT\]
Now let's add our usual expression for U.
\[K=-\frac{1}{2}U\]

\[\frac{3}{2}NkT=\frac{GM^2}{2R}\]

Simplifying we get:

\[3NkT=\frac{GM^2}{R}\]

We want to solve for pressure, so let's look at ideal gas law.

\[PV=NkT\]

We have some similar variables here.  This is a very good thing.

Let's solve both of our equations for NkT.

\[NkT=PV\]

\[NkT=\frac{GM^2}{3R}\]

Now we can set them equal to one another.

\[PV=\frac{GM^2}{3R}\]

Solving for P we get:

\[P=\frac{GM^2}{3RV}\]

V isn't in our given variables, but since the star is a sphere, we can express V in terms of R.

\[V=\frac{4}{3}\pi R^3\]

\[P=\frac{GM^2}{3R(\frac{4}{3}\pi R^3)}\]
\[\boxed{P=\frac{GM^2}{4\pi R^4}}\]
Qualitatively this makes sense, because as Mass increases pressure increases, and as Radius decreases, pressure increases.  This is the perfect recipe for an explosion.  Interestingly, (derived in AY16), the Radius of a White Dwarf Scales with Mass to the 1/3.  \(M \sim \frac{1}{R^3}\)
Thus:\[P \sim R^{-10}\]
Or:
\[P \sim M^{3.33}\]
This is cool because we can scale pressure to a single variable, so with a critical mass, we also can find a critical pressure and radius.

I worked with B. Brzycki, G. Grell, and N. James on this problem.

Blog Post 20: Hubble Tuning Fork

E0 Galaxy
A near perfectly spherical galaxy with no real defining structure.
http://www.noao.edu/image_gallery/images/d5/m89.jpg

E1 Galaxy
An elliptical galaxy with a length to width ratio of  9:10. M87 is pictured.
http://www.astronomynotes.com/galaxy/m87.gif

E4 Galaxy
An elliptical galaxy with a l/w of 6:10.  Types E2 and E3 have ratio of  8:10 and 7:10.
http://cseligman.com/text/atlas/ngc14.htm

E7 Galaxy
An elliptical galaxy with a l/w of 3:10.  Types E5 and E6 have ratio of  5:10 and 4:10.
http://www.mhhe.com/physsci/astronomy/fix/student/images/23f06.jpg

S0 Galaxy
Lenticular galaxy with a bulge and disk, but no arms or spiral.
http://bama.ua.edu/~rbuta/nearirs0/plate014c.jpg

Sa Galaxy
Spiral Galaxy with tightly bound arms.
http://www.noao.edu/image_gallery/images/d4/m81y.jpg

SBa Galaxy
Spiral Barred Galaxy with tightly bound arms.
https://en.wikipedia.org/wiki/Barred_spiral_galaxy

SBb Galaxy
Spiral Barred Galaxy with loosely bound arms.
http://nrumiano.free.fr/Images_gx2/Jalf350.gif

Sb galaxy
Spiral Galaxy with loosely bound arms.
http://www.leviathanastronomy.com/image-gallery.html

SBc Galaxy
Spiral Barred Galaxy with very loosely bound arms.
http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit4/Galaxies/

Sc Galaxy
Spiral galaxy with very loosely bound arms.
http://www2.lowell.edu/rsch/LMI/gallery/n6946_2.jpg

Blog Post 19, Free Form, Astrobites: Accretion Modes

Figure 2 of Ohsuga and Mineshige 2011: Time-averaged structure of the three models produced by the simulations. The shaded regions show the normalized density of inflowing gas. The thick lines are streamlines for the outflows. (This plot is illustrative: the simulations are in two dimensions, not three, and only calculated on one side of the equator.)

In previous free form posts we have mentioned quasars, and other black holes that consume matter through accretion. Let's take a closer look at this process with some cutting edge research. Guest Author Warrick Ball wrote a nice summary of a paper titled "Global structure of three distinct accretion flows and outflows around black holes through two-dimensional radiation-magnetohydrodynamic simulations"published by Ken Ohsuga and Shin Mineshige of the National Astronomical Observatory of Japan and Graduate of Advanced Study.  Here is a summary of that summary.  

A brief over view of accretion is that it is the process by which black holes consume matter by drawing it in.  Usually this matter is comes dust, but it also can be stars, planets, and other black holes in a cosmic food chain that is governed but the principle: "Survival of the Biggest".  

Ohsuga and Mineshige's paper addresses the energy flow and matter flow of accreting matter.  They created a computer simulation of a non-rotating black hole and placed a "doughnut" of gaseous matter around it.  The performed the simulation with gas densities of \(1, 10^(-4), \text{and} 10^{-8} g/cm^3\).  The results were wildly different every time, but seemed to conform to theoretical models fairly well, but not perfectly.  Clearly we still have plenty of work to do in this field.  Some simplifications in the model were simulating a stationary, non-rotating black hole, only simulating one half of the balks hole's equator, and using a 2D model.  This vastly simplifies the simulation and saves on the programming time and computing power needed to execute it.

Source:

http://astrobites.org/2011/06/05/the-many-modes-of-black-hole-accretion/

Blog Post 18, WS 6.1, Problem 3: Return to Virial Theorem

3. One of the most useful equations in astronomy is an extremely simple relationship known as the Virial Theorem. It can be used to derive Kepler’s Third Law, measure the mass of a cluster of stars, or the temperature and brightness of a newly-formed planet. The Virial Theorem applies to a system of particles in equilibrium that are bound by a force that is defined by an inverse central-force law (F=k/rα) (where k is a constant). It relates the kinetic (or thermal) energy of a system, K, to the potential energy, U, giving
\[K=-\frac{1}{2}U\]
(a) Consider a spherical distribution of N particles, each with a mass m. The distribution has total mass M and total radius R. Convince yourself that the total potential energy, U, is approximately
\[U\approx -\frac{GM^2}{R}\]
You can derive or look up the actual numerical constant out front. But in general in astronomy, you don’t need this prefactor, which is of order unity.

(b) Now let’s figure out what K is equal to. Consider a bound spherical distribution of N particles (perhaps stars in a globular cluster), each of mass m, and each moving with a velocity of vi with respect to the center of mass. If these stars are far away in space, their individual velocity vectors are very difficult to measure directly. Generally, it is much easier to measure the scatter around the mean velocity if the system along our line of sight, the velocity scatter σ2. Show that the kinetic energy of the system is:
\[K=N\frac{3}{2}m\sigma^2\]
(c) Use the Virial Theorem to show that the total mass of, say, a globular cluster of radius R and stellar velocity dispersion σ is (to some prefactor of order unity):
\[M\approx \frac{\sigma^2 R}{G}\]

(a) Consider a spherical distribution of N particles, each with a mass m. The distribution has total mass M and total radius R. Convince yourself that the total potential energy, U, is approximately
\[U\approx -\frac{GM^2}{R}\]
You can derive or look up the actual numerical constant out front. But in general in astronomy, you don’t need this prefactor, which is of order unity.

For this problem let's imagine a spherical distribution of particles.  For one particle, the potential energy comes from the particle's distance for the center of the distribution and the force from all other particles.  We could use calculus to solve this with integration over the distribution, but since we are ignoring numerical constants, we can think of it more intuitively and get the same result.  (For the full derivation, please see: http://ay17-ataylor.blogspot.com/2015/03/worksheet-8-problem-2-total-potential.html ). Let's instead consider a single particle in the distribution.
Generally, the potential energy of a particle of mass m related to a distribution with mass M is:
\[U_p=-\frac{GMm}{R}\]
However, we have N particles, each will mass M.  So the total potential energy is:
\[U=NU_p=-\frac{GMNm}{R}\]
However, we remember that M=Nm, thus:
\[U=-\frac{GMNm}{R}=\boxed{-\frac{GM^2}{R}}\]

(b) Now let’s figure out what K is equal to. Consider a bound spherical distribution of N particles (perhaps stars in a globular cluster), each of mass m, and each moving with a velocity of vi with respect to the center of mass. If these stars are far away in space, their individual velocity vectors are very difficult to measure directly. Generally, it is much easier to measure the scatter around the mean velocity if the system along our line of sight, the velocity scatter σ2. Show that the kinetic energy of the system is:
\[K=N\frac{3}{2}m\sigma^2\]

Let's use the same approach as before.
Commonly, for a single particle:
\[K_p=\frac{1}{2}mv^2\]
Let's modify this. This assumes motion in a single direction, for motion in three directions/dimensions, we should multiply by 3, and replace velocity (1 directional) with scatter (3 directional).
\[K_p=\frac{3}{2}m\sigma^2\]
For the total energy of the system we have N particles, so:
\[K=K_p N=\boxed{N\frac{3}{2}m\sigma^2}\]

c) Use the Virial Theorem to show that the total mass of, say, a globular cluster of radius R and stellar velocity dispersion σ is (to some prefactor of order unity):
\[M\approx \frac{\sigma^2 R}{G}\]

Let's put this all together.
\[K=-\frac{1}{2}U\]
\[N\frac{3}{2}m\sigma^2=\frac{GM^2}{2R}\]
Let's rearrange and solve for M.  We can drop all numerical constants.
\[Nm\sigma^2~\frac{GM^2}{R}\]
Earlier we mentioned that Nm=M, so:
\[M\sigma^2~\frac{GM^2}{R}\]
\[\sigma^2R~GM\]
\[\boxed{M~\frac{\sigma^2 R}{G}}\]

I worked on this problem with B. Brzycki, G. Grell, and N. James.

Blog Post 17, WS 6.1, Problem 2: Sérsic Profiles

The Hubble Classes have characteristic surface light distribution profiles. They are fairly well described by parametrized equations in the form I(R) where I = intensity and R = distance from the centre. Often each is scaled to their value at the effective radius, such that \(I_e=I(R_e)\). The most general profile is the S ́ersic Profile, given by the equation:
\[I(R)=I_e exp\left(-b_n \left(\left(\frac{R}{R_e}\right)^{1/n} -1 \right) \right)\]
The constant bn depends on the shape parameter n. n = 4 gives rise to the famous \(r^{1/4}\)-law, or the de Vaucoleur Profile, which approximates elliptical and the bulge of spiral galaxies. n = 1, on the other hand, is equivalent to a simple exponential profile, which often corresponds with the outskirts of spiral galaxies. The best fit is often obtained by a combination of the functional forms.

Another way to write the exponential profile is:
\[I(R)=I_0 exp(-R/b)\]
where I0 is the central surface brightness and b is a characteristic lengthscale, a constant.

i. Describe what b is, and suggest how one might measure it for a given spiral galaxy. Be as specific as possible about what kind of data and tools may be required.

ii. The Milky Way has an estimated b = 3.5 kpc. Plot the disk surface brightness profile relative to the central brightness and indicate the location of the characteristic lengthscale from the centre.

iii. Assuming the entire Milky Way (including the central bulge) can be described by this profile, and that it is circularly symmetric. Approximately what fraction of all its stars are interior to the Sun (at 8kpc)?

i. Describe what b is, and suggest how one might measure it for a given spiral galaxy. Be as specific as possible about what kind of data and tools may be required.

Since b is in the denominator of a term in the arguments of an exponential, when R=b, the coefficient of the exponential is multiplied by 1/e (e=2.718...).  Therefore, b is a scale length, such that every length of b that one moves away from the center of the galaxy, the surface brightness decreases by a factor of e.
\[I(R)=I_0 exp(-R/b)\]

\[I(b)=I_0 exp(-b/b)=I_0 exp(-1)=\frac{I_0}{e}\]

To determine e for a galaxy, one would first need a telescope large enough to resolve the galaxy.  By taking pictures of the galaxy with a CCD, one could use the image light counts at different radii of the galaxy to construct a surface brightness profile for the galaxy.  One could then compare their profile to the exponential profile and solve for b.

ii. The Milky Way has an estimated b = 3.5 kpc. Plot the disk surface brightness profile relative to the central brightness and indicate the location of the characteristic lengthscale from the centre. 

On this plot, radius b is at the dark vertical line, and the y-axis measures the fraction of the central brightness.

iii. Assuming the entire Milky Way (including the central bulge) can be described by this profile, and that it is circularly symmetric. Approximately what fraction of all its stars are interior to the Sun (at 8kpc)?

This is going to involve some calculus.  First, let's use this profile to get the total brightness of the galaxy.  Since we are using polar coordinates, we will need to introduce a factor of R into our integration.

\[I_T=\int^{\infty}_0 \int^{2\pi}_0 I_0 Re^{-R/b} dR d\theta\]

\[I_T=2\pi I_0 \int^{\infty}_0 Re^{-R/b} dR\]

This involves integration by parts, so we'll let Wolfram Alpha take care of that for us.

\[I_T=2\pi I_0 b^2\]

Now that we know the total brightness of the galaxy, let's solve for the part inside the sun's galactic radius.

\[I_{r_*}=2\pi I_0 \int^{r_*}_0 Re^{-R/b} dR\]

\[I_{r_*}=2\pi I_0 b(b-(r_*+b)e^{-r_*/b})\]

Taking the ratio of this with the total brightness gives us an equation for brightness fraction as a function of radius.

\[\frac{I_{r_*}}{I_T}=\frac{2\pi I_0 b(b-(r_*+b)e^{-r_*/b})}{2\pi I_0 b^2}\]

\[\frac{I_{r_*}}{I_T}=\frac{b-(r_*+b)e^{-r_*/b}}{b}\]

Now let's plug in our numbers, R=8 kpc and b=3.5 kpc.

\[\frac{I_{r_*}}{I_T}=\frac{3.5-(8+3.5)e^{-8/3.5}}{3.5}\]

\[\frac{I_{r_*}}{I_T}=.67\]

We know that brightness directly scales to number of stars, so 67% of the stars in the Milky Way are inside the sun's galactic radius.

I worked with B. Brzycki, G. Grell, and N. James on this problem.

Blog Post 16, The Great Debate

Shapley

Curtis

One of the themes for this week has been depth in astronomy.  When we look up at the night sky, we often forget that we are looking into space, not at space.  What I mean by this is that the sky is not a large sphere surrounding the Earth with all of the stars at the same distance away, it is a 3D space where each object has an RA, Dec, and Distance away form the Earth.  Before computers and the exponential improvement of autonomy equipment, observing distance was a great challenge for Astronomers, and led to many a disagreement.  On of the greater of these disagreements was The Great Debate between Harlow Shapley and Heber Curtis.

Both of these astronomers had observed the Andromeda Galaxy and other spiral galaxies and disagreed greatly upon its size, and the size of the universe.

Shapley liked to think small.  He argued that the Milky Way galaxy made up the entire universe, and as such, the spiral galaxies were simply a nebulae within the Milky Way.  He bolstered his claim using the great distance away that other galaxies must be for the extragalactic theory to work, distances unheard of in 1920.  Furthermore, respected astronomer Adriaan van Maanen claimed that he saw M101 (Pinwheel Galaxy) rotating.  If this were true, the object could not be a galaxy or else its edges would exceed the speed of light due to orbital velocity.

Curtis liked to think big.  He believed that such objects were other galaxies outside of the Milky Way.  His main arguments relied on the fact that the frequency of supernovae in Andromeda far exceeded that of supernovae in other parts of the sky, thus Andromeda must be its own galaxy to avoid an irregular distribution across the sky.  He also used the argument that the doppler shifts of the spiral objects were far greater than anything else in the Milky Way.

It is fairly clear that Curtis ultimately won.  Van Maanen's claims of eye-witnessed spiral rotation were found to be incorrect.  This vastly undermined Shapley's argument and Curtis was victorious.

And they all lived happily ever after,

THE END

Sources:
https://en.wikipedia.org/wiki/Great_Debate_(astronomy)
http://apod.nasa.gov/diamond_jubilee/debate_1920.html
http://antwrp.gsfc.nasa.gov/htmltest/gifcity/curtis.gif
http://antwrp.gsfc.nasa.gov/htmltest/gifcity/shapley.gif

Blog Post 15, WS 5.1, Problem 4: Supernovae Type Ia Distances

Some stars explode as supernovae (SNe). In particular, Type Ia Supernovae come from exploding white dwarfs in binary systems. For now, it’s not important to know how this happens. It is, however, critical to learn the consequences of this mechanism, because they too are standard candles.

(a) As with the Cepheids, we can analyze the light curve of Type Ia SNe to standardize them. Below is a set of light curves. Examine them carefully, considering quantities like light curve shape, width/timescales, relative & absolute luminosity, &c. Find a rough relation obeyed by all the Supernovae Type Ia. Note: Although the supernova light curves have many features, try to relate just one of them to the peak magnitude.

(b) Describe how you would measure the distance to a Supernova Ia.

(c) Measure the distance to SN Cornell, whose light curve is shown below.

(a) As with the Cepheids, we can analyze the light curve of Type Ia SNe to standardize them. Below is a set of light curves. Examine them carefully, considering quantities like light curve shape, width/timescales, relative & absolute luminosity, &c. Find a rough relation obeyed by all the Supernovae Type Ia. Note: Although the supernova light curves have many features, try to relate just one of them to the peak magnitude. 

In all Type Ia Supernovae, the maximum ABSOLUTE magnitude reached is about -19.3.  This is true for ANY Type Ia SNe.  Thus:

\[M_{max}=-19.3\]

This is very useful, because they can be used as distance markers.

(b) Describe how you would measure the distance to a Supernova Ia. 

This is simple using the distance modulus and a basic apparent magnitude measurement.  First one would take a light curve of the SNe in question and find the maximum apparent magnitude.

Then one would apply the distance modulus.

\[m-M=5log(d)-5\]

\[d=10^{\frac{m-M+5}{5}}\]

Plugging in for a Type Ia SNe:

\[d=10^{\frac{m+19.3+5}{5}}\]

\[\boxed{d=10^{\frac{m}{5}+4.86}}\]

(c) Measure the distance to SN Cornell, whose light curve is shown below. 

Looking at SN Cornell, it's apparent magnitude peaks at 7.  Let's find its distance.

\[d=10^{\frac{m}{5}+4.86}\]

\[d=10^{\frac{7}{5}+4.86}\]

\[d=10^{6.86}\]

\[\boxed{d=1.8\times 10^6 pc = 1.8Mpc}\]

For reference, this is roughly equivalent to 53 Milky Way Galaxies put in a line edge to edge.

Blog Post 14, WS 5.1, Problem 2: Magnitude Applications

2 (a) Suppose you are observing two stars, Star A and Star B. Star A is 3 magnitudes fainter than Star B. How much longer do you need to observe Star A to collect the same amount of energy in your detector as you do for Star B?
(b) Stars have both an apparent magnitude, m, which is how bright they appear from the Earth. They also have an absolute magnitude, M, which is the apparent magnitude a star would have at d = 10 pc. How does the apparent magnitude, m, of a star with absolute magnitude M, depend on its distance, d away from you?
(c) What is the star’s parallax in terms of its apparent and absolute magnitudes?

(a) Suppose you are observing two stars, Star A and Star B. Star A is 3 magnitudes fainter than Star B. How much longer do you need to observe Star A to collect the same amount of energy in your detector as you do for Star B? 

Let's use the recursive relation for magnitudes to solve this problem.  

\[\frac{F_B}{F_A}=2.5^{(m_A-m_B)}=2.5^3=16\]

From this, we know that Star B is 16 times brighter than Star A.  

Energy scales to \(Flux \times Time\), thus:

\[E=Ft\]

Let's let \(E_A\) be the energy received from Star A in the time \(t_A\).  We want \(E_B=E_B\), so:

\[E_A=F_A t_A=E_B=F_B t_B\]

\[F_A t_A =F_B t_B\]

We already established that \(F_B=16F_A\), so:

\[F_A t_A =16F_A t_B\]

\[\boxed{t_B=\frac{T_A}{16}}\]

Thus, one should observe Star B for one sixteenth of the time that one observed Star A in order to get the same amount of energy capture.

(b) Stars have both an apparent magnitude, m, which is how bright they appear from the Earth. They also have an absolute magnitude, M, which is the apparent magnitude a star would have at d = 10 pc. How does the apparent magnitude, m, of a star with absolute magnitude M, depend on its distance, d away from you? 

We have established in previous posts that Flux received from a star at a given distance is:

\[F=\frac{L}{4\pi d^2}\]

So, for our star at 10 pc with magnitude M, we have:

\[F_M=\frac{L}{4\pi (10pc)^2}\]

Comparatively, at another distance d, the same expression reads:

\[F_m=\frac{L}{4\pi d^2}\]

In this expressions, L is constant, so we can set the two equations equal to one another.

\[F_M 4\pi (10pc)^2 = F_m 4\pi d^2\]

With rearranging, we can get:

\[\frac{F_M}{F_m}=\frac{d^2}{(10pc)^2}\]

With arrangement, we can add another term: the magnitude relation.

\[\frac{F_M}{F_m}=\frac{d^2}{(10pc)^2}=2.5^{(m-M)}\]

We are trying to solver for magnitude in terms of distance, so let's discard the Flux comparison:

\[\frac{d^2}{(10pc)^2}=2.5^{(m-M)}\]

Next, let's solve for m in terms of d and M.

\[2log(\frac{d}{10pc})=(m-M)log(2.5)\]

\[m-M=\frac{2log(\frac{d}{10pc})}{log(2.5)}\]

\[m=\frac{2log(\frac{d}{10pc})}{log(2.5)}+M\]

\[m=5log(\frac{d}{10pc})+M\]

\[m=5(log(d)-log(10pc))+M\]

\[\boxed{m=5(log(d)-1)+M}\]

This is also known as the Distance Modulus.

(c) What is the star’s parallax in terms of its apparent and absolute magnitudes? 

We know that:

\[p=\frac{1}{d}\]

If we solve the distance modes for distance, we get:

\[m=5(log(d)-1)+M\]

\[\frac{m-M}{5}+1=log(d)\]

\[d=10^{\frac{m-M}{5}+1}\]

Plugging this in, we get:

\[\boxed{p=10^{-(\frac{m-M}{5}+1)}}\]

I worked on this problem with B. Brzycki, G. Grell, and N. James.