Introduction
Star formation is governed by the collapse of a cloud of particles into a gravitationally bound sphere which we call a star.  The radius of the could at which this occurs is called the Jeans Length, where the gravitational force of the cloud overcomes the thermal energy causing it to expand.  Here we examine the time scale of such a collapse and also calculate the Jeans Length.

Methods
In order to determine the time it takes for this collapse to occur in terms of the mass and size of the cloud, we consider a cloud of mass M and a test particle a distance away from it.  We assume the cloud has a mass given by

 

We were recently, or not so recently—I'm very good at procrastinating—assigned a multi-part blogging to find out what it truly means to be an astronomer.  I realise as a sophomore astrophysics major that I still don't understand the specifics of what being an astronomer entails or means to me.  To quote my friend Alexa,

I just want to be one. So much. “Space” is, if you think about it, everything but Earth. When we study it, we’re pausing our narcissistic tendencies for just a moment. We’re not everything; we’re part of everything. Ignoring that is shameful.

She stated in the best way possible what attracts me to astronomy, but that still doesn't mean I know what astronomy is.  Right now I just think of astronomy as some nebulous loosely defined field of Things I Would Like To Do Because They Are Amazing, but that's not an acceptable answer to the question.  So without further delay I shall attempt to synthesise my thoughts on the topic.

The point of being a professional astronomer, in my experience, is to contribute understanding of what the universe is, how it is structured, how it came to be, and what its future might hold.  Most likely this is because the first astronomer I ever met was a professor of cosmology.  I've come to accept that he's probably the reason my main research interest tends to observational cosmology.  Of course, this is a very broad and relatively unhelpful answer to the astronomer question.  Sure, that's the intention, but how do we get there?

I think it's safe to assume that the journey to becoming a professional astronomer begins as an undergrad, or if you're very lucky, as a high school student.  I think mine was a bit of both, as I did get the opportunity as a junior to do some busy work for he of blog title fame.  But that was a week long, and although it was some exposure I doubt it's how careers in astronomy start.  Careers in astronomy, at least for a Caltech student, probably start with a SURF fellowship.  I know SURF was my first real look at what an astronomer does.  I sat alone in an office 8-9 hours a day, writing code in a language I'd never seen before to analyse data I didn't understand, and I had fun doing it.  I think that enjoyment that's what sets the astronomer apart from the average person.

Then the natural course of things is to go to graduate school.  This is where you decide What You Want To Do With Your Life.  As far as I know, you don't have to decide right away.  Unless you're in the UK in which case you need to know what you want to study before you've learnt anything about it.  At least, that is what my SURF mentor who is English tells me.  Being a grad student requires doing semi-independent research under the guidance of a faculty member who works on a similar topic.  You'll probably start to hate your field at some point during this process, but hopefully you'll get over it soon.  Next is the postdoctoral fellow.  I have no idea what a postdoc does.  Don't tell my SURF mentor that because he is one.

I believe your career options then become a) professor at an academic institution, b) research scientist some place like the SAO, or c) finance.  I'm sure there are more options, I'm just uneducated in that side of things.  The first two options strike me as pretty similar except the professor track astronomer will probably have to teach at some point or another.  This part is where you get to move on to independent research in topics that interest you.  You might find out something that only you know about, and that's a rewarding experience.  Although any work in astronomy is rewarding if it's what's truly exciting and inspirational to you.

I have given my impression of what it takes to become an astronomer.  So once again, we're back to the question of what does it mean to me to be an astronomer?  It's going to take a lot of work.  Astronomy, as it turns out, is hard.  But the work will be worthwhile because I'll be learning how the universe works, or how we think the universe works.  Maybe I'll end up amending some of that knowledge.  Who knows?  Being an astronomer means getting excited about the mysteries of space and our tiny place in it.  It means realising how small we really are in the grand scheme of things, accepting that, and moving on to understand why.  Most of all, it means that when your friend starts talking about M83 and means the band, this is all you can see.

Abstract
We would like to know how the sun is being "supported".  We assume that this mechanism is hydrostatic equilibrium, but to be sure we work through the derivation.

Introduction
We know that the sun is somehow being prevented from gravitational contraction.  Our theory is that it is supported by hydrostatic equilibrium, which means that the internal pressure provides an opposing support force.  We calculate the gravitational force on a mass shell, the pressure required to balance it, and then derive the force equation for hydrostatic equilibrium.

Methods and Results
We first assume the Sun to be a spherical gas cloud with density ρ(r).  We consider a differential mass shell of this sphere with radius r.  We recall that the volume of a sphere is ⁴/₃πr³ and that the differential volume is its derivative. Then we get a differential mass dM:

We know the equation for universal gravitation:

Here we let M be the total mass enclosed by the mass shell and m be the differential mass element.  As a result, we get the differential gravitational force to be:

We know that pressure is equal to force divided by area.  So we can say:

Now dividing by dr on both sides of the equation we arrive at the equation of hydrostatic equilibrium:

Conclusions

We have derived from simple physical laws that the equation for hydrostatic equilibrium is a plausible explanation for the way the sun is supported.  A quick search shows that we are indeed correct.  Hooray!

Abstract
Considering the angular diameter of the sun and the astronomical unit, we can estimate the radius of the sun, the AU in solar diameters, and the mass of the sun using Kepler's 3rd law.

Methods
Applying basic trigonometric identities and taking the astronomical unit a to be the distances from us to the closest point of the sun to us (i.e., the centre of the circle we see from earth), we can see that:

Multiplying through by a we get a value for the radius of the sun.  It is clear from here that if we divide a by twice the solar radius we can easily determine the answer to the second part of our question.  Finally, we have Kepler's 3rd law:

Where P is the period of the earth and G = 6.7 x 10^-8 dyne-cm²/g².  From here we can solve for the mass of the sun.

Results
Solving the first equation using a = 1.5 x 10¹³ cm we get the radius of the sun equal to 6.545 x 10¹⁰ cm which is very close to the actual value of 6.955 x 10¹⁰7 s.  Dividing, we get 1 AU = 114.6 solar diameters.  Then, solving for the mass of the sun in Kepler's 3rd law with P = 3.154 x 10⁷ s, we have the mass of the sun equal to 2.007 x 10³³ g which is a surprisingly accurate number.

by Eric S. Mukherjee, Nathan Baskin, and I forget who else (sorry).


Abstract
In this problem we consider the how the temperature of the sun affects the temperature of the earth.  This is possible to estimate by assuming both the sun and the earth to behave like perfect blackbodies.

Introduction
Assuming the Earth has constant surface temperature and that it behaves like a blackbody, we can estimate the surface temperature using the energy emitted by the sun.  We also assume the sun to be a perfect blackbody.  Under these assumptions we can find the surface temperature of the Earth by knowing the temperature of the sun, the radius of the sun, the mass of the sun, the mass of the earth^*, and the radius of the earth.

Methods
We start with the equation for flux at the surface of a blackbody (σ is the Stefan-Boltzmann constant):

From this we derive the luminosity of the sun by multiplying through by the surface area:

Then the flux of the sun at the surface of the earth is (where a is the astronomical unit):

If we consider the area of the earth through which the flux passes, it is the circle of area π R_⊕².  Multiplying through by this quantity we get the power input to the earth from the sun.  We then realise that this is necessarily equal to the power output of the earth due to energy conservation which, at the surface of the earth, is equal to σT⁴π R_⊕².   Thus we have an equation of the form:

Using this equation with R_☉= 695,500 km and T_☉= 5778 K, we get T_⊕ = 279 K.

Conclusions
This temperature that we calculate is around 5.5°C which sounds reasonable for an earth without accounting for atmospheric greenhouse effects and allowing for the temperature at the poles.  The true average temperature of the earth is around 16°C but that is measured with the warming effect of the atmosphere.  The sun is not a perfect blackbody which also contributes to the difference between our calculation and the true value.

Acknowledgements
I'd like to thank the entire Ay 20 class and teachers for collective brainpower due to the fact that I can't remember who exactly worked on this problem and I'm sure we drew from the knowledge of many people in the room.  I'd also like to thank the superior computational power of Wolfram Alpha for bringing to my attention that there exponents matter when calculating ratios and that the temperature of the earth is most definitely not 1270 K.



*Note: It has been brought to my attention by Professor Johnson that the masses of the earth and sun do not actually factor into this calculation at all unless we need them to derive some of our other known constants.

As my readers may remember, a few weeks back I posted about the properties of AGN and how they affect their host galaxies.  One of these ways I listed was star formation rate.  Actually, a bit less than two weeks ago an article was reprinted from UCSD by ScienceDaily that AGN do not stop star formation as previously thought.

Prior research showed a correlation between the presence of AGN and the lack of star formation in galaxies.  This new study claims that this was a function of observational bias.  Older, more massive galaxies are easier to detect, and are also the ones with decreased star formation rate.  This study finds AGN in all types of galaxies including those in which stars are still being formed.

Read the full article here.