Worksheet 9, Problem 1: The Spacial Scale of Star Formation

The size of a modest star forming molecular cloud, like the Taurus region, is about 30 pc. The size of atypical star is, to an order of magnitude, the size of the Sun.

Taurus Region
Image Credit: http://www.usm.lmu.de/ys/tauori.gif

a) If you let the size of your body represent the size of the star forming complex, how big would the forming stars be? Come up with an analogy the describe this difference in scale.

We know that the size of a molecular cloud is about 30 pc, and the size of a forming star, which we approximate to be the size of the Sun, is about \(4\times 10^{-8}\) pc. The height of an average human is around 1.7 meters. Now we can set up a ratio to find what size a star would be if the human body was a molecular cloud:

\(\dfrac{4\times 10^{-8}\text{ pc}}{30\text{ pc}}=\dfrac{x}{1.7\text{ m}}\)

Solving for x, we find that the star would be represented as an object with size:

\(x=2.3\times 10^{-9}\text{ m}\approx 2\text{ nm}\)

Two nanometers is smaller than the size of a human cell. So this gives you a good idea of how small a forming star is in comparison to its molecular cloud. 

b) Within the Taurus complex there is roughly \(3\times 10^4M_\odot\) of gas. To order of magnitude, what is the average density of the region? What is the average density of a typical star (use the Sun as a model)? How many orders of magnitude difference is this? Consider the difference between lead \((\rho_{lead}=11.34\text{ g cm}^{-3})\) and air \((\rho_{air}=0.0013\text{ g cm}^{-3})\). This is four orders of magnitude, which is a huge difference!

Approximating the Taurus as a sphere with a radius of 15 pc, or \(4.5\times 10^{19}\) cm, we can find its volume:

\(V_{Taurus}=\frac{4}{3}\pi (4.5\times 10^{19})^3\approx 3.8\times 10^{59}\text{ cm}^3\)

We know that the mass of the gas in the Taurus is \(3\times 10^4M_\odot\). Now we can solve for density to get:

\(\rho_{Taurus} =\dfrac{M_{Taurus}}{V_{Taurus}}=\dfrac{3\times 10^4M_\odot}{3.8\times 10^{59}\text{ cm}^3}\approx 8\times 10^{-56}M_\odot \text{ cm}^{-3}\)

To find the average density of a star using the Sun as a model, we know that the radius is \(7\times 10^{10}\) cm, so its volume is:

\(V_\odot=\frac{4}{3}\pi(7\times 10^{10})^3=1.4\times 10^{33}\text{ cm}^3\)

So we calculate a density of:

\(\rho_\odot =\dfrac{M_\odot}{V_{\odot}}=\dfrac{M_\odot}{1.4\times 10^{33}\text{ cm}^3}\approx 7\times 10^{-34} M_\odot\text{ cm}^{-3}\)

There is a difference of 22 orders of magnitude between the density of a molecular cloud and the density of a star. Stars are much denser than molecular clouds, with a greater difference than even the densities of lead and air. 

I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their help in solving this problem.

Citations:
http://en.wikipedia.org/wiki/Solar_radius