Worksheet 9, Problem 1: Star Scale

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

(a) If you let the size of your body represent the size of the star forming complex, how big would the forming stars be? Can you come up with an analogy that would help a layperson understand this difference in scale? For example, if the cloud is the size of a human, then a star is the size of what? 

In an attempt to understand the extraordinary size of a star forming cloud, we aim to consider what size a newly formed star represents within its parent cloud, with respect to the size of a human being. For this problem, we will scale the star forming cloud to the size of a person. Let's examine just what size a star would be now!

The first step for approaching this problem involves drawing an appropriate figure:

From this, we know the size of an average person (not Shaq!) and how to calculate the size of a star forming complex and the size of our sun. The final piece that we will need is the size of our unknown green object. To quantify their relationship, we can establish the following ratios: \[ \frac{V_{cloud}}{V_{sun}} = \frac{V_{person}}{V_{unknown}} \] For ease of calculation, we'll consider each object as a sphere, with the volume of a sphere as: \[ V = \frac{4}{3} \pi r^3 \] Applying this to our ratio, we reach: \[ \frac{ \frac{4}{3} \pi r_{cloud}^3}{\frac{4}{3} \pi r_{sun}^3} = \frac{ \frac{4}{3} \pi r_{person}^3}{ \frac{4}{3} \pi r_{unknown}^3} \] Simplifying, we reach:  \[ \frac{r_{cloud}}{r_{sun}} = \frac{r_{person}}{r_{unknown}} \] Next, we solve for our unknown r: \[ r_{unknown} =  \frac{r_{person} \times r_{sun}}{r_{cloud}}\] With our equation set, we can now begin to plug in what we know: \[ r_{cloud} = 30 \, pc \times \frac{3 \times 10^{18} \, cm}{1 \, pc} = 9 \times 10^{19} \, cm \] \[ r_{sun} = 7 \times 10^{10} \, cm\] \[ r_{person} \approx 100 \, cm \] Solving for the unknown r, we get: \[ r_{unknown} =  \frac{(100 \, cm) \times (7 \times 10^{10} \, cm)}{9 \times 10^{19} \, cm} = 1 \times 10^7 \, cm \] \[ r_{unknown} = 1 \, nm\] We can now update our figure to show the appropriate values:

This radius corresponds approximately to the size of an amino acid, one of the simplest building blocks of life.

(b) Within the Taurus complex there is roughly \(3 \times 10^4 M_{sun}\) 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 ( \( \rho_lead = 11.34 g \, cm^{-3}\) ) and air ( \( \rho_air = 0.0013 g \, cm^{-}3\) ). This is four orders of magnitude, which is a huge difference!

First, let's calculate the density of the Taurus complex. We are given that the total mass is \(3 \times 10^4 M_{sun}\). To find the density, we can divide this by the overall (spherical estimate) volume of the complex. \[ \rho = \frac{Total \, Mass}{Volume} = \frac{(3 \times 10^4) \times ( 2 \times 10^{33} \, g)}{\frac{4}{3} \pi (9 \times 10^{19} \, cm)^3} \] \[ \rho_{complex} = 2 \times 10^{-23} \approx 10^{-22} \, \frac{g}{{cm}^3}\] Next, we can calculate the density of a star, using the sun as a model: \[ \rho = \frac{Sun \, Mass}{Volume} = \frac{2 \times 10^{33}\, g}{\frac{4}{3} \pi (7 \times 10^{10} \, cm)^3} \] \[ \rho_{sun} = 1.39 \approx 1 \, \frac{g}{{cm}^3} \] In comparing these two values, we see that there is a difference of 22 orders of magnitude. Putting this in the context of air and lead differing by only 4 orders of magnitude, it isn't surprising that as a star forms, it pulls in the massive amounts of dust and gas with its gravitational pull!

Acknowledgements: Johnathan Budd & Willie Pirc