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

The problem: 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.

Part 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?

Solve: To begin, we need to get everything on the same scale:

• Molecular Cloud Radius: \( 15 pc \frac{3 \times 10^8 cm}{1 pc} = 4.5 \times 10^{19} cm\)
• Radius of the Sun: \(7 \times 10^{10} cm\)
• Radius of a person: \( \approx 1 m = 100 cm\)

Now we can set up a ratio: \[ \frac{R_{\odot}}{R_{cloud}} = \frac{x}{R_{person}}\] Solving for x gives: \[ x = \frac{R_{\odot} R_{person}}{R_{cloud}} = \frac{7 \times 10^{10} \times 100}{4.5 \times 10^{19} } \approx 3 \times 10^{-8} cm\]

This is in the same order of magnitude of the radius of an atom! Imagine your whole body condensing into the size of an atom...

Part B: Within the Taurus complex there is roughly \(3 \times 10^4 M_{\odot}\)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 (ρlead = 11.34 g cm^-3 ) and air (ρair =  0.0013 g cm^-3 ). This is four orders of magnitude, which is a huge difference!

What we know:

• \(R_{\odot} = 7 \times 10^{10} cm \)
• \(R_T = 15 pc \frac{3 \times 10^8 cm}{1 pc} = 4.5 \times 10^{19} cm\)
• Mass of the sun: \( 2 \times 10^{33} g \)
• Mass of Taurus complex: \(3 \times 10^4 M_{\odot} = 3 \times 10^4 2 \times 10^{33} = 6 \times 10^{37} \)

Solve: To find density we know: \[ \rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi R^3} \] 

To find the density of the sun: \[ \rho = \frac{M_{\odot}}{\frac{4}{3} \pi R_{\odot}^3}= \frac{2 \times 10^{33}}{\frac{4}{3} \pi (7 \times 10^{10})^3}  = 1.4 \frac{g}{cm^3}\] And the density of the Taurus complex would be:  \[ \rho = \frac{M_T}{\frac{4}{3} \pi R_T^3}= \frac{6 \times 10^{37}}{\frac{4}{3} \pi (4.5 \times 10^{19})^3}  = 1.6 \times 10^{-22}\frac{g}{cm^3} \]

This yields a difference of 22 orders of magnitude! I can't even imagine what those densities would look like.

Acknowledgements: I worked with Dylan and Barra on this worksheet.