Worksheet 9, Problem 1- Star Formation [I'm All About that Space]

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Question: 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 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 star be? Can you come up with an analogy that would help the layperson understand this difference in scale? For example, if the cloud is the size of a human, then the star is the size of what?

Okay, so if we're trying to get an understanding of the scale of the molecular cloud in comparison to the central star, let's create an analogy to the human body. Since the height of the average human male is $175$ cm, the diameter of a hypothetical spherical human would also be $D_{human} = 175$ cm.

Let's compare the proportion of the human body to some unknown part of the human body, $D_{x}$, such that the ratio is the same as the ratio between the diameter of the central star in a molecular cloud, and the Taurus region of the cloud. The proportionality can be set up as follows:

\begin{align}
\frac{D_{human}}{D_{x}} &= \frac{D_{taurus}}{D_{star}}\\
\end{align}

We can rearrange the previous equation to solve for $D_x$. We also know the diameter of a hypothetical human body to be about $175$ cm. The problem tells us that the average star in the middle of a molecular cloud is about the same size as the Sun, which has a diameter of $1.39 \times 10^{11}$ cm. We are also given the size of the molecular cloud, which is $30$ pc. Knowing all this information, we can plug in the values directly and solve for the proportional diameter of an unknown part of the human body, $D_x$:

\begin{align}
D_x &= \frac{D_{human} \cdot D_{star}}{D_{taurus}}\\
D_x &= \frac{(175 \text{ cm}) \cdot (1.39 \times 10^{11} \text{ cm})}{30 \text{ pc} \times \frac{3.09 \times 10^{18} \text{ cm}}{1 \text{ pc}}}\\
D_x &= 2.62 \times 10^{-7} \text{ cm}
\end{align}

As you can see, if an average human, which represents the star forming molecular cloud, has a diameter of $175$ cm, then the star that in the molecular cloud will be of the size $2.62 \times 10^{-7}$ cm, or $2.62 \times 10^{-9}$ meters, which is about as big as a strand of DNA.

Other things that exist at this scale are:

Scale of the Universe 2: http://htwins.net/scale2/

(b) Within the Taurus complex there is roughly $3 \times 10^4 M_{\bigodot}$ 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 order 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!

Okay, so in order to compare the densities of the Taurus complex and a Sun-like star (the Sun has a mass of $1.99 \times 10^{33}$ g), let's try to find the density of the Taurus complex first. We know that the density $\rho$ is mass per unit volume, which can be represented by $\rho = \frac{mass}{volume}$. We can approximate the Taurus-complex as a spherical body with a radius $15$ pc to get it's volume.

\begin{align}
\rho_{taurus} = \frac{M_{taurus}}{V_{taurus}} = \frac{M_{taurus}}{\frac{4}{3} \pi r_{taurus}^3} = \frac{3 \times 10^4 M_{\bigodot}}{\frac{4}{3} \pi (15 \text{ pc} \cdot \frac{3.09 \times 10^{18} \text{ cm}}{1 \text{ pc}})^3} = \frac{3 \times 10^4 ( 1.99 \times 10^{33} \text{ g})}{\frac{4}{3} \pi (15 \text{ pc} \cdot \frac{3.09 \times 10^{18} \text{ cm}}{1 \text{ pc}})^3} = 1.4 \times 10^{-22} \frac{g}{cm^3}
\end{align}

Now that we have the density of the Taurus-complex, let's solve for the mass of a typical star, similar to the Sun. The mass of the Sun is $1.99 \times 10^{33}$ g, and the radius of the Sun is $6.97 \times 10^{10}$ cm. We can solve for the density of the Sun using the formula $\rho = \frac{mass}{volume}$ and plug the values for mass and radius directly:

\begin{align}
\rho_{\star} = \frac{M_{\star}}{V_{\star}} = \frac{M_{\star}}{\frac{4}{3} \pi r_{\star}^3} = \frac{1.99 \times 10^{33} \text{ g}}{\frac{4}{3} \pi (6.97 \times 10^{10} \text{ cm})^3} = 1.4 \frac{g}{cm^3}
\end{align}

As you can see, the density of a star is 22 orders of magnitude larger than that of a Taurus-complex. To put that in perspective, think air and lead differ in densities by an order of 4 magnitudes!