Worksheet 8, Problem 4- The Virial Theorem: The M80 Star Cluster

Question: The cluster M80 has an angular diameter about 10 arcminutes and resides about $10^4$ parsecs from the Sun. The average speed $<v> \approx 10$ $ km$ $ s^{-1}$. Approximate how much mass, in solar masses $M_{\bigodot}$, the cluster contains. 

Let's draw this scenario to see if we can get a better understanding of this situation:

In this scenario, we know the distance, $a$ from the Sun to M80 to be $10^4$. We also know that M80 has an angular diameter of 10 arcminutes, therefore, the angle $\theta = 10'$.

Since we are trying to solve for the mass of the M80 cluster, we can use the equation we derived in the previous blog post that relates mass of the cluster $M$ to the average velocity of the individual stars $<v>$ and the radius of the star cluster $R$:

\begin{align}
M = \frac{<v>^2R}{G}
\end{align}

Because we need to know the radius of the star cluster, we can refer back to the diagram and see that instead of using the angular diameter, $\theta = 10'$, we can use $\alpha = 5'$ to get the angular radius.

Using the angular radius, we have a trigonometric problem:

Since we are solving for the radius $R$, we can use trigonometry:

\begin{align}
\tan(\alpha) = \frac{R}{a}
\end{align}

Knowing that the angular radius of the star cluster, $\alpha = 5'$ is a very small angle, we can approximate using the small-angle approximation $tan(\alpha) = \alpha$ to solve for the radius of M80 $R$:

\begin{align}
\alpha &= \frac{R}{a}\\
R &= \alpha \times a\\
R &= 5' \times 10^4 parsecs\\
R &= 5' \cdot \frac{2.9 \times 10^{-4} rads}{1'} \times 10^4 parsecs \cdot \frac{3.09 \times 10^{18} cm}{1 parsec}\\
R &= 4.48 \times 10^{19} cm
\end{align}

Okay, so we now have the radius of the M80 cluster. We also know the average velocity of the individual stars in M80 cluster, so we can directly solve for the mass of the star cluster using the equation we derived earlier:

\begin{align}
M &= \frac{<v>^2 R}{G}\\
M &= \frac{(10 \frac{km}{s} \cdot \frac{10^5 cm}{1 km})^2 (4.48 \times 10^{19} cm)}{6.67 \times 10^{-8} \frac{cm^3}{g \cdot s^2}}\\
M &= \frac{(1 \times 10^12 \frac{cm^2}{s^2}) (4.48 \times 10^{19} cm)}{6.67 \times 10^{-8} \frac{cm^3}{g \cdot s^2}}\\
M &= \frac{4.48 \times 10^{31} \frac{cm^3}{s^2}}{6.67 \times 10^{-8} \frac{cm^3}{g \cdot s^2}}\\
M &= 6.72 \times 10^{38} \text{ g}
\end{align}

So the mass of the supercluster comes out to be $6.72 \times 10^{38}$ grams. However, we want this answer in terms of solar mass. Since the solar mass, $M_{\bigodot}$ of the Sun is approximately $M_{\bigodot} \approx 2 \times 10^{30}$.

Therefore,the mass of the cluster M80 is $\frac{M}{M_{\bigodot}} \approx 3.38 \times 10^5$. In other words, $M \approx 3.38 \times 10^5 M_{\bigodot}$.