Worksheet 8, Problem 4

4. The cluster M80 has an angular diameter about 10 arcminutes and resides about 104 parsecs from the Sun. The average speed of the stars in the cluster is \(<v>\:\approx\:10\frac{km}{s}\). Approximately how much mass, in solar masses, does the cluster contain?

This is a cumulative problem that allows us to incorporate the Virial Theorem and the equation for potential energy that we derived in part 2. It is asking us to compute the mass of the Messier 80 globular cluster in terms of solar masses. The M80 Cluster is located in the Scorpius Constellation, some 28,000 light years from Earth. It is named after Charles Messier, who discovered the cluster in 1781. It is one of the densest clusters of stars and is notable for having  the largest known density of "blue stragglers", unusually young and massive stars formed by the collision and merger of stars in the cluster's core.

Messier 80 Cluster (M80)

Before we begin our calculations, I would like to share some unit conversions in order to turn the given quantities into cgs units. I will hold off on converting until near the end of the calculations, but it will be useful to know up front the relative size of these new, at least in this class, units.

Now that we have our units clarified, we will follow the guidelines of the Expert Method and set up a diagram. We can draw a right triangle using the known angular diameter of M80 from a point on the Earth as well as its distance from the Earth. Note that the hypotenuse is the same as the lower side. This is clearly not exactly correct but the two distances are approximately the same, especially when considered over the massive distance (10,000 parsecs) between the Earth and M80. 

Looking at this diagram, it is clear that we want to solve for the only variable, R. We can do this using the small angle approximation. At small angles, such as \(\frac{1}{360}\) radians, \(sin\theta\:\approx\:\theta\). Using this, we can set up an easily solvable equation to solve for R. 

(Note: In the following equation, parsecs are abbreviated using "pc")

Now that we know the radius, we can start using the Virial Theorem in order to solve for the mass of the cluster.

 Here K is the kinetic energy of the system and U is the potential energy of the system. We can subsitute \(\frac{1}{2}Mv^{2}\) for K and our previously solved for \(\frac{-GM^{2}}{R}\) for U. Then we can rearrange to solve for M.

Now that we've isolated M, we just have to get every aspect of the equation into terms of known quantities. Thus, the only thing we have to change is v. The average velocity of the system, 10 \(\frac{km}{s}\), can be used here. We can then start plugging in values that we have already solved for or have been given.

We've now found the mass of the Messier 80 cluster in grams. First, let's take a moment and appreciate the fact that we were able calculate the mass of a massive cluster of hundreds of thousands of stars, thousands of light years away using just a few simple given values. However, the number that we've found is so large, it's incredibly hard for us to get a firm conceptualization of it. That number is nearly 1 duodecillion grams. That's the same mass as about 1 duodecillion paper clips. Unfortunately that doesn't help us very much because I can't even think of how big a duodecillion is let alone how much mass (and space!) a duodecillion paper clips would have (and take up). So, let's put that number in terms of the mass of our Sun, which itself has the mass of quite a bit of paper clips.

So, our final answer is that the Messier 80 cluster has a mass that is approximately 400,000 times the mass of our Sun. Though that's still hard to conceptualize, we get the idea that it's very massive and derives that mass from a large number of stars, hence the term "cluster".