WS 7, Problem 2-4: Application of the Virial Theorem

2. Consider a spherical distribution of particles, each with a mass m$_i$ and a total (collective) mass $\sum_{i}^{N}m_i=M$, and a total (collective) radius R.  Convince yourself that the total potential energy, U, is approximately 

$U=-\frac{GM^2}{R}$

We decided to solve this problem by finding the sum of all of the potential energies of all of the individual small masses, $\sum_{i}^{N}U_i$.  

Potential energy is usually given by 

$U=-\frac{GMm}{R}$

where M is the mass of the object being orbited and m is the mass of the orbiting object.  

To find the sum of all of the potential energies of each individual particle, we used the following equation:

$\sum_{i}^{N}U_i=-\frac{GM\sum_{i}^{N}m_i}{R}$

$=-\frac{GMM}{R}$

$=-\frac{GM^2}{R}$

There was some doubt in our group that we didn't have to sum over the different Rs.  We resolved the issue by assuming that the particles are uniformly distributed, so the particles at larger radii, because there would be more of them, would overpower the particles at smaller radii in the summation.


We were, however, wrong.  We really did have to take into account the different Rs, and this is how we did it:

At the beginning of the next class, we were told that we should assume the particles form a spherical distribution with uniform density and that the density can be written as

$\rho (r)=\frac{M(r)}{\frac{4}{3}\pi r^3}=\rho$           Eq. 1

Using this new information, we approached the problem as an integral with respect to the radius of the sphere.  We start with a radius barely more than 0.  At that radius, the potential energy, U, will depend only on the mass enclosed by that radius, r. Now, say we increase the radius by a tiny little bit.  We can find the mass of that new "shell" on the sphere, dM, by multiplying the surface area of the shell by its thickness, $\Delta r$.  

$dM=4\pi r^2 \rho \Delta r$

This approach leads us to the following integral:

$\int_{0}^{R}U=\int_{0}^{R}-\frac{GM}{r}dM$

At first, I was confused by the fact that $m_i$ wasn't included in the equation for potential energy of the shell.  But now I know it's not included because the point of the integral is to add up the potential energies of each shell, and the individual shells' energies only depend on mass they enclose.  

From there, the rest of the problem was pretty much algebra, with a little bit of calculus thrown in.  

First, rearrange Eq. 1 to solve for M and substitute that for M in the integral.

$-\int_{0}^{R}\frac{G\frac{4}{3}\rho \pi r^3}{r}4\pi r^2\rho \Delta r$

$-\int_{0}^{R}\frac{16}{3}G\rho^2\pi^2r^4\Delta r$

$=-\frac{16}{15}G\rho^2\pi^2r^5$                Eq. 2

This is the really complicated equation for the total potential energy of the particles.  But we know it can be simplified by dividing it by the equation provided in the problem (substituting the rearranged version of Eq. 1 for M).

$\frac{-\frac{16}{15}G\rho^2\pi^2r^5}{-\frac{GM^2}{r}}$

$=\frac{-\frac{16}{15}G\rho^2\pi^2r^5}{-\frac{G\frac{16}{9}\rho^2\pi^2r^6}{r}}$

$=\frac{3}{5}$  

This final fraction tells us that what we got when we found the integral is actually $\frac{3}{5}$ times the equation provided.  So the actual equation for the total potential energy of this distribution of particles is  $U=-\frac{3}{5}\frac{GM^2}{r}$

3.  If the average speed of a star in a cluster of thousands of stars is $\left \langle v \right \rangle$, give an expression for the total mass of the cluster.  

We started with the Virial Equation 

$K = -\frac{1}{2}U$

$\frac{1}{2}mv^2=\frac{1}{2}\frac{GM_* m}{R}$

To find the mass of the cluster, we rearranged the above equation to solve for M$_*$.  This is the result:

$M_*=\frac{Rv^2}{G}$

4.  The cluster M80 has an angular diameter about 10 arcminutes and resides about 10$^4$ parsecs from the Sun.  The average speed of the stars in the cluster is  $\left \langle v \right \rangle$=10 km/s.  Approximately how much mass, in solar masses, does the cluster contain?

This step, though it wasn't conceptually difficult, had many steps.  The first step was to convert the angular diameter provided into radians using the following conversion

$10'\left ( \frac{1^o}{60'} \right )\left ( \frac{\pi rad}{180^o} \right )$

The second step was to use the angular size and the distance to the cluster to find the clusters actual size.

where $\theta$ is the angular size (found using the above conversion) divided by 2.  

We can use geometry and small angle approximation to find R.

$\tan \theta = \theta =\frac{R}{d}$

$R=d\theta$

We can now use this R in the equation we found in part 3.  The only problem is that the $\left \langle v \right \rangle$ we were given is in km/s, not parsec/s, so we have to first convert the radius into kilometers using the fact that 1 parsec = 3x10$^13$km.  

Once that's done, we can find the mass of the cluster in grams by using the equation found in part 3.  Then, because we're asked to find the mass of the cluster in solar masses, we use the fact that 1 solar mass = 2x10$^33$ grams to find the mass of the cluster.  

We found that the mass of the cluster was $3\times 10^{5}M_\odot$.  

After googling the actual mass of M80, we found that we were well within an order of magnitude, and actually off by less than half the actual value.  Any error in our result was most likely due to rounding during our calculations.