## Info

NJi a

So for a system to which the virial theorem, in the form of equation (13.28) applies, we set {K) = - {U)/2, to give

If we take the sum of N quantities and then divide by N, the result is the average of that quantity. Therefore (1/N) 2 v\ is the average of the quantity v2. We write this average as {v2). Remembering that mN = M, equation (13.31) becomes

(We don't have to take the time average of E, since E is constant.) Remember, the gravitational potential energy, defined so that it is zero when the particles are infinitely far apart, is negative. Therefore, the total energy of a bound system is negative. This means that we have to put energy in to break up the system.

### 13.3.2 Energies

We now look at the kinetic and potential energies of a cluster. In Section 9.1, we saw that the gravitational potential energy for a constant density sphere of mass M and radius R is

Example 13.1 Potential energy for a globular cluster

Find the gravitational potential energy for a spherical cluster of stars with 106 stars each of 0.5 M0. The radius of the cluster core is 5 pc.

solution

We use the above equation to give

If we put this and the potential energy into the virial theorem, we find

The quantity {v2) is the mean (average) of the square of the velocity. If we take the square root of this quantity, we have the root mean square velocity or rms velocity. It is a measure of the internal motions in the cluster.

Example 13.2 The rms velocity in a cluster Find the rms velocity for the cluster used in the previous example.

solution

We use equation (13.34) with the given quantities:

(0.6)(6.67 X 10-8 dyn cm2/g2)(0.5)(106)(2 X 1033 g)

(5 pc)(3.18 X 1018 cm/pc) = 2.5 X 1012 (cm/s)2 Taking the square root gives vrms = 1.6 X 106 cm/s