(a) Let's think of some familiar system where there exists a long-term
equilibrium between gravitational potential energy and kinetic energy
in that system.  The solar system comes to mind.  For each object
orbiting the Sun, we know that its orbital velocity is v = sqrt(GM/r)
and its kinetic energy is E(k) = 0.5*mv^2.  Combining these equations,

  E(k) = 0.5 * m * (sqrt(GM/r)^2 = 0.5 * m * GM/r = 0.5 * GMm/r.

Now recall that the gravitational potential energy of a system is

  E(p) = - GMm/r.  

So the relationship between the kinetic and potential energies of this
system in long-term equilibrium is

  E(p) = -2 * E(k).

This relationship is the Virial Theorem.  


(b) Let's think about what it means to be in long-term equilibrium.
It makes sense that the virial theorem will be true if in a system
if it's not gaining or losing mass, if most of its energy is not being
converted into heat or lost to entropy, and if it's generally stable 
over long periods of time.  Such a system is known as a virialized system.
So, we reason:

 - our solar system is virialized.
 - a globular cluster of one million stars is virialized, as long as
   its stars remain in the cluster and do not escape.
 - our Milky Way is virialized.
 - our observable universe is probably not virialized, because it
   keeps growing in size and keeps gaining matter and energy as our 
   cosmic horizon expands.
   (BUT: the ENTIRE universe, both inside and outside our cosmic
   horizon, MAY be virialized - it keeps expanding, but it does
   not gain matter or energy, and it may stop expanding and collapse
   again, thus showing a sort of very-long-term equilibrium.)
 - an inflatable dummy orbiting a black hole is virialized.
 - an inflatable dummy falling into a black hole is not virialized,
   because it is a dynamically changing system and is continuously
   converting gravitational potential energy into kinetic energy, 
   heat, and other forms of energy.


(c) If Abell 2715 is a virialized system, then its entire E(p) will
be twice its entire E(k).  If it has 200 galaxies, each 2 * 10^42 kg,
then the mass of the entire cluster is 200 * 2 * 10^42 = 4 * 10^44 kg
So on average, each galaxy will have potential and kinetic energies

           GMm   (6.67 * 10^-11) * (4 * 10^44) * (2 * 10^42)
avg E(p) = --- = ------------------------------------------- = 1.7 * 10^52 J
            r        1 Mpc * (3.1 * 10^22 m/Kpc)

           mv^2   (2 * 10^42)
avg E(k) = ---- = ----------- * v^2  where v is the average velocity of the
            2          2             galaxies in this cluster.

For the entire cluster, we can use the Virial Theorem to solve for v.

  [200 * avg E(p)] = 2 * [200 * avg E(k)]

  200 * 1.7 * 10^52 = 2 * 200 * (10^42 * v^2)

  v = sqrt(0.85 * 10^10) = 9.2 * 10^5 m/s = 920 km/sec.

The average galaxy in this cluster would go from NY to LA in five seconds!