rj rj r. • r rj • r approaches zero. This leaves

where the { ) represents the time average of the enclosed quantity. Equation (13.28) is the simplest form of the virial theorem.

The virial theorem applies to any gravitation-ally bound system that has had sufficient time to come to equilibrium. Even simple systems, like binary stars, obey the virial theorem (see Problem 13.11). If the orbits are circular then K = -U/2 at all points. For elliptical orbits, r and v are changing, so K and U are changing, while their sum E is fixed. This means that we have to average over a whole orbit to get { K) = -{U) /2.

Remember, for any system, the total energy is

(0.6)(6.67 X 10-8 dyn cm 2/g2)[(0.5)(2 X 1033 g)(106)]2 (5 pc)(3.18 X 1018cm/pc)

If the virial theorem applies, then the total energy is E = -U/2 = 1.2 X 1051 erg.

We now look at the kinetic energy. In a cluster of stars, the kinetic energy is in the random motions of the stars. If the cluster has N stars, each of mass m, the kinetic energy is

The total mass of the cluster is M = mN; so

0 0

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