3. One of the most useful equations in astronomy is an extremely simple relationship known as the Virial Theorem. It can be used to derive Kepler’s Third Law, measure the mass of a cluster of stars, or the temperature and brightness of a newly-formed planet. The Virial Theorem applies to a system of particles in equilibrium that are bound by a force that is defined by an inverse central-force law (F=k/rα) (where k is a constant). It relates the kinetic (or thermal) energy of a system, K, to the potential energy, U, giving
\[K=-\frac{1}{2}U\]
(a) Consider a spherical distribution of N particles, each with a mass m. The distribution has total mass M and total radius R. Convince yourself that the total potential energy, U, is approximately
\[U\approx -\frac{GM^2}{R}\]
You can derive or look up the actual numerical constant out front. But in general in astronomy, you don’t need this prefactor, which is of order unity.
(b) Now let’s figure out what K is equal to. Consider a bound spherical distribution of N particles (perhaps stars in a globular cluster), each of mass m, and each moving with a velocity of vi with respect to the center of mass. If these stars are far away in space, their individual velocity vectors are very difficult to measure directly. Generally, it is much easier to measure the scatter around the mean velocity if the system along our line of sight, the velocity scatter σ2. Show that the kinetic energy of the system is:
\[K=N\frac{3}{2}m\sigma^2\]
(c) Use the Virial Theorem to show that the total mass of, say, a globular cluster of radius R and stellar velocity dispersion σ is (to some prefactor of order unity):
\[M\approx \frac{\sigma^2 R}{G}\]
(a) Consider a spherical distribution of N particles, each with a mass m. The distribution has total mass M and total radius R. Convince yourself that the total potential energy, U, is approximately
\[U\approx -\frac{GM^2}{R}\]
You can derive or look up the actual numerical constant out front. But in general in astronomy, you don’t need this prefactor, which is of order unity.
\[U\approx -\frac{GM^2}{R}\]
You can derive or look up the actual numerical constant out front. But in general in astronomy, you don’t need this prefactor, which is of order unity.
For this problem let's imagine a spherical distribution of particles. For one particle, the potential energy comes from the particle's distance for the center of the distribution and the force from all other particles. We could use calculus to solve this with integration over the distribution, but since we are ignoring numerical constants, we can think of it more intuitively and get the same result. (For the full derivation, please see: http://ay17-ataylor.blogspot.com/2015/03/worksheet-8-problem-2-total-potential.html ). Let's instead consider a single particle in the distribution.
Generally, the potential energy of a particle of mass m related to a distribution with mass M is:
\[U_p=-\frac{GMm}{R}\]
However, we have N particles, each will mass M. So the total potential energy is:
\[U=NU_p=-\frac{GMNm}{R}\]
However, we remember that M=Nm, thus:
\[U=-\frac{GMNm}{R}=\boxed{-\frac{GM^2}{R}}\]
(b) Now let’s figure out what K is equal to. Consider a bound spherical distribution of N particles (perhaps stars in a globular cluster), each of mass m, and each moving with a velocity of vi with respect to the center of mass. If these stars are far away in space, their individual velocity vectors are very difficult to measure directly. Generally, it is much easier to measure the scatter around the mean velocity if the system along our line of sight, the velocity scatter σ2. Show that the kinetic energy of the system is:
\[K=N\frac{3}{2}m\sigma^2\]
Let's use the same approach as before.
Commonly, for a single particle:
\[K_p=\frac{1}{2}mv^2\]
Let's modify this. This assumes motion in a single direction, for motion in three directions/dimensions, we should multiply by 3, and replace velocity (1 directional) with scatter (3 directional).
\[K_p=\frac{3}{2}m\sigma^2\]
For the total energy of the system we have N particles, so:
\[K=K_p N=\boxed{N\frac{3}{2}m\sigma^2}\]
Generally, the potential energy of a particle of mass m related to a distribution with mass M is:
\[U_p=-\frac{GMm}{R}\]
However, we have N particles, each will mass M. So the total potential energy is:
\[U=NU_p=-\frac{GMNm}{R}\]
However, we remember that M=Nm, thus:
\[U=-\frac{GMNm}{R}=\boxed{-\frac{GM^2}{R}}\]
(b) Now let’s figure out what K is equal to. Consider a bound spherical distribution of N particles (perhaps stars in a globular cluster), each of mass m, and each moving with a velocity of vi with respect to the center of mass. If these stars are far away in space, their individual velocity vectors are very difficult to measure directly. Generally, it is much easier to measure the scatter around the mean velocity if the system along our line of sight, the velocity scatter σ2. Show that the kinetic energy of the system is:
\[K=N\frac{3}{2}m\sigma^2\]
Let's use the same approach as before.
Commonly, for a single particle:
\[K_p=\frac{1}{2}mv^2\]
Let's modify this. This assumes motion in a single direction, for motion in three directions/dimensions, we should multiply by 3, and replace velocity (1 directional) with scatter (3 directional).
\[K_p=\frac{3}{2}m\sigma^2\]
For the total energy of the system we have N particles, so:
\[K=K_p N=\boxed{N\frac{3}{2}m\sigma^2}\]
c) Use the Virial Theorem to show that the total mass of, say, a globular cluster of radius R and stellar velocity dispersion σ is (to some prefactor of order unity):
\[M\approx \frac{\sigma^2 R}{G}\]
Let's put this all together.
\[K=-\frac{1}{2}U\]
\[N\frac{3}{2}m\sigma^2=\frac{GM^2}{2R}\]
Let's rearrange and solve for M. We can drop all numerical constants.
\[Nm\sigma^2~\frac{GM^2}{R}\]
Earlier we mentioned that Nm=M, so:
\[M\sigma^2~\frac{GM^2}{R}\]
\[\sigma^2R~GM\]
\[\boxed{M~\frac{\sigma^2 R}{G}}\]
\[M\approx \frac{\sigma^2 R}{G}\]
Let's put this all together.
\[K=-\frac{1}{2}U\]
\[N\frac{3}{2}m\sigma^2=\frac{GM^2}{2R}\]
Let's rearrange and solve for M. We can drop all numerical constants.
\[Nm\sigma^2~\frac{GM^2}{R}\]
Earlier we mentioned that Nm=M, so:
\[M\sigma^2~\frac{GM^2}{R}\]
\[\sigma^2R~GM\]
\[\boxed{M~\frac{\sigma^2 R}{G}}\]
I worked on this problem with B. Brzycki, G. Grell, and N. James.
Very nice
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