# Density of a star cluster

1. Jul 25, 2012

### AbigailM

1. The problem statement, all variables and given/known data
A large, spherically symmetric collection of point particles of mass m move in circular orbits about a common center each with the same kinetic energy. If the only force acting is the mutual gravitational attraction of the particles, find the particle density (in the continuum limit) as a function of r from the center in order that the density remain constant in time.

2. Relevant equations
$F=\frac{GMm}{R^{2}}$ ,where M is the total mass.

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

3. The attempt at a solution
$\frac{F}{m}=\frac{GM}{R^{2}}$

$\frac{4\pi F}{m}=\frac{4\pi GM}{R^{2}}$

$dM=\rho(R)R^{2}dR4\pi$

$dM=\frac{v^{2}}{G}dR$

$\rho(R)=\frac{v^{2}}{4\pi GR^{2}}$

$\frac{1}{2}mv^{2}=\frac{GMm}{r}$ solve for v and substitute into prev. equation.

$\rho(R)=\frac{GM}{2\pi R^{3}}$

Is this looking ok? Thanks for the help

2. Jul 26, 2012

### clamtrox

I'm not sure if I understand the question. I would imagine that if you have a continuous distribution of stuff on spherical orbits, then the density is always constant in time. In order to find a unique density, you need to specify the velocity of each shell, v(R).

For example, suppose you have v(R) = constant. Then you'd find
$$\frac{v^2}{R} = \frac{GM}{R^2} \rightarrow M \propto v^2 R \propto R$$
and therefore dM/dR = constant... Then
$$\frac{dM}{dR} = 4 \pi R^2 \rho \rightarrow \rho \propto R^{-2}.$$
On the other hand, suppose you want your system to rotate like a rigid body, with $v \propto R$. Then
$$M \propto R^3$$
and ρ=constant.

Also having $\rho \propto R^{-3}$ seems really suspicious to me, as this means that M diverges if you integrate to R=0.