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Density of a star cluster

  1. Jul 25, 2012 #1
    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
    [itex]F=\frac{GMm}{R^{2}}[/itex] ,where M is the total mass.

    [itex]\frac{v^{2}}{R}=\frac{GM}{R^{2}}[/itex]

    3. The attempt at a solution
    [itex]\frac{F}{m}=\frac{GM}{R^{2}}[/itex]

    [itex]\frac{4\pi F}{m}=\frac{4\pi GM}{R^{2}}[/itex]

    [itex]dM=\rho(R)R^{2}dR4\pi[/itex]

    [itex]dM=\frac{v^{2}}{G}dR[/itex]

    [itex]\rho(R)=\frac{v^{2}}{4\pi GR^{2}}[/itex]

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

    [itex]\rho(R)=\frac{GM}{2\pi R^{3}}[/itex]

    Is this looking ok? Thanks for the help
     
  2. jcsd
  3. Jul 26, 2012 #2
    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
    [tex] \frac{v^2}{R} = \frac{GM}{R^2} \rightarrow M \propto v^2 R \propto R [/tex]
    and therefore dM/dR = constant... Then
    [tex] \frac{dM}{dR} = 4 \pi R^2 \rho \rightarrow \rho \propto R^{-2}. [/tex]
    On the other hand, suppose you want your system to rotate like a rigid body, with [itex] v \propto R [/itex]. Then
    [tex] M \propto R^3 [/tex]
    and ρ=constant.

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