1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.


    3. The attempt at a solution

    [itex]\frac{4\pi F}{m}=\frac{4\pi GM}{R^{2}}[/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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook