Density of a star cluster

In summary, the discussion focuses on finding the particle density in a large, spherically symmetric collection of point particles with the same kinetic energy moving in circular orbits. The only force acting is the mutual gravitational attraction of the particles. The equation F=GMm/R^2 is used to find the density as a function of r from the center in order to keep it constant in time. The solution involves finding the velocity of each shell and considering different scenarios, such as constant velocity or rigid body rotation. The final conclusion is that the density is inversely proportional to the square or cube of the radius, depending on the scenario.
  • #1
AbigailM
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Homework Statement


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.

Homework Equations


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

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

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
 
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  • #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.
 

What is the density of a star cluster?

The density of a star cluster refers to the number of stars within a given volume of space. It is typically measured in units of stars per cubic parsec.

What factors can affect the density of a star cluster?

The density of a star cluster can be affected by several factors, such as the total number of stars in the cluster, the size and shape of the cluster, and the gravitational interactions between stars.

How is the density of a star cluster calculated?

The density of a star cluster is calculated by dividing the total number of stars in the cluster by the volume of space that the cluster occupies. This can be done using observations from telescopes or through computer simulations.

Why is the density of a star cluster important to study?

The density of a star cluster can provide valuable information about the formation and evolution of the cluster, as well as the dynamics and interactions between stars. It can also help us understand the overall structure and composition of the cluster.

Can the density of a star cluster change over time?

Yes, the density of a star cluster can change over time due to various factors such as the gravitational pull of nearby objects, interactions between stars, and the loss of stars through stellar evolution. These changes can be observed and studied through ongoing observations and simulations.

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