Rotation curve

  • #1
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How can I calculate the rotation curve, [tex]v(R)[/tex], for test particles in circular orbits of radius [tex]R[/tex] around a point mass [tex]M[/tex]?
 

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  • #2
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Ok, I guess this is just the velocity function

[tex]v(R)=\sqrt{G\frac{M}{R}}[/tex]

but how about test particles in circular orbits of radius [tex]R[/tex] inside a rotating spherical cloud with uniform density?
 
  • #3
nrqed
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Ok, I guess this is just the velocity function

[tex]v(R)=\sqrt{G\frac{M}{R}}[/tex]

but how about test particles in circular orbits of radius [tex]R[/tex] inside a rotating spherical cloud with uniform density?
Then the mass used must be the mass contained within a sphere of radius "R" where "R" is the distance from the center of the cloud (this can be proven with Gauss' law applied to gravity). Let's say you call [itex] R_0 [/itex] the radius of the cloud. Then consider a sphere of radius [itex] R < R_0 [/itex]. Find the mass contained within that sphere (here's where you will use the fact that the density is uniform) and use that mass in the above equation (of course, this mass will now be a function of R).

Hope this helps

Patrick
 
  • #4
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Yeah thanks, then

[tex]v(R) = \sqrt{\frac{4}{3} \rho G \pi R^2}[/tex].

But what if the test particle is rotating inside a spherical halo with density [tex]\rho(r) \propto 1/r^2[/tex]?
 
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  • #5
nrqed
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Yeah thanks, then

[tex]v(R) = \sqrt{\frac{4}{3} \rho G \pi R^2}[/tex].

But what if the test particle is rotating inside a spherical halo with density [tex]\rho(r) \propto 1/r^2[/tex]?
Then you proceed as before except that the mass contained within a radius R won't simply be [itex] \rho \frac{4}{3} \pi R^3 [/itex]. You will have to do a (simple) integral to find the mass contained within a radius R, namely

[tex] M(R) = 4 \pi \int_0^R dr r^2 \rho(r) [/tex]

Notice that something special happens to v(R) when the density has the radial dependence you gave....Which has some connection with observations fo rotation curves of galaxies and dark matter.

Patrick
 
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  • #6
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I assume that the speed inside the cavity is zero and that the [tex]r[/tex] here is the width of the halo and not the position from the center. Or should it be the other way around?
 
  • #7
Then you proceed as before except that the mass contained within a radius R won't simply be [itex] \rho \frac{4}{3} \pi R^3 [/itex]. You will have to do a (simple) integral to find the mass contained within a radius R, namely

[tex] M(R) = 4 \pi \int_0^R dr r^2 \rho(r) [/tex]

Notice that something special happens to v(R) when the density has the radial dependence you gave....Which has some connection with observations fo rotation curves of galaxies and dark matter.

Patrick

And is the final answer?

v^2 = 4/3 (pi) rho(r) / r
 

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