To ask a more specific one for the rotation curves of elliptical galaxies, and hope from there to later understand the dynamics of spiral galaxies.

  1. Treating the galaxy as an isothermal gravitational gas sphere, what is the equation of Density for an elliptical galaxy?

  2. Assuming the above density and modelling by the Virial Equation, what are the Velocities?

Note: I’ve drawn on a paper by Fritz Zwicky 1937 “On The Masses of Nebulae and of Clusters of Nebulae”

  • $\begingroup$ That's not angular momentum, it's centripetal acceleration times mass. Angular momentum would be mvR. $\endgroup$ – Andrew Sep 8 '11 at 19:02

The Keplerian formula assumes that only a central (point) mass exerts gravity on the orbiting mass. This is an excellent approximation for each individual planet in our Solar System, for instance, but it is totally wrong when applied to star orbits in a galaxy.

In that latter case, you consider basically all the material inside a given object's orbit. Therefore, the effective mass contributing to the orbits of test masses at various radii increases with radius, so the formula

$$ v = \sqrt{\frac{G M}{R}}$$


$$ v = \sqrt{\frac{G M_{(R)}}{R}}$$

where $M_{(R)}$ is the mass enclosed as a function of radius.

For a spiral galaxy approximated as a cylinder, $$ M_{(R)} = \int_0^R \rho_{(r)} 2 \pi r h * dr $$ $\rho$ is the density of stars, $h$ is the height of the cylinder (disk thickness), and $r$ is radius. $\rho$ is roughly proportional to $1/r$, leaving $M$ roughly proportional to $R$, leaving $v$ approximately independent of $R$.

PS: Curve A in the link is what you get for calculating rho according to visible matter only, Curve B for visible + dark matter. Or if you replace traditional gravitational calculations with MOND, of which I am personally very fond, but which is waaaaaaay beyond the scope of this answer.

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    $\begingroup$ You typeset limits of integration like \int_0^R. I've done it for you, but with my low rep it must await approval. $\endgroup$ – dmckee --- ex-moderator kitten Sep 8 '11 at 2:49
  • $\begingroup$ Does this assume all orbits are circular? $\endgroup$ – Peter Mortensen Sep 8 '11 at 18:46
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    $\begingroup$ The test mass' orbit is assumed to be circular in both the question and my answer. Masses contributing to M(r) need not be in circular orbits, so long as they don't cross the test mass' orbit much. $\endgroup$ – Andrew Sep 8 '11 at 19:04
  • $\begingroup$ So the inner part of the rotation curve before the hump is affected by viscosity (after Zwicky) which slows the orbital velocity, whereas the outer part (outside the galactic bulge) has little viscosity so the stars move freer. And then to cover the flatter part of observed velocity, Dark Matter is added to the virial equation. So that means inner part is non-Keplerian because of stellar viscosity - is this correct? $\endgroup$ – metzgeer Sep 9 '11 at 3:38
  • $\begingroup$ I'm not familiar enough with galactic core dynamics specifically to say for sure. However, since the bulge of a spiral galaxy has a lot in common with a tiny elliptical, it would make sense for the dynamics to have a different character. $\endgroup$ – Andrew Sep 9 '11 at 11:11

It looks pretty similar to me, here is the wolfram alpha plot of $$v \propto 1/\sqrt{R}$$ wolf alpha graph

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    $\begingroup$ No, it is not R^(-1/2). $\endgroup$ – Andrew Sep 8 '11 at 1:06
  • $\begingroup$ Yes it is, the predicted curve is close to just that... $\endgroup$ – Nic Sep 8 '11 at 11:33
  • $\begingroup$ You can't really eyeball inverse power law curves, even as well as positive power laws. Of course it's going to be something in the neighborhood of -1/2, because the mass density decreases outwards, but the actual predicted curve will have significant deviations from pure inverse square root. $\endgroup$ – Andrew Sep 8 '11 at 14:45

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