<p>What you do is predict the orbital velocity for a more realistic hypothesised mass distribution. There is an easy analytic solution for a disc. More complicated distributions require a numerical solution.</p>

<p>If you get far enough <em>outside</em> an arbitrary mass distribution, then that solution will asymptotically look similar to $v_{\rm circ} = \sqrt{GM/r}$.</p>

<p>Astronomers get into trouble when explaining the evidence for dark matter from galaxy rotation curves to the layperson. They often use the analogy of the solar system, where the orbital speed goes as $r^{-1/2}$. The trouble is that that a Galaxy is not like the solar system because the mass is <em>distributed</em> and in general cannot be assumed to be concentrated at a point and Newton's shell theorem does not apply.</p>

<p>In practice what you can do is posit a more realistic distribution of matter based on star counts and the gas that can be observed. You can then predict (using a computer model) what the rotation velocity should be at any point if that is all the mass that exists. This model will converge to the simple $r^{-1/2}$ law once you get well outside the bulk of the mass in the model. The model can be compared to the observed velocities of gas, globular clusters, planetary nebulae and other velocity tracers - some of which do indeed exist well outside <em>the bulk</em> of the visible galaxy matter. What you find is that they move much faster than predicted, indicating that considerably more unseen matter must exist inside their galactic orbits. You can then add that (dark) matter in various ways. Non-spherically symmetric and spherically symmetric dark matter profiles (of the same total mass) would produce different orbital velocities at different radii for precisely the reason you identify. By looking at tracers at different radii you can map out the distribution of unseen mass. Once happy with your model you can integrate the density of seen and unseen matter to estimate the total mass.</p>

<p>The bottom line is that the $\sqrt{GM/r}$ is just a "back of the envelope" demonstration of the effect. Real research uses more accurate models or works in a regime far outside the bulk of the visible matter where the "back of the envelope" approximation works.</p>

CC BY-SA 3.0
2017-10-19T09:55:47.313Z
2017-10-19T09:55:47.313Z
ProfRob
<p>The measured rotational velocities can <strong>reliably but not very accurately</strong> constrain the enclosed mass.</p>

<p>First, the deviation between $GM(r)/R$ and $R\partial\Phi/\partial R$ (note that this is a partial derivative at $z=0$) is typically small. Moreover, more careful modelling (decomposing the galaxy into disc and spheroidal components) usually reduces the resulting errors.</p>

<p>Second, even when doing so, the relation you're referring to only holds for objects on exactly circular orbits in an exactly axially symmetric gravitational potential. However, in reality even the gas does not follow exactly circular orbits, nor are galaxies exactly axially symmetric. Common deviations are spirals and bars.</p>

<p>If one includes all these effects into the modelling, the conclusions are not changed significantly.</p>

<p>When referring to the Milky Way, the rotation velocity of a circular orbit at the Solar position (known as the "Local Standard of Rest" LSR) has been measured relatively accurately to be $V_0=238\,$km$\,$s$^{-1}$ and the distance to the Galactic centre to be $R_0=8.27\,$kpc. From these numbers, the simple mass estimator gives
$$
M(&lt;R_0) \approx R_0 V_0^2/G \approx 10^{11} \textsf{M}_\odot
$$
for the mass interior to the Solar orbit. This is larger than what one sees in terms of stars and gas, but only by about a factor $\sim2$. The measurement of rotation velocities in the outer Milky Way is notoriously difficult and the evidence for dark matter in our host galaxy is therefore based on other arguments (such as the velocities and orbits of satellite galaxies).</p>

<p>For external galaxies, however, one can peruse the gas velocities measured at large distances, when the observed flatness of the rotation curve (constant $V_0$ in above formulat) demands substantial amounts of matter, roughly $M\propto R$.</p>

CC BY-SA 3.0
2017-10-20T08:39:02.510Z
2017-10-20T08:39:02.510Z
Walter
<p>If a matter distribution is spherically symmetric, the circular velocity is straightforwardly related to the mass via:</p>

<p>$v_{circ} = \sqrt{GM/r}$</p>

<p>In an axisymmetric potential, on the other hand hand, we have</p>

<p>$v_{circ}(R) = \sqrt{R\frac{d\Phi}{dR}}\neq \sqrt{GM/r}$</p>

<p>with $\Phi$ the gravitational potential.</p>

<p>What I don't quite understand is the following: the potential of the Milky Way is not spherical. How is the measured velocity then reliably related to a mass?</p>

CC BY-SA 3.0
2017-10-20T06:23:37.283Z
See also https://astronomy.stackexchange.com/questions/8172/what-are-the-arguments-against-the-feng-and-gallo-thin-disk-explanation-of-galac
ProfRob
<rotation><dark-matter><mass><velocity>
Circular velocity in axisymmetrici potential
user1991
# Circular velocity in axisymmetrici potential

(Asked by: username_0 on 2017-10-19)

If a matter distribution is spherically symmetric, the circular velocity is straightforwardly related to the mass via:

$v\_{circ} = \sqrt{GM/r}$

In an axisymmetric potential, on the other hand hand, we have

$v\_{circ}(R) = \sqrt{R\frac{d\Phi}{dR}}\neq \sqrt{GM/r}$

with $\Phi$ the gravitational potential.

What I don't quite understand is the following: the potential of the Milky Way is not spherical. How is the measured velocity then reliably related to a mass?

**Tags:** rotation, dark-matter, mass, velocity

> **Comments:**
> **username_1**: See also https://astronomy.stackexchange.com/questions/8172/what-are-the-arguments-against-the-feng-and-gallo-thin-disk-explanation-of-galac  

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## 2 answers

### Answer by username_1 on 2017-10-19. Score: 3

What you do is predict the orbital velocity for a more realistic hypothesised mass distribution. There is an easy analytic solution for a disc. More complicated distributions require a numerical solution.

If you get far enough *outside* an arbitrary mass distribution, then that solution will asymptotically look similar to $v\_{\rm circ} = \sqrt{GM/r}$.

Astronomers get into trouble when explaining the evidence for dark matter from galaxy rotation curves to the layperson. They often use the analogy of the solar system, where the orbital speed goes as $r^{-1/2}$. The trouble is that that a Galaxy is not like the solar system because the mass is *distributed* and in general cannot be assumed to be concentrated at a point and Newton's shell theorem does not apply.

In practice what you can do is posit a more realistic distribution of matter based on star counts and the gas that can be observed. You can then predict (using a computer model) what the rotation velocity should be at any point if that is all the mass that exists. This model will converge to the simple $r^{-1/2}$ law once you get well outside the bulk of the mass in the model. The model can be compared to the observed velocities of gas, globular clusters, planetary nebulae and other velocity tracers - some of which do indeed exist well outside *the bulk* of the visible galaxy matter. What you find is that they move much faster than predicted, indicating that considerably more unseen matter must exist inside their galactic orbits. You can then add that (dark) matter in various ways. Non-spherically symmetric and spherically symmetric dark matter profiles (of the same total mass) would produce different orbital velocities at different radii for precisely the reason you identify. By looking at tracers at different radii you can map out the distribution of unseen mass. Once happy with your model you can integrate the density of seen and unseen matter to estimate the total mass.

The bottom line is that the $\sqrt{GM/r}$ is just a "back of the envelope" demonstration of the effect. Real research uses more accurate models or works in a regime far outside the bulk of the visible matter where the "back of the envelope" approximation works.


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### Answer by username_2 on 2017-10-20. Score: 1

The measured rotational velocities can **reliably but not very accurately** constrain the enclosed mass.

First, the deviation between $GM(r)/R$ and $R\partial\Phi/\partial R$ (note that this is a partial derivative at $z=0$) is typically small. Moreover, more careful modelling (decomposing the galaxy into disc and spheroidal components) usually reduces the resulting errors.

Second, even when doing so, the relation you're referring to only holds for objects on exactly circular orbits in an exactly axially symmetric gravitational potential. However, in reality even the gas does not follow exactly circular orbits, nor are galaxies exactly axially symmetric. Common deviations are spirals and bars.

If one includes all these effects into the modelling, the conclusions are not changed significantly.

When referring to the Milky Way, the rotation velocity of a circular orbit at the Solar position (known as the "Local Standard of Rest" LSR) has been measured relatively accurately to be $V\_0=238\,$km$\,$s$^{-1}$ and the distance to the Galactic centre to be $R\_0=8.27\,$kpc. From these numbers, the simple mass estimator gives
$$
M(<R\_0) \approx R\_0 V\_0^2/G \approx 10^{11} \textsf{M}\_\odot
$$
for the mass interior to the Solar orbit. This is larger than what one sees in terms of stars and gas, but only by about a factor $\sim2$. The measurement of rotation velocities in the outer Milky Way is notoriously difficult and the evidence for dark matter in our host galaxy is therefore based on other arguments (such as the velocities and orbits of satellite galaxies).

For external galaxies, however, one can peruse the gas velocities measured at large distances, when the observed flatness of the rotation curve (constant $V\_0$ in above formulat) demands substantial amounts of matter, roughly $M\propto R$.


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