I am supposed to show that the orbital speed of a star is proportional with $\sqrt{r}$, where $r$ is the distance from the center of a galaxy. Suppose the galaxy's mass is equally distributed like in a disc.

I am confused since I thought orbital velocity would be defined as:

$$v=\sqrt{ \frac{G \cdot M}{r}}$$

In this case, the velocity is inversely proportional to the square of the distance. Is there something I am missing here?

Edit: I also hope that the answer does not lie simply in this algebraic manipulation:

$$v=\frac{\sqrt{r} \cdot \sqrt{GM}}{r}$$

I. The ratio between the area of the area that the star orbiting a galaxy center covers and its respective mass. Compared with the area of the disk-like the galaxy itself and its mass $\frac{\pi r_s^2}{x}=\frac {\pi r_g^2}{m_g}$

Plugging in the formula for mass of the body in orbit $m= \frac{v^2 r}{G}$, which was derived from $G\frac{mM}{r^2}=m \frac{v^2}{r}$ We get $\frac{r_g}{v_g^2}=\frac{r_s}{v_s^2}$, and isolating for the orbital speed of the star...

$v_s=v_s \cdot \sqrt{\frac{r_s}{r_g}}$

This was my desperate attempt, and I cannot figure this out it seems


2 Answers 2


Yes, you are missing something. The formula for the orbital velocity $$v=\sqrt{\frac{G M}{r}}$$ is valid only for orbits around a lumped central mass $M$ (or more exactly: where all this mass is located inside the star's orbit). This formula is derived from the equality between the gravitational force and the centripetal force $$\frac{GMm}{r^2}=\frac{mv^2}{r}$$

When you have a disc-like mass distribution (i.e. part of galaxy mass is inside the star's orbit, and part of the galaxy mass is outside the star's orbit), then you have another formula for the graviational force between star and galaxy.

  • $\begingroup$ What would that relationship look like then? I have previously calculated the mass distribution inside the star's orbit and outside it(with "outside" I mean the galaxy's area coverage mass). I compared the masses with $\frac{\pi r_s^2}{x}=\frac {\pi r_g^2}{m_g}$. Is this ratio correct and would it be relevant? $\endgroup$ Jul 23, 2022 at 13:33
  • 1
    $\begingroup$ @Aristarchus_ I think this ratio would be relevant. But I didn't check its correctness. $\endgroup$ Jul 23, 2022 at 14:17
  • $\begingroup$ Could you explain the formula for the gravitational force between a star and galaxy. That is, for disc-like mass distribution and how it can be used here $\endgroup$ Jul 23, 2022 at 14:26
  • 1
    $\begingroup$ @Aristarchus_ Since you are asking a homework question, you should show your effort first. $\endgroup$ Jul 23, 2022 at 14:34
  • $\begingroup$ Posted my attempt above... $\endgroup$ Jul 23, 2022 at 14:59

The distribution of the mass is important! The law $F=GMm/r^2$ is only true for point masses (or, through Newton's shell theorem, spherically symmetric objects).

Your derivation is close to the way I've seen it justified. You do have $v_s=\sqrt{G M/r_s}$, but $M$ is a function of $r_s$. (This is loosely justified through Newton's shell theorem). For $R$ the radius of the galaxy and $r_s$ the star's position in the galaxy, we have $M(r_s)=M_{g} r_s^2/R^2$, so $v=\sqrt{G M_{g} r_s/R^2} \propto \sqrt{r_s}$.

I'm not a huge fan of this justification because Newton's shell theorem doesn't apply. A better analysis would be to look at the full vector integral for $\vec{F}$, for a disk radius $R$ with surface density $\sigma$ and a test mass placed at $(-r_s,0)$.

\begin{align*} \frac{1}{m}\vec{F}&=\int_0^{2\pi} d\theta \int_0^{R} r dr G\sigma \frac{(r_s+r\cos(\theta),0)}{\|(r_s+r\cos(\theta),r\sin(\theta))\|^3}\\ &=G\sigma\int_0^{2\pi}d\theta \int_0^1 \ell d\ell \frac{(r_s/R+\ell\cos(\theta),0)}{\|(r_s/R+\ell\cos(\theta),\ell \sin(\theta))\|^3}\\ &=G\sigma f(r_s/R) \end{align*} Where I made the change of variables $r=\ell R$.

If we're doing a back of the envelope calculation, one idea is to toss out the unitless function $f$. Then $F=v^2/r_s=\text{const.}$ and we have the desired result $v_s\propto \sqrt{r_s}$.

If we want to be precise, the full function $f$ is very difficult and annoying to evaluate, so you can see why no one bothers with it! For example, for $y\gg 1$ we expect $f(y)\approx 2\pi/y^2$ because the force should behave like $GM/r^2$ and $M=2\pi\sigma$. For $y$ just greater than one, the force diverges (look at $\theta=0$) and we would have to add a finite thickness to the disk to get a finite result. For $0<y<1$, you have to use the Cauchy principal value to get a finite result, because you'd have to evaluate things like $\text{p.v.}\int_{-1}^1 x/|x|^3 dx=0$.

But all this only matters if we're being precise. The overall point is that the force behaves like $G\sigma$ times some unitless function of $r_s/r_g$ which, for estimation purposes, can be treated as constant.

I can't help but make plots of the velocity $v=\sqrt{r f(r)}$ for $\sigma=G=R=1$. I regularize the function by defining:

$$f_\varepsilon(r)=\int_0^{2\pi}d\theta \int_0^1 \ell d\ell \frac{r+\ell\cos(\theta)}{\|r^2+\ell^2+2\ell\cos(\theta)+\varepsilon^2 \|^3}$$

$\varepsilon\to 0$ for an infinitely thin disk, and $\sqrt{ r f_{0.1}(r)}$ is the velocity for a circular orbit for a galaxy disk with thickness $0.1 R$.

The green dashed line is the expected result for a point mass galaxy $v_s\propto r_s^{-1/2}$, the red line is $v_s\propto r_s^{1/2}$, and the other lines are the full messy truth for varying galaxy thickness.

Velocity curves for a disk with varying thickness Mathematica source code on pastebin.

Stuff like this has been pretty useless in the astrophysics courses I've taken because real galaxies have dark matter halos, central bulges, and all sorts of other things, so the issue of an infinitely thin disk of constant surface density never comes up.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.