The Oort limit

Jan Oort performed another interesting measurement: count stars as function of their height above the disk. From this he was able to estimate how much mass was in the plane of the MW. By counting stars in the plane of the disk as well, he computed a mass-to-light ratio, which was also a bit on the high side, suggesting the presence of dark matter in the disk, even at the position of the Sun.

Here is a simplified way how to do this. First let me prove that for a uniform disk, the force on a star is independent of the distance to the disk.

As an intermediate step, let's compute the gravitational acceleration on a star, at distance $d$ from the centre of a ring of matter with radius $\omega$ and thickness $d\omega$, and of surface density $\Sigma$. From symmetry, the acceleration is toward the middle of the ring, and it has magnitude

$$da(d,\omega) = -2\pi\,G\,\Sigma {\cos(\alpha)\omega\,d\omega\over \omega^2+d^2}\,,$$ (5.12)

where $\alpha$ is the angle at the position of the star, between the vertical and the ring, so $\tan(\alpha)=\omega/d$. Now the acceleration due to the whole disk is found by summing over all such rings, hence

$$a(d) =- 2\pi\,G\,\Sigma\int_0^\infty {\cos(\alpha)\omega\,d\omega... ...2} = - 2\pi\,G\,\Sigma\,\int_0^{\pi/2}\,\sin(\alpha)d\alpha=-2\pi\,G\,\Sigma\,,$$ (5.13)

independent of $d$! So the equation of motion for the star is

$${d^2z\over dt^2} = -2\pi\,G\,\Sigma\equiv -f\,.$$ (5.14)

This is just the same as that for a ball thrown vertically on the earth's surface, so the solution is $z=z_0+\dot z_0 t-(1/2)f\,t^2$. From this, one can easily obtain the fraction of stars you expect that have $z$ between $[z_1,z_1+\delta]$ and between $[z_2,z_2+\delta]$, which depends only on $f$. So, if you measure this from your data, you can determine $\Sigma$. The value Oort found is called the `Oort limit' since the MW disk needs to have at least this surface density.