Answer to Question #123326 in Astronomy | Astrophysics for frank

Question

The most mass of our Milky Way is contained in an inner region close to the core with radius R0.
Because the mass outside this inner region is almost constant, the density distribution can be
written as following (assume a flat Milky Way with height z0):
ρ(r) = (
ρ0, r ≤ R0
0, r > R0
(a) Derive an expression for the mass M(r) enclosed within the radius r.
(b) Derive the expected rotational velocity of the Milky Way v(r) at a radius r.

Answer 1

Let $z_0$ be the half of the height of the disk.

(a) If $r\le z_0$ , the sphere with radius $r$ will completely lie inside the disk, so the mass will be

$M(r) = \dfrac43\pi r^3\rho_0.$

If $r\ge \sqrt{R_0^2+z_0^2}$ then the full disk of mass lies inside the sphere, so the mass will be

$M(r) = \pi R_0^2 z_0 \rho.$

The most difficult case is for $z_0 < r < \sqrt{R_0^2+z_0^2}.$

If $z_0 < r \le R_0$ , then the volume will be spherical minus two spherical caps.

The cap has a volume of $V_1=\dfrac{\pi (r-z_0)}{6}\left(3(r^2-z_0^2)+(r-z_0)^2 \right)$ .

So the total volume will be

$V = \dfrac43\pi r^3- \dfrac{\pi (r-z_0)}{3}\left(3(r^2-z_0^2)+(r-z_0)^2 \right).$

And the mass will be $\rho_0 V.$

If $R_0 < r < \sqrt{R_0^2+z_0^2}$ , then the volume will be sphere minus two spherical caps and minus small spherical ring the Galactic plane.

The cap has a volume of $V_1=\dfrac{\pi (r-z_0)}{6}\left(3(r^2-z_0^2)+(r-z_0)^2 \right)$ . The ring is a segment minus cylindrical layer.

The cap has a volume of $V_2=\dfrac{\pi \sqrt{r^2-R_0^2}}{6}\left(3R_0^2+r^2-R_0^2 \right) = \dfrac{\pi \sqrt{r^2-R_0^2}}{6}\left(2R_0^2+r^2\right).$

The spherical ring has volume $V_3 = \dfrac43\pi r^3- 2V_2 -2\pi R_0^2\sqrt{r^2-R_0^2}$

So the total volume will be

$V = \dfrac43\pi r^3- \dfrac{\pi (r-z_0)}{3}\left(3(r^2-z_0^2)+(r-z_0)^2 \right) - \dfrac43\pi r^3 + \dfrac{\pi \sqrt{r^2-R_0^2}}{3}\left(2R_0^2+r^2\right) +2\pi R_0^2\sqrt{r^2-R_0^2} = 2\pi R_0^2\sqrt{r^2-R_0^2} + \dfrac{\pi \sqrt{r^2-R_0^2}}{3}\left(2R_0^2+r^2\right)- \dfrac{\pi (r-z_0)}{3}\left(3(r^2-z_0^2)+(r-z_0)^2 \right).$ To obtain the mass, we should multiply this volume by $\rho_0.$

(b) The potential of cylindrical bodies can be calculated using the elliptic integrals (see http://articles.adsabs.harvard.edu//full/1983CeMec..30..225L/0000227.000.html) as

$U(r,0) = -2G\rho z_0 ((R_0+r)E(k)+(R_0-r)K(k))$ , where $k = \dfrac{2\sqrt{R_0r}}{R_0+r}$ , where $K(k) = \int\limits_0^{\pi/2} \dfrac{d\psi}{\sqrt{1-k^2\sin^2\psi}}\, ; \\ E(k) = \int\limits_0^{\pi/2} {\sqrt{1-k^2\sin^2\psi}}\,d\psi.$

next, we should calculate $v(r) = \sqrt{ r \cdot \dfrac{\partial U(r,0)}{\partial r}} .$ We can see that the equation of potential contains $r$ not only as multipliers, but also in integrals, so to obtain the rotation curve we should use the numerical calculations on computer.