How much kinetic energy would a star in a galaxy have if it fell to the center?

Question

I want to calculate the speed, or equivalently, the kinetic energy of a star, if it had no rotational speed and fell from a given radius to the center of the galaxy.

I assume Newton's shell theorem works as a fair approximation.

Answer 1

A fair answer can be achieved using Newtonian physics. Assuming that the star was stationary prior to infall (as you want), we would need to find its potential energy. If it falls to the center, then the gained kinetic energy can be equated to this potential energy.

Let's assume that the star is at $r = R$. To calculate the star's potential energy, we would need to find the amount of mass within the region $r < R$. This mass will come from three sources:

(i) The massive black hole at the center say with mass $M_{\rm BH}$

(ii) The mass of the disk (I have ignored the bulge), for which we may use a power law density profile like $$\rho = \rho_0\left(r / R_{\rm in}\right)^{-n} e^{-z^2 / 2 H^2} ,$$ where $H$ is the disk scale height. This can be integrated over $z$ to give a surface density as $$\Sigma = \rho_0\left(r / R_{\rm in}\right)^{-n} \sqrt{2\pi}H$$ Here, $\rho_0$ is the density in the $z=0$ plane and $R_{\rm in}$ is the inner radius of the disk and can be approximated as $\mathcal{O}(R_s)$, $R_s$ being the Schwarzschild radius of the black hole of mass $M_{\rm BH}$. Though I haven't checked, but you might find the values of some of the other unknown parameters here.

(iii) The mass of the dark matter halo. We can assume an NFW profile here with $$\rho = \frac{\rho_0}{ \frac{r}{r_0} \left(1 + \frac{r}{r_0} \right)^2}$$ and for a Milway-like galaxy you can take $\rho_0 = 0.052 M_{\odot}/pc^3$ and $r_0 = 8.1$ kpc [Ref].

Now, we can find the total mass contained within $r<R$ by integrating over $r$ and $\phi$ for (ii) and (iii) - the $\phi$ integral will just be $2\pi$. From this, we can then directly calculate the potential energy of the star.

Answer 2

The rotation curve $v ( r )$ for the galaxy will give the centripetal force experienced at any radial position $F = \frac{mv^2}{r}$. The kinetic energy in this scenario would be $ \frac{mV^2}{2} = KE = \int_0^R F dr$. Here $V$ is the star’s velocity on reaching the galactic centre, $R$ is its initial position and $m$ is its mass.