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How can the total gravitational potential energy of a galaxy be calculated?

Lets assume for simplicity that the entire galaxy follows an exponential mass density function for an infinitely small thickness:

$\rho_r = \rho_0 e^{(-r/h)}$

with $\rho$ being mass density in $kg/m^2$, $\rho_0$ density at $r=0$, $r$ radius and $h$ scale length.

I can calculate it by imagining that it is pulled apart by successively moving ring shells to infinity, the outermost first, and finding the total energy needed for that.

The equation for gravitational potential energy is:

$dU = -GM_r m_idr/r$

with mass of ring:

$dm_i = 2\pi r\rho(r)dr$

and mass interior:

$m_i=2\pi\rho_0h^2(1-(1+r/h)e^{(-r/h)})$

I get:

$U=-G\pi^2\rho_0^2h^2(e^{(-2r/h)}(h(3-4e^{(r/h)})+2r)+h)$

which fits dimensions.

Is this derivation correct?

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  • $\begingroup$ Do you understand how to calculate the gravitational potential energy of a uniform-density ball? If not, I suggest learning that first. $\endgroup$
    – Ghoster
    Sep 20, 2023 at 21:44
  • $\begingroup$ Yes, but this is not a uniform density sphere. $\endgroup$
    – Manuel
    Sep 20, 2023 at 22:07
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    $\begingroup$ @Manuel the hint there is that the process is the same, the answer will just be slightly different because the integration is different, assuming that when you say you know the uniform sphere case that you are aware you need to perform some form of integration. $\endgroup$
    – Triatticus
    Sep 20, 2023 at 22:22
  • $\begingroup$ @Ghoster I worked out the derivation, but im not sure about my result. $\endgroup$
    – Manuel
    Sep 26, 2023 at 20:40
  • $\begingroup$ @Triatticus I worked out the derivation, but im not sure about my result. $\endgroup$
    – Manuel
    Sep 26, 2023 at 20:41

2 Answers 2

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Your method is not correct. If I understood correctly, you obtained $U$ by integrating: $$ U=Gm_R\int_0^R \frac{dm}{r} $$ with: $$ dm=2\pi r\rho dr\\ m_R=\int_0^Rdm $$ and $R$ the hard cutoff of the density. This is not the correct formula for energy. In general, when you have points mass indexed by $i$ with respective mass $m_i$ and position $r_i$, the energy is given by: $$ U=\frac{1}{2}\sum_{i\neq j}\frac{Gm_im_j}{|r_i-r_j|} $$ Similarly, for a continuous distribution, the formula is: $$ U=\frac{1}{2}\int\frac{G\rho(r)\rho(r’)}{|r-r’|}d^3rd^3r’ $$ In your modeling, it is strange that you add a hard cutoff on top of the exponential profile. I’ll just set your radius $R\to\infty$ and set $h=\rho_0=G=1$ up to change of length, mass and time scale. Your galaxy thus has total mass: $$ M=2\pi $$ The correct energy is therefore: $$ U=\pi\int \frac{e^{-r-r’}}{\sqrt{r^2+r’^2-2rr’\cos\phi}}rr’drdr’d\phi $$ I’m not sure that the integral is expressible in terms of the usual mathematical constants, but you can still evaluate it numerically.

Hope this helps.

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The mass in the disk would be \begin{align} M_{\text {disk }}(r) & =\int_{0}^{2 \pi} d \phi \int_{r_{\rm in}}^r \rho r'dr'\\ & = 2\pi \rho_0 \int_{r_{\rm in}}^{r} e^{-r'/h} r'dr' \\ & = \left[- 2\pi\rho_0 h e^{-r'/h} (h + r') \right]_{r_{\rm in}}^{r} \end{align}

In addition the infinitesimal mass element (note: this is not a ring) $$d M_{\rm disk}(r', \phi) = \rho r' dr' d\phi$$

If you combine these two, then the potential energy stored between the inner mass $M_{\rm disk}$ and the tiny mass $dM_{\rm disk}$ is $$dU (r') = -\frac{GM_{\rm disk}(r') \, dM_{\rm disk}(r')}{r'} $$

Also, the symmetry of the problem means that we only have to consider the potential due to mass in the disk $M_{\rm disk}$ at $r<r'$. Therefore $$ U (r) = -\int_0^{2\pi} d\phi\int_{r_{\rm in}}^{r} \frac{GM_{\rm disk}(r') \, \rho r'dr'}{r'} $$

I think this would result in something identical to what you have already. Also, remember that there should be mass in the halo of the galaxy as well (due to dark matter), which is not considered here.

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