I was wondering if we can predict the rotational velocity due to dark matter in this way:

Consider that we theoretically derived the rotational velocity of a galaxy ($v_{the}$). Then from experiment we got some other rotation velocity ($v_{obs}$). Then can we subtract these two velocities to obtain the rotational velocity due to dark matter ($v_{dar}$).

If not, how can we derive $v_{dar}$ by $v_{the}$ and $v_{obs}$?

  • $\begingroup$ Yes, this is essentially the method, though of course using the complete formulas. $\endgroup$ – Javier Jun 17 at 20:49
  • $\begingroup$ You mean $v_{obs}$-$v_{the}$=$v_{dar}$ just confirming $\endgroup$ – The One Eye Triangle Jun 17 at 21:11

Radial acceleration is $v^2/r$ and this is proportional to the total mass enclosed within the orbit. Adopting a rough Keplerian approximation $$ \frac{v^2}{r} = \frac{G}{r^2}\sum_i M_i,$$ where $M_i$ are the masses of the various components (baryonic, dark).

If I understand correctly, your "theoretical" rotation curve arises from considering only baryonic matter, whereas the observed speed corresponds to $v$ in the equation above. If so, then $$v^2_{\rm obs}\simeq v^2_{\rm the} + v^2_{\rm dark}\ .$$

| cite | improve this answer | |
  • $\begingroup$ Thanks I got the concept now very clear $\endgroup$ – The One Eye Triangle Jun 18 at 8:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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