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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 $v_{obs}$?

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  • $\begingroup$ 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
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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 $$v^2_{\rm obs}\simeq v^2_{\rm the} + v^2_{\rm dark}\ .$$

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

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