I am calculating galaxy rotation curves for various galaxies in Ursa Major cluster and I want distance of those galaxies from the centre of Cluster. The values referred as coordinated are RA and dec and I don't know anything about these coordinates. How/Where can I get the distances of galaxies?


If the centre of your cluster has coordinates $\alpha, \delta$, the right ascension and declination in radians$^{1}$, and you have a galaxy's coordinates $\alpha_g, \delta_g$ in radians, then the following formula gives the angular distance $\theta$ in radians.

$$ \cos \theta = \sin \delta \sin \delta_g + \cos \delta \cos \delta_g \sin (\alpha - \alpha_g)$$

Check you calculation with this calculator.

From there, to get a physical distance you need to to know the distance to the cluster $D$. The projected separation$^2$ of the galaxy from the centre of the cluster $r$ is then

$$r = D\tan(\theta) \simeq D\theta.$$

$^{1}$ To convert right ascension given in hours, minutes and seconds and declination, given in degrees, minutes and seconds, to radians you do $$\alpha = (RAh*15 + RAm/4 + RAs/240) \times \pi/180$$ $$\delta = (DEd \pm DEm/60 \pm DEs/3600) \times \pi/180,$$ where in the latter formula you use plus signs for objects in the northern hemisphere and minus signs for objects in the southern hemisphere.

$^2$ You see the galaxies on the plane of the sky; there is no way at present to get the deprojected 3-dimensional distance from the cluster centre.

  • $\begingroup$ Rob, shouldn't r be obtained as: r=D*tan(theta)? $\endgroup$ – Gabriel Dec 20 '15 at 14:09
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    $\begingroup$ @Gabriel, yes, but of course $\theta \simeq \tan \theta$ for the small angles we are talking about here. $\endgroup$ – Rob Jeffries Dec 20 '15 at 21:26
  • $\begingroup$ Thanks for the clarification Rob. I've edited your answer so this approximation is made explicit, if you don't mind. $\endgroup$ – Gabriel Dec 20 '15 at 21:29

Right Ascension and declination are angular measures, RA * 15 gives degrees &c.

Basically, RA is where the siderial zenith is at any time, declination is the distance from the zodiac. You multiply RA by 15, and get angles.

Neither of these are distance, you have to get a radial term from somewhere else. What they are are coordinates to find the points in the sky.

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    $\begingroup$ RA & declination are the celestial analogues of longitude and latitude. RA is measured in time (it derives from rotation) and only converts to degrees by multiplying by 15 at the celestial equator. $\endgroup$ – Dr Chuck Dec 19 '15 at 11:36
  • $\begingroup$ Actually, RA converts to degrees of longitude at 15 degrees per hour at all places, it's that both the hour and the degree get shorter at higher lattitudes. Siderial zenith is an accurate description of the rotation and the coordinate. 6h00m is overheat at ST: 6h00m. $\endgroup$ – wendy.krieger Dec 19 '15 at 12:01
  • $\begingroup$ Declination is the angle from the celestial equator. The zodiac is (a) ill-defined; (b) not aligned with the celestial equator. $\endgroup$ – Rob Jeffries Dec 19 '15 at 12:22

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