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


2 Answers 2


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.

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

[A previous version had a sine instead of a cosine in the last term, but that cannot be right because if you set $(\alpha, \delta) = (\alpha_g, \delta_g)$ you did not get $ \cos \theta = 1$ and thus $\theta=0$. You can check that the current expression now does satisfy this requirement: if the angles are equal, then $ \cos \theta = \sin^2 \delta + \cos^2 \delta \times \cos (0) = 1$.]

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, 2015 at 14:09
  • 2
    $\begingroup$ @Gabriel, yes, but of course $\theta \simeq \tan \theta$ for the small angles we are talking about here. $\endgroup$
    – ProfRob
    Dec 20, 2015 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, 2015 at 21:29
  • $\begingroup$ Your calculator link is not working. $\endgroup$
    – WarpPrime
    Oct 29, 2020 at 23:30

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.

  • 2
    $\begingroup$ 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, 2015 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. $\endgroup$ Dec 19, 2015 at 12:01
  • 1
    $\begingroup$ Declination is the angle from the celestial equator. The zodiac is (a) ill-defined; (b) not aligned with the celestial equator. $\endgroup$
    – ProfRob
    Dec 19, 2015 at 12:22

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