How can I get heliocentric distance to a star when I have galactic x y and z coordinates known ?

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    $\begingroup$ Can you give an example of what you mean by "galactic x y and z coordinates"? The standard Galactic coordinate system (l, b) only has two numbers, which are angular directions on the sky, as seen from the Earth. $\endgroup$ Commented Jun 29, 2018 at 18:46

1 Answer 1


The galactic coordinate system is centred on the sun with the xy plane in the plane of the galaxy and the x coordinate pointed towards the centre of the galaxy.

Since it is a heliocentric system the Euclidian distance is $\sqrt{x^2+y^2+z^2}$.

  • $\begingroup$ The Galactic coordinate system has two numbers, not three; it is an angular system, not a Cartesian system. $\endgroup$ Commented Jun 29, 2018 at 18:46
  • $\begingroup$ UVW is how cartesian coordinates are usually referred to in the galactic frame. It is possible that the OP means "xyz relative to the centre of the galaxy", if so I'll delete this. $\endgroup$
    – James K
    Commented Jun 29, 2018 at 19:55
  • $\begingroup$ UVW refers to velocities, not positions. $\endgroup$ Commented Jun 30, 2018 at 9:37
  • $\begingroup$ Having said that, yes, there are occasional uses of a "galactocentric" Cartesian system like what you describe, so in the absence of any more information from the person asking the question, that's probably a reasonable answer. $\endgroup$ Commented Jun 30, 2018 at 10:32
  • $\begingroup$ @Peter Erwin there is an alternative galactic coordinate system to the (l,b) one you mentioned above. It is a Cartesian coordinate system in which location coordinates are (X,Y,Z) and velocity vectors are (U,V,W) with the sun at (0,0,0). The X axis passes thru the sun with the positive direction being toward the center of the galaxy. The X,Y plane is the "plane" of the galaxy (not well defined and the sun is chosen arbitrarily to be "above" the plane). The Y axis is perpendicular to the X axis, positive in the direction of galactic rotation. The Z axis is perpendicular to the plane. $\endgroup$ Commented Jul 2, 2018 at 22:55

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