How do you find the distance from a star/planet/black hole to another? I know people can calculate the distance from Earth to a star, but what about from one to another?


2 Answers 2


If you know distance from Earth to both objects and the angle between them viewed from Earth, it is just a matter of trigonometry.

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    $\begingroup$ That's true if the distance are less than few millions of ligthyears. Otherwise you have to account for cosmological effects and possible cosmic curvature $\endgroup$ Commented Dec 4, 2013 at 9:22
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    $\begingroup$ @FrancescoMontesano It is still trigonometry, just not Euclid's one. $\endgroup$
    – Envite
    Commented Dec 4, 2013 at 9:29
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    $\begingroup$ its 3 dimensional trigonometry just almost similar to what we use, but far more complex when considering the fact that you are measuring distance of two objects in the vast universe and not on a sheet of paper. $\endgroup$
    – Mahe
    Commented Dec 4, 2013 at 12:38
  • $\begingroup$ The key with cosmological distances is using the appropriate distance measure depending on the purpose of the measurement. Transverse comoving distance can be used to measure distances that take the expansion of the Universe into account and don't vary with time. $\endgroup$
    – Aaron
    Commented Jun 17, 2014 at 22:42

You only need two-dimensional trigonometry if you know the distances to the two stars and their angular separation. Any two-dimensional plane can be defined by three points that lie on it so we just use the plane containing the two stars and the Earth.

You can use Earth as the origin and the closest star as a point on the $x$-axis ($x_1$,$y_1$) where $x_1$ is the distance and $y_1$ is zero. You then can use the distance of the second star and its angular separation from the first star (which is on the $x$-axis) to plot a point ($x_2 = \text{distance} \times \cos(\text{angle})$, $y_2 = \text{distance} \times \sin(\text{angle})$). The distance between those two points is $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

  • $\begingroup$ This does not apply to large distances, as noted by Francesco. $\endgroup$
    – called2voyage
    Commented Dec 5, 2013 at 21:47
  • $\begingroup$ Please explain more, I don't see why since space has been shows to be flat over most of the universe and the 'cosmological effects' aren't explained. The fact that it does not explain exactly does not make it invalid just like we can still use Newton's equations for approximations and ordinary tasks instead of invoking general relativity (they got us to the moon). $\endgroup$ Commented Dec 5, 2013 at 22:52
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    $\begingroup$ When dealing with stars in the milky-way or even galaxies in the local group you are right, and simple flat-space (euclid) works just fine. However, for the very long distances you have to account for light travel time. The universe is expanding and the rate of acceleration has changed in the past. Objects at very different (large) distances can then be thought of as being set at a different age of the universe and therefore at a different sized universe. This is what enters in the equation as 'cosmological' effects. $\endgroup$
    – Michael B.
    Commented Dec 9, 2013 at 11:03
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    $\begingroup$ I don't think that was in the question. Also, if that is the case then you have to take into account all motion of the star. For instance SDSS J091759.5+672238 have moved 400 light-years since the light we are seeing left it. I also don't hear assumed movement taken into account since there is no universal 'now' in relativity. For instance I always hear the furthest galaxies are about 13 billion light years away, not 26 billion. $\endgroup$ Commented Dec 9, 2013 at 17:40

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