<p>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.  </p>

<p>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 $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$</p>

CC BY-SA 3.0
2013-12-05T21:47:21.017Z
This does not apply to large distances, as noted by Francesco.
called2voyage
CC BY-SA 3.0
2013-12-05T22:52:55.823Z
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).
Jason Goemaat
CC BY-SA 3.0
2013-12-09T11:03:17.810Z
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.
Michael B.
CC BY-SA 3.0
2013-12-09T17:40:51.147Z
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.
Jason Goemaat
CC BY-SA 3.0
2013-12-05T21:12:32.480Z
2015-02-15T02:20:47.117Z
2015-02-15T02:20:47.117Z
Jason Goemaat
<p>If you know distance from Earth to both objects and the angle between them viewed from Earth, it is just a matter of trigonometry.</p>

CC BY-SA 3.0
2013-12-04T09:22:11.650Z
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
Francesco Montesano
CC BY-SA 3.0
2013-12-04T09:29:36.610Z
@FrancescoMontesano It is still trigonometry, just not Euclid's one.
Envite
CC BY-SA 3.0
2013-12-04T12:38:55.740Z
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.
Mahe
CC BY-SA 3.0
2014-06-17T22:42:51.587Z
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.
Aaron
CC BY-SA 3.0
2013-12-04T09:05:54.247Z
2013-12-04T09:05:54.247Z
Envite
<p>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? </p>

<star><planet><black-hole><distances>
Find distance from star to star?
Jont
# Find distance from star to star?

(Asked by: username_0 on 2013-12-04)

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?

**Tags:** star, planet, black-hole, distances


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## 2 answers

### Answer by username_4 on 2013-12-04. Score: 6 | Accepted answer

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

> **Comments:**
> **username_5**: 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  
> **username_4**: @FrancescoMontesano It is still trigonometry, just not Euclid's one.  
> **username_6**: 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.  
> **username_7**: 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.  

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### Answer by username_1 on 2013-12-05. Score: 5

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 $$\sqrt{(x\_2-x\_1)^2+(y\_2-y\_1)^2}$$

> **Comments:**
> **username_2**: This does not apply to large distances, as noted by Francesco.  
> **username_1**: 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).  
> **username_3**: 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.  
> **username_1**: 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.  

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