# Thread: What are the galactic coordinates of planet earth?

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Aug 2018
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## What are the galactic coordinates of planet earth?

Okay so I know that there is a galactic coordinate system, which uses the Sun as the center and finds other things in the universe in relation to that. Is there some calculations if it is possible to do, using the center of the Milky Way has your zero point instead of the Sun, and using the equation to figure out the galactic coordinates of our sun or Earth? sorry for my lack of technical knowledge but hopefully someone can let me know if this is possible.thanks guys

2. Assuming a polar coordinate system...

You would have to pick and arbitrary 0 radial angle.

Presumably, being Earthlings, we would pick the direction of Sol (wrt the centre of the Milky Way) as zero degrees.

You'd have to have a unit of distance from the centre - you might use light years or parsecs.
Or maybe "SU"s: Sol units - where Sol's distance from the GC is 1SU. That would mean much of the galaxy is less than one SU, so it might be too big a unit.

3. I guess we could use light years, so the sun would be at zero degrees, about 26,000 light years from the center, and say about 60 light years north of the galactic plane?

4. A bit awkward for a non-rigid body like the Milky Way.

5. Originally Posted by Glom
A bit awkward for a non-rigid body like the Milky Way.
And no immovable frame of reference, per relativity. However, we do have... the pulsar positioning system.

https://arxiv.org/ftp/arxiv/papers/1704/1704.03316.pdf

6. Originally Posted by Glom
A bit awkward for a non-rigid body like the Milky Way.
It doesn’t really matter. Wherever the sun is is defined as the Prime meridian.

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7. Originally Posted by DaveC426913
Assuming a polar coordinate system...

You would have to pick and arbitrary 0 radial angle.

Presumably, being Earthlings, we would pick the direction of Sol (wrt the centre of the Milky Way) as zero degrees.

You'd have to have a unit of distance from the centre - you might use light years or parsecs.
Or maybe "SU"s: Sol units - where Sol's distance from the GC is 1SU. That would mean much of the galaxy is less than one SU, so it might be too big a unit.

Picking the moving Sun as angle 0o degrees could be a bit confusing. How about a very bright distant object?

8. Originally Posted by swampyankee
Picking the moving Sun as angle 0o degrees could be a bit confusing. How about a very bright distant object?
Pulsars.

9. Originally Posted by swampyankee
Picking the moving Sun as angle 0o degrees could be a bit confusing. How about a very bright distant object?
Why would it be confusing?

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10. I suspect that the original poster, Emilyexploresearth, might not have been interested in or have considered the complexities introduced by the Sun's motion around the galaxy. Also, the Sun's motion is slow enough that its coordinates relative to the galactic center and the galactic plane don't change significantly over a human lifetime.

Given that, I think that one way to convert from the Sun-centered Galactic coordinate system used by professional astronomers to a new one that's relative to the Galactic center would be the following:
1. Assume that you want to have the center of the new coordinate system at the center of the Milky Way (in the vicinity of the black hole Sag A*) and be relative to the central plane of the Milky Way's "thin disk."
2. Transform (convert) the astronomical Galactic coordinates of a star from spherical to Cartesian coordinates
3. Translate (add in) the star's (or other object's) position using the Sun's distance from the galactic center along the X axis (~25,640 LY or ~7860 pc) and its height above the galactic plane along the Z axis (~112 LY or ~34.5 pc) so its position is now relative to the galactic center and the galactic plane
4. If desired, convert those Cartesian coordinates back to spherical coordinates.

The coordinate system transformations might be the difficult part for someone unfamiliar with spherical trigonometry or quaternions, but appropriate math functions are readily available in public, downloadable astronomical subroutine libraries for many programming languages. Routines also are readily available to convert between the Earth-centered Equatorial coordinate system and Galactic.

To actually answer the question posed in this Topic's title our Sun is at X=25640, Y=0, Z=112 in the Cartesian coordinate system I mention above.
Last edited by selden; 2018-Aug-10 at 12:16 AM.

11. Originally Posted by swampyankee
Picking the moving Sun as angle 0o degrees could be a bit confusing. How about a very bright distant object?
Why? It's a galactic coodinate system. It should be centred on the GC.

We already have a Sol-centred coordinate system: Right Ascension and Declination. But that would be very awkward for anyone but Sol inhabitants to use.

12. Originally Posted by selden
I suspect that the original poster, Emilyexploresearth, might not have been interested in or have considered the complexities introduced by the Sun's motion around the galaxy. Also, the Sun's motion is slow enough that its coordinates relative to the galactic center and the galactic plane don't change significantly over a human lifetime.
You wouldn't use the sun's current position as the marker - precisely because it's changing. You would affix its position as it is/was at some specific arbitrary point in time - and that angle would remain fixed forever. You'd fix it to a large number of distant, bright objects, such as quasars - which essentially serve as the unmoving background of the universe (in the same way parallax uses it).

No galactic coordinate system can hope to be at rest with any significant fraction of the galaxy. It must be arbitrary and motionless wrt to an external frame of reference, with all stars moving against it over time.
Last edited by DaveC426913; 2018-Aug-10 at 01:03 AM.

13. Still saying [extragalactic] pulsars, unless quasars move around less.

14. One problem with using pulsars to navigate over long distances is that the pulses from any particular pulsar are visible only from limited viewing angles as its "beam" passes over the observer. As the viewing angle changes, the intensity and shapes of those pulses change. The ones we've detected are only a small fraction of the total which must exist, since we can only see those which happen to have their beams pointed in our general direction at some time in their rotation. In addition to knowing all of their long-term spindown rates, a galactic traveler would also have to know which ones are visible where and how the the structure of their pulses changes with viewpoint. While traveling, for example, a sublight traveler have to keep track of how the characteristics of the visible pulsars change with viewing angle and watch for those which become newly visible. Of course, such a traveler doubtless should be constantly measuring the parallaxes of the stars they can see and could keep track of their position that way. Using extragalactic pulsars, so that their viewing angles don't change much, would be more appropriate than using pulsars that are within the Milky Way.

Quasars are billions of light years away, so they have no measurable parallax. They'd be great for determining one's orientation, but wouldn't be be usable by travelers to find out where they were in the galaxy relative to their starting point if they hadn't been keeping track (or had lost the records) of where they'd been (or who might be using some kind of discontinuous FTL transportation). Advantages of using quasars are that they do not change their positions relative to one another "on the sky" because of their great distances (which is why they're used to calibrate the measurements of positions of stars on our sky and to establish baselines for parallax measurements of "nearby" stars) and they are equally visible throughout the Milky Way (ignoring extinction problems caused by dust and gas clouds).

I think that for an FTL traveler who has gotten lost, measuring the parallaxes and orientations of nearby galaxies (like the LMC and SMC) might be a reasonable way to find one's general location.