# Lagrange Points: Where on the sky are they?

Home Forums General Discussion Lagrange Points: Where on the sky are they?

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• #574376
Grant Privett
Participant

Recently, I was reading a paper in the MNRAS and the author casually referred to pointing their telescope toward the Sun-Earth L5 point. Now, I imagine working out where that is in the sky involves offsetting from the antisolar point by a fixed (slightly varying as our orbit is elliptical) number of degrees along the ecliptic, with the L4 being on the other side of the antisolar point. I looked at The SkyX and some other bits of astronomical software I have knocking about (including AstroPy) but have found none that will tell you the RA/Dec of the L4/5 points from a given location at a given time.

Does anyone know – offhand – what that offset is for the Sun-Earth L4/5 please? Alternatively, can you point me at some software that would?

I can only imagine that finding the Earth-Moon Lagrange points must be much harder.

#581242
Dominic Ford
Keymaster

L4 and L5 are almost exactly 60 degrees away from the Sun [*], along the plane of the ecliptic, I think?

I say “almost” because as you point out, the Earth’s elliptical orbit will presumably perturb that angle slightly over the course of the year, though I imagine the perturbation is tiny.

[*] — My original forum post was incorrect, so I have edited the wording.

#581243
Grant Privett
Participant

Is that 60 degrees as viewed from the sun or the earth?

#581244
Dominic Ford
Keymaster

A correction to my previous post. Seen from the Sun, L4/5 are both 60 degrees away from the Earth. Seen from the Earth, L4/5 are both 60 degrees away from the Sun.

The Sun, Earth, and L4/5 form equilateral triangles, with all three interior angles being 60 degrees, and all the sides being 1 AU.

#581245
Grant Privett
Participant

Brilliant. Really hadnt realised you just had to look 60 degrees in front or following the sun. That makes it pretty easy, get the sun’s position on ecliptic in RA/Dec, convert to ecliptic cooords, add 60 degrees and convert back to RA/Dec again. Sure I can find something in AstroPy to do most of that. I imagine that drops it nicely into the edge of the Zodiacal light.

Thanks again.

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