<p>I'm using <a href="https://canvas.education.lu.se/courses/3801/files/277616/download?download_frd=1" rel="nofollow noreferrer">this resource</a> (will download a .pdf) to calculate the positions of all the Lagrange points in a 2-body system, but it's only concerned with the x and y components of their position vectors (2D treatment of the problem, in other words), but reality is made up of three dimensions and I would like to capture that. In the case of most of the planets I guess Z is pretty close to the ecliptic plane, but for bodies with inclined orbits like Eris, I guess that is not a good approximation.</p>
<p>Could somebody tell me how to calculate the z component of the vector that describes the positions of all the Lagrange points?</p>

CC BY-SA 4.0
2021-09-04T13:16:26.627Z
Long time no see! I'd even forgot about you, shame on me! [Looking for that cool astronomical and spaceflight simulation website that's been discussed here](https://space.meta.stackexchange.com/q/2839/12102) Lagrange points only exist in-plane. The two massive bodies have an orbital plane and the zeros of the forces (in the rotating frame) must also be in the same plane. Lagrange points are mathematical concepts and are only defined for a two-body orbit in a 2D plane.
uhoh
CC BY-SA 4.0
2021-09-04T13:30:41.297Z
Howdy! How's life treating you? Well, that's reassuring to know that we put spacecraft into orbit around mathematical concepts. So in the case of Earth, the Lagrange points are simply chilling on the ecliptic plane? And in the case of Pluto, the, ehrm, Plutonian plane? I'll get to it...
Happy Koala
CC BY-SA 4.0
2021-09-04T13:46:46.973Z
Thanks for the kind words (read the question you referred to)! You can actually create your own scenarios now, and add rings to any celestial body: https://gravitysimulator.org/misc/create-new-gravity-simulation.
Happy Koala
CC BY-SA 4.0
2021-09-04T13:53:41.883Z
The z component is identically zero.
David Hammen
CC BY-SA 4.0
2021-09-04T14:36:12.473Z
@HappyKoala In the case of Sun-Earth system with a *circular orbit*, yes. As soon as you add some reality (elliptical orbit, or other planets, or...) then the conditions under which Lagrange derived his points no longer exist. *There are still halo-like orbits around where you'd expect them to be* but the points themselves only exist *in theory* and only in the 2 dimensional orbital plane, and only for circular orbits.
uhoh
CC BY-SA 4.0
2021-09-05T17:44:06.080Z
@uhoh When I calculate the Lagrange position vectors every iteration of the simulation, I notice how, (besides moving as Earth moves), they move ever so slightly towards and away from the Sun, depending on where Earth is in its orbit. I guess that's because of the eccentricity of Earth's orbit?
Happy Koala
CC BY-SA 4.0
2021-09-05T22:39:52.573Z
@HappyKoala yes that's why, but remember that the equations you are using are themselves derived for a circular orbit (constant distance, constant speed); they're not meant to be used for elliptical orbits. So while you can do the calculation to visualize where some Lagrange-like conceptual point might be, it's not really a Lagrange point any more.
uhoh
CC BY-SA 4.0
2021-09-06T07:54:33.127Z
@uhoh Yea... Apparently, for eccentric orbits, Lagrange points are no longer Lagrange points, but Lagrange areas... I'll post a link to a scenario when I have deployed this feature to production.
Happy Koala
CC BY-SA 4.0
2021-09-06T07:57:08.800Z
@HappyKoala looking forward to it!
uhoh
CC BY-SA 4.0
2021-09-09T14:27:29.167Z
@uhoh Here you go: https://gravitysimulator.org/solar-system/jupiter-and-its-ten-largest-trojan-asteroids. Regarding the z component... Lagrange points exist in the orbital plane of the smaller mass and with respect to this plane is z is indeed 0, but looking at the rotating reference frame of the simulation, the z component of a Lagrange point can take any value.
Happy Koala
<lagrange-point>
How do you calculate the the z component of the vector that describes the position of L1, L2, L3, L4 and L5 (lagrange points)
Happy Koala
# How do you calculate the the z component of the vector that describes the position of L1, L2, L3, L4 and L5 (lagrange points)

(Asked by: username_0 on 2021-09-04)

I'm using [this resource](https://canvas.education.lu.se/courses/3801/files/277616/download?download_frd=1) (will download a .pdf) to calculate the positions of all the Lagrange points in a 2-body system, but it's only concerned with the x and y components of their position vectors (2D treatment of the problem, in other words), but reality is made up of three dimensions and I would like to capture that. In the case of most of the planets I guess Z is pretty close to the ecliptic plane, but for bodies with inclined orbits like Eris, I guess that is not a good approximation.

Could somebody tell me how to calculate the z component of the vector that describes the positions of all the Lagrange points?

**Tags:** lagrange-point

> **Comments:**
> **username_1**: Long time no see! I'd even forgot about you, shame on me! [Looking for that cool astronomical and spaceflight simulation website that's been discussed here](https://space.meta.stackexchange.com/q/2839/12102) Lagrange points only exist in-plane. The two massive bodies have an orbital plane and the zeros of the forces (in the rotating frame) must also be in the same plane. Lagrange points are mathematical concepts and are only defined for a two-body orbit in a 2D plane.  
> **username_0**: Howdy! How's life treating you? Well, that's reassuring to know that we put spacecraft into orbit around mathematical concepts. So in the case of Earth, the Lagrange points are simply chilling on the ecliptic plane? And in the case of Pluto, the, ehrm, Plutonian plane? I'll get to it...  
> **username_0**: Thanks for the kind words (read the question you referred to)! You can actually create your own scenarios now, and add rings to any celestial body: https://gravitysimulator.org/misc/create-new-gravity-simulation.  
> **username_2**: The z component is identically zero.  
> **username_1**: @username_0 In the case of Sun-Earth system with a *circular orbit*, yes. As soon as you add some reality (elliptical orbit, or other planets, or...) then the conditions under which Lagrange derived his points no longer exist. *There are still halo-like orbits around where you'd expect them to be* but the points themselves only exist *in theory* and only in the 2 dimensional orbital plane, and only for circular orbits.  
> **username_0**: @username_1 When I calculate the Lagrange position vectors every iteration of the simulation, I notice how, (besides moving as Earth moves), they move ever so slightly towards and away from the Sun, depending on where Earth is in its orbit. I guess that's because of the eccentricity of Earth's orbit?  
> **username_1**: @username_0 yes that's why, but remember that the equations you are using are themselves derived for a circular orbit (constant distance, constant speed); they're not meant to be used for elliptical orbits. So while you can do the calculation to visualize where some Lagrange-like conceptual point might be, it's not really a Lagrange point any more.  
> **username_0**: @username_1 Yea... Apparently, for eccentric orbits, Lagrange points are no longer Lagrange points, but Lagrange areas... I'll post a link to a scenario when I have deployed this feature to production.  
> **username_1**: @username_0 looking forward to it!  
> **username_0**: @username_1 Here you go: https://gravitysimulator.org/solar-system/jupiter-and-its-ten-largest-trojan-asteroids. Regarding the z component... Lagrange points exist in the orbital plane of the smaller mass and with respect to this plane is z is indeed 0, but looking at the rotating reference frame of the simulation, the z component of a Lagrange point can take any value.  

---

## 0 answers

