All I meant here is that the path is not literally an ellipse and does not even lie exactly in a plane, though that may have been a bit of a stupid thing to point out since that will never be the case, anyway. The first approximation to the trajectory will indeed be an ellipse, sure. Then, to higher accuracy, there will be a smaller elliptical epicycle on top of that, with twice the frequency. Then a third-order ellipse, and so on.
So yes it will be “roughly elliptical” around a fictitious point (under certain conditions; in our case you can judge for yourself from the numerically computed plot)
and you can write down an approximate Lagrangian function.
I have seen animations like this one (see at 0:23) that depict the JWST orbiting L2 with an orbital that is more or less perpendicular to the plane of Earth’s orbit:
This is wrong, isn’t it? if the rotation of the whole Earth-Sun system matters, then doesn’t something orbiting a LaGrange point have to be coplanar with that system?
It is correct (mostly).
See the video in post #18.
You can also see the orbit from different angles in the plot I linked to. If I understand correctly, in order for your orbit to stay periodic, there will be restrictions on the amplitude along various axes so that you absolutely have to go out of the plane of Earth’s orbit, except for the extreme case.
I couldn’t find a picture, but here is one for the Earth–Moon:
Note that only the red orbit lies in the plane; bigger ones get successively tilted.
My understanding of the particular instability of that lagrange point is that it is unstable in the radial direction – a stationary (in the rotating frame) object just barely earth-ward of L2 will fall towards earth/sun, and one just barely space-ward of L2 will escape towards deep space.
But in for objects near L2 that are neither earth-ward nor space-ward (i.e. in a plane perpendicular to the earth-sun-L2 axis), the point is stable – nearby stationary (in the rotating frame) objects will fall towards the point.
So I would presume that the orbit is being kept close to that plane* because, intuitively, in that plane, L2 looks the most like an ordinary gravitating object.
*but actually slightly earth-ward of that plane, with the falling towards earth bit corrected by thrusters, because falling towards earth is a much easier problem to solve than falling towards space.
Almost correct. An object just slightly inward or outward of the Lagrange point won’t “fall to Earth” or “fall to space”: It’ll stay in orbit around the Sun, just in a much messier orbit, rather than the nice neat Lagrange-point orbit.
And it has to stay on the inside of the Lagrange point, as it doesn’t have any maneuvering thrusters on the far side, so if it were to tip over that point, it would be difficult if not impossible to retrieve it.
At the least, they’d have to flip it over, having the telescope on the sun side of the shield, and then wait months for it to cool off again, assuming that nothing broke.
To get a feel for the orbit of this telescope, I’ve downloaded the Astronomy program Celestia and an addon representing its projected orbit,which was presumably reasonably accurate when the addon was made back in January. Perhaps I should not have been surprised to find that the Sun does not ever align with the Earth when seen from the telescope, so there will be no eclipses there. An L2 orbit would be perfectly adequate as a location to observe the Sun without interruption, but L1 has the advantage of being closer.
Another mildly interesting aspect is that the Earth appears slightly larger than the Sun, so could cover the entire stellar disc if such an eclipse were to happen.
Meaning what? This seems to mean something important, but “inside” in what sense?
Example: at Julian Day 2459730.500000000, the telescope is predicted to be 1645884 km from the Earth-Moon barycenter, but the Lagrange point will be only 1528453 km from the center.
The orbit of the telescope is huge. A good deal bigger than the orbit of the moon around the earth. So, it orbits around the shadow cast by the earth/moon. They did this so the solar power is never interrupted.
We can never perfectly orbit L2. The telescope will always be ever so slightly “inside” (closer to earth/sun) or “outside” (on the other side of L2, closer to the outer solar system).
The telescope will drift either closer to the earth or away from the earth depending on where it is (inside or outside, so to speak). As a result the telescope has maneuvering rockets which will give occasional nudges to keep it in place.
However, if it was “outside” of L2 it would need to push itself back towards earth. But, it literally cannot do that. It has no thrusters to push it “back.” So, the orbit is purposely on the “inside” of L2 so it always drifts closer to earth and the thrusters can nudge it away from earth (IIRC it will nudge itself once every 22 days or so).
This is false, or at least you do not literally mean that. Some of the time it is closer to the earth/sun than L2, and some of the time it is farther away.
I buy that it would be sent on an orbit that perturbations would tend to pull it closer. But what determines that?
I think part of your confusion is that the orbit is on the same plane as all of the planets (near enough). It isn’t. Webb orbits perpendicular to the orbital plane (near enough…you can see it in the thumbnail picture below). Perhaps it is sometimes on the “outside” of L2 but, overall, it is more on the “inside.”
I posted this video before but I will post it again, queued to the section that is relevant to how it orbits “inside” of L2. The whole video is 15 minutes but you need only watch about 1.5 minutes from where they talk about this (linked below to the correct time):
If you did take a satellite in any possible halo orbit around L2, would it not still be overall on the “inside” in your sense? How could it be outside?
That video also seems vaguely familiar, and it does not seem to discuss which, of all the possible, halo orbits will drift towards the sun, and which will go in the wrong direction in the sense that sun-oriented thrusters could not be used to stabilise them.
Take the plane of the telescope’s orbit (in the corotating frame). If the Lagrange point itself were perfectly in that plane, it’d stay balanced, and wouldn’t need course correction. But of course, perfect alignment isn’t possible, and if it’s just a little bit out of alignment, it’ll drift further out, so it needs thrusters to counteract that drift. By design, the telescope’s orbit is such that the Lagrange point is always slightly on the far side of the plane.
I think it does.
The L2 Lagrange point moves with the earth. Anything between the earth and L2 is “inside” the L2 point. Further out, past the L2 point is “outside” the L2 point. The video notes that Webb does orbit a little past the L2 point (its orbit is inclined…see pic above) BUT…the center point of its whole orbit is, crucially, “inside” (between L2 and earth) so, it will always drift towards the earth. If that center point were “outside” the L2 point then it would drift slowly off into deep space, away from the earth.
They very much want it to always drift closer to earth so its orbit is designed to ensure that is what happens. If it did otherwise the mission would be ruined.
Your (and @Chronos 's) qualitative explanations make sense, but just to be sure…
There are many theoretical halo orbits around L2, ranging from more or less in-plane, to increasingly tilted to the north or south. The center point of any such orbit will not be quite at L2; its x-coordinate will be offset slightly in the direction of the Earth. (This begs for a proof, as the equilibrium point is at L2 itself.) Is that correct, so far?
Now, to start, let’s consider a halo orbit that lies in the plane, or close to it. It will be pretty unstable and tend to drift away. You seem to be saying that this drift will always be towards, not away, from the Earth? Or are you saying that the conditions can be, and in the case of the Webb were, carefully engineered so that the drift would be in the right direction?
The center point (for some suitable definition of “center”) of a true halo orbit will be at L2. An orbit whose plane is shifted slightly off of L2 won’t be a true halo orbit. Such an orbit can be shifted either slightly inward of L2, or slightly outward. In either case, because of the instability, it will need thrust, in the direction towards L2, to keep it in that orbit. The JWST is in one of these not-quite-true halo orbits. Since the JWST has thrusters only on the inward side, it has to be a slightly inward orbit. One could also design a spacecraft that had thrusters on the other side, and so had to be slightly outward, or a spacecraft that had thrusters on both sides, and so could be off in either direction, but that’s not the design used for the JWST.
But not for long, as perturbations push it away from that point.
For many missions this would make more sense, that way, you don’t have to worry about overshooting the orbit.
But for JWST, having thrusters on the telescope side would be difficult or even impossible.