In the ongoing thread about JWST James Webb Space Telescope general discussion thread - #401 by Senegoid a side question arose–but was not satisfactorally answered of whether the telescope speeds up, resp. slows down, as it reaches the end of the end of the minor, resp. major, axis of its elliptical orbit around the L2 point. One argument was that the usual proof of Kepler’s third law is based on the gravitational attraction of the sun on a planet or a planet on a satellite, but there is no massive body at the L2 point. I would counter that it is not the mass that attracts a body, but the warpage of space/time created by the mass and, in order to allow an elliptical orbit, the same warpage must exist near the L2 point. Then I said something about conservation of angular momentum, but I have to admit that I don’t really understand what that means. A spinning ice skater speeds up but moving their arms closer to the body and there is no gravitational force involved there at all.
So what is the real answer in the case of the JWST? Does it speed up and slow down just as though there was a mass there?
I doubt it does it in exactly the same way it would if there was a mass there. It acts as it does because of the curvature of spacetime. Point masses and masses with spherical symmetry create curvatures with spherical symmetry. At least some distance away from L2, the curvature of L2 isn’t spherically symmetric. Also, as you get closer to a point mass or a mass much smaller than your distance from it, the curvature gets more intense the closer you get, whereas near L2 I think it doesn’t.
So, somewhat like there’s a mass there, but different in ways…? Not much of an answer, but the best one as of when I typed it.
There’s nothing particularly interesting going on with the spacetime at that point. Lagrange points are purely an artifact of the non-inertial coordinate system chosen.
That said, I’m unsure of the answer to the question asked, and it’s tough to settle it from first principles, because it’s unclear which of those principles relevantly applies. Sure, angular momentum is conserved, but that doesn’t help much, because there are other bodies involved to exchange angular momentum with.
IANAP, but this is precisely the question I tried to address in this post:
If any of this is wrong, as I said there, I’d be interested in understanding why. I think the “curvature of space” perspective isn’t particularly helpful, but the gravitational potential vs centrifugal force perspective seems to be compelling. This is all in reference to an elliptical halo orbit like the Webb’s which resembles an ordinary Keplerian orbit. Despite the absence of a mass at L2, the JWST is drawn there because it’s orbiting through an area of higher gravitational potential in the earth-sun system. This is similar to a weight on a string that describes an elliptical orbit because it’s drawn to the lower gravitational potential at its natural rest point, even though there’s nothing physically there.
ETA: I should add: the nature of L2 is that it’s the highest point of local gravitational potential along the sun-earth axis. It’s “downhill” on either side. But that’s not where JWST is orbiting. It’s orbit is more perpendicular to that axis. It’s downhill in the direction of its center at L2, just like in a Keplerian orbit around its barycenter.
You certainly don’t need to resort to space-time curvature to explain Lagrange points. Lagrange derived their behavior under Newton’s theory of gravity. That doesn’t mean you can’t of course, but space time curvature isn’t necessary to explain it.
I need to correct this because it was poorly stated and misleading. Yes, an object at L2 sees both sides along the sun-earth axis as “downhill”, but it’s obviously only the earthward side that has lower gravitational potential – this is the “falls back to earth if it doesn’t reach L2” scenario. On the other side, it’s pulled away by the centrifugal force of its higher orbit. L2 is the balance of the two.
Is anyone going to work it out analytically for us? I think there is no point in worrying about space-time warpage; I think what you want to do is something like, first change to a rotating system of coordinates which follows the orbit of the Earth around the Sun (the rotation rate will be given by Kepler’s law) so the Sun and the Earth are stationary. Then you obtain a pseudo-potential as the sum of the gravitational potential and the centrifugal force.
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Then you have to see whether it is bigger or smaller at points along your orbit. Note that a periodic halo orbit is actually pretty big (100000s of km radius) and very noticeably not quite elliptical. It is hard to tell from the picture, but it does look like it is slower at the ends of the “major axis” and fastest at the ends of the “minor axis”.
That works fine, if the Coriolis force and other higher-order non-potential psuedoforces aren’t relevant. I’m not at all confident that that’s the case. Honestly, without an extensive background that I don’t think any of us here have, the only approach I’d trust here would be a three-body simulation in the inertial frame.
Hey, yeah, you’re right. Didn’t think of that. It would tend to perturb the halo orbit to be smaller along the line to the Sun and larger on that perpendicular, wouldn’t it?
Anyway, there are many effects, radiation pressure, etc. we did not even think about, which is OK if you want to understand how the orbit looks to within a couple of %, but if we are going to check the numerical simulation anyway then we may as well cut to the chase; if you go to Horizons they have the orbit predicted 2 years ahead. Put in Target = -170, Coordinate center = @32. It outputs positions, velocities, range, and rate of change of range by default, but you will still need to convert it yourself to a rotating reference frame if you want to see the quasi-ellipse.
How can they be stable if it is downhill in all the directions? The Coriolis force must be taken into consideration. Note that in this figure, L2 is a saddle point.
More precisely, before you talk about perturbing a halo orbit, we need to establish the existence and properties of halo orbits in the first place. I don’t think the Coriolis force represents a perturbation so much as a term in the actual equations of motion corresponding to the simplified 3-body problem in question.
I read somewhere that a minimal halo orbit in the orbital plane has an amplitude of about 14% of the length scale along the x-axis, and 3 times that along the y-axis, so we are pretty far from hovering at the Lagrange point itself or describing a tiny ellipse around it.
The simple question arising from the Webb thread seems to have generated an amazing amount of controversy. Are we now agreed that the answer is “yes” – that an object in an elliptical halo orbit around L1 or L2 will be moving slowest at it farthest distance and fastest at its closest distance, just like in a Keplerian orbit around a center of mass?
I still maintain that one can deduce this from first principles as I described, though I may well be wrong about that. Nevertheless, it appears that the conclusion is correct, and an object in a halo orbit behaves in some ways like the elliptical orbit of a weight on a string.
Here is a simulation of the halo orbit of ISEE-3 (later renamed ICE – International Cometary Explorer) around L1, where this is evident.
It’s clear enough for the Webb as well; I was going to upload a plot but ran into some technical difficulties, but you can download the same data points yourself from the JPL web site and check.
One textbook approach, starting with assuming the two primary bodies revolve in a circle, is to work in the coordinate system where the xy plane is rotating; after appropriate normalization the square of the velocity, as a function of position, equals 2F(x,y,z)-C, where C is a constant, and
but of course that is just the beginning and you still need to analyse what happens around the Lagrangian points (to prove that you can even construct halo orbits under certain conditions, for starters)
Of course you can derive it from first principles. But to do so, you have to know what principles to use. And I reiterate that nobody without extensive experience with Lagrange-type orbits is likely to know what principles they need to start from.
The video below explains how the JWST orbits L2. It’s interesting but even simplified it is not intuitive. Rocket scientists earned their money on this one.
For those here who understand the math I think the video can help answer the OP.
I doubt the mathematical principles [classical mechanics] present much difficulty (I mean to an astrodynamicist who works this stuff out in the shower, not necessarily to me), but there is still the matter of us interpreting them correctly.
Consider, for example, the case of an orbit around a stable Lagrangian point (so not the Webb case). You can have a small orbit in that case; what does it look like? To first order, you can get something that does not merely look like an ellipse, rather an ellipse superimposed on top of another (smaller) ellipse. Is the resultant wobbling enough to screw up the smooth monotonic acceleration you get along a Keplerian orbit?
For halo orbits, as we remarked, they are not so small, but you can play the same game (maybe better done with a computer already) and expand to, say, third order.
I think it is clear, though, that
not, that is, these orbits are not simply ellipses around the Lagrange point as if there were an invisible mass there.
I’m not sure what you’re trying to say here with the word “not”. You seem to be contradicting your earlier agreement that the ISEE-3 Lagrangian orbit (I linked to an animation) as well as the Webb exhibited exactly this speedup and slowdown in an elliptical orbit. You can also see it here in a view of the WMAP orbit as seen from earth.
I think there are two separate questions here.
Q: Is a Lagrangian orbit the same as a Keplerian orbit around a center of mass?
A: No! A Langrangian orbit at L1 or L2 is generally a Lissajous or halo orbit that is metastable and needs periodic adjustments.
Q: Does an object in a roughly elliptical Lagrangian orbit speed up and slow down as in a Keplerian orbit, constantly exchanging gravitational potential energy and kinetic energy?
A: Yes. Whether or not you like my conceptual explanation, one can see from the animated models that this is true. These two screen captures from the WMAP orbit around L2 demonstrate this clearly. Look at the indicated velocity at the two points in its orbit.
I think there is no point in worrying about space-time warpage; I think what you want to do is something like, first change to a rotating system of coordinates which follows the orbit of the Earth around the Sun (the rotation rate will be given by Kepler’s law) so the Sun and the Earth are stationary. Then you obtain a pseudo-potential as the sum of the gravitational potential and the centrifugal force.
