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As L2 orbits the sun at more than 67,000 mph, JWST will approach that orbit asymptotically . In doing this, will L2 be behind JWST and catch up with the satellite or will it be ahead of JWST and JWST will catch up with it?

In either case, what is the relative velocity of these two 'objects' when JWST approaches L2?

Joe Lorimo
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    JWST will overshoot L2 a bit. then swing back gravitationally. Look at the halo orbit shape from all angles. It rotates around L2 vertically, but also wobbles in and out by a couple hundred thousand km each cycle. Thrusters finetune the orbit, with about 12 orbital adjustments per cycle (ideally 1 correction every 22 days, but more often initially) – CuteKItty_pleaseStopBArking Dec 31 '21 at 16:50
  • different but related and potentially somewhat helpful: The design of the halo orbit of the James Webb Space Telescope As CuteKitty points out it doesn't go to L2 (~1,500,000 km from Earth) but instead ends up in a really big inclined egg-shaped "orbit" around it varying between 400,000 and 800,000 from L2. In the rotating frame with Sun and Earth fixed, it moves around L2 at very roughly 0.2 km/sec. This isn't meant as an answer but it may be a little helpful. Rather than "comes to rest at L2" it "orbits around it" in the rotating frame. – uhoh Dec 31 '21 at 17:21
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    @uhoh To expand on that - it won't even orbit L2, it will always stay a bit on our side of it (energy wise) to make sure it doesn't drift away from us. – asdfex Dec 31 '21 at 17:55
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    @asdfex Good point, yes! And will manage to stay there via solar photon pressure. – uhoh Dec 31 '21 at 18:05
  • @asdfex,"it will always stay abit on our side"... I don't know how to reconcile your statement with an example of a representative trajectory of JWST given here: https://jwst-docs.stsci.edu/jwst-observatory-characteristics/jwst-orbit. If "a bit on our side" means that its X-coordinate is always kept (a bit) negative, then this contradict what is taught here. – Ng Ph Jan 02 '22 at 14:54
  • @NgPh that's why I added "energy wise". A bit on our side of the energy "hill" at L2. It can be at positive x but only with an energy that is lower than required for a stable orbit at this distance. – asdfex Jan 02 '22 at 15:06
  • @cutekitty_pleasestopbarking, perhaps it should be emphasized that the "wobbling by a couple of hundred thousands of km" around the L2 point is a design feature of JWST's science orbit. JWST's station-keeping goal is not the same as that of a GEO satellite, for example. For GEO, the goal is to keep the sat as close to the center of a nominal box as possible (with tens of km for the sides of the box). JWST has to remain in the "vicinity" of L2 (tolerance=x10^4 of GEO), but must also avoid the center area of L2 (for Sun shawowing reasons). – Ng Ph Jan 02 '22 at 15:09
  • @asdfex, Ok for that. But how is it that "it won't orbit L2" because it stays a bit down hill (energy wise). How is it that, to orbit L2, it has to be at the top of the energy hill? – Ng Ph Jan 02 '22 at 17:08
  • @NgPh Well, that gets a bit complicated, as there's no such thing as an orbit around L2, so the whole definition gets a bit hand waving... How can it orbit something that doesn't exist and that doesn't exercises any attractive force? It wobbles force and back in the vicinity of the L2 point. If you look at the orbit in detail ( https://jwst-docs.stsci.edu/jwst-observatory-characteristics/jwst-orbit ), you'll also see that it always stays on one side of L2 (plot b), which a real orbit couldn't. All we can say: At any point in time it has less energy than an object orbiting Sun directly at L2. – asdfex Jan 02 '22 at 17:31
  • @asdfex, I dispute the "no such thing as an an orbit around L2", although I agree that it's complicated. Even more so, when trying to explain with intuitive notions/graphics/animated pics. Although I understand the specific constraint of JWST's station-keeping (its thrusters can only fire in the negative X direction), it gives me headache to grasp why this means it must have "less energy than an object at L2", at any time. Thanks for the effort, BTW. – Ng Ph Jan 02 '22 at 22:41
  • @NgPh It boils down to that "moves in a circle around L2 when viewed from Earth" and "orbits L2" are two separate things. But the detailed discussion seems to be misplaced in these comments :) – asdfex Jan 03 '22 at 10:52

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Question: will JWST catch up with L2 or will L2 catch up with JWST?

Answer: To a first approximation, both. And neither. The answer depends on the frame of reference you chose.

Case 1: In a rotating frame of reference (Sun and Earth stationary), the Sun, Earth and L2 are collinear so JWST follows a radial path from Earth towards L2.

enter image description here

Rotating frames of reference produce apparent violations of the laws of motion. Try playing ping pong on a merry-go-round for a demonstration.

In a non-rotating frame of reference, things appear different.

Case 2: In the Ecliptic Coordinate System (Sun centered, non-rotating), the L2 point is in a non-Keplerian, heliocentric “orbit”. It is outside Earth’s orbit, collinear with the Sun and Earth. Its heliocentric orbital period is one year. From this frame of reference, the transfer trajectory takes JWST 1,000,000 miles radially and 49,000,0000 miles circumferentially. Since JWST is in freefall, increasing its distance from both the Earth and the Sun , it must be slowing down. So, from this frame of reference, JWST is catching up to L2. enter image description here

Case 3: In the Equatorial Coordinate System (Earth centered, non-rotating), L2 is in a 1 year, 1,000,000 mile orbit around Earth. L2 is traveling across the celestial sphere at 30$^\circ$/month. So in the (very approximately) 1 month from launch to L2 orbit, JWST appears to need to “catch up” 30$^\circ$ to J2. In this frame of reference, JWST is catching up to L2.

Fun facts:

L2 is at the meeting of Earth and Sun “spheres of gravitational influence”. So if you tried to model JWST trajectory with patched conic orbits (like Kerbal Space Program), it would fail.

The L2 point is in two different 1 year prograde “orbits” at the same time, around both Earth and the Sun.

Since L2s heliocentric “orbit” is 1% larger than the Earth’s, its orbital speed is 1% greater than the Earth’s and (roughly) 2% greater than an object in a Keplerian orbit at the same distance from the Sun.

Woody
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  • If 2 cars going in the same direction collide, the police man can say which is moving faster. If 2 asteroids collide in empty space, which one moves faster is impossible to answer w/o agreeing on the observer's position. An observer at the center of the Earth sees L2 as (quasi) fix and JWST seems moving. For him, JWST does "catch up". An observer at the center of the Sun sees both objects revolving, and at the same time one (JWST) moving radially away. How you conclude that (in Sun-centered Ecliptic Coordinate) L2 is "catching up" JWST is difficult for me to grasp. – Ng Ph Jan 02 '22 at 10:17
  • @Ng Ph I edited the Ecliptic Coordinate paragraph. See if it makes any more sense. Basically, JWST is slowing down. It eventually rendezvous with L2 which has not changed velocity. I think you are looking at the rendezvous from a rotating frame of reference (rotating around the center of the sun). From that frame, there is only apparent radial motion, but no "catch up". The Ecliptic coordinate system does not rotate, – Woody Jan 04 '22 at 00:36
  • Sorry, but I still have a problem! If the observer, sitting put at the Sun center, measures the radial distances of L2 and JWST, he will conclude that JWST is moving towards the (fix) radial distance of L2. If he observes the movements of L2 and JWST on the Celestial Sphere (hence, ignoring the radial distance), he will see that JWST leaves the L2's trajectory (at launch from Earth) and then slowly moves back, to stay in the vicinity of this point's trajectory. I still can't see how he concludes which one is catching up the other, from their paths wrt to the stars. – Ng Ph Jan 04 '22 at 11:18
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    You are describing a rotating frame of reference (Case 1) in which there is only radial motion; no “catch up” occurs. We are in agreement on this point. However, in a non-rotating inertial frame of reference (Case 2), JWSR is losing speed to make the rendezvous (while Earth and J2 maintain their speed). Therefore J2 is “catching up” to JWST. The newly edited diagram shows the initial velocity vector for JWST longer than J2’s. But the final JWST velocity vector must match J2’s. JWST must slow down during its free-fall transit because it is getting further from both the Earth and the Sun. – Woody Jan 04 '22 at 16:14
  • Your new diagram tends to demonstrate that the blue vector always has a smaller module than the red vector. Yet, you conclude that Blue(L2) catches up Red(JWST). A contradiction it seems (?). – Ng Ph Jan 04 '22 at 17:50
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    You're right. I wrote it backasswards. JWST is catching up to L2. Also, I'm dyslexic and mix up my J's and L's. I keep writing J2 instead of L2. – Woody Jan 04 '22 at 22:14
  • Thanks for the edits. – Ng Ph Jan 05 '22 at 10:27