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How precisely you need to ‘park’ on a Lagrange point? There’s obviously a singular, ideal Lagrange point, but I would assume there’s something like a practical

How precisely you need to ‘park’ on a Lagrange point? There’s obviously a singular, ideal Lagrange point, but I would assume there’s something like a practically viable sphere around it where you would benefit from the advantages. Given the vast distances, I would assume that this viable sphere is sufficiently large to accommodate more than enough man-made space infrastructure.

PartiallyTyped 3y ago
IIRC JWST is not actually at the Lagrange point, but it drifts a bit, and occasionally fixes itself.
- pests 3y ago
  A sibling comment to yours points out that it actually orbits the L2 at the same distance the moon orbits the Earth and takes 6 months to complete a revolution.
  - PartiallyTyped 3y ago
    Yes, the orbit isn't exact / perfect/ stable either [1]. > An L2 orbit is unstable, so JWST needs to use propellant to maintain its halo orbit around L2 (known as station-keeping) to prevent the telescope from drifting away from its orbital position.[186] It was designed to carry enough propellant for 10 years,[187] but the precision of the Ariane 5 launch and the first midcourse correction were credited with saving enough onboard fuel that JWST may be able to maintain its orbit for around 20 years instead.[188][189][190] [1] https://en.wikipedia.org/wiki/James_Webb_Space_Telescope https://en.wikipedia.org/wiki/James_Webb_Space_Telescope
ano-ther 3y ago
Seems to be quite roomy, at least at L2: > Webb orbits around L2; it does not sit stationary precisely at L2. Webb's orbit is represented in this screenshot from our deployment video (below), roughly to scale; it is actually similar in size to the Moon's orbit around the Earth! This orbit (which takes Webb about 6 months to complete once) keeps the telescope out of the shadows of both the Earth and Moon. https://jwst.nasa.gov/content/about/orbit.html https://jwst.nasa.gov/content/about/orbit.html
- lostmsu 3y ago
  If L2 is unstable, how can you orbit around it? Is it a point of attraction? If so, how is it not stable?
  - lm28469 3y ago
    It's a point of equilibrium, a point of equilibrium can be stable or unstable A glass bead at the bottom of a bowl is in a stable equilibrium, if you shake the bowl side to side it'll come back at the bottom. A basketball spinning on your index finger is in an unstable equilibrium, it stays there while the conditions are met and any small deviation will make it fall
    - lostmsu 3y ago
      You just basically restated the problem, and given no explanation for L2 specifics. L2 is called unstable above. So orbiting around it instead of hovering immediately at it should be like trying to spin a ball slightly off the tip of your finger. E.g. you wouldn't want to do it. Sibling replies have proper explanations.
  - deleted 3y ago
    [deleted]
  - marcosdumay 3y ago
    Wikipedia for the help: https://en.wikipedia.org/wiki/File:Lagrange_points2.svg https://en.wikipedia.org/wiki/File:Lagrange_points2.svg The idea of plainly stable and unstable points only happen if you have rotational symmetry, and doesn't really apply to most of the points. Leading the discussion on those terms isn't productive. It orbites in a plane where it's always subject to attraction, but that doesn't mean the point always attracts the satellite.
  - actinium226 3y ago
    L1, L2, and L3 have 3 "modes" associated with them, where "mode" refers to what happens to a satellite placed near those points with just the right initial conditions (position and velocity relative to the point). Since you can't place a satellite perfectly in practice, its motion will always be a combination of the modes, but in theory the 3 modes are: stable, unstable, and oscillatory. Like I said, since you can't place a satellite perfectly in practice, its motion will always be a combination of the 3 modes, and eventually the unstable mode will dominate and the satellite will move away from L1/2/3. BUT, if you get the math right you can position the satellite so that most of its motion is in the oscillatory mode (i.e. orbiting the point), and then as it drifts into the unstable mode you nudge it back to the oscillatory mode with a thruster firing. For context, the amount of thruster firings necessary to keep the ARTEMIS probes around Earth-Moon L1/2 was about 5m/s per year.
dexwiz 3y ago
L2 is unstable, so you need some amount of active thrust to stay in the orbit. The point is relatively close to the moon, and changes enough to not collect natural bodies. Same with L1 and L3. Once the James Webb runs out of fuel it will leave the point quickly, probably within 30 days. L4 and L5 are way out from the Earth Moon system and are very stable. They collect natural satellites, and once parked there you would remain indefinitely. The actual math is pretty complex, and has to do with mass ratios of bodies involved, but all the major bodies in our solar system have their own Lagrange points.
- actinium226 3y ago
  > L2 is unstable, so you need some amount of active thrust to stay in the orbit True > The point is relatively close to the moon, and changes enough to not collect natural bodies. This is awkwardly written. Let's get some background first: the Lagrange points are derived by assuming the Moon's orbit around the Earth is circular. In actual fact it's elliptical, so the distance from the Earth to the Moon changes and consequently the distance from the Moon to L2 changes. The distance from the Earth to the Moon averages 384,400km and at that point L2 is 64,000km away from the Moon. But the reason no natural bodies (like asteroids) collect around L2 is because it's unstable, not because it moves. > L4 and L5 are way out from the Earth Moon system L4 and L5 form an equilateral triangle with the Earth and the Moon, meaning they are (on average) 384,400km away from both the Earth and the Moon at any given time. > and are very stable. No, they are "meta-stable." If you drift away from L4/5 you will not be drawn back to the Lagrange point itself, but you can orbit those points without extra fuel. We see asteroids gathering at Sun-Jupiter L4/5, so that suggests they are weakly stable, but when you do the linearized stability analysis of those points you find that they are neither stable nor unstable. > The actual math is pretty complex It is. I made an effort to create some educational videos on the topic: https://www.youtube.com/watch?v=8q9in-VfpUM https://www.youtube.com/watch?v=8q9in-VfpUM
  - ttyprintk 3y ago
    My understanding is that earth-sun L4/5 also collect; not asteroids but dust https://en.wikipedia.org/wiki/Kordylewski_cloud https://en.wikipedia.org/wiki/Kordylewski_cloud I believe you’d have to avoid the point itself and orbit around it. Edit: apparently, there are two Earth Trojans: https://en.wikipedia.org/wiki/List_of_objects_at_Lagrange_points https://en.wikipedia.org/wiki/List_of_objects_at_Lagrange_po...
JKCalhoun 3y ago
I believe you can even orbit a Lagrange point.
evrydayhustling 3y ago
Seems clear from references in other comments that the usable sphere is large enough that Lagrange points can be "shared" by many parties (within most conceptions of current equipment size and interference). What about the opposite - how feasible is it to "control" a Lagrange point even if you wanted to exclude other equipment?
- calamari4065 3y ago
  At best, wildly impractical. At minimum, you need weapons coverage over a complete sphere about the point you want to protect. If your claim is larger than your spacecraft, you need multiple in a 3D perimeter. But even supposing you had enough weapons, it would take much more effort to destroy all the shrapnel before it hits you. And even then, if someone decides to throw something very heavy at you very fast, there's sweet FA you can do about it. If I throw a big satellite or a ship into a kamikaze vector at you, your one and only option is to move out of the way as fast as you can. Even if you hit it with all your weapons at once, you still have almost as much mass flying at you almost as fast, but now it's buckshot instead of a cannon ball. It really isn't practical to defend any amount of space, not even a planet. Controlling a Lagrange point would have to be a political mechanism. You can't really physically enforce it in space. You'd end up fighting wars on the ground over a point in space.
  - rocqua 3y ago
    Wouldn't a randomly moving platform with 4 gatling guns on there give full-sphere coverage and defense against passive weapons by the random movement? You have fuel issues, but that can be solved by rotating shifts, or (vulnerable) in-place refueling. If the Lagrange point is indeed unstable the random movement should be rather efficient to achieve. And kinetic weapons would need luck to hit you. For explosive weapons you might be able to rely on the gatling guns. It seems to me like the Lagrange points might be one of the few things in space you could occupy and deny an enemy. Perhaps LEO should added to that because it can be hit from the ground.

PartiallyTyped 3y ago

IIRC JWST is not actually at the Lagrange point, but it drifts a bit, and occasionally fixes itself.

pests 3y ago
A sibling comment to yours points out that it actually orbits the L2 at the same distance the moon orbits the Earth and takes 6 months to complete a revolution.
- PartiallyTyped 3y ago
  Yes, the orbit isn't exact / perfect/ stable either [1]. > An L2 orbit is unstable, so JWST needs to use propellant to maintain its halo orbit around L2 (known as station-keeping) to prevent the telescope from drifting away from its orbital position.[186] It was designed to carry enough propellant for 10 years,[187] but the precision of the Ariane 5 launch and the first midcourse correction were credited with saving enough onboard fuel that JWST may be able to maintain its orbit for around 20 years instead.[188][189][190] [1] https://en.wikipedia.org/wiki/James_Webb_Space_Telescope https://en.wikipedia.org/wiki/James_Webb_Space_Telescope

ano-ther 3y ago

Seems to be quite roomy, at least at L2: > Webb orbits around L2; it does not sit stationary precisely at L2. Webb's orbit is represented in this screenshot from our deployment video (below), roughly to scale; it is actually similar in size to the Moon's orbit around the Earth! This orbit (which takes Webb about 6 months to complete once) keeps the telescope out of the shadows of both the Earth and Moon. https://jwst.nasa.gov/content/about/orbit.html https://jwst.nasa.gov/content/about/orbit.html

lostmsu 3y ago
If L2 is unstable, how can you orbit around it? Is it a point of attraction? If so, how is it not stable?
- lm28469 3y ago
  It's a point of equilibrium, a point of equilibrium can be stable or unstable A glass bead at the bottom of a bowl is in a stable equilibrium, if you shake the bowl side to side it'll come back at the bottom. A basketball spinning on your index finger is in an unstable equilibrium, it stays there while the conditions are met and any small deviation will make it fall
  - lostmsu 3y ago
    You just basically restated the problem, and given no explanation for L2 specifics. L2 is called unstable above. So orbiting around it instead of hovering immediately at it should be like trying to spin a ball slightly off the tip of your finger. E.g. you wouldn't want to do it. Sibling replies have proper explanations.
- deleted 3y ago
  [deleted]
- marcosdumay 3y ago
  Wikipedia for the help: https://en.wikipedia.org/wiki/File:Lagrange_points2.svg https://en.wikipedia.org/wiki/File:Lagrange_points2.svg The idea of plainly stable and unstable points only happen if you have rotational symmetry, and doesn't really apply to most of the points. Leading the discussion on those terms isn't productive. It orbites in a plane where it's always subject to attraction, but that doesn't mean the point always attracts the satellite.
- actinium226 3y ago
  L1, L2, and L3 have 3 "modes" associated with them, where "mode" refers to what happens to a satellite placed near those points with just the right initial conditions (position and velocity relative to the point). Since you can't place a satellite perfectly in practice, its motion will always be a combination of the modes, but in theory the 3 modes are: stable, unstable, and oscillatory. Like I said, since you can't place a satellite perfectly in practice, its motion will always be a combination of the 3 modes, and eventually the unstable mode will dominate and the satellite will move away from L1/2/3. BUT, if you get the math right you can position the satellite so that most of its motion is in the oscillatory mode (i.e. orbiting the point), and then as it drifts into the unstable mode you nudge it back to the oscillatory mode with a thruster firing. For context, the amount of thruster firings necessary to keep the ARTEMIS probes around Earth-Moon L1/2 was about 5m/s per year.

dexwiz 3y ago

L2 is unstable, so you need some amount of active thrust to stay in the orbit. The point is relatively close to the moon, and changes enough to not collect natural bodies. Same with L1 and L3. Once the James Webb runs out of fuel it will leave the point quickly, probably within 30 days. L4 and L5 are way out from the Earth Moon system and are very stable. They collect natural satellites, and once parked there you would remain indefinitely. The actual math is pretty complex, and has to do with mass ratios of bodies involved, but all the major bodies in our solar system have their own Lagrange points.

actinium226 3y ago
> L2 is unstable, so you need some amount of active thrust to stay in the orbit True > The point is relatively close to the moon, and changes enough to not collect natural bodies. This is awkwardly written. Let's get some background first: the Lagrange points are derived by assuming the Moon's orbit around the Earth is circular. In actual fact it's elliptical, so the distance from the Earth to the Moon changes and consequently the distance from the Moon to L2 changes. The distance from the Earth to the Moon averages 384,400km and at that point L2 is 64,000km away from the Moon. But the reason no natural bodies (like asteroids) collect around L2 is because it's unstable, not because it moves. > L4 and L5 are way out from the Earth Moon system L4 and L5 form an equilateral triangle with the Earth and the Moon, meaning they are (on average) 384,400km away from both the Earth and the Moon at any given time. > and are very stable. No, they are "meta-stable." If you drift away from L4/5 you will not be drawn back to the Lagrange point itself, but you can orbit those points without extra fuel. We see asteroids gathering at Sun-Jupiter L4/5, so that suggests they are weakly stable, but when you do the linearized stability analysis of those points you find that they are neither stable nor unstable. > The actual math is pretty complex It is. I made an effort to create some educational videos on the topic: https://www.youtube.com/watch?v=8q9in-VfpUM https://www.youtube.com/watch?v=8q9in-VfpUM
- ttyprintk 3y ago
  My understanding is that earth-sun L4/5 also collect; not asteroids but dust https://en.wikipedia.org/wiki/Kordylewski_cloud https://en.wikipedia.org/wiki/Kordylewski_cloud I believe you’d have to avoid the point itself and orbit around it. Edit: apparently, there are two Earth Trojans: https://en.wikipedia.org/wiki/List_of_objects_at_Lagrange_points https://en.wikipedia.org/wiki/List_of_objects_at_Lagrange_po...

JKCalhoun 3y ago

I believe you can even orbit a Lagrange point.

evrydayhustling 3y ago

Seems clear from references in other comments that the usable sphere is large enough that Lagrange points can be "shared" by many parties (within most conceptions of current equipment size and interference). What about the opposite - how feasible is it to "control" a Lagrange point even if you wanted to exclude other equipment?

calamari4065 3y ago
At best, wildly impractical. At minimum, you need weapons coverage over a complete sphere about the point you want to protect. If your claim is larger than your spacecraft, you need multiple in a 3D perimeter. But even supposing you had enough weapons, it would take much more effort to destroy all the shrapnel before it hits you. And even then, if someone decides to throw something very heavy at you very fast, there's sweet FA you can do about it. If I throw a big satellite or a ship into a kamikaze vector at you, your one and only option is to move out of the way as fast as you can. Even if you hit it with all your weapons at once, you still have almost as much mass flying at you almost as fast, but now it's buckshot instead of a cannon ball. It really isn't practical to defend any amount of space, not even a planet. Controlling a Lagrange point would have to be a political mechanism. You can't really physically enforce it in space. You'd end up fighting wars on the ground over a point in space.
- rocqua 3y ago
  Wouldn't a randomly moving platform with 4 gatling guns on there give full-sphere coverage and defense against passive weapons by the random movement? You have fuel issues, but that can be solved by rotating shifts, or (vulnerable) in-place refueling. If the Lagrange point is indeed unstable the random movement should be rather efficient to achieve. And kinetic weapons would need luck to hit you. For explosive weapons you might be able to rely on the gatling guns. It seems to me like the Lagrange points might be one of the few things in space you could occupy and deny an enemy. Perhaps LEO should added to that because it can be hit from the ground.