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Do Lagrangian points collect micro-meteorites similar to how the Pacific trash vortex collects debris?

I was wondering if the James Webb Telescope will have to clear its Lagrange point before deploying its components. Is this a likely scenario?

Gabriel Fair
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L2 isn't a super stable point, but only quasi-stable. Things can't stay at that point for a long period of time without some work to stay there. Estimates say that number is around 5-16 m/s, depending on the object exactly. Sufficeth to say, most objects don't thrust to stay there naturally, so there isn't a lot of dust actually trapped at any particular L2 (Earth-Sun or otherwise).

PearsonArtPhoto
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    Examining maps of total potential energy (which includes both gravitational potential and Coriolis potential) is very useful. The L4 and L5 "points" are indeed local potential minima and objects orbiting there can be stably bound to those points. But L1 and L2 are "saddle points": they are minima for the radial direction, but are not minima for the tangential direction. Orbits there are not stable without some form of control, i.e. without control miniscule errors will eventually diverge resulting in escape from the region. Control was a critical part of NASA's Genesis mission. – Tom Spilker Jan 13 '20 at 20:58
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    @TomSpilker answers to Are some Halo Orbits actually Stable? say some are, but not small ones. I don't know if JWST's halo orbit is large enough to be in this regime or not. Of course those are only stable in the constraints of the CR3BP and not the real world. – uhoh Jan 14 '20 at 01:47
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    @uhoh The perturbations outside of the CR3BP scenario are sufficient to be the "miniscule errors". These include lunar perturbations and, surprisingly, Jupiter perturbations. – Tom Spilker Jan 14 '20 at 01:54
  • Any idea after running out of fuel, how long will JWST stay at L2 before it wobbles away? – matohak Dec 27 '21 at 21:55
  • @TomSpilker The L4 and L5 "points" are actually local maxima which are stabilized by the Coriolis effect. L1 and L2 are saddles (which I think can't be Coriolis-stablized) and the in-out direction is too steep to be stabilized. Of course this is in 2D, in the z-direction everything is stable so we ignore z usually. – Kevin Kostlan Oct 13 '22 at 03:33