How much margin in $\frac{m}{s}$ does JWST have for station keeping in Halo orbit?
JWST must always stay on the earth side of the L2 saddle since thrusters only point towards the sun. Also` solar radiation adds a significant force away from the sun. Also, the moon's changing orbital position relative to L2 is going to nudge JWST slightly differently at different points in the Halo orbit.
Thus determining the optimal station keeping thrust time and direction must be pretty complicated. The final MCC-2 burn on Jan24th was a mere 1.6 meters per second (5 minute burn, 3.6 miles per hour) nasa blog. Subsequent station keeping burns will be even shorter.
StackExchange: About the stability, L2 is unstable in the radial direction: if the probe is a little closer or a little further in the Sun-Earth axis it will be pushed yet further by gravitation. However L2 is stable in the perpendicular plane ...
If JWST thrusters were ever fired too long in station keeping so that JWST wound up on the far side of the saddle, then JWST would eventually drift out of the L2 orbit forever since JWST cannot correct by turning around and firing its thrusters away from the sun since the super cold mirrors and instrumentation would get fried by the direct sunlight. So I was curious how NASA figure out exactly how close the L2 saddle JWST is, and how much margin in $\frac{m}{s}$ the NASA engineers leave so that JWST always stays on the Earth side of the L2 Halo orbit, especially given the difficult to model radiation pressure and changing moon position.
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