How much margin does JWST have for station keeping in Halo orbit?

Question

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.

if this is a duplicate please post the link to where this question is answered in the comments, and can delete this question.

Answer 1

You are correct that if JWST "crosses the saddle" it would enter an unstable manifold and leave SEJ2 in an anti-sunward direction.

But if necessary, JWST could "turn around and face the sun" for the several minutes of an orbital maintenance burn without irreparable damage. The entire spacecraft would heat up to (very roughly) the average temperature of Earth and would likely take weeks to cool back down to operational temperature after correcting its orientation. Nothing would get "fried by direct sunlight".

Remember, the entire spacecraft was held at room temperature for years during fabrication and launch. It didn't start cooling down until after the sun shield was deployed. Some components must be very cold to function, but that doesn't mean they will be destroyed if unrefrigerated.

The only danger in this maneuver is if the image of an IR source (Sun, Moon, Earth) is accidentally focused by the primary mirror onto the secondary mirror. But since the angular diameter of the sun and moon are only a half a degree, this should be avoidable.

So, to answer your question, they have wiggle room since crossing the saddle would be an embarrassing loss of observation time, but not a mission killer.