My understanding of the JWST trajectory timetable (or the bits that I care about for the purpose of this question) is the following:

  1. Launched by Ariane rocket directly into escape trajectory (from earth).
  2. Unfold (for a month), with occasional trajectory modifications.
  3. More time passes, with more trajectory modifications.
  4. Arrive in operational (unstable) halo orbit around earth-sun L2.

But I recall something about a "preliminary orbit" in the broadcast. And it seems to me that, if the Ariane upper stage can fire a second time after a delay, it would make sense to perform some checks in orbit before launching the telescope onto an escape trajectory -- ranging from "deploy solar panels and check power levels" to "unfold everything" depending on how much delay the upper stage can tolerate. (Although I'm guessing that "unfold everything" would give the liquid hydrogen fuel plenty of time to boil off.)

Looking online, I have a hard time finding a direct statement answering this question; most of the information that comes up in my search is about the final orbit.

Does/did the JWST have any kind of stable earth orbit (with no rockets firing for at least, say, 60s) before being propelled into an earth escape trajectory, and if so, what was done during this time?

  • $\begingroup$ The JWST, in deployed form, would be incapable of function and indeed can easily be damaged, just by the heat from the Earth and the irregular sunlight of a low earth orbit. Additionally, once deployed it changes its center of mass, making thrusting it further using a high thrust system like the Ariane second stage impractical. So no deploying, even of the solar power system, until acceleration is done. And without power, it would die within hours, so if a checkout failled, it would just leave a nonfunctional and nonrescueable hunk of junk in LEO rather than out there. $\endgroup$ Dec 27, 2021 at 10:07

1 Answer 1


There was no parking orbit (which is I think what you're inferring), indicated by a continuous burn of the 2nd stage.

More over, it was launched not on an Earth-escape trajectory, just a really high apogee Earth orbit. This may be considered the preliminary orbit. I believe I remember hearing in the NASA Webcast the apogee was ~1.3 million km. This is pretty consistent with values from the Ariane 5 User's Manual:

Ariane 5 performance

(JWST was launched towards and will end up orbiting the Sun-Earth-L2 point)

And Brown, J. et al. "Seasonal Variations of the James Webb Space Telescope Orbital Dynamics," (2015), NTRS ID: 20150017756:

There are three possible flight programs (FP) that have been selected, each corresponding to a different apogee radius for the injection orbit: 1.02 x 10$^6$ km for FP1; 1.06 x 10$^6$ km for FP2; and 1.10 x 10$^6$ km for FP3.

The reason for not launching directly to the Earth-escape libration point orbit (LPO) is because the spacecraft is incapable of 'slowing down' or removing energy (speed) from its trajectory. This is because all of the thrusters are on the hot-side of the sunshield and thus only produce thrust in the direction of the velocity vector. The mission designers biased the launch trajectory to a lower energy to reduce the risk of launch vehicle performance uncertainties putting the spacecraft in a unrecoverable state (too much energy).

In other words, JWST is trying to climb a gravitational hill (or well) away from Earth and land on the crest$^1$ of it, without brakes to stop it if it overshoots. Mission designers aimed a little below the crest and will use JWST's thrusters to give it the final (comparatively very small) push to the LPO.

The spacecraft performed a small burn (MCC-1a) approximately 12.5 hours after launch. This burn was similarly biased shorter than required. The next burn, MCC-1b, will remove this bias and basically put JWST into the LPO$^2$ @Launch+ 2 days. This could be considered the point where Webb 'escapes' Earth, though any formal definition of this Earth-escape point is unlikely given the n-body dynamics critical to the operational orbit.

  1. JWST receives a significant (relatively speaking) force from solar radiation pressure so it is biased towards the Earth side of the gravity well with the solar radiation pressure and routine station keeping maneuvers keeping it in its LPO.

  2. Based on my interpretation of figure 9 from Brown, J. et al. and this statement from Peterson, J. et al. "James Webb Space Telescope Initial Mid-Course Correction Monte Carlo Implementation using Task Parallelism," (2014), NTRS ID: 20140008868:

MCC-2 is the smallest of the three MCC maneuvers and is effectively the first station keeping maneuver.

  • 2
    $\begingroup$ Why is it incapable of slowing down? Couldn't it perform a rotate-burn-rotate routine? I understand that there could be many reasons why they do not want to do that, but I don't see why it would be impossible. $\endgroup$
    – JohnEye
    Dec 27, 2021 at 18:19
  • 3
    $\begingroup$ @JohnEye Rotating in the direction required to do that would point the cold side of the telescope right at the Sun, frying all of the sensitive instruments. $\endgroup$ Dec 27, 2021 at 20:53

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