I am an electrical engineer and not an astrophysicist so excuse my simple question.

NASA says that the journey time for the JWST to L2 will be about thirty days. However assuming the orbit dynamics can be computed using a simple two-body case then the period of any elliptical orbit enclosing a distance of approximately one and a half million kilometres is about seventy-six days. This gives a one-way journey time to L2 from Earth of about thirty-eight days.

If all single orbits enclosing that distance must have the same period how is NASA achieving it in only thirty days? Is this where multiple orbits come into play or is the difference due to the problem being more complicated than a two-body one and the Sun and rotating frames have to be accounted for, or perhaps both?

  • 6
    $\begingroup$ I haven't tried to do the sums to prove this, but an L2 trajectory is very much not a two body problem. After all, L2 marks the point at which the a body will start orbiting the sun and not the Earth. As a 3 body problem, you'll need to solve numerically, there isn't a closed formula for the trajectory from LEO to L2. You might try playing with an Orbiter simulator. $\endgroup$
    – James K
    Commented Jan 2, 2022 at 11:27
  • $\begingroup$ @JamesK some simple calculations I did seemed to show that solar gravity is cancelled out by centrifugal force when using a rotating frame of reference, leaving the issue of the Coriolis Effect. But surely those forces are deminimus wrt the Earth’s gravity? In which case two-body kinematics would provide a reasonable approximation? $\endgroup$
    – adlibber
    Commented Jan 2, 2022 at 12:17
  • $\begingroup$ solar gravity is only cancelled by centrifugal force at 1 AU, my point is that Webb is transiting from a terrestial orbit to a solar orbit, albeit one that is in a 1:1 resonace with Earth. So to model this transition 3 body methods would be needed. As I said, I haven't checked, but I certainly don't think that it is obvious that you can just ignore solar gravity in modelling this orbit - and I rather suspect that if you include solar gravity, you'll get this orbit. $\endgroup$
    – James K
    Commented Jan 2, 2022 at 12:32
  • 2
    $\begingroup$ @JamesK The JWST can't break for orbital insertion (no way to turn the engine forward). It needs to constantly undershoot the target and requires several correction burns to just get captured by L2. $\endgroup$
    – Zac67
    Commented Jan 2, 2022 at 13:13

1 Answer 1


This is certainly a three-body problem. The transfer starts out as a planetary orbit but finishes as a combined planetary-solar orbit. There are no Lagrange points in a two-body system.

The Webb moves away from both Earth and Sun, so both are pulling back and slowing it down[*1]. Therefore, it needs to enter that transfer orbit (somewhat Hohmann transfer) with more speed than with a plain two-body calculation - making it complete the distance a bit faster. Thirty days instead of the thirty-eight you calculated sounds about right.

[*1] At that distance, the Sun's pull would actually be greater than that of the Earth, but both the Earth and the JWST are in orbit around the Sun to start with. However, the Webb gains potential energy against the Sun and needs to trade in kinetic energy.

  • $\begingroup$ I’ve just looked at the mission schedule and after the second correction burn done at day 2 there’s not another until day 29 so three weeks ballistic! I take it that the two early burns modified the shape of the orbit so that the long three week one’s apogee must be just a short way from L2. You are saying that this orbit can only be computed numerically? $\endgroup$
    – adlibber
    Commented Jan 2, 2022 at 17:41
  • $\begingroup$ There's no general solution for the three-body problem, but what's the problem with numerical computing? Computations aren't a problem and all the relevant bodies in the solar system are accounted for. $\endgroup$
    – Zac67
    Commented Jan 2, 2022 at 17:57
  • $\begingroup$ Moore’s Law has vanquished the problem of numerical computation but when I was a student in the early sixties, not so easy! Fingers crossed that all goes well for the next three weeks. Thanks. $\endgroup$
    – adlibber
    Commented Jan 2, 2022 at 18:13
  • 4
    $\begingroup$ @adlibber You are looking at a trajectory that starts overwhelmingly under the influence of one body, and ends with the other being dominant. No, you can't simplify this to a 2-body problem. $\endgroup$ Commented Jan 5, 2022 at 20:47
  • 2
    $\begingroup$ @adlibber The sole existence of a langrange point itself is expression of a 3-body problem - thus in their distance the deviation from the central gravity potential become non-negligible. On the other hand, if you are fine to ingore that for a 0th order calculation, you should also be fine with the deviation of your result to the actual result. $\endgroup$ Commented Jan 6, 2022 at 10:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .