# How can I reconcile the JWST journey time of thirty days to Lagrange L2 with simple orbital mechanics?

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?

• 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. Jan 2, 2022 at 11:27
• @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? Jan 2, 2022 at 12:17
• 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. Jan 2, 2022 at 12:32
• @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. Jan 2, 2022 at 13:13