3
$\begingroup$

I’ve been reading up on two and three body dynamics, wrt to L2 for the JWST.

Are the calculations for this L2 based on the sun/earth masses and locations, or on the sun/(earth+moon barycenter) masses and locations?

Along the same lines, how about using the solar system barycenter instead of the sun’s center of mass for M1?

Can anyone provide a order of magnitude estimate of the difference of using barycenters, vs sun/earth masses and distances in calculating the L2 for the JWST?

Perhaps the perturbations introduced by barycenter calculations are insignificant, but surely the actual location of L2 is more complicated than assuming a simplified 2-major-mass 3 body problem?

Improve this question
$\endgroup$
1
  • $\begingroup$ As I mentioned here, the point that JPL consider relevant is the L2 point of the Sun and the Earth-Moon barycentre. $\endgroup$
    – PM 2Ring
    Commented Jan 21, 2022 at 6:50

1 Answer 1

Reset to default
3
$\begingroup$

Barycenter is just another name for center of mass, although usually in the context of orbiting bodies. At a large distance from all the bodies in a system, the gravitational field can be approximated as coming from a point source positioned at the barycenter and containing the total mass of the system. However, this approximation is not useful inside the system. In this case, the individual fields (and their directions) from each body have to be calculated and summed.

Since the Sun-Earth L2 is well inside the solar system, the solar system barycenter is not useful. The Earth/Moon does not orbit around the solar barycenter. It orbits around the Sun with perturbations from Jupiter and the other planets.

The Earth/Moon barycenter is a little more relevant. L2 is at a distance four times the size of the Earth/Moon system. Between the two extremes of its orbit, the Moon is either 25% closer to L2 or 25% further from L2. The gravitational field from the Moon is either 56% stronger or 44% weaker - which seems a lot . However the Moon has only 1/80th the mass of the Earth, so its not that bad an approximation, the error is only around 0.5%.

The spacecraft trajectory from Earth to L2 is certainly calculated numerically summing the individual fields from the Earth, the Moon, the Sun, and probably Jupiter and any other significant graviational sources in the solar system. The positions of the these bodies at the time of launch will be important, particulary the position on the Earth and the positions of the Sun and Moon which are obviously very important.

The Earth/Moon barycenter as an approximate field source during stationkeeping could be helpful. But, again, my guess is that the real fields from the relevant bodies are calculated plus the estimated solar pressure and these are all summed. With the tools and computational capabilities available nowadays. these calculations are straightforward.

In conclusion, the answer is no. The solar system barycenter is not relevant at all and direct calculations of the fields are easy enough and so much more accurate that using the Earth/Moon barycenter as an approximation would not be helful either.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ thank you for this. If I understand correctly, keeping JWST correctly placed (station keeping) will always be an n-body problem requiring numerical integration, and there isn’t an analytical solution for locating L2 (since the sun, earth, moon, jupiter (and more), and solar wind pressure, etc, need to be included in the calculation). The use of centers of mass (barycenters) as approximations would not eliminate the need for numerical integrations to accommodate the perturbations due to massive planets or other forces. $\endgroup$
    – Bruce Simonson
    Commented Jan 21, 2022 at 6:41
  • 1
    $\begingroup$ @Bruce S That's right. Strictly L2 is only defined for two bodies. Part of the problem, too, is that the errors from even tiny forces accumulate given sufficient time. At L2 there's no damping mechanism and restoring force that will passively brings the orbit back to where you want it. $\endgroup$
    – Roger Wood
    Commented Jan 21, 2022 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.