What value of M should I use to calculate the speed of a satellite orbiting a Lagrangian point where there exists no mass?

  • $\begingroup$ Speculation on my part: Central and orbiting body exert easily calculated forces on a point at a given location. So called centrifugal force is also in the mix. All components of these forces parallel to the barycentric line cancel out. What you're left with is the sum of the gravity components pulling towards the L point. The halo orbit's ω^2r needs to cancel this force pulling towards the L point. $\endgroup$
    – HopDavid
    Apr 24 '17 at 18:58
  • $\begingroup$ See also physics.stackexchange.com/questions/36092/… $\endgroup$ Apr 24 '17 at 19:35
  • $\begingroup$ @HopDavid centrifugal force is used only if you are solving the problem in the rotating frame, and that's done only as a mathematical short cut. There's nothing about the three body problem that requires changing into a rotating frames or using effective potentials except the elegance, efficacy, and tradition (habit). $\endgroup$
    – uhoh
    Apr 24 '17 at 23:59
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    $\begingroup$ @uhoh Something in a halo orbit is in a rotating frame. $\endgroup$
    – HopDavid
    Apr 25 '17 at 20:38
  • $\begingroup$ @HopDavid To me a rotating frame is just an abstract concept, a tool humans use when calculating, or a check box on a visualization program. It's the same orbit when viewed in any frame one chooses. Viewed by a person in a rotating frame, it looks more "halo-ish" than viewed in an inertial frame, but the fact that a satellite is 1% closer to the sun and yet not revolving any faster around it than the Earth (on average), and wiggles strangely up and down out of the ecliptic means the halo is still there. Frames exist in our heads. It's not like different objects/orbits are "in" different frames $\endgroup$
    – uhoh
    Apr 25 '17 at 22:58

This video shows a telescope in a Halo or Lissajous orbit near the Sun-Earth L2 point. By moving slowly and tracing its path in both inertial and rotating (synodic) frames, you can see that the orbit is primarily around the sun, and the "orbit around" the libration point is a construct that comes from doing the math or thinking in a rotating frame:

Turn your audio down or off first!

Here's a simplified way to look at the situation:

Orbits are the paths that bodies follow in response to forces.

Forces are the gradients of potentials.

For a small planet around a big star, the potential is

$$\phi(r) \ = \ GM\frac{1}{r}$$

so the force felt by the planet is

$$F(r) \ = \ -GM\frac{1}{r^2}$$

pointed towards the star. When you read about Keplerian orbits, you're reading about only this situation, or a slightly modified version where the planet is bigger and you treat the two as orbiting around their center of mass.

In vector notation, you can change the force to

$$\mathbf{F}(\mathbf{r}) \ = \ -GM\frac{\mathbf{r}}{r^3}$$

If you calculate the trajectory of bodies with various starting conditions in this force field, they will all be Keplerian orbits.

In a three-body problem, and lets stick with the simple "circular restricted three body problem (CR3BP)" where there are two main bodies (a sun and a large planet for example) in circular orbits around their center of mass, the potential field and resulting force felt by the third body (so small you can treat its mass as zero) is

$$\phi(\mathbf{r}) \ = \ GM_1\frac{1}{r_1}+GM_2\frac{1}{r_2} $$

$$\mathbf{F}(\mathbf{r}) \ = \ -GM_1\frac{\mathbf{r_1}}{r_1^3}-GM_2\frac{\mathbf{r_2}}{r_2^3}$$

where $\mathbf{r_1}$ and $\mathbf{r_2}$ are the vectors which point from each of the two main bodies to the third body. Now the shape of the potential is constantly changing. Since the problem is kept simple by requiring the two bodies to be in circular orbits around each other, the potential field is just rotating.

This means that the force field is also rotating, and the orbit of the third body will be determined by its initial velocity and the acceleration due to that rotating force field.

The resulting orbit can be all kinds of crazy things, but it won't be a nice Keplerian orbit, and so none of that math applies.

To mathematically solve for orbits in this problem, mathematicians and orbital mechanics often switch to doing their math in a rotating frame, one that rotates with the two main bodies. In this frame they create a pseudopotential with a term called "centrifugal potential" which is really just a way to pretend the frame is not rotating.

In other words, in the rotating frame, along with the real gravitational potential from the two bodies (above) which is not affected by the rotation, and who's gradient produces the force field as shown above, an effective potential term is added such that its gradient produces a fictitious centrifugal force.

Resulting orbits can be all kinds of crazy shapes, but since they are not solutions of a simple 1/r potential field, they won't be Keplerian, and so you can't try to represent them with a central mass M.

  • $\begingroup$ Since the orbits aren't Keplerian, can a higher (larger) halo orbit relative to a Lagrangian point have a higher speed than a lower one? The distance to the L-point doesn't directly occur in the math. $\endgroup$
    – LocalFluff
    Apr 26 '17 at 8:18
  • $\begingroup$ If this comment didn't answer your question, better ask a new one! (note it's $2\pi$ and not $2\pi2\pi$). $\endgroup$
    – uhoh
    Apr 26 '17 at 8:20
  • $\begingroup$ @LocalFluff There are animated versions of these images somewhere but I can't find them now. When you look at these kinds of libration point orbits in an inertial frame, you see that they are just orbiting around the sun as I've describe here. This may also be helpful: ccar.colorado.edu/geryon/Slides/Slide03.html $\endgroup$
    – uhoh
    May 2 '17 at 9:58
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    $\begingroup$ I see. They just seem to appear to orbit a point. But they are just changing their inclination and Solar distance to make them look "as if" they orbit an invisible balance point. $\endgroup$
    – LocalFluff
    May 2 '17 at 10:43
  • $\begingroup$ I think that can be a good way to say it. The phrase "to orbit" something really just means go around it, and it doesn't say anything about the cause, but as @HopDavid suggests here the "going around" of the point only really happens if you look at it in a rotating (synodic) frame. In a fixed frame, it's really going around the Sun, and "wiggling" near the Lagrange point. The video I found has a nice way of showing two different traces, one that rotates, the other that doesn't. $\endgroup$
    – uhoh
    May 2 '17 at 10:53

Closed trajectories around the stable Lagrangian points (L4 and L5) are not elliptical and do not follow Kepler's laws, so there's no value of M you can use.

The same is true for the quasi-periodic Lissajous "orbits" around the unstable points L1, L2 and L3.

  • $\begingroup$ Bingo. You are no longer dealing with a two-body problem. $\endgroup$
    – Erik
    Apr 24 '17 at 19:43
  • $\begingroup$ Could a higher "orbit" have shorter period? $\endgroup$
    – LocalFluff
    Apr 24 '17 at 19:59
  • $\begingroup$ @pericynthion in the CR3BP L1, L2, L3 also have periodic, closed halo orbits, in addition to the funky Lissajous orbits. In the real n-body world, no orbit is strictly closed or strictly periodic. $\endgroup$
    – uhoh
    Apr 25 '17 at 0:32
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    $\begingroup$ @LocalFluff These halo orbits (as many do) have a period of roughly half of the main two-body orbit (which is 2π2π in this dimensionless example), and in fact become shorter as the orbit becomes "higher" (in this case farther from L2). So if you look at satellites in halo or Lissajous orbits around an Sun-Earth L1 or L2, they have, roughly speaking about five or six month orbit "periods", where period is fairly well defined for a halo, and less-so for a Lissajous. There's all kinds of other quasi-periodic ones as well. $\endgroup$
    – uhoh
    Apr 25 '17 at 0:35
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    $\begingroup$ @LocalFluff yep, yep, nope not at all, yep, yep, yep, and yep. It's not any kind of a law effect. It's just what happens in this particular kind of 3 body orbit. No rules, lots, and lots of surprises. Unless you dig deep into the math, you just have to throw up your hands and sing youtu.be/azxoVRTwlNg?t=24 $\endgroup$
    – uhoh
    Apr 26 '17 at 13:20

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