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Of the five Lagrange points, L4 and L5, as stable points, can be orbited by asteroids, satellites, and any other useful or interesting object. Assuming two-body motion however, calculating orbits with all the equations I'm familiar with relies on knowing the mass of the body being orbited. Obviously, a point in space where gravity perfectly balances out doesn't exactly have a mass, so what would be used as the "mass" of the L4 or L5 Lagrange points?

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    $\begingroup$ Read your link carefully, the two bodies defining the Lagrange points are large, the satellites at the points are small. If the mass of the satellite is not much smaller than the mass of both bodies, the theory of Lagrange is not applicable. $\endgroup$
    – Uwe
    Aug 22 at 22:23
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    $\begingroup$ The question assumes the Lagrange points act as gravitating objects...they don't. The bodies being orbited are the two that define the Lagrange points, and just can't be treated as Keplerian orbits. Here's 2010 TK7's orbit around the Earth-Sun L4 point: en.wikipedia.org/wiki/File:Animation_of_2010_TK7.gif 3753 Cruithne's orbit is even more complex, oscillating through several points: en.wikipedia.org/wiki/File:Animation_of_3753_Cruithne_orbit.gif $\endgroup$ Aug 22 at 22:47
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    $\begingroup$ Orbits around L4/L5 are neither circular nor eliptic, so Kepler laws are not even remotely applicable. In the general case, they are bean-shaped when the two main bodies are considered fixed. $\endgroup$
    – fraxinus
    Aug 23 at 11:21
  • $\begingroup$ You can see that a Lagrange point can't have the same effect as a point mass, by considering what happens to a satellite exactly at the Lagrange point. If the Lagrange point were a point mass, the satellite would be subject to an infinite gravitational attraction, and could never leave. But obviously this is not the case for a real Lagrange point. $\endgroup$
    – TonyK
    Aug 23 at 15:21
  • $\begingroup$ Duplicate? $\endgroup$
    – SF.
    Aug 26 at 10:14
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Lagrange points... can be orbited by asteroids, satellites, and any other useful or interesting object. Assuming two-body motion however...

First a quick note; the stability of a Lagrange point itself is not a predictor of the stability of an orbit associated with ("around") it.

I'm confident that the OP knows there isn't any real mass at a Lagrange point to which objects are gravitationally attracted, but it is interesting to wonder if something like an "effective mass" can be derived.

We have to keep reminding ourselves that orbits associated with Lagrange points are really orbits around the primary body that are simply in 1:1 resonance with the secondary.

See also this answer

I think it's easiest to talk about halo orbits associated with (around) Sun-Earth L1, but this will apply to any Lagrange point in a circular, restricted three-body system.

At 1.5 million km from Earth, the SE L1 is 1% closer to the Sun than Earth, so normally it should have a 1.5% shorter period and orbit the Sun 0.5% faster. But being 99 times closer to the Earth than the Sun, it is influenced by the Earth's gravity.

This is just enough to lock the object into a 1:1 resonant orbit; the object moves in a wavy orbit around the Sun with the same period as the Earth, sometimes speeding up a little then getting pulled back behind by it.

It's only when we look in the synodic frame, a rotating frame in which the Sun and Earth appear to be fixed, that it looks like the object is orbiting the Lagrange point. It's kind of an optical illusion; it really isn't.

Answer(s) to What sort of orbital elements are used to describe halo orbits? explain that there really aren't proper orbital elements about Lagrange points because these are really not proper orbits to begin with.

They are three-body dances.


Here is an animation of JWST's orbit. It's about L2 not L4 or L5, and it's not to scale, but it at least helps illustrate that orbits associated with Lagrange points are about the primary body, and when viewed in an inertial frame are definitely not "around" the Lagrange point.

Lower your volume (or turn off) before playing:

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  • $\begingroup$ Well, the orbit of the Moon is also primarily around the Sun and not the Earth ... $\endgroup$ Aug 24 at 5:17
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    $\begingroup$ @HagenvonEitzen certainly the Moon is more deeply bound to the Sun than to the Earth, but it also truly orbits the Earth 12 or 13 times every year, so I don't really know how "orbits primarily" can even be defined in an objective way for which there is general agreement. $\endgroup$
    – uhoh
    Aug 24 at 6:57
  • $\begingroup$ Well, italics or not: "Truly orbits" is an assertion rather than a reason ;-). When all will be said and done, in many billion years, the Earth's rotation will be tidally locked to its orbit, and the moon will be further out and will be locked to that rotation (assuming that's not outside Earth's Hill sphere). Its orbit will resemble an orbit around a Lagrange point. I think there is still a substantial difference but playing devil's advocate I'm wondering what exactly the difference is. $\endgroup$ Aug 24 at 11:12
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When we consider the mass something is orbiting about, we assume a gravitational potential corresponding to an inverse-square force, which corresponds to either a point mass or, by Newton's shell theorem, a spherically symmetric mass. It is this situation where any 2-body orbital dynamics can be reduced to the Kepler laws, with periods depending on the semimajor axis.

In any other potential, the dynamics will be different. The effective gravitational potential near the Lagrange points is better described by Taylor expansion as a polynomial, which is very different from the inverse-square one with its -∞ dip at zero. So it makes no sense to ascribe this potential a “mass”.

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  • $\begingroup$ +1 This is going to be the right answer. Searching "linearize" [orbital-mechanics] returns two posts $\endgroup$
    – uhoh
    Aug 23 at 13:48
  • $\begingroup$ It takes some speed to get out of gravity well, and likewise it takes some speed to get out of stable Lagrange point. I think "mass" analogy has some intuitive value. $\endgroup$
    – Vashu
    Aug 25 at 0:55
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    $\begingroup$ @Vashu but the speed required to get out of a 1/r² gravitational field is not dependent on mass only, it's also dependent on initial distance from the center. Specifically, even a very small mass can have arbitrarily large escape velocity if it's concentrated in a small enough space and you start at near-zero radius. $\endgroup$ Aug 25 at 7:17
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While asking for an "effective mass $M$" of a Lagrange point $P$ is meaningless, it still makes sense to ask about the effective field (known as little-$g$) in a neighborhood of $P$. It is (minus) the gradient (of the sum) of the gravitational and the centrifugal potential, cf. Fig. 1.

Lagrange Points

$\uparrow$ Fig. 1: The 2D orbital plane with equipotential lines and the 5 Lagrange points. (From Wikipedia.)

The effective field vanishes by definition at $P$. ($\leftarrow$ The previous sentence is essentially why it does not make to ask for an "effective mass $M$" of $P$.)

Restricting for simplicity to motion in the 2D orbital plane, a test mass $m$ (without propulsion) will not drift in the direction of the field but rather perpendicular to it (along equipotential lines) due to the Coriolis force. The drift is often superimposed with a circular motion of angular velocity $-2\Omega$, cf. Fig. 2.

$\uparrow$ Fig. 2: A possible horseshoe orbit along equipotential lines. $\star$ is the Sun, while $\bullet$ is Jupiter.

More generally, if $m$ is situated away from the 2D orbital plane, we should superimpose an oscillatory motion perpendicular to the plane.

For a derivation, explicit formulas, and more information, see e.g. part III of my Phys.SE answer here.

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