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A planet in an elliptical orbit around a star conserves angular momentum. (This is tantamount to saying that it sweeps out equal areas of the ellipse in equal times). If conditions are such that there are objects gathered at the $L_4$ and $L_5$ points of that orbit, do those objects also conserve angular momentum? They do of course in the case of the circular orbit, but that is rather trivial. What about elliptical orbits with considerable eccentricity?

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At $L_4$ and $L_5$ you are orbiting the larger body in the system (so at the Sun-Earth $L_4$/$L_5$ you're orbiting the sun). You orbit the larger body with a slightly greater semi major axis than that of the orbit of the smaller body, so typically in a two body system you would expect your object to have a greater orbital period than it does.

As you have stated from Kepler's second law, the area swept out in a given time is constant for a elliptical orbit. This still holds true for an object at $L_4$/$L_5$ since they will be moving constantly around the larger body (sun). A good way of looking at this is that if the smaller body (Earth) does it then the object does it, just with a true anomaly 60 degrees greater or lower than the smaller body (Earth).

Also by way of observation you can see that the object at $L_4$/$L_5$ will (by definition) maintain it's position relative to the smaller body (Earth) and hence will have a constant angular momentum since the smaller body (Earth) will have a constant angular momentum.

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    $\begingroup$ You should note you orbit the larger body at any Lagrangian points $\endgroup$ – ThePlanMan Jun 10 '14 at 8:35
  • $\begingroup$ Hmmm, I'm not buying it. Your comments are true for the circular case, but it is not true that the L4/L5 objects remain fixed relative to the small (or large) mass by definition. See for example this other question: space.stackexchange.com/questions/2877/… There you will find a link to a paper which finds that L4/L5 objects in an eccentric orbit will be found at the vertices of an isosceles triangle--implying that their distance from the larger masses will be changing constantly. $\endgroup$ – ben Jun 10 '14 at 18:43
  • $\begingroup$ Sorry, I mustn't have made myself clear, you stay in position relative to the smaller body, not the larger body. $\endgroup$ – ThePlanMan Jun 10 '14 at 21:21
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They do of course in the case of the circular orbit, but that is rather trivial. What about elliptical orbits with considerable eccentricity?

Lagrange points aren't really defined for elliptical orbits. They are defined only in the Circular Restricted Three Body Problem (CRTBP or CR3BP). Two bodies have significant masses and the third does not (that's the restriction; it doesn't affect the motion of the other two) and the motion of each of the two main bodies is circular and centered on their common center of mass.

So your question is a sequitur; the term Lagrange point can't apply.

What would happen is that the objects gathered near the areas we might want to call $L_4$ and $L_5$ would do whatever complicated dance they do, and their angular momentum around the Sun-Earth barycenter would vary over time, exchanging with that of the Earth and the Sun. However in the CR3BP we ignore those changes in the Earth and Sun's motion because that's the "restriction".

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With an elliptical orbit we can't have any fixed Lagrange points, not even the unstable ones aligned with the massive bodies, because the massive bodues are not fixed rekative to each other unless we make a very contrived reference frame. We can, however, conceive of trojan-like objects with the following properties:

*The mean orbit period about a primary object matches that of a more massive secondary nidy.

*$\angle SPX$, where S is the secondary mass, P the primary mass, and X the object being considered, always lies between a minimum that is strictly greater than $0°$ and a maximum that is strictly less than $180°$ (thus $X$ is locked to "one side" of $\overline{SP}$).

In a circular orbiting system the above properties are specific to trojan objects. For instance, $\text{2010 TK}_7$ in the nearly circular orbit of Earth covers a wide range of angles, but this answer shows a picture of the orbit of $\text{2010 TK}_7$ avoiding the $0°$ and $180°$angles relative to the Sun-Earth axis.

We know that for six out of the eight planets in our Solar System, the planetary orbits are close enough to circular and other perturbations are sufficiently at bay to enable the existence of objects with trojan-like properties. The real question is at what point (if any point less than 1) the eccentricity gets big enough to kill it.

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