Via a link from New Scientist, I got to a page about Neptune's Trojan asteroids. According to the author, a handful are known, all found in the last five years. One of them, newly discovered, is in an orbit tilted 25° from the ecliptic. Which is to say, 25° from the orbit of Neptune.

As I understand it, all five of Lagrange's special-case solutions to the three-body problem involve two co-planar orbits. In the real world, there's a little slack, just as not all of Jupiter's Trojan asteroids are exactly 60° ahead or behind the planet. But 25 degrees inclination seems like a lot to count as at the Lagrangian points. Does anyone know if someone has done the math for that set of orbits? I can't see how this geometry is stable: right now, 2005 TO74 is at or near Neptune's L4 point, but eventually it's going to be 25 degrees north or south of the ecliptic; with Neptune staying in the ecliptic, and at that orbital radius, we're looking at a couple of hundred millions of kilometers north or south of the L4 point. I thought the reason the Lagrange points are interesting and potentially important is that they're small areas: a body at L4 will be consistently in the same direction, and distance, from the primary.
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