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I read this Phys.SE thread which is similar Why are L4 and L5 lagrangian points stable?

but I did not want to necro that thread. It seems that most discussions of a three body problem are presented in two dimensions. I am thinking about the set of points where two cones intersect. One cone with the vertex at the sun and described by the angle sub tended by L4-sun-L5. The other cone defined by L4-earth-L5. The intersection of these two cones is a circle that includes L4 and L5. This circle is the set of locations where the gravity of the sun and earth cancel out. This circle is perpendicular to a line connecting the earth and sun. if we are considering L4 and L5 as stable points, why are we excluding all the other points on this circle?

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At L4 and L5, the object would stay in the same relative position because it is in the same orbit around the Sun as Earth. Any other points on the circle would not be in the same orbit as Earth. They would be in orbits of differing inclination and so, in the course of the orbit of that object about the Sun, it would necessarily leave those points. The L4 and L5 points are ones where the object can stay indefinitely. This is not true for any other points on the circle.

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    $\begingroup$ Thank you. Yeah orbital mechanics- the points on the circle would be a valid gateway for transferring from geocentric to heliocentric (and vice versa) but would not be a stable static location $\endgroup$ – aquagremlin Jun 4 '15 at 19:01
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This circle is the set of locations where the gravity of the sun and earth cancel out.

Uhm. No. They are neither co-linear nor the same magnitude. To get a zero "force" at the Lagrange points you have to work in a rotating frame of reference, implying a centrifugal pseudo-force which closes the triangle.

But the points on the circle you defined other than L4 and L5 have the wrong centrifugal terms. Both the wrong magnitude (because they lie a different cylindrical) radius and the wrong direction.

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