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Kepler's third law states that $$\frac{r^3}{T^2} = \frac{G (M+m)}{4 \pi^2}$$ for circular orbits. I know that all Kepler's laws work well when the smaller bodies don't have a significant gravitational interaction, otherwise there's the need to add corrections.

Now, imagine launching a satellite on a circular orbit around our planet. Its orbit lies on the Earth-Moon plane and it passes near L1 and L2. Intuitively if the distance between Earth and the satellite is slightly bigger than the distance between Earth and L1, the satellite will pass through the Moon's gravity well and that can disrupt its orbit (probably ejection or capture by the Moon). But what if it remains slightly closer to Earth than L1 and L2?

These two Lagrangian points have the same period as the Moon, i.e. 28-29 days, but the satellite will have a smaller period. So can I still rely on Kepler's third law to calculate the period knowing the distance, or vice versa, in such a situation?

For simplicity, let's consider the Moon's orbit as circular too.

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  • $\begingroup$ Any orbit that passes near L1 and L2 isn't likely to be circular anymore. The gravitational force will be very different when passing near the Moon, and will distort the orbit. $\endgroup$
    – probably_someone
    Commented Nov 22, 2019 at 13:39

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Your proposed path is one in which the gravitational interactions from the Earth, the Moon, and the Sun are all important. Kepler's Third Law assumes a two-body system where the satellite is one of the bodies, and you have a four-body system, so don't expect it to work very well, or at all.

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    $\begingroup$ This answer is correct but to emphasize as related to the original question: The Lagrange points don’t really have much to do with this. Their existence is a clear sign that you don’t have a two-body problem anymore, but you probably had significant third and fourth body effects well before you reached those points. $\endgroup$
    – Brick
    Commented Nov 22, 2019 at 14:35
  • $\begingroup$ @Brick Yes, that's what I had hoped my comment would convey. $\endgroup$
    – probably_someone
    Commented Nov 22, 2019 at 14:47

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