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The Apollo lunar missions used a free return trajectory which resembles a figure-8 when plotted in an Earth/Moon rotating frame of reference. This trajectory was chosen for safety: if the mission was scrubbed before the lunar insertion burn, the spacecraft would return to Earth “for free” (minimal delta-v).

The following diagrams are conceptual only, and far out of scale.

enter image description here

https://forum.nasaspaceflight.com/index.php?topic=16758.80

The cross-over point appears very close to the Earth/Moon Lagrange point (where total gravitational attraction equals centrifugal force in a rotating frame of reference).

httpenter image description heres://en.wikipedia.org/wiki/Lagrange_point_colonization

Contour lines connect points of equal effective gravitational potential for objects in free-fall, (such as an Apollo spacecraft in cruise phase) so a spacecraft can move along these trajectories “for free”.

Does the Apollo free return crossover point correspond to the L1 LaGrange point?

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  • $\begingroup$ Note that the drawing is not reflective of the actual trajectory, which would be curved lines. That said: mass is proportional to r^3, gravitational influence inversely proportional to r^2. If the bodies are drawn to scale, then straight lines drawn from radius to radius should therefore cross at the point where gravity would be balanced for uniform-density bodies. Earth is denser, however, so L1 should be slightly closer to the moon than the drawn crossover point. $\endgroup$ Commented Oct 17 at 17:16
  • $\begingroup$ @RussellBorogove ... L1 is not the point where net gravitational attraction cancels. It is the point where net gravitational attraction cancels centrifugal force (in a rotating frame of reference). That's why there is an L2. And an L3.... L4 and L5 are above my pay grade. $\endgroup$ Commented Oct 18 at 1:38

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From a patched conics point of view, free return trajectories are elliptical transfers with an apogee slightly overshooting the orbit of the Moon. In the flyby encounter, the exit angle is a mirror of the entry angle, and the return trajectory is therefore a mirror of the initial ellipse.

The symmetry of this means that apparent "crossover points" in a co-rotating frame will indeed fall on the line going through the Earth and the Moon, which the L1 point also lies on.

But that is where the similarities end. Free return trajectories are a continous spectrum when you change the apogee of the initial trajectory. We can for instance reason that if the overshoot is quite high, and the Moon encounter is targeted for the inbound leg, the duration of the iternary can be stretched out to multiple months. In that time the co-rotating frame has performed multiple revolutions, so there will be multiple "crossover points" on both sides of the Earth.

The crossover being at L1 can therefore be no more than a coincidence, obtained only by picking specific parameters from a range of options.

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