How much do you need to decelerate in order to be captured into a stable lunar orbit from a Trans-Lunar Ejection or TLI-type trajectory?

  • $\begingroup$ Are you asking about the minimum needed for lunar capture, or more typical insertions into low lunar orbit? $\endgroup$ Jul 1, 2018 at 13:44
  • $\begingroup$ I suspect the theoretical minimum may be zero. $\endgroup$
    – peterh
    Jul 1, 2018 at 19:40
  • $\begingroup$ The minimum won't be zero, because there's no third body to provide a braking assist to change the almost-parabolic trajectory into ellipse, but the difference between these is minimal; with the transfer maneuver performed perfectly, it would be infinitesimal; with the error common to this type of maneuvers - a couple meters per second. $\endgroup$
    – SF.
    Jul 2, 2018 at 15:36
  • $\begingroup$ It’s zero; the sun’s influence can be used. $\endgroup$ Jul 2, 2018 at 15:46
  • $\begingroup$ Let me put it this way - assuming you could be put onto a lunar flyby by piggybacking on a probe, like the 4M mission did, what would be the minimum needed to enter orbit? $\endgroup$
    – TeslaK20
    Jul 7, 2018 at 11:18

1 Answer 1


Strictly speaking, the minimum ∆v is zero: from a ballistic capture trajectory, marginally stable lunar orbits at about 20,000 km altitude can be reached for zero or nearly-zero ∆v cost; an orbiter can then descend to a lower orbit at leisure using low-thrust high-efficiency ion engines. More generally, low energy transfers allow a tradeoff between travel time and insertion ∆v.

From an Earth-Moon Hohmann transfer, it takes about 680 m/s to reach a low lunar orbit at ~100 km altitude.

The Apollo missions started from a faster transfer orbit, so they required a little more ∆v; Apollo 11's initial LOI burn was 889 m/s to enter a 113 km pericynthion, 313.4 km apocynthion orbit.


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