3
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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

8
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.