3 ms·
Because they used the free-return trajectory, which lies in the orbital plane. Also, landing near the fast-moving equator saves some delta-v.
by bronson 8y ago
Because they used the free-return trajectory, which lies in the orbital plane. Also, landing near the fast-moving equator saves some delta-v.
- vkou 8y agoUnlike the Earth, the Moon rotates very slowly - once every 27 days. It's equator is not fast-moving - it is moving at only 4 m/s. This delta-v gain is nearly negligible. The free-return trajectory, where you keep the command module's orbit in the same plane as the Moon's orbit around the Earth is more important (Since both the Moon, and the command module orbits at ~1,000 m/s - which is the delta-v that you gain, if you eject from lunar orbit, in the opposite direction to the Moon's direction of travel.)
- bronson 8y agoYou're right that delta-v is basically negligible, thanks!
- pdonis 8y ago> Because they used the free-return trajectory If you mean the free return trajectory for the Earth-Moon trip, that was done only up through Apollo 11. Apollo 12 and after did not (although Apollo 13 was put back onto one after the oxygen tank explosion).