Free Return trajectories

[IthyStPete]

The (manned) Apollo missions (except 7 and 9, ofc.) were put on a free return trajectory to the Moon, so they could coast back and reenter LEO if something went wrong after TLI, correct? But if I understand how Hohmann transfers work, wouldn't any TLI burn out them on a free return trajectory? If so, why did they need to specifically execute the TLI to put them on a free return?

 

[tblaxland]

The Apollo missions did not fly a Hohmann transfer orbit. Had they done so, they would not have had sufficient excess velocity at the moon to retrograde slingshot around it and still have a perigee close to Earth.

 

[FordPrefect]

Actually, it's impossible to fly anything but an equatorial orbit using a
free-return trajectory. In fact, Apollo 11 was the last lunar mission to
use a pure free-return trajectory. The "H" missions (12-14) started out
in free-return trajectories but shifted to non-return ones after S-IVB
separation and CSM-LM docking, with the shift done progressively earlier
in the later flights. The "J" missions injected directly into non-return
trajectories, again getting increasingly aggressive about it as time went
by -- in the event of an SM engine failure, Apollo 15 could have moved
back to a free-return trajectory using its RCS thrusters, while Apollo 17
would have had trouble doing it even with the LM descent engine. [Quoted from Henry Spencer]

 

[Tommy]

A Hohman transfer will bring you back to Earth - if nothing alters your trajectory. Trying to transfer to the Moon, you'll enter the Moon's SOI (sphere of influence) and that will alter your trajectory, and you won't return to Earth in a survivable time or velocity unless you make further course correction burns.

An Apollo style "Free Return" trajectory takes the Moons gravity into account, and relies on it to change the vessels trajectory and send it back to Earth.

A Hohman transfer will get you back home only if there are no gravity sources encountered. A Free Return Trajectory will get you back home only if a sufficient gravity source is encountered correctly.

 

[IthyStPete]

Got it. Thanks, everyone.