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When the Apollo spacecraft was en route to the Moon, was it an escape trajectory from the Earth-Moon system? If not for the burn(s) to enter lunar orbit, would it have continued in an independent solar orbit or was it at all times gravitationally bound to the Earth-Moon system?

I ask because of J002E3, originally thought to be a near-Earth asteroid, but now believed to be the S-IVB from Apollo 12.

In the case of Apollo 12, it would seem that the spent stage had to be boosted out of the Earth-Moon system, implying that up to the point of the reconfiguration maneuver, the spacecraft was still gravitationally bound to the Earth-Moon system. Was this true for all of the Apollo missions? Was a final burn of the S-IVB post-CMS separation for disposal a part of all flights?

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  • $\begingroup$ Do you mean the object en.wikipedia.org/wiki/J002E3 which turned out to be S-IVB-507? $\endgroup$
    – uhoh
    Feb 19, 2017 at 16:27
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    $\begingroup$ @uhoh Yes, exactly. $\endgroup$
    – Anthony X
    Feb 19, 2017 at 23:06

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Hmm. This question is trickier than it sounds. To know whether a body in a three-body system is always gravitationally "bound", you have to look into the future. For example, a close-enough flyby of the Moon on the trailing side would put the spacecraft on an Earth-Moon escape trajectory.

Apollo 11 was on a free return trajectory for which the lunar flyby (if it didn't go into orbit) would send it back to the Earth. So it was gravitationally bound, right?

Well, what if a spacecraft was on a very similar trajectory that just misses the Earth on the return. (I'm not going to use Apollo 11 in my hypothetical, since I don't want to kill the Apollo 11 astronauts, even hypothetically.) It would continue to orbit Earth in a lunar-crossing orbit, eventually having a close encounter with the Moon again. Then the Moon might fling the spacecraft on an escape trajectory. So was it gravitationally bound?

Exactly this happened to the Apollo 12 third stage you mentioned, which was orbiting the Earth for quite some time. It is believed to have left Earth orbit in 2003. Since it escaped three decades later, it must not have been gravitationally bound that whole time. But what if while in solar orbit it encounters the Earth-Moon system some time later, and the Moon does the opposite thing and puts it back in Earth orbit? (That third stage might do just that in the mid-2040's.) Now was it gravitationally bound the whole time? If now you think it is, might it not eventually escape again?

If its wanderings are permanently ended by impacting the Earth or the Moon, then it is now gravitationally bound in pieces on that body. So therefore it was always gravitationally bound, even when it was in solar orbit. Right?

To truly answer your question, you would need to propagate the trajectory from every state between maneuvers, potentially for a very long time, to determine its ultimate fate. There will often not be enough accuracy in the known state, as well as uncertainty in solar pressure perturbations, for that to even be deterministic.

As for that Apollo 12 third stage, they deliberately tried to have it immediately escape the Earth-Moon system, but failed. Later Apollo third stages were targeted to impact the Moon, which made for nice seismic signals.

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  • $\begingroup$ I took "gravitationally bound" to mean traveling below escape velocity -- in which case I don't believe flight path angle matters. Thus, the moon fly-bys won't matter unless they put you on a course to another body -- which is a completely different question. $\endgroup$
    – Erik
    Feb 20, 2017 at 3:20
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    $\begingroup$ "Escape velocity" only has meaning with respect to one body, e.g. the Earth, or perhaps the Earth-Moon system as a point at great distance to both. If you are not at a great distance then you would have to define escape velocity either by a) completely ignoring the Moon (in which case an "escape" trajectory past the Moon might leave you in Earth orbit, or a non-"escape" trajectory past the Moon might escape), or b) looking into the future, with the inevitable ambiguities noted in the answer. $\endgroup$
    – Mark Adler
    Feb 20, 2017 at 5:00
  • $\begingroup$ Escape velocity can be applied to the barycenter of multiple objects. $\endgroup$
    – Erik
    Feb 20, 2017 at 5:09
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    $\begingroup$ You know, I'm going to recant here and say I'm wrong. I think you can only look at the escape velocity for multiple bodies if the potential field is effectively invariant wherever you are at -- which would not be the case between or near the earth and the moon. $\endgroup$
    – Erik
    Feb 20, 2017 at 5:48
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    $\begingroup$ Escape velocity is directly bound to orbital elements; for circular orbits to orbital altitude. For a body skirting the edges of Earth's Hill Sphere, the escape velocity will be a meager 550m/s or so. When accounting for gravitational slingshots, a getting assist by the moon, never approaching LEO altitude escape velocity, will still reach velocity well excess of escape - for Moon orbital altitude. $\endgroup$
    – SF.
    Feb 20, 2017 at 10:23
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Yes. The Apollo stack was always on a free return trajectory between burns.

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  • $\begingroup$ Isn't that still "gravitationally bound to the Earth-Moon system?" There's no easy path to a heliocentric orbit from there, is there? Or am I misunderstanding something? $\endgroup$
    – uhoh
    Feb 20, 2017 at 1:16
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    $\begingroup$ I should have said "yes" as I meant that it was always bound. Fixing... $\endgroup$
    – Erik
    Feb 20, 2017 at 1:39
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    $\begingroup$ Not always. From Apollo 12 on they departed from a free-return trajectory in order to access a wider range of landing sites. $\endgroup$
    – Mark Adler
    Feb 20, 2017 at 2:36
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    $\begingroup$ True, but the midcourse burn that did this actually reduced the earth-relative velocity. $\endgroup$
    – Erik
    Feb 20, 2017 at 3:08
  • $\begingroup$ Without having looked at the data, I don't see how "the midcourse burn that [departed from a free-return trajectory] actually reduced the earth-relative velocity" implies "the Apollo stack was always on a free return trajectory between burns". Apollo 13 was most definitely not on a free return trajectory after the O2 tank explosion, despite that happening nowhere near a midcourse burn. $\endgroup$
    – user
    Feb 22, 2017 at 16:21
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The Apollo 13 trajectory looked like this.

Apollo 13 trajectory

First midcourse correction:

At 030:40:49.65, a 3.49-second midcourse correction lowered the closest point of spacecraft approach to the Moon to an altitude of 60 miles. Before this maneuver, the spacecraft had been on a free-return trajectory, in which the spacecraft would have looped around the Moon and returned to Earth without requiring a major maneuver.

Second midcourse correction:

The spacecraft was then maneuvered back into a free-return trajectory at 061:29:43.49 by firing the LM descent engine for 34.23 seconds. It then looped behind the Moon and was out of contact with the Earth tracking stations between 077:08:35 and 077:33:10, a total of 24 minutes 35 seconds.

Source

Between these two maneuvers, the spacecraft was not on a free return trajectory. I don't know yet if this section of the trajectory was bound to the Earth-Moon system.

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I'll have to look for details, but I think that on the outbound flight, the spacecraft had a high but still finite apogee, maybe 2x the lunar distance (700000km or so).

On the way back, I seem to remember that the return flight actually did exceed escape velocity, and therefore was in a hyperbolic orbit around the Earth. The periapse was low enough though that the spacecraft dipped into and entered the atmosphere instead of escaping. If the atmosphere and/or ground wasn't in the way, the spacecraft would have escaped Earth and gone into solar orbit.

Either way, the spacecraft was so close to escape velocity that tiny changes in speed result in enormous changes in orbit apoapse.

After separation, the S-IVB did a small maneuver either to impact the moon, or do a gravity assist into a heliocentric orbit, which is where the upper-stage-as-asteroid comes from. The upper stage is in orbit around the Sun, but an orbit that intersects that of the Earth and periodically re-approaches the Earth.

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    $\begingroup$ Are you commenting about Apollo 13? Because the question is not specifically about 13 and therefore your post lacks context. A simple edit could really clarify things. Sources would be nice too. $\endgroup$
    – Organic Marble
    May 26, 2021 at 3:15
  • $\begingroup$ and any answer about being temporarily bound should acknowledge that this is a three body problem and the motion is chaotic. As long as there is sufficient energy and angular momentum in the system, it may not be possible to say that it is long term bound without doing a numerical integration, and permanently bound may not be answerable at all. $\endgroup$
    – uhoh
    May 26, 2021 at 4:55
  • $\begingroup$ I recall reading that if Apollov13 had missed Earth on the return, it would have come back six weeks after launch. Not that it would have mattered, as the astronauts would have been long dead. $\endgroup$
    – Oscar Lanzi
    Feb 2, 2022 at 21:50

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