2
$\begingroup$

The NASA news item New Space Policy Directive Calls for Human Expansion Across Solar System says:

New Space Policy Directive Calls for Human Expansion Across Solar System

and the administration is working with NASA to put together a plan to get Americans back on the Moon in 2024.

The target landing site will be at the Moon's south pole, so a trajectory similar to the relatively planar Apollo era free-return-like trajectories can't be used.

Does there exist a slightly modified free-return-like trajectory that includes a polar fly-over, or will the likely orbit involve some kind of substantial plane change maneuver? Or will it be something else entirely?

Are there any other spacecraft trajectories from Earth to a lunar polar orbit that can be used for comparison?

$\endgroup$
  • $\begingroup$ “A trajectory similar to the relatively planar Apollo era free-return-like trajectories can't be used.” Just turn the plane 90 degrees - start in a polar LEO parking orbit and inject toward a polar lunar orbit. (I doubt Artemis will actually start in a 90-degree inclination polar orbit, because everything is complicated.) $\endgroup$ – Russell Borogove May 15 at 1:24
  • $\begingroup$ Supposedly they will make an interim stop at the Gateway. $\endgroup$ – Organic Marble May 15 at 1:37
  • $\begingroup$ @OrganicMarble I see, it's a mad dash to the station who's express purpose is to "retire risk"? $\endgroup$ – uhoh May 15 at 1:41
  • 3
    $\begingroup$ There's no inertia like bureaucratic inertia. $\endgroup$ – Organic Marble May 15 at 1:42
4
$\begingroup$

Will NASA put astronauts into a polar lunar obit?

Yes, but it's a bit complicated.

If so, how?

The current plan is not for an Apollo-like launch from Earth orbit to Lunar orbit, culminating in a landing descent from that orbit.

Instead, the plan is to launch a crew from Earth on the Space Launch System to the Lunar Gateway, a mini-space station of sorts, which will be in a "near-rectilinear halo orbit."

According to this plan, a "landing system" will already be docked at the Gateway. The landing system will be tripartite: "a transfer vehicle to move the lander from the Gateway to low lunar orbit, a descent stage to go to the surface, and an ascent stage to return directly to the Gateway."

Since the landing is planned to be at the south pole that orbit will necessarily be polar, but it will transfer there from the halo orbit of the Gateway, not from Earth.

enter image description here

Image source

text Source

I write this answer largely to try and clear up apparent misconceptions about what is planned, in the question itself and the other answer, not because I have any detailed knowledge of the orbits involved.

$\endgroup$
  • $\begingroup$ Thank you very much for stepping up and explaining the whole thing. For some reason I was resistant to believe that the Gateway would be up and running so soon as well. $\endgroup$ – uhoh May 16 at 0:14
  • 1
    $\begingroup$ Looks good to me! $\endgroup$ – Organic Marble May 16 at 0:36
  • 1
    $\begingroup$ @uhoh "For some reason I was resistant to believe that the Gateway would be up and running so soon as well. " which is why I kept harping on plan, planned, etc. I'm pretty skeptical of this plan. $\endgroup$ – Organic Marble May 16 at 0:40
  • $\begingroup$ based on this summary of current thinking, one could also generate a hypothesis for How many stages will be used to bring NASA astronauts to the Moon's south pole and safely back to Earth? $\endgroup$ – uhoh May 16 at 1:52
0
$\begingroup$

In terms of the question "do there exist a slightly modified free-return trajectory..." I don't know (I know that's a weird way to start an answer, but bear with me). Getting to Lunar Polar orbit is not made any easier by starting in Earth polar orbit, that is overkill.

You basically need to fly over the poles of the moon. Given the moon's size and distance, this means that it can be achieved with a relatively low inclination Earth orbit, something probably less than a few degrees inclined with respect to the moon's orbit, probably lower. It's all about angles - the moon's angular size is small, and the distance is large, so if you are inclined just a little bit, then over the distance to the moon your orbit will go over the poles.

Performing a plane change in orbit is prohibitively expensive no matter which body you are orbiting. To change just a few degrees in low Earth orbit would take several hundred m/s of delta v. To change from equatorial to polar in a lunar orbit would be insane (I can't find a good number, but there is an equation here. You'll probably find that the vast majority of the delta-v budget of a mission would be taken up by the plane change.

NASA has studied such orbits extensively, since mapping satellites are best placed in such orbits in order to get as much surface coverage as possible. Here is a NASA trajectory study of this very thing - low energy transfers.

Whether a free-return polar trajectory exists or not is an interesting question. I strongly suspect not, but I'm happy to be proven wrong here. Free-return in the sense of return to the surface without expending fuel would require a small gravitational assist from the moon, which is not possible for a purely polar orbit. Perhaps a slightly less-than-polar orbit, where some small in-plane velocity change can occur, is possible, so don't quote me on the impossibility of such an orbit yet.

$\endgroup$
  • $\begingroup$ Based on transfer times in Table 4-1 of your linked report all of those except "direct transfer" are all way too long to "put astronauts into a... lunar obit", as mentioned in the question. To my last sentence "Are there any other spacecraft trajectories from Earth to a lunar polar orbit that can be used for comparison?" why not quote a few sentences from the first paragraph of Section 4.2? $\endgroup$ – uhoh May 15 at 11:32
  • $\begingroup$ Right now this is mostly "I don't know" plus a link, but if you added a few quotes from that excellent and authoritative source and refer to a few figures there as well, I think it could be a really great answer! $\endgroup$ – uhoh May 15 at 11:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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