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I'm asking this out of curiosity, mostly as a response to Space elevator from Earth to Moon with multiple temporary anchors and some of the comments and answers there.

Suppose you're able to build space elevators on both the Earth and the Moon. The question is, can you get a capsule from Earth to Moon or vice versa without having any thrusters on the capsule?

This is an orbital mechanics question - the question is whether with suitably positioned space elevators there exists an orbital trajectory that leaves from one and arrives at another with sufficiently low delta-v for it to be launched and captured.

Note the tag - answers are expected to include calculations or references to them.

Here are some rules:

  • You can build launch/landing rails on the Moon as an alternative to a Lunar space elevator, meaning that you can leave from or arrive at the Lunar surface with a high delta-v. However, you can't leave from or arrive at a space elevator with significant delta-v, unless you can explain why the recoil wouldn't mess with the space elevator.

  • The capsule doesn't have a parachute or atmospheric shielding.

  • Unless there's some reason why it would be impossible, the capsule contains humans.

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    $\begingroup$ How can you suppose something that is demonstrably impossible, then use the hard-science label? How exactly would the science work along with something that's not scientific at all? It's like calculating heat dissipation of a fireball spell. Fireball spells don't exist. If you then suppose a fireball spell is like an exploding propane tank, then you're calculating for an exploding propane tank as if it is a fireball. There's absolutely nothing in the real world that would work as a space elevator of any sort. $\endgroup$
    – Nelson
    Jul 6 at 7:58
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    $\begingroup$ @Nelson lunar space elevators can probably be made. There probably isn't any material strong enough for an Earth-based space elevator, I agree. Nevertheless, the hard science tag is applicable, because the maths for a space elevator doesn't depend on what it's made of. The hard science tag refers to the need to back answers up with references or calculations, which it absolutely is possible to do for this question. $\endgroup$
    – N. Virgo
    Jul 6 at 9:24
  • $\begingroup$ you may want to look into skyhooks, they will be a lot more friendly to what you are tryin to do. youtube.com/watch?v=dqwpQarrDwk $\endgroup$
    – John
    Jul 6 at 13:33
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    $\begingroup$ @AlexP the question is about a transfer from one space elevator to another. You get extra kinetic energy if the Earth-based space elevator stretches further than geocentric orbit, and in addition a lunar space elevator, if built on the Earth side, stretches below the Earth-Moon Lagrange point. So you don't even need to get all that high if there's a transfer where the velocity difference is not too big. There is a potential energy difference of course but it's not obvious whether it's insurmountable. $\endgroup$
    – N. Virgo
    Jul 6 at 13:34
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    $\begingroup$ As the person who wildly asserted that this could be done on the linked question (which was admittedly not hard-science), I'm delighted to see this question on the plausibility. I was considering posting something similar. $\endgroup$
    – Zags
    Jul 7 at 18:21

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You are asking a pretty involved question that would need simulations to be properly answered.

Just conceptually, we know that if a Space Elevator tether extends beyond Geostationary orbit, the tip of the cable will move faster than orbital velocity. The velocity here is simply;

$$v = \frac {2 \pi (l+r_e)}{s}$$

Where $s$ is the number of seconds in a day (86400), $l$ is the length of the tether and $r_e$ is the radius of the Earth.

We can test this formal by plugging in the height of geostationary and comparing the resulting velocity to the actual orbiting velocity. This formal says the velocity is $3038\frac{m}{s}$. Which is right on the money. In reality it is $3070\frac{m}{s}$

We can use a very bad estimate to get ballpark figures for how much $d_v$ we need to get to the moon. By bad i mean i fired up Universe sandbox and just kind of looked how much i have to change the velocity of a probe at Geostationary to intersect the moons orbit. Obviously the moons orbit is not circular so this is an example value. From this very crude approach, i got ~$1180\frac{m}{s}$ to go from GEO to TLJ (Trans Lunar Injection).

So, we now know our Velocity at the tip of the cable has to be ~$4250\frac{m}{s}$. Which works out to a length of $l = 52786000m$. Note this is minus the 6000km from Earths radius. So the cable has to be 52780km long.

Right about now a good question would be precisely at what velocity we will intersect the moon. According to Universe Sandbox 2, we will reach Lunar orbital height at a velocity of $371\frac{m}{s}$. Which you may notice is slightly slower than the Moons orbital velocity of ~$1022\frac{m}{s}$. As a matter of fact we appear to be missing ~$700\frac{m}{s}$.

But is this actually a problem ? Well no. Nobody ever said we need a cable attached to the moon to capture us. We just need a cable that rotates with a velocity of $700\frac{m}{s}$ and is able to capture us. a Skyhook. Do i know how to do the math for that ? Absolutly not. But it should be possible to place a bit rotating skyhook in lunar orbit. On the one end it captures our spaceship coming from Earth and slows it down. And in exchange it throws another from the Moons surface to Earth. This way it stays balanced.

This entire system would require exactly 0 engine ignitions and only one half decent maglev on the moon since you need to move at $700\frac{m}{s}$ to be captured by the hook.

So yeah, aside from climbing a 52000 kilometer high elevator and one maglev on the moon, this is Energy free. Though, i would argue if your big plan is to build a 52000 kilometer elevator, just use a rocket.

EDIT; i had a smooth brain moment and wrote m/s² everywhere instead of m/s

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    $\begingroup$ There would probably need some adjustment inflight, I imagine the precission needed to let go at the right time to hit the reciever at the moon would be ... well astronomical(pun intended) $\endgroup$
    – Michael Mortensen
    Jul 6 at 9:13
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    $\begingroup$ @MichaelMortensen most certainly. I imagine you could probably do a true to god 0 adjustment transfer if the thing you transfer is a sphere of uniform density. But the moment you have any asymmetry or shifting masses going there will need to be some RCS type deal on board. On the bright side, if you miss the hook you just fall right back to the elevator xD $\endgroup$
    – ErikHall
    Jul 6 at 9:27
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    $\begingroup$ Space debris would be a huge issue with such a colossal object with a large cross section. It wouldn’t be easy to move so would be hit sooner rather than later. nasa.gov/mission_pages/station/news/orbital_debris.html Any catastrophic failure of the cable would send it spiraling Earthward through highly populous Low Earth Orbits, Geostationary orbit and others. It might be hoped that the ultra strong cable would burn up but if it didn’t then the political and financial ramifications could be very serious indeed. $\endgroup$
    – Slarty
    Jul 6 at 20:19
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    $\begingroup$ This is cool. The high velocity at the lunar surface could also be solving using a launch rail system as I proposed in the question. Arriving at the moon would be tricky, but in the other direction you'd launch from a mass driver and then arrive at the space elevator at a low enough velocity to be captured. My guess is that a 52000km elevator isn't that much harder to build than a geostationary one, which is 36000km already. If you have a counterweight in geostationary orbit then the 'up' cable doesn't need to add extra tension to the 'down' one. $\endgroup$
    – N. Virgo
    Jul 7 at 3:47
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    $\begingroup$ (I also agree that Earth-based space elevators aren't a very realistic proposition, and of course even if this proposed scheme were possible there would be very little reason to do it, except perhaps as an extreme sports type of thing. I just didn't want to let that get in the way of an interesting question.) $\endgroup$
    – N. Virgo
    Jul 7 at 3:52

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