On the other hand, I'm curious how the delta-V numbers hold up for landing on the Moon, where aero-braking is impossible. Launch and landing require by far the most delta-V between either of the two bodies, much more than the difference between lunar transfer orbit and Martian transfer.
It's about 1.7km/s delta-v to land on the moon from LLO. If you are assuming an orbital station in the lunar gateway, that's another ~0.7km/s. From LEO that's another ~3.6km/s (3.2 to TLI and 0.4 to gateway). That gives you around 6km/s delta-v flat. Of course you can optimize this a bit by using a more direct route but it's probably cheaper if you leverage existing infra like the gateway.
Compare this to anything on mars. It'll be 3.6-4.0km/s to mars injection. Then 2.0km/s to low orbit (can be optimized with clever aerobreaking down to 1.0km/s). To the surface will be a minimum of another 0.4km/s (no engine, parachute only) and closer to 1km/s if you want a soft landing. So you can expect 5km/s at the bare minimum and ~7km/s along established "routes" leveraging to-be-existing infra.
So then the question becomes whether lower delta-v returns, regular launch opportunities, and shorter comms delay are more or less important than 1km/s delta-v savings.
I'd argue the delta-v savings aren't worth it on their own. Mars will eventually be worth exploring and it's a valuable opportunity but by every measure other than raw "throw it at the orbital body with no support or return plan" delta-v savings, the moon is a better site for initial extraterrestrial habitation studies.
>> It'll be 3.6-4.0km/s to mars injection. Then 2.0km/s to low orbit (can be optimized with clever aerobreaking down to 1.0km/s).
Except that all that delta at mars is done with aerobraking/aerocapture. Inbound probes can crash strait into the mars atmosphere without slowing down for orbital insertion. That's why it takes far less fuel to get to surface of mars than the moon, at least with objects car-sized or smaller. Only with larger/heavier objects is mar's atmosphere so thin that fuel must be used for capture prior to descent.
In the following deltaV map, much of the trip to mars is a free ride so long as you follow the red arrows. There are no free rides on the way to the moon.
For the record we have not done aero capture at Mars just yet - but aerobraking has been used a lot, after first capturing into a high eccentric orbit with an engine burn.
In any case, bodies with atmosphere do enable you to effectively shave up to half of the delta-v for a transfer when you know what you are doing. :)
I was just illustrating the path along the lunar gateway (LEO -> TLI -> Gateway -> LLO -> surface). Even if you do a direct (LEO -> TLI -> LLO -> surface) route, you still have to go through LLO as you'll need to be able to pick your landing spot and potentially even delay a landing (which is not possible if LLO is not budgeted in).
Actually now that I think about it, if you want to save delta-v, you can get an extra 25% off the TLI leg of the direct route with a low energy transfer using weak stability boundary trajectories. These savings however come at the cost of a significantly longer flight time and far slimmer margins for failure.
OTOH, we can build a lunar space elevator with existing materials because of the lower gravity (even despite the longer distance to L1). Not sure, can we manage that with Mars?
The physics works for Earth too. In all cases, you make the top of the cable bigger than the bottom to support all of the weight hanging from it. From what I've read, in the case of Earth and with current materials technology, we end up with the top of the cable having a diameter comparable to that of the Earth. Clearly, that's not feasible.
For Luna or Mars, gravity is reduced and the required diameter is less. Maybe it would even be feasible to build such an elevator if it were above the Earth. But now you're building above an alien plant, so you trade one set of potentially insurmountable obstacles for another.
I mean, that's a fair point. The total volume of the elevator cable would be greater than that of the Earth. The mass might still be less since we're not building it out of iron here, but effectively we'd have a binary planet with the centre of mass well outside the Earth's surface.
I'm not sure that that system would be unstable in human timeframes since the two would be tidally locked, although it would certainly alter the engineering stresses in ways that I'm grossly unqualified to calculate. I think a portion of the cable might be under compression rather than tension? I guess it depends on the rotational speed of the whole system.
Speaking of which, substantial amounts of energy would need to be spent accelerating the spin of the Earth/space elevator system to maintain a 24-hour day/night cycle.
However, Luna's presence would perturb the whole system, either tearing it apart with tidal stresses or being ejected from the system before that could happen.
I appreciate the correction. I'm not sure where I heard that particular piece of information, nor in that case what material was being examined. Perhaps that one was steel.
On planets, a space elevator goes to (geo)stationary[0] so that the cable doesn't wind up around the planet, but you can't do that on the Moon, because luna-stationary is occupied by the Earth, which inconveniently is too massive and spinning too fast to anchor the other side of the cable. However, the L1 point is also stationery relative to the lunar surface, and is the place where the gravity of the Earth and the Moon balance out.
[0] IIRC, geostationary specifically means Earth, but there's going to be some more general term for the same idea over generic parent objects and not just Earth
Lunar stationary orbit is around 88,400 km, which would be unstable for a satellite due to the Earth's gravity, but might allow for a space elevator pointed right at Earth to efficiently launch crates of helium-3 or hydroponic grain back to the planet.
Indeed. I've not even played with this in one of the many simulators, but I believe the suggestion is to put the counterweight a tiny bit closer to Earth than the L1 to stabilise it.
Although (and I wish I could find this again), I've read that lunar He3 is so diffuse that getting it out would incidentally give us so much purified aluminium, silicon, and oxygen, that sending all that back to Earth and magnetically decelerating it on arrival would give us more energy than the He3, as would burning those ingots with that oxygen.
Lunar stationary orbit is not 88,400 km, it's 384,000 km, i.e. the distance from the Moon to the Earth. The moon is tidally locked to the Earth.
Though you could equivalently try and go for Earth-Moon L1 with a counterweight on the other side of L1. It would be significantly more unstable though.
