5 ms·

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

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.

ben_w 4y ago
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?
- randallsquared 4y ago
  Mars' moons make a stationary space elevator extremely challenging, even though known material processes are sufficient.
  - ben_w 4y ago
    Good point. They're so small I forgot about them.
- oblio 4y ago
  By lunar space elevator I assume you mean an elevator from the lunar surface to the point where lunar gravity is practically 0?
  - ben_w 4y ago
    Only by coincidence in this case. 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
    - mcbits 4y ago
      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.
      - ben_w 4y ago
        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.
      - Nzen 4y ago
        LiftPort Group's plan [0] involves actively maintaining their counterweight at Lagrange point 1. [0] https://www.youtube.com/watch?v=3ZteJmg1TfM https://www.youtube.com/watch?v=3ZteJmg1TfM (10 minute video, relevant around 4:00)
      - mlindner 4y ago
        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.
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- paledot 4y ago
  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.
  - ben_w 4y ago
    If one end of the cable is the size of the Earth, the physics can't get away with neglecting the gravity created by the cable itself.
    - paledot 4y ago
      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.
  - jodrellblank 4y ago
    Carbon Nanotubes would end up with the top being ~26% larger than the bottom: See Isaac Arthur's video about it here: https://youtu.be/dc8_AuzeYKE?t=470 https://youtu.be/dc8_AuzeYKE?t=470
CuriousCosmic 4y ago
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.
- hoseja 4y ago
  This of course doesn't mention the ~10km/s you need to actually get off Earth.
  - paledot 4y ago
    "If you can get your ship into orbit, you're halfway to anywhere." —Robert Heinlein
- sandworm101 4y ago
  >> 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. https://space.stackexchange.com/questions/2046/delta-v-chart-mathematics https://space.stackexchange.com/questions/2046/delta-v-chart...

ben_w 4y ago

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?

randallsquared 4y ago
Mars' moons make a stationary space elevator extremely challenging, even though known material processes are sufficient.
- ben_w 4y ago
  Good point. They're so small I forgot about them.
oblio 4y ago
By lunar space elevator I assume you mean an elevator from the lunar surface to the point where lunar gravity is practically 0?
- ben_w 4y ago
  Only by coincidence in this case. 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
  - mcbits 4y ago
    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.
    - ben_w 4y ago
      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.
    - Nzen 4y ago
      LiftPort Group's plan [0] involves actively maintaining their counterweight at Lagrange point 1. [0] https://www.youtube.com/watch?v=3ZteJmg1TfM https://www.youtube.com/watch?v=3ZteJmg1TfM (10 minute video, relevant around 4:00)
    - mlindner 4y ago
      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.
      - deleted 4y ago
        [deleted]
      - deleted 4y ago
        [deleted]
paledot 4y ago
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.
- ben_w 4y ago
  If one end of the cable is the size of the Earth, the physics can't get away with neglecting the gravity created by the cable itself.
  - paledot 4y ago
    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.
- jodrellblank 4y ago
  Carbon Nanotubes would end up with the top being ~26% larger than the bottom: See Isaac Arthur's video about it here: https://youtu.be/dc8_AuzeYKE?t=470 https://youtu.be/dc8_AuzeYKE?t=470

CuriousCosmic 4y ago

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.

hoseja 4y ago
This of course doesn't mention the ~10km/s you need to actually get off Earth.
- paledot 4y ago
  "If you can get your ship into orbit, you're halfway to anywhere." —Robert Heinlein
sandworm101 4y ago
>> 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. https://space.stackexchange.com/questions/2046/delta-v-chart-mathematics https://space.stackexchange.com/questions/2046/delta-v-chart...