7 ms·

> Mars is further away in travel time, but not much further in fuel costs, and it's far better than the Moon for living off the land. The prospect of an inhabi

by kibwen 4y ago

> Mars is further away in travel time, but not much further in fuel costs, and it's far better than the Moon for living off the land. The prospect of an inhabited Martian colony living off the land is dismal, for anything except extremely far-future timeframes (thousands of years or more). Any extraplanetary colony is going to be invariably tethered to Earth and highly dependent on imports for survival. The moon has the advantage of not just being closer, but having a launch window of "whenever you feel like it", in contrast to Martian launch windows that occur 26 months apart.

sobellian 4y ago
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)
  - 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
Even if I expected to be able to collect, I wouldn't give good odds on a bet that the moon or mars will physically still exist 1000 years from now. Von Neumann machines have an existence proof in ourselves, and make it fairly fast to disassemble planets into a Dyson swarm. (I think Dyson swarms are a bad idea, and will grind to dust that blows away to interstellar space in the solar wind in geologically short timescales, but that's a different problem).
- tsimionescu 4y ago
  There is absolutely no reason to beleive that any kind of machine will ever exist that can "make it fairly fast" to disassemble an entire planet. You're probably entirely ignoring any kind of wear and tear on the machine itself, which could easily be outpaced by the rate of creating new machines for anything approaching the billions of years this would actually take using any imaginable fuel.
  - ben_w 4y ago
    Assume humans are the VN machine. We can make space suits and greenhouses etc., so it's not crazy. Assume reproduction such that population doubles every 25 years. Fast, but not insane. (1000/25) = 40 doubling periods 8e9 * 2^40 * 100kg = 2.7 Mercury masses, or 12 lunar masses, or 1.3 Mars masses That's a worst-case bound using a biological anchor. Dedicated VN machines can plausibly be faster: even if it's something weird like staying with biology and uplifting dogs, that's now 75 years, a plausible but currently hypothetical self-replicating 3D printer could make the timescales even shorter. The sun provides enough power to do it in a week or two, though anything less than decades may have thermodynamics issues.
    - tsimionescu 4y ago
      I'm not sure where your 8e9 number is coming from. Population doubling every year but the parents leaving the planet after mating? In your calculation, after 1000 years, there would be 2^40 humans, each weighing 100kg - many orders of magnitude less than Mercury's mass. Still, if we give it a few thousand more years of exponential growth, you will eventually reach such masses. However, these numbers are meaningless - you are assuming that doubling the size of a population that's 2^39 individuals will take roughly the same time as a population that's 64 individuals individuals, which is not even close to plausible - especially when we reach ideas like a population weighing as much as half a planet. You're also assuming that it's even possible for a mahcine to convert a significant proportion of a planet's crust to copies of itself - which is obviously false, as the crust is mostly rock, and machines require plenty of liquids and water to be produced (whether biological or mechanical or electronic). And that doing so will not affect the growth rate at all, even as the planet starts being formed of molten magma once all of the crust has been used up. And not to mention the gigantic earthquakes and supervolcaones they would have to deal with as a significant portion of continental mass gets shifted around. Overall, you are only extrapolating some numbers to a completely absurd conclusion, and calling it plausible. There is nothing even close to realistic in your scenario, and indeed we have no idea if it's even close to possible to strip a planet down to create a Dyson swarm. I very much doubt your energy calculation as well, but that's already beside the point.

by kibwen 4y ago

sobellian 4y ago

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)
- 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

Even if I expected to be able to collect, I wouldn't give good odds on a bet that the moon or mars will physically still exist 1000 years from now. Von Neumann machines have an existence proof in ourselves, and make it fairly fast to disassemble planets into a Dyson swarm. (I think Dyson swarms are a bad idea, and will grind to dust that blows away to interstellar space in the solar wind in geologically short timescales, but that's a different problem).

tsimionescu 4y ago
There is absolutely no reason to beleive that any kind of machine will ever exist that can "make it fairly fast" to disassemble an entire planet. You're probably entirely ignoring any kind of wear and tear on the machine itself, which could easily be outpaced by the rate of creating new machines for anything approaching the billions of years this would actually take using any imaginable fuel.
- ben_w 4y ago
  Assume humans are the VN machine. We can make space suits and greenhouses etc., so it's not crazy. Assume reproduction such that population doubles every 25 years. Fast, but not insane. (1000/25) = 40 doubling periods 8e9 * 2^40 * 100kg = 2.7 Mercury masses, or 12 lunar masses, or 1.3 Mars masses That's a worst-case bound using a biological anchor. Dedicated VN machines can plausibly be faster: even if it's something weird like staying with biology and uplifting dogs, that's now 75 years, a plausible but currently hypothetical self-replicating 3D printer could make the timescales even shorter. The sun provides enough power to do it in a week or two, though anything less than decades may have thermodynamics issues.
  - tsimionescu 4y ago
    I'm not sure where your 8e9 number is coming from. Population doubling every year but the parents leaving the planet after mating? In your calculation, after 1000 years, there would be 2^40 humans, each weighing 100kg - many orders of magnitude less than Mercury's mass. Still, if we give it a few thousand more years of exponential growth, you will eventually reach such masses. However, these numbers are meaningless - you are assuming that doubling the size of a population that's 2^39 individuals will take roughly the same time as a population that's 64 individuals individuals, which is not even close to plausible - especially when we reach ideas like a population weighing as much as half a planet. You're also assuming that it's even possible for a mahcine to convert a significant proportion of a planet's crust to copies of itself - which is obviously false, as the crust is mostly rock, and machines require plenty of liquids and water to be produced (whether biological or mechanical or electronic). And that doing so will not affect the growth rate at all, even as the planet starts being formed of molten magma once all of the crust has been used up. And not to mention the gigantic earthquakes and supervolcaones they would have to deal with as a significant portion of continental mass gets shifted around. Overall, you are only extrapolating some numbers to a completely absurd conclusion, and calling it plausible. There is nothing even close to realistic in your scenario, and indeed we have no idea if it's even close to possible to strip a planet down to create a Dyson swarm. I very much doubt your energy calculation as well, but that's already beside the point.