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Speculations on a Moon Colony

Buggernut writes "As reported by the BBC, humans could be living on the Moon within 20 years, according to a leading lunar scientist."

3 of 121 comments (clear)

  1. Re:Slightly offtopic by LittleBigLui · · Score: 4, Informative

    Think about it. You correctly identified rays as going THROUGH the subject, to a hypothetic receptor on the other side. Obviously, for an image to manifest, the amount of energy that moves through the subject has to vary, otherwise the image would be all-white (or all-gray). Hence, we have some of the rays moving straight through, and the rest of the rays being scattered away. Some of those will be going straight back to where they originated from. Hence, you get just the same image (well, inversed) if you place the receptor right at the source.

    --
    Free as in mason.
  2. Re:maybe the tech is there, but show me the money by WoTG · · Score: 2, Informative

    Forget missiles and bombs, just lob big rocks. Gravity will do the rest, once you give it a small push. A moon base could be a powerful weapon in the future... let's hope our first colonies on the moon are for peaceful purposes.

  3. Re:Doing things on the moon. by barawn · · Score: 2, Informative

    Starting haphazardly.


    The delta-v quoted by your source is far lower than the delta-v needed to get into a Hohmann transfer orbit even from free space in a circular solar orbit at Earth's radius (which the C3=0 orbit is the equivalent of). As the Hohmann orbits are the lowest energy transfer orbits that don't require slingshots from other bodies, I question the values on that figure.


    Ah, there's your problem. The C3=0 orbit is NOT a circular orbit at Earth's radius. It's a parabolic orbit with Earth at its focus, which necessarily can NOT be a circular orbit about the Sun at Earth's radius. A parabolic orbit means that at infinity, it will have no velocity relative to the Earth, which means, if the craft traveled to infinity, it would then have the equivalent orbital velocity of Earth. Problem is, it never reaches infinity, as it's not a two-body system, since the Sun's there.

    If you want a spacecraft to be in a circular orbit at Earth's radius, well, it doesn't have to do anything - just stay home. It already is in one. :) After it lifts off, depending on the direction, it is doing two things - first, it is escaping from Earth's gravity, and second, it is changing its orbit. You don't have to "add" the escape velocities onto the necessary orbital delta-V. If you wanted it to actually reach a circular orbit, that takes a lot more work, actually!

    This is the problem when you're doing "you need to add an extra 5.03 + 2.38 km/s" - you're adding the Lunar escape velocity and the Martian escape velocity, which you do not need to do, because you're not exactly going to infinity. On the return, you can easily aerobrake in Earth's atmosphere as well to enter lunar orbit.

    Also don't forget about aerobraking! No matter what, any time you approach a planet (even entering lunar orbit! you can always place your perigee inside Earth's atmosphere with clever timing!) if you need to slow down, it's free.

    And if you don't like that site, how about here, which shows that Deimos is more accessible than the Moon (and shows a delta-V from Mars surface to Lunar surface of 8.0 km/s, not 13.0 km/s).

    Or here, where you'll note that "LEO to Mars" is a delta-V of 4.8 km/s, not the 5.6 km/s you're claiming - this is because, of course, it's in LEO, and therefore has some orbital velocity about Earth (and is therefore traveling at -greater- than Earth's orbital velocity at certain points).

    I can continue to give examples if you want - the point is that from the Moon, it's easier to get to Mars and back than it is to get to Earth and back.

    The easiest way to think about this is simple: You do not need to actually escape Earth orbit in order to reach Mars. A highly eccentric orbit can include both Earth and Mars (if both were stationary, obviously - they're not, so you can't orbit them, but you can of course use that path to transfer between them), and so must necessarily take less energy than the escape velocity of Earth+the escape velocity of Mars (which reaches Mars by going through infinity).

    Interestingly enough, Hohmann transfers are not lowest energy. Google for "interplanetary superhighway", which is a relatively recent discovery. Really does suck that the 3-body system isn't analytically solvable...