1
$\begingroup$

I was reading the Q/A on water as radiation shielding. The prime answer went in to detail on the cost to lift water from Earth which is cost prohibitive, but how feasible is it to get the water from the dark side of the moon? Hypothetically, including the mission cost to set up a water mining and lift off site on the Moon, what is the break even cost vs lifting it directly from Earth?

I build self selecting bioreactors for agriculture and mining and got to thinking about using the waste water/aquaponics biocyclers as additional shielding for the crew. I have a sense of the amount of water needed, about 1 meter on all sides of the vessel for a useful reduction in rads, which is a hell of a large bioreactor/propulsion fuel tank.

It might be easier to have the crew sleep cocooned in bioreactor sleeping pods. Still, you have to get a lot of water into space to do it, so, maybe the moon. Thoughts? Thanks!

$\endgroup$
  • 1
    $\begingroup$ Here's a thought: there's no water on the Moon. $\endgroup$ – AndreiROM Nov 23 '17 at 16:37
  • 1
    $\begingroup$ It should be feasible to build a space elevaoer on the moon. After its installation, energy cost can be close to zero. You may find it difficult to find the water in the first place, though. $\endgroup$ – Burki Nov 23 '17 at 16:40
  • $\begingroup$ See a link to evidence for moon ice nssdc.gsfc.nasa.gov/planetary/ice/ice_moon.html and sciencemag.org/news/2016/03/… but it should be mentioned that the "dark side" of the moon receives light, however it faces away from the earth. $\endgroup$ – P Chapman Nov 23 '17 at 16:53
  • $\begingroup$ Why do you want to get the water into LEO? Unless its going to stay in LEO it would be more efficient to move what ever is in LEO out into lunar orbit $\endgroup$ – Slarty Nov 23 '17 at 17:42
  • 1
    $\begingroup$ The cost of getting water from the moon to LEO is mostly in building a facility on the moon to mine water if it exists in any large quantity and to have a launch system on the moon that can deliver the payload. This part of the problem is much much more expensive in the short term than just sending water up from Earth. $\endgroup$ – A. C. A. C. Nov 23 '17 at 17:47
2
$\begingroup$

It is very hard to give you any exact or even a rough value as there are so many variables.

What rocket propulsion technology will be used? How much water is needed? This will affect the type of equipment needed to extract it. Extracting 10 cubic metres would need different infrastructure to extracting 1 million cubic metres. How quickly does the water need to be collected? Where will the materials for aero braking shielding come from? How much mass will they require per ton of ice returned? Will they be reusable? Are the development costs for all of the hardware included?

The only places on the moon that have any readily accessible water are deep craters at the poles. Fortunately little extra energy is required to transfer from lunar polar orbit compared to return from equatorial orbit.

The key element is the amount of velocity change require so called delta V. The delta V changes required can be found here:

9.3-10 km/s earth to LEO 2.2 km/s moon to lunar orbit with aero braking return to earth

So it is a lot more efficient to bring ice from the moon than to get it from the earth if the infrastructure is in place to do it.

As a very rough order of magnitude estimate I suggest that the breakeven point would be in the range of many hundreds of tons – a few thousand tons

$2000/kg by Falcon 9 = 200 million dollars/hundred tons in Leo from earth.

Can the lunar infrastructure be developed and put in place fro less than $200 million? I suspect not.

Could it for $2 billion - perhaps.

https://www.reddit.com/r/space/comments/62n47j/cost_per_kg_leo_for_human_spaceflight/

$\endgroup$
  • $\begingroup$ The bottom part of the answer is quite hard to read. $\endgroup$ – Oleg Lobachev Nov 23 '17 at 19:11
  • $\begingroup$ Yes I'm not sure what happened there it looks OK in the edit. Not sure where the different fonts come from... $\endgroup$ – Slarty Nov 23 '17 at 19:34
  • $\begingroup$ @OlegLobachev fixed it - it doesn't like 2 $ symbols on the same line $\endgroup$ – Slarty Nov 23 '17 at 19:39
  • $\begingroup$ Thanks for the lift cost comparison. Boy, that falcon rocket is sweet. Looking long term, we'll need a base on the moon as a springboard to other destinations. Might as well set up the means to extract resources there and ferry them to orbit as a way station. 2 billion for a moon colony sounds like a budgeting thing, not insurmountable. Once in place, the ROI is ten water lifts equivalent from Earth assuming good numbers. $\endgroup$ – Colin A Lennox Nov 24 '17 at 13:02
1
$\begingroup$

Better bet would be to snag an icy comet as it passes by, use a gravity tractor or a solar sail to redirect it and slow it down. There's more water on Mars but that's got a steeper gravity well. There's always the old SF staple of asteroid mining (Remember The Cant!).

It may not provide the plot points you're looking for, but foam metal might be the efficient solution in terms overall energy / resources spent. I guess you could make it a plot point that the best foam metal insulation can only be made in (near)weightless conditions.

$\endgroup$
  • $\begingroup$ Further out, hell yeah lets grab some comets. I was looking for the closest hypothetical "easy" water source, and an excuse to get back to the moon and stay this time. $\endgroup$ – Colin A Lennox Nov 24 '17 at 13:04
  • $\begingroup$ Well that wasn't among the questions that you asked was it ? If you're looking for reasons to stay on the moon, there's several but two off the top of my head... 1) Military base. You own the moon then you own the Earth. Being on the moon is like standing on top of a steep hill, the Earthers have gotta climb "up the hill" (their gravity well) to get to you, but to defend yourself or attack the Earth all you gotta do is throw rocks down the hill. There's lots of rocks on the moon. 2) Helium-3 minng, google it $\endgroup$ – Jack Judge Nov 24 '17 at 22:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.