View Single Post
  #18  
Old June 7th 17, 05:58 PM posted to sci.space.policy
Niklas Holsti
external usenet poster
 
Posts: 168
Default Mining the moon for rocket fuel to get us to Mars

On 17-06-05 16:09 , Greg (Strider) Moore wrote:
"Niklas Holsti" wrote in message ...

On 17-06-05 04:06 , Greg (Strider) Moore wrote:
"Jeff Findley" wrote in message
...

First, this isn't my "subject", it's the title of this article:

Mining the moon for rocket fuel to get us to Mars
May 14, 2017 8.04pm EDT ?Updated May 18, 2017 9.01am EDT
http://theconversation.com/mining-th...-to-get-us-to-

mars-76123

I saw this article (or a variation of it from another online
publication) on Twitter. I replied something to the effect that this
article glosses over all of the hard stuff, like the fact that the
lunar
soil and rock is horribly abrasive and that mining equipment isn't
anything like the lightweight rovers that NASA/JPL has flown in the
past.


[snip]

Don't get me wrong, I think *eventually* we'll be mining the moon for
water to turn into LOX and LH2 (or possibly methane) to supply a fuel
depot in lunar orbit. But, needless to say, I think the supporters of
this notion are daft if they think it's going to happen in the next 20
years or so by building a freaking factory on the moon that's
capable of
building mining equipment that's not JPL class "toys" that wear out
faster than you can build them.

Jeff

Yeah, I use a very simple first order approximation for this:
the mass of the fuel you'll get from the Moon has to be greater than
fuel used to get the mass to mine it to the Moon, otherwise it's a net
loss.

Simply put, if you're going to extract say 100 kilotons of fuel from the
Moon, you're going to have to use less than 100 kilotons of fuel getting
your mining and processing equipment there, otherwise it's a waste.


Perhaps you on purpose ignored this factor in your approximation, but
surely the *location* of the fuel must be considered?

If you use 100 kilotons of fuel to send equipment to the Moon, most of
that fuel is used up close to the Earth, and will not reach the Moon.
If that equipment then produces 100 kilotons of fuel on the Moon, that
fuel is on the Moon, which is "half-way to anywhere".

You would surely have to use *much* more than 100 kilotons of fuel to
deliver a payload of 100 kilotons of fuel from the Earth's surface to
the Moon's surface.


Not that much more really. And as I said, it's a first order
approximation. There's a lot more math involved than simply this.


No kidding :-)

But if you figure the delta V just to get to the Moon is approximately
12.52km/sec, and the delta-V from lunar transfer to surface is probably
about 2.55, and 2.55 back out, so that's less than 1/2.


What goes to the Moon (12.52 + 2.55 km/sec) is the mining and
fuel-producing equipment. If that trip uses 100 kilotons of fuel, most
of that is used near Earth (not close to the Moon) and the final part
(landing, 2.55 km/s) is for a much smaller mass.

Sure, you can probably play some numbers there (direct impact, etc.)


Not my point at all. My point is just the ratio of fuel weight to
payload weight, which is large for a trip from the Earth's surface to
lunar orbit or lunar surface, whether you are sending fuel or mining
equipment.

If you are sending fuel, to be stored in lunar orbit, there is the
advantage that it does not have to be landed onto the lunar surface,
unlike the mining equipment. But the ratio between the delta-vees (12.52
vs 2.55 km/s) means that the landing cost is relatively small.

Again as a first order approximation, figure the energy used to land
will be about the same as taking off.


What lands is the mining equipment, which must mass much less than 100
kilotons, if it took only 100 kilotons of fuel to send it from the Earth
to the Moon. What takes off is the fuel that is produced (assumed to be
100 kilotons), plus whatever containers and propulsion it needs. So the
take-off uses much more energy.

(Plus one can potentially use electromagnetic launchers for launching
the fuel from the lunar surface to lunar orbit, so the overhead in terms
of containers and propulsion could be small.)

Now, figure if you want to go to Mars, from lunar transfer to Mars is
approximately less than landing on the Moon, especially if you can use
some aerobraking.


Right, which is why fuel in lunar orbit is so valuable, compared to fuel
on the Earth.

--
Niklas Holsti
Tidorum Ltd
niklas holsti tidorum fi
. @ .