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Old June 5th 17, 02:09 PM posted to sci.space.policy
Greg \(Strider\) Moore
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Default Mining the moon for rocket fuel to get us to Mars

"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.
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. Sure, you can
probably play some numbers there (direct impact, etc.) but it doesn't change
the first order approximation much.
Again as a first order approximation, figure the energy used to land will be
about the same as taking off.

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.

So again, first order approximation, I stand by my analysis.


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