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Re: Mining on the Moon

PostPosted: Sat Aug 28, 2010 10:22 pm
by Darth Kevin
Intranaut wrote:I love the sound of that, an interplanetary gas station on the way to Mars. Will there be slushies?

No, but there will be a big sign saying "Last gas for 50,000,000 miles"!

Re: Mining on the Moon

PostPosted: Sun Aug 29, 2010 6:59 am
by Intranaut
Haha.


Has anybody on here ever seen Moon?

Re: Mining on the Moon

PostPosted: Mon Aug 30, 2010 2:04 pm
by RickLewis
Intranaut wrote:Haha good fun reading these two guys go at it about it though. Does Hop draw all his own pictures?

Yes, I wondered. But they fit so well with the context of the argument that I really think he must have drawn them just to illustrate his postings in that thread.

I think all the contributors there have produced their own diagrams, which is most impressive, but Hop's are clearly the classiest from an illustration point of view, and the funniest too!

Re: Mining on the Moon

PostPosted: Mon Aug 30, 2010 2:09 pm
by RickLewis
I do have a question I'm a bit embarrassed to ask because I'm sure I should know the answer already. But I don't so what the heck, here goes...

All the posters in that thread on the New Mars forum quote quantities of fuel in km/s. How does this measurement work? I assume the point is to use a measure of the fuel requirements which is independent of the amount of mass being lifted, bujt despite thinking about it for a bit I simply can't see how this measure works. Can anyone enlighten me? Kevin? Geoff?

Re: Mining on the Moon

PostPosted: Tue Aug 31, 2010 6:52 am
by Intranaut
Don't fear! I wasn't too sure what you meant there, so I was just as confused!
But what I think they're measuring in is amount of fuel, regardless of the thrust per kilo, needed to get to those speeds. So that once you know what fuel you're using you can plug in to a formula (I don't know it, but I imagine there is one you could come up with) the thrust to mass ratio of the fuel with the velocity that's needed to do whatever it is you have to do (like when they say it takes 3.8 km/s for low earth orbit), with mass as the subject, so you'd end up with the mass of fuel needed to carry out whatever manoeuvre you're talking about. Now that I've typed that out and that it make sense in my head, it does sound like a logical unit to use when talking about a generic fuel. I hope it made sense to you?

Re: Mining on the Moon

PostPosted: Wed Sep 01, 2010 8:52 am
by RickLewis
Thanks, but I still don't quite get it.

I can see that in deep space, away from any significant gravitational pulls, if you take an object of mass M "at rest" and attach a rocket to it, and fire the rocket so you burn a certain amount of fuel, then you will give the object a bit of a shove. By the time the burn has finished, the object will be travelling at a velocity V, and because it is a vacuum with nothing to slow it down or pull it in any direction, it will then keep on going at that velocity V (however many km/s that is) indefinitely.

So I can see that you could express the amount of fuel needed to do that as V km/s, meaning "the amount of fuel needed to speed any mass up from rest to V km/s." Is this basically it? If so, then I do understand I think.

And the amount of fuel needed to bring the object back to a state of rest by firing a rocket in the other direction? Also V km/s, obviously.

So far so good, but what if gravity is involved, as it is in most of the calculations we're talking about here? If you start with an object on the surface of the earth, fire your rocket, and burn the same amount of fuel (V km/s) then the object will accelerate up away from the earth, but gravity is pulling it back again at 9.8 m/s/s. So when you switch off the rocket, the mass will slow and may fall back to earth. This is where I still have difficulty coping with quantities of fuel in km/s.

Re: Mining on the Moon

PostPosted: Wed Sep 01, 2010 8:54 am
by RickLewis
Intranaut wrote:Haha.


Has anybody on here ever seen Moon?

No, I haven't heard of it even. Is it a film?

Re: Mining on the Moon

PostPosted: Wed Sep 01, 2010 10:49 am
by Intranaut
Ah, well that's why they refer to escape velocity on there (I think they did, if they didn't, it was inferred). Escape velocity for earth, according to wikipedia (which disputes what they were saying on there, so I'm confused now too) is 11.2 km/s. If you do a fuel burn coming up FROM the centre of the earth (not necessarily from the actual centre, but beneath the surface) so that you reach exactly 11.2 km/s as you pass the surface, you can switch off the engines and just become a projectile. From there you'll break away from the effective gravitational field of the earth. So I think what the mean when they say something like 0.6km/s burn to get into a Hohmann orbit on the way to Mars is that while you're orbiting the Earth, you only need to get to 0.6km/s to break away from it's effective gravitational field and into interplanetary space.
Hopefully clearer?


And yeah, Moon is a film:
http://en.wikipedia.org/wiki/Moon_(film)

Re: Mining on the Moon

PostPosted: Fri Oct 01, 2010 2:41 pm
by Darth Kevin
Thanks everyone for an interesting thread.

One thing that seems a little strange to me is that we've had a big thread about if it is wrong to terraform Mars, but nobody has claimed it is wrong to mine the Moon.

Re: Mining on the Moon

PostPosted: Sun Oct 03, 2010 12:07 pm
by Intranaut
A reasonable statement. For me the key difference is that Mars may have once (and may still, unlikely as it is) harbour life. If we around terraforming mars, any earth life forms may displace the Martian ones and we may never know if there was life elsewhere in this solar system.