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The answer by PearsonArtPhoto in this question got me thinking, particularly this line:

Asteroids would be cheaper to go to than the Moon

I suppose the subject here are asteroids that pass in the zone between Earth and Moon's orbit around Earth.

I imagine that such asteroids pass by Earth at high velocity and in a relatively short time. Would a near-Earth asteroid pass give enough time to complete a mining operation and generate profit?

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    $\begingroup$ Near Earth asteroids don't have to pass through cis-lunar space (within the Moon's orbit) to count as such. The just have orbits that are near the orbit of Earth - a semi-major axis of less than 1.3 AU. Many of them cross Earth's orbit, but it is rare for them to do that when the Earth is so close that they cross cis-lunar space. $\endgroup$
    – kim holder
    Jan 8, 2016 at 16:22
  • $\begingroup$ Could we nudge one into a more desirable orbit? $\endgroup$
    – PCARR
    Jan 9, 2016 at 0:55
  • $\begingroup$ "I suppose the subject here are asteroids that pass in the zone between Earth and Moon's orbit around Earth." No, it's not about distance so much as gravity wells. The moon is a larger object (deeper gravity well) than most asteroids and requires more fuel to climb up away from. $\endgroup$ Jan 9, 2016 at 2:06

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The real issue here is how much rocket propellant you have to spend to get to the location in question. In one way, asteroids are easier to get to due to the fact that landing on the Moon takes a lot of velocity change. On the other hand, an asteroid flying through the Earth-Moon system is going really fast, so catching up with it is going to take a lot of propellant.

One problem with mining NEOs is that the closer the orbit is to Earth, and therefore the velocity change needed, the longer is the synodic period. That is, you have to wait a long time before you can return. (If you do not want to waste a lot of propellant, effectively cancelling the advantage.) However, asteroids gives you the ability to choose among them, so you can target those high in ore content. Ores on the Moon are generally more equally distributed.

A better option may be to capture an asteriod in orbit, so you have more time to mine it. If mining in space is going to be profitable is difficult to answer though. This may not actually be all that difficult, see for instance: Catching an asteroid. That proposes a way to catch it with only 250m/s of delta-v.

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  • $\begingroup$ Note that the return trip doesn't necessarily have to be as low-delta-V as the initial trip. You can use the asteroid material as propellant, if you're smart enough ("sufficiently advanced technology" - but really, a mass driver to throw ferrous matter should be workable even today). This of course works for capturing the asteroid in Earth orbit as well, yummy. $\endgroup$
    – Luaan
    Jan 8, 2016 at 16:43
  • $\begingroup$ @Luaan That is of course true, but if it is any easier is still a question. Last time I checked, I had a little trouble stuffing rocks into my rocket engine. $\endgroup$ Jan 8, 2016 at 16:46
  • $\begingroup$ Sure, you wouldn't use a LHOX engine for that. However, we do have engines that work with solid fuels - for example, aluminium oxide rockets were suggested as great "return engines" for Moon shuttles (aluminium oxide being easily available on Moon's surface), and the mass driver alternative should also be workable. Of course, it would mean carrying an extra engine (not simple or cheap), but there's probably smart solutions to that as well. $\endgroup$
    – Luaan
    Jan 8, 2016 at 16:55
  • $\begingroup$ Sorry if I'm uninformed, but isn't 250m/s DeltaV of a asteroid with a high mass rather difficult to achieve? $\endgroup$ Jan 12, 2016 at 22:27
  • $\begingroup$ @SarahBourt For a high mass it is difficult, but why not catch a small one? Then you also have more asteroids to choose among. For comparison, the delta-v needed to LEO is about 9000m/s, and 3000m/s for a trans lunar injection. $\endgroup$ Jan 12, 2016 at 22:37
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That's a one trillion dollars question! Proximity of near-Earth flybys of asteroids is largely irrelevant when it comes to feasibility of matching their orbit and rendezvousing with them, what is more important is their hyperbolic excess velocity with respect to Earth, and how much delta-v is needed to do that. There are many Athen, Apollo and Amor groups of asteroids that have similar orbit to Earth's in that respect, for example, here's a nice chart of JPL's Near-Earth Asteroid Delta-V for Spacecraft Rendezvous, with many NEA listed with smaller delta-v requirement to rendezvous with them than for a lunar surface, so Pearson's answer is correct. And even prettier than a long text file, here's Asterank (Asteroid Database and Mining Rankings) that should give you a fairer impression of what it would take to rendezvous with them and their estimated worth. And you can plug names of many of these into NASA's Trajectory Search tool to also see requirements for round-trip rendezvous.

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The trick isn't to mine them when they are close, necessarily. Asteroids are cheaper to get to than the Moon because the Moon has quite a bit of gravity. If you aim correctly, you can avoid all of the loss of rocket fuel associated with gravity. Where it is doesn't matter that much (So long as it is in a nearby orbit.

The bottom line is, an Asteroid mining mission will have a lot of time to mine, not just the few hours of closest approach.

Also, asteroids make frequency close encounters. Of particular note is Apophis, which will have a close encounter in 2029, and another one in 2035. One could sent the equipment during the initial encounter, mine the asteroid, tweak the orbit slightly to return to Earth thereafter.

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  • $\begingroup$ If launcher reusability pushes launch costs towards the fuel costs, there's little motivation to mine fuel (water) in outer space. Established industries on Earth would provide fuel much cheaper. Natural gas prices are at the bottom. It's like passenger airplanes don't refuel in flight, but before departure. I think that ISRU is good for staying a long time far away, rather than for going there. $\endgroup$
    – LocalFluff
    Jan 8, 2016 at 17:04

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