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One of the cool things about the Moon is that the far side has a thicker crust that the near side.1 One theory explaining this is that the Moon was hit by an object, possibly a moonlet created by the remains of Theia as per the Giant Impact Hypothesis (while the Moon was still forming). The object hit in such a way that more material was deposited than was blown away, leading to the thicker crust.

Let's say I own an interstellar mining company, and I've come across a system with plenty to spare. There are four rocky planets orbiting a central Sun-like star. There's also an asteroid belt at between 3 and 4 AU, with a total density (i.e. number density of asteroids multiplied by the mass of each asteroid) of about three times that of the asteroid belt in our Solar System.

I don't want to mine each asteroid directly because that would require some major infrastructure problems. However, I've figured out that I might be able to divert some of the asteroids towards the planet, which lies at 2.5 AU. If I get it just right, I might be able to move some asteroids close enough to leave a lot of mineable material. Then I can swoop in and set up shop.

The issue is, too many impacts could cause a whole bunch of issues for the planet, making it worthless. And too many impacts could damage the material already there. At the same time, I want to get as much material as possible deposited onto the planet.

Assuming I can divert a large number of asteroids away via some as-yet-undetermined (but not precise) method, how many should I send in the direction of the planet to maximize my yield? Note that I'll need a large asteroid flux (in terms of number density) because many could and will completely miss.

A potentially useful paper is Asteroid and comet flux in the vicinity of Earth (Shoemaker et al, 1990).

I'd like answers to focus on how much material could be deposited on each impact, orbital perturbations of the planet, regional distribution of impacts (due to planetary rotation) and how they can impact mineability of deposited material, and asteroid flux. Answers discussing only one or two factors, however, will be much appreciated; you don't need to cover everything to write a high-quality answer.

HDE 226868
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    Why not mine the asteroids directly? that seems much easier. – John_H Mar 20 '15 at 16:05
  • Rather than shooting them at rocks and hoping they'll stick, would it not be easier to shoot them at a lake/sea of any fluid and have that absorb the impact? You could mine the rocky planets directly and have spacecraft specially designed to steer the asteroids into the right orbit to drop where you want it to. This requires some incredible calculations ofcourse. – Cronax Mar 20 '15 at 16:29
  • What's the precision and what's the accuracy? Knowing those factors would make this into a simple calculation of a multivariate Gaussian distribution. – Samuel Mar 20 '15 at 17:19
  • I hope your company is a long-term investment, because many, if not most, economically minable minerals on Earth are the result of geothermal and biological processes, which take upwards of 100 million years or so to work. – jamesqf Mar 20 '15 at 18:17
  • @John_H I have no good explanation for that. – HDE 226868 Mar 20 '15 at 21:06
  • @Cronax There's no liquids on the planet in question. – HDE 226868 Mar 20 '15 at 21:06
  • @Samuel For which factors? – HDE 226868 Mar 20 '15 at 21:06
  • @jamesqf I'm ignoring that for now. – HDE 226868 Mar 20 '15 at 21:07
  • @HDE226868 What are the margins on precision and accuracy? – Samuel Mar 20 '15 at 21:13
  • @Samuel For what values, I meant? Asteroid mass, orbital radius, etc. – HDE 226868 Mar 20 '15 at 21:15
  • @HDE226868 Ah, apologies, I mean for your aim. How well can you aim the asteroids at the planet. – Samuel Mar 20 '15 at 21:16
  • @Samuel Ah, I see what you mean. In terms of angles, a decent estimate would be that you can aim an asteroid at a target and end up being within about one arcminute of it. – HDE 226868 Mar 20 '15 at 21:18
  • @HDE226868 An arcminute is not a meaningful estimate of accuracy in orbital maneuvres; the precision of your velocity magnitude change matters a great deal too. The more I think about it, the more it seems to me that this question has way too many free variables for a straight answer. – Mike L. Mar 20 '15 at 22:20
  • @MikeL. I was referring to its initial launch angle, not where it would end up. I'll also make another thing clear, which is that I'm not looking for a numerical answer, merely the mathematical formulation of it. – HDE 226868 Mar 20 '15 at 22:23

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TL;DR: It is unlikely that this is the best option

Let's look at some numbers that describe the amounts of energy needed for some of the things you're talking about. First, let's assume that both the planet and the asteroids are in a circular orbit around the star. Also, let's say the star and the planet have the same mass as our sun and the Earth, respectively. Now, some formulas:

$$M_p = 5.972*10^{24} kg$$ $$M_s = 1.989*10^{30} kg$$ $$G = 6.67*10^{-11}\frac{m^3}{kg\cdot s²}$$ $$E_{orbit} = -\frac{GmM}{2r}$$ $$1 AU = 1.495*10^{11}m$$

These are $M_p$ the mass of the planet, $M_s$ the mass of the sun, $G$ the gravitational constant, $E_{orbit}$ the potential energy of a mass $m$ orbiting around a mass $M$ at radius $r$, and $1 AU$ in meters. Using these, we can determine how much energy it would take per kilogram to drop an asteroid onto the planet and then cost per kilogram to remove it from the system.

The potential energy of an object orbiting the sun at 2.5 AU is $-\frac{GmM_s}{2*2.5AU}=-1.775*10^8J/kg$, while at 3 AU it is $-\frac{GmM_s}{2*3AU}=-1.479*10^8J/kg$, for a delta of $2.958*10^7J/kg$. That means to move a $10^9kg$ (pretty small) asteroid you'd need $2.958*10^7*10^9=2.958*10^{16}J=8.217*10^9kWh$.

A nuclear reactor generating 1,000 megawatts (i.e full scale terrestrial nuclear reactor) would need close to a year to generate that much energy if there is no waste energy. The Tsar Bomba generated about that much energy total but most of the energy would just be lost to space, it would be very hard to aim, and it's likely that the energy would shatter the asteroid anyway, leaving very little to fall on the planet. Once you'd gotten the asteroid (or most of it) to the planet, you'd need that much energy again to move that much mass back from 2.5AU to 3AU, plus about that much again to account for escaping the gravity well of the planet.

You'd need to use a ridiculous amount of energy moving asteroids (even ones nearer the planet) down to the planet and then getting the mass back out, especially with how much loss you'd get from the asteroids missing the planet or not being entirely salvageable. It would be far easier to build some mining facility on an asteroid and then pack up and move the entire facility to the next asteroid once you're done. Even if you're using applied phlebotinum to move the asteroids, you'd have an easier time applying the phlebotinum to moving the facility.

*Note: My calculations for dropping the asteroid onto the planet assume you'd try for a relatively gentle collision with the planet. You can save some initial energy by putting it into an elliptical orbit, but then the impact will be more severe and your aim has to be much better. If those aren't an issue, then you can save 50% energy on the initial drop to the planet. Also, I don't expect that these completely correct, but it at least gives you an idea of the scale required.

Rob Watts
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  • While I agree that sending asteroids crashing into a planet is hardly the most practical way to mine them, I have to point out that you've significantly overestimated the energy necessary to bring them down. The equation you're using is for circular orbits only, but if you only want to get an asteroid on a collision course with the planet, all you need is to bring its apoapsis down to 2.5AU during the right point of its orbit. This necessitates the use of the equation for specific energy of elliptical orbits. In short, you should substitute 5.5AU (2x the semi-major axis of the transfer – Mike L. Mar 20 '15 at 20:42
  • orbit) for 5; this yields a specific energy of 1.653e8 J/kg, making the delta slightly more manageable 1.75e7 J/kg. Still a lot, but you've saved almost 50% :) Also, using a gas giant instead of a rocky planet might let you perform aerocapture and save the energy required to go back to orbit. In other news, I may have played too much KSP. (edit: that should have been periapsis above) – Mike L. Mar 20 '15 at 20:48
  • While this is a good analysis of the energy, it's not quite what I'm looking for. I'm not so interested in the orbital mechanics of it all. – HDE 226868 Mar 20 '15 at 21:08
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    @HDE226868 think of this answer as being the lead engineer of your interstellar mining company trying to explain to you why the orbital mechanics suggests that you try something else instead of dropping asteroids on the planet. – Rob Watts Mar 20 '15 at 21:16
  • @MikeL. I thought about the elliptical orbit option, but thought that it would cause more impact damage to the planet (the planet end of the orbit would be where the asteroids are going the fastest). – Rob Watts Mar 20 '15 at 21:19
  • @RobWatts (Sigh) I've been testing out the [tag:hard-science] tag and doing a very poor job of it. You're absolutely right, and you're telling me the things I need to here - you've written an answer that I might write. My original explanation was to use a gas giant to swing through the asteroid belt and accelerate asteroids towards the planet. It's a bad idea because it involves a lot less precision than I'd like. At the same time, it could happen as a natural process (e.g. planetary migration). In other words, I was trying to figure out how to mimic a natural process with manmade events. – HDE 226868 Mar 20 '15 at 21:24
  • @RobWatts A quick calculation gives me a relative approach velocity of 838 m/s, which is surprisingly sedate. You might not even lose all that much material this way. – Mike L. Mar 20 '15 at 21:35
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    @Mike L.: But isn't the 838 m/s just the difference in orbital velocities? To which you have to add velocity from gravitational attraction - same as escape velocity from the planet, but in the other direction - of ~11.2 km/sec if the planet's Earth-sized. – jamesqf Mar 20 '15 at 22:01
  • @jamesqf Under Keplerian assumptions, yes. If the asteroid is big enough, this can go down a lot, but at that point it's impossible to give a general statistical answer. – Mike L. Mar 20 '15 at 22:05
  • @Mike L.: I'm afraid I don't understand. Surely a bigger asteroid would increase the mutual attraction, and thus the final collision velocity? That is, forget the orbital velocity for a moment, and have two bodies at rest relative to each other, a large distance (say lunar orbit) apart. Then they'll mutually attract, accelerate towards each other, and collide. – jamesqf Mar 21 '15 at 05:13
  • @jamesqf Well, escape velocity is calculated on the assumption that the mass of the orbiting body is insignificant next to the mass of the central body. What you say is true, but the numbers would come out different because now both bodies are accelerating and one of them suddenly has significant inertia. Or maybe they wouldn't, we're now talking about a non-inertial frame of reference and I'm not confident in my ability to eyeball the result. – Mike L. Mar 21 '15 at 14:40