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Imagine mankind has established mining facilities on NEOs
Some of them send raw ore directly to other processing stations in space, others refine on site.

All of them use a system based on cargo 'slugs' that get flung to their destination using rail guns. Near destination 'vectors' link with the cargo, slow it down and bring it in.

This is economical as only a few vectors would be needed at destination while there may be hundreds of cargoes traveling along different trajectories, paced so that reception is possible.
The problem is there are regular launches from the NEO while mining the asteroid. This means the asteroid gets an equal impulse in the opposite direction. It's mass may be billions of times bigger than the cargo but it is reducing as it gets mined and over years of exploitation and launches the orbit is going to be altered.

Now, for some asteroids that may not be a problem, the final orbit can be calculated and be safe.
For some others that may not be the case. For instance they may be at the receiving end of cargo shipments too. Or the final altered orbit may become a hazard.

Is there another way to stabilize the orbit while keeping sending the cargoes without having to shoot an equal mass (maybe waste? but would there be enough waste? on metallic asteroids that may not be the case it seems) in the opposite direction?

Setting:

  • Our own solar system.
  • Power generation is available at low price. Firing the rail gun more than once to stabilize the asteroid would not be an issue.
  • Ion / chemical thrust engines such as those employed by vectors are available but much more expensive to use than rail guns.
  • A slight variation in the orbit is not an issue. Vectors will adjust to it if cargo needs to be brought in. A large one happening in years of work may be.
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  • $\begingroup$ @L.Dutch why editing the question? (thanks for correcting cargoes though :-) ) the mass does not have to be equal. In the edit I tried to clarify what am asking: not using shooting mass method to stabilize the orbit but something else $\endgroup$
    – Duncan Drake
    Dec 18, 2020 at 16:48
  • $\begingroup$ I have rolled back that edit because it invalidates a number of answers $\endgroup$
    – L.Dutch ♦
    Dec 18, 2020 at 16:49
  • $\begingroup$ I see but that seems to limit the chance of improving a question. I realize the problem though. I was wrong in mentioning the 'equal mass'. $\endgroup$
    – Duncan Drake
    Dec 18, 2020 at 16:59
  • $\begingroup$ Even if there was 'another way to stabilize the orbit while keeping sending the cargoes without having to shoot an equal [impulse] in the opposite direction?', you'd have to make it reliable and ship it to the asteroid, which would automatically make it less economic to do so - and the only reason to mine an asteroid rather than the planet is because of the economics. $\endgroup$
    – Pete Kirkham
    Dec 19, 2020 at 18:36

8 Answers 8

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You could in principle attach a solar sail to the asteroid to help stabilize the orbit, but I don't think it would be necessary.

In the long run you aren't going to be firing all your payloads in the same direction - even if they are all going to the same destination, that destination is presumably going to be in a solar orbit different from yours, and your relative positions will vary as you move around the sun. The changes in your orbit from the launches will therefore tend to cancel out over time, and orbital changes are unlikely to be a problem anyway; the odds of hitting anything are very small, and you'll be able to predict the orbital changes in advance and take them into account when planning incoming payloads.

What's more, there is always more than one orbital route that will take your payloads from any one point in the solar system to any other; because orbits are ellipses rather than straight lines, you can choose a direct orbit that gets you there as quickly as possible, or a less direct one that will take longer (potentially even choosing to have the payload orbit the sun multiple times before reaching the target) and that gives you a great deal of flexibility regarding which direction any particular payload is fired in. So even in the unlikely event that your orbit is threatening to become a hazard you will always have the option to give it a push in a safer direction, albeit at the cost of some of your payloads arriving at their destination at a potentially less convenient time.

Planning for all of this might not be straightforward, but it should be doable. The worst case scenario is that you occasionally need to sacrifice a payload in order to make an orbital correction that can't be otherwise accommodated, but that shouldn't happen often, and probably not at all.

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Momentum is mass times velocity. You can always use arbitrarily small reaction mass provided you shoot is off at sufficiently high speed. Just use slower guns for useful cargo and high speed guns for corrections.

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  • $\begingroup$ Right. I edited the question. $\endgroup$
    – Duncan Drake
    Dec 18, 2020 at 16:35
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Is there another way to stabilize the orbit while keeping sending the cargos without having to shoot an equal mass in the opposite direction?

You don't need to shoot an equal mass. Momentum is given by $m\Delta v$.

Let's say you are shooting a mass 1 with $\Delta v$ 100, you can neutralize the imparted momentum by shooting in the opposite direction a ballast mass 100 with $\Delta v$ 1. This means that the ballast won't leave the asteroid, since it won't reach escape velocity, and will also require less energy: remember that kinetic energy goes like $1/2mv^2$, so to shoot your load you would need $0.5\cdot 1 \cdot 100^2 = 5000$ while to shot your ballast you would need $0.5 \cdot 100 \cdot 1^2 = 50$.

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  • $\begingroup$ Right. I edited the question. $\endgroup$
    – Duncan Drake
    Dec 18, 2020 at 16:35
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    $\begingroup$ @DuncanDrake, I am not sure what you mean with it, but mind that edits cannot invalidate existing answers $\endgroup$
    – L.Dutch ♦
    Dec 18, 2020 at 16:37
  • $\begingroup$ Compare this to firing a gun. This causes a small mass (the bullet) to travel forward at high velocity, while a larger mass (the slide) moves backwards at a much smaller velocity. (Or in the case of guns without a slide, the kickback from the weapon is imparted into the arm of the shooter instead.) Obviously, you weigh considerably more than a bullet, so it doesn't knock you back. $\endgroup$
    – Darrel Hoffman
    Dec 18, 2020 at 21:24
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    $\begingroup$ If the ballast doesn't escape, but the asteroid retains its position, then you have violated conservation of momentum. The trick is that the launcher will stop imparting a force to the payload very quickly, but the ballast will continue to pull on the asteroid until it reaches apogee. Thus, the asteroid+ballast must end up with the negative momentum imparted to the payload. $\endgroup$
    – Lawnmower Man
    Dec 18, 2020 at 21:57
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    $\begingroup$ If the ballast doesn't leave the asteroid, it won't work to counteract the impulse from the cargo shoot away (or only will so shortly). $\endgroup$
    – Paŭlo Ebermann
    Dec 18, 2020 at 22:20
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move the launch site

You can actually use the launches to correct the asteroids trajectory as much as possible. You do need to be able to move the launch site though. If you can, it can be helping you instead of destabilising.

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  • $\begingroup$ The rail gun should be quite massive to be able to launch cargo slugs at a significant speed. I wonder if it would be more convenient to move it around or just rotate the whole asteroid (of course we are not talking about Ceres here). Worth some thought. $\endgroup$
    – Duncan Drake
    Dec 18, 2020 at 16:39
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Yes use a fleet of large solar sail craft to transfer the material to and from any orbit you want. It will take a long time but it would be possible. Practical perhaps not but possible yes.

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  • $\begingroup$ Can you expand on this? Would they be able to sail from an orbit farther from the sun to an inner orbit? How long would 'a long time' be? $\endgroup$
    – Duncan Drake
    Dec 18, 2020 at 16:36
  • $\begingroup$ The sail would have to be angled to the Sun in such a way that some of the force exerted on the sail was directed against it's orbital velocity, slowing the sail and letting it drop sunward. The speed would depend on the size of the sail, the distance from the Sun and the mass of the payload being carried amongst other things. I imagine it would take many years to achieve the correct orbit. The size of the sail would also be huge and it might suffer from dust impact damage over years, so it might not be that practical, but I can't be sure. $\endgroup$
    – Slarty
    Dec 18, 2020 at 18:51
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The cargos could be launched at different moments of the rotation so that the overall deviation sum up to zero. You will still be losing momentum from the asteroid because of the mass loss, but I think one could in principle compute the direction of the landings and flights in order to put the asteroid on any allowed orbit. Just to make an example. Suppose that 2 spaceships depart at the same moment and with the same amount of momentum, but at opposite sides of the asteroids and with opposite directions. Them the overall deviation will be zero.

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You also don't need an equal mass going in the opposite direction because all you need is to make sure the asteroid stays in a 'safe' orbit, rather than the 'same' orbit. It actually wouldn't surprise me if in most cases that could be accomplished even while delivering the ore to its regular destination by timing the shot. That is, sending it at a different velocity along some other-than-optimal trajectory. The ore still gets where it's needed and disaster is averted.

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You think about momentum but i think the most important issue is the gravity in this case.

I mean if we shoot the same mass to the NEO, this "shooting" action will produce a momentum but there is gravity of NEO and the Earth.

I think The Earth's gravity will be almost same in the process but NEO's gravity will change and when NEO's gravity is low, if we shoot it with anything the NEO's orbit will change with no doubt

So i don't think there is any way to 100% stabilizing

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    $\begingroup$ Please answer OP's question, this is how we work. $\endgroup$
    – L.Dutch ♦
    Dec 18, 2020 at 16:29
  • $\begingroup$ I edited the answer. Is it correct now? Sorry for my inexperience $\endgroup$
    – Ramazan Mutlu
    Dec 18, 2020 at 16:40

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