Timeline for Asteroid flux or: How to make a planet perfect for mining
Current License: CC BY-SA 3.0
14 events
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| Mar 21, 2015 at 14:40 | comment | added | Mike L. | @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. | |
| Mar 21, 2015 at 5:13 | comment | added | jamesqf | @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. | |
| Mar 20, 2015 at 22:05 | comment | added | Mike L. | @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. | |
| Mar 20, 2015 at 22:01 | comment | added | jamesqf | @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. | |
| Mar 20, 2015 at 21:35 | comment | added | Mike L. | @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. | |
| Mar 20, 2015 at 21:24 | comment | added | HDE 226868♦ | @RobWatts (Sigh) I've been testing out the 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. | |
| Mar 20, 2015 at 21:19 | comment | added | Rob Watts | @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). | |
| Mar 20, 2015 at 21:16 | comment | added | Rob Watts | @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. | |
| Mar 20, 2015 at 21:11 | history | edited | Rob Watts | CC BY-SA 3.0 |
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| Mar 20, 2015 at 21:08 | comment | added | HDE 226868♦ | 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. | |
| Mar 20, 2015 at 20:58 | history | edited | Rob Watts | CC BY-SA 3.0 |
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| Mar 20, 2015 at 20:48 | comment | added | Mike L. | 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) | |
| Mar 20, 2015 at 20:42 | comment | added | Mike L. | 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 | |
| Mar 20, 2015 at 20:20 | history | answered | Rob Watts | CC BY-SA 3.0 |