3 ms·

Isn’t identifying an unknown body as debris going to be way easier, generally, than going the other way? Especially when the body is in solar orbit, and therefo

by celoyd 15y ago

Isn’t identifying an unknown body as debris going to be way easier, generally, than going the other way? Especially when the body is in solar orbit, and therefore somewhere in a bigger volume and presumably dimmer? I don’t know much about orbital mechanics, but this line really got me: The key problem which we are taking on is a lack of solid orbital data since 1969 If it hasn’t been observed at all for 42 years, don’t minuscule errors in the initial observations give a huge uncertainty to its current position? Not just where it is on its orbit, but what its orbit is? Especially if it’s been in solar orbit at roughly 1 AU, and thus presumably interacting with the Earth-moon gravity well periodically? And perhaps accidentally venting from overheated tanks etc.? I have heaps of respect for the numerical methods that the astrophysics people have. I’ve seen little bits of them and they’re uncannily accurate and efficient. But how much can they help you if you don’t know where Snoopy was to the millimeter in 1969? Maybe this makes more sense to people with a stronger grasp of physics than mine. I just don’t see how you can go from the ephemeris data they presumably have to a realistic set of angles at which to point telescopes. Anyway, even if they don’t find it, it’s a really neat project – and, as Howes says, they’re sure to find something interesting.

kmm 15y ago
Of course it's easier, I was just trying to show that we do have the technology to identify objects of a scale of some tens of meters from a distance of hundreds of thousands of kilometres. It's a pretty amazing feat of technology, which also allows us to do geological research of the planet Venus without a probe[1]. I haven't researched the three-body problem personally, but I have had some concepts explained and as far as I know it is chaotic (like almost every dynamic system in nature). We won't be able to predict the exact position. But there are still laws similar to those of conservation of energy and momentum. In this problem they take the form of the constancy of the Jacobi integral[2]. Thus, while we don't know exactly where Snoopy is, we can exclude huge volumes where it certainly never went. It's probably still very close to the Earth-Moon system. [1] http://en.wikipedia.org/wiki/Venus#Ground-based_research http://en.wikipedia.org/wiki/Venus#Ground-based_research [2] http://en.wikipedia.org/wiki/Jacobi_integral http://en.wikipedia.org/wiki/Jacobi_integral
- celoyd 15y ago
  Yeah, real-world n-body is very hard. But astronomers can do it with said amazing integration methods. The thing is that they tend to do it for known start states. With Snoopy, our major masses are Sol, Terra, and Luna. Their dynamics are not trivial, but we have precise and accurate models. And we can drop a massless particle into the model and make really strong predictions, usually, about its position in 42 years. (Only usually because it might do complicated things like skip off our or the sun’s atmosphere, but we can overlook that.) What I’m trying to grapple with is that I don’t see how they’ll know where, and along what vector, to insert Snoopy. Say we know its start state to within ±100 m (x, y, z) and ±0.01 m/s (dx, dy, dz). I don’t think it really matters how well your methods conserve energy at that point – they’re working on too rapidly widening a volume of possible configuration space. Or is there some kind of attractor/anti-chaos effect that I don’t understand here? I’m honestly asking; I only have little bits and pieces of the knowledge here, and I’d be very interested to know the details of why my incredulity is unfounded. we can exclude huge volumes where it certainly never went. It's probably still very close to the Earth-Moon system. Granted we can exclude huge volumes – the vast, vast majority of the space in the solar system. But say we know it’s within 0.1 AU of Earth’s orbit (not Earth itself), in a solar orbit as the article suggests. Wolfram Alpha says that’s still a volume of about 3e24 km^3. And here’s what’s worse (as far as I can see). Say we’re lucky and, by ruling out slingshot effects and so forth, we know it’s in a medium earth orbit instead of random solar orbit. That’s a volume of about 2e14 km^3. Say we’re even luckier and also know it’s at lowish inclination. So in terms of the volume in the solar system, we have it basically pinned down. The thing is, that still leaves us something like half the solid angle of the sky to examine with telescopes. In other words, as long as we’re using optics that can identify it at all, its distance doesn’t really matter as much as the projected area, the area of observed sky as opposed to real volume, that it could be in. And I have trouble believing intuitively that that can be narrowed down very much when we don’t know the starting state well in any dimension other than time – but I really would like to be convinced otherwise. I think the parsimonious explanation here is probably that they had much better starting observations in 1969 than I’m imagining.
  - kmm 15y ago
    Again I don't know for sure but I fear that predicting the position of Snoopy over a period of more than some months is impossible. There's not only the gravitational interaction, but also the orientation dependent radiation pressure, tidal effects and gravity gradient torque. I would say the configuration space isn't just huge, it's probably all of the dynamically allowed space. I think the distance does matter. You have to observe an object ten times more distant a hundred times longer to receive the same amount of photons. If conservation laws dictate that it's still in the Earth-Moon system, that could mean significantly shorter observation times.

by celoyd 15y ago

kmm 15y ago

Of course it's easier, I was just trying to show that we do have the technology to identify objects of a scale of some tens of meters from a distance of hundreds of thousands of kilometres. It's a pretty amazing feat of technology, which also allows us to do geological research of the planet Venus without a probe[1]. I haven't researched the three-body problem personally, but I have had some concepts explained and as far as I know it is chaotic (like almost every dynamic system in nature). We won't be able to predict the exact position. But there are still laws similar to those of conservation of energy and momentum. In this problem they take the form of the constancy of the Jacobi integral[2]. Thus, while we don't know exactly where Snoopy is, we can exclude huge volumes where it certainly never went. It's probably still very close to the Earth-Moon system. [1] http://en.wikipedia.org/wiki/Venus#Ground-based_research http://en.wikipedia.org/wiki/Venus#Ground-based_research [2] http://en.wikipedia.org/wiki/Jacobi_integral http://en.wikipedia.org/wiki/Jacobi_integral

celoyd 15y ago
Yeah, real-world n-body is very hard. But astronomers can do it with said amazing integration methods. The thing is that they tend to do it for known start states. With Snoopy, our major masses are Sol, Terra, and Luna. Their dynamics are not trivial, but we have precise and accurate models. And we can drop a massless particle into the model and make really strong predictions, usually, about its position in 42 years. (Only usually because it might do complicated things like skip off our or the sun’s atmosphere, but we can overlook that.) What I’m trying to grapple with is that I don’t see how they’ll know where, and along what vector, to insert Snoopy. Say we know its start state to within ±100 m (x, y, z) and ±0.01 m/s (dx, dy, dz). I don’t think it really matters how well your methods conserve energy at that point – they’re working on too rapidly widening a volume of possible configuration space. Or is there some kind of attractor/anti-chaos effect that I don’t understand here? I’m honestly asking; I only have little bits and pieces of the knowledge here, and I’d be very interested to know the details of why my incredulity is unfounded. we can exclude huge volumes where it certainly never went. It's probably still very close to the Earth-Moon system. Granted we can exclude huge volumes – the vast, vast majority of the space in the solar system. But say we know it’s within 0.1 AU of Earth’s orbit (not Earth itself), in a solar orbit as the article suggests. Wolfram Alpha says that’s still a volume of about 3e24 km^3. And here’s what’s worse (as far as I can see). Say we’re lucky and, by ruling out slingshot effects and so forth, we know it’s in a medium earth orbit instead of random solar orbit. That’s a volume of about 2e14 km^3. Say we’re even luckier and also know it’s at lowish inclination. So in terms of the volume in the solar system, we have it basically pinned down. The thing is, that still leaves us something like half the solid angle of the sky to examine with telescopes. In other words, as long as we’re using optics that can identify it at all, its distance doesn’t really matter as much as the projected area, the area of observed sky as opposed to real volume, that it could be in. And I have trouble believing intuitively that that can be narrowed down very much when we don’t know the starting state well in any dimension other than time – but I really would like to be convinced otherwise. I think the parsimonious explanation here is probably that they had much better starting observations in 1969 than I’m imagining.
- kmm 15y ago
  Again I don't know for sure but I fear that predicting the position of Snoopy over a period of more than some months is impossible. There's not only the gravitational interaction, but also the orientation dependent radiation pressure, tidal effects and gravity gradient torque. I would say the configuration space isn't just huge, it's probably all of the dynamically allowed space. I think the distance does matter. You have to observe an object ten times more distant a hundred times longer to receive the same amount of photons. If conservation laws dictate that it's still in the Earth-Moon system, that could mean significantly shorter observation times.