3 ms·
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 stat
by 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 agoAgain 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.