6 ms·

Great post, thank you! Where does the uncertainty (1%) come from? For example, is it more from our ability to precisely determine the orbit based on limited ob

• renjimen 2y ago
  Parent mentioned Monte Carlo simulations, which allow you to simulate across a range possible scenario parameters and see what % result in some outcome (like a collision with Earth or the moon).
  • VierScar 2y ago
    I'm sure there's a reason, but it seems like an unusual use of Monte Carlo - it's all deterministic and there is no opposing player making choices. Must have something to do with uncertainties in projected orbits or imperfect simulations maybe?
    • deleted 2y ago
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    • heikkilevanto 2y ago
      Uncertain observations, I'm sure
    • deleted 2y ago
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    • zacharycohn 2y ago
      Right - we only got a short glimpse of it, not enough to get a high confidence of its trajectory.
    • Projectiboga 2y ago
      The guesswork is uncertainity about the object's exact paraneters. Because of this they have to use informed estimates (scientiffic guesswork).
    • basementcat 2y ago
      The observations are not 100% certain. There are a variety of body states and configurations that might result in the same (few) pixels being lit up in the few measurements collected so far. As additional measurements are collected, some possibilities may be eliminated and the uncertainty of the trajectory can be reduced. This usually results in the impact probability converging toward 0%.
      • davrosthedalek 2y ago
        ...or 100%. But yeah, the MC comes in this way. You have a current most probable value for the position and some distribution around it, depending on the precision of the measurement device etc. That can be a high-dimensional space. You draw some (many) random points from this space and propagate them all deterministically. Taking into account how likely a certain random point was in the first place, you can then estimate the hit probability. MC is numerically approximating an integral. Here it replaces the high-dim integral over the start parameters.
    • cozzyd 2y ago
      It is very common to use Monte Carlo for deterministic problems. It's just an integral over complicated PDFs
    • NL807 2y ago
      >it's all deterministic and there is no opposing player making choices It's not deterministic, it's chaotic. That is the nature of the N-body problem. We can only approximate trajectories in such a system using numerical methods, within a certain margin of error. In principle, the object is gravitationally interacting with everything else in the solar system. But for the most part, most interactions are negligible and could be ignored (eg, other small objects far away), except of the large bodies. But there are many unknowns (as stated before), where initial conditions will affect the outcome of the trajectory simulation, and errors will certainly amplify over time. I'm guessing Monte Carlo is used to "fuzz" the simulations with randomised initial conditions to account for the range of unknowns, and see what the outcome is under these different scenarios.
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      • jaggederest 2y ago
        At the very least you can use monte carlo to provide examples from the expected distribution. It's hard to visualize a probability cloud.
      • coderenegade 2y ago
        Chaotic doesn't mean non-deterministic, it just means that small changes in initial conditions result in a large change in the trajectory. The system itself can be both chaotic and deterministic. It's also a reasonable question to ask, because the simulations are deterministic. It's just that because the system is also chaotic and there's noise in the measurement, that can result in a large spread of deterministic trajectory simulations.
        • NL807 2y ago
          It's only deterministic in the sense of the mathematical constructs that models the system, like differential equations that drive the simulations at each finite time step. But the information or the state which the simulation is applied on is always chaotic. That is because delta at each time step is an approximation with some error. It's impossible to make the state in the system behave deterministically, because that requires time deltas to approach to zero (or infinite amount of infinitely small differential steps).
    • refulgentis 2y ago
      Right: - "Since we saw it so briefly, our knowledge of its orbit is not that great" - "[for example, in 2016 the data shows] a large chunk of sky where it could have been, and [the object is quite small." - "Our knowledge of the diameter of this object is a bit fuzzy, because of surface reflectivity,"
    • baq 2y ago
      Looks like alpha zero stole the original meaning of Monte Carlo… it is not required to have players, or a game, to use the method: https://en.m.wikipedia.org/wiki/Monte_Carlo_method https://en.m.wikipedia.org/wiki/Monte_Carlo_method
      • bananaflag 2y ago
        It wasn't stolen, it was a well-established particular case well before alphago https://en.wikipedia.org/wiki/Monte_Carlo_tree_search https://en.wikipedia.org/wiki/Monte_Carlo_tree_search
        • baq 2y ago
          I only meant the total mind share :) looks like folks have never heard of MC in other contexts than game bots.
    • RossBencina 2y ago
      I would assume that it is because we have imperfect knowledge of the state of the asteroid (i.e. mass and current position/velocity/...). This imperfect knowledge is characterised by a probability distribution. Similarly the state of all other objects in the solar system is only known up to some distribution. To propagate the information forward in time to impact requires a complicated function f(state of solar system; state of asteroid). If all of the data was known (and expressible numerically) with perfect accuracy, and f were computable with perfect accuracy then all would be good. But as noted, (state of solar system; state of asteroid) is a probability distribution, and there are very few distributions and very few types of maps f that are amenable to analytic transformation. For example if the state was a normal distribution with mean x and covariance P, and f were a linear transformation, then x,P mapped through f is also normally distributed with mean y and covariance P_y, you can get the mean of the transform as y=fx, and P_y = fPf' (where ' indicates transpose). Needless to say our knowledge of the state of the asteroid and the solar system is probably a rather complicated distribution, and the n-body problem is not a linear transformation. Monte-carlo simulation is often used to propagate probability distributions through non-linear transformations.
    • andrepd 2y ago
      It's "deterministic" the same way the weather is.
      • brewtide 2y ago
        Exactly. It will do exactly this, as long as nothing changes.
• sdenton4 2y ago
  There's going to be some degree of measurement error, which will likely be greater for objects which have not been observed many times. Multiple observations should allow both better estimation of the object attributes (average out the noise), and allow some judgement of the quality of predictions given what you think you know about it.
• SR2Z 2y ago
  My guess is that small objects like this suffer greatly from the 3-body problem, and multiple trajectories are generated from various starting points inside our measured error bars for the current states of these objects. Small inaccuracies compound over the years.
  • mmooss 2y ago
    > My guess is that small objects like this suffer greatly from the 3-body problem What bodies? My impression is that the only objects around Earth with enough gravity to significantly impact trajectories are the Earth and Moon. Will the other small objects have any significant gravitational impact on this body? I also understand that in cislunar space, the Earth-Moon dynamic does create a three-body problem and trajectories are fundamentally unpredictable, with some exceptions. I wonder how that affects objects such as this one if they pass through the Moon's gravitational well.
    • SR2Z 2y ago
      Clicking the link, I found this visualization of the approximate orbit: https://ssd.jpl.nasa.gov/tools/sbdb_lookup.html#/?sstr=2024%20YR4&view=VOP https://ssd.jpl.nasa.gov/tools/sbdb_lookup.html#/?sstr=2024%... This orbit is around the sun (as asteroids tend to) and the apoapsis is closer to Jupiter's orbit than Mars.
      • pclmulqdq 2y ago
        Also, the gravity of asteroids or small planetary bodies like moons that it passes close to will have some small effects that can add up over a long time period.
    • slashdev 2y ago
      The Sun, Jupiter, Saturn are the main ones. Depends on the orbit as to how much influence they have.
    • baq 2y ago
      Gravity has unlimited range, the patched conics method you think of is a good approximation on short time spans, but breaks down surprisingly quickly. Keep in mind the Sun moves all the water in Earth’s oceans all the time…
• bagels 2y ago
  Mesurement uncertainty when propagated over long periods of time leads to very large uncertainties, imperfect gravity models, space weather
• ddahlen 2y ago
  It's a bit of both, observing has uncertainty in a lot of places, if you are on the ground you get atmospheric effects, imprecision of timing, imprecision of optics, etc etc. You are also observing an object where you dont know how far away it is. That distance has to be solved by basically doing a sort of triangulation, which requires either the observer or the object to move enough. So if you observe over a short time (hours for example), you can see it is moving, but it is hard to tell distance. Once you have an estimated orbit, if it has any interactions with planets (IE: flyby of Earth), small differences in positions during the close encounter make LARGE differences decades later. Add to this the effects of photons from the sun pushing on the smaller asteroids or dust, or out-gassing /dust from comets cause these objects to slightly drift from just the basic gravitational forces. Generally inner solar system asteroids (inside mars) are very chaotic over hundreds of years, though typically predictable less than a century. Note that I am not an expert on impact calculations, I just know a bit about and and can do back of the envelope ones. There are a number of ways to get to the ~1%, the orbit fits have uncertainties on them and those can be propagated forward in time. However there are all sorts of complexities with doing that, and often the easiest method is to sample the uncertainty region a few hundred thousand times (Monte-carlo), and propagate those and see what hits.
  • coderenegade 2y ago
    Very cool. How are samples drawn from the uncertainty region? MCMC or does it simplify down? I'm guessing that this would drive the final percentage values that you guys determine, since the orbital dynamics would be deterministic.
    • ddahlen 2y ago
      I can tell you how I do it, but again I am not an impact study person. It helps to understand a bit of the background of how we fit orbits in general: 1) someone with a telescope sees something moving (typically these days these are bigger surveys) 2) These observations are submitted to the Minor Plant Center (MPC), the clearinghouse of all asteroid/comet observations. 3) Several groups pull observations from the MPC to fit orbits, including JPL Horizons (MPC also fits orbits) 4) You now have a pile of observations which you have to figure out which observation links to other observations, which is a complex math problem on its own. Solve that. 5) JPL Horizons for example then fits the orbits to the observations, and since the observations may be 100 years of data of wildly varying quality, from hand written notes in the 1920s through to modern data, this is very difficult. They publish a covariance matrix with the associated fit (IE: basically a gaussian error fit for the parameters). 6) I grab that covariance matrix and sample from it using some pretty vanilla statistics to build orbits. 7) Propagate and see what happens. Here is an example of an observation from 1950: https://caltech-ipac.github.io/kete/tutorials/palomar.html https://caltech-ipac.github.io/kete/tutorials/palomar.html The image was developed on a glass plate, this one was never even sent in to the MPC, the guy taking the observation just wrote down "Asteroid" on the cover slip for the image. It was not formally discovered until the 1980s. We now know its orbit very well, so this particular observation is not that interesting other than as a curiosity. Here is an example of an orbit fit by JPL Horizons: https://ssd.jpl.nasa.gov/tools/sbdb_lookup.html#/?sstr=c%2F2010%20l5 https://ssd.jpl.nasa.gov/tools/sbdb_lookup.html#/?sstr=c%2F2... Note the "condition code" on the right, which is a score of how good their orbit fit matches the data, 0 means we know the orbit with high precision. This one is an 8, meaning we have a fit, but its not that great. Most likely because we only have 31 days of observations.