3 ms·

Stupid question: if the impact time is +/- 10 minutes, doesn't that translate into a lot of uncertainty for where it will hit? How do they know it's going to be

Stupid question: if the impact time is +/- 10 minutes, doesn't that translate into a lot of uncertainty for where it will hit? How do they know it's going to be Rouen?

dredmorbius 4y ago
My (lay, nonexpert) understanding is: - The centroid of probability is near Rouen. - Earlier/later impact probability would be along a path along the movement/orbit of the impactor. I'm going to guess that that's largely east-west track, though with a possible orbital inclination. - The probability distribution across that range is likely far higher near the centre than at the extremes. - Relative motion of Earth's rotation (~1,600 km/hr at the equator, less than that at the latitude of Rouen) is a much smaller component than the likely relative velocity of the meteorite (~20 km/s). - 20 km/s +/- 10 minutes ... gives 12,000 km as a possible range around the impact site. There's also the question of accurately noting the original position of the impactor. That said: good question and I'd love to hear from someone who knows what they're talking about. Thought also comes to mind of Aristotle's future contingents, "there will be a sea battle tomorrow" ...
- cshimmin 4y ago
  Hmmm, interesting analysis, but 12,000km is 3x wider than continental US. Based on the plot from ESA's projection, the track over the +-10m period is much smaller than France. I think the issue with your reasoning is that when you are dealing with orbits, you cannot really talk separately about position and (relative) velocity. Instead, for a given orbital trajectory, the position completely defines the speed. Moreover, your estimate might tell us how far _in space_ the impact would be given +/- 10min, but since both earth and asteroid are moving though space that doesn't really tell us much about how far apart it would be in an earthbound reference frame. In practice, I don't think they can accurately measure the position OR speed of an object like this... Instead, they make multiple observations over time and can fit the "shape" of the orbit, which can be done even with a fair amount of noise since we know all orbits are elliptical. If the object's orbit doesn't intersect Earth's, there's no impact. If the orbits do interest, there _may or may not_ be an impact event. To determine this, you have to figure out if both the object and the earth will be at that intersection point, _at the same time_. I'm guessing this is why the quoted uncertainty is given in minutes.
  - dredmorbius 4y ago
    The 12,000 km estimate is straight maths assuming a (fairly typical) relative orbital velocity of ~20,000 km/s for the meteroid, and looking at distance travelled in 10 minutes. Mind that the geometry of the collision might truncate that considerably. Again, I'm not claiming expertise here, and the "straight maths" rubric isn't a justification for correctness so much as explaining how I arrived at the number.
    - cshimmin 4y ago
      Understood! My point was that, while dimensional analysis almost always works (including for this problem), you have to make sure you are using the right dimensionful quantities. In this case that there are constraints here due to orbital mechanics that were not respected, and hence the 12,000km is way off. To successfully apply DA in this problem, we should start by noting that since we are talking about impact, it is a foregone conclusion that the earth and the asteroid will be in the same place at the same time (your application instead was related to figuring out where earth and asteroid would be in relation to each over at a given time). Once you've specified that they will be at the same place, asking where it lands _on earth_ amounts to asking which part of the earth will be facing the asteroid at that time. Since there is a quoted uncertainty of +- 10min, the range of outcomes is almost entirely defined by the angular distance that the earth with travel in those 20min -- about half a time zone! This gets us much closer to the actual projection from ESA. The actual central location at the nominal time is defined almost entirely by the direction the asteroid is approaching from -- i.e. the relative inclination and phase anomaly of the orbits. So, the rest of the projected range can be chalked up to uncertainty on the geometric fit of the object's orbit based on astrometric observations. Which we can't apply DA to without further information.
      - dredmorbius 4y ago
        Thanks, appreciate the clarifications.
cshimmin 4y ago
Yes that's exactly right. You can see this information in the plot attached to the linked Twitter post. There is a range of times (indicated by color) and the corresponding projections for impact location. But note that meteors can be seen from quite far away. Large fireballs can be heard for dozens of miles around! Also, they posted a more recent update that the estimate is zeroing in on Rouen.
kijin 4y ago
Think of a bell curve, more or less. It's much, much, much more likely for the asteroid to come down near the center of the predicted range than some place 5 standard deviations away.

Stupid question: if the impact time is +/- 10 minutes, doesn't that translate into a lot of uncertainty for where it will hit? How do they know it's going to be Rouen?

dredmorbius 4y ago

My (lay, nonexpert) understanding is: - The centroid of probability is near Rouen. - Earlier/later impact probability would be along a path along the movement/orbit of the impactor. I'm going to guess that that's largely east-west track, though with a possible orbital inclination. - The probability distribution across that range is likely far higher near the centre than at the extremes. - Relative motion of Earth's rotation (~1,600 km/hr at the equator, less than that at the latitude of Rouen) is a much smaller component than the likely relative velocity of the meteorite (~20 km/s). - 20 km/s +/- 10 minutes ... gives 12,000 km as a possible range around the impact site. There's also the question of accurately noting the original position of the impactor. That said: good question and I'd love to hear from someone who knows what they're talking about. Thought also comes to mind of Aristotle's future contingents, "there will be a sea battle tomorrow" ...

cshimmin 4y ago
Hmmm, interesting analysis, but 12,000km is 3x wider than continental US. Based on the plot from ESA's projection, the track over the +-10m period is much smaller than France. I think the issue with your reasoning is that when you are dealing with orbits, you cannot really talk separately about position and (relative) velocity. Instead, for a given orbital trajectory, the position completely defines the speed. Moreover, your estimate might tell us how far _in space_ the impact would be given +/- 10min, but since both earth and asteroid are moving though space that doesn't really tell us much about how far apart it would be in an earthbound reference frame. In practice, I don't think they can accurately measure the position OR speed of an object like this... Instead, they make multiple observations over time and can fit the "shape" of the orbit, which can be done even with a fair amount of noise since we know all orbits are elliptical. If the object's orbit doesn't intersect Earth's, there's no impact. If the orbits do interest, there _may or may not_ be an impact event. To determine this, you have to figure out if both the object and the earth will be at that intersection point, _at the same time_. I'm guessing this is why the quoted uncertainty is given in minutes.
- dredmorbius 4y ago
  The 12,000 km estimate is straight maths assuming a (fairly typical) relative orbital velocity of ~20,000 km/s for the meteroid, and looking at distance travelled in 10 minutes. Mind that the geometry of the collision might truncate that considerably. Again, I'm not claiming expertise here, and the "straight maths" rubric isn't a justification for correctness so much as explaining how I arrived at the number.
  - cshimmin 4y ago
    Understood! My point was that, while dimensional analysis almost always works (including for this problem), you have to make sure you are using the right dimensionful quantities. In this case that there are constraints here due to orbital mechanics that were not respected, and hence the 12,000km is way off. To successfully apply DA in this problem, we should start by noting that since we are talking about impact, it is a foregone conclusion that the earth and the asteroid will be in the same place at the same time (your application instead was related to figuring out where earth and asteroid would be in relation to each over at a given time). Once you've specified that they will be at the same place, asking where it lands _on earth_ amounts to asking which part of the earth will be facing the asteroid at that time. Since there is a quoted uncertainty of +- 10min, the range of outcomes is almost entirely defined by the angular distance that the earth with travel in those 20min -- about half a time zone! This gets us much closer to the actual projection from ESA. The actual central location at the nominal time is defined almost entirely by the direction the asteroid is approaching from -- i.e. the relative inclination and phase anomaly of the orbits. So, the rest of the projected range can be chalked up to uncertainty on the geometric fit of the object's orbit based on astrometric observations. Which we can't apply DA to without further information.
    - dredmorbius 4y ago
      Thanks, appreciate the clarifications.

cshimmin 4y ago

Yes that's exactly right. You can see this information in the plot attached to the linked Twitter post. There is a range of times (indicated by color) and the corresponding projections for impact location. But note that meteors can be seen from quite far away. Large fireballs can be heard for dozens of miles around! Also, they posted a more recent update that the estimate is zeroing in on Rouen.

kijin 4y ago

Think of a bell curve, more or less. It's much, much, much more likely for the asteroid to come down near the center of the predicted range than some place 5 standard deviations away.