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?
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
It wasn't stolen, it was a well-established particular case well before alphago
I work on the Near Earth Object Surveyor space telescope (NEO Surveyor) writing simulation code which predicts which objects we will see. This one has drummed up a bit of interest due to its (relatively) high chance of impact. I actually spent quite a bit of time yesterday digging through archive images trying trying to see if it was spotted on some previous times it came by the Earth (no luck unfortunately). Since w…
What are the implications of a moon impact? Would it affect life on Earth?
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 t…
Energy drift doesn't make the system non-deterministic, it just means that the time evolution has some error. The error is still deterministic. If you simulate a deterministic but chaotic system like n-body orbitals with a non-symplectic integrator, you'll always get the same result for the same initial conditions. The drift created by the finite timestep will also be the same.
It’s the error with the ground truth that you can’t predict. Otherwise you would just be able to cancel it out. You can only predict probability distributions..
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).
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?
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.
I work on the Near Earth Object Surveyor space telescope (NEO Surveyor) writing simulation code which predicts which objects we will see. This one has drummed up a bit of interest due to its (relatively) high chance of impact. I actually spent quite a bit of time yesterday digging through archive images trying trying to see if it was spotted on some previous times it came by the Earth (no luck unfortunately). Since w…
I wonder if we have a chance to catch it still on radar this pass. I know Arecibo used to do that and is now sadly gone, but Goldstone also has capability. Anyone know more?
The mass of such an object is likely be miniscule compared to the mass of the earth. So even a very close approach isn't likely to make much difference to the earth's orbit. E.g. an asteroid 100m across with the same density as the earth is going to have ~0.0000000000001 the mass of the earth.
Mass is relative to speed. If the object has a very high speed, it can reach the mass of the earth or even higher. That's why my understanding is that if it can escape the gravity of the earth, it means its own gravitational field is as or more powerful.
The mass of an object increases appreciably with velocity only at relativistic velocities. This is completely irrelevant at the speed an asteroid is likely to be travelling.
If an asteroid travelling at relativistic velocity passed through the atmosphere, a change in the earth's orbit would be the least of our worries!
Ah, I understand now. For one thing, you could weaponize asteroids as a second-strike response, if there's an imbalance where one superpower has asteroid deflection and one does not. Even if it's delayed by months, the certainty that the asteroid will eventually hit establishes a deterrent against attacking the asteroid-controlling country. If there's defenses against ballistic missiles, this circumvents those. For a…
...boomer submarines exist. Nuclear second strike has always been from the submarines, which guarantee a response within minutes. Versus your proposal here of launching a highly visible rocket, to manipulate a highly visible object, with a flight time measured in days to weeks that's mostly ballistic. Frankly: this is strategic nonsense.
It's nonsense today, but technical factors change over time. It's unsafe to try to predict too far ahead.
The mass of such an object is likely be miniscule compared to the mass of the earth. So even a very close approach isn't likely to make much difference to the earth's orbit. E.g. an asteroid 100m across with the same density as the earth is going to have ~0.0000000000001 the mass of the earth.
Mass is relative to speed. If the object has a very high speed, it can reach the mass of the earth or even higher. That's why my understanding is that if it can escape the gravity of the earth, it means its own gravitational field is as or more powerful.
That's only relevant at relativistic speed, an Asteroid isn't going to come in at 10% of the speed of light.
Is there a "sweet spot" for asteroids where they don't impact the earth, but come within like 20 meters of it? Like is it possible for an asteroid to give a city a "buzz cut" and then continue on into outer space?
realistically any scenario is possible if the asteroid is going fast enough
...and dense enough and with enough shear strength and yield strength. Air blast disintegration is making a lot of scenarios impossible.