I Probability distribution of angle of asteroid entry to the atmosphere

Used to play with gravitational attraction simulations ages ago. One thing I noticed it was difficult to get a small object to collide with a bigger spherical one vertically and far more likely to hit at an angle far from 0. Has the math of this been worked out for asteroids entering the earth's atmosphere? Maybe the distribution of the angle of entry has a peak at 45 degrees or whatever? Maybe observation data shows where the peak is and the shape is not symmetrical ?
 
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andrewkirk

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If we assume an asteroid can approach from any direction and that the Earth is perfectly spherical, the calculation is straightforward, because we can model it as a particle approaching a hemisphere head-on. It would get much more complicated if we relax those assumptions, but my feeling is that they would not make a huge difference to the probability.

With the hemisphere, we can calculate the probability of it striking at a particular angle theta by considering the closest distance r at which it would pass the Earth's centre, if the Earth were not there. The probability of the distance being between r and r+dr is ##p(r)dr = \frac{2\pi r\,dr}{\pi R^2}## where ##R## is the Earth's radius. The probability of the distance being less than ##r## is ##\int_0^r p(u)du = \left(\frac rR\right)^2##.
We can then do a change of variable and calculate the CDF and PDF for the angle theta, given that ##r=R\cos\theta##.

The result is that the probability density for angle ##\theta## is ##\sin 2\theta## and the probability of the angle being less than ##\theta## is ##\sin^2\theta##.

The probability density has a maximum when theta is 45 degrees, so your supposition is correct. The probability density is zero for theta equal to zero (brushing the surface) or ninety degrees (head-on impact).

This calculation also ignores the effect of gravity, which will make the object's trajectory curve inwards as it approaches Earth. However, if the object is travelling very fast relative to Earth, that effect should be minor.
 

stefan r

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the rotating surface would distort the model. The asteroid has to be flying through space angled eastward in order to leave a vertical trail in the atmosphere.

This calculation also ignores the effect of gravity, which will make the object's trajectory curve inwards as it approaches Earth. However, if the object is travelling very fast relative to Earth, that effect should be minor.
I thought most asteroids are prograde and part of the solar system. Earth's gravity would create a very large fraction of the total speed. The original post just said a "gravitational attraction simulations ages ago ". It is possible that it was only simulating low velocity Earth crossing asteroids.
 
Wow. I haven't understood yet how you got this result but it is impressive that you ended up with 45 degrees. Haven't pictured your geometry yet. Why is r = R cos theta?

Those simulations allowed you to give each object any initial position, any initial velocity and any mass. So you could make the meteorite orbit the earth object if you wanted. But you could not make the meteorite hit the atmosphere vertically without shooting it exactly towards the center of the earth, given an initially motionless earth.

In fact I think the meteorite cannot pass through the center of the earth (if the earth were not there) unless it is aimed at exactly the center of the motionless earth, thinking in terms of point masses. Is this true?
 

davenn

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just for your learning :smile:


Those simulations allowed you to give each object any initial position, any initial velocity and any mass. So you could make the meteorite meteor orbit the earth object if you wanted. But you could not make the meteorite meteor hit the atmosphere vertically without shooting it exactly towards the center of the earth, given an initially motionless earth.
Meteorite, once it is on the ground :smile:
 

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