On Wed, Jan 16, 2002 at 06:09:46PM -0600, Dan Minette wrote: > Sorry I took so long, but its at a point where it looks as though we > need to be at a blackboard together.
No problem. > Let us consider 200,000,000,000 asteroids that are isotropically > distributed at 40,000,000 miles. Of them, using the numbers I gave > above, 1000 will pass within 4000 miles of the center of the earth and > 10,000,000 will pass within 400,000. > > OK, I bet you are saying so far so good. Yes. > So, now let us just consider the 10,000,000 that will pass within > 400,000 miles of the earth and your origional arguement, substituting > 400,000 for R and 4,000 for r: Those are different asteroids than the ones I was considering (they are not isotropic at R, for example). Anyway, I still was not sufficiently convinced after reading your explanation. Although I expected that your answer was correct, I still could not clearly understand the flaw in my approach. But you pointed me in the right direction (pun intended). I should not have considered an isotropic source at distance R. The proability distribution is, in fact, cos(theta), where theta is the angle of the asteroid's velocity vector with respect to a plane tangent to the sphere of radius R centered on the earth (in spherical coordinates, so theta is between 0 and 90 degrees and the distribution is normalized over this range). This is apparent if you consider asteroids as being distributed homogeneously and having isotropic velocity distribution, with average velocity v. Then consider the asteroids crossing a given plane from the left per unit area per second. The number crossing the plane is proportional to cos(theta). And due to symmetry of the assumptions made so far, every point on the plane is equivalent. So, consider the tangent point of the plane with the sphere of radius R centered on the earth, and the distribution is cos(theta). The equivalent solid angle probability distribution is 2*sin(theta)*cos(theta)*d(theta), and noting that the maximum angle resulting in an earth collision is sin(theta)= r / R, integrating from 0 to arcsin( r / R ) gives the probability of collision as exactly ( r / R ) ^ 2, as you obtained in a much less tedious fashion with a much more elegant method :-) -- "Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.com/
