----- Original Message ----- From: "Erik Reuter" <[EMAIL PROTECTED]> To: "Brin-L" <[EMAIL PROTECTED]> Sent: Thursday, January 10, 2002 10:34 PM Subject: Re: another asteroid near miss
> On Tue, Jan 08, 2002 at 10:09:20AM -0600, Dan Minette wrote: > > > ----- Original Message ----- From: "Erik Reuter" > > <[EMAIL PROTECTED]> > > > > > > > > > Therefore, the probability P that the asteroid hits the earth is > > > > > > 2*Pi / (Pi*r^2/R^2) > > > > > > 1/P = 2 ( R / r ) ^2 or 20,000:1 (actually, 19,999:1 if we are > > > picky). > > > > > > Is there an error in my thinking? > > > > > > Let me use that same calculation twice to show that, together with > > the assumption made above, it comprises a system with internal > > inconsistency. Let us assume that R is 40,000,000 miles. Let us > > assume that there is indeed an isotropic source of asteroids in a thin > > spherical belt at that distance. Let us also neglect the effects of > > gravity (at 40 million miles this is certainly invalid, but it does > > help illustrate the difficulty in the technique). > > > > Let us look at two cases, where r1=400,000 miles, and where r2=4,000 > > miles. For r1, we have 1/P =2*100^2=20,000, for r2, we have > > 1/P=2*10000^2=200,000,000. The ratio of these two numbers is 10,000, > > not 20,000. > > Maybe I am just dense, but I don't see how this is inconsistent. You've > shown that if 1/P = a * ( R / r ) ^2 , then the ratio of P's for > different r's cancels out the constant 'a'. Please fill in the missing > steps from here to inconsistency. > Sorry I took so long, but its at a point where it looks as though we need to be at a blackboard together. 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. 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: For every asteroid that passes within 400,000 miles of the earth, consider the instant when the asteroid first intersects the sphere, centered at the earth, with radius 400,000 miles. Assume the radius of the earth is 4000 miles. Then the earth subtends a solid angle of Pi*4000^2/400,000^2 for the asteroid at that instant. The asteroid is equally likely to be heading in any solid angle of 2*Pi steradians (since it is going into the sphere, not out, it is half of 4*Pi). Therefore, the probability P that the asteroid hits the earth is 2*Pi / (Pi*4000^2/400000^2) 1/P = 2 ( 40000/ 4000 ) ^2 or 20,000:1 (actually, 19,999:1 if we are picky). > > Since you did the math OK, I think the problem is assuming a half > > isotropic source. > > Unfortunately, if I assume a full isotropic source, then I get > > 1/P = 4 ( R / r ) ^2 > I'm not saying its fully isotropic. I'm saying the asteroids who's closest approach to the earth in R cannot be replaced by either an isotropic source at R or by a half isotropic source at R. It is really a biased sample. Dan M. > which is even further from your answer. I suspect your answer is > correct, but I'd like to know exactly why my thinking is wrong here. > > > -- > "Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.com/ >
