Worlds in collision (Posted on 2008-02-16)

	Submitted by FrankM
Rating: 3.0000 (1 votes)
Solution:	(Hide)
We want 1/R1(t) = 1/R2(t+D) for some t. That is: p2(1 + e1cos(t)) = p1(1 + e2cos(t+D)), so [p1e2cos(D) - p2e1]cos(t) - p1e2sin(D)sin(t) = p2 - p1 But the expression Asin(t) + Bcos(t) assumes all values between +- sqr(A^2+B^2). Thus the intersection condition is met iff the sum of the squares of the cos(t) and sin(t) coefficients is greater than (p2 - p1)^2, i.e. p1^2e2^2 - 2p1p2e1e2cos(D) + p2^2e1^2 => p2^2 + p1^2 - 2p1p2 Rearranging terms gives the proposed inequality. ________________________________ Bonus problem Results from the previous problem can be reapplied to determine the point of intersection. Let R^2 = p1^2e2^2 - 2p1p2e1e2cos(D) + p2^2e1^2 and sin(H) = p1e2sin(D)/R. Then, the ellipse points are equidistant from the focus when cos(t* + H) = (p2 - p1)/R (if this has absolute value no greater than 1) We construct the vector w extending from the focus to the intersection point (in the ellipse plane an angle t* removed from the periapsis of ellipse 1). The planes' of the ellipses intersect along the line L1 x L2. The intersection condition is then that w must be parallel to one of the cross products +-L1 x L2.

	Subject	Author	Date
	computer exploration	Charlie	2008-02-16 12:48:43