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Message: <blockquote data-quote="Ovinomancer" data-source="post: 7495222" data-attributes="member: 16814">Yep, I missed that Excel was using radians.  It was late, I wasn't paying attention, and it should have jumped up and smacked me.  So, yes, between 6 and 7 degrees (I see you edited your original claim to include this update, which is, oddly again, one of the reasons I went to check, because ~.12 degrees doesn't solve as you initially claimed, seems we both had issues mistaking radians for degrees).  Sadly, I originally had this worked out in degrees using a calculator, which helped confuse me on the Excel sheet because, in radians, there's also a point between 6 and 7 radians were the value crosses 1.  Okay, here's the issue -- your formula does work, but not how you originally said it does -- in the Earth frame.  Your formula works in the Solar frame, and only works in the Earth frame in a case where the path of the asteroid goes through the center of Earth.  All the shifting of the scenarios made me miss this when we did shift to this assumption.  As I documented above, the assumption that the path of the Asteroid goes through the center of Earth is erroneous in all cases except a head-on collision.  Even when we make the flat disc assumption for simplification of the problem, the change in frame to the Earth frame means the asteroid is not approaching the disc from the perpendicular anymore, and so any calculated angle from the path of the asteroid in the Earth frame is incorrect (although it may cause a miss).  The Solar frame doesn't have this problem, so your formula works.  It also works in this problem, because we've changed the assumption to a head on collision where the path of the asteroid lies through the center of the Earth/approaches the disc of Earth from the perpendicular.One of the reasons I was making the 'forces don't change direction on frame change' argument so strongly is because of this -- the apparent approach vector doesn't matter, it's the force applied that matters, and this becomes apparent if you actually don't use a point mass for Earth and have the asteroid impact on the surface.  When you shift frames, the impact point being the same means the asteroid's path doesn't lie through the center of Earth (it actually doesn't in the Solar frame, either, but the path does align with the center of Earth at the time of impact in the cases laid out above -- it doesn't have to, but that changes the math which is still easier to deal with in the Solar frame).</blockquote>

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