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Science: asteroid vs. hero physics
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<blockquote data-quote="Ovinomancer" data-source="post: 7488500" data-attributes="member: 16814"><p>You <em>don't </em>shift the wind. Unless and until you rotate the diagram, in which case you shift the wind, but rotation doesn't work well in this case because Earth is so big and the impact point, while appearing to be in line with Earth's center in one frame isn't in another. Here's a to scale diagram showing Earth as the circle on the left of a diameter including 150km of atmo (so it rounds up to a nice 13,000km diameter) and three asteroids (20km/s, 30km/s, 50km/s) at one hour from impact. This shows the travel of Earth in that time and the travel of the asteroids in that time in both the Solar frame (Earth moving up, asteroids moving left) and the Earth frame (diagonal asteroid paths). The grey arrow is the direction of an example push so that it's apparent that the shift in frame doesn't rotate the push. As you can see, you can't just rotate the Earth frame because the asteroids aren't moving towards Earth's center but instead a point on the circumference, and that circumference is so large you can't ignore it for the purposes of the example.</p><p></p><p>[ATTACH]100973[/ATTACH]</p><p></p><p>You must mean angle from the vector of the asteroid's travel, because otherwise a 90 degree push would be sin(0)=0.</p><p></p><p>That accounted for, the issue here is obvious when you consider it. The maximum dp occurs at sin(+/-90), but it would obviously require a much, much harder push if you went past 90 to generate a miss if you pushed from any angle behind the asteroid (ie, speeding it up while you generate a lateral velocity) but your formula has dp dropping off, not increasing. Also, it's possible to generate a miss by accelerating the asteroid "ahead" of the Earth, but your formula says needed dp there is 0.</p><p></p><p>I agree, there's an angle that minimizes pushed needed, but your assumption doesn't do it. Momentum is the wrong frame of thinking, as the asteroid's mass is constant so it falls out of the equations and you're just dealing with velocities. This is how most orbital problems work -- you deal in delta-v not momentum change. The challenge here is the point of impact being on the surface of a large circle instead of the center in the Earth frame, so the math gets messy. In the solar frame, it's two quickly moving objects. The math doesn't resolve into a simple sine equation.</p></blockquote><p></p>
[QUOTE="Ovinomancer, post: 7488500, member: 16814"] You [I]don't [/I]shift the wind. Unless and until you rotate the diagram, in which case you shift the wind, but rotation doesn't work well in this case because Earth is so big and the impact point, while appearing to be in line with Earth's center in one frame isn't in another. Here's a to scale diagram showing Earth as the circle on the left of a diameter including 150km of atmo (so it rounds up to a nice 13,000km diameter) and three asteroids (20km/s, 30km/s, 50km/s) at one hour from impact. This shows the travel of Earth in that time and the travel of the asteroids in that time in both the Solar frame (Earth moving up, asteroids moving left) and the Earth frame (diagonal asteroid paths). The grey arrow is the direction of an example push so that it's apparent that the shift in frame doesn't rotate the push. As you can see, you can't just rotate the Earth frame because the asteroids aren't moving towards Earth's center but instead a point on the circumference, and that circumference is so large you can't ignore it for the purposes of the example. [ATTACH=CONFIG]100973._xfImport[/ATTACH] You must mean angle from the vector of the asteroid's travel, because otherwise a 90 degree push would be sin(0)=0. That accounted for, the issue here is obvious when you consider it. The maximum dp occurs at sin(+/-90), but it would obviously require a much, much harder push if you went past 90 to generate a miss if you pushed from any angle behind the asteroid (ie, speeding it up while you generate a lateral velocity) but your formula has dp dropping off, not increasing. Also, it's possible to generate a miss by accelerating the asteroid "ahead" of the Earth, but your formula says needed dp there is 0. I agree, there's an angle that minimizes pushed needed, but your assumption doesn't do it. Momentum is the wrong frame of thinking, as the asteroid's mass is constant so it falls out of the equations and you're just dealing with velocities. This is how most orbital problems work -- you deal in delta-v not momentum change. The challenge here is the point of impact being on the surface of a large circle instead of the center in the Earth frame, so the math gets messy. In the solar frame, it's two quickly moving objects. The math doesn't resolve into a simple sine equation. [/QUOTE]
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