But it's not really, is it? It's about a preferred coordinate system for measurement -- the angle is the same referenced to the same vectors in either frame. Only if you add the asteroid's solar speed and Earth's solar speed when you change to Earth's frame does the apparent vector of the asteroid's momentum change, but the push is still in the same direction with regards to the asteroid. Measuring that push from a new vector is changing the coordinate system, not the observer frame. I brought up the difference between an observer frame shift and a coordinate shift long ago. The car example, for instance, deals only with an observer frame shift, not a coordinate shift. You're doing both now with the asteroid and claiming it's analogous -- it's not.
Shifting the asteroid's velocity (not speed) by the earth's velocity (both measured in the solar frame) is the mathematical representation of switching to the earth's frame. That's what changing frames means. If someone taught you otherwise, they did you a disservice. The little formula I wrote down uses the angle between the asteroid's initial momentum and the change in momentum it experiences when Pierce hits it. I have said all along that I have applied that in the earth frame. You therefore can't just apply my argument in the solar frame without adjustment.
This is exactly the problem you've been having with the car and the wind. The velocity of object B as measured in object A's rest frame is by definition the velocity of B relative to A. In a "lab frame" this relative velocity is vB-vA (subtraction done vectorially), and that's vB as measured in A's rest frame. I don't know how to make this clearer given how much we've discussed it already. If you can't understand or accept this fact, I don't see much point in continuing. I'll answer the rest below, but this is a key point.
Yes, but it's also impossible to cause a miss by pushing in that same direction in the Solar frame. No one's arguing that there are bad angles of push. Why this is a sticking point, I'm not sure.
As you pointed out all the way back in post 14 of this thread, in the solar frame, the asteroid is aiming where the earth will be, and the earth is moving. Therefore, you can slow the asteroid down without changing its direction and still get it to miss the earth without changing its direction. In the solar rest frame, not the earth's rest frame. Are you changing your mind on this? Because you were right then. The only problem then was that you insisted that the earth frame is invalid.
Um, yes? I wasn't confused about that once I realized my error in radians vs degrees.
Right, and, as I've said before, I'm not just writing for you but for other readers who maybe aren't quite as familiar with the math.
I understand that. I've challenged that it doesn't work except in the head-on case because Earth is off-center and it might not generate a miss depending on approach path. You acknowledge this in point 3, and offer a solution of not measuring from the path of the asteroid, which is what I've been saying.
I'm glad you understand. However, you've clearly misunderstood my point 3. I'll address that below.
Yeah, your formula doesn't create a miss in some geometries (it can fail in minimum dp cases for all but the head-on case, frex). I'm definitely not confused that the objective is to cause a miss. Explaining that to me is rather condescending. Yes, I made a few blunders because I haven't dusted off my trig outside of narrow applications for about a decade, but I've picked up everything you've laid down and caught a number of my own errors on doublecheck. I'm an electrical engineer by trade, so, yeah, I may be rusty but I don't need to be explained to that we're trying to get an asteroid to miss the Earth at this point.
The head-on, path through the center of the earth case actually requires a greater deflection angle and therefore harder push dp than most other cases.
As for the explanation, I'm sorry you found it condescending since that wasn't the intent. I was trying to be methodical and clear about why I am choosing to apply the results of point 1 in the earth frame. You haven't responded to that part. In any case, you clearly have a fine grasp of the math and basic physical laws. I just want to straighten out some confusion on working in reference frames.
But, if we're talking about tone of posts, please go back and look at yours, particularly post 27 in the thread where you called my prior post "not even wrong." I am well aware of the origin of the quote and its less than complimentary use in physics discussions.
I'm not sure what you mean by 'equatorial plane.' I'm going to assume you mean the 'East-West' in the non-rotated coordinate scheme (where Earth's movement relative to the Sun is 'up'), yeah? Okay, I'll buy that for the exact reason that it's the same plane as the Solar frame case under discussion, so I know that works. But, point in fact, this means that you're now agreeing with me that the optimum deflection angle is NOT from the perpendicular of the asteroid's path with sin(x)=dp/p, but instead from a different reference (excepting the head-on case)? Hallelujah! I'm confused, though, that you started this point with a refutation of this.
No, I am not agreeing with you at all. Once again, in the solar frame, you can get the asteroid to miss just by slowing it down, so deflection angle isn't the only thing to consider there. Let me explain what I said in my third point:
a) First off, I am now working exclusively in the earth's reference frame, so everything is measured relative to the earth.
b) I defined the equatorial plane as the plane through the center of the asteroid perpendicular to its velocity vector (see my point 1). If we break dp up into its components, it has one component along the asteroid's velocity and another component in the equatorial plane. The magnitude of the asteroid's deflection angle --- we called this angle a --- is unaffected by the orientation of that second component (all that matters are the magnitudes of the components), but which direction that deflection goes in does depend on the orientation of the component in the equatorial plane.
c) If our hero can intercept (and push) the asteroid at a certain time before collision with the earth, she has to deflect it by a minimal angle, which we have called m. To determine this angle, we draw a line tangent to the earth through the point where she intercepts the asteroid. The required minimal deflection angle is the angle between that tangent line and the asteroid's original path. The tangent line is a grazing trajectory where the asteroid just skims the top of the earth (or the atmosphere, however you like). Actually, there is a whole family of such tangent lines, but typically one will require a smaller deflection angle than the others. So the push should then be oriented properly in the asteroid's equatorial plane to take advantage of the smallest required deflection angle.
d) Note that the earlier Pierce can intercept the asteroid, the smaller the minimal required deflection angle will be, which I think we all agree on already.
I'm not going to reproduce your example and go through the whole argument, since it doesn't get to what I want to explain and what you seem to be confused about. In fact, you seem to be confusing what we've been calling angle x (the angle of the push relative to the asteroid's equatorial plane and which is equal to arcsin(dp/p) if we want to maximize deflection angle a given a fixed magnitude dp) with the minimal required deflection angle m. I will instead draw everything that I am talking about:
The earth is the big hollow circle. The filled one is the asteroid at the point where Pierce intercepts it. The black line is the asteroid's initial trajectory, and it's continuation through the earth is dotted. The green lines are tangents to the earth, indicating the minimal deflection needed to miss the earth (realistically, Pierce should give some leeway because we're ignoring the earth's gravity, etc). The solid one is the one with the smallest deflection angle, which is labeled m. The dashed one is another tangent which will cause the asteroid to miss but requires more deflection. I've labeled dp as a red vector hitting the asteroid along with angle x. On the other side of the asteroid is a red dashed vector indicating a push that could get the asteroid to deflect along the dashed green trajectory. None of this has numbers worked out, and it's not to scale, since that's not the point. But we know from previous work that the minimum push Pierce has to exert has dp/p = sin(m) oriented such that sin(x)=dp/p, or x=m. Please note that I very specifically did not use a head-on collision with the asteroid going through the center of the earth. Please also be forgiving if not all the tangents are precisely tangent, etc, since I did this in a hurry.
OK, last thing. I haven't wanted to use the "appeal to authority" fallacy because it's, well, a fallacy, but I would just like to ask you to consider that I know what I'm talking about even if you still find this confusing. As you may know from upthread or other EN World discussions, I am a physicist professionally. Not only that, I have taught a course on relativity, including a large section on changing reference frames in Newtonian mechanics every academic year for the past 7 years. So I hope that clears things up. I am going to try to get myself to stay out of this thread, since I can't really take credit for this at work.
