Right, the direction of the change in momentum of the asteroid (or the force on the asteroid, assuming a sharp push) remains the same in all frames. But the direction of the asteroid's initial motion is different in different frames. The discussion has been about the angle between these two vectors, which is different in the different frames.
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.
That is why it can be impossible to cause a miss in the earth's frame by pushing straight back on the asteroid and just slowing down (again, unless you can push hard enough to stop the asteroid completely) but possible to cause a miss by just slowing the asteroid but not changing its direction in the solar frame.
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.
Yes, it's hard to remember to use degrees, which are a weird unit. But please note that I edited it about an hour and a quarter after the initial post.
The solutions between 6 and 7 radians are just the two solutions we've been talking about shifted by 2 Pi because sine and cosine are periodic.
Um, yes? I wasn't confused about that once I realized my error in radians vs degrees.
Let me explain what my formula is for again and how it might be useful for the asteroid problem Janx posed again, since I must not have made it clear enough.
1) Ignore our discussion about reference frames for a moment. In any arbitrary frame, some object (in our case, the asteroid) is moving with some initial momentum vector p. We are able to strike it and change its momentum by vector dp. Now imagine a plane drawn perpendicular to the vector p (the equatorial plane, so to speak) and define x as the angle between vector dp and that plane where positive x corresponds to the component of dp parallel to p pushing against p (that is, slowing the object down some). Then choosing x such that sin(x)=dp/p maximizes the deflection angle (the angle between the object's initial velocity and final velocity) in the frame we're working in for a fixed magnitude of dp. In other words, if you can only push so hard, how do you get the most bang for your buck?
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.
2) The problem we are considering is how to get an asteroid to miss the earth. I would like to consider this problem in terms of the deflection angle of the asteroid. What frame lets us do that? The earth frame. In the rest frame of the earth, the earth is sitting still. The question is just how to get the asteroid not to be aimed at the earth, in other words, how to deflect the direction of the asteroid's motion. That is not true in the solar frame. As you have told us in fact, it is possible in the solar frame to get the asteroid to miss the earth by slowing it down or speeding it up without changing its direction. This may or may not be the easiest thing to do, but it can work. As stated before, all this rests on the assumption that the asteroid is no more than a couple of hours out, so we can safely ignore the fact that the earth is accelerating in its orbit.
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.
3) Nowhere is it necessary to assume that the asteroid hits the earth at the perpendicular or that its path will go through the center of the earth. However the asteroid is going to hit, if we're looking at this in the earth frame, there is a minimal deflection angle needed to cause a miss. Now, if the asteroid's trajectory isn't straight down the center of the earth, that minimal deflection angle may not be the same in all directions. So then you also have to choose the direction of vector dp in the equatorial plane. It might help if you sketch this out. I'd really rather not have to take the time to draw pictures.
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.
If you want to know whether the asteroid hits the earth, the approach vector (initial asteroid velocity) always matters in any frame. Think about it. If the asteroid isn't initially aimed to hit some point of the earth, who cares? (Extreme example, I know.) Whether it hits the center of the earth is not really the point. I would also argue that the earth frame has simpler math. How can it make the problem easier if the earth is moving? That's an extra motion we have to keep track of. And, even if we wanted the earth to move, why choose the solar frame and not, say, the rest frame of Jupiter or something? As long as things are happening fast enough that we can ignore the acceleration of the earth's and asteroid's orbits to a reasonable first approximation, the sun doesn't affect the physics at all. So let's not introduce an extra confusing factor. Reducing a calculation to only the necessary elements is an early lesson in physics, or at least it should be.
I disagree on the frame for the example case. For others, sure. In the solar frame, I can very easily calculate the forces that will cause a miss in both the lateral and the parallel directions. Determining optimum angle of push with a force in between those two somewhat straightforward. In the Earth frame, I now have a lopsided target and non-right angles to calculate the minimum miss angles above and below. Let me present the drawing I did earlier:
As you see above, each speed (red was 20km/s, purple 30 km/s, and yellow 50 km/s lateral velocity) creates a different angle because of the displacement due to the size of Earth. Even if you use the disc Earth simplification, the angles of approach in the Earth frame don't change. This makes figuring the optimum angle harder -- it's not symmetrical and it's not sin(x)=dp/p measured from the perpendicular to the asteroid's path.
Let's take what we've already determined for a 30km/s asteroid in the above Solar frame (purple line). There exists a dp such that it will cause a miss if the angle of the push is perpendicular to the path of the asteroid. We solved for that, it was an instantaneously applied 1.81 km/s (downwards). That force can be applied to 0 degrees from perpendicular or ~6.12 degrees counterclockwise from perpendicular. We've agreed, here.
Now, if we switch to the Earth frame, the angle of the asteroids apparent path from the observer on the surface is 45 degrees. We've also agreed that until we change coordinates, the force we figured out above is still in the same direction, or perpendicular to the original path. This means that force is still applied downwards even with the new path. If we rotate the coordinate frame to match the current path of the asteroid, we rotate our measurements 45 degrees counterclockwise. Going from the vertical of the asteroid's path now, the original force is
clockwise by 45 degrees -- ie, behind the perpendicular. If we use your formula for the optimum angle of push now, sin(x)=dp/p where dp=1.81km/s and p=sqrt(30km/s^2+30km/s^2) = 42.4 km/s. x = 2.43 degrees clockwise from perpendicular. We don't have matching answers here.
If, instead, you use the 'equatorial' plane, then at least your optimum is between the two values for 1.81 (0 < 2.43 < 6.12). I'm not sure how valid that is, though.
To test the above, 1.81 km/s applied downwards means the down velocity component in the Earth frame is now 31.8 km/s and the 'left' component velocity is still 30km/s. It takes 1 hour to approach (this was the time assumption for the above graph, and the one we've been using recently), so it will take 3600s to reach the disc. At which point it will be 1.81 km/s * 3600 s = 6516 km further 'down', which is a miss. Barely, as predicted. If applied at 2.43 degrees, the left component of that is -0.0767 km/s and the 'down' part is 1.808. Left speed is 29.9233, so it will take 3609 seconds to approach, which means it will be 6,526.4 km further down. A bigger miss. Your formula does optimize on dp for push angle when used from the 'equatorial' plane (given the asteroid approaches on this plane in the solar frame, not sure of other cases). Notably, these angles work when mirrored across the horizontal as well (ie, 180 degrees and 177.57 degrees).
If we use the angle generated from the path of the asteroid in the Earth frame, it doesn't miss. I'm going to rotate it back to the initial coordinate plane, which means that push is now 47.43 degrees from the vertical in the 'equatorial' plane. Left component is -1.3256 km/s and down component is 1.2244 km/s. Asteroid approach time is 108,000 km/(30 km/s - 1.3256 km/s) = 3766.4 s. Down distance is then 4611.58. This is not a miss.
And that's all without getting into the psychology of it all. Humans have a strong tendency to think of the earth as still, and Pierce probably isn't going to overcome that kind of visceral intuition in the short time she has to deflect this thing. After all, how often do you think "the earth is moving past me at 60 mph" rather than "I'm driving at 60 mph" when you're out in your car? Furthermore, Pierce is presumably leaving on her mission from somewhere on earth, so she is providing force/thrust to herself to give herself a velocity relative to the earth. Why should she get out into space and then have to think to herself, "OK, now how fast am I going relative to the sun?"
I don't have a problem with it. But then, I like orbital mechanics stuff. If you can go into space, you figure it out pretty quickly. It's, heh, a different frame of reference.
I hope this clears things up.
Not sure. I'm still saying what I have been, and you've moved a bit in my direction. I've agreed your optimization formula works, at least in some scenarios. We're drifting towards agreement, but I think the big issue to overcome is the frame of reference issue -- forces don't change direction on a frame change, and this has pretty big repercussions. Also, assuming sin(x)=dp/p maximizes x in all cases isn't warranted -- you have to be careful where you measure from. The frame change to Earth needs to account for the non-perpendicular impact on Earth, which adds additional complexity in the Earth frame. We're closer, but not there.
