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Message: <blockquote data-quote="freyar" data-source="post: 7489763" data-attributes="member: 40227">We must somehow be talking about different things.  Let me just give an example on the earth.  Start in the rest frame of some city streets laid out on a rectangular grid with the wind blowing directly toward the right on the grid.  Suppose there is a car driving "up" on the grid, perpendicular to the wind in the frame of the streets.  In the rest frame of the car, the grid is moving "down," and the wind is moving at the diagonal down and right.  It has clearly shifted.  The same is true in this case: Pierce's velocity will be an apparently different vector when measured in the earth or sun rest frames.  You agree, right?No, I mean the angle from the perpendicular.  So if the asteroid is infinitely massive and therefore has infinite momentum p for finite momentum transfer dp, you push perpendicular to the asteroid's travel to get the maximum change in the direction of the trajectory.I must not have made my assumptions clear.  (1) I am working in the earth's rest frame. (2) I am assuming that Pierce can transfer a maximum total amount of momentum dp to the asteroid; this fits with the idea of Pierce ramming into the asteroid once, but it also works with a continuous push over some time (EDIT: to be clear, the continuous push would need to end sufficiently before the asteroid passes the earth that it's on a straight-line path again as it passes earth, so this really is best for a single big push).  Therefore, in vector terms, the initial asteroid momentum is vec(p), and the final asteroid momentum is vec(p)+vec(dp).  (3) In the earth's frame, assuming Pierce can't ram the asteroid enough to turn it around, we are concerned about the change in direction of the asteroid's path, which we can give by the angle y between the initial and final asteroid momenta.  (5) If vec(p) is oriented at angle x compared to the perpendicular to vec(p) (and with parallel component oriented opposite vec(p) in order to slow down the asteroid), the deflection angle y is given by tan<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44d.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(y)" title="Thumbs up    (y)"  data-smilie="22"data-shortname="(y)" /> = dp cos(x)/ (p-dp sin(x)). (6) Because tangent is monotonic, y is maximized by maximizing the tangent.  We have assumed dp is fixed, so we are maximizing with respect to x.  Differentiating, we find maximization occurs when -dp sin(x)/ (p-dp sin(x))+dp^2 cos(x)^2/(p-dp sin(x))^2=0, which is equivalent to cos(x)=0 (which minimizes the deflection angle at y=0) or-sin(x) (p-dp sin(x))+dp cos (x)^2 =0, which solves to sin(x) =dp/p (remembering cos^2+sin^2=1).(7) All this ignore's the earth's gravity, but I think we've all agreed that the asteroid is probably moving fast enough that gravitational attraction is unimportant as a first approximation.Since the asteroid's mass is constant, the change in its momentum and its velocity tell us the same thing.  But what's important is that we know Pierce has some max momentum she can transfer to the asteroid, so that's what we have to work with.  The important thing is that, no matter how the asteroid is going to hit the earth if left alone, all that matters to whether it will hit the earth is how much Pierce can deflect it in the earth's frame.  This we can figure out using conservation of momentum.  What I've told you is, given how much momentum Pierce can gather and transfer to the asteroid, what's the best angle for her to take in order to deflect it the most.</blockquote>

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