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Message: <blockquote data-quote="freyar" data-source="post: 7495303" data-attributes="member: 40227">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.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, 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.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?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.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.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.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 hope this clears things up.</blockquote>

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