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Science: asteroid vs. hero physics
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<blockquote data-quote="freyar" data-source="post: 7493309" data-attributes="member: 40227"><p>Sorry I haven't been able to respond to this for a week. But it's important enough to get things right for the public that I don't want to let it pass. There is a lot correct here, but it's missing enough that the overall message is incorrect.</p><p></p><p></p><p></p><p>In both cases, the force of an object from the wind is in the direction of the air's velocity <em>relative to the object</em>. Let me just address your second example, since the logic holds in the first example also. You are right that the car should experience the same force in both the city frame and the car frame. You're also right that the force in the car frame is "down-right" because the wind velocity is at an angle with respect to the grid in this frame. Now look at the frame of the city. You are right that the wind blows across the grid, exerting a force to the right on the car. You have forgotten that the car is moving "up" in this frame and pushing through the air, which by Newton's 3rd law is pushing back -- "down." So in this frame, the car still experiences a "down-right" force from the air.</p><p></p><p>The simplest example of this I can think of at the moment is the following. Suppose you're standing outside with no wind, ie, the air is still with respect to the ground. Now you start running. You will feel wind in your face. This is because, in your frame, the air is moving "backward" and exerting a force (what you feel) on your skin. In the ground's frame, the air exerts a force on you because you are pushing through it.</p><p></p><p></p><p></p><p></p><p>Yes, that is what I've done. You have to change the velocity of <em>everything</em> when changing frames, including the wind. You're also right that the force experienced by an object is the same in both frames, and I've explained why that happens in your example.</p><p></p><p></p><p></p><p>I'm afraid I don't quite follow what you're setting up here. But you're right that the force doesn't change directions --- but the wind has to change velocities because the force of a wind on an object is in the direction of the <em>difference</em> of the object's and wind's velocities.</p><p></p><p></p><p>Again, you're right, but you've forgotten parts of the force, as I've said above, so your overall conclusions above are incorrect. And I'm not entirely sure what you're talking about integrating in the last sentence here.</p><p></p><p></p><p>No, I have made no such assumption. No matter the impact location on the earth, at a given time before impact, there is a minimum angle that the relative velocity (that is, the asteroid's velocity in the earth's frame) must be changed to avoid hitting the earth somewhere. The calculation I gave found the direction of push that Pierce can make to maximize the deflection angle, assuming Pierce can impart a fixed (or maximum) magnitude of momentum to the asteroid. To be fair, this calculation is axially symmetric about the relative velocity, so it doesn't tell you which direction around that circle to push in order to get clear of the closest edge of the earth, but the calculation does what I said it does.</p><p></p><p></p><p></p><p>As with the pictures you've uploaded previously, you seem to be making some assumption about the direction of the asteroid's approach in the solar frame. For example, if the collision is head on, slowing down or speeding up the asteroid won't help (unless of course you can slow it enough that it just precedes the earth around its orbit).</p></blockquote><p></p>
[QUOTE="freyar, post: 7493309, member: 40227"] Sorry I haven't been able to respond to this for a week. But it's important enough to get things right for the public that I don't want to let it pass. There is a lot correct here, but it's missing enough that the overall message is incorrect. In both cases, the force of an object from the wind is in the direction of the air's velocity [I]relative to the object[/I]. Let me just address your second example, since the logic holds in the first example also. You are right that the car should experience the same force in both the city frame and the car frame. You're also right that the force in the car frame is "down-right" because the wind velocity is at an angle with respect to the grid in this frame. Now look at the frame of the city. You are right that the wind blows across the grid, exerting a force to the right on the car. You have forgotten that the car is moving "up" in this frame and pushing through the air, which by Newton's 3rd law is pushing back -- "down." So in this frame, the car still experiences a "down-right" force from the air. The simplest example of this I can think of at the moment is the following. Suppose you're standing outside with no wind, ie, the air is still with respect to the ground. Now you start running. You will feel wind in your face. This is because, in your frame, the air is moving "backward" and exerting a force (what you feel) on your skin. In the ground's frame, the air exerts a force on you because you are pushing through it. Yes, that is what I've done. You have to change the velocity of [I]everything[/I] when changing frames, including the wind. You're also right that the force experienced by an object is the same in both frames, and I've explained why that happens in your example. I'm afraid I don't quite follow what you're setting up here. But you're right that the force doesn't change directions --- but the wind has to change velocities because the force of a wind on an object is in the direction of the [I]difference[/I] of the object's and wind's velocities. Again, you're right, but you've forgotten parts of the force, as I've said above, so your overall conclusions above are incorrect. And I'm not entirely sure what you're talking about integrating in the last sentence here. No, I have made no such assumption. No matter the impact location on the earth, at a given time before impact, there is a minimum angle that the relative velocity (that is, the asteroid's velocity in the earth's frame) must be changed to avoid hitting the earth somewhere. The calculation I gave found the direction of push that Pierce can make to maximize the deflection angle, assuming Pierce can impart a fixed (or maximum) magnitude of momentum to the asteroid. To be fair, this calculation is axially symmetric about the relative velocity, so it doesn't tell you which direction around that circle to push in order to get clear of the closest edge of the earth, but the calculation does what I said it does. As with the pictures you've uploaded previously, you seem to be making some assumption about the direction of the asteroid's approach in the solar frame. For example, if the collision is head on, slowing down or speeding up the asteroid won't help (unless of course you can slow it enough that it just precedes the earth around its orbit). [/QUOTE]
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