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.
Here’s a thought experiment for you: in your city frame, there’s a flagpole ahead of the car with a flag blowing in the wind. The wind blows left to right, so the flag blows left to right. The car is approaching the stationary flagpole. Now, let’s shift the frame to that of the car. Now, it appears that the car is stationary and the flagpole is approaching the car. But the flag – the flag is still blowing left to right with the wind. It doesn’t swivel to blow diagonally when you change the frame.
Here’s a second though experiment. Let’s use your setup and agree that the wind shifts direction when you change frames. Agreeing to that, can we also agree that the wind exerts a force on the car? In the city frame, that force is left to right, ie, it pushes the car to the right. Perhaps the car is steering to overcome or the friction of the tires to lateral movement is sufficient, but, in any case, the wind is a force acting to push the car to the right in the city frame, yes? Now, shift to the car frame and assume, for the sake of the argument, that the wind is now blowing diagonally down and right towards the car. There’s now some part of the force pushing the car backwards. If this is true, the car must compensate against the wind in two directions rather than one, and that means it must have some forward force even in the car stationary frame or the rate that the city moves past must slow with the added force. This is clearly not the case – the car isn’t dealing with a different force when you change the frame.
In both cases, the force of an object from the wind is in the direction of the air's velocity
relative to the object. 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.
What you’ve done is a translation on velocity. In this translation, you’ve subtracted the velocity of the car and also subtracted that velocity from the city, resulting in the car having a zero velocity and the city gaining a velocity exactly opposite the car’s original velocity. But the wind is a force, ma, and ma+v doesn’t change the vector of the ma any more than moving a building would affect it’s velocity in the translation. Any given particle of the field of particles comprising the wind will, indeed, adopt the car’s velocity when you shift frames, but the force applied is still left to right.
Yes, that is what I've done. You have to change the velocity of
everything 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.
To make an even more clear example, let’s go back to the ball and the car. In the field’s frame, the car has a velocity v(car) upwards and the ball has a velocity v(ball) leftwards. There’s a wind applying a force to both blowing from right to left (ie, accelerating the ball leftwards). The car, due to steering or friction, resists this leftward force of the wind. Now, let’s switch frames. The car’s velocity is subtracted so it now has a relative velocity of 0. The car’s velocity is also subtracted from the ball, so it now has the velocity of the car downward and now appears to move left and down towards the car (in the lower right-hand corner). The wind still blows. If the wind’s direction doesn’t change, then the ball is being accelerated leftwards and will, increasingly, appear to veer to the left as viewed by the car. But, if what you say is true, and the wind also shifts to blow left and down, then the ball merely accelerates towards the car in a straight line. But, if we sum the forces, we have a problem. In the field frame, the ball is experiencing a net force to the left and the car is experiencing the same force with a balancing force (steering or friction) to the right. Now, if we shift and use the former (my) assumption that the wind does not shift, the forces on everything are the same – the car is still at 0 net force and the ball is still being accelerated to the leftwards. If, however, we use your assumption that the wind shifts, the car is now experiencing a down and left force and must compensate by pushing back right (steering or friction) and forward (ie, using the motor) while the ball is experiencing a net acceleration down and right. Your assumption means the car has to press the gas to counter the force of the wind when you shift from the field frame to the car frame. The ball is there to illustrate that the real effect is that the force doesn’t change vectors and will still be accelerated leftwards and no left and down.
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
difference of the object's and wind's velocities.
I may have just stumbled on a better way to say this: slope doesn’t change if you move the intercept. Y= mx+b. m is independent of b. Changing b is like the frame shifts we’re talking about – the force vector doesn’t change if you move around the velocities, just like the slope of a line doesn’t change if you move it’s y-intercept. To put it in calculus terms, you’re changing the +c, which doesn’t affect the integral.
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.
The thing that jumps out at me here is that you’ve assumed that the translation to the Earth frame centers the line of approach on Earth’s center. It doesn’t. Look at the graphic I posted above. The size of the Earth isn’t negligible in this calculation – it’s a sizable fraction of the distance traveled in 1 hour for all the cases discussed. Further, the impact being on the circumference of a 6500km radius circle means that the approaching vectors will not be coming straight down for an observer at that point but will instead by steeply angled with respect to the vertical (again, reference the diagram). This affects your formula which assumes a point mass for Earth such that a perpendicular to the asteroid’s path is parallel to the tangent line at the point of impact. This isn’t so, and dramatically affects the outcomes. For instance, your formula now has a ‘right’ and a ‘wrong’ side for pushing, where a hypothetical deflection from one side’s perpendicular causes a miss but the opposite force on the other perpendicular still results in a hit. For that reason, your simplification fails a first approximation due to bad assumptions.
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.
In the Solar frame, I can pretty easily figure what’s needed to generate a miss with either a lateral push or a slowdown push. The formulas are dependent on the asteroid speed and the time to impact.
V(slow) = V(ast)(1-t(i)/(t(i)+217)) -- Slowdown velocity is equal to the velocity of the asteroid times 1 minus the time of impact divided by the time of impact plus 217 seconds. The 217 comes from the time it takes Earth at 30km/s to clear it's own 6500km radius. That doesn't change.
V(lateral) = r(E)/t(i) -- lateral velocity is equal to the radius of the Earth divided by the time to impact.
This approximation treats Earth as a flat disc. For a sphere, you have to calculate a T(m) which is equal to r(E)/V(asteroid), or the time it takes the asteroid to travel a radius of the Earth. This is the time to clear the widest part of Earth from the impact point (assuming impact at the closest point). You add this to t(i) in the above. It complicates things a bit for a bit more fidelity. I'm going with the easy here.
So, for an asteroid going 20km/s 1 hour our, the V(s) is 20km/s(1-3600s/3817s) = 1.14 km/s. The V(l) is 1.81 km/s. Either of these will generate a miss. And optimal push would be at some angle between 0 and 90 degrees and would have a value of less than 1.14.
I've run out of time at this point. I might get back to the optimization in the Solar frame problem so that it could be checked against your formula.
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).
