Science: asteroid vs. hero physics

freyar

Extradimensional Explorer
But in order for a moving flagpole to still have its flag blowing straight out sideways, wouldn't the wind have to be blowing diagonally? After all, with no prevailing wind at all, the natural propensity for a flag on a moving pole would be to stream straight backwards from it, opposite to the direction of motion.
Yes.
Sum the forces.
He did. See my last post.
 

freyar

Extradimensional Explorer
In the Earth frame, do you have the asteroid approaching the disc at an angle or did you have it approach perpendicularly? Because the latter isn't correct -- the angle of appoach is 45 degrees for a 30km/s asteroid. It makes a big difference in the deflection vector.
You're clearly making some assumption about the approach of the asteroid and the earth in the solar frame --- in genera, you can have an asteroid approach with any angle and any speed. Did I miss something in the thread where Janx said how the asteroid is approaching?
 

Ovinomancer

No flips for you!
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 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.




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.



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.
Yes, I see that you don't get what I've set up, and it goes directly to your complaints above. Let's try again.

Let's set up the situation without the wind. In city frame, the car's motor is applying force to the car to move it along the road. The air, which is stationary, is applying resistance to that force in equal measure (the car is at constant velocity, so the forces are equal). What you have is:

Up: force applied by car's motor = down: force applied by air resistance.

Now, let's switch this to the car's frame. The car is not stationary and it appear that there's a strong headwind (air resistance) pushing the car backwards. The car's motor is applying force to counter that headwind (air resistance) and keep the city moving past at a constant rate. You have:

down: headwind (air resistance) = up: car's motor,

Good so far? Okay, let's add the wind.

Now there's a cross wind of some force F. First the city frame. The car's motor hasn't changed force, so the air resistance also hasn't changed force. Up and down forces are still equal. But, we've added a lateral force, the wind(F). This will push the car in the direction of the wind unless countered. Let's assume that the friction of the tires on the surface counter the wind, here, such that the lateral forces are balanced. You not have:
Up: car's motor = down: air resistance.
Right: wind (F) = Left: tire friction.

All forces balance. Now, let's shift to the car's frame, and follow your suggestion that the wind also shifts direction when we shift frames. We run into a problem. If the wind shifts by some angle x, the lateral force of that wind(F) is now cos(x)*F, and there's now a up/down componenet of sin(x)*F. If we sum the forces now, we have:

Down: air resistance + sin(x)*F = up: car's motor
Right: cos(x)*F = Left: tire friction

So, you're arguing that the tires exert less lateral friction force in the car frame than in the city frame, and that the car's motor must push harder to combat the increase in downward force because some part of the wind is adding to the air resistance. This doesn't add up, though, as the forces on the car DO NOT CHANGE with a frame shift -- they still must be the same forces as in any other frame.

The error in your thinking above is that the up/down component of the air's movement was already accounted for in the frames by the declaration of constant velocity -- meaning that the force of the car's motor exactly counters the force of the air resistance/wind in the up/down direction. You have the lateral and up/down forces changing in your frame change, and that just doesn't happen.


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.
Um, okay, let's look at your equation as the force of the push goes to zero. When this happens the limit of angle x goes to 0 degrees, which is nonsensical. IE, the less you push, the more perpendicular you should push. Further, if you increase your push to the velocity of the asteroid, sinx goes to 90. That kinda makes sense in that completely stopping the asteroid relative to Earth will cause a miss (although gravity now becomes a dominating factor). But what happens if you exceed the push? There exists some dp greater than p where your formula says that pushing the asteroid even faster directly at Earth generates a miss.

Your error here was assuming some dp and then finding a formula that created an x that seemed right to you. You're not testing other dp to find if the formula works. It doesn't.

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).
Yes, those assumptions were established well upthread as plausible given the launch point of the asteroid and how it should interact with Earth's orbit. A head on example in the Solar frame requires an extra-solar origin for the asteroid or an orbital period of much greater than 100 years (ie, the ellipse of the asteroid's path would have to be at Earth's orbit close to perihelion, and the combined speeds would be YUGE. Further, for your purposes, a head on in the Solar frame is exactly identical to the Earth frame, with the Earth's 30km/s transferred to the asteroid's velocity. At that point, your formula breaks very badly, as it suggests that the weaker your push, the more towards the perpendicular you should push but the stronger you push the more towards opposing the asteroid's path your should push. That doesn't work.

In the Earth frame, at a given distance, and assuming an asteroid path aimed straight at Earth, there exists a minimum angle that will cause a miss. This angle doesn't change based on the speed of the asteroid because the extended path will still need to be outside that angle. This angle is easy to determine if you know the distance (d) and Earth's radius (Re) as it's simply Tan-1 (Re/d). At sufficient distance, this angle is small. To generate a miss, you must create an instantaneous dp such that the resulting path exceeds this miss angle. This generates some math, but it's not too bad. Let's call the miss angle (m), the angle of applied dp (x), and the angle the asteroid takes as (a); all angles are measured clockwise from the path. Tan (a) is going to equal the lateral portion of dp, or dpcos(x) divided by p minus the vertical portion of dp, or dpsin(x). This is tan(a) = dpcos(x)/(p-dpsin(x)). We can set Tan(a) equal to Tan(m) and get that dpcos(x)/(p-dpsin(x))=Re/d. From this, we can see a few things. One, dp cannot go to 0, and two, x also cannot go to 0 (or 180). This eliminates a push in the direction of the asteroid from generating a miss, as I stated above. Now, let's solve for a dp if x = 90, ie, a perfectly lateral push. IN this case, the formula goes go dp/p=Re/d. Assuming p is known, we have a dp(lat) that generates a miss. Now, can this dp applied to any other angle x, generate a miss (ignoring 270)? Doing some substitution from above, this would mean that dp(lat)cos(x)/(p-dp(lat)sin(x))=dp(lat)/p. There is no solution for x for a known dp(lat) in this scenario where x can be anything other than 90. For a dp that exceeds dp(lat), you can have angles that aren't 90, because the resultant angle of miss can still be greater than m.

If we apply your forumla to this, though, where your sin(x)=dp/p, a known dp(lat) would result in an angle that isn't 90. Your construction doesn't work even if we use your assumption that the path of the asteroid is directly through Earth's center. In the realistic case, where it is not, it also doesn't work.
 
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Ovinomancer

No flips for you!
Yes.

He did. See my last post.
Nope! Flagpole in city frame. No up/down forces. Lateral forces are wind - being attached to the flagpole = 0.

Car frame: forces don't change. The 'head wind' the car feels isn't present for the flag, which is moving precisely along with and at the same speed as this 'headwind'. So, up/down forces are still zero. Side forces are also the same, ie: wind - being attached to the flagpole = 0. If the wind shifted to the down and right, the flagpole is now pushing up and left, but a change of frame doesn't affect the sum of the forces, so that's impossible.

Sum. The. Forces.
 

MarkB

Hero
Nope! Flagpole in city frame. No up/down forces. Lateral forces are wind - being attached to the flagpole = 0.

Car frame: forces don't change. The 'head wind' the car feels isn't present for the flag, which is moving precisely along with and at the same speed as this 'headwind'. So, up/down forces are still zero. Side forces are also the same, ie: wind - being attached to the flagpole = 0. If the wind shifted to the down and right, the flagpole is now pushing up and left, but a change of frame doesn't affect the sum of the forces, so that's impossible.

Sum. The. Forces.
The headwind does exist for the flagpole, it's just precisely countered by its movement. Being precisely balanced out doesn't make it non-existent. And since the headwind does exist for the car, when you add in the lateral wind those two factors combine into a diagonal wind. It is impossible that they would not.

Forget the flagpole for a moment, and consider instead a crisp packet blowing across the car's path, moved purely from the force of the wind. From the city's perspective, if the crisp packet starts blowing left-to-right across the car's path, it is moving precisely perpendicular to the car, while the car is moving forwards to intersect its path. Start them at the right moment, and they'll collide.

But switch to the car's perspective. The car is perfectly still, and the crisp packet is carried by the wind, starting forward and to the left of the car. If the wind is still directly perpendicular to the car's facing, and the car is stationary, then the car cannot possibly ever intersect with the crisp packet's path. The only way in which it can do so is if the packet is travelling diagonally, from a position forward and to the left of the car to a position behind and to the right of the car.
 

Ovinomancer

No flips for you!
The headwind does exist for the flagpole, it's just precisely countered by its movement. Being precisely balanced out doesn't make it non-existent. And since the headwind does exist for the car, when you add in the lateral wind those two factors combine into a diagonal wind. It is impossible that they would not.

Forget the flagpole for a moment, and consider instead a crisp packet blowing across the car's path, moved purely from the force of the wind. From the city's perspective, if the crisp packet starts blowing left-to-right across the car's path, it is moving precisely perpendicular to the car, while the car is moving forwards to intersect its path. Start them at the right moment, and they'll collide.

But switch to the car's perspective. The car is perfectly still, and the crisp packet is carried by the wind, starting forward and to the left of the car. If the wind is still directly perpendicular to the car's facing, and the car is stationary, then the car cannot possibly ever intersect with the crisp packet's path. The only way in which it can do so is if the packet is travelling diagonally, from a position forward and to the left of the car to a position behind and to the right of the car.
How does the headwind exist for the flagpole? Let's look at a no wind situation. The flag-pole in relation to the city frame has no movement. There's no wind, not push, it's entirely stationary. Now, shift that to the car's frame. The flagpole now appears to be moving toward the car along with the city. The air is also moving towards the car at the same speed. The flagpole doesn't experience any forces from the headwind, as it's moving along with it and not opposing it in any way.

When you add the wind, the above doesn't change for the up/down direction. The wind blows sideways against the flagpole exactly the same way in both frames. You can even see this if you shift to the flagpole's frame (hint: it's the same as the city frame).

As for the crisp packet, what does it look like with no wind? The crisp packet, alongside the road, is still moving 'downward' at the speed of the frame. When the wind blows to the right, and pushes it in front of the car, the speed at which it's moving 'downward' does not increase. IE, the wind acts to push the packet to the right, but doesn't increase it's speed 'downward'. The packet is still moving with all of the air at the same speed 'downward' with or without the wind. It's speed laterally increases with the wind, but the wind adds no speed 'downward'.
 

MarkB

Hero
How does the headwind exist for the flagpole? Let's look at a no wind situation. The flag-pole in relation to the city frame has no movement. There's no wind, not push, it's entirely stationary. Now, shift that to the car's frame. The flagpole now appears to be moving toward the car along with the city. The air is also moving towards the car at the same speed. The flagpole doesn't experience any forces from the headwind, as it's moving along with it and not opposing it in any way.
Why does that matter? The car is experiencing the headwind, and we're viewing things from the car's frame of reference. If the headwind exists for the car, then it exists for every other object in the model. The other objects simply happen to be travelling at a velocity that matches it.
 

MarkB

Hero
As for the crisp packet, what does it look like with no wind? The crisp packet, alongside the road, is still moving 'downward' at the speed of the frame. When the wind blows to the right, and pushes it in front of the car, the speed at which it's moving 'downward' does not increase. IE, the wind acts to push the packet to the right, but doesn't increase it's speed 'downward'. The packet is still moving with all of the air at the same speed 'downward' with or without the wind. It's speed laterally increases with the wind, but the wind adds no speed 'downward'.
Bear in mind that from the car's frame of reference, the car is stationary but the city around it is moving. For there to be no wind from the car's frame of reference, there would need to be an upward-blowing wind from the city's frame of reference, blowing at precisely the same velocity as the car.
 

Ovinomancer

No flips for you!
Why does that matter? The car is experiencing the headwind, and we're viewing things from the car's frame of reference. If the headwind exists for the car, then it exists for every other object in the model. The other objects simply happen to be travelling at a velocity that matches it.
Um, no, no it doesn't exist for everything else in the model. That's a big conceptual error, there. The point is that the flagpole is at rest with the city in all frames. That doesn't change. The car is not at rest with the city frame, so when you change frames, either the car is moving relative to the city frame or the city is moving relative to the car frame. But the flagpole remains at rest with the city regardless of the frame. The headwind the car experiences in the car frame is also at rest with the city -- it's blowing exactly at the same speed the city is moving in any frame. In the city frame, the air resistance is stationary with the city while the car moves against it. In the car frame, the air moves against the car with the city while the car remains motionless. But all of the air is moving with the city. When you add the wind, it creates a field effect to the right for all of that air, which is still at rest with regards to the city in the up/down direction in all frames.
 

Ovinomancer

No flips for you!
Bear in mind that from the car's frame of reference, the car is stationary but the city around it is moving. For there to be no wind from the car's frame of reference, there would need to be an upward-blowing wind from the city's frame of reference, blowing at precisely the same velocity as the car.
Yes... you're almost there!
 

MarkB

Hero
Um, no, no it doesn't exist for everything else in the model. That's a big conceptual error, there. The point is that the flagpole is at rest with the city in all frames.
But the city itself is not at rest in the car's frame. That's the part you're missing.
 

Ovinomancer

No flips for you!
But the city itself is not at rest in the car's frame. That's the part you're missing.
Where am I missing that? I just said that in the post you editted down to the above! Come on, now. The flagpole is part of the city frame, as is the air. It ALL is at rest with respect to each other and ALL in exactly the same motion with respect to the car. What they do not do is change motion with respect to each other.

Look, the forces are the same no matter what frame you're looking in. If you add up the forces in one frame, they are exactly the same in a different frame. What you're confusing here is trying to account for things partially. There are no forces operating on the flagpole in the up or down direction regardless of the frame. In the city frame, the air isn't moving against the flagpole in the up or down direction. In the car's frame, the air isn't moving against the flagpole in the up or down direction. The flagpole AND the air appear to be moving in the down direction, at exactly the same speed, from the car's perspective. But the actual forces on the flagpole do. not. change. Sum the forces yourself and see.

A force does not change direction with a frame shift of perspective. What you're proposing is that the car's motor would have to work harder if you change how you think of it's motion. Clearly this is false, so the error has to be in how you're conceiving of this. Make the jump.
 

MarkB

Hero
Where am I missing that? I just said that in the post you editted down to the above! Come on, now. The flagpole is part of the city frame, as is the air. It ALL is at rest with respect to each other and ALL in exactly the same motion with respect to the car. What they do not do is change motion with respect to each other.

Look, the forces are the same no matter what frame you're looking in. If you add up the forces in one frame, they are exactly the same in a different frame. What you're confusing here is trying to account for things partially. There are no forces operating on the flagpole in the up or down direction regardless of the frame. In the city frame, the air isn't moving against the flagpole in the up or down direction. In the car's frame, the air isn't moving against the flagpole in the up or down direction. The flagpole AND the air appear to be moving in the down direction, at exactly the same speed, from the car's perspective. But the actual forces on the flagpole do. not. change. Sum the forces yourself and see.

A force does not change direction with a frame shift of perspective. What you're proposing is that the car's motor would have to work harder if you change how you think of it's motion. Clearly this is false, so the error has to be in how you're conceiving of this. Make the jump.
You are seriously missing the point here. In one frame the city is stationary and the car is in motion. In the other the car is stationary and the city is in motion. Whether the car is working to move forwards across a stationary city, or to remain stationary as the city tries to carry it backwards, its motor exerts the same effort.

The error is in how you are conceiving of this. Regardless of the frame of reference, you are insisting upon picturing things only from the perspective of the city.
 

Ovinomancer

No flips for you!
You are seriously missing the point here. In one frame the city is stationary and the car is in motion. In the other the car is stationary and the city is in motion. Whether the car is working to move forwards across a stationary city, or to remain stationary as the city tries to carry it backwards, its motor exerts the same effort.
YES!!!!

The error is in how you are conceiving of this. Regardless of the frame of reference, you are insisting upon picturing things only from the perspective of the city.
No. Sigh.

Okay. Again. No wind. Car moves through city in city frame at 20 kph. It therefore feels a 20 kph headwind due to this movement, yes? Okay. Now, go to the car frame, the headwind is still 20 kph, right? Agreed?

Good.

Now, add in a 5 kph crosswind in the city frame. So, now car is experiencing a 20 kph headwind AND a 5 kph crosswind. Combined, this feels like a 20.6 kph wind coming in at about 14 degrees counterclockwise from the direction of the car's travel. Now, let's switch to the car's frame -- the wind is still 20.6 kph from 14 degree off of dead ahead. The car doesn't feel a different combined wind when you change the frame.

Now, that 20.6 kph wind at 14 degrees? It's still a 20 kph headwind and a 5 kph crosswind combining. IE, the crosswind doesn't change direction when you shift the frame.
 

MarkB

Hero
Okay. Again. No wind. Car moves through city in city frame at 20 kph. It therefore feels a 20 kph headwind due to this movement, yes? Okay. Now, go to the car frame, the headwind is still 20 kph, right? Agreed?

Good.

Now, add in a 5 kph crosswind in the city frame. So, now car is experiencing a 20 kph headwind AND a 5 kph crosswind. Combined, this feels like a 20.6 kph wind coming in at about 14 degrees counterclockwise from the direction of the car's travel. Now, let's switch to the car's frame -- the wind is still 20.6 kph from 14 degree off of dead ahead. The car doesn't feel a different combined wind when you change the frame.

Now, that 20.6 kph wind at 14 degrees? It's still a 20 kph headwind and a 5 kph crosswind combining. IE, the crosswind doesn't change direction when you shift the frame.
But we're not just talking about the crosswind. We're talking about the sum of the crosswind and the headwind - the latter of which only exists in the car's frame. So, as you've so elegantly demonstrated, the wind does change depending upon which frame you're in.
 

Ovinomancer

No flips for you!
But we're not just talking about the crosswind. We're talking about the sum of the crosswind and the headwind - the latter of which only exists in the car's frame. So, as you've so elegantly demonstrated, the wind does change depending upon which frame you're in.
Gah. Okay, once more unto the breach.

NO CROSSWIND CASE:
City frame -- car drives at 20 kph. Air resists. Force of the car's motor is exactly cancelled by the force of the air's resistance.
Car's frame -- headwind blows at 20 kph. Car resists. Force of the wind's push is exactly cancelled by the force of the car's motor.

CROSSWIND CASE:
5 kph crosswind at 90 degrees to the path of the car.
City frame -- car drives at 20 kph. Air resists. Force of the car's motor is exactly cancelled by the force of the air's resistance. Wind blows from left. Pushes car to right. Force of push exactly cancelled by tire friction.
Car frame -- headwind blows at 20 kph. Car resists. Force of the wind's push is exactly cancelled by the force of the car's motor. Wind blows from left. Pushes car to right. Force of push exactly cancelled by tire friction.

In both cases, the forces in the direction the car is pointing DO NOT CHANGE. In the crosswind case, there's a new force pushing the car to the right. This force doesn't change IN EITHER FRAME.

What you're doing is saying that the headwind is part of the crosswind. It isn't. It exists exactly the same whether the crosswind exists or not. If the crosswind exists, it exists exactly the same in either frame. The vector of adding the headwind to the crosswind in the car frame DOES NOT MEAN THE WIND CHANGES DIRECTION. It means you can add them if you want, like all vectors. But the forces to the front of the car and to the side of the car do not change when you change the frame. The crosswind applies the exact same force to the side of the car in either frame. It doesn't change direction.

Honestly, when I first saw this example, I was unexcited because of this exact problem -- the presence of air would lead to people making this exact mistake of conception with 'headwinds'.
 

freyar

Extradimensional Explorer
Yes, I see that you don't get what I've set up, and it goes directly to your complaints above. Let's try again.

Let's set up the situation without the wind. In city frame, the car's motor is applying force to the car to move it along the road. The air, which is stationary, is applying resistance to that force in equal measure (the car is at constant velocity, so the forces are equal). What you have is:

Up: force applied by car's motor = down: force applied by air resistance.

Now, let's switch this to the car's frame. The car is not stationary and it appear that there's a strong headwind (air resistance) pushing the car backwards. The car's motor is applying force to counter that headwind (air resistance) and keep the city moving past at a constant rate. You have:

down: headwind (air resistance) = up: car's motor,

Good so far? Okay, let's add the wind.

Now there's a cross wind of some force F. First the city frame. The car's motor hasn't changed force, so the air resistance also hasn't changed force. Up and down forces are still equal. But, we've added a lateral force, the wind(F). This will push the car in the direction of the wind unless countered. Let's assume that the friction of the tires on the surface counter the wind, here, such that the lateral forces are balanced. You not have:
Up: car's motor = down: air resistance.
Right: wind (F) = Left: tire friction.
So far, you are right. But you need to remember that the air resistance and the "horizontal" wind force you've labeled F are just two components of the force of the air on the car. The problem you are having is below.
All forces balance. Now, let's shift to the car's frame, and follow your suggestion that the wind also shifts direction when we shift frames. We run into a problem. If the wind shifts by some angle x, the lateral force of that wind(F) is now cos(x)*F, and there's now a up/down componenet of sin(x)*F. If we sum the forces now, we have:

Down: air resistance + sin(x)*F = up: car's motor
Right: cos(x)*F = Left: tire friction

So, you're arguing that the tires exert less lateral friction force in the car frame than in the city frame, and that the car's motor must push harder to combat the increase in downward force because some part of the wind is adding to the air resistance. This doesn't add up, though, as the forces on the car DO NOT CHANGE with a frame shift -- they still must be the same forces as in any other frame.

The error in your thinking above is that the up/down component of the air's movement was already accounted for in the frames by the declaration of constant velocity -- meaning that the force of the car's motor exactly counters the force of the air resistance/wind in the up/down direction. You have the lateral and up/down forces changing in your frame change, and that just doesn't happen.
Actually, no. I don't have the up/down or lateral forces changing. The problem you're having is that you are double counting the air resistance and "vertical" component of the wind force. You still have the same force of the air on the car --- magnitude F "horizontally" and the same magnitude "vertically" as in the city frame. It's just now that you say both components are from the air moving, rather than saying in the city frame that the horizontal component is from the wind moving and the vertical component is from the car moving through the air. Remember, the force of the air on the car is due only to the relative velocity of the air and the car, which is the same in any reference frame.



Um, okay, let's look at your equation as the force of the push goes to zero. When this happens the limit of angle x goes to 0 degrees, which is nonsensical. IE, the less you push, the more perpendicular you should push. Further, if you increase your push to the velocity of the asteroid, sinx goes to 90. That kinda makes sense in that completely stopping the asteroid relative to Earth will cause a miss (although gravity now becomes a dominating factor). But what happens if you exceed the push? There exists some dp greater than p where your formula says that pushing the asteroid even faster directly at Earth generates a miss.

Your error here was assuming some dp and then finding a formula that created an x that seemed right to you. You're not testing other dp to find if the formula works. It doesn't.
Yes, as dp goes to zero, if you want to deflect the asteroid at all, you need to be pushing it as much "to the side" of its initial motion as possible in order to change its direction of motion.

Yes, x goes to +90 degrees as dp matches the initial momentum of the asteroid. This is pushing directly against the asteroid's motion, so the final asteroid momentum along its line of motion is p-dp->0 in this limit, meaning the asteroid stops. That's a "total deflection," or the best you can do. Note that pushing the asteroid forward is a negative value of x. That never optimizes the deflection angle.

I may have forgotten to note that I've assumed dp<p (which should be clear from the formula) because otherwise Pierce could just easily turn the asteroid around, and we don't need any dramatic "just missing the earth" bit.



Yes, those assumptions were established well upthread as plausible given the launch point of the asteroid and how it should interact with Earth's orbit. A head on example in the Solar frame requires an extra-solar origin for the asteroid or an orbital period of much greater than 100 years (ie, the ellipse of the asteroid's path would have to be at Earth's orbit close to perihelion, and the combined speeds would be YUGE. Further, for your purposes, a head on in the Solar frame is exactly identical to the Earth frame, with the Earth's 30km/s transferred to the asteroid's velocity. At that point, your formula breaks very badly, as it suggests that the weaker your push, the more towards the perpendicular you should push but the stronger you push the more towards opposing the asteroid's path your should push. That doesn't work.
You are giving very specific numbers, like saying "the angle of appoach is 45 degrees for a 30km/s asteroid" in a response to tomB above. While I admit I didn't read the whole thread, that seems more specific than the range of "reasonable" launch points. But it doesn't really matter.

Yes, head-on in solar and earth frames looks the same but with different asteroid speeds. My formula doesn't break at all. Look, line up the asteroid's initial velocity/momentum (remember, these are proportional) along the horizontal axis. Pierce then pushes. The asteroid now has a final momentum with a vertical component --- entirely due to Pierce's perpendicular push -- and a horizontal component, which is its initial momentum minus Pierce's parallel push slowing it down. The slope of the new velocity is the vertical part divided by the horizontal part. This slope can get larger if one of two things happens: the numerator gets bigger, or the denominator gets smaller. But if Pierce's push has some fixed total magnitude, you reduce the amount you can shrink the horizontal component of the asteroid momentum by making the numerator bigger, so these are competing effects. My calculation tells you how to balance them to get the biggest slope.

I will go through the next bit in more detail...

In the Earth frame, at a given distance, and assuming an asteroid path aimed straight at Earth, there exists a minimum angle that will cause a miss. This angle doesn't change based on the speed of the asteroid because the extended path will still need to be outside that angle. This angle is easy to determine if you know the distance (d) and Earth's radius (Re) as it's simply Tan-1 (Re/d). At sufficient distance, this angle is small. To generate a miss, you must create an instantaneous dp such that the resulting path exceeds this miss angle.
Yes!
This generates some math, but it's not too bad. Let's call the miss angle (m), the angle of applied dp (x), and the angle the asteroid takes as (a); all angles are measured clockwise from the path. Tan (a) is going to equal the lateral portion of dp, or dpcos(x) divided by p minus the vertical portion of dp, or dpsin(x).
You are getting your trigonometry wrong. If x is the angle from the initial velocity, dp sin(x) is the amount perpendicular to the initial velocity, and dp cos(x) is the amount along the initial velocity, which you have to subtract from the initial momentum. You are effectively choosing x to be measured from the perpendicular to the initial asteroid velocity, with the push parallel to the initial velocity directed backward against the asteroid's velocity. That's what I did also, but this may be part of your confusion.

This is tan(a) = dpcos(x)/(p-dpsin(x)). We can set Tan(a) equal to Tan(m) and get that dpcos(x)/(p-dpsin(x))=Re/d. From this, we can see a few things. One, dp cannot go to 0, and two, x also cannot go to 0 (or 180). This eliminates a push in the direction of the asteroid from generating a miss, as I stated above. Now, let's solve for a dp if x = 90, ie, a perfectly lateral push.
A perfectly perpendicular push is x=0. Cos(90 degrees)=0, which means there is no deflection at all.

IN this case, the formula goes go dp/p=Re/d. Assuming p is known, we have a dp(lat) that generates a miss. Now, can this dp applied to any other angle x, generate a miss (ignoring 270)? Doing some substitution from above, this would mean that dp(lat)cos(x)/(p-dp(lat)sin(x))=dp(lat)/p. There is no solution for x for a known dp(lat) in this scenario where x can be anything other than 90.
This is true until the last sentence. You can plot this on some software like maple or mathematica and see that there is in fact another solution for x even when dp=dp(lat). You can also see that there is a range of x in which tan(a) is greater than dp(lat)/p. The smaller dp(lat) is, however, the more that range is crammed toward x=0, which means you have to push more perpendicularly.

For a dp that exceeds dp(lat), you can have angles that aren't 90, because the resultant angle of miss can still be greater than m.
This is true, but you can also have dp<dp(lat) which will get tan(a)>tan(m). There is still some minimum necessary value of dp, however.

If we apply your forumla to this, though, where your sin(x)=dp/p, a known dp(lat) would result in an angle that isn't 90. Your construction doesn't work even if we use your assumption that the path of the asteroid is directly through Earth's center. In the realistic case, where it is not, it also doesn't work.
You're missing the point of what I'm doing. If you have any fixed dp (and p), this formula maximizes tan(a). But I didn't look at all about the necessary deflection tan(m), so I didn't say anything about what you call dp(lat) until this post. Furthermore, there is nothing in my calculation that talks about the asteroid hitting the center of the earth.


Nope! Flagpole in city frame. No up/down forces. Lateral forces are wind - being attached to the flagpole = 0.

Car frame: forces don't change. The 'head wind' the car feels isn't present for the flag, which is moving precisely along with and at the same speed as this 'headwind'. So, up/down forces are still zero. Side forces are also the same, ie: wind - being attached to the flagpole = 0. If the wind shifted to the down and right, the flagpole is now pushing up and left, but a change of frame doesn't affect the sum of the forces, so that's impossible.

Sum. The. Forces.
The wind in the car's frame is down and right. However, the flag is also moving down, so it is pushing additionally through the air. That means there is an additional force from the air on the flag that cancels the push "down" from the wind. In other words, the total force from the air on the flag is to the right, just like the relative velocity of the air and flag. You have forgotten about a force, so you are not summing the forces correctly.

Look, I doubt I can convince you, so I am mostly talking to other readers, if anyone else still cares. But I'm not sure it's productive continuing this discussion if you can't consider what I'm saying carefully.
 

freyar

Extradimensional Explorer
I just saw more discussion came up. I will look at it later, but I doubt there's much point participating. It seems like everyone else gets it.
 

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