Science: asteroid vs. hero physics

[Nagol]

"Yeah, we feel really bad about destroying the birthplace of humanity. But there was serious cash involved...."



The only really relevant point is that it didn't go down *accidentally*. There is no, "Whoops! We threw a mountain-sized rock across interplanetary distances, and *just happened* to hit the Earth. So sorry!" Whoever did it, meant it. Consider what would be necessary for that to place those responsible on the right side of history.

They needed to win? Then they're on the right side of history much like Rome was on the right side compared to Carthage. Unfortunately, they missed and their descendants and the descendants of their enemies now have to deal with the leftover threat.

 

[Janx]

"Yeah, we feel really bad about destroying the birthplace of humanity. But there was serious cash involved...."



The only really relevant point is that it didn't go down *accidentally*. There is no, "Whoops! We threw a mountain-sized rock across interplanetary distances, and *just happened* to hit the Earth. So sorry!" Whoever did it, meant it. Consider what would be necessary for that to place those responsible on the right side of history.

and that is the makings for another story.


I haven't decided on the details, as it doesn't matter for the current story, from Earth's perspective long after the event and knowledge has been lost.

But in my mind, a number of elements exist to work with:
that was 100 years ago, what does being angry at Mars now matter?
Why do we assume Mars was a unified people in agreement and responsible for the act?
What if you dropped the first bomb over Hiroshima and as it's falling, regretted it?
Who says the people responsible weren't punished already?
How is Mars doing these days?

Except for the first point, none of these factors play into the current story about people who've scavenged the heck out of things to survive and don't know somebody was mad at them 100 years ago, let alone have the means to retaliate. To what end?

 

[Janx]

They needed to win? Then they're on the right side of history much like Rome was on the right side compared to Carthage. Unfortunately, they missed and their descendants and the descendants of their enemies now have to deal with the leftover threat.

Could be. I was being cheeky when I said taxation without representation.

Consider what Babylon 5 or The Expanse shows us of Earth's treatment toward settlements/colonies.

 

[tomBitonti]

This is the problem when you have independent folks out in the belt with the capability to adjust orbits (as in The Expanse). Individuals piloting asteroid tenders would have the capacity to do tremendous harm. That seems unstable, with a terrible disaster just a matter of time.

This could be the basis for a decent set of stories -- the decentralization of control from individuals as a necessary step for surviving into a solar system spanning civilization. I can't say I've read stories that use this as a basic idea.

This actually seems to be where we are headed with just an Earthbound civilization: Moving control over high energy means -- large vessels and large airplanes, then to smaller vessels and trucks, and eventually to cars -- from people to automation. We are building that technology right now, and I am thinking that one of the motivations is exactly to remove control from non-approved actors.

Thx!
TomB

 

[Tonguez]

"Yeah, we feel really bad about destroying the birthplace of humanity. But there was serious cash involved...."



The only really relevant point is that it didn't go down *accidentally*. There is no, "Whoops! We threw a mountain-sized rock across interplanetary distances, and *just happened* to hit the Earth. So sorry!" Whoever did it, meant it. Consider what would be necessary for that to place those responsible on the right side of history.

Humans have this great capacity to rationalise things in hindsight, so I'm sure that should a society with the capacity to launch asteroids be established on Mars then relations between them and Earth will balance out eventually

 

[freyar]

You don't shift the wind. Unless and until you rotate the diagram, in which case you shift the wind, but rotation doesn't work well in this case because Earth is so big and the impact point, while appearing to be in line with Earth's center in one frame isn't in another. Here's a to scale diagram showing Earth as the circle on the left of a diameter including 150km of atmo (so it rounds up to a nice 13,000km diameter) and three asteroids (20km/s, 30km/s, 50km/s) at one hour from impact. This shows the travel of Earth in that time and the travel of the asteroids in that time in both the Solar frame (Earth moving up, asteroids moving left) and the Earth frame (diagonal asteroid paths). The grey arrow is the direction of an example push so that it's apparent that the shift in frame doesn't rotate the push. As you can see, you can't just rotate the Earth frame because the asteroids aren't moving towards Earth's center but instead a point on the circumference, and that circumference is so large you can't ignore it for the purposes of the example.

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?

You must mean angle from the vector of the asteroid's travel, because otherwise a 90 degree push would be sin(0)=0.

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.

That accounted for, the issue here is obvious when you consider it. The maximum dp occurs at sin(+/-90), but it would obviously require a much, much harder push if you went past 90 to generate a miss if you pushed from any angle behind the asteroid (ie, speeding it up while you generate a lateral velocity) but your formula has dp dropping off, not increasing. Also, it's possible to generate a miss by accelerating the asteroid "ahead" of the Earth, but your formula says needed dp there is 0.

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 = 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.

I agree, there's an angle that minimizes pushed needed, but your assumption doesn't do it. Momentum is the wrong frame of thinking, as the asteroid's mass is constant so it falls out of the equations and you're just dealing with velocities. This is how most orbital problems work -- you deal in delta-v not momentum change. The challenge here is the point of impact being on the surface of a large circle instead of the center in the Earth frame, so the math gets messy. In the solar frame, it's two quickly moving objects. The math doesn't resolve into a simple sine equation.

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.

 

[Eltab]

Consider what Babylon 5 or The Expanse shows us of Earth's treatment toward settlements/colonies.

"The War of Earth's Aggression" is a sci-fi trope. You are not required to invoke it for your Future History if you don't want to.
It would make your universe different from so many other authors', if you didn't.

 

[Ovinomancer]

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?

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.

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.

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 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.

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 = 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.

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.

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.

 

[Eltab]

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.

Presuming the wind lines up with the streets (which I think you are describing), then

Have the moving car make a right-hand turn, and the wind will be coming from behind it.
So from the car's point of view, when it goes into motion down the original perpendicular street, the wind IS coming from another direction (not from its due left any more).

 

[Ovinomancer]

Presuming the wind lines up with the streets (which I think you are describing), then

Have the moving car make a right-hand turn, and the wind will be coming from behind it.
So from the car's point of view, when it goes into motion down the original perpendicular street, the wind IS coming from another direction (not from its due left any more).

Yes, but then you're not translating to a different reference frame, you're rotating the coordinate frame. Changing the coordinate frame is a different exercise.

This is apparent if you set that coordinate frame prior to the turn of the car -- the wind won't change direction. Obviously, if the wind is blowing in the +y direction but you then invert the y-axis, the wind will now be blowing in the -y direction. You've switched your coordinates, not changed the direction of the wind.