Science: asteroid vs. hero physics

[Ovinomancer]

Changing from the earth’s Frame of reference and a sun centered frame of reference, a vector perpendicular to the velocity of the moon in the earth frame is no longer perpendicular in a solar frame. The shift is from 30-60 degrees, with the earth moving perpendicular to the moon in the solar frame, and varying the relative speeds from 1/2 to 2.

Huh? I can't make heads or tails of that. You seem to be talking about a soecific scenario, but haven't presented it. I'm completely lost on your relative speed thing, though.

Over the course of a day, the earth’s motion shifts by one degree. That doesn’t seem to change the situation much, in the sense of how you would want to adjust the asteroid’s movement.

Well, that 1 degree is about 2.5 million km (per day), so you tell me if that seems like a lot.

In all of this, is the deflection caused by the earth’s gravity big enough to matter? The effect in the earth frame puts the asteroid on an arc instead of a line, with (I’m thinking) more curvature closer to the earth.

Thx!
TomB

Not to the impact, actually. At the speeds involved, the acceleration due to Earth's gravity is mostly going to be seen in the outgoing trajectory. Near Earth Orbital speed is low, comparitively. The ISS is going about 7 km/s, so the asteroid is an order of magnitude faster. It will be bent, but the fastest acceleration will come when it's passing Earth, so it matters little for the intercept.

Go slower, and it matters more.

 

[Janx]

So, didn't bother reading all 27 posts, so forgive me if this has been said.

How big is your asteroid? Cause if you're looking at something, say, the size of Texas (like in Armageddon), then what you have done is create a scenario that you are unable to win. You are constraining your hero to be unable to simply push it, and simply "flying real fast around the earth and hitting it perpendicularly so it goes off course" wouldn't work. She's too small. Even a few tons of mass would simply be too small. It wouldn't push something massive like that off course, it would puncture a hole through it and go through the other side.

I think we're going with these stats:
the planet-killer asteroid has a mass of 1EE14 kg and is traveling at 50km/s (same size as Chicxulb)

I don't know if it could be smaller. I don't know how big that is, it's just a number, and it killed a lot of good dinosaurs.

Oviniomancer says it's less energy (still alot) to push it faster, past its intercept point, than to hit it perpendicular (another order of magnitude of fig newtons needed and in a post-disaster world, fig newtons are rare).

The hero can scoop up and reconfigure matter in a localized area (let's say withing 30 feet of her suit). So she can break a number of physics rules, up to a point. Let's say she can gather and push up to 2 tons of matter (could lift a car). It's more like like telekinesis, but yeah.

You raise a good question, that might enable other tactics. What's a typical asteroid that big made of? Can a 2 ton bullet driven by her, pierce and shatter the asteroid.

In other fun facts, her name is Pierce. So I kinda had that in mind way back when I started the story...

 

[Nagol]

I think we're going with these stats:
the planet-killer asteroid has a mass of 1EE14 kg and is traveling at 50km/s (same size as Chicxulb)

I don't know if it could be smaller. I don't know how big that is, it's just a number, and it killed a lot of good dinosaurs.

Oviniomancer says it's less energy (still alot) to push it faster, past its intercept point, than to hit it perpendicular (another order of magnitude of fig newtons needed and in a post-disaster world, fig newtons are rare).

The hero can scoop up and reconfigure matter in a localized area (let's say withing 30 feet of her suit). So she can break a number of physics rules, up to a point. Let's say she can gather and push up to 2 tons of matter (could lift a car). It's more like like telekinesis, but yeah.

You raise a good question, that might enable other tactics. What's a typical asteroid that big made of? Can a 2 ton bullet driven by her, pierce and shatter the asteroid.

In other fun facts, her name is Pierce. So I kinda had that in mind way back when I started the story...

1EE14 is 100,000,000,000,000 kg or 100,000,000,000 tonnes or 100,000 mega tonnes. You don't need that much speed with something that size: typical asteroid collision speeds of 11 km/s is more than sufficient.

A 2 tonne impactor isn't going to cause it to wobble let alone impart enough velocity to cause it to miss. Momentum transfer is necessary and momentum is proportional to both mass and velocity. My original calc of 1,500 kg vs. 600,000 tonnes was off by 3 orders of magnitude (I swapped kg for tonnes when I did the math).

Shattering the asteroid is only helpful if there is sufficient time for the debris to scatter and actually miss the Earth. The energy released from the pieces striking is the same as if the object was one piece -- whether it is released in the atmosphere or on impact. Air release may be less damaging, but at these scales it'll still be devastating.

There are 3 main types of asteroid: nickel-iron, rock, and more rarely carbonaceous. Chicxulb was probably the latter. Since this impactor is actually launched by an alien force, nickel-iron probably a better option -- the volume is smaller and they can use electromagnetic forces to change its orbit. 100,000,000,000 tonnes is about 12,000,000,000 cubic metres or a cube 2.3 km on a side.

 

[tomBitonti]

Re: what angle to push the asteroid. If the earth has motion perpendicular to the asteroid, the angle appears different if you examine it the frame of reference which is moving with the earth vs the angle in a sun centered frame of reference.

Put it like this: Earth towards the bottom, moving upwards. Asteroid to the right, moving to the left. Paths intersect in the top left. That’s the solar view. The view moving with the earth simply has the earth to the left and the asteroid to the right, with the asteroid moving directly leftwards towards the earth.

In the earth centric view, a perpendicular push on the asteroid points directly down. Translating that into the sun centric view, the push is no longer perpendicular to the motion of the moon. If the earth is moving at the same speed as the asteroid, the angle in a solar frame seems to be 45 degrees. If the earth is moving at half the speed, 30 degrees, and if at twice the speed, 60 degrees. As a check, if the earth is still, the angle is the same, and if the earth is moving a lot (10x) faster the angle gets close to 90 degrees. In which case you really are mostly speeding up or slowing down the asteroid — just in the sun centric view.

This is meant to unify the two perspective: Both are correct.

Thx!
TomB

 

[Ovinomancer]

Re: what angle to push the asteroid. If the earth has motion perpendicular to the asteroid, the angle appears different if you examine it the frame of reference which is moving with the earth vs the angle in a sun centered frame of reference.

Put it like this: Earth towards the bottom, moving upwards. Asteroid to the right, moving to the left. Paths intersect in the top left. That’s the solar view. The view moving with the earth simply has the earth to the left and the asteroid to the right, with the asteroid moving directly leftwards towards the earth.

In the earth centric view, a perpendicular push on the asteroid points directly down. Translating that into the sun centric view, the push is no longer perpendicular to the motion of the moon. If the earth is moving at the same speed as the asteroid, the angle in a solar frame seems to be 45 degrees. If the earth is moving at half the speed, 30 degrees, and if at twice the speed, 60 degrees. As a check, if the earth is still, the angle is the same, and if the earth is moving a lot (10x) faster the angle gets close to 90 degrees. In which case you really are mostly speeding up or slowing down the asteroid — just in the sun centric view.

This is meant to unify the two perspective: Both are correct.

Thx!
TomB

Took me a few readings, but I follow you here, I just don't see the point. Your example here shows that treating this from the Earth's perspective and applying a perpendicular push to the apparent motion of the asteroid means that you're being very inefficient because you aren't fully slowing the asteroid nor are you committing fully to pushing it to the side. It'll work, but you're borrowing the worst of both for no benefit. Also, Earth doesn't change speeds, so... maybe you meant the asteroid? Yes, if the asteroid approaches infinite speed, the inefficient push approaches the line of motion, but that's not helpful anywhere.

This whole translation doesn't illuminate the issues, so I'm confused as to what your purpose is?

 

[freyar]

This is... not even wrong. To attempt to explain, again, both the Earth and the asteroid are orbiting the Sun. The Earth is in constant motion along it's orbital path, as is the asteroid. The point of impact is where those two lines cross. In other words, the asteroid isn't moving towards the Earth, it's moving towards where the Earth is going to be.

You missed what I said about being in the rest frame of the earth over a short time. For someone sitting on the earth, in the last little bit before the asteroid hits, it's going to look like the asteroid is coming straight at the earth. Over longer periods of time, yes, you're right that you have to take the earth's acceleration around the sun into account. But not over the course of a few hours. Looking at the numbers, the difference between the earth's actual orbit and a constant velocity path is less than the earth's radius for about a 12-hour period, so that's about how long we can neglect the acceleration (sorry, I was just guessing when I said a day or two).

Imagine trying to hit a moving car with a baseball from the field next to the road the car is travelling along. If you throw the baseball at the car, you will miss because by the time the baseball arrives at the road, the car will have already traveled past that point. Instead, you throw the baseball ahead of the car so that when the baseball gets to the road, it meets the car there.

That's in the rest frame of the baseball field. In the rest frame of the car --- assuming no acceleration, which would change your scenario as well --- the baseball has to come straight at the car, or it will miss. You and the baseball field are, however, moving compared to the car. This is a simple example of change of reference frame from introductory physics.

In this scenario, the hero is like a wind pushing the baseball after you throw it. The car is moving so fast that it takes less wind to slow the thrown ball so that the car zooms past the rendezvous than it does to push the ball to the side so that it gets there at the same time but far enough off to one side. Largely this is because, for reference, the car is huge and moving really fast so that a small change in speed leverages the speed of the car to cause the miss rather than having to push the baseball off target by half the length of the car from zero lateral speed.

Depends on what frame you're measuring the wind's speed from. In the earth's frame (the natural one for our hero), slowing down the asteroid won't stop it from hitting the earth. It will just make it hit later.

If you insist on treating Earth as the center point, then the asteroid is going to appear to move under constantly changing acceleration (because it has to include Earth's orbital movement, which is elliptical, alongside it's own highly elliptical orbit, which is going to do weird things -- other planets appear to occasionally do loop-de-loops in the sky, for instance). You're essentially recreating the problem of predicting planetary motions in an Earth-centric universe, a field of rather complicated mathematical modelling. There's a reason everything got a lot easier (but not exactly easy) to predict when we moved to a heliocentric model. Don't ignore Kepler!

That's only if you're talking about longer time periods than this problem. The OP was talking about a 5-minute time frame, which seems to have been extended maybe to a few hours. If you want to talk about a day, you have to worry about this a little. More than a day or two, yes, I agree you have to worry about the full orbital mechanics. Kepler is certainly important over the course of a year, but one thing we learn in physics is how to know when different effects are actually important.

 

[freyar]

Took me a few readings, but I follow you here, I just don't see the point. Your example here shows that treating this from the Earth's perspective and applying a perpendicular push to the apparent motion of the asteroid means that you're being very inefficient because you aren't fully slowing the asteroid nor are you committing fully to pushing it to the side. It'll work, but you're borrowing the worst of both for no benefit. Also, Earth doesn't change speeds, so... maybe you meant the asteroid? Yes, if the asteroid approaches infinite speed, the inefficient push approaches the line of motion, but that's not helpful anywhere.

This whole translation doesn't illuminate the issues, so I'm confused as to what your purpose is?

The point is that the push needs to be perpendicular to the asteroid's motion in the earth's rest frame (again assuming short time periods). In the heliocentric frame, it will be angled differently, depending on the orientation of the asteroid's motion with respect to the earth's.

 

[Ovinomancer]

Hmm. Earth is moving pretty quickly in space, so, depending on approach vectors, speeding the asteroid up or slowing it down a bit should generate enough 'miss' if you're far enough out.

For example, if the asteroid is crossing the orbit of Earth perpendicularly, the window to strike Earth is only 13,000 km / 30 km/s or about 435 seconds (about 7 and a quarter minutes). Assuming the asteroid aimed precisely at the center of Earth, you'd need to speed up or slow down the approach by half that 7 minute time, or about 4 minutes to be safe (240 seconds). A 4 minute delta in arrival time depends on the velocity of the asteroid and how long you take to start the acceleration. Assuming a 50 km/s speed on the asteroid, 4 minutes at 1 hour distance is a ratio: needed speed over current speed = needed time over original time. Or needed speed = original speed x (new time/old time). In this case, needed speed = 50km/s x (56 minutes/60 minutes) = 50 km/s x (.933) = 46.67 km/s. You'd need to generate a delta-v of 3.33 km/s at exactly 1 hour out (or earlier) to cause a complete miss. I think this is lower than the necessary delta-v to accelerate the mass sideways 6500 km over an hour (6500km/360 seconds = 18 km/s, so, yep, 6 times less).

Now, if your geometry has the asteroid coming in much closer to the orbital direction, this changes, as earth's relative movement is much less with respect to the path of the asteroid. However, a rock launched from the asteroid belt some time in the past should have a high angle of intercept to Earth's orbit, as it would be on a steep elliptical around the Sun. But, assuming a near head on collision, say at about a 20 degree approach, combined delta-v would be 30km/s (Earth) plus cosine(20)x50km/s or about 77 km/s. Earth relative speed on the perpendicular would be sin(20)x30km/s or about 10km/s. This puts a Earth on-target time window of 1300 seconds or 21.7 minutes. Again assuming a center strike, that's a time window of 11 minutes you have to modify. Needed speed up/slow down delta-v at 1 hour would be 14.7 km/s. This is still less than the needed perpendicular delta-v to cause a miss, although not much.

Orbital mechanics are funny. Often speeding up or slowing down along your path causes huge changes in orbit, while orthogonal thrusts change orbits more slowly. For the purposes of your story, gathering up a bunch of debris and smacking the asteroid just enough to slow it down would be sufficient for a high angle of intercept. This is the most likely scenario given the orbital geometries of a rock launched from the belt that's coming around again.

A neat way to tell this story might be that the angle of impact is such that meeting the asteroid and speeding it up is the right call (this would be the case if the asteroid was going to hit on the forward half of the Earth). Timing could be such that the push means the ateroid scrapes the atmo and skips, causing an awesome light show and placing the heroine at risk.

I reviewed this math, and I made an error that changes the analysis. I made a magnitude error in the step where I calculated the deflection speed at 1 hour when I used 360 seconds instead of 3600 seconds. This error was propogated. At 50km/s, it's better to generate velocity lateral to the movement of the asteroid in the solar frame, not slow it. There is a breakpoint in speed where slowing is better, but it's dependent on time. At 1 hour, that breakpoint is slightly less than 30km/s.

Mea culpa, [MENTION=8835]Janx[/MENTION].

 

[Ovinomancer]

The point is that the push needs to be perpendicular to the asteroid's motion in the earth's rest frame (again assuming short time periods). In the heliocentric frame, it will be angled differently, depending on the orientation of the asteroid's motion with respect to the earth's.

I disagree. The problem here is the same one you make about -- when you shifted the frame from the road/field to the car, you DON'T shift the vector of the wind -- it remains the same. Ergo, if you go with an Earth centric reference frame, the asteroid's motion is actually comprised of it's vector and the Earth's speed vector, creating the apparent straight line movement towards Earth. If you restart the Earth, you don't change the Asteroid's original vector, you just recover the Earth's vector back to Earth. The push that was perpendicular in the Earth frame is still in the same direction, which is now diagonal to the asteroid's path. This means you're both slowing the asteroid AND pushing it to the side. At 5 minutes out, the breakpoint speed where slowing becomes more efficient than pushing it to the side is:


Speed of Earth = 30 km/s; radius of Earth (with atmo at a bit of slop): 6500km
Time needed for Earth to move out of the way: 6500km/30km/s = 216.7 seconds. This is the time you need to generate by slowing the asteroid for Earth to get out of the way.

S1 * T1 = S2 * T2. Rearranging, we find S2 = S1 * T1/T2. T2 = T1 + 216.7s. T1 = 300 s (five minutes). So, simplifying, S2 = S1 *(300s/516.7s) -> S2 = 0.581*S1.

21.667 km/s - speed needed to push perpendicular to the asteroid's path to cause a miss (6500km/300 seconds <-- right this time!). Note the similarity in the time for Earth to move and this. Just a fun fact, no importance.

Setting S2 equal to 21.667 km/s, we get an S1 of 37.3 km/s.

So long as the asteroid's speed isn't exactly 37.3 km/s, your perpendicular push in the Earth reference frame will be less efficient than a different angle at generating a miss.

Here's a graphic showing the Earth centric frame of three scenarios: a 20km/s asteroid 5 minutes out in green, a 30km/s asteroid 5 minutes out in purple, and a 50km/s asteroid 5 minutes out in yellow-orange. The big things here is that the graphic is to scale. The Earth is the proper size, and the path of the asteroids is what they will appear to travel over the 5 minutes.

 

[tglassy]

So....I’m going to say go with the Rule of Cool.

Find the solution that sounds the coolest, and go with that. Go watch One Punch Man. He’s awesome. Completely invulnerable, can jump to the moon and back, and can, literally, kill anything with one punch. He’s freaking amazing. Nobody cares about the science behind it.

Now, it’s great to have some science to make things make sense. That way, you can create one or two “fictional” things, like a super suit, and use the science to know how it would interact with the rest of the world.

But if you stick with the Rule of Cool, you can’t go wrong.