Shut the fuck up. :icon14:Turbine wrote:Good point. :icon14:[xeno]Julios wrote:I love applying physics and math whenever possible. It's amazing what you can do with highschool physics and grade 9 math. It's a good intellectual exercise.Turbine wrote:What's with the arguing, when this rock is going to miss us by a long margin?
Using the Orbiter ( http://www.orbitersim.com ) you can get a greater perspective of how things actually work in space. Including what it takes, and how to maneuver in real life space. Not sci-fi space.
asteroid strike 100% likely, but... (warning - physics post)
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Nightshade
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Nightshade[no u]
We don't really have the resources and technologies yet and the time, may be. So, the best I am thinking at the moment would be to just sting it with types of parachutes to decelerate its fall and thus, a lot less destruction on collision with earth.
May be, we could even change its path with the help of robotic or teleguidance from the ground with the help of the parachutes.
Have you seen the big chip that can be sailed partly by sails to save on fuel even though the wind is in the opposite direction?
Same thing.
May be, we could even change its path with the help of robotic or teleguidance from the ground with the help of the parachutes.
Have you seen the big chip that can be sailed partly by sails to save on fuel even though the wind is in the opposite direction?
Same thing.
Jules I don't know a whole lot about celestial physics/mechanics, but to my layman's eye it seems that you're oversimplifying the problem and over complicating it at the same time.
Over simplifying by using elementary 2D physics to extrapolate results to a more complex model with several different forces at work.... And over complicating by not accounting for the fact that all you need to do is adjust the asteroid's vector, which doesn't require nearly the same amount of force it would actually take to PUSH the asteroid the distance of the earth's radius.
It seems like you are trying to calculate how much force it would take to push the asteroid the whole way - when only slightly altering its course far away would have a huge impact on its position by the time it got to earth.
Again, I'm certainly not a physics or mechanics expert and I may be misunderstanding you.
Over simplifying by using elementary 2D physics to extrapolate results to a more complex model with several different forces at work.... And over complicating by not accounting for the fact that all you need to do is adjust the asteroid's vector, which doesn't require nearly the same amount of force it would actually take to PUSH the asteroid the distance of the earth's radius.
It seems like you are trying to calculate how much force it would take to push the asteroid the whole way - when only slightly altering its course far away would have a huge impact on its position by the time it got to earth.
Again, I'm certainly not a physics or mechanics expert and I may be misunderstanding you.
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[xeno]Julios
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nope, both you and menkent have misunderstood the approach. Nightshade seems to be the only one that has understood it so far.
Go back and read my long post with the diagrams:
the forces at work here are actually not that complicated, and all that is needed is a 2d analysis (and I believe this would be the case even if we were assuming earth's gravity).
As for the pushing the asteroid the distance of the earth's radius, remember this:
we're dealing with a predetermined force (since weight of asteroid and tractor craft is already given), and a predetermined timelimit.
For lower level feasibility, I'm assuming that the force is being applied the entire time.
If it turns out to be not enough force if force is applied the entire time, then it will certainly not be enough force if you only did a momentary push.
A lot of you are still not understanding this, menkent included.
Again, go back and read the diagram post carefully.
A key conceptual difficulty you may be having is in failing to recognize that the orthogonal movement of the asteroid can be treated on a completely independent basis. This is what orthogonal ultimately means.
That's why you could solve the problem even if the asteroid was completely motionless to start with. So long as you had a timelimit, you could calculate how much force is required to displace the asteroid orthogonally in that amount of time.
Yes you could achieve the same effect (displacement over that period of time) with a "momentary push", rather than a sustained one, but you'd need a larger force.
Remember, I'm looking at the lower end of the feasibility.
In other words, if it's not feasible under my conditions, then it will certainly not be feasible in any other conditions.
(all these conditions assume asteroid is aimed for centre of earth, and that the earth is not moving).
So:
if it's not feasible with a sustained push, in the absence of earth's gravity, it is most certainly not feasible with a momentary push, in the presence of earth's gravity.
I've made this really clear a number of times, and people keep making the same objections without taking this into account.
I guess I should also ask: do you all understand what orthogonal means, and it's significance?
(if not, I can go through some simple projectile motion to explain it)
Go back and read my long post with the diagrams:
the forces at work here are actually not that complicated, and all that is needed is a 2d analysis (and I believe this would be the case even if we were assuming earth's gravity).
As for the pushing the asteroid the distance of the earth's radius, remember this:
we're dealing with a predetermined force (since weight of asteroid and tractor craft is already given), and a predetermined timelimit.
For lower level feasibility, I'm assuming that the force is being applied the entire time.
If it turns out to be not enough force if force is applied the entire time, then it will certainly not be enough force if you only did a momentary push.
A lot of you are still not understanding this, menkent included.
Again, go back and read the diagram post carefully.
A key conceptual difficulty you may be having is in failing to recognize that the orthogonal movement of the asteroid can be treated on a completely independent basis. This is what orthogonal ultimately means.
That's why you could solve the problem even if the asteroid was completely motionless to start with. So long as you had a timelimit, you could calculate how much force is required to displace the asteroid orthogonally in that amount of time.
Yes you could achieve the same effect (displacement over that period of time) with a "momentary push", rather than a sustained one, but you'd need a larger force.
Remember, I'm looking at the lower end of the feasibility.
In other words, if it's not feasible under my conditions, then it will certainly not be feasible in any other conditions.
(all these conditions assume asteroid is aimed for centre of earth, and that the earth is not moving).
So:
if it's not feasible with a sustained push, in the absence of earth's gravity, it is most certainly not feasible with a momentary push, in the presence of earth's gravity.
I've made this really clear a number of times, and people keep making the same objections without taking this into account.
I guess I should also ask: do you all understand what orthogonal means, and it's significance?
(if not, I can go through some simple projectile motion to explain it)
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[xeno]Julios
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here - i'll try to prove my approach:
A spacecraft, weighing 1000 kg has lost engine control, and is headed directly towards the centre of a spherical body.
The sphere has a radius of 7,000,000 metres.
There are some auxilliary thrusters on the lower surface of the spacecraft (activating them will apply a force in a direction orthogonal to the current direction the craft is heading).
Calculations have shown that if nothing is done, the craft will collide with the sphere in 3600 seconds.
The thrusters are capable of imparting a force of 1000 newtons.
Question: Will this be enough to clear the craft from a collision?
Here's how I would do it, using 2d kinematics:
force = mass*acceleration
acceleration = force/mass
= 1000N/1000Kg = 1 m/s/s
Now that we have the acceleration that the thrusters are capable of generating, we can calculate the distance that the craft can be displaced, in the amount of time given.
Initial velocity of craft, along orthogonal direction (in direction that the thrusters impart the force - remember, force is imparted along a direction): 0 m/s
time = 3600 seconds
acceleration = 1 m/s/s
So:
distance = v1*t+0.5*a*t^2
= 0 + (0.5)(3600)(3600)
= 6480000m = 6480km
Answer: No, the craft will not clear the sphere in time - by the time it reaches the sphere, it'll be 520 metres short of clearing it.
We can therefore safely say that it would certainly not clear it if the sphere had a gravitational field.
And we can safely say that it would certainly not clear it if we only turned on the thrusters for a short period of time.
Now let's see someone else use a vector analysis with polar coordinates and see what answer they get.
A spacecraft, weighing 1000 kg has lost engine control, and is headed directly towards the centre of a spherical body.
The sphere has a radius of 7,000,000 metres.
There are some auxilliary thrusters on the lower surface of the spacecraft (activating them will apply a force in a direction orthogonal to the current direction the craft is heading).
Calculations have shown that if nothing is done, the craft will collide with the sphere in 3600 seconds.
The thrusters are capable of imparting a force of 1000 newtons.
Question: Will this be enough to clear the craft from a collision?
Here's how I would do it, using 2d kinematics:
force = mass*acceleration
acceleration = force/mass
= 1000N/1000Kg = 1 m/s/s
Now that we have the acceleration that the thrusters are capable of generating, we can calculate the distance that the craft can be displaced, in the amount of time given.
Initial velocity of craft, along orthogonal direction (in direction that the thrusters impart the force - remember, force is imparted along a direction): 0 m/s
time = 3600 seconds
acceleration = 1 m/s/s
So:
distance = v1*t+0.5*a*t^2
= 0 + (0.5)(3600)(3600)
= 6480000m = 6480km
Answer: No, the craft will not clear the sphere in time - by the time it reaches the sphere, it'll be 520 metres short of clearing it.
We can therefore safely say that it would certainly not clear it if the sphere had a gravitational field.
And we can safely say that it would certainly not clear it if we only turned on the thrusters for a short period of time.
Now let's see someone else use a vector analysis with polar coordinates and see what answer they get.
I've barely read any of this topic aside from your premise on the first page.
But consider this...
The amount of force you're talking about (force required to move the body orthogonally by the distance of the radius of the earth) would be massively greater than the force we're talking about (force required to slightly change its vector in motion).
What you're essentially saying in your premise, is that if we can't apply enough force to push an object this size the length of the earth's radius, orthogonally, then there is no way we could possibly apply enough force to alter its path enough to prevent it from hitting earth.
That doesn't hold up, because much less force is required to alter its path slightly than is required to push its entire mass sideways by several thousand kilometers.
The force it takes to alter its course is a single instance of momentary pressure (x newtons). The force you require in your premise is x newtons multiplied by the amount of time you are trying to exert that force.
So your model requires a much greater amount of force than NASAs model -- which means your experiment does not bear on the success of NASAs model in any way.
Well, I guess it does, just not in the way you are offering: if your model works, then NASAs would easily work.
Right?
But consider this...
The amount of force you're talking about (force required to move the body orthogonally by the distance of the radius of the earth) would be massively greater than the force we're talking about (force required to slightly change its vector in motion).
What you're essentially saying in your premise, is that if we can't apply enough force to push an object this size the length of the earth's radius, orthogonally, then there is no way we could possibly apply enough force to alter its path enough to prevent it from hitting earth.
That doesn't hold up, because much less force is required to alter its path slightly than is required to push its entire mass sideways by several thousand kilometers.
The force it takes to alter its course is a single instance of momentary pressure (x newtons). The force you require in your premise is x newtons multiplied by the amount of time you are trying to exert that force.
So your model requires a much greater amount of force than NASAs model -- which means your experiment does not bear on the success of NASAs model in any way.
Well, I guess it does, just not in the way you are offering: if your model works, then NASAs would easily work.
Right?
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[xeno]Julios
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Have you read my last two posts carefully?R00k wrote:I've barely read any of this topic aside from your premise on the first page.
No. By changing the vector, you're implicitly displacing it along an orthogonal axis.R00k wrote: The amount of force you're talking about (force required to move the body orthogonally by the distance of the radius of the earth) would be massively greater than the force we're talking about (force required to slightly change its vector in motion).
My analysis is only concerned with the orthogonal component of this vector - you can analyze it independently.
almost: if my model doesn't work, then NASA's most certainly won't.R00k wrote:
Well, I guess it does, just not in the way you are offering: if your model works, then NASAs would easily work.
If mine does, then not much can be said.
Read the last post, where I give the spacecraft with faulty engines example.
Are you able to follow that one?
btw thanks for replying and actually engaging the issue
Yes, I read your last two posts.
But you are assuming that the amount of force required to change the vector of an object in motion, is the same amount of force required to move that object orthogonally the same distance (y) over the same amount of distance (x) or time. That simply isn't true.
You played pool last weekend didn't you? You should know this.
If an object is already traveling in one direction, its current inertia is the acting force. Also keep in mind there is no friction, for all practical purposes.
Once you apply an orthogonal force to that object (even momentarily), its path changes completely. It has a new flight vector. The amount of force required to change its flight path is infinitesimal -- the force required to change that path by any measurable degree is much higher and can be calculated.
For instance, if a 16 lb. bowling ball is rolling down a lane toward the pins, and you give it a very light punch, it will not move orthogonally according to the amount of force you applied, and then correct its path to be parallel to its original one. Its new path can be measured in degrees of deviation from the original.
So it doesn't take much force at all to change a bowling ball's course and make it miss all the pins in the lane.
However, it would take a much greater force to actually move the ball orthogonally all the way to the gutter - which you would only have to do if the ball weren't in motion, or if the ball needed to be moved that distance instantaneously.
Is that a little clearer?
But you are assuming that the amount of force required to change the vector of an object in motion, is the same amount of force required to move that object orthogonally the same distance (y) over the same amount of distance (x) or time. That simply isn't true.
You played pool last weekend didn't you? You should know this.
If an object is already traveling in one direction, its current inertia is the acting force. Also keep in mind there is no friction, for all practical purposes.
Once you apply an orthogonal force to that object (even momentarily), its path changes completely. It has a new flight vector. The amount of force required to change its flight path is infinitesimal -- the force required to change that path by any measurable degree is much higher and can be calculated.
For instance, if a 16 lb. bowling ball is rolling down a lane toward the pins, and you give it a very light punch, it will not move orthogonally according to the amount of force you applied, and then correct its path to be parallel to its original one. Its new path can be measured in degrees of deviation from the original.
So it doesn't take much force at all to change a bowling ball's course and make it miss all the pins in the lane.
However, it would take a much greater force to actually move the ball orthogonally all the way to the gutter - which you would only have to do if the ball weren't in motion, or if the ball needed to be moved that distance instantaneously.
Is that a little clearer?
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[xeno]Julios
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you still aren't getting the orthogonal bit.
In the spacecraft example i gave above, the craft would never move orthogonally, so long as the thrusters are acting in the direction that they are.
The actual path of the craft would be a curved path.
but you can analyze the orthogonal component independently.
Here let me give you a down to earth example: projectile motion.
Notice that Vx is constant (ignoring air resistance)
Vy changes as a function of time, due to the force of gravity.
You can analyze the vertical displacement without worrying at all about Vx, because the force of gravity is acting in a direction orthogonal to the direction of x.
Think of it this way:
If I roll a ball off a cliff, into a chasm of infinite depth, the ball is going to take a curved path downwards and to the right (assuming the ball rolled rightwards off the cliff).
Let's say that on the other side of this chasm, there is another cliff whose edge is on the same level as the edge of the first cliff.
Now for the important part:
It doesn't matter how much gravitational pull there is on the ball - it will always hit the second cliff at the same time.
In fact, if you knew in advance what this time was, and you knew the force due to gravity, you could easily calculate how far down the second cliff the ball hits.
This distance is the orthogonal displacement due to the force of gravity.
Replace the second cliff with the sphere, and the force of gravity with thrusters, and you've got the exact same situation.
bbl off to lab.
In the spacecraft example i gave above, the craft would never move orthogonally, so long as the thrusters are acting in the direction that they are.
The actual path of the craft would be a curved path.
but you can analyze the orthogonal component independently.
Here let me give you a down to earth example: projectile motion.
Notice that Vx is constant (ignoring air resistance)
Vy changes as a function of time, due to the force of gravity.
You can analyze the vertical displacement without worrying at all about Vx, because the force of gravity is acting in a direction orthogonal to the direction of x.
Think of it this way:
If I roll a ball off a cliff, into a chasm of infinite depth, the ball is going to take a curved path downwards and to the right (assuming the ball rolled rightwards off the cliff).
Let's say that on the other side of this chasm, there is another cliff whose edge is on the same level as the edge of the first cliff.
Now for the important part:
It doesn't matter how much gravitational pull there is on the ball - it will always hit the second cliff at the same time.
In fact, if you knew in advance what this time was, and you knew the force due to gravity, you could easily calculate how far down the second cliff the ball hits.
This distance is the orthogonal displacement due to the force of gravity.
Replace the second cliff with the sphere, and the force of gravity with thrusters, and you've got the exact same situation.
bbl off to lab.
It's not the same situation at all though. There is no gravity acting on the asteroid that we are talking about -- the only force working on it is inertia (its own mass and velocity).
The entire point of our exercise is to make sure that earth's gravity NEVER works on the asteroid.
In that example, you are assuming that the asteroid is being pulled by earth's gravitational field, so it constantly has a force pulling it toward earth. The bowling ball example is much more apt for this scenario, since the only force acting on it before you hit it (aside from friction), is its own mass and velocity.
Take this scenario:
A sphere is traveling through a vacuum under its own inertia (let's assume it's in a giant vacuum tube in a lab and we are looking at it from the top, for the sake of simplicity).
If you apply a small amount of force at a perpendicular angle to its flight path, then the vector/angle of the sphere's inertial force will have changed. It now has inertia acting on it that will carry it to the edge of our vacuum tube, whereas before it would have gone in a perfectly straight line to the end of the tube. The amount of force required to do this is relatively small, because the force we applied does not have to be sufficient to carry it all the way to the edge of the tube -- it only has to be enough to change the inertial vector. From that point forward, inertia (in its new direction) will carry the sphere to the edge of the tube, and no more external forces are required.
The entire point of our exercise is to make sure that earth's gravity NEVER works on the asteroid.
In that example, you are assuming that the asteroid is being pulled by earth's gravitational field, so it constantly has a force pulling it toward earth. The bowling ball example is much more apt for this scenario, since the only force acting on it before you hit it (aside from friction), is its own mass and velocity.
Take this scenario:
A sphere is traveling through a vacuum under its own inertia (let's assume it's in a giant vacuum tube in a lab and we are looking at it from the top, for the sake of simplicity).
If you apply a small amount of force at a perpendicular angle to its flight path, then the vector/angle of the sphere's inertial force will have changed. It now has inertia acting on it that will carry it to the edge of our vacuum tube, whereas before it would have gone in a perfectly straight line to the end of the tube. The amount of force required to do this is relatively small, because the force we applied does not have to be sufficient to carry it all the way to the edge of the tube -- it only has to be enough to change the inertial vector. From that point forward, inertia (in its new direction) will carry the sphere to the edge of the tube, and no more external forces are required.
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[xeno]Julios
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No - the inertia is acting in the direction towards the planet. The tugging force of the tractor craft has the same effect on the motion of the asteroid as gravity has in the ball-cliff example.R00k wrote:It's not the same situation at all though. There is no gravity acting on the asteroid that we are talking about -- the only force working on it is inertia (its own mass and velocity).
No - please read more carefully. The force is due to the tractor craft (or in the other example i gave, the lateral thrusters)In that example, you are assuming that the asteroid is being pulled by earth's gravitational field, so it constantly has a force pulling it toward earth.
Correct.Take this scenario:
A sphere is traveling through a vacuum under its own inertia (let's assume it's in a giant vacuum tube in a lab and we are looking at it from the top, for the sake of simplicity).
If you apply a small amount of force at a perpendicular angle to its flight path, then the vector/angle of the sphere's inertial force will have changed.
Correct.It now has inertia acting on it that will carry it to the edge of our vacuum tube, whereas before it would have gone in a perfectly straight line to the end of the tube.
No, it absolutely does have to be sufficient to carry it all the way to the edge. This does not mean the force has to be applied for the entire time, but the displacement due to that force has to be sufficient. An understanding of Newton's 1st and 2nd laws of motion will help here.The amount of force required to do this is relatively small, because the force we applied does not have to be sufficient to carry it all the way to the edge of the tube -- it only has to be enough to change the inertial vector.
Do you acknowledge that it's possible to exert a momentary force which isn't enough to change its vector such that it will not reach the sides of the vacuum tube?
If you understand this, then you must understand that a sustained force will have an even larger effect.
Remember, because we're doing a lower level feasibility assessment, we're looking to see whether it's possible to displace the object given a sustained force.
Last edited by [xeno]Julios on Thu Feb 22, 2007 11:23 pm, edited 2 times in total.
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[xeno]Julios
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Here Rook, this post is for you
Read it carefully.
Look at the following diagram:
Imagine this is taking place on a frictionless surface - a circular disk, weighing 1 kilogram, on an air-hockey table is moving upwards towards the top edge of the table (by upwards i don't mean up against gravity - it's a top down view of the table).
The width of the table is D = 1000 metres.
We are told in advance that the puck will reach the top edge in 10 seconds.
There is a jet thruster on the right side of the puck, exerting a constant force of 10 newtons.
The question is: how far will the puck have been displaced to the right, by the time it reaches the top of the table.
(assume that it doesn't bounce off the edge of the table if it clears the edge before reaching the end, and can rather continue rightwards).
Here is how you solve the problem.
Fx = M*A (newtons second law of motion)
Ax = F/M = 10/1 = 10 m/s/s
(the x refers to the direction along the x axis, i.e. rightwards in this case)
Now that we have acceleration, and time, we can calculate distance:
Dx = V1x*t + 1/2at^2
V1x is 0, since it is initially not moving at all in the direction of the x axis.
so Dx = 1/2*a*t^2
= 0.5*10*10^2
= 0.5*1000 = 500 metres.
If we knew that the width of the table was in advance, we could then say whether or not the puck would clear the side of the table before it reached the top (would have to clear half the width).
Do you follow this example? Are you skeptical of the way I figured out the solution?
I can probably find online examples where the exact same approach is used. This is really basic highschool physics - it would be an extremely easy test question.
Read it carefully.
Look at the following diagram:
Imagine this is taking place on a frictionless surface - a circular disk, weighing 1 kilogram, on an air-hockey table is moving upwards towards the top edge of the table (by upwards i don't mean up against gravity - it's a top down view of the table).
The width of the table is D = 1000 metres.
We are told in advance that the puck will reach the top edge in 10 seconds.
There is a jet thruster on the right side of the puck, exerting a constant force of 10 newtons.
The question is: how far will the puck have been displaced to the right, by the time it reaches the top of the table.
(assume that it doesn't bounce off the edge of the table if it clears the edge before reaching the end, and can rather continue rightwards).
Here is how you solve the problem.
Fx = M*A (newtons second law of motion)
Ax = F/M = 10/1 = 10 m/s/s
(the x refers to the direction along the x axis, i.e. rightwards in this case)
Now that we have acceleration, and time, we can calculate distance:
Dx = V1x*t + 1/2at^2
V1x is 0, since it is initially not moving at all in the direction of the x axis.
so Dx = 1/2*a*t^2
= 0.5*10*10^2
= 0.5*1000 = 500 metres.
If we knew that the width of the table was in advance, we could then say whether or not the puck would clear the side of the table before it reached the top (would have to clear half the width).
Do you follow this example? Are you skeptical of the way I figured out the solution?
I can probably find online examples where the exact same approach is used. This is really basic highschool physics - it would be an extremely easy test question.
rook, you're oversimplifying now.
my point from the beginning is that you don't need the thing to miss earth for a lower limit of the feasibility or however julez is phrasing that. that's a worst case scenario. all you have to do is get it to miss that 400m keyhole.
in a head-on collision scenario julez' model is probably pretty optimistic, since the direction of gravity's pull on the asteroid would continually change direction... it's always pulling down (which is why the ball rolling off a cliff isn't entirely accurate depending on how far from earth we're talking about). you'd have to make the towing ship capable of repositioning itself to constantly exert a force normal to that of earth's gravity. logistical nightmare!
that all said, i still think julez is missing something here. gravity is the real solution to this because we're really wanting to use angular momentum against this thing. the most effective way to make it miss might not be to push it "away" from earth or orthoginally to its current motion. if you push it straight forward it might have enough momentum to blast right past earth or slingshot around is like comets do with the sun. certainly the last thing you'd want to do is slow its approach down... remember that demonstation where you swing a bucket of water in a circle? don't want to slow it down when it's above your head
my point from the beginning is that you don't need the thing to miss earth for a lower limit of the feasibility or however julez is phrasing that. that's a worst case scenario. all you have to do is get it to miss that 400m keyhole.
in a head-on collision scenario julez' model is probably pretty optimistic, since the direction of gravity's pull on the asteroid would continually change direction... it's always pulling down (which is why the ball rolling off a cliff isn't entirely accurate depending on how far from earth we're talking about). you'd have to make the towing ship capable of repositioning itself to constantly exert a force normal to that of earth's gravity. logistical nightmare!
that all said, i still think julez is missing something here. gravity is the real solution to this because we're really wanting to use angular momentum against this thing. the most effective way to make it miss might not be to push it "away" from earth or orthoginally to its current motion. if you push it straight forward it might have enough momentum to blast right past earth or slingshot around is like comets do with the sun. certainly the last thing you'd want to do is slow its approach down... remember that demonstation where you swing a bucket of water in a circle? don't want to slow it down when it's above your head
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[xeno]Julios
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True, and that's a dynamic I hadn't considered previously. We still agree that if it's not feasible without that dynamic, then it certainly wouldn't be feasible with it, right?menkent wrote: in a head-on collision scenario julez' model is probably pretty optimistic, since the direction of gravity's pull on the asteroid would continually change direction... it's always pulling down (which is why the ball rolling off a cliff isn't entirely accurate depending on how far from earth we're talking about). you'd have to make the towing ship capable of repositioning itself to constantly exert a force normal to that of earth's gravity. logistical nightmare!
That's the main point I've been making from post 1. i.e. this whole exercise only has the potential of demonstrating infeasibility rather than feasibility.
Interesting point - I hadn't considered that.that all said, i still think julez is missing something here. gravity is the real solution to this because we're really wanting to use angular momentum against this thing. the most effective way to make it miss might not be to push it "away" from earth or orthoginally to its current motion. if you push it straight forward it might have enough momentum to blast right past earth or slingshot around is like comets do with the sun.
After reading the nature paper which sparked this whole thing, though, it seems they were interested in displacing the asteroid, rather than speeding it up. Hard to tell because they referenced a formula from a book which I don't have electronic access to.
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Anhedoniac
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I'm going to gracefully bow out of this thread after this post, in the hopes of not highlighting my ignorance.
I should have read the article first. This is the path we're talking about:
The white bar indicates uncertainty in the range of possible positions. (Percent error)
(Jules, I'm not really sure which article you were reading, since I didn't see anything about a 100% chance of the asteroid hitting earth in that Wiki entry.)
At any rate, I now see what you mean about the "keyhole."
So the mass of the asteroid is around 1.012x10^11 lbs (a hundred billion lbs) and we essentially have to displace its orbit by 400 meters or so, to miss the keyhole. This isn't much longer than the asteroid itself.
But what has to be taken into account here (and sort of ties in with what menkent said) is the time frame in which these astronauts are talking about doing this.
The idea is to use this gravity tractor for a year in order to change its path to miss us. Interestingly enough, the asteroid's orbit around the sun is nearly the same as earth's -- 323 days.
That means that, if we were to send the tractor to it while it is near earth, then that tractor would be affecting it for an entire revolution of its orbit. Between that, and not knowing the exact rotational speed of the asteroid, it would be next to impossible to position an object to hover near the asteroid that would affect its orbit in a predictable/controllable manner.
I imagine the asteroid is far from spherical as well, so that adds more difficulty to controlling it with an external source of gravity.
Not only that, but Jules, your models don't seem to have much to offer in this context, since simply changing the vector or position of the object itself can have widely varying degrees of impact on the rest of the orbital path, depending on where the object was in its orbit when the change was made.
You were talking about changing the asteroid's direction if it were on a direct course to the center of the planet. You justified this because, since it is a low-level feasibility test, then if you can change it on a straight course, then it should be even easier to change if it were on an indirect/glancing course. But this leaves out the fact that the changes you will be making would occur around a 360 degree arc - and not a perfect circle.
So I'm just going to declare myself ignorant on the subject -- the only thing I'm sure of is menkent's earlier statement, that the system is far too complex to test with simple physics in a 2D model.
Sorry - if I had read the article, I would not interjected as much as I have. :icon32:
I should have read the article first. This is the path we're talking about:
The white bar indicates uncertainty in the range of possible positions. (Percent error)
(Jules, I'm not really sure which article you were reading, since I didn't see anything about a 100% chance of the asteroid hitting earth in that Wiki entry.)
At any rate, I now see what you mean about the "keyhole."
So the mass of the asteroid is around 1.012x10^11 lbs (a hundred billion lbs) and we essentially have to displace its orbit by 400 meters or so, to miss the keyhole. This isn't much longer than the asteroid itself.
But what has to be taken into account here (and sort of ties in with what menkent said) is the time frame in which these astronauts are talking about doing this.
The idea is to use this gravity tractor for a year in order to change its path to miss us. Interestingly enough, the asteroid's orbit around the sun is nearly the same as earth's -- 323 days.
That means that, if we were to send the tractor to it while it is near earth, then that tractor would be affecting it for an entire revolution of its orbit. Between that, and not knowing the exact rotational speed of the asteroid, it would be next to impossible to position an object to hover near the asteroid that would affect its orbit in a predictable/controllable manner.
I imagine the asteroid is far from spherical as well, so that adds more difficulty to controlling it with an external source of gravity.
Not only that, but Jules, your models don't seem to have much to offer in this context, since simply changing the vector or position of the object itself can have widely varying degrees of impact on the rest of the orbital path, depending on where the object was in its orbit when the change was made.
You were talking about changing the asteroid's direction if it were on a direct course to the center of the planet. You justified this because, since it is a low-level feasibility test, then if you can change it on a straight course, then it should be even easier to change if it were on an indirect/glancing course. But this leaves out the fact that the changes you will be making would occur around a 360 degree arc - and not a perfect circle.
So I'm just going to declare myself ignorant on the subject -- the only thing I'm sure of is menkent's earlier statement, that the system is far too complex to test with simple physics in a 2D model.
Sorry - if I had read the article, I would not interjected as much as I have. :icon32:
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[xeno]Julios
- Posts: 6216
- Joined: Fri Dec 10, 1999 8:00 am
I forgot to link to the article i read - the wiki quote was not where it was mentioned.R00k wrote:
(Jules, I'm not really sure which article you were reading, since I didn't see anything about a 100% chance of the asteroid hitting earth in that Wiki entry.)
No - the opposite: If it were on a direct collision course, and it turned out to be infeasible without earth's gravity to consider, then it would certainly be infeasible with earth's gravity to considerYou were talking about changing the asteroid's direction if it were on a direct course to the center of the planet. You justified this because, since it is a low-level feasibility test, then if you can change it on a straight course, then it should be even easier to change if it were on an indirect/glancing course.
(all the while assuming that it's aimed for centre of earth, and yes, in a straight line).
And about the simple 2d model, it seems that they used a similar model in the nature paper. The calculations are remarkably simple - you don't always need complex equations to solve real life issues, especially in the relatively controlled environment that space is.
Well, not necessarily just the fact that it's 2D -- you can do some pretty advanced orbital simulations in 2D.
But what about considering changing the complete orbit of an object that probably has multiple gravitational forces acting on it at different times? That doesn't seem like something your models could account for.
But what about considering changing the complete orbit of an object that probably has multiple gravitational forces acting on it at different times? That doesn't seem like something your models could account for.
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[xeno]Julios
- Posts: 6216
- Joined: Fri Dec 10, 1999 8:00 am
even the nature paper didn't account for those multiple gravitational influences. In fact, I'm not even sure they accounted for the earth's gravitational field - they may have done, but only in the sense that the deflection would have to transcend the gravity well.
But the size of this well, as menkent points out, depends on the inertia of the asteroid - and they used a stable equation for all asteroids at all speeds - so they made simplifications.
In fact, they used a very similar approach to mine, in basing their calculations on lead time as the primary variable.
You also gotta realize that these events are taking place years before collision, where the influence of earth's gravity isn't a factor worth considering.
ultimately you'd have to account for every object in the universe if you wanted it perfectly accurate - but you don't have to - there's a lot of space in space.
But the size of this well, as menkent points out, depends on the inertia of the asteroid - and they used a stable equation for all asteroids at all speeds - so they made simplifications.
In fact, they used a very similar approach to mine, in basing their calculations on lead time as the primary variable.
You also gotta realize that these events are taking place years before collision, where the influence of earth's gravity isn't a factor worth considering.
ultimately you'd have to account for every object in the universe if you wanted it perfectly accurate - but you don't have to - there's a lot of space in space.