Asteroid word problem.

[DaShen]

Originally posted by: Tom

Originally posted by: DaShen

Originally posted by: HomeBrewerDude
rotational speed is highest at the equator. go south and your chances of avoiding the collision improve.

negligible. If it hits you, it hits you, but otherwise, the best thing to do is be prepared.

For this problem, there is no such thing as negligible. If there is a mathmatical difference.

This isn't about asteroids, it's about the mathmatics involved. I'm having trouble figuring out if there are mathmatical differences..

HomebrewerDude is the first person to grasp what I'm getting at, but I don't know if his assertion, which is where I started from, is correct or not.

if the improvement in chances is within the standard deviation of mathematical error, it is negligible. It would be the same as me throwing a baseball 100 feet away from a minature globe and having an ant travel to the equator of the spinning globe. The chances are pretty much equal for me to hit the ant.

 

[Tom]

Originally posted by: joshsquall
Mathematically speaking, every 100 x 100 block on the face of the Earth has the same chance of being hit.

I think I'm beginning to grasp that this is correct, if we assume the impact is instantaneous and the Earth and the asteroid are both flat.


However, given my original parameters and common knowledge, we know the Earth is basically a sphere, we also know the velocity and diameter of the asteroid.

So if I add one more fact, that the asteroid can be assume to be a perfect sphere, does that change anything ?

Given that info, isn't it true that depending on the angle and location of the intersection, the impacted area could be larger or smaller than 100 feet ? I suppose that this doesn't matter, because all the effects of this cancel each other out in terms of releative risk..

Or does the rotation of the Earth being relatively constant and in a particular direction mean that Todd can use that info to reduce his risk ?

 

[Gunslinger08]

Originally posted by: Tom

Originally posted by: joshsquall
Mathematically speaking, every 100 x 100 block on the face of the Earth has the same chance of being hit.

I think I'm beginning to grasp that this is correct, if we assume the impact is instantaneous and the Earth and the asteroid are both flat.


However, given my original parameters and common knowledge, we know the Earth is basically a sphere, we also know the velocity and diameter of the asteroid.

So if I add one more fact, that the asteroid can be assume to be a perfect sphere, does that change anything ?

Given that info, isn't it true that depending on the angle and location of the intersection, the impacted area could be larger or smaller than 100 feet ? I suppose that this doesn't matter, because all the effects of this cancel each other out in terms of releative risk..

Or does the rotation of the Earth being relatively constant and in a particular direction mean that Todd can use that info to reduce his risk ?

That's just the thing. We don't know the velocity of the asteroid. Velocity is a vector, meaning we know the direction of the force as well. In this case, we only know the speed. There is absolutely no way to predict where it will land with just the scalar of speed. All spots on Earth have an equal chance.

 

[Tom]

Originally posted by: DaShen

Originally posted by: Tom

Originally posted by: DaShen

Originally posted by: HomeBrewerDude
rotational speed is highest at the equator. go south and your chances of avoiding the collision improve.

negligible. If it hits you, it hits you, but otherwise, the best thing to do is be prepared.

For this problem, there is no such thing as negligible. If there is a mathmatical difference.

This isn't about asteroids, it's about the mathmatics involved. I'm having trouble figuring out if there are mathmatical differences..

HomebrewerDude is the first person to grasp what I'm getting at, but I don't know if his assertion, which is where I started from, is correct or not.

if the improvement in chances is within the standard deviation of mathematical error, it is negligible. It would be the same as me throwing a baseball 100 feet away from a minature globe and having an ant travel to the equator of the spinning globe. The chances are pretty much equal for me to hit the ant.

You are talking about practical applications, I am trying to determine if there is a mathmatical difference. Not error, but a real mathmatical difference.

btw, as to your analogy, a 100 foot asteroid relative to the Earth, is much much smaller than a baseball relative to the size of most globes I've seen. Your globe would have to be a half mile in diameter. Then your chance of hitting the ant may vary more than you said, although I'm not convinced it is true.

 

[Tom]

Originally posted by: joshsquall

Originally posted by: Tom

Originally posted by: joshsquall
Mathematically speaking, every 100 x 100 block on the face of the Earth has the same chance of being hit.

I think I'm beginning to grasp that this is correct, if we assume the impact is instantaneous and the Earth and the asteroid are both flat.


However, given my original parameters and common knowledge, we know the Earth is basically a sphere, we also know the velocity and diameter of the asteroid.

So if I add one more fact, that the asteroid can be assume to be a perfect sphere, does that change anything ?

Given that info, isn't it true that depending on the angle and location of the intersection, the impacted area could be larger or smaller than 100 feet ? I suppose that this doesn't matter, because all the effects of this cancel each other out in terms of releative risk..

Or does the rotation of the Earth being relatively constant and in a particular direction mean that Todd can use that info to reduce his risk ?

That's just the thing. We don't know the velocity of the asteroid. Velocity is a vector, meaning we know the direction of the force as well. In this case, we only know the speed. There is absolutely no way to predict where it will land with just the scalar of speed. All spots on Earth have an equal chance.

Well, I should have specified it, but the velocity of the asteroid is relative to the Earth, at impact, if that clarifies things at all.

 

[Chewy]

A. speed up - if his girlfriend is in NY
B. slow down - if his girlfriend is ugly
C. turn around and head back to LA - if his girlfriend is REALLY ugly
D. Head North - if he likes them top heavy
E. Head south - if he likes to give more than to receive
F. It doesn't matter what he does - he has his own gravitational pull
G. Stop where he is - his girlfriend has her own gravitational pull

😛