4 ms·

How come they are unable to know precisely where it will hit? Isn't it "just" lots of math to go through?

by joering2 13y ago

gibybo 13y ago
I'm guessing they don't know exactly how the wind currents are going to change over the course of it's flight.
- joering2 13y ago
  Yes, however winds would apply only to earth atmosphere. Most of its traction happens above earth atmosphere, right?
  - gibybo 13y ago
    The time it is affected by wind might still be enough to qualify its projected landing location as 'not exact'. Maybe they have a 200 mile radius projection or something. I could also imagine its non-spherical shape might present some challenges predicting its path as it rotates and 'flies' in slightly different directions.
  - sp332 13y ago
    What would cause drag besides the atmosphere?
  - maaku 13y ago
    > Most of its traction happens above earth atmosphere, right? What do you think is pulling it in?
parham 13y ago
It might also involve the shape of the satellite
- bbllee 13y ago
  And its orientation.
  - negamax 13y ago
    Assuming it's not rotating continuously.
bbllee 13y ago
Yes, it is "just" lots of math-- IF you know what the atmospheric conditions will be like. Having worked with a few atmospheric density models, I have the distinct impression that they are meant to give a mean density estimate but that variations are significant and velocity variations are not even modeled.
InclinedPlane 13y ago
The degredation of the orbit is due to atmospheric drag. Drag at such high altitudes is tricky to determine. If the spacecraft were already on a re-entry trajectory it'd be easy, the uncertainty in the exact conditions in the upper atmosphere wouldn't be significant. However, those conditions are dominant now when the spacecraft is still in orbit, also they are very non-linear. If the spacecraft happens to hit a small "bump" in the outer atmosphere which significantly slows it down that will lower its altitude, dramatically increasing its drag and hastening the re-entry.
omegant 13y ago
It hasn't an aerodinamic shape that behaves in a predictable way. Also it can and will break in different pieces, that will fall in different places due to their different weight and shape. Look at the Columbia debris pattern. It was scattered through half the USA. I don't think that the wind could do more than a docen miles of difference in the impact point. It will spend too little time at the lower atmosphere.
Tloewald 13y ago
The rest is "mere algebra" as one of my classmates used to say ;-)
blahedo 13y ago
The drag is bad enough---somewhat chaotic, hard to compute due to atmospheric variation---but I bet they could give a reasonably good estimate as long as it remains in one piece. The problem is, the moment it starts to break up, all bets are off, because we don't know which pieces will break off first, how large the respective sub-pieces will be, and so on; and each break could send one or more of the pieces into a tumble and/or heat up some other part more than another, which affects chances on further breakage, etc, etc, etc. Basically, once it starts breaking up, the system goes from "sort of unpredictable" to "wildly chaotic".

How come they are unable to know precisely where it will hit? Isn't it "just" lots of math to go through?

gibybo 13y ago

I'm guessing they don't know exactly how the wind currents are going to change over the course of it's flight.

joering2 13y ago
Yes, however winds would apply only to earth atmosphere. Most of its traction happens above earth atmosphere, right?
- gibybo 13y ago
  The time it is affected by wind might still be enough to qualify its projected landing location as 'not exact'. Maybe they have a 200 mile radius projection or something. I could also imagine its non-spherical shape might present some challenges predicting its path as it rotates and 'flies' in slightly different directions.
- sp332 13y ago
  What would cause drag besides the atmosphere?
- maaku 13y ago
  > Most of its traction happens above earth atmosphere, right? What do you think is pulling it in?

parham 13y ago

It might also involve the shape of the satellite

bbllee 13y ago
And its orientation.
- negamax 13y ago
  Assuming it's not rotating continuously.

bbllee 13y ago

Yes, it is "just" lots of math-- IF you know what the atmospheric conditions will be like. Having worked with a few atmospheric density models, I have the distinct impression that they are meant to give a mean density estimate but that variations are significant and velocity variations are not even modeled.

InclinedPlane 13y ago

The degredation of the orbit is due to atmospheric drag. Drag at such high altitudes is tricky to determine. If the spacecraft were already on a re-entry trajectory it'd be easy, the uncertainty in the exact conditions in the upper atmosphere wouldn't be significant. However, those conditions are dominant now when the spacecraft is still in orbit, also they are very non-linear. If the spacecraft happens to hit a small "bump" in the outer atmosphere which significantly slows it down that will lower its altitude, dramatically increasing its drag and hastening the re-entry.

omegant 13y ago

It hasn't an aerodinamic shape that behaves in a predictable way. Also it can and will break in different pieces, that will fall in different places due to their different weight and shape. Look at the Columbia debris pattern. It was scattered through half the USA. I don't think that the wind could do more than a docen miles of difference in the impact point. It will spend too little time at the lower atmosphere.

Tloewald 13y ago

The rest is "mere algebra" as one of my classmates used to say ;-)

blahedo 13y ago

The drag is bad enough---somewhat chaotic, hard to compute due to atmospheric variation---but I bet they could give a reasonably good estimate as long as it remains in one piece. The problem is, the moment it starts to break up, all bets are off, because we don't know which pieces will break off first, how large the respective sub-pieces will be, and so on; and each break could send one or more of the pieces into a tumble and/or heat up some other part more than another, which affects chances on further breakage, etc, etc, etc. Basically, once it starts breaking up, the system goes from "sort of unpredictable" to "wildly chaotic".