This morning GOCE was at an altitude of around 160 km. As expected, the drag levels have increased very much, with the average now around 90 mN (milliNewton).
Re-entry of GOCE into Earth’s atmosphere is predicted to occur during the night between Sunday and Monday, 10/11 November. Break-up of the spacecraft will occur at an altitude of approximately 80 km.
Q: Do we expect any of these to reach the surface?
HK: Most of these fragments will completely burn up. A small fraction of the initial spacecraft mass – about 20% or 200kg – is expected to reach ground, distributed across dozens of fragments, spread over a sizable re-entry ground swath.
Q: Is there any risk to anyone on ground?
HK: The risk to the population on ground will be minute. Statistically speaking, it is 250,000 times more probable to win the jackpot in the German Lotto than to get hit by a GOCE fragment. In 56 years of space flight, no man-made space objects that have re-entered into Earth’s atmosphere have ever caused injury to humans.
Hmm.. is that statistic calculated for each individual or for the entire population. The changes are to hit are very low in either case, but there is still a large difference mathematically.
Let's try to get an order of the probability... Let's say the parts land on earth with an an angle of more than 45 degrees. I'd take 1 sqm for each human, understanding the angle would reduce that number. Which makes around 10000 sqkm, counted generously. The earth total surface is around 510 millions sqkm. So that's one in 51000 chances for any human to be hit (calculated high). So quite makes sense that no one was impacted yet, but this seems not impossible as well - of course depending on how many satellites impact earth each year. Not counting that people are often in buildings, but one impact could kill more than 1, etc.
Personally, I take this number for negligeable still. It would rather be hard finding anything build by humans killing less people.
Please correct me if I'm wrong but, your calculations don't take into account the population's distribution over the surface of the earth; most importantly water masses.
> HK: The risk to the population on ground will be minute. Statistically speaking,
Its nice to see some common sense and statistics to play key role in "preparedness" in this situation.
To the contrary, the "War on Terror" continues with 1/18th of US debt being spent [1], even though the odds of dying due to terrorist attack are minimalist [2].
<NSA>The odds of dying in a terrorist attack are minimalist because we spend 1/18 of the US budget to keep you civilians safe! Also, your wife just emailed you to remind you to pick up milk at the store.</NSA>
> The odds of dying in a terrorist attack are minimalist because we spend 1/18 of the US budget
Yes, I agree. Terrorism in US falls under basic "create problem and offer solution" category.
The odds of dying in a terrorist attack would be even smaller, had we never gone to Iraq, never started Afghanistan war and did thousands other things throughout the history of US that got most of the world pissed off, starting from hard core terrorist from Afgan mountains all the way down to the WW2 veterans (american citizens) being pissed off because Gov closed national parks.
So why don't you stop all those shady operations around the world and eventually, as a result, keep that 1/18th of the budget that you so desperately need to spend on "keeping civilians safe" in your/citizens own pockets.
It is dropping altitude faster (40 metres/sec) as it nears the equator and is moving at almost constant velocity (between 7.89-7.91). Looks like it will continue to slowly spiral around the earth and crash in the Atlantic Ocean/N.America based on simple extrapolation. Except that the it is probably not going to be that simple. Just hit the equator. Alt: 156, Spd: 7.93, Long: 128.8, Lat: 0. Now altitude is increasing again.
The altitude is rising and falling because the satellite is in an (approximately) elliptical orbit around Earth. Go play you some Kerbal Space Program! :)
I'd think there'd be somewhat of a chance that parts could break off during reentry which might have a much greater chance of hitting a lower terminal velocity and therefore not burn up, can anyone explain why this doesn't seem to be the case? Is it that the high velocity in the upper atmosphere causes much greater heat and destruction than if it was falling through a more dense atmosphere?
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.
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".
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.
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.
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.
Yeah, hard bounce off the atmosphere trades angular momentum for radial velocity and increases eccentricity, meaning next time you hit the atmosphere is gonna be even worse.
AFAIU it doesn't go up, per se. Because our planet is not a perfect sphere, but rather one that is squeezed on the Poles, its actually the planet distance itself longer while rotating..
There's only a 31km difference between the "lowest" and "highest" surface point on earth so I doubt that's what's going on here. More likely is its orbit eccentric and while it seems to go up it continually loses speed due to the atmosphere slowing it down and its average altitude keeps going down continually (although obviously more when it's close to the periapsis, the point of closest approach).
Could you explain that figure to me please? Everest isn't nearly that high, so are you starting at the bottom of a trench? Surely the meaningful low point is the lowest point (be it the sea or some other low point) that the inbound fiery ball could hit?
The earth is not spherical. The diameter of the earth through the equator is more than the diameter through the poles, because of its angular velocity. See http://en.wikipedia.org/wiki/Figure_of_the_Earth
This morning GOCE was at an altitude of around 160 km. As expected, the drag levels have increased very much, with the average now around 90 mN (milliNewton).
Re-entry of GOCE into Earth’s atmosphere is predicted to occur during the night between Sunday and Monday, 10/11 November. Break-up of the spacecraft will occur at an altitude of approximately 80 km.
Q: Do we expect any of these to reach the surface?
HK: Most of these fragments will completely burn up. A small fraction of the initial spacecraft mass – about 20% or 200kg – is expected to reach ground, distributed across dozens of fragments, spread over a sizable re-entry ground swath.
Q: Is there any risk to anyone on ground?
HK: The risk to the population on ground will be minute. Statistically speaking, it is 250,000 times more probable to win the jackpot in the German Lotto than to get hit by a GOCE fragment. In 56 years of space flight, no man-made space objects that have re-entered into Earth’s atmosphere have ever caused injury to humans.