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Tiangong-1 Re-entry


GluttonyReaper

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The re-entry site is unpredictable.

An orbit takes about 90 minutes, which means it flies completely around the world 16 times per day.

You can only predict the impact area when you have a precise re-entry date down to a few minutes, which simply isn't possible yet. At this stage we don't even know which week it will come down since the margin is plus or minus 3 days.

You only get to predict the actual date a few days before, and the actual time (along with an approximate impact area) a few hours before it comes down.

So at this stage, the estimated re-entry site is anywhere between 42.7°N and 42°7 S. That's it.
 

Edited by Nibb31
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2 minutes ago, Nibb31 said:

You can only predict the location when you have a precise re-entry date, down to a few minutes. At this stage we don't even know which week it will come down. The estimate is plus or minus 3 days. You only get to predict the exact date à few days before, and the time (and location) a few hours before.

This.  Any prediction of the reentry footprint at this point isn't even a SWAG or informed speculation.  It's pure guesswork.

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https://phys.org/news/2018-03-track-chinese-space-station-falls.html

March 31 +/- 3 days.
Window will start to narrow considerably tomorrow.

Anyone here live in Oregon? (I think im kidding, but who knows).

Quote

“It’s much more common to be hit by lightning,” said Dr. William Ailor, principal engineer for the Center for Orbital and Reentry Debris Studies at Aerospace. - http://www.kgw.com/article/news/local/space-station-falling-to-earth-could-be-headed-for-oregon/283-531393318

Seeing how many times Chinese launch vehicles hit houses, lightning is going to get a run for its money.

Edited by PB666
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We have already explained why any predictions as to where it will reenter are pure guesswork. As long as no one can predict the exact time when it will come down then nobody can say where it will come down, since the two are obviously linked. And the exact time can only be pinpointed a few hours beforehand.

Currently, we have a plus or minus 3 day window, which means that it can fall down anywhere during 96 full orbits of the Earth. So basically any location between 42.7°N and 42.7°S has exactly the same chance as Oregon to be hit by a space station.

Edited by Nibb31
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4 hours ago, Nibb31 said:

We have already explained why any predictions as to where it will reenter are pure guesswork. As long as no one can predict the exact time when it will come down then nobody can say where it will come down, since the two are obviously linked. And the exact time can only be pinpointed a few hours beforehand.

Currently, we have a plus or minus 3 day window, which means that it can fall down anywhere during 96 full orbits of the Earth. So basically any location between 42.7°N and 42.7°S has exactly the same chance as Oregon to be hit by a space station.

I was joking about a local News story. 

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Two things the people are afraid of:

1. To get a falling truck-sized object right onto the head. One per whole Earth surface area.
2. To miss the falling truck-sized object visible as a weak meteor somewhere at horizon.

People, make your mind up! Are you afraid to catch or not to catch?

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13 hours ago, PB666 said:

Seeing how many times Chinese launch vehicles hit houses, lightning is going to get a run for its money.

In terms of risk, there's a great deal of difference between living in a village or town close to the Chinese launch site (which puts you under the falling boosters for every launch, meaning your risk of getting one on your house is like living on a target face at a shooting range), and living anywhere between 42.7 North and 42.7 South during the uncontrolled deorbit.

FWIW, though, the article I read was correct that there's a slightly increased risk of the impact falling close to the limits vs. anywhere else in that belt because Tiangong 1 spends proportionally more of its orbit time close to the "turnaround points" at the northern or southern parts of the orbit than it does near the equator.  For an illustration of how this works, look at the motion of a pendulum bob -- it passes center at maximum velocity, but it slows way down at the ends of the arc, and spends more time (per millimeter of arc length) near the ends than in the middle.  That means the least likely place for Tiangong 1 to land is close to the equator; the most likely is close to the northern or southern limits of the orbital ground path.

Conversely, Oregon takes up a tiny fraction of the land under the northern limit, and the bulk of the state is too far north anyway (the 45th parallel runs through the northern part of the state; that's 2.3 degrees, or almost 140 miles (225 km) outside the orbital path coverage).  Statistically, the station is more likely to hit Colorado (100% within the path and similar area to Oregon), California (further south, but 2-3 times the area), Arizona, New Mexico, or Texas (same or a bit further south on average than California, but even larger).  I live in North Carolina; we're 100% within the area at risk, but significantly further south than Colorado and much smaller area than Colorado or Oregon.  Honestly, China probably has the highest risk here; most of their country is under the path, and that's much more land area than the portion of the US at risk.  Brazil, India, and nearly all of Africa are at risk, too -- and more than half of the population of Oregon lives near Portland, too far north for concern.

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16 hours ago, Zeiss Ikon said:

FWIW, though, the article I read was correct that there's a slightly increased risk of the impact falling close to the limits vs. anywhere else in that belt because Tiangong 1 spends proportionally more of its orbit time close to the "turnaround points" at the northern or southern parts of the orbit than it does near the equator.   For an illustration of how this works, look at the motion of a pendulum bob -- it passes center at maximum velocity, but it slows way down at the ends of the arc, and spends more time (per millimeter of arc length) near the ends than in the middle.


o.0  A pendulum bob behaves as it does because it's being slowed down by gravity as it travels from the center to the end of the arc, and speeded up by gravity when it travel in the reverse direction.  Tiangong 1 however, isn't a pendulum bob and doesn't have any "turnaround points".  It isn't being slowed down as it heads towards the northern or southern extremities of its orbit and sped up as it heads towards the equator. It's traveling at a more-or-less constant angular velocity through it's entire orbit.

I'd have to see the article to judge, but based on your description it's hogwash.

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4 minutes ago, DerekL1963 said:


o.0  A pendulum bob behaves as it does because it's being slowed down by gravity as it travels from the center to the end of the arc, and speeded up by gravity when it travel in the reverse direction.  Tiangong 1 however, isn't a pendulum bob and doesn't have any "turnaround points".  It isn't being slowed down as it heads towards the northern or southern extremities of its orbit and sped up as it heads towards the equator. It's traveling at a more-or-less constant angular velocity through it's entire orbit.

I'd have to see the article to judge, but based on your description it's hogwash.

lol, you know that the oregon news article was linked as a joke. I can think of oth r places it would be funny to crash into if you like, assuming it stays together during reentry (it wont so dont write five posts about it). What if it landed on OCISLY, would it be considered a landing failure at spaceX? What if it landed at an airport, would it be considered as a crash landing. What if it landed in the Indian ocean and crashed into Zuma? What if it landed onto N Kor guided missile launchers. What if it crashed one of Putins publicity stunts. 

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2 hours ago, DerekL1963 said:


o.0  A pendulum bob behaves as it does because it's being slowed down by gravity as it travels from the center to the end of the arc, and speeded up by gravity when it travel in the reverse direction.  Tiangong 1 however, isn't a pendulum bob and doesn't have any "turnaround points".  It isn't being slowed down as it heads towards the northern or southern extremities of its orbit and sped up as it heads towards the equator. It's traveling at a more-or-less constant angular velocity through it's entire orbit.

I'd have to see the article to judge, but based on your description it's hogwash.

No, an inclined orbit doesn't "slow down" in terms of orbital velocity, but the velocity vector (relative to the surface) turns eastward near the inclination limits, so the north-south component is reduced.  In the limit (that moment when the satellite is exactly as far north or south as it will ever get) the north-south velocity is zero, just as it would be with a pendulum at the end of its swing.

In fact, the north-south component of a satellite's ground track is a better approximation to true "simple harmonic motion" than a pendulum.  Straighten out the ground track (that is, map it to a flat projection) and assuming the eccentricity is low (a good assumption in this case, since orbital decay tends to reduce eccentricity -- the drag impulse applied near periapsis lowers apoasis preferentially), the plot will be a sine wave, which is also the displacement/time plot of a system in true simple harmonic motion (characterized by a linear restoring force).  In the simple case of a circular orbit, the north-south component is simple harmonic motion.

I keep coming back to sine waves and simple harmonic motion because you can easily see, in the plot of a sine wave, that the half of the time centered on the peaks of the waves has much lower slope (analogous to north-south velocity in an orbit plot) than the half centered on the zero crossings.  Those peaks, with their reduced slope, show how an inclined circular orbit spends more time near the north and south limits than in a similar distance from the equator -- because the limits are approached with north-south velocity dropping to zero, while the crossing are approached with north-south velocity approaching its maximum value.  This is why any given square kilometer near 42.7 degrees north or south is at greater risk than a square kilometer near the equator.

 

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