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74

The ISS is not designed to be run unmanned, entirely. The staff on board, when there are 6 astronauts, between exercise, sleeping, and maintenance get a single person-day of science work completed. (That is an 8 hour days' worth). The ISS is at a fairly low orbit, so that Soyuz, Dragon, CST-100, Cygnus, and the Shuttle can reach it. De-orbiting it will take ...


42

The extent of Earth's atmosphere and thus the amount of drag on the station varies greatly with solar weather; the relationship between drag and time-to-deorbit is very nonlinear. Combined, this means that reentry time is not very predictable on the scale of hours, which means the ISS could come down anywhere on its ground track and we wouldn't know where it ...


38

Let's explore the options: Try to keep it flying: due to structural and other stress, you would need to replace parts of the ISS. You might end up with a situation like Mir where more time is spent patching the station than actually making use of it and waste a lot of money and time. Just abandon the ISS: Not a good idea, its orbit decays and the ISS would ...


36

This question was partly addressed in a 1994 report (warning: not peer reviewed) by the Environmental Management of the Space & Missile Systems Command in the United States. Their focus was to consider the impact of deorbiting space debris on ozone at the time, and their conclusion was that deorbiting space debris has very little impact on stratospheric ...


33

The mass of Earth's atmosphere is 5E+18 kg and the Troposphere alone has 3/4's of that. With an average height of 13 km that makes its volume $4 \pi r^2 h$ or about 6.6E+18 m^3. If we break up one thousand 100 kg satellites into semi-porous PM2.5 particles that works out to be 1.5E-08 micrograms per cubic meter, and we generally worry about tens of ...


18

The problem with the ISS is that it cannot simply be left in its current orbit, because there is a (miniscule) amount of atmospheric drag acting on the structure to continually slow it down. Since the first module was placed in orbit, we've had to periodically boost the ISS back up to its nominal orbit (e.g. using the space shuttle's RCS engines while it was ...


10

Yes! The current status (end of June 2019) according to a SpaceX statement via Michael Sheetz is: 45 in final orbits 5 still raising, in final orbits shortly 5 paused during raise for adjustments, will continue 2 intentionally being deorbited to show debris disposal 3 stopped communicating, "passively" deorbiting


9

No, the reflector hasn't been deployed. Project head, Alexander Shayenko reported about it (RUS) (ENG) today. Early report information On July 17th the team reported about possible success (RUS) (ENG). They analyzed the TLE of orbits of 73 satellites in cluster. Based on braking factor(#9 element) and ballistic ratio(#11 element) they found 2017-042F, ...


9

Why was MESSENGER deorbited? It wasn't. (The wikipedia page on MESSENGER is wrong in this regard.) MESSENGER was inserted into an orbit about Mercury with a high inclination, a high eccentricity, and a rather low periapsis. The Kozai mechanism would have naturally resulted in this orbit evolving toward an even greater eccentricity, with the inevitable ...


8

Probably not. To control the point of reentry, you need to be able to adjust from a perigee high enough to not promptly reenter (i.e. above 200km) to one low enough to promptly reenter (i.e. below 80km) in significantly less than the time it takes to complete a single orbit -- otherwise, the unpredictable effects of drag in the variable density upper ...


8

It doesn't, not really, but there will come a time when it will simply not be worth keeping it up. Eventually the solar panels will not produce enough power, some airtight joint will break, or something similar. Right now the station is planned to be used until 2024. There is a study going on to see if it can last until 2028. The things that they are the ...


6

Assume that the 0.379 m/s² / km is a unit error, and the factor is supposed to be 0.379 m/s / km. I believe this is the fundamental mistake in the problem statement. Step 1: The delta-v required is equal to the change in altitude in km, multiplied by the conversion factor. ∆v is measured in meters per second. Step 2: Compute the acceleration of the ...


6

There is no need to have a steeper angle, and in fact, that's probably counter-productive. As explained in this thorough answer; the quickest deorbit would be to have a roughly circular orbit that is stable for a few months (270 km was proposed), and then lower the perigee to below 100 km, the lower the better. This lowering should happen such that the ...


5

This is not a complete answer as I won't be including the exact calculation needed to find out your burn time, but at least I will address the direct return vs bi elliptic approach. For a return from orbit of a manned spacecraft, you want to balance two factors: On one side you want to minimise the amount of fuel required for the operation; on the other, ...


4

Assuming a 400km circular starting orbit(and disregarding drag), how much delta V would be required to bring the perigee down to 0km altitude, or what equation could I use to find this out? The vis-viva equation is the go-to equation for a lot of things: $$ v^2(r)=GM\left(\frac{2}{r}-\frac{1}{a} \right)$$ $$a = \frac{r_{peri} + r_{apo}}{2} $$ You start ...


4

Some breaking news on this, according to this source (not the most reputable, to be sure) the reflectors have failed to deploy. They link to this source which appears to be from the creators of Mayak, saying that technical failures have occurred and they are still trying to diagnose them. I don't speak Russian and Google Translate only makes it partially ...


4

While the question could be marked as duplicate of Through what process does MESSENGER undergo orbital decay?, there's some hesitation to do that so I'll post an answer and also link to @DavidHammen's answer. edit: and now also link to @DavidHammen's other answer! MESSENGER was not specifically sent crashing into Mercury by direction of NASA, in fact they ...


3

I made an Excel spreadsheet to look at different scenarios. You are welcome to download it. To answer your question I entered 100 into cell F38 (Periapsis altitude) and 400 into cell F39 (Apoapsis altitude). I didn't enter 0 into F38 because getting the periapsis down into the upper atmosphere suffices to de-orbit a satellite. Over in cell J40 is the ...


3

Just to complement uhoh's answer, note that your spacecraft would reach the surface of the Earth at a speed $$v=\sqrt{2GM\left(\frac{1}{r_{Earth}}-\frac{1}{2a}\right)},$$ a little higher than 8 km/s = 28,889 km/h (not counting the rotation of the planet). To make the spacecraft land smoothly you'd need to brake to zero, and spend almost as much propellant ...


3

Because it's using Earth's magnetic field to create drag. It's one of several passive deorbiting systems. An electromagnetic tether uses a conductive tether to generate an electromagnetic force as the tether system moves relative to Earth’s magnetic field. NASA: State of the Art of Small Spacecraft Technology, 12. Passive Deorbit Systems In-Space ...


2

It doesn't matter when, as the system is rotationally symmetric. A retrograde burn is almost certainly the most efficient method (excluding any very long time period wait-for-perturbations-to-become-significant ones). It is uncommon to de-orbit geostationary object so you might not find much about it directly (the common approach is to increase the orbital ...


1

Neither NASA nor Xinhua news, in their detailed descriptions of the overall mission, mention any role for the orbiter after the returner separates from it in Earth orbit to land. Any extended mission for the orbiter that requires equipment beyond what is needed to retrieve Lunar samples would have eaten into the payload budget and probably reduced the size ...


1

In the linked AIAA article, the bottom of page 4 (excerpted below) estimates numbers for a 250 m x 0.28 m tape. When 700 km high, electrodynamic drag and aerodynamic drag are both about 15 μN. Higher up, electrodynamic drag dominates. To generalize this, into equation (4) plug in values for tape width w, a bunch of numbers ∆V me mi that I don't know how ...


1

If you have trajectory, then you have position and velocity over time, so what you need next is aero heating versus velocity. If you can calculate that over the trajectory, you can easily find the peak. If all you care about is the point of maximum heating, you could get pretty close by looking at acceleration times velocity, which should be a good proxy for ...


1

I've posted this answer to make sure there is some information here. If someone would like to use this as a starting point for an additional answer, that's great! The comment by @Hobbes seems to be correct. I've just found the YouTube video A tour through the Mayak project and I show some screenshots below. The original plan was to use ammonium carbonate ...


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