I'm a noob in this subject and know little about space exploration, but I wonder about this every time I read the news:

Wouldn't drag make all space junk fall back to Earth after some time?

I'm reading about several projects to get rid of space junk but I was wondering if all those pieces of junk aren't doomed to fall to Earth anyway.

Or maybe it's impractical in terms of waiting time. How long will take for a ~1 cm piece of junk in LEO to fall back to Earth on its own?

  • $\begingroup$ Complementary questions for whoever interested in the debris issue. How many artificial objects of size >1cm are orbiting the Earth now and won't decay by themselves in our's and our grandchildren's lifetimes? How many of such objects existed before mankind first set foot on the Moon? Hint: visit stuffin.space. $\endgroup$
    – Ng Ph
    Dec 2 '21 at 13:44
  • $\begingroup$ What does drag have to do with debris in a vacuum? Or is the act of gravity on an object considered drag? $\endgroup$ Dec 2 '21 at 16:39
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    $\begingroup$ @JOHNKEEBLE the vacuum in LEO is far from perfect. There are still lots of atmosphere particles, there is solar wind, so you constantly collide with something. Even though the density and the resulting drag are tiny, the effect adds up over time. See the answers. $\endgroup$
    – Frax
    Dec 3 '21 at 1:14
  • $\begingroup$ @JOHNKEEBLE While drag has nothing to do with debris in a vacuum, that's not relevant to this question. Many objects in LEO orbit entirely within the thermosphere. While the air there is very thin, it's enough to drag objects out of orbit with enough time. Also, look up the exosphere -- it extends well past the area we designate as LEO (10,000km vs 2,000km). $\endgroup$
    – Brian
    Dec 3 '21 at 1:20
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    $\begingroup$ @JOHNKEEBLE There is air in orbit. Very thin air, as thin as what you would find in a vacuum tube or thinner. But it's air nonetheless. Heck, there's even "air" (if you can call it that) in deep space. In deep space you can expect there to be around 100000 atoms/molecules of "stuff" per cubic meter. At the International Space Station there is around 10 trillion molecules per cubic meter in "vacuum". Earth atmosphere at sea level is 10 trillion trillion molecules per cubic meter. So "vacuum" in orbit is just 10 trillion times less dense than normal air but the density is not zero $\endgroup$
    – slebetman
    Dec 3 '21 at 2:27

It depends on the altitude. Here is a chart from ESA and UNOOSA. Basically, anything under 500 km will fall relatively quickly, maybe 25 years. Everything under 800 km should fall within a century or so. 1200 km will take almost 2000 years to fall, and anything higher than that will take a REALLY long time to fall.

enter image description here

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    $\begingroup$ Thank you. This makes a lot of sense. I understand these figures are approximate and many other parameters should be taken into account like surface, weight, drag, density, etc. but this really gives a good idea that it's usually impractical to wait for them to fall on their own. $\endgroup$ Dec 1 '21 at 2:59
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    $\begingroup$ ... and objects higher than geostationary don't even fall. The Earth drags them higher. $\endgroup$
    – fraxinus
    Dec 1 '21 at 7:27
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    $\begingroup$ I want to see that Roman satellite though! Imagine if the Romans had cleared up everything they left behind, that would be quite bad for the Italian tourist industry ;-) $\endgroup$
    – gerrit
    Dec 1 '21 at 12:37
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    $\begingroup$ It would be great if you can repeat the essential info from the image as text to aid those using screen readers. $\endgroup$
    – Tim
    Dec 2 '21 at 0:00
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    $\begingroup$ @Nelson: Except that the higher up they are, the slower they are moving. Mean orbital speed scales in proportion to the inverse square root of the orbit's semi-major axis. (Objects in higher orbits do, however, have more total energy and angular momentum that they need to lose before they fall. But they still move slower. Orbital mechanics is weird.) $\endgroup$ Dec 2 '21 at 11:07

Space debris poses a real risk for spacecraft in LEO. From the Technical Report on Space Debris UN Committee on the Peaceful uses of Outer Space (Table 5), a satellite in orbit can expect to collide with a small piece of debris every few decades:

Technical Report on Space Debris UN Committee on the Peaceful use of Outer Space

Table 5. Mean time between impacts on a satellite with a cross-section area of 10 square metres

Height of circular orbit Objects 0.1–1.0 cm Objects 1–10 > cm Objects > 10 cm
500 km 10–100 years 3500–7000 years 150,000 years
1,000 km 3-30 years 700-1400 years 20,000 years
1,500 km 7-70 years 1000-2000 years 30,000 years

According to NASA https://www.nasa.gov/news/debris_faq.html ,

“The higher the altitude, the longer the orbital debris will typically remain in Earth orbit.

  • Debris left in orbits below 370 miles (600 km) normally fall back to Earth within several years.
  • At altitudes of 500 miles (800 km), the time for orbital decay is often measured in decades.
  • Above 620 miles (1,000 km), orbital debris normally will continue circling Earth for a century or more.”

The time to de-orbit depends on a large number of variables:

Since the peak concentration of space debris is at an altitude of 1000km, it will take many decades for spontaneous reduction of centimeter size debris and centuries or millennia for larger chunks

enter image description here


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    $\begingroup$ The table 5 you cited appears to be very incorrect. There is no way that objects in a 500 km altitude orbit remain in orbit for a longer period of time (sometimes a lot longer period of time) than objects in a 1000 km or a 1500 km altitude altitude orbit. Please add a citation, and double check that you have the numbers correct. $\endgroup$ Dec 1 '21 at 13:55
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    $\begingroup$ @DavidHammen: If I'm reading that correctly, the table is not giving the lifetime of an orbiting object, it's giving the average rate of impacts on a satellite at that altitude. It doesn't seem unbelievable to me that there would be factors that would lead to lower impact probabilities at both higher and lower altitudes, with the intermediate range being the "worst of both worlds". (And of course, the figures in the table don't answer the question the OP's question, which is how long a given piece of debris takes to de-orbit.) $\endgroup$ Dec 1 '21 at 14:55
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    $\begingroup$ The table 5 brings confusion because it doesn't answer the question of how long it takes for debris to deorbit but uses the same units as you'd expect from an answer to that question. Consider adding a an explanation of the data, because the text in the image is grainy and many people skip to the reading just the numbers and assume the table is meant to serve as an answer the question rather than just give context to the answer. $\endgroup$ Dec 1 '21 at 16:16
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    $\begingroup$ I've proposed an edit that (I hope) makes the relevance of the first table clearer. $\endgroup$ Dec 1 '21 at 17:20
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    $\begingroup$ I suspect it is a square/cube issue. Drag increases with the square of size, but mass increases with the cube. $\endgroup$
    – Woody
    Dec 2 '21 at 14:51

Your question leaves out two rather important variables to give accurate answers. You've given us a size, but not a density or shape. Because orbital decay is related to the loss of an orbiting objects potential energy and its potential energy depends on its mass and how fast the energy is lost will be related to its shape.

For a more step by step explanation you can read through the math of the Wikipedia explanation here (https://en.wikipedia.org/wiki/Orbital_decay), its actually pretty good explanation which avoids getting too deep into the weeds.

This is pretty abstract stuff talking about dimensionless values and such, but it gets a lot more obvious what's going on when you focus just what aerodynamic drag is caused by. Atmospheric drag on an object, regardless of velocity, and even for the tenuous wisps of molecular gasses above 400km altitude, is determined by the ballistic co-efficient (https://en.wikipedia.org/wiki/Ballistic_coefficient) which can be understood as (( how dense is the object * how long is the object ) / an special adjustment factor for the shape of the object ).

So lets get to how this affects your question. You've asked about a 1cm piece of debris, the easy way to show how this is affected is probably to show a little table for comparison. Using average density from wikipedia and the basic drag coefficient values https://en.wikipedia.org/wiki/Drag_coefficient lets get some rough ballistic coefficients for various hypothetical bits of debris of about 1cm in size.

Material Shape Approximate Ballistic Co-efficient
Frozen Water Sphere 18.3
Glass Cubic 23.8
Mylar Insulation Square (Flat) 0.0029
Steel Square (Flat) 7870
Steel Sphere 175
Steel Cube 78
Steel Short cylinder (Imagine the sheered off end of a broken bolt 1cm long) 71.5

Now these are extremely rough numbers. But you can see from how large the range is depending on the density of the materials when you look at the flat square and how massive the difference is between light mylar insulation and steel, and then how simply changing the shape of the steel debris in the last few lines from our square to sphere, the cube, and then to our hypothetical chunk of a broken bolt. The changing shape has a large difference in the final ballistic coefficient and this is what drives how quickly our object will shed its kinetic energy and thus how fast its orbit will decay. When you look at how this plugs into the math on orbital decay you can see this parameter means that for a given size of debris, 1cm in the case of your question, without knowing what its made of and having a better idea of the shape, you can't analytically predict ahead of time how long it will take to decay. When you add the additional factor of how different the mass will be between a thin flat piece of broken of sheet, compared something like a cube or sphere or cylinder this further magnifies how large of a range you are looking at.

When people talk about approximate times for things to decay they are often just looking at the statistical averages of real world decay times. One of the best examples is cubesats which because they have a known shape and all fit in a fairly common total mass can be approximated rather well, most 1U 2U or 3U Cubesats orbiting under 500km should re-enter in under 25 years based on modelling such as https://iopscience.iop.org/article/10.1088/1742-6596/641/1/012026/pdf But random space junk is not uniformly shaped and its material is usually not well known for the smallest pieces, consequently people just fall back to taking the tracking information we do have and building statistical approximations which can be used as a best guess, leading to approximations like the one put out by UNOOSA, referenced by one of the other answers here https://space.stackexchange.com/a/55995/19695

  • $\begingroup$ Particles undergoing orbital decay are actually losing potential energy, dropping lower and lower, as they speed up..., $\endgroup$
    – DJohnM
    Dec 3 '21 at 19:58
  • $\begingroup$ Thanks @DJohnM for the reminder, fixed that. So easy to get them swapped when your thinking local interaction with a streaming medium versus the overall energy of the orbiting object . $\endgroup$
    – Techdragon
    Dec 4 '21 at 12:52

Well 1 cm items are different. But I remember the falling out of orbit of a rather LARGE item. Skylab was launched into orbit in 1973, and there was a bit of a scare in the world about it's descent just 6 years later in '79. Their original plan was that the Space Shuttle would soon be available to help boost it's orbit.


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