If something is "dropped" from ISS altitude, do we need to worry about it burning up on re-entry

Factual Questions
1

I know that asteroids, spacecraft, etc. are subject to burning up when entering the atmosphere due to the friction of the object against the atmosphere – which is why great care is taken to safeguard spacecraft and the occupants within.

Thing is, these things are moving at thousands, if not tens of thousands, of kilometers per hour. If we had a space elevator (say it’s ~400 kilometers high, roughly the same as the ISS) and I dropped something, I imagine the highest speed it would achieve would be terminal velocity for that object’s mass and shape – likely not more than a few hundred kilometers per hour. At these speeds, is burning up on re-entry possible? Or would it simply fall to the Earth?

2

That something would have to be huge.

Also, my understanding is that anything “dropped” would be in orbit for a long time before we had to worry if at all.

Again this is just my understanding. I could have it wrong.

3

If “dropped” from the ISS, yes, because it’s moving laterally at thousands of kilometers per hour. You’re not going to get it to go straight down to the Earth without attaching a rocket to it. But if dropped from a space elevator, both the ground and the thing being dropped are moving laterally at the same speed, relative to each other, so there’s nowhere for it to go but down.

4

The terminus of a space elevator would by necessity be at geostationary altitude (35,786 km AMSL at the equator). Anything released at that altitude would remain in order for many thousands of years.

Items released from the ISS would eventually fall to Earth (within a few years, depending on ballistic coefficient). The ISS Is moving in orbit at around 27,600 km/hr or about 7.7 km/sec. While it will slow slightly to fall down below orbital speed it will still enter with something pretty close to that speed, and the dynamic pressure when it interacts with and compresses the tenuous upper atmosphere will erode most materials except for very high temperature metals and ceramics such that it will be mostly reduced to component atoms long before it reaches the stratosphere and “terminal velocity” (the speed at which the force due to drag equals that of acceleration due to gravity) becomes a limiting factor.

Meteorites (objects from space that make it to Earth’s surface) either have a dense mass of refractory elements (typically dominated by iron and nickel along with ‘rare earth’ elements), or they are very large such that the erosion while going through the atmosphere does not vaporize the core (1 km and larger water-ice objects).

Stranger

5

Way to avoid answering the question. Yes, the terminus of a space elevator must necessarily be above (not at) geosynchronous height. So don’t drop it from the terminus. Ride the elevator up only partway, to ISS height (which would not be in the ISS’s orbit, because you’re going way too slow), and drop from there.

It’s sort of true that you would be limited by terminal velocity, but that’s not the whole picture. Terminal velocity depends on the thickness of the atmosphere, so very high up where the atmosphere is only just barely there, the terminal velocity is much higher than it is at ground level. I think (though I haven’t done the calculations) that a dropped object would fall through the layers of atmosphere more quickly than it would decelerate due to air resistance, so that, for most points on its fall, it would be moving at faster than the terminal velocity for that point.

In any event, the simple answer is that you can get, at most, the gravitational potential energy of the object, and that’s not going to be very much. A kilogram dropped from 200 km up is going to have approximately 470 kcal of energy. If that kilogram is all water, it’ll take 540 kcal to evaporate all of it (plus more to raise the temperature first, plus more to melt it if it started out as ice). And of course, water is about the most volatile material you could be doing this with: Metal or rock would just shrug it off.

6

I think this is correct but, even at that altitude, there is still a teeny bit of “stuff” that would cause some friction and slow the object down. Unless it can boost itself on occasion, like the ISS, it will eventually fall back to earth. It may take a while but it will happen.

7

I did answer the question regarding terminal speed and burning up, but since you want to take issue with everything I say for some reason, here’s a thread for you to stop beating around and just spew out whatever your issue is with me.

Stranger

8

It isn’t actually friction, it is ram pressure:

Meteoroids enter Earth’s atmosphere from outer space traveling at hypersonic speeds of at least 11 km/s (7 mi/s) and often much faster. Despite moving through the rarified upper reaches of Earth’s atmosphere the immense speed at which a meteor travels nevertheless rapidly compresses the air in its path, creating a shock wave. The meteoroid then experiences what is known as ram pressure. As the air in front of the meteoroid is compressed its temperature quickly rises. This is not due to friction, rather it is simply a consequence of many molecules and atoms being made to occupy a smaller space than formerly. Ram pressure and the very high temperatures it causes are the reasons few meteors make it all the way to the ground and most simply burn up or are ablated into tiny fragments. Larger or more solid meteorites may explode instead in a meteor airburst.[17][18]

Compressing air heats it like, for example, in a diesel engine.

Diesel Engine Additive | Diesel Engine Basics | E-ZOIL

9

The only part of a traditionally-envisioned space elevator moving at orbital velocity is the part at the level of geosynchronous orbit (being the one height where the rotational speed of the orbit natches the rotational speed of the ground is what makes it geosynchronous.) Everything below that is moving slower than the orbital velocity at that height and isn’t actually in orbit. The traditional space elevator would be a cable hanging down from the anchor at geosynchronous orbit but it works like if you could build a tower up from the ground to geosynchronous distance: the top of the Empire State building isn’t moving at the speed of a 1,454 foot orbit (and the top of your head isn’t moving at the speed of a roughly 6 foot orbit).

10

Yes, but not in the way the OP was asked. I read the OP as “A car is going up a space elevator and about 400 km up, something goes overboard and falls back to Earth. Will it burn up in the atmosphere?” That is not a question you answered.

11

So trying to parse out my extensive confusion…

The question specifies ISS height space elevator so zero velocity relative to earth ground.

But some velocity relative to upper atmosphere since it isn’t traveling exactly at ground velocity?

So not a drop from ISS or meteorite circumstance?

Initial drop not slowed down much by friction or ram pressure because less dense atmosphere that high. But initial drop real slow because far away. Offsetting?

And if I am reading right it’s the ram pressure that would be the concern not the friction? Does that limit of gravitational potential energy limit in that case?

12

Earth’s gravity at 400 km up is still around 90% of surface gravity, so the fall would start only slightly slower.

13

I think I still don’t see the answer for something dropped at ISS altitude at geosynchronous speed (say, from a space elevator). My instinct is that it would basically just fall to earth, without any significant heating. It would go faster than ground-level terminal velocity at first, but I imagine it would adjust back to terminal velocity smoothly as the atmosphere thickened up.

14

Obviously, you don’t “drop” stuff from the ISS since it’s in orbit, the object would just float along at 28,000kph.

A space elevator? It’s rotating with the earth, so a circuit in 24 hours rather than the 90 minutes the ISS does. at 400km up, it’s 4400km from the center of the earth, so it’s travelling (2 xpi x 4400)/24 = 1151 kph. Presumably that is nowhere near the speed required to remain in orbit, so the dropped object would curve downward, attempting to perform an elliptical orbit where its starting point is the high point. It would intersect the atmosphere fairly soon. I would suspect it would not be too different from the simplest case.

Simplest case, no forward velocity (which I assume is what you are really asking). Straight drop down.

Basic back-of-envelope calculation… Standard acceleration formula is v^2 = 2ad. Take a=9.81m/s^2 (gravity) so say 10, and ignore the difference for being 400km further from the center of the earth. The object will fall 300km, let’s say, before friction becomes a seroious impediment. (at 100km, 60 mi altitude)

v^2 = 2 (10)(300x1000) (we’re working in meters not km in this one)

v^2 = 6,000,000

v = 2,450m/s or 8,818km/h.

And if we return to the space elevator scenario, the most it’s going to add the velocity is the extra 1151kph. (Actually, less)

Note the stuff that hits earth’s atmosphere from orbit is going sideways, at roughly the same speed as the ISS’s 28,000kph. Mostly, unless it’s big, it burns up. But your dropped object is only going about one third that speed, and less if you argue that maybe atmospheric friction starts to impeded acceleration before 100km altitude. And whether something burns up depends on what it’s made of, combustible, mass, melting point, etc. etc.

So the answer would be… maybe.

(My superpower in college was messing up calculations, so feel free to check my math)

15

Nevermind, ninja’d by md-2000 who showed their work.

16

The rotation is the same as the Earth’s ground (360 degrees in 24 hours) but not the speed because it is further out. Think of it like putting coins on a spinning record, one coin placed near the spindle and one near the edge. Both travel at exactly the same angular speed (such as 33.3 rotations per minute) but the coin on the outer edge travels a much greater distance.

A spot on the Earth’s surface at the equator travels at around 1,670 km/h. A spot on a space elevator/tower 400 km up would travel at around 1,774 km/h, so the difference in “wind speed” even between 400 km and the surface is only around 100 km/h. (Something orbiting at 400 km has an orbital velocity of around 27,600 km/h, so more than 25,000 km/h difference from surface “wind speed”.) And of course the nearer the air is to 400 km the less that difference is going to be. So the friction or ram pressure from the “sidewise moving” velocity is much, much less than even the friction or ram pressure from traveling at terminal velocity. Maximum pressure if it somehow instantly moved from 400 km up to 400 m up would be the amount you feel on your hand sticking out the window of a car moving at 100 km/60 miles per hour. What it would actually encounter over the length of the fall would barely qualify as a gentle breeze.

(Disclaimer: this is off the top of my head and I could be missing something that makes it completely wrong.)

17

This is wrong. As I mentioned in my post (written as yours was being posted) a spot on the Earth’s surface at the equator travels at around 1,670 km/h. A point 400 km further up from the surface would travel a larger diameter, therefore the speed would have to be higher than the surface speed.

18

I think you are confusing people by posing the question this way. The only part of the space elevator that is actually going at geosynchronous speed is the counterweight, which is way up at around 36,000 km and traveling at about 11,399 kph. This is required to maintain the tension on the cable so that it’s useful. At the point along the elevator cable that happens to be at ISS altitude, the relative velocity is much lower than the actual ISS, which is untethered and actually has to travel at orbital velocity to stay in orbit.

19

Just what happened to Felix Baumgartner when he jumped from an altitude of 39 km (24 mi): first he accelerated to almost supersonic speed:

  • Maximum vertical speed (without drogue) of 1,357.6 kilometres per hour (843.6 mph)

He slowed down considerably before opening the parachute a couple of minutes later. The same would happen from 400 km height, your top speed would just be even faster. The speed of sound varies with altitude/air density: Baumgartner did not fall faster than sound at his respective levels, but from 400 km height you probably would, only to slow down again before the parachute opens. Hopefully. Pro tip: stretch your arms and legs out, try not to spin, always look down to the ground. Do not forget to pull the cord. Plan some mechanism for automatic release of the parachute in case you lose consciousness.

20

Well, I didn’t say geosynchronous orbit, I said geo speed. For example, I’m sitting in a chair in my house right now traveling at geosynchronous speed.

And, he didn’t burn up. I think I understand now, thanks!