Relative to atmosphere it is contact with? To a meaningful degree? Why?
It is rotating too. From the best I can find possibly faster than ground is but by a third if that “super rotation” does happen. There are also updrafts daytime from solar heating causing expansion and opposite at night … but are they of importance at these scales?
Looks like the outside burns, but so quickly that the inside stays cold, at least if you’re dropping a steak:
Great link!
So key is if it starts to tumble or not …
From ISS height:
The steak reaches a top speed of Mach 6, and the outer surface may even get pleasantly seared. The inside, unfortunately, is still uncooked. Unless, that is, it goes into a hypersonic tumble and explodes into chunks.
Because a space elevator will be in tension, held in place by the counterweight. Without that counterweight, the whole thing will move sideways and wrap itself around the Earth. An object released part way up the elevator will also move sideways. Someone upthread calculated the velocity it will have, but I don’t have time now to hunt down the post.
That’s because of drag on the whole line through the whole atmosphere. At ISS height the drag is present but small enough that the ISS just needs small boosts every so often.
If anyone has answered what the speed of the object would be relative to the atmosphere in the horizontal plane I have missed it. I’m guessing minor in the scale of these calculations. (Compared to top speed before thicker atmosphere slows the object down or even terminal velocity.) But happy to read an expert answer that is more informed than my guess!
In one of my posts, I pointed out:
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.
If air movement was in lockstep with ground movement all the way up, the horizontal speed of the atmosphere where it reaches appreciable density relative to the falling object would be between those two numbers, so it would be somewhere less than 100 km/h.
My math:
Radius of Earth: 6,371 km
Circumference of Earth at Equator: 40,030 km
Speed: 1,668 km/h
Radius of Earth+400 km: 6,771 km
Circumference of Earth+400 km: 42,543 km
Speed: 1,772 km/h
(This is simply 2r*π for the circumference, and circumference divided by 24 for the speed.)
If you consider the atmosphere being thick enough to matter at high speeds being at 100 km, then
Radius of Earth+100 km: 6,471 km
Circumference of Earth+100 km: 40,658 km
Speed: 1694 km/h
So the difference in horizontal speed at 100 km would be 78 km/h
(But of the air speed won’t be locked relative to the ground speed at that height, so that math isn’t exactly useful.)
I found myself wondering why there should be any horizontal displacement at all since from the reference frame at the surface of the earth, the space elevator cable is stationary so one might think a heavy object would come straight down. But the earth’s rotation is relevant due to the object’s higher angular velocity than at the surface.
Calculations from ChatGPT (so take it for what it’s worth) say that, ignoring air resistance, an object dropped from 400 km on a space elevator tethered at the equator would fall for 301 seconds and land 5.8 km east of the base of the cable, so there’s a horizontal displacement but it’s not huge. In the real world, winds and air resistance could significantly change those numbers.
Correct. Sorry. Does the upper atmosphere in general follow in lockstep of a radian/s basis? Faster? Slower? I’ve read faster by a third but also read doubt about that. And other articles that says it varies seasonally.
Would that matter much?
For this I have to rely on the University of Google. This old study from satellite observations in 1966 says that the atmosphere at 200-300 km is rotating around 1.3 times faster than Earth:
(That’s a scanned PDF but worth reading at least a bit of it.)
I dug out a spreadsheet I created to calculate reentry ballistics in a course on rocket propulsion I took a few years ago. With a few minor “space elevator” adjustments, here’s a summary of an object’s behavior being released from a space elevator’s tether at IIS altitude. It’s essentially your basic 2-D projectile motion calculations, but it also calculates dynamic pressure as the object falls deeper into the atmosphere (along with terminal velocity and max Q). For simplicity, the payload is a solid steel ball 1ft in diameter. While q is calculated, I ignore heat load in these calculations (assume an indestructible object)
Inertial (space) frame: the payload has an eastward tangential speed of ~497 m/s at 402 km.
Air/drag (Earth-rotating) frame: the local air co-rotates at the same speed, so the payload’s relative wind is ~0 while attached.
Implication: at the instant of release, horizontal relative wind ≈ 0, vertical = 0. However, after release, angular momentum is ~conserved.
Frame transition: We now compute drag in the air/drag frame using relative velocity to the air. Since angular momentum is conserved, the initial velocity is 497 m/s
Dynamics: gravity accelerates the sphere; speed rises like vacuum free-fall.
Milestone: 90 km → 2422 m/s, Mach ~8.9, dynamic pressure q ~16 Pa.
Density grows fast; drag still modest.
60 km → 2359 m/s, Mach ~8.5, q~2.0×10³ Pa.
Drag < gravity for a bit longer in this still-thin layer.
Peak speed 2604 m/s at 34.05 km, t = 4.79 min, Mach ~8.49.
Density ramps by orders of magnitude; drag ramps waaaaay up.
30 km: 2409 m/s (M~8.2), q~7.1×10⁴ Pa
20 km: 2208 m/s (M~7.5), q~2.7×10⁵ Pa
Max q: 6.93×10⁵ Pa at 9.67 km, t = 4.96 min, with speed ~1817 m/s, Mach ~6.0.
Velocity collapses but remains supersonic due to huge ballistic coefficient (~3300 kg/m²).
5 km: 1202 m/s (M~3.75), q~5.39×10⁵ Pa
1 km: 654 m/s (M~1.95), q~2.38×10⁵ Pa
Time to impact: ~5.11 min
Impact speed: ≈ 501 m/s (~Mach 1.5–1.6 at sea level)
Note that due to the high ballistic coefficient, it will smack into the ground long before reaching terminal velocity. It never achieves terminal velocity because the required density for that would occur only below where impact happens.
Now, in reality, our steel ball won’t make it anywhere close to the ground. I have a separate spreadsheet that calculates static air temp (using the same atmospheric model for density), stagnation temp, and adiabatic wall temp at each point along the trajectory. Here are the highlights:
90 km: M~8.9, T∞ ∼190–220 K, Taw∼2600–2800 K
60 km: M~8.5M, Taw∼3000 K (near peak)
35 km: Taw around 3,100–3,200 K
10 km (near max-q): M∼6, Taw∼2,700–2,900 K
1 km: M∼2, Taw∼900–1,100 K
Steel’s melting point is ~1,800 K. These Taw values exceed that above ~40–60 km, implying the sphere would melt/ablate long before reaching the lower atmosphere
Darren has already answered, but I just want to make this point. The tension in a space elevator, for the most part, has nothing to do with atmospheric drag. Even if the Earth had no atmosphere, there would still be tension, and it would still wrap itself around the Earth if the counterweight were released. Atmospheric drag might add some more tension, but it’s not the main source.
The ISS is different. It’s at orbital velocity, which a space elevator is not except at the geosynchronous altitude. That is, the altitude a satellite would be at if it were in geosynchronous orbit (38,000 km above ground level).
Can we all at least agree that the object be a spherical cow?
Where the atmosphere ‘starts’ depends upon the kind of phenomena you are looking at but in terms of aeroelastic behavior (where you can treat the atmosphere as a fluid continuum rather than a free molecular flow where the particles don’t really interact with each other in a mechanical sense) is at the Kármán line in the middle mesosphere, defined at 100 km (62 cmi). This is where you will see meteors getting hot enough to visibly glow mostly via radiative heating as ablated material interacts with the shock boundary producing ionized gas; as pointed out above, the interior of meteors isn’t really affected and even surface isn’t that hot because it is actually protected by this layer, which his the same principle by which ablative heat shields work to protect a spacecraft upon reentry.
At the point that drag becomes significant enough to place a limitation on how fast a body accelerated only by Earth’s gravity can go (excluding massive bodies with momentum beyond that of what an orbiting body would have) is even lower, essentially starting at the stratopause, about 48-50 km (about 26-27 nmi or ~160 kft). At that height, so much energy is lost to form drag and at supersonic speeds to the shockwave that a free body will quickly slow to a speed limited by drag. The specifics are determined by the aerodynamic properties (ballistic coefficient) of the body.
Stranger
Sure, a Solid steel spherical cow 1-ft in diameter 
Okay. Thinking about it for a second I get that and why. But my question remains - how velocity differential relative to entry of thick enough atmosphere to matter for the ram pressure concerns does that translate to? Enough to matter or rounding error amounts?
As Darren said, around 100 km/h. Closer to rounding error than enough to matter.
That’s below the breed standard for the Princeton Spherical Cattle used as the reference cow for standard bovine-related aerospace simulations. Maybe you are referring to the Stanford Ellipsoidal Dwarf Cow?
Stranger
It’s absolutely 100% definite that anything dropped would heat up. The question is just how much it would heat up. Certainly not enough to vaporize the entire object (there isn’t enough energy for that), but maybe enough to vaporize a thin outer layer of it.
OK, now that is a thorough answer.
I presume that’s just the surface temperature?
The key here is ablation. The surface heats up very quickly, loses structural integrity (i.e, softens/melts), and is torn away by hypersonic airflow. Exposing a new surface, which heats up very quickly, loses structural integrity, and is torn away by hypersonic airflow. Rinse and repeat.
That’s how everything that “burns up” in the atmosphere is destroyed. Nothing heats up uniformly and vaporizes all at once. The 1 kg ice ball you mentioned earlier in the thread (actually you said 1 kg of water) won’t completely vaporize, but a very large part of it will never reach the ground.
So going back to the OP, the answer is “do we need to worry about it burning up on re-entry” is yes, for any reasonable definition of “burning up”.