You have two separate posters who calculated that the object will be traveling at least 2.5 km/sec vertically when it hits the atmosphere. Survival will depend on the shape and composition of the object - you could design an object to survive, but a random object like a baseball wouldn’t have a chance.
Frame of reference is key.
Geosynchronous speed is that speed which is “appearing stationary with respect to a point on the ground.” A point on the ground is stationary compared to itself and is geosynchronous … It has a speed relative to the center of rotation and the object at ISS height and the object at geosynchronous orbit height have different ones.
I think this conversation gets very confusing without clear specification of which frame of reference is being used and switching between them.
Would something dropped off the edge of a space elevator at ISS height from earth surface drop straight down relative to earth surface? I don’t think so. I think the atmosphere is moving at a different speed.
An object at ISS height is already in the atmosphere. Thin enough to be called a vacuum, but thick enough to be constantly dragging at the ISS enough that it requires periodic boosting. Every moment that object is falling towards Earth the very thin atmosphere it is passing through is getting slightly thicker. So how far has it fallen when it is considered to have “hit the atmosphere”, and how much does that thin atmosphere effect the speed calculations? (Meteors are around 80 to 120 kilometers up when atmospheric pressure is high enough for them to become incandescent.)
Right, I think that just about any physicist would interpret the term “geosynchronous speed” as the speed of an actual geosynchronous orbit. I don’t believe I’ve ever seen it used any other way until today.
Ignoring winds, I would assume that the upper atmosphere rotates with the Earth. In other words, if you take an arbitrary chunk of it, this will stay over the same piece of surface, other things being equal. The whole Earth plus atmosphere system has had a LONG time to settle into equilibrium.
I don’t understand them but nah. There are huge circulation cells and currents partly created by the earth’s rotation, the Coriolis effect, but also by solar heating ionization tidal effects and more. The atmosphere is not geostationary.
Ah yes… I was forgetting things like the jet streams and other circulation cells, you’re quite right.
On the global scale, all integrated together, I would assume the whole system rotates together, just by conservation of momentum.
But there are a lot of local variations. Almost all of these are below the tropopause though, I think?
There’s certainly some air movement relative to the surface, but that’s still a heck of a lot slower than orbital speed.
And on thinking about it some more, how fast an object is falling at various heights, and how long/far before it ends up at (close to) terminal velocity, would depend on what the objet was. Certainly on its sectional density (mass per cross sectional area), and also on its shape. A very large or dense object might have a terminal velocity (even at ground-level air densities) greater than the velocity it’d ever reach by falling, wile a very small or diffuse object would be slowed much more by air resistance.
I’m not sure what all the debate here is.
The OP question was obviously along the lines of “the object falls vertically, but does not have orbital velocity” and implies the only velocity would be from falling vertically.
The question is not phrased well -wheels within wheels complexity - since a space elevator is in fact also rotating around the center of the earth. The speed at any point up and down the elevator is simply the speed needed to complete a circuit of the earth in 24 hours. Nowhere near orbital velocity untill you get to 22,300 miles from the surface.
At ISS level, 250 miles up, you are 4,250 miles (!) from the center of the earth, and so to complete a circuit of (2 x pi x 4250) = 26,690 miles in 24 hours you are going a direction tangent to the earth’s surface at 1,112mph or 1,631 feet per second or about 652m/s.
I did the calculation for a direct drop, no horizontal component. That 1,112mph horizontal velocity of the space elevator at 400km/250mi would have to be added to the direct vertical acceleration. Additional factors - yes, there would be some air resistance, but it is irrelevant generally until you get close to the 60mi/100km which is generally said to be the edge of space (according to Bezos, who wants you to pay for the privilege of getting up that high).
Note that Starlink satellites are launched to roughly 260km and use internal thrusters to slowly climb to operational orbits. One recent launch, a number of satellites were lost - atmospheric friction slowed them out of orbit over a week or two - because recent solar flares had heated and expanded the upper atmosphere so it was denser than usual at that altitude and the thrusters weren’t powerful enough. To get an idea how much resistance - Something going 17,000mph needs a week or two to slow enough to hit thick atmosphere. I’m guessing for our drop calculations, upper air friction is negligible.
Also, my calculation neglected the fall-off that gravity, hence acceleration, is slightly less with altitude. The actual velocity will be slightly lower.
The 1000 MPH lateral velocity isn’t relevant, just the difference between the lateral velocity at drop height and at the Earth’s surface, which isn’t much.
Regarding the Blue Origin spacecraft, which only goes up to the Karman Line (the arbitrary boundary between atmosphere and space, 100km up), not to ISS height 4x higher. Nevertheless, it does have a heat shield. Sort of.
“The white and blue rocket is partially coated in “Comet,” a thermal protection system developed by Blue Origin. It was applied to New Glenn’s fins, forward module, strakes, tank tunnel and the aft section, including the rocket’s legs.”
The answer is already above.
I really would have thought the OP was perfectly unambiguous. (Grumble.) A space elevator was specified. They only work in one way. They are geosynchronous from the ground all the way up. It is baked into the definition of what they are. The niceties of the need for the top to extend way past the height of geosynchronous orbit doesn’t matter. The entire structure is geosynchronous. Geosynchonous speed really only makes sense as defining an angular speed. 0.0000727 radians/sec.
So you are climbing the elevator, and just as your height crosses the height of the orbit of the ISS, you drop something. What happens to it? It accelerates to a seriously large speed before hitting a thickening atmosphere, where the radiative heating from the entry shock will heat up the outside, it gets seriously buffeted, most likely resulting in the fragmentation and general destruction of the object. Unless it has enough intrinsic strength or heat resistance to stay together, in which case it will hit the ground quite, but not ridiculously, fast.
One notes that meteorites that reach the ground are usually a solid rock, and counter-intuitively, are, whilst hot on the outside, are actually still frosty cold on the inside. Our dropped object probably doesn’t have enough potential energy when dropped to vaporise it, so if it survives the onslaught of heating and buffeting in the supersonic part of its journey, it will likely make it to the ground largely intact, albeit a little diminished. Drop a personal object, like say a camera or phone, nothing is going to make it. Drop a rock, some of it will. Cube of tungsten? Please, just don’t.
Or a tungsten cylinder, aka Rod from God.
They aren’t even hot on the outside. During ablation the molten rock/metal flows off the surface as soon as it melts. Heat doesn’t have time to get more than millimeters deep at a time. Then it slows down to too slow to melt at around 30 to 60 (ish) miles up. This is known as the “dark flight” time. Falling at a terminal velocity of around 200 to 400 mph. Call it a 30 mile fall at 300 miles per hour and you get 6 minutes of falling through strong “wind” to remove whatever surface heat is there.
Wait, so now we’re saying that something would heat up when dropped from an elevator at LEO?
Yes. And we have been for most of this thread.
I see this:
Shrugging it off isn’t burning up. I see a few people mentioning Baumgartner, for example:
He didn’t burn up.
Escape velocity from the earth’s surface is about 11 km/s, which means that something falling from infinitely far away towards earth in a straight line, without orbiting, would crash on the surface at 11 km/s if we ignore air resistance.
Meteorites falling towards earth (and satellites/space stations/lunar modules) already have a significant fraction of those 11 km/s when they are at ISS height. If we start from 0 km/h speed at 400 km height, which could be done with a vertical parabolic flight that ends at that height: how fast would an object get
a) with air resistance as it is, with the layered atmosphere we have, denser at the bottom, less dense at the top, and
b) without air resistance
Say the object is a balloon filled with water. Or a can of beer. Would it burst before reaching the ground?
My WAG as to the differences - he didn’t have as far to fall through less thick atmosphere, so less chance to accelerate before hitting the thicker portion that slowed him down to his terminal velocity. And even without parachute not a very aerodynamic object.
Absolutely. As someone quoted above, Baumgartner fell from 39 km. That’s one tenth the altitude the question here calls for. And that 360 km extra fall is going to be with virtually no air resistance to slow the object down. There’s also going to be some additional horizontal velocity from something falling from a space elevator at 400 km that Baumgartner wouldn’t have experienced. That’ll add to the velocity the object will have when it does meet air resistance.