I have a question about which forces cause alignment of objects with the rotation of the Earth and how:

It is known to me from public sources that objects up to a height of about 36,000 meters above the surface of the Earth move (if we ignore their own speed) completely in line with the surface of the Earth, because they are attracted by the Earth's core. When, for example, an airplane flies at an altitude of 10,000 m, it is not decisive for the earth's rotation whether it flies east or west, north or south.

Furthermore, the speed of the Earth's rotation at the equator is approximately 1670 km/h, and the core of the earth - to which objects are said to be attracted, according to scientists, rotates differently - there are even theories that the other way around - than the earth's shell.

That's what I've studied.

And my question is - what forces, and how, cause the fact that an object at a height of up to 36,000 m is still attracted to the same place on the Earth's surface despite all the rotations, i.e. that it exactly copies the rotation of the Earth's surface when it is 36,000 m above it (or 10,000 m for a normal aircraft).

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    $\begingroup$ do you know what the conservation of angular momentum is? It has nothing to do with the atmosphere. If something leaves from the earths surface it has angular momentum, changing that requires an input of energy. $\endgroup$
    – John
    Commented Aug 26, 2023 at 4:26
  • $\begingroup$ @f.thorpe not sure whether your changes match the intent of the question? Unfortunately the user never returned... may need to close the question as unclear? To me, they seem to be asking why the air 36km above the Earth matches with the location below it (of course it doesn't exactly... wind... but a rough idea). unfortunately they never gave the public source of such a specific number, which makes the question excessively precise. But removing the word atmosphere from the title seems it would lead others to answering with consideration that the topic is aloft-focused. $\endgroup$ Commented Sep 28, 2023 at 7:58
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    $\begingroup$ (And indeed, the 36 [k]km is likely a number they got from geostationary orbit. But they also jump to the topic of planes, showing it's not just about that) $\endgroup$ Commented Sep 28, 2023 at 8:00
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    $\begingroup$ The question was misinformed so I tried to open it up to an actual answer other than "none". Feel free to edit further, I think the intent of the OP is less important than a good question, especially when they don't follow-up. $\endgroup$
    – f.thorpe
    Commented Sep 28, 2023 at 14:01

1 Answer 1


I think what you are referring to is a geostationary orbit, judging from your statement:

is still attracted to the same place on the Earth's surface despite all the rotations

However, I think your distance is out by a factor of 1000: 36,000 km not 36,000 m.

For an object to orbit the Earth and still be above a particular location on the Earth's surface, the object needs to be at least 36,000 above the Earth.

An object in such an orbit has an orbital period equal to Earth's rotational period, one sidereal day, and so to ground observers it appears motionless, in a fixed position in the sky. The concept of a geostationary orbit was popularised by the science fiction writer Arthur C. Clarke in the 1940s as a way to revolutionise telecommunications

Objects in such orbits are usually weather or communications satellites and sometimes spy satellites.


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