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This seems like a banal question, but the more I think about it, the less I understand the mechanics of it. Does the progressively denser atmosphere pick up the spacecraft and carries it along the direction of Earth rotation, eventually bringing the spacecraft in synchronization with Earth rotation when it reaches the surface? (assuming the spacecraft is eventually descending vertically, effectively making the relative lateral speed between Earth and spacecraft zero)

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    $\begingroup$ Yes, atmospheric drag is the entirety of the effect. Earth’s atmosphere is mostly still relative to the surface of the Earth, so the relative velocity of something entering the atmosphere at orbital speed is quite a lot. At lower speeds, though, drag significantly falls off. That’s when parachutes or propulsive landing systems are needed. $\endgroup$ – CourageousPotato Sep 26 at 5:18
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Background

Earth's $2 \pi \times 6378 = 40074$ km circumference rotates every 23h 56m 04s = 86164 seconds, so the equator rotates at about 0.465 km/s. At a very low 200 km altitude, orbital velocity is $\sqrt{GM/(6378+200)} = 7.784$ km/s (using the vis-viva equation where GM is Earth's standard gravitational parameter of about 398600 km^/s^2.)

Launching prograde uses the Earth's rotation to an advantage, it takes significantly less fuel to reach orbit than it would launching retrograde, though it's done sometimes for reasons.

The same idea works for atmospheric reentry. The Earth's atmosphere is fairly well "stuck" to the planet due to friction, so it rotates near the equator at roughly the same speed as Earth rotates. When coming in at about 7.8 km/s you would prefer to be reentering prograde so that relative to the atmosphere you are only moving 7.3 km/s rather than 8.3 km/s if you reentered in a retrograde direction.

While that sounds like a +/- 6% effect, it's much more important. Per answers to What aspects of reentry heating 'scale as the 8th power'? the heat generated during reentry is a very strong function of the relative velocity. Maybe not exactly to the 8th power, but is's still huge.

Question

What forces a spacecraft, returning from Earth orbit, to synchronize with Earth spin?

Answer

@CourageousPotato's answer as comment is right:

Yes, atmospheric drag is the entirety of the effect. Earth’s atmosphere is mostly still relative to the surface of the Earth, so the relative velocity of something entering the atmosphere at orbital speed is quite a lot. At lower speeds, though, drag significantly falls off. That’s when parachutes or propulsive landing systems are needed.

When reentering and preparing to land you are in intimate contact with the Earth's atmosphere and it's that that sets your final sideways velocity. You slow down and match the motion of the atmosphere (in the transverse or sideways direction) so if there is a strong cross-wind you'll hit the surface with some sideways velocity.

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    $\begingroup$ Thank you! If I were to visualize it, would it be akin to stirring a pot of soup, and then dropping a noodle into it? Noodle plops in and starts rotating along with other noodles in the soup? Except, the density transition from air to liquid is a lot more abrupt than the density transition for vacuum to air - correct? $\endgroup$ – Mitch99 Sep 26 at 18:16
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    $\begingroup$ @Mitch99 I think that's a good way to think about it, with the following caveat. Since coming from orbit the spacecraft's speed is 16x larger than the Earth rotates, it's like throwing your noodles into the soup at a fairly fast clip, much faster than the soup rotates. The outside of the soup is the equator, and the center (which is not spinning with appreciable speed) is like the poles, roughly speaking. Enjoy your space soup! ;-) $\endgroup$ – uhoh Sep 26 at 23:41

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