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Given that everything on earth is rotating at the same speed as Earth, it is simple for an item leaving the earth surface to return to the earth surface.

However, once an object has left the earth's atmosphere and entered space, how can it return to earth and rejoin the rotational speed of the planet?

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  • $\begingroup$ What do you mean it is simple? It costs $100,000/kg, that sounds hard to me! Seriously I can't understand your first sentence. Do you mean if you throw a ball in the air, it will return to the ground nearby because it already has the Earth's rotational speed when it leaves your hand? $\endgroup$ – uhoh Feb 24 '17 at 11:15
  • $\begingroup$ No. Not even remotely. I am asking how a spacecraft can rejoin the earth as it rotates and at what point in the return journey it picks up the rotational speed of the body it is landing on. $\endgroup$ – Venture2099 Feb 24 '17 at 11:17
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    $\begingroup$ I'm just asking about your first sentence, not the second sentence. The reason is that I'm trying to get an idea what it is that you don't understand yet that you would like to, so starting with your assumptions is helpful (for me and other people who will stop by). Rockets have literally "tons of thrust" and can attain a wide variety of velocities by using it on return. Air friction does a lot of the work too. Is that roughly the kind of information you'd like in an answer? $\endgroup$ – uhoh Feb 24 '17 at 11:22
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Your question here is more about the frame of reference than relative speeds. To make things easier to understand, let us consider those points:

  • you have a launch site exactly on the equator
  • earth equatorial rotation velocity is 1,674.4 km/h, approx 465m/s
  • you want to join the ISS which is in our case also in a equatorial orbit and with a 7.67 km/s speed (7670m/s)
  • there is no wind, never.

As earth is rotating from west to east, you "only" need to add 7205m/s to your vessel, because of ground speed and the orbit you want is west-east oriented (fyi, it would require 2x465 = 930m/s of additional delta-v to have an orbit east-west oriented). So:

  • when on launchpad, you are already at a speed of 465m/s in the frame of reference of space.
  • when in orbit, you go at 7670m/s in the frame of reference of space, but "only" 7205m/s in the frame of reference of ground.

Returning to earth only means slowing in both frame of reference, and to land safely you want a ground speed of 0m/s. Here is where atmosphere helps a lot. As we previously stated there is no wind, we can consider the atmosphere as a outer layer of earth rotating the same speed as the ground, and so:

  • going through this layer will make you brake and heat, and you may need some protection.
  • in order to low your ground speed to 0, you then use parachute (for example on an Apollo capsule)
  • you then only wait to touch down, and voila.
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  • $\begingroup$ Perfect. Thanks :-) The voila summed it up amazingly. $\endgroup$ – Venture2099 Feb 24 '17 at 13:58
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Oh, at reentry.

There's no magical change of frame of reference or other weird stuff. The Apollo Command Module returns from its trip to the Moon. It moves in trajectory that makes it move at 11 km/s relative to Earth surface, while rotation of Earth near Equator makes the surface move at about 465.1 m/s, at poles - 0, in between - in between.

So how does the craft slow down from 11,000m/s to some 400m/s when it lands somewhere near Hawaii? Air drag. It plunges into the atmosphere at full speed, at a rather shallow angle, air gets compressed under the heatshield, heating immensely and creating huge plasma trails, and reducing the craft's horizontal speed, then, when safe, parachute opens, and the craft begins falling down, its "horizontal speed" bound (through the parachute) to local wind speed which is tightly bound to Earth's rotation speed.

If you look at suborbital flights, their landing sites are often quite close to launch sites - they don't magically gain or lose speed in horizontal direction when leaving the atmosphere and entering space - if you want to enter the orbit, instead of falling directly where you launched from, like in that ball example, you have to burn lots and lots of fuel in horizontal direction, to get the orbital speed, much higher than Earth rotation speed.

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  • $\begingroup$ It's possible the OP is including stuff not orbiting fast, but coming in radially - on a straight line, with no rotation about the Earth's center at all. As a thought experiment, imagine hovering over the equator in a heliocentric orbit around the sun ahead or behind the Earth, and dropping a paper airplane. Then the 0.46 km/s rotational speed of the Earth and it's atmosphere would spin-up the paper airplane as it entered. It would literally start dragging it along w/ drag. I think that's the kind of thing meant by "how can it return to earth and rejoin the rotational speed of the planet" $\endgroup$ – uhoh Feb 24 '17 at 14:23
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    $\begingroup$ @uhoh: Nevertheless, the answer remains the same: air drag. $\endgroup$ – SF. Feb 24 '17 at 14:51
  • $\begingroup$ Yes indeed, didn't mean to imply otherwise. But maybe it's possible to supplement "So how does the craft slow down..." with a "(or in some cases speed up)" or something similar to help address one potential interpretation of the question, that's all. $\endgroup$ – uhoh Feb 24 '17 at 14:55

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