Is there a way, or an aircraft that can actually perform this? The F35B fighter jet has hovering capabilities. The only "supposed" images of this 1000 mph spin is from NASA, but the clouds do not move, so it is, in most logic; a computer-animated clip.

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    $\begingroup$ If it is hovering, it does not see the earth rotate, it would defy the definition of hovering. $\endgroup$ – Federico Jul 11 '15 at 10:31
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    $\begingroup$ The definition of hovering is not height-dependent. $\endgroup$ – Federico Jul 11 '15 at 10:36
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    $\begingroup$ A spin/rotational motion is measured in angle unit per time unit (e.g. degree per second), not in length unit per time unit. See this video. $\endgroup$ – mins Jul 11 '15 at 12:27
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    $\begingroup$ It also seems to me (although this one might get me flagged as rude) that the OP is starting to show a habit of asking 'interesting' questions which have problems of factuality (i.e. military over antarctica, planes hovering in space) $\endgroup$ – CGCampbell Jul 11 '15 at 14:22
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    $\begingroup$ You should take this question to Physics SE for a full explanation (which will expand on why it is impossible to define fixed point and stationary and why, in order to see the rotation of the Earth, you must also be moving) and to Sceptics SE for the "supposed rotation" (I don't even know what you're saying here - do you doubt that the Earth rotates?). $\endgroup$ – Simon Jul 11 '15 at 15:15

At an absolute fixed position you would see the earth move at 1 inch altitude. But you would need to traveling 1000 mph relative to the surface of the earth to hold a fixed position relative to the North star.

At lower altitude the atmosphere moves with the earth. Otherwise we would experience 1000 mph wind on the surface.

There is a spirited debate on fixed position. And this is physics. To me the intent from the OP was clear - take rotation out of the equation.

  • At the equator hold a position such that you can draw a straight line between you the rotational axis of the the earth and the center of the Sun. To achieve that you would need to travel apx 1000 mph relative to the surface of the earth and should be able to see that at any altitude.
  • Or directly above the axis (north or south) point directly at the Sun. How high would you need to be see the earth spin below you? I don't think you would see it spin as it relatively low angular velocity (one rotation every 24 hours).

I agree this is not an aviation question. I did not think it would into this level of physics. I am good with it getting migrated. If an moderator wants to delete my answer I am good with that.

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    $\begingroup$ [physicist] an "absolute fixed position" does not exist [/physicist] $\endgroup$ – Federico Jul 11 '15 at 11:26
  • $\begingroup$ @Federico And that is why I added hold a fixed position relative to the North star. $\endgroup$ – paparazzo Jul 11 '15 at 11:28
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    $\begingroup$ @Federico: While it is known the linear speed at the equator is about 1000 mph (relative to the center), the Earth is not fixed relative to Polaris, hence the linear speed of a point on the equator relative to Polaris is unlikely to be the same. For a start the Earth moves around the Sun (66,000 mph relative to the Sun). Also I doubt we know what is our velocity relative to Polaris, as we barely know the distance of Polaris, in the range of 400 l.y. Or am I wrong? $\endgroup$ – mins Jul 11 '15 at 14:28
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    $\begingroup$ @Blam The intent of a "fixed position" is only ever clear when the reference frame is also specified. If you say "fixed position above the Earths surface", then you are only fixed in respect of that point of the Earth. To everyone else, you are not in a fixed position. It is just not possible to say a fixed position is clear. The North Star is also moving, moving really fast, so if you are fixed relative to the North Star, then you are also moving really fast relative to an observer in a different reference frame. $\endgroup$ – Simon Jul 11 '15 at 15:22
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    $\begingroup$ OK, this has nothing to do with aviation. Please take it to physics SE. $\endgroup$ – Simon Jul 11 '15 at 16:01

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