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This question already has an answer here:

If I jump and stay in the air for a long enough time, will my position be changed since the Earth is rotating?

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marked as duplicate by Kyle Kanos, Ilmari Karonen, rob Jun 11 '18 at 16:36

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  • $\begingroup$ If you jump inside a moving train does your position change due to the movement of the train? $\endgroup$ – lurscher Jun 11 '18 at 15:44
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You will land ever so slightly behind where you started from, as the earth rotates under you. Yes, in general, at all times you are moving at the same tangential velocity (with respect to the center of the earth) as the ground and the atmosphere. However, when you jump, you attain a higher elevation and therefore a lower angular velocity compared to the ground you jumped off.

Say you are at the equator where everything is moving the fastest with respect to the center of the planet. If you jumped 1m, your angular velocity drops to 99.999984% of the rest of the earth. So while the surface of the earth zooms past at 463m/s, you lag behind by 73nm for each second you spend at 1m altitude. That's a thousandth of the width of a human hair. Now to land far back to even a measurable degree, we're talking either crazy altitude jumps, or long air time. That's some crazy air time! Or should I say..hair time...

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    $\begingroup$ Big jumps: you may even want to use a rocket for even bigger jumps, and once out of that pesky atmosphere, you haven't got the wind blowing you around and having far more of an influence on you than the movement of the Earth $\endgroup$ – Baldrickk Jun 11 '18 at 13:39
  • $\begingroup$ Why would the angular velocity drop? Maybe there is a simple explanation (some kind of conservation law or something), but I'd like your answer to explain this, because at the moment, it is not clear to me that the angular velocity would drop: $\endgroup$ – Quantumwhisp Jun 11 '18 at 13:48
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    $\begingroup$ @Quantumwhisp Angular momentum is conserved, so if you move further from the centre of the Earth your angular velocity must fall. $\endgroup$ – tfb Jun 11 '18 at 13:51
  • $\begingroup$ @tfb : Thanks, that makes it clearer. The previous explanation didn't consider the coriolis force, (which gave me the suspicions that it wouldn't work). $\endgroup$ – Quantumwhisp Jun 11 '18 at 13:54
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    $\begingroup$ Your simple calculation is also forgetting that the air around you is also rotating with the earth's surface, and will make the displacement even less! $\endgroup$ – rubenvb Jun 11 '18 at 14:27
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When you jump, you're already travelling in roughly the same direction as the Earth so obviously the Earth is not going to move along a few metres. This is the same as if you jump in a train - you don't land a metre behind where you took off.

Having said that, there is a slight effect due to the Coriolis Force. This is an apparent force that arises due to motion in a rotating reference frame. The maths is quite complex (you'd usually study this two or three years into a university degree). The strength of the effect depends on the latitude, the direction of the jump and how fast you go.

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    $\begingroup$ The exact math of the coriolis force may be complex but the underlying issue is trivial: On a rotating platform the inner parts move slower (in terms of velocity), so that objects moving inwards or outwards on that platform experience a difference in ground velocity; the ground moves slower (inwards) or faster (outwards) and hence they need to decelerate or accelerate if they don't want to start moving tangentially relative to the rotating ground. The phenomenon is subjectively puzzling because all locations on the platform rotate at the same angular speed, appearing (wrongly) equally fast. $\endgroup$ – Peter A. Schneider Jun 11 '18 at 14:46
  • $\begingroup$ Also note that any jump (except at the poles; but especially at the equator, jumping vertically) will experience the Coriolis effect because the jumper will increase their distance to the rotation axis. $\endgroup$ – Peter A. Schneider Jun 11 '18 at 14:53
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While on ground, you are moving with the Earth and hence have some horizontal momentum. If you jump upwards, you don't lose this horizontal momentum just because you have lost contact with the ground. So, you will continue to move horizontally approximately in step with the Earth. Also, the atmosphere is moving (approximately) in step with the Earth and it will help you along.

Consider that you are on a train. The train is moving fast at a constant speed on a smooth, straight track. There is a table in front of you and a coffee cup on it. If you push the cup off the table, will it fall to the fall near to you or shoot to the back of the carriage? You are facing forwards and there is a passenger opposite you. If he drops his knife, does it fall on the table below or shoot backwards and into you?

With sufficiently accurate calculation or measurement, you might find a slight variation in your position but much less than you would expect from the rotational speed of the Earth.

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No, there won't be any significant change in your position. To make a significant shift in your position, you might have to jump as high as some orbit of a satellite revolving around earth.

Because the speed of rotation of earth and your horizontal velocity do not have comparable magnitude, the displacement produced will be negligible.

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  • $\begingroup$ Still there'll be some displacement? $\endgroup$ – Theoretical Jun 11 '18 at 12:01
  • $\begingroup$ @AsifIqubal Yes, of course. See gregsan's answer. $\endgroup$ – Peter A. Schneider Jun 11 '18 at 14:43
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Simple, the highest you jump (related to the time counted), the more mm(s), cm(s) or meter(s) you've got. In other words, the final position is based on how high you jump. Try to jump up to the sky :)

Perhaps, that could be described as follows:
Df = Differential Position in mm(s);
Hi = Height (jumping) in meter(s);
Ti = Constant of Time = 24 hours

Note: (24 hours x 60 minutes = 1440 minutes)

If you jump 1 meter (1000 mms), then;

Df = Hi * Ti/1440
Df = 0.0166666666666667 mms

Note: you also must think the interference of wind blows :D more complex. Or, for instance, the interference of atmosphere is ignored (or null). but, That's only illustration! :)

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    $\begingroup$ That the math here is incorrect can be seen just from the failure of the units to work out. $\endgroup$ – dmckee Jun 11 '18 at 16:33

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