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Ignoring statistical variation in wind, will a helicopter that only applies force perpendicular to a perfectly sphere earth remain in the exact same spot relative to the earth?

I am arguing with globe earthers who claim the helicopter will remain in the exact same spot because somehow the helicopter maintains the same angular velocity and thus increase its tangential speed to compensate for the longer distance it has to travel, while I claim the helicopter would slowly deflect away from it's original position due to

  1. The helicopter loses angular momentum and air cannot perfectly conserve its angular momentum.
  2. There is a net force pushing the helicopter towards the equator if it isn't on the equator to begin with.

This is not a duplicate because I am asking for more in-depth info on the forces acting on the helicopter. Those links do not address the question at all.

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marked as duplicate by HDE 226868, Qmechanic May 6 at 16:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/1193/2451 , physics.stackexchange.com/q/58154/2451 and links therein. $\endgroup$ – Qmechanic May 6 at 13:42
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    $\begingroup$ Possible duplicate of Can a hovering helicopter travel half the globe in 12 hours? $\endgroup$ – HDE 226868 May 6 at 14:23
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    $\begingroup$ Note: This question will be difficult because you are using a helicopter. The dynamics of helicopters are tremendously complicated and simply hand-waving away a lot of those dynamics with "only applies force perpendicular to the earth" is tricky. I believe that JMac's answer correctly captures the spirit of what you were asking, without getting bogged down in the helicopter details, but just know that the choice of vehicle might cause confusion when you use it to explain the motion of objects on our Earth. $\endgroup$ – Cort Ammon May 6 at 15:06
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I think you are technically correct, that it will deflect from it's original position, but it seems like it is not for the reasons you are describing.

The Coriolis effect for example will cause some drift for the helicopter that would not be prevented by normal force. This means that ultimately, there would be some drift. I don't think that is the effect you are thinking of, or the other person is discussing though.

Let's assume we are close to the surface of the Earth, and as you said, we are ignoring the wind, and only considering the air that is moving along with the surface of the Earth. As a consequence of viscosity and the no-slip condition, the air touching the surface of the Earth is being dragged by the Earth, and will have no slippage between the air and the Earth. This no slippage forms what is known as a boundary layer, and on the rotating curved surface of Earth, this causes a wind gradient where in the higher you are, the greater the angular velocity; which is necessary to prevent the layers below from "slipping" on one another. This is equivalent to saying the air above a point on the surface approximately will stay above that point on the surface, ignoring the inertial effects such as the Coriolis effect.

If you are trying to argue that the angular speed of the wind due to height will be less, and this will cause the helicopter to fall behind, that would be incorrect. If you are saying the Coriolis effect is causing the helicopter to change relative position with the surface, that would be correct, and is something that the other side does not seem to be considering.

To address the edit a bit more directly: The Coriolis effect would be pushing the helicopter towards the equator, although not with much force.

As far as angular momentum goes, you asked about a helicopter that is hovering above the ground. If the helicopter were lifting off from the ground, it would likely require a slight adjustment to stay above the same location; but this would only be as you are changing height. The airflow would also assist with this increase in angular momentum, because the angular velocity of the air is increasing with height as well (see the boundary condition and no-slip condition as explained above). Once you're floating in the air hovering above the surface, maintaining that does not require any more force, because the air around you is also travelling across the surface at the same speed the Earth is rotating.

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  • $\begingroup$ So if the helicopter is hovering within the boundary layer, it will maintain the same angular velocity as it did when it was on the ground after it reaches its height? And if it hovers above that then it will actually have a faster angular velocity? $\endgroup$ – Flat Earther May 6 at 19:14
  • $\begingroup$ @FlatEarther No, I'm assuming it's always within the boundary layer, where the air is essentially "stuck" to the surface and moves with it without slipping. To move without slipping, you can imagine a column of air stretching upwards; to prevent slipping of air layers, this means that this column always has to move with the surface at the same pace. The higher up on the column you are, the greater the angular velocity, but your apparent velocity to the surface, and all the air sticking to the surface, is 0. You travel further due to the bigger arc, but the angular velocity compensates. $\endgroup$ – JMac May 6 at 19:18
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Both of those answers are correct. If the helicopter were simply hovering at the equator it would maintain its position over the planet. If you tried to ascend or descend, applying thrust only perpendicular to the Earth, it would no longer maintain that same spot over the Earth. The atmosphere is basically moving along with the planet at the equator. Look at a geostationary satellite, it has a much higher tangential speed than the surface of the Earth yet it maintains the same angular speed as the Earth, the same height above the Earth, and position over a spot on the Earth.

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  • $\begingroup$ Also, if the helicopters were hovering directly over the North and South Pole they would also maintain their spot directly over the ground with simply a downward thrust. And there wouldn't be all these questions of atmospheric movement. $\endgroup$ – James Montagne May 7 at 14:50

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