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First imagine a pilot takes off from a North South runway situated on a specific northern latitude, the circumference of which is 888 miles, where the angular spin velocity of the earth (and the plane) would naturally be calculated at approximately 37 miles per hour in an Easterly direction. Then imagine that he attempts to land on a North South runway at the equator where the angular Easterly spin velocity is 1037 mph.

Explain how this 1000 mph difference of angular spin velocity is gained with enough exactitude that his plane will not flip over upon touchdown?

For extra points explain how such a plane under these conditions could lose the greater angular spin velocity if the trip were reversed and started out at the equator and landed at the northern runway?

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    $\begingroup$ The pilot controls the speed their plane travels at. If there was any build-up of the effect you speak of, they could just increase the drag of the plane, and slow down. $\endgroup$ – CDCM Sep 3 '17 at 3:16
  • $\begingroup$ The speed in which the plane travels in a southerly (or northerly direction} is irrelevant as pertains to to the gaining {or losing} of angular spin velocity, which should remain the same as it was upon takeoff, (37mph) absent some additional angular force to modify it with the necessary exactitude to match the spin velocity of 1037 mph. $\endgroup$ – Lee Sep 3 '17 at 3:49
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/16390/2451 , physics.stackexchange.com/q/173074/2451 and links therein. $\endgroup$ – Qmechanic Sep 3 '17 at 8:08
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    $\begingroup$ This question might be better suited to Aviation SE where related questions crop up like this one.. $\endgroup$ – StephenG Sep 3 '17 at 8:08
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If the pilot were flying in a vacuum, then it would take an enormous amount of energy to catch up to a landing site near the equator. One way of thinking about this is that the vehicle has to overcome the Coriolis force that comes from moving across a sphere in a direction parallel to the sphere's axis of rotation (i.e., North-South). Artillery and battleships guns have to account for this deflection in the trajectory of their shells in order to hit targets to the North or South.

This extra velocity near the equator is why rockets that put spacecraft into orbit are located as close to the equator as possible. In this list of American rocket launch sites, 10 out of 12 are in the southern half of the continental Unites States. The most famous launch sites, Cape Canaveral and the Kennedy Space Center, are in Florida since it is the state that is closest to the equator. This location takes advantage of the rotation speed of the Earth to give rockets an extra boost to get to orbit. This is also why nearly all satellites orbit west-to-east, and nearly none orbit east-to-west.

Luckily, the problem for airplanes is much reduced because they travel in Earth's atmosphere. Due to air being a fluid with non-zero viscosity, it gets dragged by the Earth's surface and ends up rotating with the Earth in the same direction as its rotation. The air's circulation speed does not exactly match Earth's rotational speed and lags behind it, as evidenced by the existence of trade winds, which blow east-to-west near the equator. Since aircraft are subject to the movement of air around them, their movement over the ground is a combination of their movement through air and the air's movement over the ground. For example, if an airplane flies with an airspeed of 100 miles per hour flies towards the East while the local wind is 100 miles per hour towards the west, the airplane with have zero ground velocity--essentially hovering (note the very steep landing angles starting at 0:42). If the wind is constant, airplanes do not feel it as a crosswind that pushes them. It only affects their velocity over the ground. Aside from the force from the engine, an airplane will match the local air movement due to its drag. This is similar to how a swimmer in a river does not feel the river pushing them downstream, they only notice that they have a downstream velocity with respect to land.

So, as the airplane flies South, the air at each latitude that is rotating with the Earth drags the plane with steadily increasing velocity to the East so that, by the time it reaches the equator, it already has the 1000-mph sideways velocity needed to land. This sideways speed is measured from an inertial reference frame in which the Earth rotates, not with respect to the ground. The airplane will have to expend fuel to counter the trade winds, but these winds are nowhere near 1000 mph. At the destination runway, the airplane, the surrounding air, and the ground are all moving with the same speed in the same direction, so the airplane can easily land with no threat of flipping over due to sideways velocity. In the reverse direction, the slower-moving air to the North puts a drag on the initial sideways velocity of the airplane so that it matches the northern runway's slower rotational speed by the time it lands.

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Presumably this is because of air resistance. As the plane travels south, the air would also be moving at roughly same speed as the ground so the plane will be forced into moving the same speed as the ground (in the east-west direction). Otherwise the plane would be flying against 1037mph crosswinds.

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  • $\begingroup$ We know that the air or atmosphere moves in many and all directions as is shown by numerous meteorological examples. Unless it could be shown that the atmosphere was being uniformly and constantly dragged along with the spin of the earth, then it would appear we lack the necessary facts to make such a presumption as you mention. It would require us to depend solely upon the absence of uniform atmospheric forces in the hope that as the plane travels it will either increase or decrease angular velocity until at the moment of touchdown it will match the angular spin of the earth at that point. $\endgroup$ – Lee Sep 3 '17 at 5:47
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    $\begingroup$ The atmosphere has an overall motion that matches that of the earth. The motion of winds is superimposed on that. The difference in angular momentum between equator and poles is the cause of the Coriolis force. Unless you want to claim that there is a constant 1600 km/h wind at the equator. Or are you trying to claim the earth is flat and stationary? $\endgroup$ – hdhondt Sep 3 '17 at 6:06
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    $\begingroup$ @Lee I don't understand your comment. There's no place on Earth were there is a 1000-mph wind at the surface due to Coriolis effects (though there are such winds at higher altitudes). What other "necessary facts" do you have in mind? $\endgroup$ – rob Sep 3 '17 at 6:13
  • $\begingroup$ @ hdhondt, I am not claiming that there is a constant 1600 km/h wind at the equator. Nor am I trying to claim the earth is flat and stationary. $\endgroup$ – Lee Sep 3 '17 at 6:28
  • $\begingroup$ @ Rob and hdhondt, what I am asking is please explain how this 1000 mph difference of angular spin velocity is either gained or lost with enough exactitude that his plane will not flip over upon touchdown? If atmospheric drag is in fact responsible for the gain or loss then it would appear that if one is flying either North or South then they would at all times experience continuous crosswinds in an amount sufficient to make the necessary gains or losses. I have never heard of the existence of such a phenomenon when flying North or South. So does a pilot consider the angular spin? $\endgroup$ – Lee Sep 3 '17 at 6:54
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Explain how this 1000 mph difference of angular spin velocity is gained with enough exactitude that his plane will not flip over upon touchdown?

This question seems to presume that planes commonly move around with little change in their sideways velocity, and that controlling such is difficult. The non-physics but simple answer is that the pilot uses the airplane controls to adjust its velocity so that a landing is permitted. Whether the earth's rotation or a thunderstorm is responsible for any unwanted velocity is irrelevant.

The craft is piloted to a useful point above the earth's surface pointed in the correct direction to approach the runway via constant manipulation of the controls and constant references to the desired flight path. Rather than precompute the effect of all possible items that could affect the path (wind, fuel and craft balance, Coriolis/inertial effects, etc.), any deviations that appear can be corrected without needed to identify the specific source of the deviation.

If atmospheric drag is in fact responsible for the gain or loss then it would appear that if one is flying either North or South then they would at all times experience continuous crosswinds in an amount sufficient to make the necessary gains or losses. I have never heard of the existence of such a phenomenon when flying North or South.

A crosswind is not required. In fact, an atmosphere that is exactly fixed over the surface below it would work as well. We don't have that atmosphere (instead wind with respect to the surface is constantly present), but let's see how it would work and then just imagine wind as a correction to this model.

As the airplane in the scenario flies south, it has relatively too little eastward speed to match the surface of the new,more southerly latitude. Therefore relative to the earth (and the fixed atmosphere in this case), it would appear to be skidding to the west as it flew south. This sideways skid creates drag on the fuselage, which accelerates the plane to the east. If the drag is large enough, it will create some complex motions that the pilot would notice. But if it is small enough, it will be lost in the noise of all the other turbulent forces.

At no time would anyone describe the air as having a crosswind (wind moving sideways to the direction of flight). But even so the plane would constantly be accelerated to the east because the various parts of the atmosphere have different rotational speeds.

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Flat-earthers ask this all the time, and they never like the answer, which is simple.

Suppose the aircraft is flying due south from near the north pole (as in your example) at a speed of 500 miles per hour (or whatever units you like). So after one hour of flight the surface of the earth is traveling to the east faster than it was at the beginning of the hour.

How much faster? It's traveling in a circle 500 miles bigger in radius, which means it is 500 x 2pi or 3,141.6 miles further around. Since it has to cover that in 24 hours, that means it is traveling eastward 130.9 miles per hour faster than the departure point was.
(And so is the air, in case you were worried about a big wind.)

So the plane has to accelerate eastward at 130.9 miles per hour in one hour to keep up with the ground.
Let's convert that to feet per second per second. Multiply by 5280 and divide by 3600 twice.

I get .0533 feet per second per second. How does that compare to gravity, which is about 32 feet per second per second? Well .0533/32 = .00167, or about two tenths of one percent of gravity. For the plane to do this, it has to continually bank to the left by that number of radians. Multiply by 57.3, and that's about one tenth of one degree of bank, in order to get to their desired landmarks.

Can you see how they would never notice?

And that's near the pole. The effect gets even smaller, down to nothing as they get closer to the equator, because they are no longer getting as far away from the earth's axis of rotation, in every hour.


And for extra credit, just reverse it :)

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  • $\begingroup$ I'm not a FlatEarther but can understand why they would not like this answer. The math as well as the analysis is incorrect. $\endgroup$ – Lee Sep 7 '17 at 17:12
  • $\begingroup$ The circumference of the earth from the North Pole to the equator is approx 6225.25 miles. Distance along the spherical plane of the circumference does not increase the radius by the same amount. If that were the case then by the time you got to the equator flying at 500 mph the radius of the earth would be 6225.25 miles. As a Globe Earther I am sure you know that cannot be true. $\endgroup$ – Lee Sep 7 '17 at 17:25
  • $\begingroup$ @Lee: When they are near the pole it is correct, because they are traveling almost directly away from the spin axis. That's why I said "And that's near the pole. The effect gets even smaller, down to nothing as they get closer to the equator". I'm trying to simplify it for the high-school-level reader. $\endgroup$ – Mike Dunlavey Sep 7 '17 at 20:35
  • $\begingroup$ Mike wrote"So after one hour of flight the surface of the earth is traveling to the east faster than it was at the beginning of the hour. How much faster? It's traveling in a circle 500 miles bigger in radius, which means it is 500 x 2pi or 3,141.6 miles further around." Math is incorrect and shouldn't be taught to "high-school-level reader[s]". It is incorrect even "near the pole". Ignoring the plane's altitude as a factor, 500 miles N/S around the circumference causes one to descend to a 89.54 degree latitude, a cir. of 1992.5 miles, spinning at 83mph, if flight began at the North pole. $\endgroup$ – Lee Sep 8 '17 at 7:03
  • $\begingroup$ @Lee: If you leave the north pole and travel 500 miles radially outward, that's 500/3959 = .1263 radians (about equal to tangent and sine), times 57.3 = 7.24 degrees. Subtract that from 90 and you get a latitude of 82.8 degrees, not 89.54. Besides, I think we may have a confusion between miles and kilometers. $\endgroup$ – Mike Dunlavey Sep 8 '17 at 14:45

protected by Qmechanic Sep 3 '17 at 8:07

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