Earth's Rotation and Atmosphere

[Iseous]

How does the atmosphere rotate with the Earth on its axis? There are no forces acting on it that would be strong enough to keep it moving with the ground. Gravity acts perpendicular to the direction of rotation, so it would not be able to cause this motion. It would only be able to keep the atmosphere from drifting outside of Earth's gravitational field. Viscosity would not be responsible, or else moving anything through the atmosphere would pull the entire atmosphere above it with it. That is, in order for the atmosphere to be pulled with Earth, each point on the ground would be dragging the atmosphere above it. However, if this were the case, then anything like a car or plane would be able to do the same, but they obviously cannot. And even if the atmosphere somehow got forced into this motion by some other means, viscosity/friction would have slowed this motion until there was only a small boundary layer near the surface that was affected at all, while the rest would be moving at high speeds relative to the ground. Imagine water flowing through a pipe, or reverse it so the pipe moves relative to the water to make it more similar to the Earth/atmosphere. If the pipe is moving, then imagine adding pressure to get the water up to the same speed as the pipe, and then remove that pressure, would the water keep moving with the pipe forever like our atmosphere?

Furthermore, even if we were to assume that somehow there was a force keeping the atmosphere relatively stationary to the ground, this would create an atmosphere that does not match what we have around us. Each point on the Earth is not the same distance from the axis of rotation, so the speed of rotation is 0 mph at the poles, and increases to over 1000 mph at the equator. This means the atmosphere would have to be moving at different speeds depending on where it was, which would result in a pressure gradient with highest pressure at the poles and the lowest pressure at the equator. This would cause the air to want to flow from high to low pressure, creating constant winds that went from the poles to the equator. This flow could potentially stabilize as the density increased the closer you got to the equator, but we do not see this density variation either.

Lastly, since the atmosphere would have to be at different speeds depending on how far you were from the poles, traveling in an airplane would be largely influenced if the destinations were at very different latitudes. They could either gain or lose 100's of mph, but this does not seem to happen or have to be taken into account.

 

[Drakkith]

The atmosphere formed with the rest of the Earth and thus has been rotating with it from the very beginning.

 

[Astronuc]

There is certainly drag by the Earth on the atmosphere. At the Earth's surface, the air speed is very low, but aloft the air speed is much higher. Look at the jet stream.

Gravity affects the atmosphere through buoyancy effects vis-à-vis differences in density in regions of air, and by the Coriolis affect. The sun provides plenty of energy to the atmosphere to keep things moving.

An example of the effect of drag -

 

[russ_watters]

Furthermore, even if we were to assume that somehow there was a force keeping the atmosphere relatively stationary to the ground, this would create an atmosphere that does not match what we have around us. Each point on the Earth is not the same distance from the axis of rotation, so the speed of rotation is 0 mph at the poles, and increases to over 1000 mph at the equator. This means the atmosphere would have to be moving at different speeds depending on where it was, which would result in a pressure gradient with highest pressure at the poles and the lowest pressure at the equator. This would cause the air to want to flow from high to low pressure, creating constant winds that went from the poles to the equator.

The atmosphere is stationary with respect to the earth, so there are no pressure gradients. And even if there were, they'd quickly equalize, as just a little air flowing in one direction would result in a pressure change, making the air stop flowing. Obviously, air can't continuously flow from the poles to the equator because of conservation of mass (once the air is gone from the poles, there would be no more air to flow).

 

[Iseous]

The Earth is rotating about its axis, which means each point on the Earth is not rotating at the same speed since they are at different distances from the axis of rotation. So if it were stationary with the Earth, that does not mean each point on the Earth is moving at the same speed. If you are at the equator, the atmosphere would be moving 1000 mph since the Earth would be rotating that fast at that location. If you are at a pole, the atmosphere would not be rotating since it is on the axis of rotation (or close to it). Thus the atmosphere would be ranging from 0 to 1000 mph from the poles to the equator. So there would be a pressure gradient since each part of the atmosphere would be moving at different speeds to be "stationary" with the respective part of Earth it was over.

Furthermore, the very next sentence I said it would probably stabilize as the flow continued, but that would mean the density would increase in order to equalize the pressures. In either case we don't see that. Since there is no flow, that means it had to stabilize, but if it stabilized, then the density of the atmosphere would increase as you approached the equator. That doesn't happen.

 

[berkeman]

The Earth is rotating about its axis, which means each point on the Earth is not rotating at the same speed since they are at different distances from the axis of rotation. So if it were stationary with the Earth, that does not mean each point on the Earth is moving at the same speed. If you are at the equator, the atmosphere would be moving 1000 mph since the Earth would be rotating that fast at that location. If you are at a pole, the atmosphere would not be rotating since it is on the axis of rotation (or close to it). Thus the atmosphere would be ranging from 0 to 1000 mph from the poles to the equator. So there would be a pressure gradient since each part of the atmosphere would be moving at different speeds to be "stationary" with the respective part of Earth it was over.

Furthermore, the very next sentence I said it would probably stabilize as the flow continued, but that would mean the density would increase in order to equalize the pressures. In either case we don't see that. Since there is no flow, that means it had to stabilize, but if it stabilized, then the density of the atmosphere would increase as you approached the equator. That doesn't happen.

It would help if you could show us some math. What is the percent change in density that you calculate? The Earth is a pretty big object.

 

[russ_watters]

Since there is no flow, that means it had to stabilize, but if it stabilized, then the density of the atmosphere would increase as you approached the equator. That doesn't happen.

Right: which means the entire first paragraph of your post was wrong. There is no pressure gradient like you describe because the air is not moving with respect to the Earth's surface. Think about it this way: take a parcel of air just above the equator and a parcel of air on the equator. Is the distance between them changing? The reality is that at different latitudes, different points on the Earth and in the air are moving parallel to each other. As a result, there are no forces (other than centripetal acceleration, discussed below) caused by this motion. IE, if you are trying to apply Bernoulli's equation, that only works along a streamline, which means a parcel of air must be flowing from one place to another. If the parcel of air at the pole never flows to the equator, then it isn't a streamline and you can't compare the pressures using Bernoulli's equation.

Now, what you do have because of the rotational motion is a centripetal acceleration. That's what makes the Earth a slightly flattened sphere. The same effect exists on the atmosphere.

Back to the OP for a sec:

Viscosity would not be responsible, or else moving anything through the atmosphere would pull the entire atmosphere above it with it. That is, in order for the atmosphere to be pulled with Earth, each point on the ground would be dragging the atmosphere above it. However, if this were the case, then anything like a car or plane would be able to do the same, but they obviously cannot.

Have you never felt the wind rushing by from a train or a car passing you? Obviously, they can and do pull the atmosphere with them.

The atmosphere formed with earth, so there was no need to get it to spin-up to the Earth's speed. But even if it hadn't, the drag force you describe and we feel every time there is a breeze would fairly quickly bring the atmosphere to a near halt with respect to the Earth's surface.

 

[Iseous]

The air is not moving relative to the surface, but that's not where the gradient would come from. Imagine moving in a car and rolling your window down. The air inside your car is not moving relative to the car, but it is moving at a different speed than the air outside it. Thus there is air flow when the window goes down. The atmosphere would have to be moving at different speeds relative to itself at different points on Earth, just as different points on the ground are rotating faster depending on their distance from the axis of rotation.

I said the train or car does not pull the entire atmosphere above it and that there would only be a small boundary layer that would go with it. Hence why if you were far away from the car you would not feel the wind rushing by you. So if you were far from the surface of the Earth, the ground would not be pulling it as you suggest.

The atmosphere was pulled toward the Earth by gravity, so it moves around the sun with it, but that doesn't mean it would have been forced to rotate with it. Those are two different things.

 

[berkeman]

The air is not moving relative to the surface, but that's not where the gradient would come from. Imagine moving in a car and rolling your window down. The air inside your car is not moving relative to the car, but it is moving at a different speed than the air outside it. Thus there is air flow when the window goes down. The atmosphere would have to be moving at different speeds relative to itself at different points on Earth, just as different points on the ground are rotating faster depending on their distance from the axis of rotation.

I said the train or car does not pull the entire atmosphere above it and that there would only be a small boundary layer that would go with it. Hence why if you were far away from the car you would not feel the wind rushing by you. So if you were far from the surface of the Earth, the ground would not be pulling it as you suggest.

The atmosphere was pulled toward the Earth by gravity, so it moves around the sun with it, but that doesn't mean it would have been forced to rotate with it. Those are two different things.

Your posts are bordering on crackpottery and misinformation. The atmosphere obviously moves with the Earth. Either you post your calculations, or this thread will be closed.

 

[Iseous]

Well, in a simple calculation using Bernoulli's Principle, we have:
v^2/2 + gz + p/rho = constant

v = speed of fluid
g = gravitational acceleration
z = altitude
p = pressure
rho = density

Since we would deal with points at equal altitudes, gz is negligible.

So, assuming the flow stabilized, all of the pressures would be the same, which is the air pressure around us.

We could compare two different points, one where the atmosphere was moving at 0 m/s (pole), and another where the atmosphere was moving at 100 m/s (up to 465 at equator).
v1^2/2 + p/rho1 = v2^2/2 + p/rho2
v1 = speed of atmosphere at pole (0 m/s)
rho1 = density of the atmosphere at pole (sea level)
v2 = speed of atmosphere at point on Earth that would rotate at 100 m/s
rho2 = density of the atmosphere at the point where Earth is rotating at 100 m/s (sea level)
p = pressure at sea level (101325 Pa)

0^2/2 + p/rho1 = 100^2/2 + p/rho2
p/rho1 = 5000 + p/rho2
Divide by pressure:
1/rho1 = 5000/101325 + 1/rho2
Multiply by rho2:
rho2/rho1 = 0.05*rho2 + 1

Thus, the ratio of rho2/rho1 would be greater than 1, meaning the density would increase as you increased in velocity or moved closer to the equator. Using 1.2 kg/m^3 as rho1, rho2 would be 1.27 kg/m^3 or about 6% more. This would increase further as you moved to the equator.

 

[Drakkith]

The atmosphere would have to be moving at different speeds relative to itself at different points on Earth, just as different points on the ground are rotating faster depending on their distance from the axis of rotation.

Someone correct me if I'm wrong, but wouldn't the relative velocity between two points on the Earth's surface be zero? It's like being on a spinning merry-go-round with me in the center and by buddy at the edge. Neither of us are moving relative to one another. Or have I misunderstood how things work in this non-inertial frame?

 

[Nugatory]

Someone correct me if I'm wrong, but wouldn't the relative velocity between two points on the Earth's surface be zero?

That is not right. Think about how the Coriolis effect sets a hurricane to spinning...

 

[Nugatory]

Well, in a simple calculation using Bernoulli's Principle...

Bernoulli's Principle applies within a flow line, not between flow lines, so it doesn't apply here. The Wikipedia article on Bernouilli's Principle has a section on misapplications of the principle that you might want to look at; you're basically presenting a more sophisticated version of the widespread but bogus classroom demonstration in which blowing across a sheet of paper causes it to rise.

 

[Nugatory]

Lastly, since the atmosphere would have to be at different speeds depending on how far you were from the poles, traveling in an airplane would be largely influenced if the destinations were at very different latitudes. They could either gain or lose 100's of mph, but this does not seem to happen or have to be taken into account.

It most certainly does happen and is taken onto account. An aircraft in the northern hemisphere experiences a slight crosswind from the west as it flies south into air that is moving more rapidly to the east. The effect of this crosswind is to increase the eastward component of the aircraft's speed relative to the fixed stars, matching it to the eastward speed of the ground underneath it.

 

[russ_watters]

It most certainly does happen and is taken onto account. An aircraft in the northern hemisphere experiences a slight crosswind from the west as it flies south into air that is moving more rapidly to the east. The effect of this crosswind is to increase the eastward component of the aircraft's speed relative to the fixed stars, matching it to the eastward speed of the ground underneath it.

That's not the effect the OP is referring to.

That is not right. Think about how the Coriolis effect sets a hurricane to spinning...

In a stationary frame at Earth's center, the atmosphere is moving at 1000 mph at the equator, but in the rotating frame of Earth's surface - not including the wind - the atmosphere is stationary. That means there are no shear (or any other fluid dynamic) forces between adjacent air molecules even though in the stationary frame they rotate around each other.

It most certainly does happen and is taken onto account. An aircraft in the northern hemisphere experiences a slight crosswind from the west as it flies south into air that is moving more rapidly to the east. The effect of this crosswind is to increase the eastward component of the aircraft's speed relative to the fixed stars, matching it to the eastward speed of the ground underneath it.

The OP is not referring to the wind as we know it, he's wondering why a plane flying at the equator isn't going 1500 mph in when flying west and barely moving if it flies east.

 

[Iseous]

Someone correct me if I'm wrong, but wouldn't the relative velocity between two points on the Earth's surface be zero? It's like being on a spinning merry-go-round with me in the center and by buddy at the edge. Neither of us are moving relative to one another. Or have I misunderstood how things work in this non-inertial frame?

Isn't this motion the basis for a centrifuge, however? As it rotates, the more dense fluids get pulled further away from the axis of rotation, just as I am describing. For the merry-go-round, that is an interesting point. I think the rotation makes it a bit confusing, as an outside observer would see one moving and the other not moving, so aren't they moving relative to each other even if they aren't getting further away from each other? That seems odd.

Bernoulli's Principle applies within a flow line, not between flow lines, so it doesn't apply here. The Wikipedia article on Bernouilli's Principle has a section on misapplications of the principle that you might want to look at; you're basically presenting a more sophisticated version of the widespread but bogus classroom demonstration in which blowing across a sheet of paper causes it to rise.

True, but does the basic idea still not apply? That a faster flow would result in lower pressure? Or any difference between the two fluids? For instance, imagine a car driving. The air inside the car is not moving relative to a person inside it, but the air outside the car is moving relative to that same person. How would you compare the two different fluids? If you roll down the window, there obviously seems to be a difference, so how would you represent those differences mathematically?

It most certainly does happen and is taken onto account. An aircraft in the northern hemisphere experiences a slight crosswind from the west as it flies south into air that is moving more rapidly to the east. The effect of this crosswind is to increase the eastward component of the aircraft's speed relative to the fixed stars, matching it to the eastward speed of the ground underneath it.

This is exactly what I am talking about. So as it moves south, the air would be moving more rapidly to the east.

The OP is not referring to the wind as we know it, he's wondering why a plane flying at the equator isn't going 1500 mph in when flying west and barely moving if it flies east.

No I was referring to what Nugatory was talking about. If you stay at the same latitude, you will have the same rotational speed that you started with. If you are at the equator, you would already be moving at 1000 mph with the rotation, so it would be like walking in a plane traveling at 1000 mph; it wouldn't be harder to move forward, back, or side to side because you were already moving with it. So going east or west would not have any effect. It would, however, have an effect if you moved north or south because those points aren't moving as fast as the equator. For instance, a point north of the equator might be rotating at only 900 mph because it is not as far out on the curvature of the Earth (not as far from axis of rotation). Thus, you would have a 100 mph speed difference if you moved to that point.

 

[russ_watters]

Bernoulli's Principle applies within a flow line, not between flow lines, so it doesn't apply here. The Wikipedia article on Bernouilli's Principle has a section on misapplications of the principle that you might want to look at; you're basically presenting a more sophisticated version of the widespread but bogus classroom demonstration in which blowing across a sheet of paper causes it to rise.

I disagree with everything the wiki says about the paper demonstration. I don't think it is relevant here though, because the entire issue of flowing air doesn't exist in the example we are discussing.

Edit: ehh, thinking about it more, the first and key issue is correct, that the static pressure in the jet is equal to the static pressure of the surrounding air. Still don't like the other criticisms though.

 

[PeterDonis]

in the rotating frame of Earth's surface - not including the wind - the atmosphere is stationary. That means there are no shear (or any other fluid dynamic) forces between adjacent air molecules even though in the stationary frame they rotate around each other.

In the rotating frame of the Earth's surface, there are fictitious forces present because the frame is non-inertial. The key one for this discussion is Coriolis force, as Nugatory said. If every parcel of air on Earth were exactly stationary all the time, then Coriolis force would not come into play, since it is velocity-dependent. But in the real atmosphere, parcels of air do move relative to each other, just from random fluctuation if nothing else, and as soon as they do, Coriolis force comes into play.

does the basic idea still not apply? That a faster flow would result in lower pressure? Or any difference between the two fluids?

Yes, it applies. The differences in pressure between air parcels in the atmosphere are a major contributor to weather and wind patterns. But for those differences in pressure to exist, there has to be some relative motion between the air parcels. In your car example, the car provides the relative motion. In the actual atmosphere, the relative motion is ultimately due to random fluctuations in the motion of air molecules, plus differences in density and temperature due to differential heating. See below.

If you stay at the same latitude, you will have the same rotational speed that you started with. If you are at the equator, you would already be moving at 1000 mph with the rotation, so it would be like walking in a plane traveling at 1000 mph; it wouldn't be harder to move forward, back, or side to side because you were already moving with it. So going east or west would not have any effect. It would, however, have an effect if you moved north or south because those points aren't moving as fast as the equator. For instance, a point north of the equator might be rotating at only 900 mph because it is not as far out on the curvature of the Earth (not as far from axis of rotation). Thus, you would have a 100 mph speed difference if you moved to that point.

Yes, this is the Coriolis force that Nugatory referred to. But, as I said above, there needs to be some initial motion of a parcel of air for it to come into play; if every parcel of air in the atmosphere were exactly stationary with respect to the Earth, then there would be no Coriolis force observed. This idealized case (all parcels of air motionless with respect to the rotating Earth) is what russ_watters is talking about in post #15: in this idealized case, he is correct that there are no shear forces present. If you really don't understand how that idealized case is possible, look up the equations describing a rotating frame, and note how the Coriolis force depends on velocity relative to the rotating frame: an object at rest in the rotating frame experiences no Coriolis force, so the entire atmosphere sitting at rest with respect to the rotating Earth is a possible state.

But in the real atmosphere, all parcels of air are never motionless with respect to the rotating Earth; there are random fluctuations, plus different parts of the atmosphere are heated differently because of day/night differences, differences in sun angle, differences in ground temperature, etc., and this differential heating drives convection. And once the air is moving, the Earth's rotation does affect how the air moves (an obvious example is the circulation of air in cyclones, which is driven by the Coriolis force and is different in the Northern and Southern hemispheres). And obviously, if you are in an airplane and you move relative to the rotating Earth, you are introducing motion yourself.

 

[russ_watters]

Iseous, your original question asked why the atmosphere rotates with the Earth's surface. That has the simple, two-pronged answer:
A. It always has.
B. If it didn't, drag would quickly make it.

But you don't seem to like that and it looks to me like you are jumping around between several other explanations, hoping one might apply. That makes it hard to follow. So far, we have:
1. Bernoulli's principle.
2. Centrifugal/centripetal force.
3. Coriolis effect.

#2 and #3 have an impact on the atmosphere (and objects) due to Earth's rotation. #1 does not. But none of these have anything to do with your original question. This jumping-around makes the thread confusing and unfocused. Please try to keep the separate issues separate in your head and be clear that you understand that they are separate when discussing them.

Isn't this motion the basis for a centrifuge, however?

Yes. Earth's rotation causes the Earth to be flattened as a result of those forces (#2, above). But this has nothing to do with aerodynamic forces and the rotation (or not) of the atmosphere with Earth. Also, my understanding is that because the Earth is flattened, the gravitational potential is equalized at the Earth's surface. As a result, the atmosphere would not have its own, additional bulge. I'm not certain of that, though.

True, but does the basic idea still not apply? That a faster flow would result in lower pressure? Or any difference between the two fluids? For instance, imagine a car driving. The air inside the car is not moving relative to a person inside it, but the air outside the car is moving relative to that same person. How would you compare the two different fluids? If you roll down the window, there obviously seems to be a difference, so how would you represent those differences mathematically?

A car is not well sealed, so the pressure situation is not clear/simple. However, if you seal a car air-tight and then set it in motion and measure the absolute pressure inside and outside (for someone stationary), they will be equal, at atmospheric pressure.

When you open a window, you create a pathway for air to flow into and out of the car and the orientation of that path determines the pressure change. For a window perpendicular to the direction of motion, no air (net) should be flowing into or out of the car and the pressure should not change (like a "static port"). If you angle an air-scoop forward, that directs air into the car and acts like a pitot tube, making the pressure inside rise (up to, potentially, stagnation pressure)

This is exactly what I am talking about. So as it moves south, the air would be moving more rapidly to the east.

The "crosswind" is the Jet Stream, it's a weather effect caused by the coriolis effect and regular wind. If the Earth was not heated by the sun and thus had no wind, there would be no jet stream. This is totally different from other things you were discussing.

Note that there is a southern jet stream and it flows in the opposite direction of the northern one.

No I was referring to what Nugatory was talking about. If you stay at the same latitude, you will have the same rotational speed that you started with. If you are at the equator, you would already be moving at 1000 mph with the rotation, so it would be like walking in a plane traveling at 1000 mph; it wouldn't be harder to move forward, back, or side to side because you were already moving with it. So going east or west would not have any effect. It would, however, have an effect if you moved north or south because those points aren't moving as fast as the equator. For instance, a point north of the equator might be rotating at only 900 mph because it is not as far out on the curvature of the Earth (not as far from axis of rotation). Thus, you would have a 100 mph speed difference if you moved to that point.

That's the coriolis effect and it has nothing to do with wind/the atmosphere: it applies in a vacuum, causing a projectile to appear to curve as it moves north or south.

 

[russ_watters]

In the rotating frame of the Earth's surface, there are fictitious forces present because the frame is non-inertial. The key one for this discussion is Coriolis force, as Nugatory said. If every parcel of air on Earth were exactly stationary all the time, then Coriolis force would not come into play, since it is velocity-dependent. But in the real atmosphere, parcels of air do move relative to each other, just from random fluctuation if nothing else, and as soon as they do, Coriolis force comes into play.

That's all fine, as long as we are no longer discussing the question in the first sentence of the OP. It isn't clear to me that the OP recognized that that's not the same issue.

 

[PeterDonis]

my understanding is that because the Earth is flattened, the gravitational potential is equalized at the Earth's surface.

This is true, at least as a good approximation. (It would be exactly true if the Earth were a fluid, but since the Earth is solid, its rigidity can resist, to some extent, the gravitational effects that try to make its surface exactly equipotential. This effect is very small, however.)

As a result, the atmosphere would not have its own, additional bulge.

This is not true. The gravitational potential as a function of height is different at the equator than at the poles; this effect does not stop at the Earth's surface. To put it another way, the fact that the Earth is an oblate spheroid means that the gravitational potential it produces is not a simple spherically symmetric $1 / r$ function; there is an additional term due to the Earth's quadrupole moment, which is different at different latitudes. This affects the shape of the atmosphere.

 

[russ_watters]

This is not true. The gravitational potential as a function of height is different at the equator than at the poles; this effect does not stop at the Earth's surface.

Yeah, ok, I can see that. I wonder how big an additional oblateness that is. The Earth's oblateness is caused by the interaction of gravity and centrifugal force over the entire radius of 6400 km. Any additional oblateness would be due to the additional few tens of km of atmosphere, not the entire radius. So it may be a few orders of magnitude smaller of an effect than the 21km of the surface oblateness.

 

[PeterDonis]

The Earth's oblateness is caused by the interaction of gravity and centrifugal force over the entire radius of 6400 km. Any additional oblateness would be due to the additional few tens of km of atmosphere, not the entire radius.

The atmosphere's oblateness is not just caused by the gravity of the atmosphere itself. It's caused by the gravity of the entire Earth being oblate. The oblateness of the Earth's gravity doesn't just stop at the Earth's surface; the variation of potential with distance from the Earth's center does not suddenly start being the same at all latitudes when you get beyond the Earth's surface. That is, if I am, say, 100 km above the North Pole, the difference in potential between that and the surface at the North Pole is not the same as the difference in potential between 100 km above the equator, and the surface at the equator.

 

[Iseous]

your original question asked why the atmosphere rotates with the Earth's surface. That has the simple, two-pronged answer:
A. It always has.
B. If it didn't, drag would quickly make it.

I do not understand how B could be true. So the ground is essentially moving eastward, and the atmosphere above each point on the ground is being "dragged" with it. If there is a treadmill on the ground, then the atmosphere above is being pulled by each point on the treadmill. But if you turn it on so that the treadmill (ground) is essentially moving faster, the entire atmosphere above does not move faster with it. Even if it were a smaller closed system only a small layer close to it would be pulled with it (boundary layer). Where else would you see a moving or rotating object take a static fluid and get it to move with it over a long distance? So how could drag be the answer? So we are left with A, which is a non-answer. It doesn't actually explain anything.

 

[Bandersnatch]

I do not understand how B could be true.

It looks like you're under the misapprehension that the only two possible states to consider are drag having no effect at all and drag that is causing the whole column of air to move simultaneously and immediately.
Consider drag accelerating the layer closest to the ground first, then that layer transferring its kinetic energy to the layer above, and that one to the next one up. The ground keeps adding kinetic energy to the atmosphere until it is all moving at the same angular speed.

This is the same effect that can be seen when you take a cup of tea and stir it. At first you're only accelerating the volume elements in contact with the spoon, but the rest of the liquid soon starts rotating as well - you don't have to 'hit' every bit of volume of the liquid with the spoon to make it rotate.
Once the tea is rotating at the same angular velocity, you again can see the same effect as the walls of the cup begin to slow it down through friction. Again, it is only the layer directly in contact that is decelerated by the cup, yet in a finite time you observe the tea to match the rotation (or lack thereof) of the cup.

 

[russ_watters]

I was too busy this morning to answer, but Bandersnatch answered in nearly exactly the same way I would have, including using exactly the same analogy: stirring a cup of tea. Doublelike.

 

[Iseous]

I didn't say drag has no effect at all; I said it affects a boundary layer, which is nowhere near the size of the entire atmosphere. I didn't say it would do it immediately either. But even if you let the motion of a treadmill continue for a very long time in a closed system, the entire column of air would not eventually move with it unless the system only enclosed the size of the boundary layer. But even then, the boundary layer is not uniformly moving together as you seem to suggest it would over time.

Your cup of tea example isn't really an accurate depiction of what is going on. For one, you would have to be rotating the cup, not a spoon inside of the cup. But if you did that, the tea would rotate with the cup at a constant angular speed (one rotation per rotation of the cup), but the tea itself (and even the cup) would be rotating at different speeds depending on the location away from the center. For another, you are saying that the atmosphere would be moving with the ground, yet your example clearly shows a relative motion between the tea and the cup... That is the complete opposite of the atmosphere/ground system we are talking about. Furthermore, your spoon is large compared to the size of the cup. Put your spoon in a lake and stir it and tell me that the entire lake will be affected (and it would have to be affected evenly) by your stirring if you keep going. You just cleverly chose a small enough object such that the entire fluid would move.

 

[russ_watters]

I didn't say drag has no effect at all; I said it affects a boundary layer...

That's not what a boundary layer does/is. And in fact, because of fluid viscocity, a boundary layer is theoretically infinite in size. Boundary layers are often arbitrarily defined at 99% of freestream velocity, but as with any hyperbolic function, they can never reach 100%: https://en.wikipedia.org/wiki/Boundary_layer

But even if you let the motion of a treadmill continue for a very long time in a closed system, the entire column of air would not eventually move with it unless the system only enclosed the size of the boundary layer. But even then, the boundary layer is not uniformly moving together as you seem to suggest it would over time.

This is not correct. The larger the container, the longer it will take for the enclosed fluid to spin-up to uniform speed, but it would eventually.

Your cup of tea example isn't really an accurate depiction of what is going on. For one, you would have to be rotating the cup, not a spoon inside of the cup.

Same issue, different reference frame, and you can indeed do either if you feel like it. You could take a cup of liquid, put it on a turnable, and spin it or you could set the liquid spinning and wait for it to stop spinning. Both work the same way.

Think about this on a much smaller level. If an arbitrarily small element of viscous fluid is moving past another at any speed, it must be exerting a viscous force on it. If one element exerts a force on the next, the next moves and exerts a force on the next after that...and next after that, etc., forever.

 

[Iseous]

From your page "a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant". Key words there are "immediate vicinity" and "significant". So sure the boundary layer goes on forever but it becomes negligible outside of the "immediate vicinity".

The energy each of those molecules transfers to one another decreases as you get further out, hence why the boundary layer is only significant in the immediate vicinity. That's also why you don't feel a car going by if you are sufficiently far away. There's only so far that layer goes in any significant way. I don't understand how you could think stirring a lake would get the entire lake rotating, unless you are imagining a giant spoon that is the same proportion as the one with the cup of tea. If you took a spoon and did a small circular stir just as you would in the cup of tea, but do it in a lake, there is no way the entire lake would eventually rotate at the same speed. You would just get a little vortex.

 

[Drakkith]

I don't understand how you could think stirring a lake would get the entire lake rotating, unless you are imagining a giant spoon that is the same proportion as the one with the cup of tea. If you took a spoon and did a small circular stir just as you would in the cup of tea, but do it in a lake, there is no way the entire lake would eventually rotate at the same speed. You would just get a little vortex.

That's because the lake bed and shore have friction. There is nothing to steal energy from the atmosphere, so as the bottom layers begin to rotate the effect is transferred outwards to the upper layers. Besides, the Earth is enormous compared to the atmosphere. It's analogous to the effect a spinning basketball has on a layer of air with a thickness equal to a thin film of plastic (so less than a mm).