Earth's Rotation and Atmosphere

[Bystander]

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.

You understand drag. You understand boundary layer. You understand the difference between angular and tangential speeds as functions of radius.

Yet, you ask the question,

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.

 

[russ_watters]

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".

What is negligible in one situation may not be quite so negligible in another.

Specifically, have a look at the equation for the thickness of the boundary layer:
https://en.wikipedia.org/wiki/Boundary_layer_thickness#99.25_Boundary_Layer_thickness

Do you see that there is a term in it for the length of the surface? Because of the viscocity/drag effects we are discussing, the longer a surface is, the thicker the boundary layer gets (the previous wiki article has graphs of this). The further along a surface the wind blows, the more time the the drag effects have to build-up and move outward in the way I described previously.

So, how long is the relevant length of the surface: the distance downstream from which you want to calculate the boundary layer's thickness? On Earth or in the coffee cup?

 

[Iseous]

For laminar boundary layers over a flat plate, the Blasius solution gives:

For turbulent boundary layers over a flat plate, the boundary layer thickness is given by:

where

is the overall thickness (or height) of the boundary layer

is the Reynolds Number

is the density

is the freestream velocity

is the distance downstream from the start of the boundary layer

is the kinematic viscosity

is the dynamic viscosity

This calculation will be done for the equator with the assumption that the distance is the circumference of the Earth. I'm not sure if that would really be accurate since these equations are for a flat plate. It may be more accurate to do a section of the Earth with a small amount of curvature to essentially make it flat. However, using the entire circumference would give it the largest possible boundary layer since it is proportional to the square root of that distance. Thus, this would be the best case scenario to show if this drag could actually cause this.

= 1.2 kg/m³

= 465 m/s
x = 40,233,600 m (circumference of Earth)

= 1.8 * 10^-5 kg/m/s (http://www.engineeringtoolbox.com/dry-air-properties-d_973.html)
Re_x = 1,247,241,600,000,000

With that Reynolds Number, the flow is turbulent (> 4000), but here is the value using laminar flow:

= 4.91*x/(Re_x)^½

= 5.6 meters

With turbulent:

= 0.382*x/(Re_x^1/5

= 14,704 m = 9 miles

Whether those equations would work for this is questionable. But it would seem to be the best case scenario. However, according to http://www.space.com/17683-earth-atmosphere.html, "Earth's atmosphere is about 300 miles (480 kilometers) thick, but most of it is within 10 miles (16 km) of the surface". Since there is still a mile where there are a decent amount of particles to encounter, this would be an interesting situation for high altitude balloons that can go much higher than 9 miles. Those particles would be moving at 1000 mph or several hundred mph most other places around the world.

Additionally, this distance is the distance to where it would reach 99% of the freestream velocity. But this does not mean the speed would be 0% of that velocity up until 9 miles and then sharply increase to 99% at 9 miles. If you look at the graphs of the previous page you provided, the speed increases exponentially from 0, so the atmosphere would still be moving a good fraction of that up until that point. Just that graph alone shows that the size of the boundary layer is almost irrelevant. The velocity changes very rapidly from the ground up, which means we would see 100+ mph winds even at relatively low altitudes.

Even a balloon rising through 20% of that speed would encounter pretty significant and noticeable wind (200 mph). And even if this increase was gradual so that the balloon slowly increased in speed and started moving with the rotation of the Earth, this type of traveling with the rotation of the Earth has been said not to happen. If it did, couldn't we just put balloons up to a high enough altitude and let them rotate around Earth like a satellite? Furthermore, this Reynolds Number would indicate turbulent flow everywhere just from the rotation of the planet alone, ignoring weather patterns caused by temperature and other variables. So even with a 9 mile boundary layer, this doesn't seem to describe what we see.

 

[Bandersnatch]

In your analysis of the motion of the atmosphere, what is it that maintains the constant flow over the surface that you need for Blasius boundary layer?

 

[Iseous]

The constant rotation of the Earth essentially provides the relative motion.

 

[Bystander]

the relative motion.

"Relative" to what? What is moving relative to what else?

 

[Bandersnatch]

The constant rotation of the Earth essentially provides the relative motion.

And you think this relative motion remains constant why exactly? Would you say there is no friction between molecules in the air?

@Bystander: the scenario is the atmosphere begins stationary over rotating Earth.

 

[Iseous]

The Earth is rotating at a constant speed, so the molecules outside of the boundary layer will be moving at the freestream velocity (speed of rotation) relative to the Earth. And I'm not saying there is no friction between the molecules. The viscous forces are causing the boundary layer.

 

[OmCheeto]

Ok. So what is the new question?

 

[Iseous]

Ok. So what is the new question?

Russ linked me to a page about boundary layers and I used those equations to get a very rough estimate of the size of that layer for Earth. Those equations weren't really for a sphere rotating, so I'm not sure if they really mean anything.

 

[Bandersnatch]

The Earth is rotating at a constant speed, so the molecules outside of the boundary layer will be moving at the freestream velocity (speed of rotation) relative to the Earth.

No they will not. All they have is inertia, and there is a constant force acting on them through their viscosity. Unless you provide another force to keep the freestream velocity constant with respect to the ground, the atmosphere will accelerate and eventually match the rotation as per the Newton's laws.

Perhaps approach this this way: I'm sure you do recognise that for every action there is a reaction. The lower layer of the atmosphere in contact with the ground will exert a force on the surface in the opposite direction to the drag the surface exerts on the atmosphere. This force will produce torque through the Earth's radius acting to slow down the rotation.
Under your understanding, the atmosphere will then eventually stop the rotation of the Earth, or any other planet, but the planet will somehow not manage to accelerate the atmosphere.

 

[Iseous]

So if you were to take a ball, put it in the center of a large enclosed room, and keep it spinning constantly, the layer of air being rotated by the ball would increase in size over time?

 

[russ_watters]

Your error is here:

This calculation will be done for the equator with the assumption that the distance is the circumference of the Earth. I'm not sure if that would really be accurate since these equations are for a flat plate.

The Earth has no beginning or end. If a bundle of air has been circling Earth for four billion years, the distance is the entire distance it has covered in that time.

 

[Iseous]

The Earth has no beginning or end. If a bundle of air has been circling Earth for four billion years, the distance is the entire distance it has covered in that time.

I started to think about that and I think that would be correct. However, that's assuming those equations would even apply to a rotating sphere. I found a paper about boundary layers of a rotating sphere in a still fluid, and I'll try to look into it more to see what it says when I get a chance. http://www2.leicester.ac.uk/departm...f-profiles/sjg50/Garrett - Peake 2002 JFM.pdf

However, there was an interesting tidbit in the introduction:
"When a sphere rotates in still fluid a flow is induced in which the fluid moves over the outer surface from the poles to the equator and is ejected radially from the equator." That's what I was saying in the beginning.

 

[Bystander]

When a sphere rotates in still fluid

What "still" fluid?

 

[Iseous]

In a fluid that was initially stationary, if that's what you mean.

 

[Bystander]

What are the boundaries of this stationary fluid volume/mass?

 

[PeterDonis]

"When a sphere rotates in still fluid a flow is induced in which the fluid moves over the outer surface from the poles to the equator and is ejected radially from the equator."

What happens if the sphere has enough gravity to keep the fluid from escaping when it rises radially at the equator?

 

[olivermsun]

What happens if the sphere has enough gravity to keep the fluid from escaping when it rises radially at the equator?

Presumably you develop some sort of closed circulation

 

[jim hardy]

Dad's 1940-ish meteorology textbook "Climate and Man" had a simplistic explanation that i kept in a crevice of my alleged mind:

Air near the equator gets heated by the sun because the ground is warm there.

It picks up a lot of moisture which lightens it further because molecular weight of water is only 18 compared to air's 29...
So it rises and is replaced by cooler air flowing generally toward equator over the surface.
As it rises above the "boundary layer" it carries aloft its eastbound momentum
so the atmosphere rotates because it's well stirred. Somebody mentioned teacup...

Warm equatorial air aloft travels away from equator to where Earth's surface velocity is lower,
so it has excess eastbound momentum compared to surface air at same latitude..
At around 30 deg it sinks again and some of it becomes the aforementioned equator-bound surface air,
some of it becomes poleward-bound surface air.
The sinking air that continued away from equator overtakes Earth's surface due to its excess eastbound momentum giving us the Westerlies .
Near the equator Earth rotates out from under the south equator-bound air giving us the Tradewinds. (edit - sorry , had to edit it to be hemisphere-neutral jh)
See picture below.

So the thought model is a rotating sphere covered with a viscous liquid that's unevenly heated.

Way more thorough explanation here:
http://nc-climate.ncsu.edu/edu/k12/.atmosphere_circulation

Some folks think of it as a Carnot engine moving heat from equator toward poles.

 

[PeterDonis]

Presumably you develop some sort of closed circulation

Yes. Which, as jim hardy's post points out, is exactly what happens in the Earth's atmosphere.

 

[OmCheeto]

In a fluid that was initially stationary, if that's what you mean.

I think Drakkith answered this in post #2

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

I'm still not sure what we are discussing, so I'll reference back to some research I did back in March, and my commentary on why the atmosphere is so weird, that I can't begin to explain it.

I was diddling around with the http://earth.nullschool.net simulator yesterday, and things were no better.
From sea level to about 1500 meters, the atmosphere is consistently random, and very much traveling at the same speed as the Earth.
Above that elevation, things get very strange, and I suspect it is the tilt of the Earth which is causing it.
The northern and southern hemispheres, atmospherically, look like they are from two different worlds.

 

[olivermsun]

Yes. Which, as jim hardy's post points out, is exactly what happens in the Earth's atmosphere.

Keep in mind, though, as Jim Hardy's post also points out, the Earth system is being heated at the equator, so there's quite a different set of mechanisms causing the large-scale circulation that we actually observe.

 

[rootone]

It has to be said a well that the atmosphere always has been a part of the planet. just as the ocean is.
It' s overall angular momentum has been in synch with the rocky planetary body from the beggnning, it did not need to be accelerated to match the rocky stuff,
and there is nothing at the outer edge of the atmosphere which is applying a braking force to it.

 

[PeterDonis]

the Earth system is being heated at the equator, so there's quite a different set of mechanisms causing the large-scale circulation that we actually observe.

Yes, good point; what we should be doing is comparing the relative numbers for convection and the rotation effect.

 

[D H]

I started to think about that and I think that would be correct. However, that's assuming those equations would even apply to a rotating sphere. I found a paper about boundary layers of a rotating sphere in a still fluid, and I'll try to look into it more to see what it says when I get a chance. http://www2.leicester.ac.uk/departments/mathematics/extranet/staff-material/staff-profiles/sjg50/Garrett - Peake 2002 JFM.pdf

However, there was an interesting tidbit in the introduction:
"When a sphere rotates in still fluid a flow is induced in which the fluid moves over the outer surface from the poles to the equator and is ejected radially from the equator." That's what I was saying in the beginning.

That article is about flow of air around a small object in a large, uniform gas. It has nothing to do with the topic at hand. I suggest you read about general circulation models instead.

 

[Iseous]

That article is about flow of air around a small object in a large, uniform gas. It has nothing to do with the topic at hand. I suggest you read about general circulation models instead.

So a spherical Earth rotating with an initially still atmosphere is nothing like a sphere rotating in an initially still gas?

 

[rootone]

But it didn't have an initially still atmosphere, (and what exactly does 'still' mean here anyway?, still in relation to what?)
The whole planet including the atmosphere originally formed out of a condensing cloud of material including gases, and all of it had angular momentum.

 

[Iseous]

Still to an outside observer who is not rotating with Earth. Regardless, once the sphere in that experiment started rotating and getting the gas to rotate with it, then it is no longer "still" either. So whether it started off not rotating, or you looked at a point once the sphere and gas around it was already rotating (fast forward into the experiment), you would get the same steady state eventually unless there was an outside energy source influencing the gas. But if you're going to claim it all had angular momentum, then where did that come from? These clouds were the result of gravitational forces acting on one another. But how would gravity cause something to rotate on its own axis? It acts at the center of mass, which would not do that. That would require a force not acting on the center of mass (i.e. not gravity).

 

[PeterDonis]

But how would gravity cause something to rotate on its own axis? It acts at the center of mass, which would not do that.

You are taking an approximation (that the gravity of a spherically symmetric body can be viewed as being produced at its center of mass) and using it where it is not valid. The solar system formed from a diffuse cloud of gas and dust. That cloud was not exactly spherically symmetric; the forces between individual pieces of the cloud did not average out to a single "force of gravity" acting from the center of mass of the cloud.

The above is actually another way of saying that the cloud started out with angular momentum. Other clouds that formed other stars also started with angular momentum. In order to average everything out to get zero angular momentum and spherical symmetry (to a good approximation), you need to look at much larger distance scales, comparable to the entire observable universe. At that scale, the universe as a whole has (as far as we can tell) zero angular momentum. But much smaller regions of it, like stellar systems, do not.