Is the movement of winds due to coriolis or centripetal force?

[Order]

Is the movement of winds due to coriolis or centripetal force? Or is there no way to tell, like for example for tidal forces, there is no way to tell wether they are caused by centripetal forces or gravity. (At least that is how I have understood the problem.)

Earth is like a huge turntable with the difference that there are no centripetal forces if you are standing still (and the Earth is rotating), because Earth is slightly flattened towards the equator. So the gravitational forces balance out the centripetal forces, if you are standing still.

But when wind is moving east there is an extra movement in the rotational direction, so there is a force towards the equator. Now, looking from an inertial frame the force is $$F_{r}=\frac{mv^{2}}{r}=\frac{mv_{earth}^{2}}{r} + \frac{mv_{wind}^{2}}{r}+\frac{2mv_{earth}v_{wind}}{r}$$ The first term cancels to gravity. The second term is the centripetal force and the third term $2mv_{wind}\omega$ is the coriolis force. My problem is: if the coriolis force is constant and the centripetal force obviously is not constant, since the wind's speed in the east west direction changes. How comes winds are moving in circles and not in elipses?

 

[MrNerd]

I always thought winds resulted from pressure differences in the atmosphere resulting from differences in temperature.

In spot A, the temperature is higher. In spot B, the temperature is lower. Therefore, there is a difference in temperature. A difference in temperature will also result in differences in pressure. The difference in pressure causes a flow of air from high pressure to low pressure. Thus, wind.

As a disclaimer, I don't think I've ever had any formal education in this, but I do start community college this week with meteorology(oddly enough, it's in the geography or geology class group, I don't remember which) as a class.

 

[BruceW]

the coriolis effect is the same thing as the centrifugal force.

Low pressure systems in the north hemisphere rotate anticlockwise and low pressure systems rotate clockwise in the southern hemisphere due to the coriolis effect.

In general, the weather is explained by temperature, pressure and the rotation of the earth.

 

[A.T.]

the coriolis effect is the same thing as the centrifugal force.

No, it's not. While the centrifugal force also affects objects at rest in the rotating frame of reference, the Coriolis force affects only objects that are already moving in the rotating frame. So you need something to make the air move in the first place. Coriolis force alone cannot create any wind.

 

[Order]

Thanks for answers and good luck MN with your new class. I would like more proofs of what you are saying instead of just claims, though.

I was looking at my equation above (which is not completely correct I can tell, but still good enough for reasoning about the issue) and I can see in this equation that the coriolis term is much larger than the centripetal term, since the velocity of the wind is much smaller than the rotational speed of the earth. Is this right? If this is right, it would explain a lot.

I can see also that much of the difficulty here arises from dealing with pseudo forces. Therefore the answer to my questions might be, kind of, frame dependent. No wonder I am so confused!

 

[BruceW]

Oh I see now. Thanks for pointing that out A.T. I got it wrong because I thought Coriolis force and Centrifugal force both meant the total fictitious force. But actually the total fictitious force equals the sum of the Centrifugal force, Coriolis force and Euler force.

In the case of the earth, the Euler force is practically zero. and the Coriolis force is:
$$-2m \Omega \times v$$
And the Centrifugal force is:
$$-m \Omega \times ( \Omega \times x)$$

where omega is the vector angular frequency of the earth, and the velocity and position are of some object with respect to the rotating reference frame.
So if we consider an air parcel which is stationary with respect to the rotating earth, then the Coriolis force on it is equal to zero, as A.T. said.

 

[rcgldr]

There's also the jet stream near mid latitudes and the opposite flowing prevailing winds nearer the equator. Also the Earth approximates a rotating sphere:

http://en.wikipedia.org/wiki/Coriolis_effect#Rotating_sphere

 

[RedX]

I'm pretty confused by the post - I think you might have a lot of misconceptions.

First, tidal forces have nothing to do with centripetal acceleration. Tidal forces are due to gravity.

Second, if you're standing still on the earth, unless you're at the poles, there is a centripetal force on you.

Third, the reason your equation says the wind moves in circles is because you wrote it that way: at the very beginning you said assume the wind blows from the west to the east, so of course you'll get that they move in circles from west to east. I'm also not sure how you can say that the Coriolis force is obviously constant and the centripetal force is not.

Here is a picture of global surface winds, taken from the reference frame of a stationary observer fixed to the earth:

http://www.colorado.edu/geography/blanken/GEOG%206181%20Fall%202003/noble/images/06_tradewind_1.jpg

The Earth is hotter at the equator, causing warm air to rise high in the air where it's at greater pressure than colder air at the same height, so the warm air moves north and south of the equator where it cools down and sinks at 30 degrees latitude. The air here is now high pressure at the surface, so winds flow back towards the equator on the surface of the Earth but due to the Coriolis force (which pushes things right in the Northern hemisphere and left in the Southern hemisphere) it gets bent westward. These are the trade winds. Similarly, above 30 degrees you have the Westerlies until you reach the polar front (when the cold air drops at 30 degrees, it can flow north or south, so if it flows north in the northern hemisphere the coriolis force pushes it east).

 

[chrisbaird]

I simplified way of thinking about it is that in general local winds are caused by temperature/pressure differences and global winds are caused by the Coriolis force. I don't think the centrifugal force has much effect on winds since it applies equally outwards at all points and essentially just weakens the effect of gravity.

 

[chrisbaird]

By the way, you have to be careful to define what frame you are in. If your frame is a stationary spot in space, watching the Earth spin, then there is no coriolis force or centrifugal force. The global winds are described directly by the rotation of the earth. There is only gravity which is a centripetal force holding everything in orbit and your inertia. If your frame of reference is fixed to the surface of earth, so that you are rotating with the earth, but from your frame you do not appear to be rotating. The rotational effects are then treated as fictitious forces: The coriolis force and the centrifugal force. So, to use "coriolis force" and "centripetal force" in the same sentence is to confuse together the two different frames.

 

[A.T.]

I don't think the centrifugal force has much effect on winds since it applies equally outwards at all points and essentially just weakens the effect of gravity.

Really? What does "outwards" mean?

 

[russ_watters]

I simplified way of thinking about it is that in general local winds are caused by temperature/pressure differences and global winds are caused by the Coriolis force.

No, that is not correct. As said previously, temperature and density/pressure variations cause the winds to blow. The coriolis force only affects the direction in which they blow.

 

[olivermsun]

As said previously, temperature and density/pressure variations cause the winds to blow. The coriolis force only affects the direction in which they blow.

But of course motions (the winds) are caused by forces (pressure gradients)!

I wonder if the OP is asking in a sense if and how the winds as we know them are governed by the Coriolis force.

 

[K^2]

Earth surface is a (nearly) equipotential surface. Since centrifugal effect merely adds to effective potential, it makes no contribution to the weather.

The two forces that give you weather are the pressure gradients and Coriolis force.

 

[Order]

By the way, you have to be careful to define what frame you are in. If your frame is a stationary spot in space, watching the Earth spin, then there is no coriolis force or centrifugal force. The global winds are described directly by the rotation of the earth. There is only gravity which is a centripetal force holding everything in orbit and your inertia. If your frame of reference is fixed to the surface of earth, so that you are rotating with the earth, but from your frame you do not appear to be rotating. The rotational effects are then treated as fictitious forces: The coriolis force and the centrifugal force. So, to use "coriolis force" and "centripetal force" in the same sentence is to confuse together the two different frames.

Yes, that is enlightening. Good explanation.

Thinking about it, I think I really meant centrifugal force (rotation of the earth) and not centripetal force (gravity).

 

[Order]

Thanks for a good post.

I'm pretty confused by the post - I think you might have a lot of misconceptions.

Sure I have a lot of misconceptions since I have never studied this seriously.

First, tidal forces have nothing to do with centripetal acceleration. Tidal forces are due to gravity.

Maybe I should start a new post just about this then.

Second, if you're standing still on the earth, unless you're at the poles, there is a centripetal force on you.

Oh, I thought this was exactly balanced by gravity. I better watch out when I am going outside then; I might fall down to the equator. Right? Or all the air might go from where I live and down to Africa. Then it will be really hard to breath. Right? I am just surprised, that’s all.

Third, the reason your equation says the wind moves in circles is because you wrote it that way: at the very beginning you said assume the wind blows from the west to the east, so of course you'll get that they move in circles from west to east. I'm also not sure how you can say that the Coriolis force is obviously constant and the centripetal force is not.

But look at the equations that Bruce posted. To begin with the coriolis effect is given by $$-2m \Omega \times v$$ Here $\Omega$ is perpendicular to v no matter how v moves so it is always constant. Now look at the equation for centrifugal force $$-m \Omega \times ( \Omega \times x)$$ Here $\Omega$ will be larger when something is moving in east west direction than if it is moving in north south direction. The north south components of the equation cancels out effectively as far as I can see.

Here is a picture of global surface winds, taken from the reference frame of a stationary observer fixed to the earth:

http://www.colorado.edu/geography/blanken/GEOG%206181%20Fall%202003/noble/images/06_tradewind_1.jpg

The Earth is hotter at the equator, causing warm air to rise high in the air where it's at greater pressure than colder air at the same height, so the warm air moves north and south of the equator where it cools down and sinks at 30 degrees latitude. The air here is now high pressure at the surface, so winds flow back towards the equator on the surface of the Earth but due to the Coriolis force (which pushes things right in the Northern hemisphere and left in the Southern hemisphere) it gets bent westward. These are the trade winds. Similarly, above 30 degrees you have the Westerlies until you reach the polar front (when the cold air drops at 30 degrees, it can flow north or south, so if it flows north in the northern hemisphere the coriolis force pushes it east).

Obviously it is a lot more complicated than I thought. Unfortunately I don't have time to study meteorology in depth. But it is good to know a little bit about it. Good thing you posted me this.

 

[Order]

But of course motions (the winds) are caused by forces (pressure gradients)!

I wonder if the OP is asking in a sense if and how the winds as we know them are governed by the Coriolis force.

Obviously the Coriolis force does not make the wind move from rest. But just to be clear, my original question was due to the fact that in popular explanations of the global winds one says that centrifugal forces causes wind to bend, which confused me a lot. The consensus on this forum seems to be that this is wrong and that the coriolis effect alone is the cause.

 

[olivermsun]

I guess the way it's being discussed is kind of funny anyhow. If you look at the equations of motion in the rotating (earth) frame, there aren't special terms which distinguish between Coriolis and centrifugal forces. As one earlier posted pointed out, the Earth is close to an equipotential surface, so our "up" which is perpendicular to the ground happens to point the opposite way from the sum gravity + centrifugal (which is outward from the axis of rotation, not the center of the earth!). Therefore, winds which blow along "horizontal" (equipotential surfaces) aren't affected by centrifugal forces, and only Coriolis is left.

But this sounds like a bit of a definitional trick, right? I mean, suppose the wind starts blowing "poleward" from the equator. It pretty much has to bend to stay with the curve of the earth, right? And that curve is essentially defined by gravity + centrifugal...

But yes, if you are talking about "horizontal" cyclonic motions around around an axis sticking straight out of the Earth (wherever you happen to be), then that's Coriolis.

 

[MrNerd]

Alright, I've got my book here now, Essentials of Meteorology, 5th edition. It says, and I quote:

"An object will always accelerate in the direction of the total force acting on it. Therefore, to determine in which direction the wind will blow, we must identify and examine all of the forces that affect the horizontal movement of air. These forces include:
1. pressure gradient force
2. Coriolis force
3. centripetal force(not centrifugal, which is technically imaginary and outward. Centripetal means inward, thus causing the rotation. The outward illusion results from the tendency of you to keep forward, but the object accelerating you directionwise is pushing in on you, thus you push out, thus the illusion of an outward force)
4. friction"
the centrifugal/petal thing was me, not the book, by the way.
Pressuse gradient force is given by (difference in pressure)/distance(essentially the second law of thermodynamics, high to low)
The Coriolis Force/Effect results from the Earth's rotation, also only affecting direction, never speed.
The centripetal force is involved in spinning winds, such as tornadoes and hurricanes.
Friction is self-explanatory(I assume), it slows down the winds(presumably).

 

[RedX]

Thanks for a good post.

Sure I have a lot of misconceptions since I have never studied this seriously.

Maybe I should start a new post just about this then.

For some information on tides, here's a website:

http://www.lhup.edu/~dsimanek/scenario/tides.htm

For some mathematical detail (just algebra):

http://mb-soft.com/public/tides.html

You only need gravity to explain the tides.

But look at the equations that Bruce posted. To begin with the coriolis effect is given by $$-2m \Omega \times v$$ Here $\Omega$ is perpendicular to v no matter how v moves so it is always constant. Now look at the equation for centrifugal force −mΩ×(Ω×x) . Here Ω will be larger when something is moving in east west direction than if it is moving in north south direction. The north south components of the equation cancels out effectively as far as I can see.

Omega is a constant, equal to the angular velocity of the earth. In those two equations, you plug in different x and v to calculate the centrifugal and coriolis force, but omega is fixed at the angular velocity of the earth.

 

[klimatos]

A difference in temperature will also result in differences in pressure. The difference in pressure causes a flow of air from high pressure to low pressure. Thus, wind.

The genesis of winds is an extremely complex field of study. Differences in pressure is the laboratory explanation. In the free atmosphere winds can be caused by Bernoulli effects, vaporization, condensation, differences in temperature, differences in humidity, and--of course--differences in pressure.

Strangely enough, many winds blow from areas of lower pressure toward areas of higher pressure. Gravity winds are a good example. Also any area where air masses are sinking represents air moving from lower pressure toward higher pressure (I'm talking real pressures here, not the fictional pressures obtained by reducing everything to sea level).

 

[BruceW]

Hold on. Is the OP asking about centripetal forces due to the circular movement of air in a tornado or similar? Because in that case, centripetal force is much more important than coriolis force.
In the case of large-scale low pressure systems, it is the coriolis force that dominates.
I think the Rossby number gives an indication of whether coriolis force or local inertia is more important.

But maybe the OP was asking about the centripetal force due to the Earth spinning?

 

[Order]

For some information on tides, here's a website:

http://www.lhup.edu/~dsimanek/scenario/tides.htm

For some mathematical detail (just algebra):

http://mb-soft.com/public/tides.html

You only need gravity to explain the tides.

Ok, good, now I won't make a fool of myself if I try to explain this to someone. The book that explained tidal forces with centrifugal forces was Feynmans lectures on physics part 1. Obviously even a genious such as Feynman can be wrong.

 

[olivermsun]

The book that explained tidal forces with centrifugal forces was Feynmans lectures on physics part 1. Obviously even a genious such as Feynman can be wrong.

What did Feynman actually say in his book, though?

 

[Order]

The cause of tidal forces

What did Feynman actually say in his book, though?

Thanks for asking Oliver. It is in chapter 7-4. First he talks about gravitational pull, but then these lines follows:
What do we mean by "balanced"? What balances? If the moon pulls the whole Earth toward it, why doesn't the eart fall right "up" to the moon? Because the Earth does the same trick as the moon, it goes in a circle around a point which is inside the eart but not at its center. The moon does not just go around the earth, the Earth and the moon both go around a central position, each falling toward this common position, as shown in Fig. 7-5. This motion around the common center is what balances the fall of each. So the Earth is not going in a straight line either; it travels in a circle. The water on the far side is thrown out more by the "centrifugal force" than the average for the center of the earth, which is just balanced by the moon's attraction. The moon's attraction on the far side is weaker and the "centrifugal force" is stronger. The net result is an imbalance of the water in a direction away from the center of the earth. On the near side, the attraction from the moon is stronger, and because the radius vector is shorter, the "centrifugal force" is also weaker and the imbalance is in the opposite direction in space, but again away from the center of the earth. The net result is that we get two tidal bulges.

So he uses quotation marks, which does not make things clear. But the important thing is to prove if there really is an effect such as this. I don't know how to do the actual calculation, but I think that a period of one month would not lead to any significant variation in the height of the sea level. I can be wrong of course, and then Feynman is right.

 

[tiny-tim]

Hi Order!

Winds:

https://www.physicsforums.com/library.php?do=view_item&itemid=84" give you the direction of acceleration of the wind, not the direction of the wind itself.

Centrifugal force will always provide an acceleration directly towards the equator (and partly up also). The magnitude of the acceleration depends only on the latitude, not on the speed of the wind.

Coriolis force will always provide an acceleration perpendicular to the wind (and purely horizontal) … so it makes the wind go in a circle. The magnitude of the acceleration is proportional to the speed of the wind and the sine of the latitude.

Obviously, both these forces are connected with the rotation of the Earth.

Tides:

Feynman is referring to centrifugal force connected with the rotation of the Moon.

An observer standing still on the Moon regards the Earth as stationary, suspended in space.

Applying good ol' Newton's second law, he knows that there are two forces on the Earth, gravitational and centrifugal.

(there's no Coriolis force, because the Earth is stationary! )

They have to be equal and opposite at the centre of the Earth to keep the Earth suspended there!

But closer to the Moon, the Moon's gravitational force becomes stronger while the (Moon-related) centrifugal force becomes weaker.

So the overall Moon-related force is stronger towards the Moon on the Moon side of the Earth, and weaker towards the Moon on the opposite side of the Earth.

 

[olivermsun]

Thanks for asking Oliver. It is in chapter 7-4.

In my cursory reading of the quoted passage, I don't see anything wrong with Feynman's explanation.

On the other hand, I do see a fairly major problem with (or at least a misleading conclusion being drawn from) Simanek's explanation which is being cited in this thread (and in the other, related thread).

 

[olivermsun]

Feynman is referring to centrifugal force connected with the rotation of the Moon.

I believe he was talking about the "centrifugal force" related to the revolution of the Earth (and moon) around the common COM.

 

[K^2]

Centrifugal force will always provide an acceleration directly towards the equator (and partly up also).

No, it will never, ever do that. Centrifugal force does not affect the winds.

 

[JeffKoch]

There is so much wrong in this thread that it's mildly amusing, but Re: centrifugal force, you can calculate it if you like using standard formulas and you'll find that you weigh about 0.35% less at the equator than you do at the pole, neglecting the oblate nature of the Earth's surface (which makes you weight even a bit less at the equator, total variation is around 0.5%).

The centrifugal force is not directed towards the equator, I don't know where this idea comes from but at least two people seem to think this is true.

It's a headshaking moment when someone claims on a physics forum that they are correct and Feynman is wrong. OP might head to Africa where the air has gone, and carefully study what Feynman had to say.

 

[A.T.]

It's a headshaking moment when someone claims on a physics forum that they are correct and Feynman is wrong.

I think those who say "centrifugal force doesn't affect winds" mean only the horizontal component (parallel to surface). Since the Earth in not a perfect sphere, gravity is not acting perpendicular to the surface everywhere. The horizontal component of the centrifugal force is canceled by the horizontal component of gravity, so that the vector sum of gravity & centrifugal force is acting perpendicular to the surface:

The resulting vector g* has less magnitude on the equator than on the poles, so the air is lighter on the equator. But the effect on winds should be tiny, compared to the temperature difference between equator and poles.

 

[K^2]

Centrifugal force doesn't affect winds in any way. Centrifugal force field is conservative and is constant for every point on the planet. Therefore, all it can do is affect the equilibrium pressure. That's even if it did have a component along the surface.

 

[BruceW]

I already gave the equation for the centrifugal force:
$$-m \vec{\Omega} \times ( \vec{\Omega} \times \vec{r} )$$
(I've written it with r instead of x in case that was causing confusion).
And it can be re-written as:
$$m \Omega^2 ( \vec{r} - \hat{z} ( \hat{z} \cdot \vec{r} ) )$$
So the magnitude depends on how far you are in the z direction. For example, on the north pole, the force is equal to zero. And on the equator, the force is maximum.

 

[BruceW]

In fact, the magnitude is $m \Omega^2 r sin(\theta)$
And the radial component is:
$$m \Omega^2 r sin^2(\theta) \hat{e_r}$$
and the component along the surface is:
$$m \Omega^2 r \frac{1}{2} sin(2 \theta) \hat{e_{\theta}}$$

Edit: to be clear, $\hat{e_r}$ and $\hat{e_{\theta}}$ are the radial and polar angle unit vectors.

 

[A.T.]

Centrifugal force doesn't affect winds in any way. Centrifugal force field is conservative and is constant for every point on the planet. Therefore, all it can do is affect the equilibrium pressure.

So the equilibrium air pressure at the same temperature at the equator is less than at the pole.

That's even if it did have a component along the surface.

What do you mean by "if"? Aside from poles and equator the centrifugal force does have a component along the surface. But that component is exactly canceled by the gravity component along the surface. See diagram:
http://paoc.mit.edu/labguide/images_new/solid4.jpg