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

[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

 

[BruceW]

I don't understand why the centrifugal force along the surface of the Earth is canceled by gravity... Gravity is radial..

 

[BruceW]

from a rough calculation, I get the maximum radial component of the centrifugal acceleration to be 0.03 m/s^2. So this would effectively cancel roughly 1/1000 of the effect of gravity for someone standing on the equator compared to someone standing on the north or south pole.

And the max component along the surface due to the centrifugal force is half that.
I'm guessing the reason that we don't notice this slight acceleration is because pressure gradient only needs to change slowly over the surface of the Earth to be able to cancel out the acceleration.

 

[K^2]

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.

Sorry, that's what I meant. I should have said "unbalanced component along the surface". In general, though, the surface doesn't have to be perfectly smooth. (I mean, it's not, if you go down to fine detail.) But just like slope of the mountain doesn't result in a wind always blowing downwards, any leftover projection of the centrifugal force won't result in a wind, either.

I don't understand why the centrifugal force along the surface of the Earth is canceled by gravity... Gravity is radial..

Gravity IS radial, but Earth isn't spherical.

Edit: Proof of this is simple enough. Earth shape attempts to minimize energy locally at every point on the surface. That means that the entire surface has the same potential energy. The potential in question is a sum of gravitational and centrifugal potentials. So the total force, which is the sum of gravitational and centrifugal force, is the gradient in that potential, and the gradient is always perpendicular to the surface of constant potential. So the sum of gravitational and centrifugal forces is always perpendicular to the surface.

 

[BruceW]

I finally see what you're talking about. The Earth isn't spherical, But its shape is such that the total centrifugal+gravitational force along its surface is equal to zero.
My problem was that I was working under the assumption that the Earth is spherical.

 

[Order]

So is Feynman right or wrong...let's see

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.

Maybe I will. But let me first ask about your calculation. I am very confused about all this, so let me at least start somewhere. There is a way to investigate if Feynman is right.

Let us suppose that the Earth is revolving around a point at the surface of the earth. Then the centrifugal acceleration at the other side of the Earth (subtracted with the acceleration of the center of the earth) should be a=r\omega^{2}=2R_{earth}\omega^{2}-R_{earth}\omega^{2}=R_{earth}\omega^{2}=\frac{6,378 \cdot 10^{6} \cdot (2 \pi)^{2}}{(60 \cdot 60 \cdot 24 \cdot 30)^{2}}m/s^{2}=3,7 \cdot 10^{-5}m/s^{2} This calculation might not be correct, but it shows that you should be lighter at a degree 3,7 \cdot 10^{-5}/9,81=3,8 \cdot 10^{-6}.

This can also be compared with the gravitational effect of the moon subtracted by the gravitation on the center of the Earth (from http://mb-soft.com/public/tides.html)which is 1,129 \cdot 10^{-6}m/s^{2}.

There is a difference here by a factor of 30. And all this although the smaller factor is said to be the cause of tidal forces. This leaves me very confused. Is there something wrong with my calculations? And what is the cause of tidal forces? Gravitation or centrifugal forces?

 

[A.T.]

Let us suppose that the Earth is revolving around a point at the surface of the earth. Then the centrifugal acceleration...

Which reference frame are you looking at? It sounds like you use the frame where that revolving point is at rest, but the frame is rotating around it (hence centrifugal forces). This frame is analyzed here:
http://www.vialattea.net/maree/eng/index.htm
Under: Case 3

 

[Order]

Which reference frame are you looking at? It sounds like you use the frame where that revolving point is at rest, but the frame is rotating around it (hence centrifugal forces). This frame is analyzed here:
http://www.vialattea.net/maree/eng/index.htm
Under: Case 3

Thanks for link!

Case 3 is indeed tricky to me. I don't see how the radial terms are canceled in equation 12 and 13. It seems to me the coriolis force just puts the accelerations in the "other" direction.

I can understand case 2 though, that the forces are equal on all points if you choose this referance frame. And if this is so, then I don't understand why Feynman is talking about "centrifugal forces".

 

[olivermsun]

So supposing that everyone agrees that the magnitude of the Earth's rotation around its own axis is not important to the explanation of the tidal bulges (since a uniform equatorial bulge is created), why don't we propose a slightly different setup. Instead of zero rotation in the inertial frame, let's use a model where the Earth rotates at exactly ω, the orbital angular frequency around the common center of mass of the earth-moon system. In this case, the same two points on the Earth face toward and away from the moon at all times and the Earth as a whole does rotate around the COM.Does this change the qualitative description of the system at all?

 

[A.T.]

So supposing that everyone agrees that the magnitude of the Earth's rotation around its own axis is not important to the explanation of the tidal bulges (since a uniform equatorial bulge is created), why don't we propose a slightly different setup. Instead of zero rotation in the inertial frame, let's use a model where the Earth rotates at exactly ω, the orbital angular frequency around the common center of mass of the earth-moon system. In this case, the same two points on the Earth face toward and away from the moon at all times and the Earth as a whole does rotate around the COM.

This scenario is nice, because there a no Coriolis forces in the co-rotating frame. You have to be a careful though when you compare it with the no-rotation case (Earth & Moon falling linearly towards each other). When you consider only the near and far point, you should find greater relative accelerations between the points in the rotating case. But this might be solely due to the uniform equatorial bulge from the spin at ω, and doesn't necessarily mean that the tidal bulges would be higher.

I have not done any math on this myself, so I'm interested what you might find.

 

[olivermsun]

This scenario is nice, because there a no Coriolis forces in the co-rotating frame...I have not done any math on this myself, so I'm interested what you might find.

Well, what you find is, unsurprisingly, the same as every other way of looking at the system (up to a change in reference frame). The argument seems to center around (no pun intended!) the choice of how to decompose the total motion into linear and angular parts, which therefore affects what you believe are parallel fields and "centrifugal" fields, respectively.