# Coriolis from our orbit around the Sun

1. Nov 7, 2009

### YellowTaxi

Can we detect any coriolis forces induced from our circular orbit around the Sun ?

We're going in a big circle so we should be able to detect some coriolis if we deviate from that circle - ie if we move around a bit instead of following a perfect circular path.

Considering the Earth is spinning once per day we should be able to detect coriolis effects (forces or accelerations) at any single point on Earth's surface if we use accurate enough accelerometers. :-)

True or false ?

2. Nov 7, 2009

Staff Emeritus
The solar Coriolis force is quite weak - it's proportional to the angular velocity, which is 365 times smaller than that of the earth's rotation.

3. Nov 7, 2009

### Cleonis

Good question.
An instrument located on the equator has a velocity relative to the Earth's center of 465 meters per second.
The angular velocity of Earth in its orbit around the sun is 2*pi radians per year. A year is 31 556 926 seconds, so the orbital angular velocity is in the order of 1/10million radian per second
The magnitude of the Coriolis factor is:

$$2 \Omega v$$

Where $\Omega$ is the system angular velocity, and $v$ is velocity with respect to the rotating system.

The acceleration comes out in the order of 1/10thousand meter per second squared. Modern gravimetric instruments have a sensitivity that is smaller than that, I'm sure. But gravimetric instruments cannot differentiate, they measure a single gravitational acceleration.

- For instance, at the equator some of the Earth's gravitation is "spent" on providing required centripetal force. So just from that alone you expect that at the equator you will register a smaller gravitation than at the poles.
- Then the gravity from the Sun tugs harder or less hard at you depending on whether you are on the part of Earth that is facing the Sun (the point during the Earth day that you are closest to the Sun) or whether you are on the opposite side. These are the tidal effects.

It's only after the above two effects have been theoretically computed and subtracted from the record that you are in a position to recognize the Earth-orbital Coriolis effect. I'm really curious whether in gravimetry the Earth-orbital Coriolis effect is taken into account or considered negligable. I expect it's taken into account. If I find out I will write about that on my website, my website is largely dedicated to the Coriolis effect.

Cleonis
http://www.cleonis.nl

4. Nov 7, 2009

### D H

Staff Emeritus
What coriolis force? From the perspective of what frame of reference? See below.

False.

Some key facts about the coriolis force:
1. It is an inertial, or fictitious force. One cannot build a sensor that directly detects the coriolis force because in a very real sense it is not real.
2. It appears as a result of force-fitting Newton's second law to a rotating frame of reference.
3. Newton's third law does not apply to inertial forces.
4. The coriolis force on an object is proportional to the velocity of that object as observed from the perspective of the rotating frame.

Second question first:
From the perspective of an Earth-fixed frame, the coriolis force at a point on the surface of the Earth is zero because the point has zero velocity in an Earth-fixed frame. The coriolis force only kicks in when some object has a non-zero velocity in the rotating frame. An moving airplane or a packet of air, for example, are subject to the coriolis force. A rock at rest on the surface of the Earth is not.

Regarding the first question:
From the perspective of an Earth-centered, Earth-fixed (ECEF) frame, the Sun is subject to several inertial forces: The inertial force due to the acceleration of the frame's origin, centrifugal force due to the Sun's non-zero distance from the origin, and a coriolis force due to the Sun's non-zero velocity in this frame. There is however no third law interaction pair to these forces.

The Sun does of course exert a force on us. From the perspective of an ECEF frame, the most convenient thing to do is to determine the total force from the Sun at some point in the frame is to combine the acceleration of the frame origin due to the Sun's gravity with the Newtonian gravitational acceleration at the point in question. The result is the tidal gravitational acceleration at that point. This tidal acceleration does not depend on the Sun's velocity. There is no coriolis force from the Sun from the perspective of an ECEF frame.

Instead of the Earth-centered, Earth-fixed frame, consider things from the perspective of a frame rotating with the Earth's mean orbital angular velocity (one rotation per sidereal year) and with origin at the center of the Sun. The Earth would be stationary in this frame if the Earth truly did have a circular orbit about the Sun. The Earth's orbit is not circular. It is an elliptical orbit (e=0.0167). The Earth orbits about its mean position with respect to the Sun. A coriolis term does arise in this frame because the Earth is moving in this frame.

5. Nov 7, 2009

### Cleonis

It seems to me you are focussing on the non-interesting aspects of the question.

Let give an example.
A gravimeter located on the equator registers less gravitational acceleration than one at the poles. Part of that difference is due to the fact that some of the Earth's gravity is "spent" in providing required centripetal force.
The observation in itself, that a gravimeter at the equator registers less gravitational acceleration, is not dependent on the perspective. Depending on the coordinate system that is used the terms in the calculation come out differently, but the physical effect that is at play is always the same physical effect.

If you have measured the Earth's gravitational acceleration at the poles, and you know the Earth rotation rate, then you can predict the Earth's equatorial bulge, predict theoretically the Earth's gravitational potential, and then proceed to find out whether the measurement at the equator matches the prediction.

Something analogous can be done for the Coriolis effect due to the velocity of an object with respect to the Earth's center, for the case of a perfectly circular orbit of the Earth around the Sun. How will it affect the registered acceleration, can you predict it and find out whether the observations match the prediction? I estimate that that's possible.

As you point out, if you push even harder than that the eccentricity of the Earth's orbit comes into view.

Cleonis

6. Nov 7, 2009

### D H

Staff Emeritus
Perspective is in my mind the most important aspect of the question. From the perspective of an inertial frame, there is no such thing as coriolis force, or centrifugal force, or any inertial force. Inertial forces result from pretending that Newton's second law works in a non-inertial frame.

7. Nov 7, 2009

### Cleonis

Let me make sure, and repeat my earlier remark:

The setup
A gravimeter located on the equator registers less gravitational acceleration than one at the poles. Part of that difference is due to the fact that some of the Earth's gravity is "spent" in providing required centripetal force.
The observation in itself, that a gravimeter at the equator registers less gravitational acceleration, is not dependent on the perspective. Depending on the coordinate system that is used the terms in the calculation come out differently, but the physical effect that is at play is always the same physical effect.

The question:
From a physics point of view the question is: why does the accelerometer at the equator register a smaller gravitational acceleration than at the poles?

The answer to that question is well understood, and it does not depend on perspective.

Well, there could be babylonian confusion as to how profound the word 'perspective' is taken. I use the word 'perspective' only to refer to how the calculation is set up, the coordinate system of the calculation. For physical explanation I use as elements the phenomena gravitation and inertia. Some things that have their use in calculations, such as a centrifugal term or a coriolis term, are not elements for physical explanation.

Cleonis

Last edited: Nov 7, 2009
8. Nov 7, 2009

### D H

Staff Emeritus
The answer to that question most certainly does depend on perspective. Look at it this way: From the perspective of an inertial observer there is no such thing as a centrifugal force. An inertial observer cannot invoke centrifugal force to explain this phenomenon.

There is an centrifugal force from the perspective of an Earth-fixed. Per this observer, the time it takes an object to fall from rest a short height h above the surface of the Earth at the equator involves solving for t

$$h = \frac 1 2 (g_e - R_e\omega_e^2)t^2$$

where $$g_e$$ is the gravitational acceleration toward the Earth at the equator and $$- R_e\omega_e^2$$ is the centrifugal acceleration away from the Earth at the equator.

Since there is no centrifugal force from the perspective of an inertial observer, how does an inertial observer explain this same phenomenon? One simple answer arises from the assuming a uniform gravity field but accounting for the fact that that the Earth is not flat. From the perspective of an inertial observer, the object does not start falling from at rest. It instead has a velocity of $$v_h = R_e\omega_e$$ parallel to the surface of the Earth. In the time t it takes the object to fall to the surface of the Earth, the object will have traveled horizontally, to first order, by a distance of $$y = v_h t = R_e\omega_e t$$.

The Earth is not flat. In terms of a uniform gravity field, the point at which the object hits the Earth is a bit lower than the point directly beneath the initial location of the object. How much lower is simple geometry. The angular displacement corresponding to that horizontal travel is $$\theta = \approx y/R_e = \omega_e t$$. The extra vertical distance that the object must fall is $$\Delta h = R_e(1-\cos\theta) \approx 1/2 R_e \theta^2 = 1/2 R_e \omega_e^2 t^2$$. The object falls with vertical acceleration ge, and thus the time taken to fall to the surface of the Earth is given by

$$h+\Delta h = h + \frac 1 2 R_e \omega_e^2 t^2 = \frac 1 2 g_e t^2$$

or

$$h = \frac 1 2 (g_e - R_e \omega_e^2)t^2$$

This is the same result obtained by the Earth-fixed observer, but without invoking centrifugal force.

9. Nov 7, 2009

### Cleonis

What I agree with of course is that the same thing can be calculated in different ways.

The thing is: I'd like to argue against stating things metaphorically when there is no need to do so.

I think it's unpractical to use a phrasing such as "there is a centrifugal force from the perspective of [...]". It is more straightforward to say for instance: "When the motion is mapped in a coordinate system that is co-rotating with the Earth then a centrifugal term is added to the equation of motion."

That is what you actually do when using a co-rotating frame: you add the appropriate terms to the equation of motion. No physicist actually thinks that the centrifugal term in the equation represents a physically real centrifugal force. Adding the centrifugal term represents the fact that a co-rotating coordinate system is used.

For explanation of physics the following elements are admissible:
- The four fundamental interactions of nature (Gravitation, Electromagnetism, strong nuclear force, weak nuclear force)
- Inertia

If we consider only macroscopic physics then the list is of course shorter:
- The two long range fundamental interactions of nature: Gravitation, Electromagnetism
- Inertia.

Example involving EM and inertia:
An electric car that is designed for regenerative braking. When the car's driver switches to braking the energy flow in the drive-train reverses, and the batteries are recharged.

Regenerative braking works because of the existence of inertia. A force is required to reduce the relative velocity between the car and the road to zero. During the braking kinetic energy is converted to chemical potential energy in the batteries.

In the case of the electric car with regenerative braking you can opt to map the motion with respect to the road, or you can map the motion in a coordinate system that is co-moving with the initial velocity of the car. Either way it's the same physics taking place, and it's actually important to emphasize the physics taking place is independent of the choice of coordinate system.

Cleonis

10. Nov 7, 2009

### Cleonis

YellowTaxi, my apologies for hijacking the thread you started.

Recapitulating:
The Earth orbiting the Sun constitutes a rotating system. (The central axis is the axis of the Earth's orbit.) An object located on the equator, co-rotating with the Earth, has a velocity with respect to that rotating system.
That will give rise to a rotation effect, and I have estimated the magnitude of the effect to be in the order of 0.00001 m/s2

That is, if you predict the value of gravitational acceleration theoretically, and you correct for other effects, (such as tidal effects) then in the end you will find a remaining effect, that is accounted for in terms of the spinning-in-orbit effect.

However, I think that in geology the flow of information is the other way round.
The geologist starts with registering local gravitational acceleration, and then all effects known to influence the reading are corrected for. What remains is used as input for subsequent calculations. For instance, the denser the rock right underneath the surface the stronger the local gravitational acceleration. The density of the rock right underneath the measuring spot is inferred from the gravimetric reading. That is what geologists do, gravimetry is used to find out about the rocks underneath the surface.

Cleonis

11. Nov 7, 2009

### D H

Staff Emeritus
You are mixing frames, and because coriolis force is a frame-dependent quantity, your calculation is nonsense.

In a frame with the Sun at the origin and rotating with the Earth's orbital angular velocity, there is indeed a coriolis acceleration at points on the surface of the Earth due to the Earth's rotation about its axis. This coriolis acceleration vanishes in an Earth-fixed frame (e.g., a frame in which a gravimeter is based). The coriolis force is frame dependent. You cannot simply calculate its value in one frame and assume that this calculation applies to all frames.

12. Nov 7, 2009

### Cleonis

What I calculated was the magnitude of the rotation effect.

In this particular case it so happens that the expression for the rotation effect is the same as the expression for the fictitious Coriolis force. But they are in fact distinct concepts. I realize of course I won't be convincing you here and now; I'm aware how much it goes against expectation.

For background I refer to my website.
The applets there show a dual perspective: one panel displays the motion as seen from an inertial point of view, the other panel displays the motion as seen from a co-rotating point of view. The Java applets were created with the http://www.um.es/fem/EjsWiki/index.php/Main/WhatIsEJS?userlang=en"; the mathematical setup is open to inspection.

The rotation-of-Earth-effect that is taken into account in Meteorology is really cool physics, well worth looking into.

Cleonis
http://www.cleonis.nl

Last edited by a moderator: Apr 24, 2017
13. Nov 7, 2009

### D H

Staff Emeritus
What you did was invalid. The coriolis effect as calculated in one frame of reference is not the same as the coriolis effect in another frame of reference. In particular, the coriolis effect due to the Earth's orbit in the case frame of a gravimeter is identically zero.

I fully understand what fictitious forces are. I don't know if the same applies to you given that you did something that is completely invalid.

14. Nov 7, 2009

### Cleonis

I also understand what the concept of fictitious force is, and I don't use it; I don't need it.

You can check out the http://www.compadre.org/osp/" [Broken] site.
If you search on the OSP-site with the URL 'www.cleonis.nl' you'll find the simulations that have been included in the Open Source Physics library.

Cleonis

Last edited by a moderator: May 4, 2017
15. Nov 7, 2009

### D H

Staff Emeritus

16. Nov 7, 2009

### Cleonis

What? That must be the worst babylonian confusion ever.

You must be using the concept 'fictitious force' in a totally different manner than I do. I can't imagine.
There's something about the physics of rotation. Different physicists can have mutually contradicting ideas on how to one should think about it.

Cleonis

17. Nov 7, 2009

### D H

Staff Emeritus
18. Nov 8, 2009

### Cleonis

My point of view is that it's unnecessarily complicated to use the expression 'fictitious force' (or any of those other names for it that are in circulation.)

When inertia is at play then simply call it 'inertia'.

By the way, in the entry in Scientific American David Politzer refers to the everyday observation that if you have stirred a cup of tea, and the tea is slowing down again the tea leaves are gathering at the center of the bottom surface.

K Muldrew reproduced an article by a well known scientist that may have been where Politzer got the example from:
http://www.ucalgary.ca/~kmuldrew/river.html

Let's take a closer look at the tea leaves example.

Initially, as the fluid is actively swirled around, the fluid is in solid body rotation (or close enough to solid body rotation to regard it as so.) When the spoon is lifted the fluid starts to slow down again, due to friction.

Near the bottom the fluid experiences friction both from the side and from the bottom, so there the fluid will slow down most. As the fluid near the bottom is circumnavigating a bit slower than the overall column it needs less centripetal force. However, since the concave shape of the fluid's surface hasn't changed the centripetal force is still the same and because of that there is a surplus of centripetal force on the fluid near the bottom.

The surplus of centripetal force induces convection: near the bottom fluid is pushed to the center, along the walls fluid moves down, at the surface fluid moves out and along the center a column rises. The convection current is very slow, it will pull the tea leaves along the bottom of the tea glass, but it doesn't carry the tea leaves up, and so the leaves gather at the center of the bottom.

Cleonis

Last edited: Nov 8, 2009
19. Nov 8, 2009

### YellowTaxi

I imagined that if it existed at all then it would be difficult to detect. It's why I suggested using a very accurate accelerometer. Do you know if a measurement has ever been attempted ?

btw it's going to depend on instantaneous velocity of the lab on Earth's surface right?
'.' Centre of the Earth is ~fixed in the reference frame, but the lab on the surface certainly isn't fixed. And the direction of velocity is going to rotate once per day. Therefore the coriolis force/acceleration should be rotating too (If it even exists...)

I also wondered if gravity masks the effect entirely, because we don't feel the pull of the Sun do we. Or do we ? :-)

20. Nov 8, 2009

### Cleonis

A planet that is free floating in interstellar space would be free of tidal effects.
Earth has tidal effects, both from the Sun and the Moon, and the contributions from the Sun and the Moon are comparable in strength.
In other words, we are affected by the Sun's gravitational field: it gives rise to tidal effects.

Cleonis