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My question is pretty much all in the title. I was always told that our planet is flattened at its poles due to the centrifugal force generated by its own rotation. However I don’t see how centrifugal force can explain anything at all, it not being a real force, but rather a mathematical expedient (together with Coriolis force) to make Newton’s laws of motion hold in a frame of reference that is rotating at constant angular speed (therefore a non-inertial one). Am I missing something? Is centrifugal force a good explanation of this phenomenon? And if not, why is Earth flattened at the poles?

Qmechanic
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Federico
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    If it's the centrifugal force argument that causes problem for you, perhaps look into Newton's bucket experiment. It involves a similar method, but the calculations are a lot easier. – LPZ Mar 21 '23 at 14:24
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    Possible duplicates: Why is the Earth so fat? and links therein. – Qmechanic Mar 21 '23 at 14:33
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    I think the all confusion for you comes from the statement "that centrifugal force is a pseudo-force". That is totally different from characterizing it as non-force. The prefix "pseudo" really means that although this effect is caused by mass inertia the end result behaves effectively like a real force where acceleration and force vectors point toward the same direction. – Markoul11 Mar 21 '23 at 14:46
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    I think people completely misunderstand the "not a real force" part. They take it that centrifugal effect itself is not real, when what is meant is that it's not a force, because basically it's inertia. But the effect is very, very real. Just sit on a rollercoaster once, and you'll see how real it is. – Davor Mar 21 '23 at 21:53
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    @Davor I concur. I prefer the following definition of force: a mutual interaction, such that Newton's third law is applicable. For short: it's a force when there is a third law force pair. Inertia is in a category of its own. Categorizing inertia as a force is not an option; there is no third law force pair. Since inertia is in a category of its own: when something happens due to inertia it should be explicitly attributed to inertia. – Cleonis Mar 21 '23 at 22:15
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    Anecdotal Experiment: Take a sack of oranges or grapefruits with you and have someone drive down an empty highway at 65 mph or so. Drop an orange or grapefruit out the window. When it hits the pavement it will start to roll very fast. As it spins it will flatten at the poles. Would such an illustration help explain centrifugal forces on spherical objects? – zundarz Mar 22 '23 at 12:50
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    Obligatory xkcd: https://xkcd.com/123/ – Apoorv Potnis Mar 22 '23 at 13:28
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    @Cleonis But centrifugal force can be part of Newton's third law pair. Imagine rotating a heavy ball using a long rope. We can imagine the rope applying a centripetal force on the ball and the ball applying the equal and opposite force on the rope because of the centripetal force. I think in this case considering cetnrifugal force to be a real force that balances out the centripetal force is completely reasonable. – vats dimri Mar 22 '23 at 15:04
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    @vatsdimri I disagree that centrifugal and centripetal forces are a third law pair, since they can't even be defined in a common reference frame. In the frame of the person, there is no fictitious force pulling them outward, there is the very real force of tension. What you're describing isn't centrifugal force. In the frame of the ball, there is the completely fictitious force centrifugal pushing the ball outward which has no physical explanation in that frame. – Nuclear Hoagie Mar 22 '23 at 17:34
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    @vatsdimri About the reason why inertia is in a category of its own: we have that inertia acts in opposition to change of velocity. We don't know how inertia does that, but we know this: in order for the opposition to manifest itself at all there must be change of velocity; inertia is inherently an phenomenon that acts in response. The larger the rate of change of velocity the larger the magnitude of the opposition to change. Hence a larger acceleration requires a larger accelerating force. – Cleonis Mar 22 '23 at 19:27
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    You can find in an answer to my own question a detailed calculation showing that the section through a plane containing the axis of rotation has form given by the equation$$x^2+z^2=\frac{R^2}{(1-\varepsilon x^2)^2}$$ – მამუკა ჯიბლაძე Mar 23 '23 at 13:31

9 Answers9

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The description from an interial frame is that the bits of the Earth at the equator are moving faster than those near the poles, so they have more kinetic energy, and thus they are a bit farther away from the center of mass. It's like they are every so slightly trying to escape the gravitational field of the Earth.

But note something important: this explanation and the centrifugal force are the same phenomenon. One is not "fake" and the other "real". Both are exactly as valid as the other, and not only that: they are different ways of saying the same thing. And this is what happens every time inertial forces are involved: what looks like inertia from one frame looks like a force from another. "Real" is a very tricky word in physics.

Javier
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    Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Buzz Mar 23 '23 at 21:04
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(modified ans after corrective comment)

The ground at the equator is moving fast around a circle as the Earth rotates. So there has to be a force on it, directed inwards towards the axis of rotation. Gravity can provide this force, and so can elastic deformation forces in the rocks of the Earth.

For elastic deformation forces, the object in question (here the material making up the Earth) has to be deformed out of its equilibrium shape. If it is squeezed it will push back out and if it is stretched it will pull back in. This certainly happens for spinning objects, but in the case of planets another fact comes into play: the timescale is so long that the planet can slowly deform until its equilibrium shape is no longer a sphere. It eventually settles to a shape such that the surface is perpendicular to the local effective gravity at each point, where by effective gravity I mean the combination $$ m{\bf g} - m {\bf a} $$ where to good approximation ${\bf g}$ is in the radial direction towards the Earth's centre, and ${\bf a}$ is directed towards the axis of rotation. Since this combination is not exactly in the radial direction at most places, the Earth's surface is not a sphere.

In the above I did not need the idea of centrifugal force so I did not mention it. But one can always do the analysis in a frame of reference which is itself moving in a circle. It takes some technical expertise to do the analysis that way, so I would not recommend that approach to a beginner. But in the accelerating reference frame (i.e. the one going around in a circle) the $m{\bf a}$ term is an inertial force in the outwards direction (a centrifugal force) and this force is combined with the gravitational one, with the net result of no acceleration of the ground relative to the rotating frame. But, as I already said, I would not recommend this second type of explanation to a learner.

Andrew Steane
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    Except that the rotating frame is directly relatable to real world experience. It's physical. That it's (perhaps) more difficult using symbols on a whiteboard shows the limitations of mathematics. – John Doty Mar 21 '23 at 14:20
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    This is - I think - not correct. The earth doesn't bulge like a bouncy ball would if it were spinning. It is not elastic forces that at all that cause the bulge - and the force of gravity is entirely responsible for keeping the spinning mass at the equator from flying away, not "a small contribution". Consider that a planet sized spinning ball of water with no elastic forces whatsoever would have the same equatorial bulge. – AXensen Mar 21 '23 at 17:30
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    @AndrewChristensen thanks for your thoughtful comment. I wrote too hastily and now I think you might be right. I will modify accordingly. – Andrew Steane Mar 21 '23 at 17:47
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    I'm guessing that in the back of your mind is/was a tabletop demonstration of equatorial bulge. We have of course that with a tabletop demonstration the device starts with zero angular velocity, and then it is demonstrated how the device deforms as you spin it up. The Earth, on the other hand, started as a proto-planetary disc, so it arrived at the equilibrium state coming from a configuration that was (way) more flattened than the final equilibrium state. – Cleonis Mar 21 '23 at 18:32
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    I disagree with the end of the first paragraph - elastic deformation doesn't provide any of the inward-pulling centripetal force that keeps the earth spinning. The earth is effectively a loose pile of gravel held together entirely by gravity, not a solid rock. Elastic forces aren't at work at all here. Not sure what you mean with "the planet can slowly deform until its equilibrium shape is no longer a sphere" - equilibrium states are by definition unchanging. The earth was never a sphere, the equilibrium state was always an oblate spheroid. – Nuclear Hoagie Mar 22 '23 at 15:26
  • @NuclearHoagie The rocks (more generally, the whole interior whether it can be called rock or not) are in compression and this compression force certainly squeezes them. But at the equator the compression is less than it would be for a non-spinning Earth. This adds to the result one would get on a model of incompressible gravel. – Andrew Steane Mar 22 '23 at 19:00
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    @AndrewSteane I think you are describing the normal force as an elastic force which I suppose is correct in a way, but the bulk elastic deformation properties of rock are entirely irrelevant. All large spinning cosmic bodies are oblate spheroids, including gas giants and stars. You can compress a gas and have the pressure normal force increase, but it doesn't really make sense to talk about elastic deformation of a fluid. Oblateness doesn't require an elastic material. – Nuclear Hoagie Mar 22 '23 at 19:51
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Centrifugal acceleration is a real acceleration. It is an acceleration term which appears when converting Netwon's laws of motion from an inertial frame into a rotating one -- the difference between standing on the ground in the middle of a carousel and standing on the carousel while it whirls around. When you do this mathematical transformation, an acceleration appears, pointing outwards (centrifugal - center fleeing). This is as real as any other thing one might model.

We say centrifugal force is a fictitious force because it's actually just an acceleration multiplied by the mass of an object. If you model things as-if there was a centrifugal force you get the right answer. Pedantically we claim it isn't a force, because it doesn't have other properties associated with forces. For instance, there is no equal and opposite reaction to it. There can't be, because there's no other object applying the force.

Intuitively one might think the Earth is the other object for Netwon's third law, but that's just because we chose an unusually convenient rotating frame. You can define a rotating frame around any arbitrary point in space and the laws of physics will still work.

Regardless, you can either solve the problem in a rotating frame, in which case there is a centrifugal force-ish thing which is opposing gravity more near the equators near the poles. Or you can solve the problem in an inertial frame, in which case we find the particles on the surface are moving rapidly enough near the equator to naturally move outward. Either view is valid, its just a question of which frame you find the most convenient to operate in. Just make sure that, if you treat this centrifugal effect as a force, you don't book-keep too carefully... for you'll never find the equal and opposite force associated with it. There is none.

Cort Ammon
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    We have: in the equation of motion for a rotating coordinate system the centrifugal term and the coriolis term are added. The magnitude of the centrifugal/coriolis term depends on the angular velocity of the rotating coordinate system with respect to the inertial coordinate system. That is: the act of using a rotating coordinate system is dependent on an prior established reference: the inertial coordinate system. If one doesn't use the inertial coordinate system as underlying reference one doesn't have a theory of motion. This shows that the assertion of equal validity is untenable. – Cleonis Mar 21 '23 at 21:14
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    @Cleonis By "equally valid" I mean "solve the equations in whichever system you like, because both will yield the same motion." If we wanted to get deep into the philosophy of things, I'd say that there is an isomorphism between the inertial equations of motion and the rotating equations of motion. Inertial just happens to be a description of the frames used in Newton's three laws. One could have come up with a theory of motion that operates in a very particular rotating frame, and it would have been tenable. It's just not what Newton proposed. – Cort Ammon Mar 21 '23 at 21:19
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    Well, the question is about attributing causation. Of course: attributing causation is a judgment call; there is no such thing as a mathematical proof for a particular attribution of causation. In the case of inertia the evidence is abundant. I judge the case to be proved beyond reasonable doubt. For the solar system there is one coordinate system such that the interacting celestial bodies move relative to it so that their motion is described by an inverse square law of gravity: the inertial coordinate system. The equivalence class of inertial coordinate systems is preferred reference. – Cleonis Mar 21 '23 at 21:42
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    I don't think the notion that rotation is necessary is correct. You get the centrifugal term immediately when transforming into spherical coordinates, which are non-inertial. However that does not imply rotation per se. That's only added on top later, when we use said spherical coordinates to describe motions on spinning spheres. – AtmosphericPrisonEscape Mar 22 '23 at 09:02
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    @AtmosphericPrisonEscape How does transforming into spherical coordinates yield a centrifugal term, unless said coordinates are rotating? Typically one thinks of centrifugal acceleration as a product of a frame transformation (from inertial to rotating), which can actually be explained and even calculated in a coordinate-free manner as the positions and rotations can all be described with vectors, not just coordinate vectors. – Cort Ammon Mar 22 '23 at 17:28
  • @CortAmmon The centrifugal term is the second term in $\dot{\vec r} = \dot{r} \vec e_{r} + r \dot{\phi} \vec{e}_{\phi}$. That just comes out of the definitions of geometry, and doesn't involve frame acceleration. True, for some variants of this derivation you involve the 'dot', but that's just a helper. – AtmosphericPrisonEscape Mar 22 '23 at 18:04
  • @AtmosphericPrisonEscape That formulation is new to me. I'd love to see an answer expanding on it. Every time I have come across centripetal/Coriolis effects it has always been rotating-frame dependent and coordinate-system agnostic. It'd be interesting to see a centripetal effect that is frame-agnostic and coordinate-system dependent. – Cort Ammon Mar 22 '23 at 22:17
  • @CortAmmon I'd be doing nothing else than setting $x e_x + y e_y + z e_z = r e_r$, and using $x=rsin(\theta)cos(\phi)$ etc. to obtain the relations $\dot{e_r} =\dot{\phi} e_{\phi}$, from which follows the above statement for $\dot{r} = \dot{(r e_{r})}$. Then, $\dot{\phi}$ is just the general coordinate velocity. Coriolis comes into this after we choose to rotate at a certain $\Omega \equiv \dot{\Phi} \equiv \dot{\phi} - \Delta\dot{\phi}$. This procedure also correctly reproduces the quadratic/nonlinear terms in the equation of motion for $\ddot{r}$. What procedure were you thinking of? – AtmosphericPrisonEscape Mar 22 '23 at 22:47
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Take a rock of mass $m$ on the equator of the earth. Imagine it stops rotating with the rest of the earth, stops being affected by gravity, and just continues on with its current velocity. It will move away from the earth - it will start at a radius R (radius of the earth) and move away like $r(t)=R+t^2R\omega^2/2$ (where $\omega$ is the angular velocity of the earth). To counteract this acceleration, you need to apply a force to the rock $mR\omega^2/2$.

Let's start with the right framework - the surface of a planet is an "equipotential" - meaning that the the gravitational potential energy of something sitting on the equator is the same as the gravitational potential of something sitting on the poles. This ensures that "downhill" isn't in any particular direction - if the poles were "downhill" of the equator, mass would move from the equator to the poles to fix that (especially in a gas giant, where mass is free to move in any direction). So we'll add the gravitational potential to the centrifugal potential and find that the new equipotential surfaces are wider at the equator. And you can see this yourself. Let your arms relax then spin really fast. Your arms will move away from your body for the same reason - and they move away until they're in equilibrium (gravity pulling them down balances the centrifugal force pulling them out).

With nothing else around, a body will be spherical, and the potential at the surface will be $-GM/R$. If it spins, we add our first most significant correction, the centrifugal potential $-\omega^2r^2\cos^2(\theta)/2$, where $\theta$ is the latitude and $\omega$ is the angular velocity of the rotation. This (small) perturbation makes the planet slightly wider at the equator than at the poles. To see how, let $R$ be the radius of the planet at the poles, then the potential of the surface is $-GM/R$. The potential of the gravity of the planet is still $-GM/r$, where $r(\theta)$ will be the radius of the planet as a function of latitude. Then we need to find $r(\theta)$ such that $$-GM/r-\omega^2r^2\cos^2(\theta)/2=GM/R$$ Assuming $r=R+\delta(\theta)$ (the correction is small) we get $$GM\delta/R^2-\omega^2r^2\cos^2(\theta)/2=0$$ $$GM\delta/R^2-\omega^2(R^2+2R\delta)\cos^2(\theta)/2=0$$ $$\delta=\frac{\omega^2R^2\cos^2(\theta)}{2(GM/R^2-2\omega^2R\cos^2(\theta))}$$

NOTE: this calculation is off by a factor of two due to the effect described here Why is the Earth so fat?

JosephDoggie
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AXensen
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    The word longitude is used but is it really latitude that is needed in the above. I.e., the equator is always 0 degrees latitude, but has every value of longitude (for instance, meridian, etc.) – JosephDoggie Mar 24 '23 at 20:27
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    Ha I got mixed up by a diagram that was showing "lines of constant longitude" and not "the direction that longitude changes in" – AXensen Mar 25 '23 at 10:04
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To understand why the Earth's rotation makes it shape non-spherical we can use the following simpler case: a (shallow) cilindrical tank, with a layer of fluid in it, and the tank is spinning.

There is a series of pages, created by John Marshall and R. Alan Plumb

That series of pages accompanies a textbook on atmospheric and oceanic dynamics, written by Marshall and Plumb.

The cilindrical tank demonstration is described on the page Construction of parabolic turntable

So:
A (shallow) cilindrical tank, with fluid in it.
When the students spin up that tank the fluid becomes concave. We all know that of course: when you have a fluid in a cup, and you make it swirl around the surface becomes concave.

Gravity tends to pull the fluid down the slope, and that pull is providing the required centripetal force for the fluid to be in circumnavigating motion.


Another way of thinking about slope:
Look up pictures of the Brooklands racing circuit. Built in 1907, all the corners were banked corners.

Back then they had the engine power to go fast, but the tires on the cars were flimsy, so they couldn't go fast through level corners.

On a circuit with banked corners the slope of the banked corners makes gravity provide the required centripetal force.

The banking is steeper and steeper as you go to the outside of the corner. Depending on the speed of the car the driver can drive the line where gravity provides just the amount of centripetal force that is required.

Assuming the tires had desparately little grip: if the driver steers too far to the inside of the corner for his speed: the slope there is too shallow for the speed, not enough centripetal force: the car, losing grip, will swing wide. If the driver is too far up the banked corner then there is too much centripetal force for the speed, and the car will slump down.

Additional remark about the case of not enough centripetal force being available:
As we know: if no force is acting (or if the forces balance out) then objects move in a straight line, with uniform velocity. If a car is going through a corner, and there isn't enough centripetal force, then it is the car's inertia that makes it not follow the corner


Now to the Earth.

The formation of the planets of the solar system:
Prior to coalescing there were proto-planetary discs.

As the proto-Earth contracted more and more the proto-Earth became more and more spherical in shape.

For any rotating system we have: as the rotating system contracts the angular velocity increases.

The Earth contracted towards spherical shape and that process of contraction came to a halt when an equilibrium was reached.

That equilibrium is not a static equilibrium, but a dynamic equilibrium.

The equatorial bulge means the Earth is not in the shape of lowest possible gravitational potential energy. If there would be opportunity to contract all the way to spherical shape there would be a corresponding release of gravitational potential energy.

During the stage of contracting from proto-planetary disc to planet gravitational potential energy that was released converted to rotational kinetic energy.

As long as the rate of release of gravitational potential energy was larger than the rate of increase of rotational kinetic energy the process of contracting towards spherical shape continued. (The energy that was not converted to kinetic energy converted to heat; a large portion of that heat radiated away.)

The process of contracting towards spherical shape halted at the point were the rate of increase of rotational kinetic energy matched the rate of release of gravitational potential energy.

(Actually, there are estimates that when it was just formed the Earth had a day of about 6 today's hours. Throughout the Earth's history: a process of tidal interaction with the Moon has been decreasing the Earth's rotation rate. The equatorial bulge has decreased in accordance with that; the dynamic equilibrium is a self-adjusting equilibrium.)

Cleonis
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Another way to approach this question is continuous mechanics. All the different materials of the Earth are being compressed by gravity. Due to the size and time of existence of the planet, it is reasonable to suppose that the shear forces can be neglected and we can treat it as a fluid, where only pressure matters. I use Cartesian instead of spherical coordinates to make the argument simpler, so the equations are approximations for a thin shell (say $\frac{1}{100}$ of the Earth's radius). Because it is in equilibrium, a small cubic portion of the Earth near the surface in the poles must have the sum of all forces zero:$$F_r = \rho \Delta V g + F_{r+\Delta r} \implies P_{r+\Delta r}\Delta A - P_r \Delta A + \rho \Delta V g = 0$$ Dividing by $\Delta V$ and letting $\Delta$'s go to zero: $$\frac{\partial P}{\partial r} = -\rho g$$

But a cubic element near the surface of the equator is not in equilibrium due to the rotation, and the net force is not zero, but $F_{net} = ma$. The equation is now: $$P_{r+\Delta r}\Delta A - P_r \Delta A + \rho \Delta V g = \rho \Delta V \omega^2 r$$ Repeating the same approach, we have:

$$\frac{\partial P}{\partial r} = -\rho g + \rho \omega^2 r$$

The pressure at the surface is zero (suppose no atmosphere to simplify). So it increases slowly in the equator (compared with the poles) with depth . Supposing a initial state of a perfect sphere, the material would flow from high pressure poles to low pressure equator. This flow would bulge the equator with time.

Claudio Saspinski
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Centrifugal forces are "fictitious" in the sense that their causes are fictitious, not their effects. Note that no matter what reference frame you view some physical setup, the motion isn't actually different, each observer sees the exact same thing from a different point of view(assuming no weird relativistic effects). The difference is in how they explain the motion - a viewer in an inertial frame can explain motion with centripetal forces which have real physical causes, while a viewer in a non-inertial frame must resort to centrifugal forces which seemingly arise from nowhere.

Consider the familiar case of a passenger in a turning car. When viewed from outside the car in the inertial frame, we see the car apply a centripetal force inward that makes the passenger turn. This force is readily explained from the motion of the car and its interaction with the ground. Now view the scenario from the frame of the passenger - they are spontaneously pressed against the door of the car by a centrifugal force that has no physical explanation in that frame. The centrifugal force is fictitious since its cause is not real, but the passenger certainly feels its effect.

As you can see, whether we choose to define a centrifugal force or not depends entirely on the reference frame. From the reference frame of observers on the earth, centrifugal force does indeed explain the upward push at the equator which causes the bulge. From the reference frame of an observer outside the earth, there is no such thing as a centrifugal force.

Nuclear Hoagie
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    If centrifugal force has no physical cause, then the force of gravity also has no physical cause. – John Doty Mar 24 '23 at 15:23
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    @JohnDoty From the perspective of a passenger in a turning car, what is the origin of the force that pushes them into the wall? Imagine a windowless car heading down a windy road - the person is flung against the walls seemingly randomly, as the centrifugal force varies in a manner unrelated to anything the passenger can observe. Centrifugal forces are merely the bookkeeping needed to use a non-inertial frame and are dependent entirely on the choice of non-inertial frame, not the physical apparatus being observed. – Nuclear Hoagie Mar 24 '23 at 15:36
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    The passenger can observe the car's acceleration relative to a free-fall trajectory. That's no different than the force of gravity, which depends on the choice of frame just as profoundly as other inertial forces. – John Doty Mar 24 '23 at 15:48
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    The choice of an inertial frame is still a choice, mathematically convenient for pedagogy, but no different physically than other choices. – John Doty Mar 24 '23 at 15:51
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    @JohnDoty You lost me - the force of gravity between two objects doesn't depend at all on choice of reference frame. The apparent weight of an object depends on what frame you observe it in. That's kind of my point, that apparent observed forces can take any value depending on your choice of frame. Centrifugal force is an apparent force that can take any value depending on reference frame - certainly there is not one physical cause that explains a force of arbitrary magnitude across different frames. – Nuclear Hoagie Mar 24 '23 at 16:10
  • Weight is what you measure, not a whiteboard theoretical concept. It's measured in the frame of the measuring instrument. – John Doty Mar 24 '23 at 16:14
  • @JohnDoty Right. When in an elevator accelerating upward, you measure your weight as greater than the force of gravity on you. In your reference frame, there is no explanation for your "extra" weight. There is some additional force driving you into the floor which is indistinguishable from extra gravity as far as you can tell, but the physical explanation in reality is not actually gravity. Fictitious pseudoforces do not arise from any kind of physical interaction like gravity, contact, or electromagnetism, and have no observable cause in the frame they're defined (otherwise they'd be real). – Nuclear Hoagie Mar 28 '23 at 17:53
  • Since a force gauge can measure it, it is an actual physical force. "Interactions" are part of a mathematical model. The measurement is fundamental, the model is not, and indeed we have different models we can use here: Newtonian and GR. The physics doesn't care what your model is: the gauge measures force regardless. – John Doty Mar 28 '23 at 18:22
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I think your confusion lies in thinking that as "centrifugal force is not a real force" then it can be ignored.

For example, think of holding a bowling ball in each hand and spin round. Your arms will feel the centrifugal force. It is there and just as important as a "real" force.

colin rowe
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  • That force-in-centrifugal-direction is a force exerted by the bowling ball on your hand. For comparison, in athletics there is the discipline of Hammer throw. You see the athlete and the hammer going around the Common Center of Mass of the two masses; the hammerhead and the athlete. The athlete is exerting a centripetal force upon the hammerhead, the hammerhead is exerting a centripetal force upon the athlete. The athlete's hands are holding on to the hammer handle. The hammer handle exerts a force in centrifugal direction on the athlete's hands, but that is just transfer of force. – Cleonis Apr 01 '23 at 08:21
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The poles are 50km closer to the center of the Earth than the equator. That's why sending rockets closer to the equator is cheaper (it's closer to space)

Apply centrifuge force to something spheric and elastic. You'll see it becoming more oval

Matheus Lacerda Bezerra
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