Is Earth really flattened at the poles because of centrifugal force?

Question

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?

Answer 1

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.

Answer 2

(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.

Answer 3

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.

Answer 4

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?

Answer 5

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.)

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

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