# Does gravity cause the Earth's equatorial bulge?

The way I understand centrifugal force, I don't see how Earth's daily rotation alone would cause equatorial bulges to form. The usual explanation is that the centrifugal force increases with distance from the rotation axis, and since this distance varies from the poles to the equator, the equator experiences a greater centrifugal force and bulges. My understanding of centrifugal force is a little different. Consider two observers, $$1$$ who is on a space shuttle in orbit around the Earth, and $$2$$ who is stationary relative to the distant stars. From $$1$$'s perspective, it is not the shuttle that is orbiting the Earth, but instead the Earth that is orbiting the shuttle. $$2$$ sees gravity providing the centripetal acceleration required for the shuttle's orbit in the $$1$$-to-Earth direction, while $$1$$ sees a pseudo centrifugal force providing the centripetal acceleration required for the Earth's orbit in the Earth-to-$$1$$ direction. The two always point in opposite directions (they are 180° out of phase), just like any other pseudo force. If $$1$$'s orbital radius was bigger (but the angular rate was the same), then $$1$$ would indeed see a greater centrifugal force, but only to the extent that the Earth's orbit "balances" to make up for its increased orbital circumference. The point is that $$1$$ would not expect the Earth to stretch in any way due to a greater centrifugal force, only for it as a whole to accelerate faster to make up for the fact that they themselves are accelerating faster because their orbital radius is bigger. If we pretend $$1$$ was a small volume of mass at the equator, then the same argument applies and $$1$$ should not expect the rest of the Earth to stretch, only to accelerate fast enough to balance its orbit around $$1$$.

If the argument is to make an analogy to the radially-outward force you feel when you are on a spinning amusement park ride (such as Night Mares at Canada's Wonderland), then my understanding is that the force you feel is the force between your feet, which are connected to the ride by friction, and the rest of your body, which is only dragged along by your feet.

The only way I can see bulges forming under rotation is by the fact that gravity points towards the Earth's centre, not its axis of rotation. Thus, gravity is not purely perpendicular to the axis, but has a parallel-to-axis component which makes up 0% of the total at the equator and 100% of the total at the poles. Because of the fact that Earth is spinning, it does not compress under its own gravity, but this is only true at the equator; there is no centripetal force at the poles, so the poles compress under gravity, and the excess mass finds its way to the equator and bulges.

Is this the correct line of reasoning, that gravity along with rotation causes the equatorial bulge?

(I'm sorry if this makes no sense at all!)

while 1 sees a pseudo centrifugal force providing the centripetal acceleration required for the Earth's orbit in the Earth-to-1 direction

This doesn’t work. The fictitious centrifugal force (as its name suggests) accelerates things outward, not inward. It cannot provide a centripetal acceleration. The force that provides the centripetal acceleration in the non-inertial frame is the Coriolis force.

If 1's orbital radius was bigger (but the angular rate was the same), then 1 would indeed see a greater centrifugal force, but only to the extent that the Earth's orbit "balances" to make up for its increased orbital circumference.

There is no need to introduce any additional balancing. The standard centrifugal and Coriolis forces explain the motion.

The point is that 1 would not expect the Earth to stretch in any way due to a greater centrifugal force, only for it as a whole to accelerate faster to make up for the fact that they themselves are accelerating faster because their orbital radius is bigger.

This is not correct. Again, you are missing the Coriolis force. Suppose, for simplicity, that you have fixed 1’s orbit as a geosynchronous equatorial orbit. Then the combination of the centrifugal and Coriolis forces can be treated as a potential, and this potential has the same form as the centrifugal potential in the Earth’s frame. It this leads to the same equatorial bulge.

• The idea was that flipping the direction of the centripetal force vector from 1 to the Earth at any point in time would create a new vector that is centrifugal to Earth (and hence the rotation axis) but centripetal to 1. This would be the exact acceleration needed to describe Earth's rotation from 1's perspective, considering 1 always sees the Earth 180° ahead of where the Earth sees 1. I'm still puzzled as to where I'm going wrong.
– arz
Sep 10, 2022 at 3:31
• @arz I told you where you are going wrong. From 1’s perspective, where the earth is orbiting around 1, the centrifugal force is outward and it is the Coriolis force that is inward. Your reasoning in the question and your reasoning in the comment is incorrect both in identifying the direction of the centrifugal force and in neglecting the Coriolis force. If you correct both of those errors then you get the usual potential
– Dale
Sep 10, 2022 at 4:41

There is a analogon for the Earth's equatorial bulge that is helpful for understanding what is going on: the dynamics of a liquid mirror telescope.

For simplicity take the case of a flat disk with a rim, holding liquid.

The friction between the surface of the disk and the liquid transfers velocity from the disk to the liquid. As described by Newton's first law, objects in motion tend to continue moving in a straight line. So as the disk is spun up liquid tends to start moving to the rim.

When the disk has reached the intended angular velocity, and this velocity is maintained: the liquid will then settle into a dynamic equilibrium.

The cross section of the surface is in the shape of a parabola then.

As we know, the amount of centripetal force required to sustain circumnavigating motion along a circle is proportional to the distance to the center of rotation.

A parabola has the following geometric property: the slope of that curve is proportional to the distance to the midline.

The surface of a rotating liquid settles on a parabolic shape because with that surface profile we have the following property: at each distance to the center of rotation the slope of the surface is such that the required amount of centripetal force is provided.

Close to the perimeter the liquid is at a larger height than at the center. For a given angular velocity there is a corresponding height to which the liquid will climb. It cannot climb higher. Climbing higher would make the slope of the surface steeper, which would increase the amount of centripetal force, and that surplus of centripetal force would then bring the height down again. Conversely, when the liquid is not yet high enough there is insufficient centripetal force, and then the radius of circumnavigation increases.

In the case of the Earth:
As the solar system formed the Earth started as a protoplanetary disk, all loose objects.

As the Proto-Earth formed gravity made everything contract towards spherical shape. This process of contraction continued until a dynamic equilibrium was reached.

When a spinning ellipsoid contracts towards a more spherical shape there is conversion of potential energy to kinetic energy: there is an increase of angular velocity of the contracting system.

The process of contraction of the spinning system stops at the point where the amount gravitational potential energy that is released is equal to the amount of kinetic energy that is gained. The contraction does not proceed beyond that point, because that would require more potential energy than is available for it.

There are two obvious ways you can analyze the situation:

• From an inertial frame of reference in which the the center of the Earth is stationary
• From an accelerated frame of reference in which the surface of the Earth is stationary

Let's take the inertial frame first.

If you are standing on the Earth or floating in a boat, once per day you travel in a circle perpendicular to the Earth's axis. There are two forces acting on you: gravity and the reaction force of the land or water. These two forces add up to a centripetal acceleration that moves you in the circle.

If you are standing at the pole, there is no centripetal force. Gravity and the reaction from the ground exactly cancel. You have no net acceleration. You are stationary in this inertial frame (aside from rotating on your own axis once per day.)

If you are standing on the equator, gravity and the reaction force almost cancel. They add up to a small centripetal acceleration toward the center of the Earth. The reaction force is a tiny bit less than gravity.

$$a_{centripetal} = \omega^2r = \bigg(\frac{2 \pi}{T}\bigg)^2r = \frac{4 \pi^2}{(86400 \space sec)^2}(6.4 \cdot 10^6 \space m) = 0.034 \space m/s^2 = 0.0034 \space g$$

The reaction force is important. It determines

• How much you weigh
• Which direction is vertical, and which are horizontal.

If you fall off a building, you are weightless. The weight you feel is caused by the ground pushing up against your feet. If you put a bathroom scale between the ground and your feet, the ground pushes it against your feet. This compresses a spring in the scale. The amount of compression measures your weight. You weigh less on the equator than at the pole.

To balance a stick on the ground, you must exactly align the stick parallel to the reaction force. When the stick is balanced, we say it is vertical.

Likewise a bowl of water will have a level, horizontal surface. If any point is higher than another, water fill flow downhill until all is level. A horizontal surface is perpendicular to the vertical.

Now lets look at a point between the pole and the equator. Any will do, be let's choose one at a latitude of $$45^o$$.

Gravity and the reaction force almost cancel, leaving a tiny centripetal force toward the axis of rotation. This is at $$45^o$$ to the direction of gravity. The reaction force is not quite anti-parallel to gravity. It tilts slightly toward the axis of rotation. The reaction force is vertical. Horizontal is perpendicular to the reaction force. The surface of the ocean is horizontal. So the surface of the ocean does not follow a spherical shape.

Instead, the ocean surface everywhere is slightly tilted compared to a sphere. As you travel toward the pole, you get farther below a spherical surface. The ocean has an equatorial bulge.

The Earth is also fluid on a geologic time scale. The center is molten and can flow. But even solid rock flows over millions of years. The Earth slowly flows and maintains an equatorial bulge too. This explanation was from an inertial frame of reference. No pseudo forces were needed.

But you can use an accelerated frame. If you are motionless in an accelerated frame, it is because you are accelerated with respect to an inertial frame and there is a force providing that acceleration. But you measure your acceleration with respect to your own frame. In your frame, you are motionless, not accelerated.

This sounds like nonsense, but it is often a very useful thing to do. Here you are sitting on the Earth as it rotates. A while later, you are sitting in the same place. You haven't moved. A car drives by you at a uniform velocity. It is easy to describe the car's motion with respect to you. It would be much harder from an inertial frame, where you are moving one way and the care is moving somewhat differently.

It does get confusing when you talk about forces. $$F = ma$$ is a very useful law. People want to use it in an accelerated frame. But how can you do that if you and the car have a true inertial acceleration of a, and you are using a frame where it is measured as $$0$$?

There is a way. There is a force that makes you match the acceleration of the frame. So you add it in to the law. You add a pseudo force to $$F = ma$$

$$F = ma_{measured} + F_{pseudo}$$

In this case, the pseudo force is call centrifugal force.

Names like pseudo force, fictitious force, and centrifugal force are unfortunate. They make you think that these forces are not real. They make it harder to understand that they are a different way of keeping track of ordinary forces like centripetal forces.

For a detailed example of how this works, see Coriolis Force: Direction Perpendicular to Rotation Axis Visualization

• Thank you so much for the detailed response, this is what I was looking for.
– arz
Sep 10, 2022 at 21:27
• In answer like this (comparing inertial and accelerating perspective) I think it is worthwhile to point out the following: in the case of a rotating coordinate system the centrifugal and coriolis vector are functions of the angular velocity of the rotating coordinate system wrt to the inertial coordinate system. That is: the ability to use a rotating coordinate system is dependent on using the equivalence class of inertial coordinate systems as underlying reference. This is fundamental: to have theory of motion at all: for reference of acceleration one cannot not use inertial coordinates. Sep 17, 2022 at 10:04
• @Cleonis - You are right. Position and velocity are relative, but acceleration is absolute. That is in one inertial frame an object can have a velocity of $0$ and keep on occupying the same position . In another, it can have a non-$0$ velocity and keep changing the point it occupies. But if it is accelerated in one inertial frame, it is accelerated in all of them. Sep 17, 2022 at 15:03
• The nature of acceleration is, in my opinion, one of the most interesting aspects of theory of motion. It raises a question: in mathematical physics, what does it mean to take a derivative? In mathematics the operation of taking a derivative cannot add information. In mathematics, whatever information content the derivative represents is inherited from the entity that it is the derivative of. But in theory of motion we have that position and velocity are relative quantities, and acceleration, while being the time derivative of velocity, is absolute. Sep 17, 2022 at 15:51