# Why does the Earth have an equatorial bulge?

1. Dec 4, 2013

### jouncey

I understand there are two factors: the gravitational forces from the sun and the moon and rotation. Rotation has a greater effect than the gravitational forces, so rotation is what I'd like to focus on.
When the Earth spins, a centripetal force acts on the Earth's crust and points to the core (centripetal force points inward), right? But that seems to contradict the fact that the Earth bulges at the equator. Sources online day that centripetal force cancels out the force of gravity but wouldn't they be pointing the same way...?

2. Dec 4, 2013

### xAxis

I doubt any online source says that. Even the unreliable source won't claim that centripetal force cancels out the gravity, because centripetal force is in the same direction as gravity.
You should concentrate on the study of rotational motion. Hyperphysics web site has good articles on centrifugal and centripetal forces.

3. Dec 4, 2013

### xAxis

Sorry, I see now that you know that these two forces act in the same direction and naturally it confuses you. Can you cite any source that says that they cancel out?

4. Dec 4, 2013

### Drakkith

Staff Emeritus
Gravity IS the centripetal force. The bulge is caused by centrifugal force.

5. Dec 4, 2013

### Bandersnatch

Hi, jouncey. Welcome to PF!

It might be handy to realise that the centripetal force is not a fundamental force of nature(of which exist four: gravity, electromagnetic, weak, and strong). There is always some other force providing or functioning as, the centripetal force.
For example, it could be the tension of a string when you swing a ball on a tether(i.e., electromagnetic interactions between the string's molecules), it could be gravity in many other cases such as orbits and the one in question.

So then, imagine two points on the surface of the Earth - one somewhere on the equator, and the other on one of the poles. For now, let's imagine they're both equidistant from the centre of perfectly spherical Earth.

Both points experience the same force, of equal magnitude, pulling them inward - the force of gravity. Both are pushed back by the reaction force of the Earth's bulk(again, electormagnetic interactions, fundamentally), which gets stronger the more the material gets compressed, until an equlibrium state emerges, with both forces equal to each other, and the point on the pole stationary.

However, the point on the equator also has some non-zero tangential velocity due to Earth's rotation.

According to Newton's laws, in the absence of any force, a mass will either stay at rest or continue moving with constant velocity. If there is a force, then it will act to accelerate the mass.

If not for gravity, the point on the equator would fly away into space in a straight line. With gravity, that point is accelerated inwards.

But since some of the force of gravity is "wasted" on accelerating the point inwards, there's less of it left to compress the material below. That's why, as compared to the poles, the points on the equator reach equilibrium further away from the planet's centre.

6. Dec 4, 2013

### xAxis

Bandersnatch, I don't think gravity gets wasted. Equatorial bulge is simply there because the radius of rotation increases when you move from poles to the equator and there is more mass.

7. Dec 4, 2013

### Bandersnatch

Can you elaborate, xAxis? I didn't get what you mean.

8. Dec 4, 2013

### xAxis

Centrifugal force is proportional to mass and speed squared, and reciprocally proportional to radius r. Points close to poles have small radius, but also small speed and mass. Approaching equator, the speed and radius increase linearly, so the centrifugal force is greater there because of the squared speed. Also, there is more mass rotating further from the axis of rotation in the equator than on poles

9. Dec 4, 2013

### Bandersnatch

Thanks.

The question asked why was there more mass around the equator(the bulge), so I don't think that simply restating the fact helps solve the OP's dillema.

My post was using an inertial frame, as that's what the OP appears to have chosen. I'm not sure if introducing non-inertial frame and centrifugal force will help here or just confuse more.

You're right that in a non-inertial, rotating frame of reference, there will be two forces: gravity, and centrifugal(disregarding Coriolis). The net force compresses the Earth's bulk, and since the centrifugal force varies from maximum at the equator to 0 on the poles, the globe is compressed the most on the poles.

In an inertial frame there are no centrifugal forces, so we need to describe the situation only with what we do have - namely gravity.
Perhaps writing "wasted" earlier was a poor word choice on my part. Of course it's not wasted in the sense of being somehow less than it should according to Newton's law of gravity. It's wasted in the sense that part of it goes towards curving the path of the rotating object, and only the remainder compresses the material underneath. So if you want to figure out with how much force an object will act on the Earth's surface, you can write $F_{net}=F_G-F_{centripetal}$, even though it is FG that provides Fcentripetal. The centripetal force here is simply the amount of force that goes toward curving the path of the object.
In the extreme case of $F_{centripetal}=F_G$ we'd have 0 net force, and the object would be in free-fall in a circular orbit near the surface. It would exert no force on other objects beneath it, so it could not compress the Earth, and whatever was below it would be able to relax its shape.

10. Dec 4, 2013

### xAxis

I think this is typical example where using rotating frame of reference explains the phenomenon better. Besides, why do you think that how much force an object will act on the Earth's surface is important, and what is Fcentripetal in that formula?

11. Dec 4, 2013

### Staff: Mentor

The centripetal force is provided by gravity. If some of the force has to be used to keep you from flying off the earth, it has to be subtracted from the total to get what a scale reads.