Antagonism in earth physics looking to be solved

the Interceptor

Premium
Messages
4,191
Belgium
BEL / GER
Messages
theInterceptor77
Dear GTP members,

I am facing an interesting problem which I was not able to solve yet. Since I know there are some very bright heads here, I am certain you can help me.

This is the problem: there is a contradiction in the following logical chain. I am trying to find that contradiction, and why it is there.

Okay, imagine the planet Earth. We know that the earth has the shape of a sphere. Yet, if you look into the physics behind our home planet some more, you will discover that it actually isn't a perfectly shaped sphere. In reality, Earth is a socalled oblate spheroid. The reason is the rotation, which leads to centrifugal forces pushing the area around the equator outwards.

Regarding the gravity on earth, its intensity is directly dependent on the distance to the center of the Earth. You know this from spaceflight: the further you get away from the Earth's center, the lighter your body becomes, because the Earth's mass loses hold of you. That said, it is a known and understandable fact that the gravity at the equator is slightly lower than at the poles, because due to the shape of the Earth, you are closer to the center and thus experiencing a more intense gravity at the poles. Others say you are lighter at the equator because of the centrifugal force. Whichever way it is, gravity is slightly lower here. Following me so far? Good! 👍

Now, let's talk about the sea. The water on Earth of course also follows gravity. Where there's more gravity, there will be more water. Only few people know (nevertheless it is a fact) that the mean sea level actually isn't perfectly level due to local gravity differences. The water doesn't level out, because it follows gravity. When there's more gravity at the poles compared to the equator, the water will naturally flow away from the equator.

Now, we take a different look at the water level. As long as you stay close to the Earth's surface, there is a simple equation for potential energy:

(potential) E(nergy) = m(ass) * g(ravity) * h(eight)

Put simply, this says that given a constant mass (note the difference between mass and weight. The mass of a body is constant, its weight depends on the gravity), the potential energy solely depends on the local gravity and the height the mass sits at. To visualize that, let's think of a bungee jumper that jumps from a platform that's 30 meters above ground. That jumper will have a different amount of potential energy at the north pole compared to the equator. To reach the same amount of potential energy, he would have to use a higher platform when at the equator, because the local gravity is lower. Still with me?

Now, back to the water. As we know, water tries to engange the lowest possible state of potential energy. That's why it always levels out. Water never has a gradient, because the water higher up would then have more potential energy. Thus, the water higher up will flow towards the water which is lower down until the surface is level. This, local influences aside, does also apply to the seas. Thus, one can say that all seas combined will engage the height of a specific equipotential surface, which is an imaginary level that states that the potential energy on this level is the same at every single place.

However ... coming back to the above equation, you end up with the antagonism I stumbled over. Gravity on Earth says that the water level should be higher at the poles compared to the equator regions. The equation however says that a lower gravity is followed by a larger height to ensure the same amount of potential energy. That means that water levels should be higher where gravity is lower. This contradicts the above statement, and only one of them can be correct. I'm sure there's a major fault in my thinking, but as you probably have noticed, this matter is quite complex.

Can you spot the error?
 
Last edited:
The water is also externally influenced by the gravity of the moon, which is what causes tides. ;)
Water, unlike the land is not fixed and is therefore free to "flow" towards the gravitational attraction of the moon and because the Earth rotates and the moon orbits the Earth the water "follows" the pull of the moon, leading to high and low tides in different areas of the globe at different times.
 
I think you might be missing kinetic energy, or energy or rotation. An object (in this case, the sea) wants to follow a tangent path to the earth's surface. So you have a conflict between the acceleration due to the earth's rotation and the acceleration of gravity.

Also, the difference in gravity between the poles and the equator isn't going to be that much. For problems like these, the equation for gravitational potential energy is going to be V=-g(M1*M2)/r, where g is the gravitational constant, m1 and m2 are the masses of the object and r is the distance apart. Using the V=mgz formula works only because other one is a log curve, but for small distances (interactions on the earth's surface), you can approximate.

But anyway, you're going to see that the difference in V due to distance form the center of the earth is going to be pretty small, so the difference in sea levels due to the acceleration due to the rotation of the earth is going to have a much greater effect on sea levels.
 
Okay, I think I can understand that. Yet, the problem I have is a very basic one: the same physics model says two opposite things, and only one can be right.
 
Okay, I think I can understand that. Yet, the problem I have is a very basic one: the same physics model says two opposite things, and only one can be right.

I don't know if this makes any sense, but what sticks out to me is that gravity is based on the distance from the center of the earth where as potential energy has the variable of height that is relative to the earth's surface.
 
I've put some more thought into this. As I understand it, it boils down to a very general kind of mathematical (or logical) problem:
  1. we know water follows gravity
  2. we know the sea level remains at a certain height (an equipotential surface) equal to said gravity worldwide
  3. we can calculate that said height is higher where gravity is lower (Epot = m*g*h)
This is a contradiction. If water follows gravity to where it is strongest, and gravity defines an equipotential surface which the sea level levels out at, why is that surface and therefore the sea level supposed to be higher where gravity is lower?
 
Are you saying that it doesn't add up that the water would "bulge" because the water "should" spread out evenly over a surface to achieve a uniform gravity?
 
No. I am saying that water follows gravity where it is stronger, such as at the poles of the Earth. Also, water levels out at an equipotential surface. Said surface can be calculated by Epot = m*g*h (at low heights). Since water follows Epot all over the world, it should stick to that equation. The equation however says that when you take a mass to a place with a stronger gravity, the height must be lower to get the same amount of potential energy. That means that the sea level would have to be lower where gravity is stronger, which is contradictory to the fact that water follows gravity and thus is higher where gravity is stronger.
 
Last edited:
Maybe it's not the water's density that is changing, and not the actual level. Water seeks a uniform level, and it does, but the density of the water could be different at the poles and equator because gravity is effecting the water differently.

Let me write that so I understand it better. Your equation says that the water level should be lower at the poles because there is more gravity. But it is not, because water always levels itself out. The reason the water is level is because there is a larger amount at a higher density at the poles in order to make up for the supposed lower level, and at the equator there is a lesser amount at a lower density in order to make up for the supposed higher level.

Good idea, bad idea? I've got no math to go on here, just logic.
 
That said, it is a known and understandable fact that the gravity at the equator is slightly lower than at the poles, because due to the shape of the Earth, you are closer to the center and thus experiencing a more intense gravity at the poles.

Nope. Right conclusion, wrong reason.

You weigh more at the poles because you're not being thrown off the ground by the earth's rotation - something which the acceleration of gravity must counteract. There is actually less "gravity" at the pole because there is less mass under your feet.
 
First major mistake - you are looking at relative terms. Height is relative to the system you are checking.

Followed by the fact that gravity is not consistent X feet from the middle of the Earth, as variable density, plate mechanics, et cetera come into play.

But for this, you've just over simplified things. Water doesn't follow gravity, it simply attempts to achieve the lowest potential energy. Which is exactly what the math you keep throwing back out there is saying.
 
Actually, water does follow gravity, because gravity defines the lowest possible state of potential energy for every place.

http://en.wikipedia.org/wiki/Geoid
The gravity field of the earth is neither perfect nor uniform. A flattened ellipsoid is typically used as the idealized earth, but even if the earth were perfectly spherical, the strength of gravity would not be the same everywhere, because density (and therefore mass) varies throughout the planet. This is due to magma distributions, mountain ranges, deep sea trenches, and so on. If that perfect sphere were then covered in water, the water would not be the same height everywhere. Instead, the water level would be higher or lower depending on the particular strength of gravity in that location.
As it looks right now, the problem is that F=m*g*h can not be used in this case. Also, there are a lot of factors to be considered when examining Earth, yet, the problem I stated does also exist in a much more basic model of a planet.
 
/refrain from facepalming.

Water flows for lowest potential energy.

If gravity is stronger in a certain place, to maintain the same energy across the globe, it will closer to the core of the earth, or "less deep."

The water will then move to various spots to make sure the energy remains distributed evenly, not follow gravity. If follows energy, to the lowest potential state.

If water followed gravity like you keep suggesting, all the water would pool at the poles and leave the equator dry. It doesn't, it just seems to follow gravity on a smaller scale because it is moving to the lowest potential energy. On a larger scale, you cannot simply to such an extreme.

PE = m * g * h is still very much valid. While G is a variable based on location, H is fixed as the distance from the center of mass of the Earth, and M is the mass of whatever you are checking.
 
Last edited:
Cody, your equation there shouldn't be force. Force due to gravity is F=m*a, the equation you have there is for potential energy.

EDIT: And the water is going to follow the net acceleration, which in this case is directly down (due to gravity), and a little toward the equator and "out" (due to the rotation of the earth and Newton's first law of motion). Therefore, the water "follows" this acceleration to the point it is now with a bulge at the equator.
 
Last edited:
If the effective downward acceleration at the equator is less than the poles (which it is, due to the larger gravity being canceled out by the rotational acceleration of the earth), then you would expect a larger height of water at the equator given a constant potential energy.

g goes down at the equator, h must go up to compensate.
 
/refrain from facepalming.
Relax man, we're just trying to figure out a problem here! No need for facepalms, I just want to understand the reality as much as you. :)

If the effective downward acceleration at the equator is less than the poles (which it is, due to the larger gravity being canceled out by the rotational acceleration of the earth), then you would expect a larger height of water at the equator given a constant potential energy.

g goes down at the equator, h must go up to compensate.
Yes, that's exactly what the formula says. Yet, we know that the sea level responses to local gravity in the way that it rises where gravity is higher.

The question is: does this contradict the above examination, are these two independent events, or is the formula itself not to be used for this question?

EDIT:
I may have spotted the problem:

The simplified formula for the potential energy is Epot = m * g * h
... with m being the mass, g being gravity and h being the height.

The actual formula for two masses in a distance is Epot = -G * (m1 * m2) / R (or R²)
... with G being the gravitational constant, m1 and m2 being the two masses and R being their distance to each other.

Now, just looking at this formula, you will notice that the distance in the simple formula (h) is a multiplier, while the distance in the complex formula (R) is a divisor. That means that a growing distance will increase Epot with the simple formula, but decrease it with the complex formula. Now we just need to sort out why. To me, it looks like the simple one must not be used for this case, because it was made to calculate the potential energy for objects on the surface, discarding other factors.
 
Last edited:
You're typing "Fpot" the whole times, but doesn't that stand for potential force, which is comething very different from Ep, potential energy? No complaint or anything, just something I was confused by 👍
 
Yes, that's exactly what the formula says. Yet, we know that the sea level responses to local gravity in the way that it rises where gravity is higher.

The effective gravity (downward acceleration) is lower at the equator - thus equatorial bulge.
 
This is a contradiction. If water follows gravity to where it is strongest, and gravity defines an equipotential surface which the sea level levels out at, why is that surface and therefore the sea level supposed to be higher where gravity is lower?

I remember reading something to the effect that the mean sea level of the Pacific Ocean is, on average, about 8 inches higher than the mean sea level of the Atlantic Ocean...on either end of the Panama Canal. The higher elevation of the Pacific Ocean is primarily due to currents, and the fact that the tidal variation is much higher on the Pacific side than on the Atlantic side....and these two points are probably less than 50 miles apart.

There are other factors besides gravity that affect sea level.
 
Back