Sea-Level on Oblate Spheroid

[arkie87]

So i was watching a flat earth video (because its fun to laugh at crazy people), and one of their questions got me thinking: given that the radius of the earth at the equator is 20km longer than at the poles, how does sea level (as a radius from center) work from the equator to the poles? Obviously, 20km of potential energy is a lot, so where is that energy harnessed? Does it have something to do with the m*v^2/R term changing as you go from the equator to the pole (since this mv^2/R term is what creates the oblateness to begin with), or is the gravitational well itself of an oblate spheroid different to account for this 20km of potential energy.

TIA (and people who know what they are talking about only, please)

[m4v]

Sea level is literally the sea level, it isn't the radius of a imagined sphere, it's also oblate. The rotation of the planet produces centrifugal forces that cancels a bit of the gravitational potential, this force is strongest at the equator so stuff weight less there, thus the planet's bulge.

[arkie87]

3 minutes ago, m4v said:

Sea level is literally the sea level, it isn't the radius of a imagined sphere, it's also oblate. The rotation of the planet produces centrifugal forces that cancels a bit of the gravitational potential, this force is strongest at the equator so stuff weight less there, thus the planet's bulge.

Sea level is also level, meaning constant potential energy. So how does that work, given that as you travel to the pole, you are getting closer to the center.

[Superfluous J]

19 minutes ago, arkie87 said:

Sea level is also level, meaning constant potential energy. So how does that work, given that as you travel to the pole, you are getting closer to the center.

Potential energy isn't "height" like most people approximate. It's much more complicated than that.

Basically, the extra energy you get from rotating around Earth at the equator means you have the same potential energy 20km higher than someone not rotating around the Earth at the poles.

And yes, that's also an approximation of the situation but hey I'm not a physics teacher.

[arkie87]

Just now, 5thHorseman said:

Potential energy isn't "height" like most people approximate. It's much more complicated than that.

Basically, the extra energy you get from rotating around Earth at the equator means you have the same potential energy 20km higher than someone not rotating around the Earth at the poles.

And yes, that's also an approximation of the situation but hey I'm not a physics teacher.

Well yeah, potential energy is complicated so lets just use equations.

What you are describing is something like this:

-GMm/R_pole = -GMm/R_equator + mV^2/R_equator*(R_equator - R_pole)

If you plug in the numbers, they are not equal, so my guess is the gravitational potential of an oblate sphere is slightly different than a perfect sphere in an amount that would cause an energy balance. 

2 minutes ago, arkie87 said:

Well yeah, potential energy is complicated so lets just use equations.

What you are describing is something like this:

-GMm/R_pole = -GMm/R_equator + mV^2/R_equator*(R_equator - R_pole)

If you plug in the numbers, they are not equal, so my guess is the gravitational potential of an oblate sphere is slightly different than a perfect sphere in an amount that would cause an energy balance. 

Oh, you said kinetic energy, which comes out to 101 kJ/kg, which is about half of the deficit (but still much more important than centripetal forces)

[m4v]

33 minutes ago, arkie87 said:

Sea level is also level, meaning constant potential energy. So how does that work, given that as you travel to the pole, you are getting closer to the center. 

Sea level is at constant potential energy, nobody said that that has to be spherical, it is a spheroid. At the poles the centrifugal forces are lower thus the gravity is stronger, at equator the centrifugal force is maximum thus the gravity is weaker and the potential level you have at the poles ends up higher.

Edited June 11, 2018 by m4v

[arkie87]

Just now, m4v said:

Sea level is at constant potential energy, nobody said that that has to be spherical, it is a spheroid.

I'm sorry if my wording was poor/unclear, let me rephrase:

Sea-level is at constant potential energy. Sea level at the poles is 20 km closer to earth's center than at the equator. Considering gravitational potential energy alone, as PE = -GMm/R, we can see they are at different energy potentials due to the different radii (i mention radii to indicate different gravity potentials, not to imply that they are spherical or anything to do with a sphere).

Quote

At the poles the centrifugal forces are lower thus the gravity is stronger, at equator the centrifugal force is maximum thus the gravity is weaker and the potential level you have at the poles ends up higher.

If i do a calculation like what you are describing, i dont get an energy balance, so that explanation doesnt work (counter-intuitively):

-GMm/R_pole = -GMm/R_equator + mV^2/R_equator*(R_equator - R_pole)

[m4v]

5 minutes ago, arkie87 said:

Considering gravitational potential energy alone, as PE = -GMm/R, we can see they are at different energy potentials due to the different radii.

When you add centrifugal forces it deforms that potential field into a spheroid. You cannot consider gravitational potential alone when in a rotating frame such as Earth, this is why we usually talk about gravity and not gravitation.

[arkie87]

1 minute ago, m4v said:

When you add centrifugal forces it deforms that potential field into a spheroid. You cannot consider gravitational potential alone when in a rotating frame such as Earth, this is why we usually talk about gravity and not gravitation.

Arg. We don't seem to be communicating well. 

 

Anything to say about the second half of my response? 

[m4v]

45 minutes ago, arkie87 said:

Anything to say about the second half of my response? 

no, I don't know what this term "mV^2/R_equator*(R_equator - R_pole)" is supposed to mean, is the integral of the centrifugal force for put it in units of energy? And the centrifugal force depends of the distance from the rotating axis, not the center of Earth, at the poles it would be zero.

Edited June 11, 2018 by m4v

[YNM]

"(Mean) Sea Level" is just one of the conveniently chosen standards of an equipotential surface. For one, they are (and were) very easy to measure - all you need to do is to mark the high tide and the low tide and see where they fall, then make rough average - and seas are all connected across the world (apart from two or three).

The theoretical basis of "why does the sea level even represents equipotential surface ?" would be the rather simple reason that "water finds it's own level". The entire ocean is a big bowl of water - they should "level" themself out.

It is to be noted, however, that Mean Sea Level is only one reference method for an equipotential surface. Other methods includes atmospheric pressure equivalence or other geometric-equipotential methods (see how we made them for Mars).

Now, the question is : "well, the entire thing led itself to a pretty deviation from 'perfect' - so how does that work out ?"

It might be interesting to note that there are parts of the 'imperfect' method that actually is in itself still imperfect.

So much so, some parts differ by ~200 m even on the same latitude (ie. Equator - look the area around Indonesia / Papua archipelago vs Laccadive sea).

So even by our standards, these seems bloody absurd.

But guess what - these are reality.

And again, the flat worlders are free to take what they think, but reality is here to stay.

[arkie87]

6 hours ago, m4v said:

no, I don't know what this term "mV^2/R_equator*(R_equator - R_pole)" is supposed to mean, is the integral of the centrifugal force for put it in units of energy? And the centrifugal force depends of the distance from the rotating axis, not the center of Earth, at the poles it would be zero.

You should notice that i have not put a centripetal term at the pole, but there is at the equator... i never said it had anything to do with radius. 

The problem is that this term is orders of magnitude off (approx 600 J/kg compared to 200 kJ/kg).

 

3 hours ago, YNM said:

"(Mean) Sea Level" is just one of the conveniently chosen standards of an equipotential surface. For one, they are (and were) very easy to measure - all you need to do is to mark the high tide and the low tide and see where they fall, then make rough average - and seas are all connected across the world (apart from two or three).

The theoretical basis of "why does the sea level even represents equipotential surface ?" would be the rather simple reason that "water finds it's own level". The entire ocean is a big bowl of water - they should "level" themself out.

It is to be noted, however, that Mean Sea Level is only one reference method for an equipotential surface. Other methods includes atmospheric pressure equivalence or other geometric-equipotential methods (see how we made them for Mars).

Now, the question is : "well, the entire thing led itself to a pretty deviation from 'perfect' - so how does that work out ?"

It might be interesting to note that there are parts of the 'imperfect' method that actually is in itself still imperfect.

So much so, some parts differ by ~200 m even on the same latitude (ie. Equator - look the area around Indonesia / Papua archipelago vs Laccadive sea).

So even by our standards, these seems bloody absurd.

But guess what - these are reality.

And again, the flat worlders are free to take what they think, but reality is here to stay.

I'm not sure what to do with this...

[Snark]

11 hours ago, arkie87 said:

or is the gravitational well itself of an oblate spheroid different to account for this 20km of potential energy.

^ Pretty much this.

A few things to bear in mind:

• The Earth is not a sphere.
• It also does not have uniform mass distribution.
• The old standby GM/r² is an approximation that works pretty well... but it assumes a perfectly spherical mass distribution and is just that, an approximation.

The above considerations become relevant when you try to start "solving" this problem with just a few simple equations, like GM/R² for gravity or v/r² for centripetal acceleration and the like. Those equations are great, as an approximation-- but given that the actual Earth is lumpy and bulgy and irregular in various ways, solving it perfectly is a lot more complicated than that.  It'll need complex mathematics at the least, a data-intensive numerical computation at worst.

One important thing to remember is, don't try to separate the water from the planet.  Remember that the Earth is not a rigid body.  Sure, it feels "solid" enough under our feet... but the square-cube law works heavily against anything as big as a planet, which is why all sizable planets are pretty close to being spheres.  On the planetary scale, it's easiest to think of the Earth as being liquid-- certainly, it behaves that way on the macro level.  It has dense liquid at the center (the iron core), and less dense liquid at upper layers (rock), and really less dense liquid sitting on top (water)... but it's basically a big gobbet of density-stratified liquid, drawn together by its own gravitation and altered by its rotation.

So when the Earth spins, the liquid does what liquid is wont to do, and settles to an equipotential surface.  Liquid flows "downhill".  So the oceans are simply following the shape of the Earth's gravitaitional field-- just as the rocky part does-- which is oblate to match the mass distribution coupled with the rotating reference frame.  The oceans bulge for the same reason that the rest of the planet does.

If, when you're doing your math, you try to think of the Earth as being a rigid, perfect sphere with perfectly spherical GM/r² gravity, and with a liquid ocean sloshing around on top, then you're going to go astray and your numbers won't make any sense because that's not what the situation is.

Here's a thought experiment that may help:

Step 1

1. Imagine that you have a planet-sized lump of some dense material.  It's perfectly rigid.  However, it's possible to melt it, in which case it becomes perfectly liquid.  It's just sitting there, stationary in space.
2. Now melt it.  It will settle into a perfect sphere as it self-gravitates.
3. Now freeze it solid.  Imagine you can land on it, and walk around.  No matter where you walk on it, the ground is perfectly "horizontal"-- e.g. if you drop a ball bearing, it'll just sit there and not roll.  This "perfectly horizontal" surface is guaranteed because it formed from a liquid, and liquid flows downhill, and will settle to a state of minimum potential energy, i.e. perfectly flat.
4. Imagine that you now have some liquid of very low density.  You've got a volume of this liquid, equal to the area of the planet times one centimeter:  in other words, enough to cover the whole surface of the planet to a depth of 1 cm.  Now you dump this onto the surface of the planet.  In our thought experiment, the liquid is of such low density, and so small relative to the planet's size, that the liquid itself doesn't affect the planet's gravitational field.  It simply reacts to the planet's gravity.  What happens?
5. Answer:  It'll flatten and spread out until it has covered the whole surface of the planet to a perfectly uniform depth of 1 cm.  The whole surface is covered with a very shallow "ocean", one centimeter deep everywhere.
6. So far, so good-- simple, right?

Step 2

1. Okay, next step.  We start the whole (perfectly spherical, perfectly rigid) planet rotating on its axis, fairly rapidly.  Not so fast that it's trying to fling itself apart, but enough that the perceived "gravity" at the equator is much less than at the poles.  But the shape doesn't change because it's magically perfectly rigid.
2. If you walked around on the surface now, it wouldn't seem "flat" anymore.  The poles would feel "higher" than the equator.  Walking from the pole to the equator, you'd be walking downhill (according to your local perception of gravity).  This, in spite of the fact that the surface is geometrically "horizontal" (i.e. perpendicular to a vector to the planet's center) everywhere.  It's horizontal geometrically, but not gravitationally.  And gravitationally is what matters, here.
3. Okay, now we dump our imaginary low-density, enough-to-cover-whole-surface-1-cm-deep liquid onto the planet.  What happens?
4. Answer: The liquid flows downhill.  You end up with a ring-shaped ocean around the planet's equator.  It'll be deepest at the equator, getting shallower until you have a shoreline somewhere.  So, the liquid is "bulgy" at the equator, due to the planet's rotation-- and it's much bulgier than the planet is (since the planet isn't bulgy at all, it's a perfect sphere).

Step 3

1. Okay, now imagine that we have our rapidly-spinning, perfectly spherical, perfectly rigid planet.  Now we melt it to a liquid again.  It will alter its shape due to the rotation.  The thing will settle into a flattened, highly oblate shape.  Depending on what the internal density distribution is, the actual shape could be a variety of possibilities (it won't necessarily be a simple ellipsoid), but in any case it'll be cylindrically symmetric and its equatorial radius will be a lot bigger than its polar radius.
2. Wait until it settles down and isn't wobbling around.  Then freeze it again, in its new shape.
3. Now land on it again.  Once again:  no matter where you land on it, the surface always appears to you, locally, to be perfectly "horizontal".  Drop a ball bearing, it'll just sit there.  If you stand at the pole, you'll have very high gravity, on a perfectly horizontal surface.  Stand on the equator, you'll have low gravity, still with a perfectly horizontal surface.  If you walk from the pole to the equator, the surface will seem to be horizontal to you, locally, the whole way.
4. Okay, now we dump our imaginary low-density, enough-to-cover-whole-surface-1-cm-deep liquid onto the planet.  What happens?
5. Answer:  It will distribute itself evenly.  You'll get a 1-centimeter-deep "ocean" covering the whole planet.  Uniformly 1 centimeter deep everywhere.  The planet's radius may vary by many kilometers over its surface-- even many hundreds of kilometers-- but the ocean will stay at a consistent, uniform 1 cm depth everywhere.

Why is that?  Why the uniformity?  Answer:  because the surface is "horizontal" at all locations.  It was established thus when the planet settled to gravitational equipotential.  Imagine the opposite assumption, that the "ocean" isn't uniformly deep.  It would mean that it's mounded higher in one spot than a neighboring spot.  Well, since the surface is gravitationally horizontal, that means you can't have a hill of liquid there any more than you can have a hill of water in a bucket.  It'll collapse under its own weight until the surface of the liquid is perfectly horizontal, too.  So, when you have a perfectly horizontal liquid surface, over a perfectly horizontal planetary surface... you get constant depth.

Now, the Earth is a more complicated system than the highly idealized thought experiment above, of course.  The Earth isn't perfectly liquid, the crust has solid irregularities (contintents, ocean floors), the oceans are subject to tidal forces and atmospheric pressure, the water is of varying density with temperature and salinity, there are ocean currents, the water itself has gravitation, yadda yadda.  But broadly speaking, it'll act approximately like "Step 3" in the thought experiment above.

So, if you're trying to do simple GM/r² and v²/r calculations... they won't work.  And the reason they won't work is that you're trying to model the Earth as if it were a "Step 2" situation per the above thought experiment-- i.e. you're trying to treat it as a rigid sphere.  However, really it's much closer to a "Step 3" situation, and that's why the math doesn't work.

Use the wrong model, and you'll get the wrong math. 

[YNM]

4 hours ago, arkie87 said:

I'm not sure what to do with this...

Simple - this is what taking the red pill allows you to know.

Taking the blue pill is fine and is about as nice, but you'll lose the opportunity of a red pill.

 

Joke aside, in short, the Earth is very lumpy. So much so, even an ellipsoid approximation (WGS84) still fall short of actualy providing the correct mean sea level in places where there are more gravitational pull and where there are less gravitational pull, and the errors could mean that despite standing near the sea shore, let's say in India, the ellipsoid might tell your altitude to be a hundred meters beneath sea level.

So coping with a 20 km difference in average figures is just the same stuff.

Edited June 11, 2018 by YNM

[arkie87]

11 minutes ago, YNM said:

Simple - this is what taking the red pill allows you to know.

Taking the blue pill is fine and is about as nice, but you'll lose the opportunity of a red pill.

 

Joke aside, in short, the Earth is very lumpy. So much so, even an ellipsoid approximation (WGS84) still fall short of actualy providing the correct mean sea level in places where there are more gravitational pull and where there are less gravitational pull, and the errors could mean that despite standing near the sea shore, let's say in India, the ellipsoid might tell your altitude to be a hundred meters beneath sea level.

So coping with a 20 km difference in average figures is just the same stuff.

The errors you showed were on the order of 80m, compared to 20,000m. Before we can understand the 80m, we need to understand the 20,000m error. 

[m4v]

4 hours ago, arkie87 said:

You should notice that i have not put a centripetal term at the pole, but there is at the equator... i never said it had anything to do with radius. 

But you have R_pole in there, it shouldn't be there if there's just the centrifugal term at the equator (centripetal points to the center, centrifugal outwards) so what it is? how did you get it? what values are you using? I did the integral myself, and I get 0.5*m(w R)^2, (w being angular velocity in radians, V=wR), calculating that for the equator you get 108kJ/kg so half of the 200kJ/kg you said before. Since is in the order of magnitude I would say that's good enough, we aren't going to get exact numbers with the approximations we're using.

We would need to know the math for calculating the shape of bodies at hydrostatic equilibrium for get a complete answer.

Edited June 11, 2018 by m4v

[arkie87]

2 hours ago, Snark said:

^ Pretty much this.

A few things to bear in mind:

• The Earth is not a sphere.
• It also does not have uniform mass distribution.
• The old standby GM/r² is an approximation that works pretty well... but it assumes a perfectly spherical mass distribution and is just that, an approximation.

The above considerations become relevant when you try to start "solving" this problem with just a few simple equations, like GM/R² for gravity or v/r² for centripetal acceleration and the like. Those equations are great, as an approximation-- but given that the actual Earth is lumpy and bulgy and irregular in various ways, solving it perfectly is a lot more complicated than that.  It'll need complex mathematics at the least, a data-intensive numerical computation at worst.

One important thing to remember is, don't try to separate the water from the planet.  Remember that the Earth is not a rigid body.  Sure, it feels "solid" enough under our feet... but the square-cube law works heavily against anything as big as a planet, which is why all sizable planets are pretty close to being spheres.  On the planetary scale, it's easiest to think of the Earth as being liquid-- certainly, it behaves that way on the macro level.  It has dense liquid at the center (the iron core), and less dense liquid at upper layers (rock), and really less dense liquid sitting on top (water)... but it's basically a big gobbet of density-stratified liquid, drawn together by its own gravitation and altered by its rotation.

So when the Earth spins, the liquid does what liquid is wont to do, and settles to an equipotential surface.  Liquid flows "downhill".  So the oceans are simply following the shape of the Earth's gravitaitional field-- just as the rocky part does-- which is oblate to match the mass distribution coupled with the rotating reference frame.  The oceans bulge for the same reason that the rest of the planet does.

If, when you're doing your math, you try to think of the Earth as being a rigid, perfect sphere with perfectly spherical GM/r² gravity, and with a liquid ocean sloshing around on top, then you're going to go astray and your numbers won't make any sense because that's not what the situation is.

Here's a thought experiment that may help:

Step 1

1. Imagine that you have a planet-sized lump of some dense material.  It's perfectly rigid.  However, it's possible to melt it, in which case it becomes perfectly liquid.  It's just sitting there, stationary in space.
2. Now melt it.  It will settle into a perfect sphere as it self-gravitates.
3. Now freeze it solid.  Imagine you can land on it, and walk around.  No matter where you walk on it, the ground is perfectly "horizontal"-- e.g. if you drop a ball bearing, it'll just sit there and not roll.  This "perfectly horizontal" surface is guaranteed because it formed from a liquid, and liquid flows downhill, and will settle to a state of minimum potential energy, i.e. perfectly flat.
4. Imagine that you now have some liquid of very low density.  You've got a volume of this liquid, equal to the area of the planet times one centimeter:  in other words, enough to cover the whole surface of the planet to a depth of 1 cm.  Now you dump this onto the surface of the planet.  In our thought experiment, the liquid is of such low density, and so small relative to the planet's size, that the liquid itself doesn't affect the planet's gravitational field.  It simply reacts to the planet's gravity.  What happens?
5. Answer:  It'll flatten and spread out until it has covered the whole surface of the planet to a perfectly uniform depth of 1 cm.  The whole surface is covered with a very shallow "ocean", one centimeter deep everywhere.
6. So far, so good-- simple, right?

Step 2

1. Okay, next step.  We start the whole (perfectly spherical, perfectly rigid) planet rotating on its axis, fairly rapidly.  Not so fast that it's trying to fling itself apart, but enough that the perceived "gravity" at the equator is much less than at the poles.  But the shape doesn't change because it's magically perfectly rigid.
2. If you walked around on the surface now, it wouldn't seem "flat" anymore.  The poles would feel "higher" than the equator.  Walking from the pole to the equator, you'd be walking downhill (according to your local perception of gravity).  This, in spite of the fact that the surface is geometrically "horizontal" (i.e. perpendicular to a vector to the planet's center) everywhere.  It's horizontal geometrically, but not gravitationally.  And gravitationally is what matters, here.
3. Okay, now we dump our imaginary low-density, enough-to-cover-whole-surface-1-cm-deep liquid onto the planet.  What happens?
4. Answer: The liquid flows downhill.  You end up with a ring-shaped ocean around the planet's equator.  It'll be deepest at the equator, getting shallower until you have a shoreline somewhere.  So, the liquid is "bulgy" at the equator, due to the planet's rotation-- and it's much bulgier than the planet is (since the planet isn't bulgy at all, it's a perfect sphere).

Step 3

1. Okay, now imagine that we have our rapidly-spinning, perfectly spherical, perfectly rigid planet.  Now we melt it to a liquid again.  It will alter its shape due to the rotation.  The thing will settle into a flattened, highly oblate shape.  Depending on what the internal density distribution is, the actual shape could be a variety of possibilities (it won't necessarily be a simple ellipsoid), but in any case it'll be cylindrically symmetric and its equatorial radius will be a lot bigger than its polar radius.
2. Wait until it settles down and isn't wobbling around.  Then freeze it again, in its new shape.
3. Now land on it again.  Once again:  no matter where you land on it, the surface always appears to you, locally, to be perfectly "horizontal".  Drop a ball bearing, it'll just sit there.  If you stand at the pole, you'll have very high gravity, on a perfectly horizontal surface.  Stand on the equator, you'll have low gravity, still with a perfectly horizontal surface.  If you walk from the pole to the equator, the surface will seem to be horizontal to you, locally, the whole way.
4. Okay, now we dump our imaginary low-density, enough-to-cover-whole-surface-1-cm-deep liquid onto the planet.  What happens?
5. Answer:  It will distribute itself evenly.  You'll get a 1-centimeter-deep "ocean" covering the whole planet.  Uniformly 1 centimeter deep everywhere.  The planet's radius may vary by many kilometers over its surface-- even many hundreds of kilometers-- but the ocean will stay at a consistent, uniform 1 cm depth everywhere.

Why is that?  Why the uniformity?  Answer:  because the surface is "horizontal" at all locations.  It was established thus when the planet settled to gravitational equipotential.  Imagine the opposite assumption, that the "ocean" isn't uniformly deep.  It would mean that it's mounded higher in one spot than a neighboring spot.  Well, since the surface is gravitationally horizontal, that means you can't have a hill of liquid there any more than you can have a hill of water in a bucket.  It'll collapse under its own weight until the surface of the liquid is perfectly horizontal, too.  So, when you have a perfectly horizontal liquid surface, over a perfectly horizontal planetary surface... you get constant depth.

Now, the Earth is a more complicated system than the highly idealized thought experiment above, of course.  The Earth isn't perfectly liquid, the crust has solid irregularities (contintents, ocean floors), the oceans are subject to tidal forces and atmospheric pressure, the water is of varying density with temperature and salinity, there are ocean currents, the water itself has gravitation, yadda yadda.  But broadly speaking, it'll act approximately like "Step 3" in the thought experiment above.

So, if you're trying to do simple GM/r² and v²/r calculations... they won't work.  And the reason they won't work is that you're trying to model the Earth as if it were a "Step 2" situation per the above thought experiment-- i.e. you're trying to treat it as a rigid sphere.  However, really it's much closer to a "Step 3" situation, and that's why the math doesn't work.

Use the wrong model, and you'll get the wrong math. 

Thanks for this post. I'll need some time to digest this...

[VaPaL]

36 minutes ago, arkie87 said:

The errors you showed were on the order of 80m, compared to 20,000m. Before we can understand the 80m, we need to understand the 20,000m error. 

The 20km is the difference between a sphere and an ellipsoid (better aproximation of actual Earths shape). The 200m is the difference from a perfect ellipsoid and actual Earths shape. He's saying that Earth is even more irregular than just not been a sphere.

 

EDIT: missed a word

Edited June 11, 2018 by VaPaL

[arkie87]

7 minutes ago, m4v said:

But you have R_pole in there, it shouldn't be there if there's just the centrifugal term at the equator (centripetal points to the center, centrifugal outwards) so what it is? how did you get it? what values are you using? I did the integral myself, and I get 0.5*m(w R)^2, (w being angular velocity in radians, V=wR), calculating that for the equator you get 108kJ/kg so half of the 200kJ/kg you said before. Since is in the order of magnitude I would say that's good enough, we aren't going to get exact numbers with the approximations we're using.

We would need to know the math for calculating the shape of bodies at hydrostatic equilibrium for get a complete answer.

The first half is the centripetal acceleration; the second is the height, so analogous of g*h

g = V^2/R

h = (R_equator-R_pole)

What you are computing is not centripetal acceleration, but kinetic energy. 

3 minutes ago, VaPaL said:

The 20km is the difference between a sphere and an ellipsoid (better aproximation of actual Earths shape). The 200m is the difference from a perfect ellipsoid and actual Earths shape. He's saying that Earth is even more irregular than just not been a sphere.

 

EDIT: missed a word

Yeah. It's not a perfect sphere, but can be approximated as an ellipse. Yes, its not a perfect ellipse, but it has minor deviations. Each subsequent approximation gets more and more accurate.

But that has no bearing on the question. 

[m4v]

26 minutes ago, arkie87 said:

The first half is the centripetal acceleration; the second is the height, so analogous of g*h

g = V^2/R

h = (R_equator-R_pole)

What you are computing is not centripetal acceleration, but kinetic energy. 

I applied the definition of potential field, which is the amount of work needed to move a mass m from a zero potential (this case the center of Earth) to the point of interest, that the result is equal to kinetic energy makes total sense, since the centrifugal force is a fictitious force that arises due to the angular speed of an object.

Now I recognize you're using the  potential energy of the near earth gravity equation but instead of the gravity acceleration you're putting centrifugal acceleration. Yeah, I wouldn't mix them like that since they aren't analogous, for one the gravitational force decreases as you increase R while the centrifugal increases. 

Edited June 11, 2018 by m4v

[VaPaL]

8 hours ago, YNM said:

I think this ^ together whit this:

Resumes pretty much what @Snark said.

In your model, you are approximating gravity as a constant and Earth as a ellipsoid, this wont match. Your model is incoherent because you are using to different approaches. If you want to use real Earth shape, you should user real Earth gravity, or use matching approximations. Ellipsoid Earth and Ellipsoid gravity as Snark said:

[sevenperforce]

If the Earth were non-rotating, it would take an overall spherical shape, and its gravitational equipotential would share this shape, although there would be mass concentrations corresponding to the continental plates and local deviations. 

Since the Earth is rotating, there is a small centrifugal (not centripetal) force working against gravity at the equator. This causes the Earth to bulge at the equator.

Because the Earth bulges at the equator, it is lifted farther away from the iron core, and thus experiences slightly less gravitational force. Which makes it bulge more. Which means less gravity, and so forth.

 So there's no simple equation.

[mikegarrison]

13 hours ago, m4v said:

Sea level is literally the sea level, it isn't the radius of a imagined sphere, it's also oblate. The rotation of the planet produces centrifugal forces that cancels a bit of the gravitational potential, this force is strongest at the equator so stuff weight less there, thus the planet's bulge.

Sea level is literally the sea level, but "mean sea level" is a datum. And since the sea level itself has been changing over time, then the datum either needs to also change over time or else be identified as, for example, being MSL in 1950.

[Kerbart]

Imagine a water filled cup perfectly centered on turntable.

as the turntable spins faster, water on the outside of the cup will climb up against the wall of the cup. Since the water is coming out of the center of the cup, water levels there will drop.

The water is in an equilibrium with gravity and centripetal (imaginary) forces. As long as the cup is spinning it doesn’t want to “flow down”to the center (at least not anymore than the spinning forces it up to the rim), so while there’s a difference in waterlevel there’s not really a potential to unlock that energy.

Earth is just an inverted cup, but the same principle applies.

[arkie87]

44 minutes ago, Kerbart said:

Imagine a water filled cup perfectly centered on turntable.

as the turntable spins faster, water on the outside of the cup will climb up against the wall of the cup. Since the water is coming out of the center of the cup, water levels there will drop.

The water is in an equilibrium with gravity and centripetal (imaginary) forces. As long as the cup is spinning it doesn’t want to “flow down”to the center (at least not anymore than the spinning forces it up to the rim), so while there’s a difference in waterlevel there’s not really a potential to unlock that energy.

Earth is just an inverted cup, but the same principle applies.

I understand the physics. I want a calculation that would show what "level" is or equivalently, the surface of constant potential energy.