Sea-Level on Oblate Spheroid

[arkie87]

1 hour ago, sevenperforce said:

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.

I really dont think mass concentrations compare to the 20km radius difference from equator to pole. 
 

I get the governing physics. I want some math to quantitatively show where this seeming 20km height difference went, since 20km of potential energy is a lot. 

[arkie87]

2 hours 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.

Ok, i think i see what you were implying here. Since at the pole, you are zero distance from the axis of rotation, the distance covered is not (R_equator - R_pole), but rather, just R_equator. Interestingly enough, your equation is exactly equal to kinetic energy, which itself is also equal to half the potential energy difference. Thus, the centrifugal and kinetic energy terms sum to ~100% of the potential energy difference due to height, which implies that the gravitational potential of an oblate spheroid has nothing to do with it. Sorry,  Snark.

Does that seem right to everyone?

[sevenperforce]

1 hour ago, arkie87 said:

Ok, i think i see what you were implying here. Since at the pole, you are zero distance from the axis of rotation, the distance covered is not (R_equator - R_pole), but rather, just R_equator. Interestingly enough, your equation is exactly equal to kinetic energy, which itself is also equal to half the potential energy difference. Thus, the centrifugal and kinetic energy terms sum to ~100% of the potential energy difference due to height, which implies that the gravitational potential of an oblate spheroid has nothing to do with it. Sorry,  Snark.

Does that seem right to everyone?

Yep.

I get 191 kJ for the gravitational potential energy of a 1 kg mass dropped from a height of 20 km, and 108 kJ for the kinetic energy of an object at the equator traveling at the speed of the Earth's rotation. So it's approximately half, but not exactly half.

Edited June 11, 2018 by sevenperforce

[arkie87]

32 minutes ago, arkie87 said:

Ok, i think i see what you were implying here. Since at the pole, you are zero distance from the axis of rotation, the distance covered is not (R_equator - R_pole), but rather, just R_equator. Interestingly enough, your equation is exactly equal to kinetic energy, which itself is also equal to half the potential energy difference. Thus, the centrifugal and kinetic energy terms sum to ~100% of the potential energy difference due to height, which implies that the gravitational potential of an oblate spheroid has nothing to do with it. Sorry,  Snark.

Does that seem right to everyone?

On second thought, if we imagine someone holding onto a rigid rod spinning around at a constant angular velocity about a central axis floating in space. If the person decided to pull themselves closer to the center, that would take energy. At the same time, they would lose kinetic energy (since the rod does not speed up). The kinetic energy formula and the centrifugal energy formula are equivalent, thus one is exchanged for the other. In this case, there is no gravitational potential, and centrifugal and kinetic energy are exchanged, thus, it appears my explanation above is incorrect.

Thus, half of the energy is due to centrifugal forces/kinetic energy (its the same thing in this case), and the other half could be due to the gravitational well of an oblate spheroid. 

 

Edited June 11, 2018 by arkie87

[YNM]

3 hours 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 20000 m error is just due to centripetal forces. Or presumably.

The deviation for that is the one that's even more baffling.

And again, as I've said, these have nothing to do with whatever they were caused of. You could live on a planet that has a weird massive mountain ridge just around the equator - and their geoid will deviate from any sphere or ellipsoid. We aren't dictating what the shape of the Earth is, we are observing what the shape of the Earth is.

Hence it all depends on whether you want to take the red pill and always stay true to what the truth is, or the blue pill and be comforted by the "consensus" and the fact that further scrutiny would have made you took the red pill.

Edited June 11, 2018 by YNM

[sevenperforce]

1 hour ago, arkie87 said:

On second thought, if we imagine someone holding onto a rigid rod spinning around at a constant angular velocity about a central axis floating in space. If the person decided to pull themselves closer to the center, that would take energy. At the same time, they would lose kinetic energy (since the rod does not speed up).

Incorrect. As they pull themselves closer to the center, they will decrease the rod's moment of inertia, and conservation of angular momentum will cause the rod's RPM to increase.

Same reason a neutron star can spin hundreds of times per second -- the parent star's modest rotation rate went through the roof when most of its mass collapsed down to a tiny radius.

[arkie87]

7 minutes ago, sevenperforce said:

Incorrect. As they pull themselves closer to the center, they will decrease the rod's moment of inertia, and conservation of angular momentum will cause the rod's RPM to increase.

Same reason a neutron star can spin hundreds of times per second -- the parent star's modest rotation rate went through the roof when most of its mass collapsed down to a tiny radius.

Oh god. I specifically said the rod is rotating at constant angular velocity (i.e. the mass of the rod is much larger than the mass of the human). Also, you clearly missed the point of the thought experiment.... 

Edited June 11, 2018 by arkie87

[sevenperforce]

13 hours ago, arkie87 said:

Oh god. I specifically said the rod is rotating at constant angular velocity (i.e. the mass of the rod is much larger than the mass of the human). Also, you clearly missed the point of the thought experiment.... 

If a person is hanging on to the end of a rod rotating at constant angular velocity, then pulls herself in toward the center of the rod, then the rod will no longer be rotating at constant angular velocity. Angular momentum must be conserved. Change the moment of inertia, and unless you have a way to change angular momentum, the angular velocity WILL change.

The fact that the mass of the rod is much greater than the mass of the person means this angular velocity change will be very slight, but it will nonetheless be there. If you take a scoop of dirt from the equator and carry it to the North Pole, you increase the rotation rate of the Earth. 

[K^2]

15 hours ago, arkie87 said:

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

When you are looking at surface of the water as equipotential surface, you have to consider frame of reference where water is at rest. It's the only case where it makes sense to talk about equilibrium. And since the Earth is rotating, the relevant frame of reference is a rotating one. In a rotating frame of reference, potential energy picks up the mω²(r sinθ)²/2 term. So the total specific (per unit mass) potential energy of fluid on the surface of a rotating sphere is given by

U = ω²(R sinθ)²/2 - μ/R

For Earth, θ is the latitude, so r has to be smaller near poles for the same potential energy.

Formal derivation is via Canonical Transformation of the Hamiltonian. It's a lot easier to work with cylindrical coordinate system here, so I'm going to ignore θ. The generating function for azimuth angle is then given by

F(φ, L, t) = (φ - ωt)L

This generates a contribution to Hamiltonian (total energy) equal to

K - H = ∂F/∂t = ωL

For an object at rest in a rotating frame, kinetic energy is exactly ωL/2, so we have

U - U₀ = K - H - ωL/2 = ωL - ωL/2

Finally, for an object at rest in a rotating reference frame, L = ωr², so we can simplify in one of two ways

U - U₀ = L²/(2r²) = ω²r²/2

Which is exactly the effective potential of a unit mass in a rotating frame, and with r = R sinθ it matches up with the equation at the top.

[arkie87]

2 hours ago, sevenperforce said:

If a person is hanging on to the end of a rod rotating at constant angular velocity, then pulls herself in toward the center of the rod, then the rod will no longer be rotating at constant angular velocity. Angular momentum must be conserved. Change the moment of inertia, and unless you have a way to change angular momentum, the angular velocity WILL change.

The fact that the mass of the rod is much greater than the mass of the person means this angular velocity change will be very slight, but it will nonetheless be there. If you take a scoop of dirt from the equator and carry it to the North Pole, you increase the rotation rate of the Earth. 

I clearly stipulated in the OP that the angular velocity doesnt change. Since there is a way for this to happen in real life--if the mass of the rod is much greater than the mass of the body, your minute technical point is both obvious and trivial.

Being technically correct is cancer. 

[YNM]

5 minutes ago, arkie87 said:

if the mass of the rod is much greater than the mass of the body,

The mass retraction would cause an even greater change.

(assuming the person is in the center of the rotation.)

 

If otherwise then otherwise.

 

Still, an highly interesting explanation made by @K^2 , but some question : Even if the body wasn't spinning, but is innately oblate, would it still makes an oblate geoid ? Like, would we still have sun-synchronous orbits and so ?

Edited June 12, 2018 by YNM

[sevenperforce]

4 minutes ago, arkie87 said:

I clearly stipulated in the OP that the angular velocity doesnt change. Since there is a way for this to happen in real life--if the mass of the rod is much greater than the mass of the body....

No. You can't stipulate away conservation of angular momentum.

5 minutes ago, arkie87 said:

...your minute technical point is both obvious and trivial.

Being technically correct is cancer. 

You are the one who said, "If the person decided to pull themselves closer to the center, that would take energy. At the same time, they would lose kinetic energy (since the rod does not speed up)."

That is false. Kinetic energy is not lost; it is compensated for by a minute change in the angular velocity of the entire system. You cannot use the change in rotational kinetic energy of a small mass to set up your physics while handwaving the change in angular velocity of the system.

Is the difference minute? Yes. But if you are pretending it doesn't exist, you are setting up your body diagram incorrectly and will get a fallacious result.

This is not me being a physics pedant (though I'd gladly accept such blame when deserved); this is me stopping you from creating an incorrect construct of a physical system. If you make the wrong assumptions, you won't get a good outcome.

[arkie87]

9 minutes ago, sevenperforce said:

No. You can't stipulate away conservation of angular momentum.

You are the one who said, "If the person decided to pull themselves closer to the center, that would take energy. At the same time, they would lose kinetic energy (since the rod does not speed up)."

That is false. Kinetic energy is not lost; it is compensated for by a minute change in the angular velocity of the entire system. You cannot use the change in rotational kinetic energy of a small mass to set up your physics while handwaving the change in angular velocity of the system.

Is the difference minute? Yes. But if you are pretending it doesn't exist, you are setting up your body diagram incorrectly and will get a fallacious result.

This is not me being a physics pedant (though I'd gladly accept such blame when deserved); this is me stopping you from creating an incorrect construct of a physical system. If you make the wrong assumptions, you won't get a good outcome.

I'm not stipulating away conservation of momentum. It just that angular velocity for all practical purposes doesnt change for that case. Stop wasting our time with trivial distinctions.

And yes, on a system level, the energy goes into a minute increase in angular momentum. But our reference frame is clearly the person, not the system as a whole. On the level of the person, the energy require to "climb" towards the center is used to reduce his kinetic energy. I never claimed energy was destroyed.

11 minutes ago, YNM said:

The mass retraction would cause an even greater change.

(assuming the person is in the center of the rotation.)

 

If otherwise then otherwise.

What? The mass doesnt retract. The body does...

Edited June 12, 2018 by arkie87

[VaPaL]

7 minutes ago, arkie87 said:

I'm not stipulating away conservation of momentum. It just that angular velocity for all practical purposes doesnt change for that case. Stop wasting our time with trivial distinctions.

You cannot ignore something and argue that something doesn't made sence or is missing. From the start, if ignore something that shouldn't your analysis is doomed or you will have to make for it somewhere else.

 

10 minutes ago, arkie87 said:

And yes, on a system level, the energy goes into a minute increase in angular momentum. But our reference frame is clearly the person, not the system as a whole.

But you cannot ignore the system as a whole. Your just said that the energy goes into a increase in angular velocity, so if you just consider the person, it's energy will no be the same from the start, because part of it, as you said, went to increase angular velocity. Ignoring part of a system is the easiest way to get unbalanced energy. You only have energy conservation in a closed system, in a open one, there's no problem with energy being "lost" or "gained".

It's the same if you said that by stepping on a cars trhottle you as a person gain knetic energy and you energy increase so there's must be something... You can't do that

[sevenperforce]

14 minutes ago, arkie87 said:

Stop wasting our time with trivial distinctions.

On the level of the person, the energy require to "climb" towards the center is used to reduce his kinetic energy.

Yes, I understand that is the system construction you are proposing. If you want to set it up that way, then do the maths and see if it comes out.

Spoiler

E.g., "Suppose a person of mass 100 kg is clinging to one end of a 1000 m rod, of mass 10,000 kg, with an initial rotation rate of 1 RPM. The centrifugal force perceived by the person is _______ Newtons. The tangential kinetic energy of the person is initially _______ kJ. If the person climbs 250 meters up the rod, the work performed will be _____ kJ and their new tangential kinetic energy will be ______ kJ."

Remember that you're gonna have to take the integral of the apparent force because the centrifugal force term is a dependent variable of distance, which is the thing being covered.

 

[arkie87]

13 minutes ago, VaPaL said:

You cannot ignore something and argue that something doesn't made sence or is missing. From the start, if ignore something that shouldn't your analysis is doomed or you will have to make for it somewhere else.

I wasnt ignoring it. There are real-world situations where the term is negligible, which do not violate physics. 

Quote

But you cannot ignore the system as a whole. Your just said that the energy goes into a increase in angular velocity, so if you just consider the person, it's energy will no be the same from the start, because part of it, as you said, went to increase angular velocity. Ignoring part of a system is the easiest way to get unbalanced energy. You only have energy conservation in a closed system, in a open one, there's no problem with energy being "lost" or "gained".

I'm not ignoring the system as a whole, i am just not presently analyzing it. You can apply conservation of energy at any level you choose, and it should be satisfied at all levels. Here, i am interested with the energy of the astronaut/person. As he climbs towards the center, he inputs work and loses kinetic energy.

13 minutes ago, sevenperforce said:

Yes, I understand that is the system construction you are proposing. If you want to set it up that way, then do the maths and see if it comes out.

I already did the math, and presented it here (above), and it works out.

Here it is again:

The energy required for the astronaut to climb to the center is:

integral(F,dR,R,0)=integral(w^2*R,R,0)=1/2 w^2 R^2 = 1/2 V^2, which is the same form as kinetic energy, so the energy required to pull inwards is equal to the change in kinetic energy.

Now, granted, in reality, the angular velocity changes proportional to the ratio of masses. If we computed the change in angular velocity, and then summed energy from the person and the rod, we would also have energy balance.

Edited June 12, 2018 by arkie87

[VaPaL]

3 minutes ago, arkie87 said:

There are real-world situations where the term is negligible, which do not violate physics. 

Negligible =/= nonexistent, if you punch the numbers with enough precision you will get an inbalance. Also, some time it's ignored because others errors (measuring, manufacturing, control, etc.) are grater than the imbalance, therefore more important.

 

5 minutes ago, arkie87 said:

You can apply conservation of energy at any level you choose

Providade that you add or subtract the energy the entered or left the system (for open systems) or the system is closed. YOU CANNOT apply it to open systems, bacause, by deffinition, energy can enter or exit the system.

[arkie87]

12 minutes ago, VaPaL said:

Negligible =/= nonexistent, if you punch the numbers with enough precision you will get an inbalance. Also, some time it's ignored because others errors (measuring, manufacturing, control, etc.) are grater than the imbalance, therefore more important.

My integrals are analytical. If w doesnt change, we have an analytical match, so i dont know what you are talking about.

12 minutes ago, VaPaL said:

Providade that you add or subtract the energy the entered or left the system (for open systems) or the system is closed. YOU CANNOT apply it to open systems, bacause, by deffinition, energy can enter or exit the system.

What? Ever heard of transport equation? You can apply conservation of energy for open and closed systems, it just has slightly different form. sum(delta_energy) = 0 vs sum(delta_energy) = IN - OUT

[VaPaL]

4 minutes ago, arkie87 said:

What? Ever heard of transport equation? You can apply conservation of energy for open and closed systems, it just has slightly different form. sum(delta_energy) = 0 vs sum(delta_energy) = IN - OUT

That's what I said here:

4 minutes ago, arkie87 said:

Providade that you add or subtract the energy the entered or left the system

EDIT: Above I queted myself from you reply, so the quote is attributed to you, and I'm not able to add quotes in a edit for some reason, so the post I quoted myself was:

 

Edited June 12, 2018 by VaPaL

[arkie87]

13 minutes ago, VaPaL said:

That's what I said here:

EDIT: Above I queted myself from you reply, so the quote is attributed to you, and I'm not able to add quotes in a edit for some reason, so the post I quoted myself was:

 

You said you cannot apply it to an open system. What are you talking about then? 

[VaPaL]

3 minutes ago, arkie87 said:

You said you cannot apply it to an open system. What are you talking about then? 

I was talking about delta energy = 0 only, I should have clarified better

[arkie87]

3 minutes ago, VaPaL said:

I was talking about delta energy = 0 only, I should have clarified better

I'm not sure when i said that delta_energy needed to be zero... just that energy needed to be conserved. 

[YNM]

1 hour ago, arkie87 said:

What? The mass doesnt retract. The body does...

... And rotational movements are dependent of rotational inertia, which refers to mass distribution. Not just mass.

We're not doing free energy here. A tiny tyre does take less energy to roll than a large one, even if they have the same mass.

Edited June 12, 2018 by YNM

[arkie87]

1 minute ago, YNM said:

... And rotational movements are dependent of rotational inertia, which refers to mass distribution. Not just mass.

We're not doing free energy here. A tiny tyre does take less energy to roll than a large one, even if they have the same mass.

What's your point? I am talking about A and you are talking about B and C.

[YNM]

4 minutes ago, arkie87 said:

What's your point? I am talking about A and you are talking about B and C.

"Retracting" is a form of mass distribution change. Which means the moment of inertia will change, and so the rotational speed will change to compensate for it.

Do tell me if this isn't what it's about though.

Edited June 12, 2018 by YNM
Bloody WYSIWG ! real slow...