Why don't I weigh more at the poles

[GrapesOfWrath]

[post]427794[/post]:
The two effects should nearly cancel each other, making the surface of the Earth an equipotential.

EDIT : Googling, I've found that I'm wrong...but I can't see how.

Just because the surface is an equipotential, does not mean that the force or acceleration due to
gravity is the same at all points of the surface.
In order for the surface to be an equipotential, no energy should be gained in going from one point to another--there would be no flow. The force perpendicular to the surface may vary.

[post]428223[/post]:
But gravity is not the biggest anywhere on the surface of the Earth. So where would that be instead?

I'm not sure I understand this question. Can you rephrase it?

[post]428483[/post]:
The Earth's rotation rate is slowing (ever so slightly), but the redistribution of mass to match the rotation rate is even slower. A lot of the motion of tectonic plates, etc is the Earth trying to redistribute its mass to match the current rotation rate.

The Earth has mostly adjusted to the decrease in rotation rate. There was a theory, forty years ago, that the excess bulge at the equator (about 100 meters) was the result of a delay in the response of the earth, but that resulted in such a high value for the viscosity of the Earth mantle, that plate tectonics would have been impossible.

More than likely, the plate motions are caused by convection of the Earth's mantle.

[post]428563[/post]:
The origin of this expression may appear mysterious but it's not really, though it's a bit involved.

It's basically the result of a series expansion. There is an unfortunately not very legible derivation online

<http://www.apl.ucl.ac.uk/lectures/3c37/3c37-6.html>

When you said "not very legible" I didn't expect you to mean it literally. How often do you see penmanship on the web? :)

[post]428608[/post]:
I was wondering about the changing shape of the earth. as its rotational speed changes it needs to readjust its shape to maintain equilibruim.

Hence the plate shift that recently killed so many.

Most of the plate shifts are not caused by adjustments to the slowing of the Earth's rotation, as I said above.

[post]428858[/post]:
Yes I realize the Earth does wobble. in a slow deliberate way thus we have seasons.

The Earth's wobble does not produce the seasons. The tilt of the axis relative to the Earth orbit does. The Earth wobbles, but even without a wobble there would still be seasons.

[post]428875[/post]:
Anyways, the force is greater at the poles and an underwater pipeline run from the pole to the Equator would indeed flow to the equator. Even if there is friction and coriolis forces involved. Any bets as to what would happen?

The force of gravity is greater at the poles, but as many people have pointed out, the (sealevel) surface is an equipotential surface--which means that water would not flow. After all, the ocean is free and unconstrained to flow from the north pole to the equator now.

 

[pervect]

For what it's worth, it's my understanding that it's actually the Hamiltonian / energy function that's constant on the Earth's surface at sea level.

If we use cartesian coordinates that co-rotate with the earth, assumed to rotate with an angular frequency w, with the x-y plane being the equatior, the energy function h for a small test mass of mass m in terms of x,y,z and their derivatives xdot,ydot,zdot is:

h = .5*m*(xdot^2+ydot^2+zdot^2) - .5*m*w^2*(x^2+y^2) + V(x,y,z)

here V(x,y,z) is the actual gravitational potential. The expression

V(x,y,z) - .5*m*w^2*(x^2+y^2)

is sort of an "effective potential" that includes the work done by centrifugal forces to yield a conserved quantity in the rotating coordinate system. (The detailed argument for it being conserved requires Hamiltonian mechanics, unfortunately - well, there's probably some way of reformulating it without using Hamiltonian mechanics with enough effort). It is the above effective potential which is constant for the Earth's surface AFAIK.

Because it's expressed in terms of x,y,z and their derivatives, the expression for h is called the energy function. If we substitute for the conjugate momenta

px = m*(xdot - w*y)
py = m*(ydot + w*x)
pz = m*zdot

we get the usual expression for the Hamiltonian H as a function of position and the conjugate momentum in a rotating coordinate system, which is

H = px^2/2m + py^2/2m + pz^2/2m + w*(px*y-py*x) + V(x,y,z)

Real enthusiasts who want to derive the above from first principles using Hamiltonian mechanics can start with

x_inertial := x(t)*cos(w*t)-y(t)*sin(w*t)
y_inertial := y(t)*cos(w*t)+x(t)*sin(w*t)
z_inertial := z


compute

L = .5*m*((d x_inertial/dt)^2 + (d y_inertial/dt)^2) + (dz/dt)^2) + V(x,y,z)

using the chain rule to find the total derivatives,

find the conjugate momenta

px = dL/dxdot
py=dL/dydot
pz=dL/dzdot

and compute the energy function via

h = xdot*(dL/dxdot) + ydot*(dL/dydot) + zdot*(dL/dzdot) - L

The last step in the computation of the Hamiltoniain, changing the variables to compute H, is optional in this case, as it's mostly the energy function we're interested in.

 

[GrapesOfWrath]

For what it's worth, it's my understanding that it's actually the Hamiltonian / energy function that's constant on the Earth's surface at sea level.

It does depend upon what we mean by sea level--usually, that is a datum, and the tides are high or low with respect to it. At the level of detail of the tides (or wind and shore conditions), it's not constant at sea level. But you're right, there are potentials other than gravitational that come into play.

 

[errorist]

Nonetheless, the gravity potential at sea level at the poles is greater than the gravity potential at the equator because of the 14 mile difference in elevation. Therefore, the poles have a greater downward net force in this example.

 

[notsureanymore]

Actually, we have seasons because the Earth's axis is tilted relative to the Earth's orbit plane.

It is the wobble that gives us the seasons not the tilt, if it was just the tilt then it would be always summer in one hemisphere and always winter in the other. which is not 'seasonal'

And that gives rise to another question which is unrelated and so should be in another thread.

Why is that wobble aligned so closely aligned to the period of rotation around the sun.
(1 year) coincidence or something else? As I said another thread so no answers to this musing here.

 

[pervect]

It does depend upon what we mean by sea level-

By sea level I actually mean "on the geoid".

http://www.ngs.noaa.gov/GEOID/geoid_def.html

or better yet

http://geophysics.ou.edu/solid_earth/notes/geoid/Earth's_geoid.htm

Note that dynamic effects such as tides are specifically excluded.

 

[pervect]

I found a reasonable reference on-line for the "effective potential" that I mentioned in my previous post, though it takes careful reading to find all of the definitions of the terms used in the following URL:

http://stommel.tamu.edu/~baum/reid/book1/book/node42.html

Note that the "force" (by which I mean generalized force because the Earth is not an inertial frame of reference) of gravity on the Earth's surface is the sum of two separate forces - a centrifugal force component fc, and a gravitational force component fg.

These forces are each the gradient of a potential function, in the notation of the web page above

\Phi_a is the potential whose gradient gives the gravitational force, and

\Phi_e is the potential whose gradient gives the centrifugal force.

The sum of these two potential functions \Phi = \Phi_a + \Phi_e is what I called the "effective potential". The Earth's geoid is an equipotential function in terms of this effective/total potential \Phi, which is the sum of \Phi_a + \Phi_e

Thus it is incorrect to say that the _potential_ is different at the Earth's poles and it's equator. The Earth's surface, the geoid, is an equipotential surface (with a partricular defintion of potential, one that includes terms due both to gravity and the Earth's rotation).

It is correct to note that the gravitational force at the poles is greater than the gravitational force at the equator, however. Not only will fg, the force due to gravity, be greater at the pole, but at the equator fc will oppose fg, making the total force f=fg+fc even lower at the equator (fc is of course zero at the poles).

 

[errorist]

"It is correct to note that the gravitational force at the poles is greater than the gravitational force at the equator, however. Not only will fg, the force due to gravity, be greater at the pole, but at the equator fc will oppose fg, making the total force f=fg+fc even lower at the equator (fc is of course zero at the poles)."

Correct.
So a straight pipeline run underwater 50 feet below the surface and fully primed would flow all the way to the Equator from the poles if both open ends are 50 feet below the surface.

 

[chroot]

So a straight pipeline run underwater 50 feet below the surface and fully primed would flow all the way to the Equator from the poles if both open ends are 50 feet below the surface.

This is not correct. Sea level is an equipotential surface. The two points, both 50 feet under sea level, are at identical gravitational potential.

- Warren

 

[pervect]

"It is correct to note that the gravitational force at the poles is greater than the gravitational force at the equator, however. Not only will fg, the force due to gravity, be greater at the pole, but at the equator fc will oppose fg, making the total force f=fg+fc even lower at the equator (fc is of course zero at the poles)."

Correct.
So a straight pipeline run underwater 50 feet below the surface and fully primed would flow all the way to the Equator from the poles if both open ends are 50 feet below the surface.

Nope.

Consider a tank of water in a uniform gravitational field for simplicity. The potential at 50 feet depth in the tank and at 51 feet depth in the tank is different, the potential energy being given in this simple case by -m*g*h. Yet a pipe going from 50 feet down to 51 feet will not experinece a net flow of water. The reason for this is very simple, the pressure at 51 feet of depth is greater than the pressure at 50 feet of depth. When the system is in equilibrium there is no net flow through a vertical pipe, because the potential difference between 50 feet and 51 feet is exactly compensated by the pressure difference.

The situation with your pipe from the equator to the poles is similar. The surface of the water has the same potential at both places (with the previously mentioned definition of potential, which includes a term due to the Earth's rotation as well as a term due to the Earth's gravitation.)

Let's call the potential at the surface of the Earth 0 for convenience. The gravitational field at the poles is g1, so the potential 50 feet below the surface is -m*g1*h, where h=50. The gravitational field at the equator is g2, so the potential 50 feet below the surface is -m*g2*h, where h=50. Since g1>g2, the potential is indeed different 50 feet below the surface. But the pressure 50 feet below the surface of the water is also different at the poles and the equator. The difference in pressure prevents any net flow, just as it does in the simple example of a tank of water with a pipe going from 50 feet down to 51 feet.

 

[Andrew Mason]

It is the wobble that gives us the seasons not the tilt, if it was just the tilt then it would be always summer in one hemisphere and always winter in the other. which is not 'seasonal'

The period of the Earth's 'wobble' is about 26,000 years. I don't think wobble can explain the seasons.

Why is that wobble aligned so closely aligned to the period of rotation around the sun. (1 year) coincidence or something else?

The Earth's axis changes its direction relative to a radial vector to the sun as it travels in its orbit around the sun. This is because that radial vector changes direction, not because the Earth's axis changes direction. The Earth's axis points in the same direction (ie. to the same star) at all times, subject, as I said, to the 26,000 year period of wobble.

AM

 

[BobG]

It is the wobble that gives us the seasons not the tilt, if it was just the tilt then it would be always summer in one hemisphere and always winter in the other. which is not 'seasonal'

And that gives rise to another question which is unrelated and so should be in another thread.

Why is that wobble aligned so closely aligned to the period of rotation around the sun.
(1 year) coincidence or something else? As I said another thread so no answers to this musing here.

I would pretty much echo Andrew Mason's post. The Earth's axis always points the same direction. The only thing that changes is the Earth's location. In the summer we're on one side of the Sun (the Northern Hemisphere is tilted towards the Sun). In the winter, we're on the opposite side of the Sun (the Southern Hemisphere is tilted away from the Sun).

Or, even easier to visualize, http://www.enchantedlearning.com/subjects/astronomy/planets/earth/Seasons.shtml

 

[errorist]

Let's call the potential at the surface of the Earth 0 for convenience. The gravitational field at the poles is g1, so the potential 50 feet below the surface is -m*g1*h, where h=50. The gravitational field at the equator is g2, so the potential 50 feet below the surface is -m*g2*h, where h=50. Since g1>g2, the potential is indeed "different 50 feet below the surface. But the pressure 50 feet below the surface of the water is also different at the poles and the equator. The difference in pressure prevents any net flow, just as it does in the simple example of a tank of water with a pipe going from 50 feet down to 51 feet."

Now throw in rotational velocity. What happens?

 

[chroot]

errorist,

The surface of the water is a gravitational equipotential surface. (As you might be aware, water flows downhill until it can flow no longer.) The pressure is the same 50 feet underwater everywhere in the world.

- Warren

 

[errorist]

Sorry to tell you but the radius is different at the pole by 14 miles thus the gravity is different when compared to the equatorial radius.

 

[JasonRox]

It is totally obvious.

There is no real transportation to get to the poles. All the work that you put into getting there results in losing weight, therefore that counter balances what you think would happen.

 

[chroot]

errorist,

You already started a thread about this more than a month ago: https://www.physicsforums.com/showthread.php?t=56857. Your arguments were shown to be false many times in that thread. Several people went to great lengths to explain the simple physics to you. You are now simply repeating the same arguments again, as if you have not listened to a word anyone has said to you. This kind of behavior is not welcome here.

- Warren

 

[russ_watters]

Sorry to tell you but the radius is different at the pole by 14 miles thus the gravity is different when compared to the equatorial radius.

This is still wrong. It doesn't matter how many times you say it. The relevant datum is sea level. If water would flow through your pipe, it would also flow without your pipe.

 

[notsureanymore]

The Earth's axis changes its direction relative to a radial vector to the sun as it travels in its orbit around the sun. This is because that radial vector changes direction, not because the Earth's axis changes direction. The Earth's axis points in the same direction (ie. to the same star) at all times, subject, as I said, to the 26,000 year period of wobble.

Silly me. and I always believed the Earth rocked on its axis. and this was the cause of seasons, I wonder what gave me that idea?

Oh well another lesson learnt. Thanks.

 

[russ_watters]

Silly me. and I always believed the Earth rocked on its axis. and this was the cause of seasons, I wonder what gave me that idea?

Oh well another lesson learnt. Thanks.

I'm not being sarcastic here: if it helps, hold up a pair of oranges and see for yourself. It can be counterintuitive and its similar to the reason we see the same side of the moon all the time, yet it rotates. Another simple piece of evidence: Polaris is always the north star (with the already stated caveat).

 

[pervect]

errorist,

The surface of the water is a gravitational equipotential surface. (As you might be aware, water flows downhill until it can flow no longer.) The pressure is the same 50 feet underwater everywhere in the world.

- Warren

Errorist actually got this from my post. I believe that I am correct in stating that the pressure 50 feet under the surface is (ever so slightly!) different at the poles than at the equator.

The formula for static fluid presure is

P = \rho * g * h

where \mbox{\rho} is the density of the fluid, g is the acceleration of gravity, and h is the height

reference:
http://hyperphysics.phy-astr.gsu.edu/hbase/pflu.html#fp

Because 'g' is ever so slightly different at the poles and the equator, the pressure is ever so slightly different 50 feet under the surface of the ocean at the poles than it is at the equator. (A purist might also point out that the density of water is different at the poles and the equator, because the temperature is different.)

NONE of these comments should be taken to support errorists strange idea that water will flow from the poles to the equator. This is simply untrue.

The main point I'd like to make is that surfaces of constant pressure on the Earth will be equipotential surfaces.

The secondary point I'd like to make is that the potential function defining the shape of the Earth includes both gravitational terms, and terms due to the rotation of the Earth, as described in

http://stommel.tamu.edu/~baum/reid/book1/book/node42.html

The third point I'd like to make is that equipotential surfaces will not occur at a constant spacing - because the Earth is oblate, equipotential surfaces will be closer together at the poles than they are at the equator. (Hence, a surface that is uniformly 50 feet below the geoid will not be an equipotential surface).

The fourth point I'd like to make is that because the Earth is reasonably close to a state of hydrodynamic equilibrium, a pipe connecting any two points of the ocean will not transport water (Small exceptions might occur due to tides and weather - the Earth is only approximately in a state of hydrodynamic equilibrium). A vertical pipe will not transport water, even though it connects region of differing potential, because of the pressure differences. A horizontal pipe along an equipotential surface will also not transport water. A combination of a vertical and a horizontal pipe will also not transport water. No combination of pipes will transport water! It won't happen!

 

[Andrew Mason]

The fourth point I'd like to make is that because the Earth is reasonably close to a state of hydrodynamic equilibrium, a pipe connecting any two points of the ocean will not transport water (Small exceptions might occur due to tides and weather - the Earth is only approximately in a state of hydrodynamic equilibrium). A vertical pipe will not transport water, even though it connects region of differing potential, because of the pressure differences. A horizontal pipe along an equipotential surface will also not transport water. A combination of a vertical and a horizontal pipe will also not transport water. No combination of pipes will transport water! It won't happen!

The height of water surface at the equator, relative to the centre of the earth, is greater than the height at the north pole. If the surface is an equipotential surface, which I agree must be the case, I don't see how that equipotential could extend to any depth. It would be complicated to work out because as you go deeper, the pressure gradient, which is a function of water depth and gravity and temperature, would differ at the equator and at the poles. This results in ocean currents.

So I don't think it is true that the Earth is in a state of hydrodynamic equilibrium. How would you explain the ocean currents if that is the case?

If the equator was not warm we could see a huge movement of water away from the equator that would not be returned.

AM

 

[russ_watters]

So I don't think it is true that the Earth is in a state of hydrodynamic equilibrium. How would you explain the ocean currents if that is the case?

If the equator was not warm we could see a huge movement of water away from the equator that would not be returned.

AM

Temperature variation and the rotation of the Earth cause the currents, they don't reduce them.

 

[pervect]

The height of water surface at the equator, relative to the centre of the earth, is greater than the height at the north pole. If the surface is an equipotential surface, which I agree must be the case, I don't see how that equipotential could extend to any depth. It would be complicated to work out because as you go deeper, the pressure gradient, which is a function of water depth and gravity and temperature, would differ at the equator and at the poles. This results in ocean currents.

So I don't think it is true that the Earth is in a state of hydrodynamic equilibrium. How would you explain the ocean currents if that is the case?

If the equator was not warm we could see a huge movement of water away from the equator that would not be returned.

AM

OK, I'll agree that the Earth is not in a perfect state of thermodynamic equilibrium - ocean currents are a good example.

If you want to work out the exact solution that takes into account the density variation of water with temperature and pressure, the problem would indeed be complicated. But one can gain a lot of insight from the solution where the density is constant.

The equations for hydrodynamic equilibrium are fairly simple

\rho F = \nabla P

http://astron.berkeley.edu/~jrg/ay202/node6.html

here \mbox{\rho} is the density, and F is the force, and P is the pressure.

Because the force is a gradient of a conservative potential function one can write

\rho \nabla \Phi = \nabla P

This has a particularly simple solution (derived in the above URL)

\rho \Phi + P = constant

where \mbox{\Phi} is the potential function

From this equation it's obvious that contours of constant pressure are contours of constant potential.

The non-constant density case is more complicated, the above URL demonstrates that in the most general case for a fluid in equilibirium, the surfaces of constant density must be surfaces of constant potential, and that this implies that the surfaces of constant pressure are also surfaces of constant potential,

This is quite a mouthful, but what it means is that if one ignores the issue of heat transport, a big blob of fluid or fluid covered rock would reach an equilibrium condition where the surfaces of constant potential, constant temperature, constant pressure, and constant density all coincided (assuming that denisity of the fluid is a function of pressure and temperature alone).

Our Earth departs slightly from this equilibrium condtion because the sun heats the water, air, and ground, causing non-equilbirum flows such as ocean currents, air currents, winds, etc etc. Without the sun, or some other mechanism involving generated heat/heat transport/heat flow, these currents would not exist.

 

[pervect]

I found a much better link that goes through and calculates the actual flattening of the Earth for anyone really interested.

http://quake.mit.edu/ftp/12.201-12.501/ch2.pdf

It's quite a long paper, but it discusses how to correct the gravitational field of the Earth for the oblateness parameter J2 (going into the details of the Legendre polynomial expansion) and derives the expression for the geopotential U - the solution of which U=constant gives the overall shape of the Earth.

It also gives some useful formulas for the acceleration of Earth's gravity as a function of latitude (including some accepted international standard formulas), and seems to have just about anything anyone would want to know on the topic of the Earth's figure.

 

[notsureanymore]

Excellent find pervect. I have not fully read it, but it seems to cover my original question precisely.


... and a lot more

 

[bligh]

why I don't weigh more at the poles

When someone posed this to me I reasoned that mass is equal to energy and that the distance to the center of Earth would be shorter at the poles, therefore
there would be less potential energy in my mass at the poles and if it were possible to measure (someone mentioned a strain gauge) I would weigh less at the poles. Less total energy equal to less total mass, at least in the universe as a whole. This may not answer the question however if it is limited to Newton's concept of gravity alone?
Bligh

 

[rcgldr]

The constant for gravitational acceleration is considered to be 9.822 meters / second^2 when dealing with spacecraft , because the rotation of the Earth is not a factor of an orbiting satellite. I'm not sure what "radius" is used as the constant for spacecraft .

Regarding the tides, because the water in the oceans are pulled towards the moon, the center of mass of the Earth moves in a orbit similar to the moon (but out of phase). The result is the Earth's rate of rotation slows down and the distance between the moon and Earth increase at about 3.8 cm per year.

http://en.wikipedia.org/wiki/Moon

 

[russ_watters]

When someone posed this to me I reasoned that mass is equal to energy and that the distance to the center of Earth would be shorter at the poles, therefore
there would be less potential energy in my mass at the poles and if it were possible to measure (someone mentioned a strain gauge) I would weigh less at the poles. Less total energy equal to less total mass, at least in the universe as a whole. This may not answer the question however if it is limited to Newton's concept of gravity alone?
Bligh

Mass, gravitational potential energy, and weight aren't the same things - you're mixing three separate concepts.

The mass/energy equivalency has to do with nuclear reactions and the conversion from one to the other. It does not apply here. For this discussion, mass is a completely constant physical property of an object.

For potential energy vs weight, as two objects get closer together, the potential energy decreases because it is the integral of force over distance (shorter distance, less energy to gain by letting the two objects be pulled together), but weight (that force) is determined by Newton's theory of gravity, and increases as distance decreases.

 

[bligh]

Thank you Russ!
I thought I was mixing concepts
BUT, do we live in an Einsteinian Relativity world or not?
Do we measure weight on Earth differently than in space?
I guess I still don't understand why we don't factor in Potential Energy and Kinetic energy etc in questions of this sort.
I guess that Relativity doesn't come in unless we are talking of things nearer the speed of light or nearer the scale of nuclear processes?
Bligh