Why don't I weigh more at the poles

  • Thread starter notsureanymore
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  • #26
notsureanymore
Yes I realise the earth does wobble. in a slow deliberate way thus we have seasons.

What I meant was that some event which causes the earth to become imbalanced
although the chance of an imbalance being so large as to throw us out of orbit ever occuring is negligable. it was just a passing comment. We would be dead long long before that.

Any 'sudden' change in the earths rotation, even relativaly minor would be disastrous. (think of the winds) but again the mass of the earth is such that it is extraordinarilly unlikely.

My original thought was of the oceans acting as a balancer. smoothing the rotation of the earth so that minor imbalances could not occur.

As you know even a very minor imbalance can cause a tyre to shred or a washing machine to 'walk'. thats why modern washing machines have balancing mechanisms (usually a fluid ring) and car tryes are regularly checked and rebalanced.
 
  • #27
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I can make a correlation to the harmonic balance on a cars engine.It balances things out, also.
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?
 
  • #28
russ_watters
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errorist said:
I can make a correlation to the harmonic balance on a cars engine.It balances things out, also.
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?
We've discussed that before and it still wouldn't work.
 
  • #29
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Show me the mathmatical formula as to why it won't. Then I will believe you!
 
  • #30
BobG
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notsureanymore said:
Yes I realise the earth does wobble. in a slow deliberate way thus we have seasons.

What I meant was that some event which causes the earth to become imbalanced
although the chance of an imbalance being so large as to throw us out of orbit ever occuring is negligable. it was just a passing comment. We would be dead long long before that.

Any 'sudden' change in the earths rotation, even relativaly minor would be disastrous. (think of the winds) but again the mass of the earth is such that it is extraordinarilly unlikely.

My original thought was of the oceans acting as a balancer. smoothing the rotation of the earth so that minor imbalances could not occur.

As you know even a very minor imbalance can cause a tyre to shred or a washing machine to 'walk'. thats why modern washing machines have balancing mechanisms (usually a fluid ring) and car tryes are regularly checked and rebalanced.
Actually, we have seasons because the Earth's axis is tilted relative to the Earth's orbit plane.

The oceans are a little more volatile than the surface. The Moon pulls them back and forth all the time. That's why folks on the coast have to keep track of tides. It also changes the Earth's moment of inertia constantly, which actually adds to the imbalance, not decreases it. Not only is there a second gravitational force acting on us, but our own mass tries to realign itself with the axis between the Moon and Earth.

Your idea about the tyre is a good point, but there's a little more to the tyre problem. If it were just rotating in free space instead trying to support and steer your car, it wouldn't shred. And if the floor weren't in the way, your washing machine would be perfectly happy rotating in an unbalanced manner.
 
  • #31
Gokul43201 said:
[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.
Andre said:
[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?
BobG said:
[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.
pervect said:
[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> [Broken]
When you said "not very legible" I didn't expect you to mean it literally. How often do you see penmanship on the web? :)
notsureanymore said:
[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.
notsureanymore said:
[post]428858[/post]:
Yes I realise 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.
errorist said:
[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.
 
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  • #32
pervect
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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.
 
  • #33
pervect said:
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.
 
  • #34
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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.
 
  • #35
notsureanymore
BobG said:
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.
 
  • #36
pervect
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GrapesOfWrath said:
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/earths_geoid.htm [Broken]

Note that dynamic effects such as tides are specifically excluded.
 
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  • #37
pervect
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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

[tex]\Phi_a[/tex] is the potential whose gradient gives the gravitational force, and

[tex]\Phi_e[/tex] is the potential whose gradient gives the centrifugal force.

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

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).
 
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  • #38
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"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.
 
  • #39
chroot
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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
 
  • #40
pervect
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errorist said:
"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.
 
  • #41
Andrew Mason
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notsureanymore said:
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
 
  • #42
BobG
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notsureanymore said:
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
 
  • #43
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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?
 
  • #44
chroot
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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
 
  • #45
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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.
 
  • #46
JasonRox
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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.
 
  • #47
chroot
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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
 
  • #48
russ_watters
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errorist said:
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.
 
  • #49
notsureanymore
Andrew Mason said:
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.
 
  • #50
russ_watters
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notsureanymore said:
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).
 

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