Why don't I weigh more at the poles

[notsureanymore]

Hi all

Another thought I have been pondering is why don't I weigh significantly more at the poles than I do at the equator

The Earth rotates at n kilometers per hour, I think around ~1300 kmph

Would this not create a centrigual effect countering Earth's gravity. thus making me lighter at the equator than the poles where no real rotation is evident.

I know the Earth has an equatorial bulge. but to me. it seems to be a very small bulge, does this bulge increase the Earth's mass at the equator enough to counter the centrifugal forces.

I read elsewhere pole to pole the Earth is 22 km's less in diameter than at the equator. this seems marginal.

And.. will the Earth eventually flatten ?

Chris H.

 

[vincentchan]

good question... the Earth rotation speed is about
2 \pi R / T
6 378100 m (radius of earth) * 2 pi / 1 day ( period of the Earth's rotation)
=2*3.14*6378100m/(60*60*24)sec
=464m/s at the equator
the centrifugal force is mv^2/R
for a 70kg man, the force will be
70*464^2/6378100
=2.36N...
F=mg
m=F/g=2.36/9.8=0.2kg
so, at the equater.. your weight will be about 0.2 kg less than the weight at the pole

 

[rachmaninoff]

There were several similar discussions recently, the major point of which was that the gravitational potential at sea level is the same everywhere.

 

[notsureanymore]

so, at the equater.. your weight will be about 0.2 kg less than the weight at the pole

Ok the cetrifugal force is less than I anticipated. so the smallish bulge is enough to balance the gravitational forces.

This of course occurs at all points on the surface.

The upshot is that if the Earth was spherical and I wanted to lose weight I would move to the equator.

but as the Earth is suffering middle age spread I will have to go back to the gym.

Damn shame that .

Chris H.

 

[errorist]

Actually, the force of gravity is a little different at the pole. The pole is about 14 miles below sea level when compared to the Equator do to rotational velocity forcing the mass at the equator to bulge out. The difference is due to the altitude difference from the center of the Earth. Gravity changes with altitude.
This means there is more downward force at the poles.So, this must be added to the effect of rotational velocity.

 

[DaveC426913]

"And.. will the Earth eventually flatten ?"

No. The two forces ('centrifugal' and gravitational) reach equilibrium at some point and the flattening does not continue.

I expect that happened during the formation of the planet. I don't know of any evidence to suggest that the bulging is changing in present day.

 

[errorist]

The gravity is different due to the altitude difference. You can't get away from that. More downward force is there due to the elevation.

 

[BobG]

The force of gravity is more at the poles both due to being closer to the center of the Earth and due to no centrifugal force. The use of 9.8 m/sec^2 is used as an approximation regardless of your latitude, or whether you're standing on Mt Everest or at sea level, because the difference is so small.

Btw, you wouldn't 'weigh' .2kg less at the poles because kg is a measure of mass, not weight. Regardless of how a particular scale uses or misuses units, you won't notice any difference on a balance scale since that's really measuring mass - you might notice a difference on a spring scale, if it's sensitive enough.

 

[errorist]

The use of 9.8 m/sec^2 is used as an approximation regardless of your latitude, or whether you're standing on Mt Everest or at sea level, because the difference is so small.

Correct. But, there is still a gravity difference because you are closer to the center of the Earth at the poles. No way around that.

 

[motswac]

Hi all

Another thought I have been pondering is why don't I weigh significantly more at the poles than I do at the equator

The Earth rotates at n kilometers per hour, I think around ~1300 kmph

Would this not create a centrigual effect countering Earth's gravity. thus making me lighter at the equator than the poles where no real rotation is evident.

I know the Earth has an equatorial bulge. but to me. it seems to be a very small bulge, does this bulge increase the Earth's mass at the equator enough to counter the centrifugal forces.

I read elsewhere pole to pole the Earth is 22 km's less in diameter than at the equator. this seems marginal.

And.. will the Earth eventually flatten ?

Chris H.

hi, there,
i was just reading your message. are you suggesting that Earth's gravity has something to do with the Earth's rotational speed. if so, since we would not be attached by anything to Earth when it is stationary, wouldn't we be pushed away from Earth when it start's rotating? Newton's law of motion."we will continue moving on a straight line unless an external force acts upon us". If you suggest that this gravitaional force is created by our Earth's rotation, what force (external) is changing our course (from straight line to circular, along with the Earth's rotation?)


Cyp Mot

 

[JesseM]

hi, there,
i was just reading your message. are you suggesting that Earth's gravity has something to do with the Earth's rotational speed. if so, since we would not be attached by anything to Earth when it is stationary, wouldn't we be pushed away from Earth when it start's rotating? Newton's law of motion."we will continue moving on a straight line unless an external force acts upon us". If you suggest that this gravitaional force is created by our Earth's rotation, what force (external) is changing our course (from straight line to circular, along with the Earth's rotation?)


Cyp Mot

He wasn't saying gravity is caused by rotation, he was just saying that at the equator the centrifugal force would oppose the gravitational force, decreasing your weight slightly (assuming we define 'weight' in terms of the total force you feel, not solely in terms of the force due to gravity).

 

[notsureanymore]

The force of gravity is more at the poles both due to being closer to the center of the Earth

Ok this is interesting you are saying gravity is not a function of mass so much as a function of the centre of mass.


The following totally ignores other forces...
Wouldnt gravity be less at the poles due to less mass between you and the centre of the earth?

Wouldn't this also mean gravity is less at sea level over a deep portion of the ocean. than at the same level over land due to the fact that sea water is less dense than rock and magma?

What if you climbed a high mountain. Wouldnt gravity be more for the mountain climber than for the photographer hovering nearby in a helicopter. due to the large mass of the mountain below the mountaineer? (assuming the helicopter is not over the mountain)

OR

Is gravity the same at all points of equal radius from the centre of the Earth regardless of intervening mass?

 

[Gokul43201]

The gravitational equipotentials are spherical shells centered on the center of the earth. The field increases outwards as a linear function of the radial distance from the center.

E = \frac {GM}{r^2} = \frac{G\rho}{r^2}~\cdot~ \frac {4}{3} \pi r^3 = Kr

This expression holds from r=0 to r=R(poles). For R(poles) < r < R(local surface), you would expect the increase in the field to be less than linear, because the spherical shells are no longer completely filled with earth. Nevertheless, there must be an increase in the field.

But due to the centrifugal effect, the net field sees a reduction as described in an earlier post here.

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.

 

[errorist]

Is gravity the same at all points of equal radius from the centre of the Earth regardless of intervening mass?

In theory yes. But, in actuality no.

http://www.csr.utexas.edu/grace/gallery/animations/ggm01/


At the poles there is a fourteen mile difference in elevation than at the Equator. The amount of missing land mass at the pole does have an effect of the gravity. It is not enough to change or equal out the total difference.

 

[Andre]

OK Gokul was very close, I think.The most simple expression of Newtons law only accounts for point masses and for homogeneous spheres. Earth is not a homogeneous sphere by far. The most remarkable abaration is the equatorial bulge. This means that if you are at the equator, three things must be accounted for gravity correction.
First you're farther away of the centre oif the Earth this tends to decrease the gravitational force w/Newtons universal law. Next you're spinning, creating an apparent centrifugal force that is opposing gravity as well. But the excess mass is directly beneath you and this tends to increase gravity. The nett result is a very slight decrease.

At the poles the reverse is true. Closer to the centre, less R more gravity. No centrifugal force opposing gravity, but the excess mass is off to the sides instead of beneath you, that would decrease the gravitational force, compared to the sphere.

Therefore the gravity at sea level is not equal everywhere but there is a relation between gravity, and total pressure on the surface area, balancing the Earth.

But gravity is not the biggest anywhere on the surface of the Earth. So where would that be instead?

 

[BobG]

The gravitational equipotentials are spherical shells centered on the center of the earth. The field increases outwards as a linear function of the radial distance from the center.

E = \frac {GM}{r^2} = \frac{G\rho}{r^2}~\cdot~ \frac {4}{3} \pi r^3 = Kr

This expression holds from r=0 to r=R(poles). For R(poles) < r < R(local surface), you would expect the increase in the field to be less than linear, because the spherical shells are no longer completely filled with earth. Nevertheless, there must be an increase in the field.

But due to the centrifugal effect, the net field sees a reduction as described in an earlier post here.

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.

You would think so, but the only time the spherical shells model truly works for mass 'outside' your radius is if your mass really is a perfect sphere.

The perfect sphere model makes the real problem simpler. There is a certain amount of mass in front of you (towards the center of Earth) and a certain amount of mass behind you. Instead of calculating the amount of mass in front of you and the amount behind you, you can figure out how much mass way out on the opposite side is canceled by the mass behind you.

For non-spherical Earth, even though you're closer to the center of mass at the poles, all of the mass is still in front of you ... And you're closer to the mass on the opposite side of the sphere (and all the rest of the mass), as well.

Edit: Actually, not all of the mass is closer. Some of the mass near the equator is slightly further away. Your average distance from every point in the oblate sphere is closer and, more importantly, all of the mass is still 'in front' of you.

 

[errorist]

Seems to me if the Earth stopped rotating it would all equal out??

 

[BobG]

Seems to me if the Earth stopped rotating it would all equal out??

Eventually.

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.

So, yes, if the Earth stopped rotating, eventually the Earth would form a virtually perfect sphere.

 

[pervect]

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

The Earth isn't spherical, so integrating out the spherical shells won't give the right answer.

The Newtonian potential of the non-spherical Earth is used in the GPS calculations, for instance.

http://xxx.lanl.gov/abs/gr-qc/9508043 has an expression for it, which is their eq 2)

<br /> V = -(\frac{GM}{R})(1- \frac{J2}{2}(R/r)^2\,(3 cos^2(\theta)-1)))<br />

Theta here is the lattitude.

J2 is a commonly used measure of the Earth's departure from sphericity. It's definitinon is given in the NASA planetary data sheet

http://nssdc.gsfc.nasa.gov/planetary/factsheet/fact_notes.html

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

Some keywords if you want to go Google-fishing "Gravitatioanl Potential Theory, Legendre polynomials"

Goldstein also derives the gravitatioanl potential for an approximately spherical body in "Classical mechanics", pg 226.

I think I wrote a little about this in a thread here somewhere before, but I can't find it :-(.

 

[notsureanymore]

Slightly OT

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.

When a thought entered my cranium. the oceans act like a balancer. Now I am sure its only slight but without it wouldn't there be a risk of the Earth developing a wobble like an out of balance washing machine.

This could increase to the point of throwing us out of orbit (unlikely to ever get that severe of course), but without the oceans rebalancing the Earth everytime a major event occurs we would most likely be in a major p**.

now my thoughts wander to other planets without oceans. how do they cope? perhaps the atmosphere is dense enough to do the job?

Any thoughts?
(perhaps should this go to another thread.)

 

[errorist]

I think I wrote a little about this in a thread here somewhere before, but I can't find it :-(.

I think I started a thread about this a few days ago but they locked it for some reason. Why, I don't know. I think we were arguing to much or something.

 

[Integral]

now my thoughts wander to other planets without oceans. how do they cope? perhaps the atmosphere is dense enough to do the job?

Hang onto that thought, and realize that neither the moon nor Mercury have any atmosphere to speak of.

It is not clear to me that there is any way to disrupt the orbit of a planet by rearranging its mass.

The total energy of the system will remain constant. To change a planetary orbit requires that energy be added or removed from the system. A earthquake does not change the total energy.

Recall that the Earth's crust is pretty thin wrt to the radius of the Earth (roughly a few parts in a thousand), displacements of the crust, as in a major earthquake are on the order of a few meters, this is pretty insignificant on a planetary scale. It is said that the Indonesian earthquake caused a hitch in the rotation of the earth, note that this is rotation about the Earth's axis, NOT revolution about the sun.

 

[Gokul43201]

You would think so, but the only time the spherical shells model truly works for mass 'outside' your radius is if your mass really is a perfect sphere.

The Earth isn't spherical, so integrating out the spherical shells won't give the right answer.

Duh ! I knew there was something funny about my dinner last night. I blame it all on spicy chicken !

 

[BobG]

... I was wondering about the changing shape of the earth. as its rotational speed changes it needs to readjust its shape to maintain equilibruim. ...

This could increase to the point of throwing us out of orbit (unlikely to ever get that severe of course), but without the oceans rebalancing the Earth everytime a major event occurs we would most likely be in a major p**.

Integral's correct. You can't throw a planet out of orbit by changing it's rotation about it's axis.

but without it wouldn't there be a risk of the Earth developing a wobble like an out of balance washing machine

A risk? The Earth does wobble, thanks to torque from the Earth's axis being tilted relative to the Sun (about 23.4 degrees), thanks to the Earth's axis being tilted relative to the the Moon's orbit (the Moon's orbit has almost dropped into the same plane as the Earth's orbit around the Sun - it's only 5 degrees off or, it ranges from around 18 to 28 degrees relative to the Equator), and due to the other planets.

It's minor, but it is tracked by the International Earth Rotation Service ( http://www.iers.org/). They also track the Earth's rotation. You can peruse their bulletins to find out how fast the Earth was rotating a couple of months ago and how far the poles have moved (relative to celestial space). It takes them awhile to analyze the data considering the small magnitude of the variations their analyzing.

Check the Web site map. They have more Earth data than you ever imagined.

 

[BobG]

The Earth isn't spherical, so integrating out the spherical shells won't give the right answer.

The Newtonian potential of the non-spherical Earth is used in the GPS calculations, for instance.

http://xxx.lanl.gov/abs/gr-qc/9508043 has an expression for it, which is their eq 2)

<br /> V = -(\frac{GM}{R})(1- \frac{J2}{2}(R/r)^2\,(3 cos^2(\theta)-1)))<br />

Theta here is the lattitude.

J2 is a commonly used measure of the Earth's departure from sphericity. It's definitinon is given in the NASA planetary data sheet

http://nssdc.gsfc.nasa.gov/planetary/factsheet/fact_notes.html

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

Some keywords if you want to go Google-fishing "Gravitatioanl Potential Theory, Legendre polynomials"

Goldstein also derives the gravitatioanl potential for an approximately spherical body in "Classical mechanics", pg 226.

I think I wrote a little about this in a thread here somewhere before, but I can't find it :-(.

To really get a feel what they're doing with J2 and Legendre polynomials, check this site (http://www.floridageomatics.com/publications/gfl/toc.htm ), particularly the Appendix on spherical harmonics (http://www.floridageomatics.com/publications/gfl/spherical-harmonics.htm ). This is the best explanation I've seen on the internet and it's been copied, word for word, in many places. Unfortunately, the figures look worse and worse each time someone tries to copy them, so it's nice to read the original article.

This is really a sine regression where they created an equation to model the observed shape using Legendre polynomials.

 

[notsureanymore]

Yes I realize 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 occurring is negligable. it was just a passing comment. We would be dead long long before that.

Any 'sudden' change in the Earth's 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'. that's why modern washing machines have balancing mechanisms (usually a fluid ring) and car tryes are regularly checked and rebalanced.

 

[errorist]

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?

 

[russ_watters]

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.

 

[errorist]

Show me the mathmatical formula as to why it won't. Then I will believe you!

 

[BobG]

Yes I realize 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 occurring is negligable. it was just a passing comment. We would be dead long long before that.

Any 'sudden' change in the Earth's 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'. that's 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.

 

[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).