Why don't I weigh more at the poles

[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

 

[russ_watters]

BUT, do we live in an Einsteinian Relativity world or not?

Yes we do, but it isn't always relevant. For situations where the energy involved isn't nuclear, the equivalence of matter and energy produces immeasurably tiny changes in mass. Recently, there was a thread asking about the relativistic mass change in burning oil, and I calculated it was something like 1/100,000,000th %.

Some nuclar reactions have mass conversions to energy on the order of .1%. Still pretty small, but a lot more than in a chemical reaction. That tells you that nuclear reactions are really powerful compared to chemical reactions.

Do we measure weight on Earth differently than in space?

Weight is properly measured with a spring-scale, anywhere.

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

Yes, that's correct. If you are unsure, do the calculations yourself. They are easy. Here's that other thread I was talking about, with an example for the energy released in burning oil (just plug the energy into e=mc^2 to find the equivalent mass): https://www.physicsforums.com/showthread.php?t=170280

 

[D H]

Weight is properly measured with a spring-scale, anywhere.

There is no way to measure "weight" if one defines "weight" as the "force exerted on an object by gravity". Balance scales measure mass. Spring scales measure the total of all forces acting on a body except for gravity: in short, everything but weight. Using a spring scale in a jet pulling 6Gs would yield a markedly different result than using the same scale on the ground, even though the object's "weight" is the unchanged.

This whole thread has been talking about many different things: mass, weight (force of gravity), weight (what a scale measures), the Earth's gravitational potential, and geopotential. Note: the potential fields are typically represented in units of Joules/kg.

The geopotential includes the Earth's gravitational and a potential resulting from the Earth's rotation. Both the geopotential and gravitational potential are "real". The Earth's rotation affects the shape of the Earth and of the oceans that cover the Earth. The mean sea surface is an equipotential surface of the geopotential, not the gravitational potential.

Taking the gradient of the gravitational potential yields the acceleration due to gravity. Multiplying this by the mass of some object yields its "weight". As noted above, this is not what a scale measures. Taking the gradient of the geopotential yields an acceleration as well. The additive inverse of this acceleration is what a spring scale measures: the force that must be applied to an object to make it stationary with respect to the rotating Earth.

 

[MadMax]

Didn't read the whole thread, so not sure if it was mentioned... but at the start people were mentioning "centrifugal force". There is no such thing. There is only a centripetal force which pulls us towards the centre of the earth. This is countered by a reaction force from the surface of the earth, and the net force is exactly the force needed to keep us in our geostationary orbit on the surface of the earth.

 

[AlephZero]

All this was sorted out pretty well by Newton - see Principia Book 3, Proposition 20, Problem 4.

This was "big science" at the time, because sailors were trying to use clocks as navigation tools to find longtitude, and had discovered that pendulum clocks ran at different speeds at different latitudes. There was a big debate about whether this was caused by changes in "gravity" (i.e. what we would now call the resultant of gravity and centripetal effects) or by the clock pendulums expanding at the equator because the climate was hotter. There was no agreed scale of temperature and no accurate thermometers, so nailing down the temperature effects was a serious experimental problem.

Newton takes about 3 pages to list all the experimental data he has on the subject, gives a table showing the length of a pendulum with period 1 sec at various latitudes between the equator and the poles according to his theory, and concludes that "if the experimental observations are correct, the Earth is higher under the equator than at the poles, by an excess of about 17 miles".

Been there, done that.

 

[laurelelizabeth]

I have wondered this sort of thing too. When they calculated the gravity of Earth (F=Gm1m2/r^2), they must have taken into account the spinning of the Earth (F=mv^2/r) so the Fnet would really be Gm1m2/r^2-mv^2/r

I'm assuming they took that into account otherwise they wouldn't have been able to send spaceships anywhere, right?

 

[bligh]

Laurel. I was trying to think in terms of Potential E and Kinetic E of an isolated mass (me) at Equator and Pole with Gravity one force, Centrifug force another etc. I think the moderator told me there is no way to have a strain gauge in space to measure weight of an isolated mass there, or something similar.
Anyway, I am an amateur physicist. One year college 1958 or so.
I do enjoy reading physics, however. Lay type Quantum stuff. Enjoyed Bagott's "the meaning of quantum mechanics"
Anyway, with enough energy you could launch almost anyone?
B

 

[laurelelizabeth]

Sorry i didn't make myself clear -- i meant that they must have taken into account centripetal force from the Earth spinning when figuring out the gravitational constant (G), because otherwise their calculations would be wrong about other things -- like the distance to the moon, in which case they couldn't have orbited the moon. :)