# Earth gravity

Andre
( I'm not making a poll of this. It no opinion.) But a simple question, where would you be weighting the most?

A - On the equator
B - On the North pole because you are closest to the centre of the Earth
C - On the North pole because there is no centrigufal force of the spinning Earth counteracting gravity
D - Indifferent. Gravity is not depending on lattitude.

So what is it?

Both B&C are true, but I believe the centrifugal effect is much greater than the effect from the non-spherical Earth.

Andre
Are you sure? Everybody agree?
[?]

Creator
Originally posted by Andre
( I'm not making a poll of this. It no opinion.) But a simple question, where would you be weighting the most?
.

Actually B & C .

The value of g at different latitudes is well established and well measured. At sea level, the the effective (measured) value of g:

g (At 0* -at equator) = 9.78039 m/sec^2

g (90* -at North pole) = 9.83217 m/sec^2

Thus you weigh more on the pole...
The difference in g values is about 0.05178 m/sec^2

Of this difference it is easily calculable that a little over half is due to the centripetal acceleration provided by Earth's rotation: (about 0.0336 m/sec^2).

A 100 kg. man on the pole could loose over 5 kg. simply by moving to the equator (provided both were measured at sea level).

Creator
P.S. We're thinking of starting a weight loss program for overweight Eskimos. We fly you to Indonesia and guaranteed a 5 % weight loss by the time you get off the plane.

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Andre
But why not "D"

I wonder if that eskimo "g" was weighted or calculated. The textbook answer, right? But I think it's not that easy. I was intending to have "D" as the correct aswer. gravity should not be depending on lattitude

Since the the Earth is spinning the centrifugal force at the equator are the strongest it tend to make you lighter. It also makes the Earth bulge outwards and it hence that bulge also put you are farther from the Earth's center of gravity at the equator so gravity is weaker, according to Newtons Universal Law of Gravitation. Of course the effects of both centrifugal force and reduced gravity add together so you seem to be much lighter. right? Wrong.

There is an error in that assumption. We assumed without saying that masses were the same as if concentrated in the centre of gravity. This is exactly true for spheres but we just deviated from the assumption that the Earth is a sphere. Hence this is not true anymore. We overlook another essential effect of the equatorial bulge. It puts more mass directly underneath you at the equator. This compensates exactly (theoretically) for that centrifugal force and the longer distance between the COG's.

Consequently if you are on the poles, you are closer to the COG, hence a stronger gravitation force and there is no centrifugal force left, so you should be a lot heavier. Again not true (theorically), because there is a lot less mass directly underneath you and more mass to the left and the right in the equatorial bulge. This compensates again exactly for the other effects.

Look at an extreme example and make the Earth flat like a pancake. If you're on the Northpole and hence in the center of the pancake, all the mass is to the left and right. There's almost nothing underneath and you're almost weightless. When standing on the the edge of the pancake (the equator) all the mass is below you and you will weight quite a bit.

Also consider this: the shape of the Earth (geodoid) is exactly like that because of equalizing all forces. If somewhere the gravity would be less, the Earth shape would correct for it automatically. (isostacy?).

That concept is not fully operational, because the Earth is not an ideal model by all means. There are a lot of local gravity variations. You seem to weight more in Indonesia and less on the indian ocean. http://www.Earth'sky.com/2002/esmi020629.html [Broken] are trying to find out how and what

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Staff Emeritus
Gold Member
Can you prove any of that?

Your pancake example is extremely pathological; assuming an actual finite mass is uniformly concentrated into a disk, an observer on the rim of the disk will feel infinite gravitational attraction, and the observer standing on the center of the disk will actually feel a gravitational force 3.414 times as strong as if his height along the axis was equal to the radius of the disk.

The latter fact is quite familiar (think electric field lines near a large flat plate). The former is due entirely to the infinitessimal piece of the disk upon which that observer is standing.

Clearly intuition unreliable in this matter! Also, there does not appear to be any restriction that the effects you mention would be exactly compensatory.

Hurkyl

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Andre
OK Hurkyl,

There are some differences between electrical and gravitational fields.

Lets do more "pathological" experiments. Suppose that Earth was a water drop. It would be a perfect sphere if there were no forces or movements. Because all the molecules are subject to perfect symmetrical gravity. Now, we spin the drop and due to the centrifugal forces something happens. Normally the gravitational force on all water molecules at the surface was equal. Now, the total force of the spinning molecules decreases due to the opposite centrifugal force.

The static not spinning molecules of the poles are pushing down stronger now than those at the equator. Hence the Earth starts to deform, the poles are coming down and the equator is moving up, so the bulge around the equator starts to form. When does this process stop? This is only possible when equilibrium is reached. Hence when the polar molecules come down, their gravity pull decreases and when the equatorial molecules move up, their gravity increases until polar and equatorial gravity is equal again.

If that equilibrium was not reached, the deformation would have to continue until the water world was a disk again .

Again, as I tried to point out, this equal gravity is caused by the location of the mass. The poles have less mass directly underneath compared to the equator. Consider that the poles have the gravity of a smaller Earth and the equator has the gravity of a bigger Earth according to their respective radius.

Now back the pathological plate (alliteration intended ).
Exactly in the centre of the Earth you are weighless (big difference with electrical fields). Why? Because all the mass is equally distrbuted around you and it is pulling you in all directions. The sum of all forces is zero. Now cut Earth in half and cut a slice through the middle (perhaps a few inches thick, not zero). Now you have the disk again. In the centre of that disk you are still weigtless because we still have the symmetrical forces in all directions albeit in two dimensions now (well almost). At the rim or the previous Eart surface there is still some gravity left caused by the finite mass of the disk but it certainly is not infinite.

This was just to illustrate that textbooks can be wrong on very basic things. That differential gravity is just about the most annoying.

Staff Emeritus
Gold Member
On the spinning water droplet:

I still see no reason why the surface pressure needs to be equal... if all of the "external" force was acting on the surface of the water (such as surface tension or in hydraulics), I could see how the surface must be an equipotential surface, but since gravity and centrifugal forces act mostly on the interior of the droplet and not the surface, this argument doesn't work.

And things, of course, can be more complicated!

Elasticity of the Earth would also seem to have a role in the equilibrium shape, one that would not extend to acting on those on the surface of the earth.

And, of course, density variation makes things icky too. One can contrive an example where 90% of the Earth's mass is concentrated underneath the equatorial observer's feet, so he'd obviously feel heavier. What effect would natural density gradients have on the solution?

On the "more mass underneath" approach:

That's not a very good proof! It gives a heuristic reason why it could be possible for them to be equal, but is far from actually proving it so. As a sample example why its not proof, I constructed this simple model:

Scenario 1: The "planet" is composed of 4 point masses: (+/-1, 0) and (0, +/-1). Observer A is at (0, 2) and observer B is at (2, 0).

Scenario 2: Flatten the planet by scaling the x-axis a factor of 2 and the y-axis by (1/2). Compute the forces and you'll find that the force A feels is tremendously greater than what B feels!

Heuristically, one may argue that while the equatorial observer has more mass "beneath" him, he is on average further from that mass than the polar observer, and distance is more important!

On the pancake:

For an observer at a height d above the disk on its axis, if you integrate C d s^3 dA over the disk where s is the distance from the observer to a given area element and C is a constant involving masses and gravitational constant, the resulting formula is:

F = 2 pi C (1 - d / sqrt(r^2 + d^2))

for the magnitude of the gravitational force the observer feels, where r is the radius. While its true that there is 0 net gravitational force on the geometric point in the center of the disk, that point is a discontinuity in the field, an observer approaching the disk along its axis will find the gravitational force steadily increasing, approaching 2 pi C. Also, deviation from this is second order in (d / r), so near the disk it will very much appear as a uniform field, so for a real disk with nonzero thickness t, the gravitational force near the surface of the disk will simply be 2 pi C t. Of course, from there, it would drop nearly linearly to 0 as you go past the surface to the actual center of the disk.

As for the observer on the rim, there's a significant difference between zero thickness and small thickness! Taking all of that material on the outer edge that was originally a small nonzero distance away and compressing it into that point 0 distance away from the observer is sufficient to introduce an infinity. (Again computed by integrating Newton's law). I can't manage to produce a closed form solution to this integral and I can't get the TI-89 to give me a numeric answer either, so I can't give any actual figures for nonzero thickness.

Hmmm, this is actually a really interesting question... I'm willing to buy that under certain idealized conditions our spinning drop would be an equipotential surface -- but that doesn't necessarily mean gravitational force, dU/dh, would be equal at all points. Hmm.

Hydr0matic
-> Hurkyl

Your "More mass underneath" analysis is incorrect. Distance is not more important, distribution is ! An observer that is very close to the center of mass does not necessarily experience more attraction. In the case with the disc, the mass is distributed fairly uniformed around the observer at the axis so he would barely experience any attraction at all.

...

I'm with Andre on all points.

Andre
Final blow.

Have a look at this exagarated gravity map Not much evidence of lattitude related gravity. Just strong local anomalys. You'd be as heavy in Indonesia as in Norway.

Staff Emeritus
Gold Member
In the case with the disc, the mass is distributed fairly uniformed around the observer at the axis so he would barely experience any attraction at all.

That's just not so.

Assuming Newtonian style gravity:

Suppose the gravitational source is a disk in the x-y plane centered on the origin, with a uniform density &rho and radius a.

Suppose we have a point observer on the positive z axis measuring gravity. Let z represent his z-coordinate, and let his mass be m.

For an area element of the disk dA located a distance r from the origin, its mass is &rho r dr d&theta and its distance from the observer is sqrt(z^2 + r^2), so the z-component of the gravitational force on the observer by this area element is:

dF = -G m z / (z^2 + r^2)^(3/2) &rho r dr d&theta

Integrating over the disk:

&int dF = &int -G m z &rho r / (z^2 + r^2)^(3/2) dr d&theta
= (-G m z &rho) &int d&theta &int r / (z^2 + r^2)^(3/2) dr
= (-G m z &rho) (2&pi) (-1) (r^2 + z^2)^(-1/2) | r = 0..a
= (-2&pi G m z &rho) (1 / z - 1 / sqrt(z^2 + a^2))
= -2&pi G m &rho (1 - z / sqrt(z^2 + a^2))

Due to symmetry, the x and y components of the gravitational force are 0, so the magnitude of the gravitational force experienced by our axial observer is:

F = 2&pi G m &rho (1 - z / sqrt(z^2 + a^2))

Notice that:

F -> 2&pi G m &rho as z -> 0

so that it is not true that F -> 0 as z -> 0.

Plugging in some numbers for comparison:

F(z = a) = 2&pi G m &rho (1 - 1 / sqrt(2)) = 2&pi G m &rho * 0.2929
F(z -> 0) = 2&pi G m &rho

So we see that as our observer approaches the disk's surface, the gravitational force he feels increases up to about 3.414 times as strong as the force he felt when his distance from the disk was equal to its radius.

Taylor expanding around z = 0 shows that near the surface, deviation from the uniform field F = 2&pi G m &rho is second order in z, so when you're sufficiently close to the surface, you appear to be in a uniform gravitational field.

This is all exactly the same as the case for a uniformly charged plate in electrostatics.

For a disk with thickness t, because of the aforementioned near-uniformity, an observer near the surface of the cylinder (or on the surface of the cylinder, as long as he's not sinking into the cylinder) will experience a gravitational force:

F = 2&pi G m &rho t

For the exact solution for an observer who is a height h above the surface of the cylinder, integrate F(z) for z in h .. h + t yielding

F(h) = 2&pi G m &rho (t + sqrt(h^2 + a^2) - sqrt(h^2 + (t + a)^2))

Plugging in h = 0 yields:

F(h = 0) = 2&pi G m &rho (t + a - sqrt(t^2 + a^2))

So again we see that an observer near (or on) the surface experiences a nonvanishing gravitational force. The total mass of the cylinder is &pi a^2 t &rho, so we can rewrite the above equation using M as the total mass of the cylinder:

M = &pi a^2 t &rho
&pi &rho = M / (a^2 t)

F(h = 0) = 2 G m M / (a^2 t) * (t + a - sqrt(t^2 + a^2))

Plugging in some numbers for comparison, suppose the thickness of the disk is 1 unit and the radius is 1000 units. Then, the gravitational force is approximately:

F(h = 0) = G m M (2 * 10^-6)

Which is the same gravitational force felt by an observer 707 units from a point gravitational source with the same mass as the cylinder. Hardly "barely any attraction at all".

For completeness, another example would be a thickness of 1 unit and a radius of 10^6 units. In this case, the equivalent observer would be 7*10^6 units from the point mass. Clearly the ratio between thickness and radius has a dramatic effect on things, but as pointed out before the force will never vanish, and an observer standing on (not in) the disk will always experience a greater gravitational attraction than an observer who is some distance away from the disk along the axis.

When a is much larger than t, the equation reduces to:

F = 2 G m M / a^2

An observer standing on a disk of negligable thickness and radius a will experience a gravitational force twice as much as an observer standing on a sphere with the same mass and radius.

This all assumes uniform thickness; in reality we would expect the disk (or cylinder) to be more dense in the center, this increasing the force felt by our axial observer.

Hurkyl

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Nigel
Very interesting! A lecturer once told me that first people thought the Earth to be flat, then like a circular plate, then spherical.

So far so good. But they Newton came up with a calculation showing that its a flattened sphere or "oblate spheroid". Next we were told that more accurate determinations show it to be "pear shaped". I can't remember whether the big bulge is supposed to be near the north or the south pole, or whether it is big enough to have a significant effect.

The problem is, when you look at the NASA pictures of the Earth from space, they show a very spherical shape. It may be true that a minor distortion can have a measurable effect.

Originally posted by Creator
Of this difference it is easily calculable that a little over half is due to the centripetal acceleration provided by Earth's rotation: (about 0.0336 m/sec^2).

Wait, are you saying that the centripetal accelration makes you heavier or lighter at the equator?

Creator

Originally posted by climbhi
Wait, are you saying that the centripetal accelration makes you heavier or lighter at the equator?

Lighter on the surface of Earth (at sea level) at the equator relative to sea level at the poles. The figures I gave are MEASURED VALUES of surface gravity acceleration (g) at the equator vs. that of g at the poles. Half of that reduction is due to centripetal accel.

Creator

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I really must be missing something, wouldn't the increased centripetal acceleration at the equator make you heavier?

Creator
Sorry, climbhi; I should have said 'centrifugal' accel., meaning in a direction opposite to that of gravity.
So over 1/2 of the reduction in gravity at the equator is due to Earth rotation.

The other part of the reduction in weight at the equator is due to being at a greater distance from the Earth's center.

Creator

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Rockazella
Quick question... Is centrifugal force even factored into the accepted 9.8m/s^2 ? Now that I think about it, I'm not so sure.

When an object is outside of the Earth's atmosphere and 'falling' twords Earth (not orbiting, but just going right 2 the center) it is not affected by centrifugal force. For objects that are in the Earth's atmosphere and are falling twords earth, they would be spinning with the Earth and would be affected by centrifugal force.

So back my question, Is the 9.8m/s^2 for objects that are affected by centrifugal force or for objects that are not?

Mentor
Originally posted by Rockazella
Quick question... Is centrifugal force even factored into the accepted 9.8m/s^2 ? Now that I think about it, I'm not so sure.

When an object is outside of the Earth's atmosphere and 'falling' twords Earth (not orbiting, but just going right 2 the center) it is not affected by centrifugal force. For objects that are in the Earth's atmosphere and are falling twords earth, they would be spinning with the Earth and would be affected by centrifugal force.

So back my question, Is the 9.8m/s^2 for objects that are affected by centrifugal force or for objects that are not?

Andre
Creator,

I'm puzzled. Reading your post you seem to persist in being lighter at the Equator, while I'm trying to promote that the total force (gravity and centrifugal) is not depending on lattitude. So we actually seem to have a disagreement. May I assume that your calculations are based on a simplification that all the mass of the Earth is working from it's centre. But again that is only valid for spheres and the Earth is no sphere. This link shows a neat real gravitation model of the earth.

http://antwrp.gsfc.nasa.gov/apod/ap011113.html

Creator
Originally posted by Andre
Creator,

... Reading your post you seem to persist in being lighter at the Equator, while I'm trying to promote that the total force (gravity and centrifugal) is not depending on lattitude.

Reading your post,Andre, you persist in promoting a theory that is in total contradiction to the well known measured values of surface gravity (which I originally presented); which is the reason we have refused to accept what you are promoting. I thought given enough time you would be able to figure out your error without argument.
We will contiue to present that which is KNOWN physical evidence.

It does no good to continue to entertain someone who develops a gravity theory based on speculation that is contary to the known measured values of g at Earth's surface. You can present all sorts of propositions to try to justify why gravity SHOULD be a certain way BUT IN THE LEAST it must reflect the known MEASURED VALUES, otherwise no one will give it the time of day. Hurkyl has tried to point to the fallaciousness of your arguments; why should I continue to do so when you have made apparent you want to ignore the value of physical data, especially when you still have presented no empirical or mathematical derivations of your theory?

You continue to ask how the data was "calculated" when you have been told repeatedly these values of g are "MEASURED". Gravitometers are regularly used to gain very accurate g measurements at the surface to within Mgals. There are very slight variations locally, which are regularly used to spot ore deposits, density variations, etc.; but this is far below the 5 percent variation from pole to equator.

It will be very easy to show the erroneous assumptions you are making but FIRST you will need to acknowledge that we have presented the actual 'measured' surface g values? The measured values of surface g don't change simply because you want to adopt some theory that negates them. You ought to be changing your theory to conform to data, not trying to negate the data to fit your theory! If you continue to question the veracity of data in order to substitute your own theory, then there is no use arguing any further. We will be forced to consider you another Baghdad Bob from the Iraqi Ministry of (Mis)Information.

Creator

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Rockazella

russ,
I reall don't know if I did...?

Creator
Originally posted by Rockazella
Quick question... Is centrifugal force even factored into the accepted 9.8m/s^2 ? Now that I think about it, I'm not so sure.

Actually, Rockazella, it is an excellent question.
First, the commonly used value of 9.8 m/sec^2 is only an average value. It actually varies from about 9.83 at the poles to about 9.78 m/sec^2 at the equator (if measured at sea level).These are the actual measured values.

The object falling vertical at the surface is actually rotating with the surface. Therefore, yes, the centrifugal force is included in the measurement.

When an object is outside of the Earth's atmosphere and 'falling' twords Earth (not orbiting, but just going right 2 the center) it is not affected by centrifugal force.

I believe if I understand correctly, you are correct. If it is shot at the center from space (not orbiting), there is no centrifugal component.
More importantly, it is a serious error to try to use satellite measurements of gravity variations ABOVE the Earth (like GRACE satellite) to prove that DIFFERENT VARIATIONS don't exist at the Earth's SURFACE.

DON"T BE DECEIVED BY THOSE WHO ARE POSTING SATELLITE RELIEF MAPS TO TRY TO DENY THE LARGE LATITUDE VARIATION ON THE SURFACE.

This is the obvious ERROR made by Andre. The reason is because SURFACE gravity measurements vary with the distance from Earth's center, getting smaller at equator which is farther from the center. But satellite measurements measure the variations "felt" by the satellite at a certain distance in space which is FIXED RELATIVE TO THE CENTER OF EARTH.
Notice; to a good approximation these satellites are in a 'circular' orbit, meaning they orbit at a FIXED distance FROM THE EARTH'S CENTER. Their orbital variations determine gravity fluctuations AT THAT POINT IN SPACE.

In effect, they measure the very small and subtle global variations that exist while the satellite REMAINS AT AN (almost) FIXED DISTANCE FROM THE CENTER OF EARTH; and so they do not measure the relative variations that occur on the surface as a result of VARYING DISTANCES from the center due to oblateness.

ALSO, these satellite g variations are VERY small compared to SURFACE variations with latitude.

Hope that clarifies some of the misconception that has been posted by those trying to prove there is no latitude surface variation simply by posting a site showing a satellite relief graph !

Creator

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Andre
Thanks Creator, you are really too kind. A very elegant way of discussing. I did not know that the war of the boldface capitals has started.

There seem to be more practical situations that have adjusted to theories. First of all, I'm not the only one thinking of equality of gravity:
http://www.em-bed.com/ee/02/02/gravity/ [Broken]

And as far as the gravity on the disk is concerned:

http://www.em-bed.com/ee/02/02/gravity/math.html [Broken]

Finally if the average gravity on higher lattitudes is measured to be greater than on the equator, this little hypothesis would indicate that Earth is not balanced perfectly right now which I never denied BTW:

http://www.gsfc.nasa.gov/topstory/20020801gravityfield.html [Broken]

Moreover it would be very nice to hear from the expert where the balance hypothesis is wrong other than plainly stating that it is wrong. It seems to be obvious that the pulsating Earth is unable to show its equilibrium at the moment.

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Creator
Originally posted by Andre
Thanks Creator, you are really too kind. A very elegant way of discussing. I did not know that the war of the boldface capitals has started.

Ha Ha that site you posted is laughable..
If that's your response to the facts then no wonder no one believes your theory.

Anyone can hide from the facts by posting a reference to some obscure site that openly says it is trying to prove both NASA and Scientific American to be wrong, which is what you just did.
I hope you realize that certainly doesn't make me believe your theory MORE...quite the contrary..it confirms the absurdity of it!

**If you are going to continue to refuse to address the real issue at hand>>>>THE FACT THAT YOU ARE WRONG ABOUT THE MEASURED g VALUES OF SURFACE GRAVITY WHICH VARY WITH LATITUDE, there is nothing further to discuss. Your refusal to admit to the valid physical evidence from valid physic's texts nails your theory into a coffin!
All your speculation and quoting weird web sites will do no good, Baghdad Bob...

You can find the physical (MEASURED)values of surface gravity variations with latitude in any good physics book or possibly in Handbook of Physics. OR you can simply get a gravitometer and do the measurements yourself. It seems as though that's the only thing that will convince a skeptic and amateur like yourself... but it won't convince willful ignorance.
If that doesn't work, then I suggest you can find good employment taking Baghdad Bob's place at the Iraqi Ministry of Mis-Information; they are looking for people expert in self denial who refuse reality.

Creator

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NileQueen
Haha yourself, Creator, because Andre is right, you're wrong and no amount of rudeness will strengthen your teetering position. Someone who engages in personal attacks and ridicule is obviously on the offensive to cover up their shortcomings.

Oh yes, just let me get out my handy dandy gravitometer and check my local gravity...great suggestion. I suppose you have one there with you O world enlightener

Gravity maps are tools used by geologists to see variations in gravity over the planet, and over time. How is this data collected? GRACE (Gravity Recovery And Climatic Experiment) twin satellites
http://news.bbc.co.uk/1/hi/sci/tech/1668872.stm
I don't get why you are ridiculing this data.

And we don't really care about your personal promotion of jobs in Iraq. Just take yourself off to Iraq.

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I'm sorry to bring this back up but I really must be missing something, wouldn't the spinning of the Earth make you heavier at the equator not lighter? F = mv^2/r ; v = &omega;r ; F = m&omega;^2r , thus since &omega; is constant for anyone on the Earth the further you are away from the center of the Earth the heavier you are ... or am I doing something wrong here?

FZ+
Yes. F here give the effective centrifugal force. This should be DEDUCTED from the gravitational attraction.

Originally posted by FZ+
Yes. F here give the effective centrifugal force. This should be DEDUCTED from the gravitational attraction.
Why please, physically? I think I'm beginning to see why, you see my high school physics teacher once said that you'd be heavier at the equator for the reason I stated above, so I'm having a hard time kicking it. Perhaps an analogy or something would do me good.

Creator
Originally posted by NileQueen

I don't get why you are ridiculing this data.

Well, Cleopatra,
Noone is ridiculing the GRACE data; we are laughing at the inapplicability of trying to use it to prove that there is no SURFACE variation in g with latitude, when all physics texts clearly show the MEASURED value of g at the surface DOES vary PREDOMINATELY with latitude. Go back and read my post to Rockazila again...Maybe you will understand it this time ...

The question is, "Why is Andre (and apparently you too) ridiculing the MEASURED DATA presented in many standard physics sources"? And trying to get us to believe speculative theory CONTRARY to established physical data.

The actual MEASURED g variation with latitude on the surface of the Earth is given,for example, in Halliday & Resnick (Standard) Physics text:

"Variation of g with latitude at sea level:
0* ...9.780 m/sec^2 (equator)
40*... 9.80 m/sec^2
60*...9.819 m/sec^2
90*...9.83 m/sec^2 (pole) "

These MEASURED VALUES prove that Andre's theory (and you) are flat wrong.

Example 2:
"Gravity varies PRDOMINATELY with latitude FROM 983 Gal at the poles TO 978 Gal at the equator..." (A Gal is 0.01 m/sec^2)
--- Source: Encyclodedia of Physics, Lerner & Trigg 2nd ed.

UNLESS YOU HAVE "MEASURED VALUES" OF g THAT ARE DIFFERENT FROM THESE FROM AN EQUALLY REPUTABLE SOURCE, DON'T EXPECT ME OR ANY KNOWLEDGEABLE PERSON IN PHYSICS TO BELIEVE YOU (OR ANDRE'S SPECULATIVE THEORY)

Creator

P.S. "Denial is not a river in Egypt"

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Creator
Originally posted by climbhi
I'm sorry to bring this back up but I really must be missing something, wouldn't the spinning of the Earth make you heavier at the equator not lighter? F = mv^2/r ; v = &omega;r ; F = m&omega;^2r , thus since &omega; is constant for anyone on the Earth the further you are away from the center of the Earth the heavier you are ... or am I doing something wrong here?

Climbhi,
No problem; Its an honest inquiring question, obviously with a willingness to find the truth.

You are correct again with the equation, but you are simply pointing the force in the wrong DIRECTION. The centrifugal force is OPPOSITE to the DIRECTION of acceleration due to gravity. Thus it must be subtracted from the gravitational force.

If the Earth were rotating fast enough, objects on the equator would have a 'effective' value of g = 0 and the objects would seem to be weightless.

...Which brings us around again to the OTHER reason why GRACE satellite measurements don't give you the effective g value equivalent to that meassured at a specific surface location;-- the satellite doesn't take into account the lower measured g value due to the object's ROTATION on the surface.

You and Rockzella both deserve credit for bringing that up.

Creator

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Andre
Thanx, your deity, creator, for your sublime posts, how delicately and patiently put. I clearly see my omission now thanks to your elaborated masterpieces.

Perhaps it is permitted to return to my water droplet world again, let's build one, orbiting in space around the sun having an average radius of 6371km. Let's assume rather heavy water, with a mean density of 5515kg/m^3. Now let's rotate this droplet with one revolution per 86400 seconds. We see that results in flattening and the equatorial radius increases to 6378,1 km while the polar radius decreases to 6356,8 km. We measure the gravity to be 9,78 m/s^2 at the equator and 9,83 m/s^2 at the poles, a difference of 0,5%.

We calculate that the rotation gives us a centrifugal force of rω^2 =0,034 m/s^2 on the equator surface. We also calculate a delta in the gravity due to the difference of the pole and equator surface to the center of gravity to be a ratio of 0,67% or about 0,0677 m/s^2. Both values added together is just a bit more than the actual difference in the measured values (0,05 m/s^2). But we assumed a sphere here, which is a invalid simplification since the mass distribution is not spherical and mass is pulling down more effective at the equator, the basic message of the whole thread. This should make up for the difference.

Another question is: do we have a balanced situation now?

For balance we still require equilibrium, but –and here is my omission- the equilibrium is not dependable on force but on pressure. It is the total pressure from every angle in every position that needs to be balanced and of course, pressure is force per area. So for more area, less force is required to maintain equilibrium.

I would ramble that on every plane through the center of gravity the value of the sum of all pressures must be equal for all possible disks, while the vector sum of all forces should be zero. I figure this would require a great deal of calculus to catch this. But perhaps we can make some very simplified assumptions to get an approximation.

Lets concentrate on two disks, one disk cut in the plane of the equator and one disk cut in a polar plane. In two dimensions the total pressure of gravitational forces of the rims for instance should be equal. Hence the sum of the forces times the circumference. This proportionality equals to the ratio between the circumferences: More circumference, less force. And obviously, the equatorial rim is longer than the polar rim (67 kilometers).

But not only the rim is acting as a mass, the whole disk is. Integration for the disk would yield the same logic for the ratio of the areas of both disks, hence more area, less force. This should equal the difference in relation between both extreme gravity values or 0,5%.

Now the equatorial disk has a area of 127,80 million km square while the polar disk has a area of 127,37 million km square, a difference of only 0,33%, not enough to match the gravity difference. Hence it seems that the equatorial bulge of the water droplet is insufficient to form equilibrium. I see that I have to increase the equatorial bulge with some 4 kilometers and reducing the poles likewise, to having the difference in areas to match the 0,5% gravity difference, but this also would affect the ratio in gravity differences. The centrifugal force would only increase 0,07% for instance, the differential gravity however would increase 34%. It seems that we need to calculate the effect of the uneven mass distribution to find a more accurate result.

It seems that all the mentioned effects are working, including the effect of uneven mass distribution. But instead of forces, pressures should be in equilibrium. This leads to different values for gravitational forces but the sums don’t add up due to the mass distribution not being spherical. And in my example, the water droplet seemed not to be flattened enough to be in equilibrium. So I wonder about the Earth. http://www.gsfc.nasa.gov/topstory/20020801gravityfield.html [Broken] But then again, Earth is much more complicated than a water droplet.

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