SULPHUR

The earth is oblate FACT. Why are you making it so complicated .GO talk to a geophysicist.

Aquila ka Hecate

I pop in and out of here and I knew something was bothering me about wiploc's formulation.

It's this:
No, I don't think so.
The earth's mass is the same wether you stand at the poles or at the equator.

Just at the equator you are further away from the geometric centre of mass-so you weigh less.

Sulphur, I still contend however that the centrifugal force weight-loss is comparable in effect to the greater distance weight loss.

I contend just to be argumentative, obviously-I'm at work and don't have time to figure them.
Both effects are fairly small,but I think they have almost equal contributions.
After all, both depend upon the full radius of the earth squared, don't they?
(I'm thinking vaguely of kinetic energy here where the linear velocity is a square and depends upon the radius?)

SULPHUR

It has been proved using a unit mass that in areas of greater mass the weight icreases no matter where you are if the gravitational force caused by say, massive sulphide deposits you are going to weigh more
THE GREATEST FACTORS ARE MASS AND DISTANCE

SULPHUR

The earth is oblate therefore the the radius is not constant.

SULPHUR

I did not read your comment fully. If abody is on the equator its KE matches all other particles around it and therefore have no effect.

Wiploc

I was doing that reductio ad absudam type stuff.

I still think that that weight is better increased by mass underneith you than by mass off to the side.

Imagine you are standing on a manhole cover. It will attract you more strongly if you stand on the top than on the edge, for exactly the reason you give: it is, on average, closer to you when you stand on the top. But now imagine it is a giant manhole cover, a hundred miles across. If you stand on the face of that, most of the mass pulls you very inefficiently, off at an angle; whereas if you stand on the edge of the disk, the mass is all well situated to increase your weight. It's farther away on average, but it's all pulling down.

Let me confess that last night I was discussing this with a physics student who surprised me by having some sympathy for your position, though he granted that I would be right in the extreme case of the coin-shaped world that I used as illustration. He said he'd do the math and get back to me.

So I've got a new theory that a perfectly spherical world may be the most attractive. That is, people on the poles and the equatorial bulge may both weigh less than they would on a spherical world.
crc

Wiploc

The mass is the same, but the mass isn't downward, see?

You intuitively recognise that you would weigh less if you dug a hole and went into the earth, because some of the earth would be pulling up, and some sideways; and those parts wouldn't contribute to your weight. Well what if, instead of digging down into the earth, you got closer to the center by just sanding off a big flat spot? Same effect. You are moving earth from underneith you to off-to-the-side, and the effect is to reduce your weight. The equitorial bulge, in effect, I suspect, produces such a flat spot at the poles.
crc

Bumble Bee Tuna

SULPHUR, I'd understand you better if you kept it all in a single post and actually went in depth.

-B

SULPHUR

For me to in depth would take several hours. sorry

Yggdrasill

If you start digging a hole downwards, you would at first become heavier, because the increase in gravity as a result of getting closer to the mass that is below you would be greater than the decrease in gravity as a result of the mass above you plus the decreased gravity as a result of not having the mass directly below you. Then, as you get deeper you would reach a point where the effects of the mass above you and the decreased gravity as a result of not having the mass directly below you would equal the effect of getting closer to the mass. Here you would weigh as much as you do on the surface. From this point and below you would start weighing less until you reach the centre of the earth.

Increase in weight as a result of getting closer to the mass below you: X
Decrease in weight as a result of the mass above you: Y
Decrease in weight as a result of not having the mass directly below you: Z
Gravity at the surface: G0
No acceleration/No gravity: G1

X,Y and Z is different in each situation.

Situation 1 (the surface): X-(Z+Y)=G0
Situation 2 (below the surface): Y+Z<X
Situation 3 (radius A): Y+Z=X -> X-(Z+Y)=G0
Situation 4 (below radius A): Y+Z>X
Situation 5 (the center): Y+Z=X -> X-(Z+Y)=G1

Now imagine a perfectly spherical planet, you are standing on the north pole. As you are standing there the planet slowly becomes disc-shaped (with smooth edges). The same thing would happen, you would at first become heavier, then return to your previous weight, then become lighter until you are weightless. The only difference is that you don't have effect Y.

Situation 1 (the surface): X-Z=G0
Situation 2 (below the original surface): Z<X
Situation 3 (radius B (below radius A)): Z=X -> X-Z=G0
Situation 4 (below radius B): Z>X
Situation 5 (the center): Z=X -> X-Z=G1

If you were at the equator when the sphere started becoming a disk, it would be much like climbing a mountain, you would have the mass beneath you to a larger degree, but you would have less of it near you, so you would weigh less. If the sphere flattened without ever being interrupted, the gravity would go towards infinitely small.

Because the gravity is higher on the poles (at least until the surface is below radius B) on a disc shaped planet, and the earth is slightly disc-shaped (and the surface is nowhere near being below radius B) it is logical to conclude that the gravity is higher on the poles of the earth.

Oh gawd I hate planets, they hurt my brain. I might try to make it more coherent later.