Force of gravity at poles and equator

Ladewig · 14th February 2003, 09:00 PM

I'm having a bit of trouble fully understanding the replies I got over in Banter so I am jumping over to Science. I tried searching the net and came up with values between 0.1 and 1.0 % which seemed like a large range.

Quote:

20) Gravity has a stronger pull at the Earth's poles than it does at the equator. As a result, a person who weights 150lbs at the equator, would weigh almost a pound heavier if they stood at the North Pole.
True

The earth is not perfectly round. It flattens out somewhat at the poles. Therefore, a person standing at the poles is actually 13 miles closer to the center of the earth than they are when standing at the equator. The pull of gravity increases as you move closer to the center of a gravitational mass, making objects heavier (just as when you move away from the center of a gravitational mass, such as when you fly into outer space, gravity weakens). In addition, the centrifugal spin effect at the equator slightly counteracts the pull of gravity. All of this translates into a difference of almost a pound between what a person would weigh at the equator versus the poles. See this article for more info.

So if I had a uniform cube large enough to easily measure the gravitational force, and I stood at the center of a face, I would weigh more than if I stood at a corner? Part of my confusion comes from the (perhaps mistaken) belief that as one descends below the surface of the earth, one's weight decreases.

Keneke · 14th February 2003, 10:51 PM

I think the crux here is that gravity is created by the whole of the mass. As you stand on the surface of the earth, some of the mass is pulling you slightly down and to the left, right, etc. However, almost all the mass pulls you below the horizontal axis. The sum of the forces of all points equals to straight down. However, if you go underground, some of those vectors start to point upward, negating the downward pull. If you go to the center of the earth, you have an average force vector of approximately zero, since the mass is equally distributed around you. So yes, as you go underground, you lose weight, and if you go up above the surface of the earth, you lose weight because of the distance from the gravitating object. The reason you gain weight at the poles and not at the equator is because, though you are closer to the center, none of the mass is above your head and cancelling out gravity vectors like going underground does. Apparently the distance to the center of gravity is more important to the final weight than the fact that the mass of the earth, being slightly flattened, means there is roughly more mass out on the edges than directly below you.

I'm not too sure about the square question, I'm only versed in spherical bodies, but thinking about the problem makes me say yes, the corner position would make you weigh less because you are farther from the center, and none of it is above your head.

Soapy Sam · 15th February 2003, 06:39 AM

I'm puzzled too. It seems to me (naive view) that what is keeping me on the ground is the attractive force due to the mass of the Earth acting on the mass of me (and vice versa).

I accept that at the centre of the Earth, that force averages to zero. I also accept Newton's conclusion that the mass distribution can be treated as equivalent to a point at the centre. (ie No matter where you stand, the whole planet is below you). So the force is the same (inverse square law) at any point on a sphere centred on the centre of gravity. Now if that imaginary sphere coincides with the Earth's surface at the pole (radius RP), it will be below ground at the equator (Radius RP + bulge). So if I want to weigh the same at the equator, I have to go down a mine till I reach a point on the sphere. (So I was wrong in my comment in the banter thread).

The above holds even if Earth is not rotating. So there must be a second decrease in g due to my 1000+mph rotation at the equator, compared with zero at the pole. This would be proportional to the vertical component of my tangential momentum. Assuming the Earth actually was spherical, how fast would it need to rotate before effective escape velocity was reached at the equator? 1 Rev per hour is about 25200 mph, which is close to actual escape velocity. Would that be enough or is this far too simplistic?

Now, it seems to me that we can speed the thing up by decreasing angular momentum. First we remove all the mountains, then we get rid of the moon, then we get all the bicycles in China lined up...

Walter Wayne · 15th February 2003, 08:43 PM

Originally posted by Soapy Sam
.,. I also accept Newton's conclusion that the mass distribution can be treated as equivalent to a point at the centre. (ie No matter where you stand, the whole planet is below you)....

I think this only hold if you are a a significant distance from an object. For instance, if we measure the distance vector to every point of the sun from here each of them is very close to the same distance and same angle, so the direction of the force and r squared term is the same for all of the sun and us.

However, this doesn't hold for the earth below us as the distance vector varies a lot, so you can calculate the mass of an object simply by measuring g at its surface. One would have to know its geometry (and hope its density is relatively constant).

The above holds even if Earth is not rotating. So there must be a second decrease in g due to my 1000+mph rotation at the equator, compared with zero at the pole.

This just adds another force, but doesn't change g itself.

Walt

Keneke · 15th February 2003, 10:00 PM

Quote:

Originally posted by Soapy Sam
So if I want to weigh the same at the equator, I have to go down a mine till I reach a point on the sphere. (So I was wrong in my comment in the banter thread).

No, if you go underground at any point, some of the mass begins to pull you upward, offsetting the gravity.

So, at any point of X latitude and X longitude, the elevation that will give you the maximum gravity is exactly at ground level. So, if you are at the equator, you can't be as heavy as if you were at a pole, no matter how far up or down you went.

garys_2k · 16th February 2003, 07:18 AM

Quote:

Originally posted by Keneke

No, if you go underground at any point, some of the mass begins to pull you upward, offsetting the gravity.

So, at any point of X latitude and X longitude, the elevation that will give you the maximum gravity is exactly at ground level. So, if you are at the equator, you can't be as heavy as if you were at a pole, no matter how far up or down you went.

Right. If you want to weigh the same at the pole as you do at the equator, the place to dig the hole is at the pole. Since you are heavier at the pole than the equator, dropping into a mine at the pole will reduce your weight. You could also climb onto a tall tower at the pole and get the same effect.

Soapy Sam · 16th February 2003, 10:48 AM

Walter- granted, the rotation effect (I'm trying to avoid the disgusting term "centrifugal force" does not affect g, but it does reduce my effective weight. If we can get the old girl spinning fast enough, free launches are the norm. Indeed we have to glue everything down. Oceans for instance. I'm just curious about how many rpm are required.)

On Kenneke & Gary's points, let's see if I follow:- The gravitational gradient from core to pole is higher then that from core to equator. Surface gravity is always maximum, but not everywhere the same. Absolute maximum is at the pole,where the polar radius puts you nearest the core. There is no sub-surface point where g exceeds surface gravity at the pole (SgP).

However as SgP is higher than surface gravity at the equator,(SgE), there are sub surface points where g is higher than SgE. Presumably these lie on an ellipsoid surface which coincides with the Earth's surface at the equator and dips to a fixed depth under the poles.

I like threads like this. Forty years you carry a vague, wrong notion in your head and then you are forced to think it through. Nice one , Ladewig.

bjornart · 16th February 2003, 11:10 AM

Quote:

Originally posted by Soapy Sam
Walter- granted, the rotation effect (I'm trying to avoid the disgusting term "centrifugal force" does not affect g, but it does reduce my effective weight. If we can get the old girl spinning fast enough, free launches are the norm. Indeed we have to glue everything down. Oceans for instance. I'm just curious about how many rpm are required.)

Since orbital speed equals the root of a*r, where a is gravity in that orbit and r is the radius of the orbit, the speed needed to stay in orbit right on the equator is the root of 6378km * 9,81m/s*s, which is 7910m/s. Run at that speed and a jump would be enough to launch you into space, if it wasn't for air resistance.
(in rpms the answer is approximately 0.012, good luck)

Soapy Sam · 17th February 2003, 09:47 AM

.012 * 60 =0.72revs / hour. Rather less than the 1 I had guesstimated. Even better. Call the Chinese embassy and get them bikes lined up. Here we go...

Cecil · 18th February 2003, 12:38 AM

Quote:

Originally posted by Keneke
So, at any point of X latitude and X longitude, the elevation that will give you the maximum gravity is exactly at ground level. So, if you are at the equator, you can't be as heavy as if you were at a pole, no matter how far up or down you went.

*cough*

That seems very arbitrary. If I dig 6 feet underground, then although there is 6 feet of Earth now pulling me up, every other point on the Earth is now 6 feet closer, and pulling that much harder. I was taught in Physics that your point of maximum gravity was about 1000 kilometers under the surface, or about 10% of the way to the center.

Iconoclast · 18th February 2003, 01:06 AM

Quote:

Originally posted by Cecil
*cough*

That seems very arbitrary. If I dig 6 feet underground, then although there is 6 feet of Earth now pulling me up, every other point on the Earth is now 6 feet closer, and pulling that much harder. I was taught in Physics that your point of maximum gravity was about 1000 kilometers under the surface, or about 10% of the way to the center.

I was thinking along the same lines, and this is indeed a fascinating topic. It seems someone's going to have to bite the bullet and come up with a function for gravitational attraction versus distance from the core for distances less than the radius of the sphere. I'd do it but.. er.. my integrator broke and I haven't bought another one yet. Stimpy?

Iconoclast · 18th February 2003, 01:46 AM

OK, here's a thread on the subject from the sci.physics.research newsgroup from a few years ago, the thread in question is entitled "Baltimore Sun Says Gravity Inside Sphere is Uncalculable" (the thread title is misleading, that's not what the article was implying). From this discussion it appears that:

- For a homogeneous sphere, gravity is always at a maximum at a point on the surface, it falls off above and below that point. Gravity falls off faster as we move outwards from the surface than than it does as we move inwards.

- It is believed that the earth has a much higher density at the core than at the surface, thus gravity on earth will increase for quite a few hundred miles as we descend.

I didn't know either of those things.

DrMatt · 18th February 2003, 06:57 AM

No matter where you stand on the cube, the whole cube is "below" you...

Imagine the Earth--or the whole cube--divided into smaller subparts. The total gravitational force of the earth on you is the vector sum of the gravitational forces supplied by the parts. Divide the thing into a greater number of smaller pieces without bound, and you see that this is an integral.

It's fair to assume that, within bounds of measurablility, the Earth has bilateral symmetry both about a polar axis and about a diameter through the equator. Therefore, for a person standing on it in either of those two places, all components of gravitation perpendicular to the diameter have a vector sum of zero, and only the components parallel to the diameter extending through the earth to the person count for anything. In other words, at those locations, the earth only pulls down.

From there, it's a matter of grinding through the arithmetic and seeing whether the figures differ...

boooeee · 18th February 2003, 08:06 AM

Just to clarify a potential misunderstanding, if you dug a hole six feet deep, the mass shell above you would actually exert no force at all.

It is one of the happy consequences of the inverse square law that a spherical mass shell exerts no force at all on you if you are anywhere inside the shell, and if you are outside the shell, the force acts as if all the mass were concentrated at a single point in the center.

I'm pretty sure that no exact expression exists for the force exerted by an ellipsoid on any arbitrary point, but it is possible that an exact expression might exist for certain special points, like the poles, or along the equator. But, like DrMatt and Iconoclast said, the only way to know is to bite the bullet and start integrating.

Soapy Sam · 18th February 2003, 10:12 AM

Or, of course, start digging. Anyone got a gravitometer they aren't using?

Great how the "simple" questions turn out more complicated than you would think.

Next question. Does a pendulum clock (well insulated and at standard T & P) run at the same rate at pole and equator. (Ignoring relativistic effects of equatorial rotation velocity)?

CurtC · 18th February 2003, 02:01 PM

As I was digging into this subject a few years ago, I started to wonder, if you were at a point not at the pole and not at the equator, which direction gravity pulls. For example, let's say you're on the ocean at 45 degrees north latitude, due to the flattened shape of the earth, the direction perpindicular to the surface is *not* pointing at the center of the earth. My question was whether the force you feel at that point is directed towards the center, or along the surface's normal vector, or somewhere between. After a few non-answers, one smart person finally responded with a simple and inituitive answer. Anyone care to pin it down?

fishbob · 18th February 2003, 03:24 PM

The direction gravity pulls is "down". Down is toward the center of mass of the earth, no matter where on the earth you are. The center of mass = center of gravity = the point where all the down directions around the planet intersect.

Down is also perpendicular to the surface of a body of water.

The center of mass of the earth is not necessarily the same point as the geometric center of the earth.

I hope this helps.

garys_2k · 18th February 2003, 04:54 PM

Interesting about the effects of the Earth's increasing density causing a weight INCREASE for a while when descending into the hole. I'd assumed a homogeneous sphere which would decrease weight.

True, too, that the "shell" of earth above you (where "above" is defined as farther from the center than you are) has no net force on you, all of it cancels out. As you descend into this uniform sphere the effect is that you are "standing on" a smaller sphere. The deeper you go, the smaller the effective planet you are on. In the center you're effectively floating in space.

Regarding the local vector of "down," that varies a lot from mountain ranges, mineral deposits and nearby bodies of water. In fact, if you followed a "down arrow" from most spots on earth all the way in it would not follow a straight line, being pulled toward higher density areas at different depths.

Iconoclast · 18th February 2003, 05:52 PM

Quote:

Originally posted by boooeee
It is one of the happy consequences of the inverse square law that a spherical mass shell exerts no force at all on you if you are anywhere inside the shell.

How come everyone except me seemed to already know this? As an electircal engineer I have this sinking feeling that I was supposed to already know this rule, I'm sure it has applications in EE.

garys_2k · 18th February 2003, 05:54 PM

Quote:

Originally posted by Iconoclast

How come everyone except me seemed to already know this? As an electircal engineer I have this sinking feeling that I was supposed to already know this rule, I'm sure it has applications in EE.

You already know about it, in a way. Think Faraday cage.

Ladewig · 18th February 2003, 08:03 PM

So the cube thing...

Quote:

So if I had a uniform cube large enough to easily measure the gravitational force, and I stood at the center of a face, I would weigh more than if I stood at a corner?

Yes?

And a follow-up. If I were to walk from the center of a face to a corner, then would I be walking uphill? The gravitational vector would not be perpendicular to the surface.

Iconoclast · 18th February 2003, 08:04 PM

Quote:

Originally posted by garys_2k
You already know about it, in a way. Think Faraday cage.

Well I knew the field inside a faraday cage was not effected by any field outside it, but I was anaware that it was because of this shell rule. I feel so inadequate.

CurtC · 18th February 2003, 08:17 PM

Quote:

fishbob wrote:
The direction gravity pulls is "down". Down is toward the center of mass of the earth, no matter where on the earth you are. The center of mass = center of gravity = the point where all the down directions around the planet intersect.

BZZZZZZZZZT!

On a perfect ellipsoid, which is extremely close to the shape of the earth, the surface is *not* normal to the line through the center at all points. Only at the poles and the equator. All all the other points on the surface, the vector that's normal to the surface will *not* be pointed toward's the ellipsoid's center. So does gravity pull towards the center, or is it perpindicular to the surface?

69dodge · 18th February 2003, 10:48 PM

Quote:

Originally posted by CurtC
As I was digging into this subject a few years ago, I started to wonder, if you were at a point not at the pole and not at the equator, which direction gravity pulls. For example, let's say you're on the ocean at 45 degrees north latitude, due to the flattened shape of the earth, the direction perpindicular to the surface is *not* pointing at the center of the earth. My question was whether the force you feel at that point is directed towards the center, or along the surface's normal vector, or somewhere between. After a few non-answers, one smart person finally responded with a simple and inituitive answer. Anyone care to pin it down?

Water flows downhill.

The water of the ocean moves until its surface becomes perpendicular to the force of gravity.

(Well, not "the force of gravity", exactly. Rather, "the net force at that point", due to anything: gravity, centrifugal force, tidal forces from the sun and moon, etc.)

69dodge · 18th February 2003, 11:23 PM

Quote:

Originally posted by Iconoclast

Well I knew the field inside a faraday cage was not effected by any field outside it, but I was anaware that it was because of this shell rule. I feel so inadequate.

Faraday cages don't have to be spherical to work. And spherical mass shells don't shield gravity; they just don't create any gravity inside themselves. So I don't see the connection either. On the other hand, I don't know enough to say there definitely isn't one. Can anyone elaborate?

I'm pretty sure that both Faraday cage shielding and the lack of field inside a spherical shell are consequences of the inverse square law, and I think they might even be equivalent to it. So that's some sort of connection. But is there a more direct connection that doesn't go through the inverse square law?

fishbob · 19th February 2003, 12:23 AM

CurtC sez:

Quote:

On a perfect ellipsoid, which is extremely close to the shape of the earth, the surface is *not* normal to the line through the center at all points. Only at the poles and the equator. All all the other points on the surface, the vector that's normal to the surface will *not* be pointed toward's the ellipsoid's center.

You appear to be thinking geometry. The geometric center of the earth is not the same as the center of mass (= center of gravity).

As garys_2k said:

Quote:

In fact, if you followed a "down arrow" from most spots on earth all the way in it would not follow a straight line, being pulled toward higher density areas at different depths.

Soapy Sam · 19th February 2003, 07:45 AM

Last I heard, Earth is an oblate ellipsoid- slightly pear shaped with the big end in the S. hemisphere. Given convection in the mantle (and possibly parts of the core??), and tidal effects from Sun and Moon, rivers, trucks, plate tectonics and migrating lemmings, can the Earth be said to have one " permanently fixed" centre of mass anyway? (The lemmings may be small, but they are way out on the edge. High angular momentum, lemmings- even more so, migrating continents.)

As for the argument that gravity doesn't act upwards inside a sphere- As an idiot, I demand more than just an assertion that this is so. I need an explanation.

If I'm a mile under Australia I reckon Oz exerts the same pull upwards as the Australia sized mile thick slab below me does downwards. It's mass can't just vanish. Granted there's a similar slab on the far side of the world, but that's 8000miles away (Inverse squares) and furthermore that antipodean slab was there below me before I went down the shaft as well, whereas Oz wos below me too and now is above me.

Something does not ring true here. Upward attraction may well average out to zero at the centre of mass, but I can't believe it does so at every point. After all, where is the surface of the Earth? If I sit under a table, is the upward gravity of the table exactly balanced by the pull of the whole Earth? If so, why is the table not lighter when lifted from beneath? Should it not be weightless?

I need a cup of strong tea.

CurtC · 19th February 2003, 08:01 AM

fishbob, I accept that since the earth is not homogeneous, or even has symmetric density, that the geometric center is not exactly the same as the center of mass. But you stated that all the "down" directions all over the surface point towards that common point, and this is not true. They don't intersect. As 69dodge said, "down" is always the normal to the surface (with water, anyway), and I'm sure you can see that all those normal vectors don't intersect.

Soapy Sam, if you're inside a even-density shell, then that shell exerts no net gravity on you. Even though that little piece of land that might be closer to you has a higher gravity per unit volume, there is a lot more volume on the other side doing its part to pull you away, and it just so happens that they exactly cancel. The proof is a classic excercise in freshman college physics.

If you're a mile under Oz, then Oz does indeed pull you upwards, but the first one mile in depth of the Atlantic Ocean (on the opposite side of the earth) pulls you the other direction, pretty closely negating the effect of Oz pulling you up. As has been pointed out, there are errors due to the fact that the earth isn't even density, but that's basically how it's supposed to work.

boooeee · 19th February 2003, 08:16 AM

Quote:

Originally Posted By Soapy Sam
As for the argument that gravity doesn't act upwards inside a sphere- As an idiot, I demand more than just an assertion that this is so. I need an explanation.

The best way to prove this is to use Gauss' law, but short of proving Gauss' law, this just replaces one mystery with another. The best way I can think of to explain it is as follows:

Let's say you're at some arbitrary point within a sphere, say 5 miles below the surface of a 50 mile radius sphere. You could reason as follows:

Obviously, the portion of the sphere "above" me is going to pull up, and the portion of the sphere "below" me is going to pull down. So, you could say: "The portion of ther sphere 'below' me is going to exert a greater force because there is more mass in the 'below' portion. But then you might say: "But the portion 'above' me, although less massive, is closer to me on average, so the 'above' portion should exert a greater force".

It turns out that these two effects (mass and distance) completely balance, leaving no net force on you no matter where you are within the sphere. Hopefully I haven't just muddied the waters further.

As far as CurtC's question goes, it was running through my head all throughout my evening commute yesterday, and I'm still not sure. I am pretty sure that the force does not necessarily have to point towards the center of mass, but it very well may be the case on the surface of an ellipsoid.

Soapy Sam · 19th February 2003, 01:45 PM

CurtC , Boooeee. Thanks for the attempt gents. If I can't quite credit your argument, I'm sure it's my fault, not yours. I UNDERSTAND the principle but find it hard to believe in a real case.

In a sphere of uniform density gradients I can happily accept the averaging argument, but not in a real planet. After all, if the inner average is uniform at every point, then surely that must hold for every point on a polar axis- including the pole itself, where we have already agreed g is higher than at the equator.

My confusion is increased by the fact that I have been involved in enough gravimetric surveys to know that there are few things less constant on the Earth's surface than gravity.

And Earth is far from isotropic in density. (We know most about the crust, which I grant is pretty insignificant in terms of the mass distribution of the whole planet. But lacking data, we have no proof that mantle or core densities are any less anisotropic locally than those of the crust. )

However, putting that aside for a minute, what about Ladewig's cubic world? ARE the corners downhill from the face centres / edge middles??

Skeptoid · 19th February 2003, 03:05 PM

Think of the corners as tall mountains on a sphere where "sea level" is the center of a face of the cube.

Soapy Sam · 20th February 2003, 09:14 AM

What a very interesting image! Thanks for that. Immediately makes me think of Larry Niven's Mount Lookitthat.

Skeptical Greg · 20th February 2003, 10:09 AM

Quote:

Originally posted by Soapy Sam
What a very interesting image! Thanks for that. Immediately makes me think of Larry Niven's Mount Lookitthat.

Interesting!!!

My rough calculations tell me, such a mountain on earth (assuming a sphere with diameter of approx 8,000 miles),
would be about 1,200 miles high!!!!

garys_2k · 20th February 2003, 06:23 PM

Quote:

Originally posted by 69dodge

Faraday cages don't have to be spherical to work. And spherical mass shells don't shield gravity; they just don't create any gravity inside themselves. So I don't see the connection either. On the other hand, I don't know enough to say there definitely isn't one. Can anyone elaborate?

I'm pretty sure that both Faraday cage shielding and the lack of field inside a spherical shell are consequences of the inverse square law, and I think they might even be equivalent to it. So that's some sort of connection. But is there a more direct connection that doesn't go through the inverse square law?

Well, to be honest, a Faraday cage can work from two ways. First, all the charges will tend to the outside surface because of mutual repulsion, leaving the inside neutral (assuming there's nothing inside to attract them back). But, if a body with an opposite charge was placed inside a cage, then the inverse-square law would negate any net force in one particular direction, even though the inside would show a charge.

I haven't thought about a cubic cage, or one of an arbitrary shape. I suspect the net force would still be zero, but I definitely could be wrong on that.

pupdog · 23rd February 2003, 09:48 AM

It might be worth mentioning the GEOID--this is an equipotential surface of the Earth's gravity field (taking into account both mass and centrifugal force) that corresponds (approximately) to mean sea level. That is, it's a surface of equal gravimetric force. It has some topography (relief of almost 200 meters), so it deviates from a sphere, and can be described by vertical distances from a reference spheroid. At any point on the geoid, the pull of gravity is perpendicular.

Lemastre · 23rd February 2003, 01:14 PM

Considering that an object's mass is concentrated in a central point, as we do in assuming that earth's gravity emanates from the planet's center, is a convenience in mathematical theories and calculations but has limited value in reality. As already pointed out, earth's gravity varies in strength and direction as you move over the surface, as does the magnetic field. I'd think that the problem with studying gravity in a controlled way here on earth is that earth's gravity overwhelms any experiment we might set up. I've seen some rather simple demonstrations of gravity in a vacuum chamber where a couple of lead balls were shown to attract one another. Maybe gravity is something to be investigated on the space station.

garys_2k · 23rd February 2003, 03:25 PM

Quote:

Originally posted by Lemastre
I'd think that the problem with studying gravity in a controlled way here on earth is that earth's gravity overwhelms any experiment we might set up. I've seen some rather simple demonstrations of gravity in a vacuum chamber where a couple of lead balls were shown to attract one another. Maybe gravity is something to be investigated on the space station.

Nah, no need to be that exotic. You need some lead weights, monofilament fishing line, a ladder, styrofoam and a few other things. This would be a great science fair experiment:

http://www.fourmilab.ch/gravitation/foobar/

Lemastre · 24th February 2003, 04:38 AM

Demonstrating that gravity exists via simple experiments seems redundant –– we've known that it's there for a long time. So far, though, exactly what it is seems a mystery. As far as I know, no one has figured out how to duplicate it using electrical forces or anything else. If that could be done, then you probably could counteract it, which might lead to a levitation scheme that would reduce the need for rocket power in escaping from the earth, etc. My mention of using the space station in this was partly to provide a locale free of earth's ubiquitous pull and partly to suggest something for the space station to do.

garys_2k · 24th February 2003, 08:21 AM

Quote:

Originally posted by Lemastre
Demonstrating that gravity exists via simple experiments seems redundant –– we've known that it's there for a long time. So far, though, exactly what it is seems a mystery. As far as I know, no one has figured out how to duplicate it using electrical forces or anything else. If that could be done, then you probably could counteract it, which might lead to a levitation scheme that would reduce the need for rocket power in escaping from the earth, etc. My mention of using the space station in this was partly to provide a locale free of earth's ubiquitous pull and partly to suggest something for the space station to do.

Some research is going on right now, ref. this link from Slashdot a while ago:

http://www.wired.com/wired/archive/6...igravity.html?

My mind instantly screamed "CRACKPOT!," but on reflection, with no solid theoretical underpinning for gravity, we can't rule out anything. There is also research on capacitors being cyclically charged while being accelerated -- this, too, may cause "shielding" of gravity.

As long as nobody is claiming to be running something "over unity," or otherwise breaking a law of thermodynamics, I've got an open mind for this stuff.

24th February 2003, 05:29 PM

Quote:

Originally posted by Lemastre
Demonstrating that gravity exists via simple experiments seems redundant –– we've known that it's there for a long time. So far, though, exactly what it is seems a mystery. As far as I know, no one has figured out how to duplicate it using electrical forces or anything else. ...

Well, the same can be said for almost anything. How well do we "understand" electrical forces?

All of current expiremental evidence is in very close agreement with Einstein's General Theory of Relativity, which in some sense "explains" gravity very very very well. There are some VERY small effects that the astronomers think might not agree with GR, but the jury is still out on those. GR, or any of the "improvements" to GR being proposed to extend the theory to cover these early observations, does not allow for any sort of "sheilding" or any practical way of getting anti-gravity - it seems unlikely in the extreme that any ever will.

Attempts at making GR fit in with quantum theories also so very little hope in allowing for nifty flying cars or zippy anti-gravity space ships.

But we can dream I suppose...

(One of the difficulties in getting anti=gravity is that gravity is (via GR) closely tied in with space and time and when you start to imagine ways of blocking it you often also end up with wierd other effects such as time travel and cause following effect and faster than light travel and stuff like that. Very fun to think about but probably clues that what you are thinking about is utter bunk.)

14th February 2003, 09:00 PM	#1
Ladewig Hipster alien Join Date: Dec 2001 Location: not measurable Posts: 16,798	Force of gravity at poles and equator I'm having a bit of trouble fully understanding the replies I got over in Banter so I am jumping over to Science. I tried searching the net and came up with values between 0.1 and 1.0 % which seemed like a large range. Quote: 20) Gravity has a stronger pull at the Earth's poles than it does at the equator. As a result, a person who weights 150lbs at the equator, would weigh almost a pound heavier if they stood at the North Pole. True The earth is not perfectly round. It flattens out somewhat at the poles. Therefore, a person standing at the poles is actually 13 miles closer to the center of the earth than they are when standing at the equator. The pull of gravity increases as you move closer to the center of a gravitational mass, making objects heavier (just as when you move away from the center of a gravitational mass, such as when you fly into outer space, gravity weakens). In addition, the centrifugal spin effect at the equator slightly counteracts the pull of gravity. All of this translates into a difference of almost a pound between what a person would weigh at the equator versus the poles. See this article for more info. So if I had a uniform cube large enough to easily measure the gravitational force, and I stood at the center of a face, I would weigh more than if I stood at a corner? Part of my confusion comes from the (perhaps mistaken) belief that as one descends below the surface of the earth, one's weight decreases.
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14th February 2003, 10:51 PM	#2
Keneke Muse Join Date: Jan 2003 Posts: 980	I think the crux here is that gravity is created by the whole of the mass. As you stand on the surface of the earth, some of the mass is pulling you slightly down and to the left, right, etc. However, almost all the mass pulls you below the horizontal axis. The sum of the forces of all points equals to straight down. However, if you go underground, some of those vectors start to point upward, negating the downward pull. If you go to the center of the earth, you have an average force vector of approximately zero, since the mass is equally distributed around you. So yes, as you go underground, you lose weight, and if you go up above the surface of the earth, you lose weight because of the distance from the gravitating object. The reason you gain weight at the poles and not at the equator is because, though you are closer to the center, none of the mass is above your head and cancelling out gravity vectors like going underground does. Apparently the distance to the center of gravity is more important to the final weight than the fact that the mass of the earth, being slightly flattened, means there is roughly more mass out on the edges than directly below you. I'm not too sure about the square question, I'm only versed in spherical bodies, but thinking about the problem makes me say yes, the corner position would make you weigh less because you are farther from the center, and none of it is above your head.
Keneke Muse Join Date: Jan 2003 Posts: 980

15th February 2003, 06:39 AM	#3
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885	I'm puzzled too. It seems to me (naive view) that what is keeping me on the ground is the attractive force due to the mass of the Earth acting on the mass of me (and vice versa). I accept that at the centre of the Earth, that force averages to zero. I also accept Newton's conclusion that the mass distribution can be treated as equivalent to a point at the centre. (ie No matter where you stand, the whole planet is below you). So the force is the same (inverse square law) at any point on a sphere centred on the centre of gravity. Now if that imaginary sphere coincides with the Earth's surface at the pole (radius RP), it will be below ground at the equator (Radius RP + bulge). So if I want to weigh the same at the equator, I have to go down a mine till I reach a point on the sphere. (So I was wrong in my comment in the banter thread). The above holds even if Earth is not rotating. So there must be a second decrease in g due to my 1000+mph rotation at the equator, compared with zero at the pole. This would be proportional to the vertical component of my tangential momentum. Assuming the Earth actually was spherical, how fast would it need to rotate before effective escape velocity was reached at the equator? 1 Rev per hour is about 25200 mph, which is close to actual escape velocity. Would that be enough or is this far too simplistic? Now, it seems to me that we can speed the thing up by decreasing angular momentum. First we remove all the mountains, then we get rid of the moon, then we get all the bicycles in China lined up...
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885

15th February 2003, 10:00 PM	#5
Keneke Muse Join Date: Jan 2003 Posts: 980	Quote: Originally posted by Soapy Sam So if I want to weigh the same at the equator, I have to go down a mine till I reach a point on the sphere. (So I was wrong in my comment in the banter thread). No, if you go underground at any point, some of the mass begins to pull you upward, offsetting the gravity. So, at any point of X latitude and X longitude, the elevation that will give you the maximum gravity is exactly at ground level. So, if you are at the equator, you can't be as heavy as if you were at a pole, no matter how far up or down you went.
Keneke Muse Join Date: Jan 2003 Posts: 980

16th February 2003, 07:18 AM	#6
garys_2k Muse Join Date: Jan 2003 Posts: 756	Quote: Originally posted by Keneke No, if you go underground at any point, some of the mass begins to pull you upward, offsetting the gravity. So, at any point of X latitude and X longitude, the elevation that will give you the maximum gravity is exactly at ground level. So, if you are at the equator, you can't be as heavy as if you were at a pole, no matter how far up or down you went. Right. If you want to weigh the same at the pole as you do at the equator, the place to dig the hole is at the pole. Since you are heavier at the pole than the equator, dropping into a mine at the pole will reduce your weight. You could also climb onto a tall tower at the pole and get the same effect.
garys_2k Muse Join Date: Jan 2003 Posts: 756	__________________ - Gary

15th February 2003, 08:43 PM	#4
Walter Wayne Wayne's Words Join Date: Jul 2002 Location: Ottawa, ON Posts: 2,439	Originally posted by Soapy Sam .,. I also accept Newton's conclusion that the mass distribution can be treated as equivalent to a point at the centre. (ie No matter where you stand, the whole planet is below you).... I think this only hold if you are a a significant distance from an object. For instance, if we measure the distance vector to every point of the sun from here each of them is very close to the same distance and same angle, so the direction of the force and r squared term is the same for all of the sun and us. However, this doesn't hold for the earth below us as the distance vector varies a lot, so you can calculate the mass of an object simply by measuring g at its surface. One would have to know its geometry (and hope its density is relatively constant). The above holds even if Earth is not rotating. So there must be a second decrease in g due to my 1000+mph rotation at the equator, compared with zero at the pole. This just adds another force, but doesn't change g itself. Walt

16th February 2003, 10:48 AM	#7
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885	Walter- granted, the rotation effect (I'm trying to avoid the disgusting term "centrifugal force" does not affect g, but it does reduce my effective weight. If we can get the old girl spinning fast enough, free launches are the norm. Indeed we have to glue everything down. Oceans for instance. I'm just curious about how many rpm are required.) On Kenneke & Gary's points, let's see if I follow:- The gravitational gradient from core to pole is higher then that from core to equator. Surface gravity is always maximum, but not everywhere the same. Absolute maximum is at the pole,where the polar radius puts you nearest the core. There is no sub-surface point where g exceeds surface gravity at the pole (SgP). However as SgP is higher than surface gravity at the equator,(SgE), there are sub surface points where g is higher than SgE. Presumably these lie on an ellipsoid surface which coincides with the Earth's surface at the equator and dips to a fixed depth under the poles. I like threads like this. Forty years you carry a vague, wrong notion in your head and then you are forced to think it through. Nice one , Ladewig.
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885

16th February 2003, 11:10 AM	#8
bjornart Master Poster Join Date: Nov 2002 Posts: 2,395	Quote: Originally posted by Soapy Sam Walter- granted, the rotation effect (I'm trying to avoid the disgusting term "centrifugal force" does not affect g, but it does reduce my effective weight. If we can get the old girl spinning fast enough, free launches are the norm. Indeed we have to glue everything down. Oceans for instance. I'm just curious about how many rpm are required.) Since orbital speed equals the root of ar, where a is gravity in that orbit and r is the radius of the orbit, the speed needed to stay in orbit right on the equator is the root of 6378km 9,81m/s*s, which is 7910m/s. Run at that speed and a jump would be enough to launch you into space, if it wasn't for air resistance. (in rpms the answer is approximately 0.012, good luck)
bjornart Master Poster Join Date: Nov 2002 Posts: 2,395	__________________ Well, I DON'T CARE WHAT YOU LIKE TO BELIEVE, GODDAMMIT! I DEAL IN THE FACTS! -Cecil Adams

17th February 2003, 09:47 AM	#9
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885	.012 * 60 =0.72revs / hour. Rather less than the 1 I had guesstimated. Even better. Call the Chinese embassy and get them bikes lined up. Here we go...
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885

18th February 2003, 12:38 AM	#10
Cecil Muse Join Date: Oct 2002 Location: Vancouver Posts: 991	Quote: Originally posted by Keneke So, at any point of X latitude and X longitude, the elevation that will give you the maximum gravity is exactly at ground level. So, if you are at the equator, you can't be as heavy as if you were at a pole, no matter how far up or down you went. cough That seems very arbitrary. If I dig 6 feet underground, then although there is 6 feet of Earth now pulling me up, every other point on the Earth is now 6 feet closer, and pulling that much harder. I was taught in Physics that your point of maximum gravity was about 1000 kilometers under the surface, or about 10% of the way to the center.
	__________________ "Great spirits have always encountered violent opposition from mediocre minds." --Albert Einstein "The common man marvels at the uncommon; the wise man marvels at the commonplace." --Confucious "The whole problem with the world is that fools and fanatics are always so certain of themselves, and wiser people so full of doubts." --Bertrand Russell

18th February 2003, 01:06 AM	#11
Iconoclast Breaker of Icons Join Date: Sep 2001 Posts: 1,797	Quote: Originally posted by Cecil cough That seems very arbitrary. If I dig 6 feet underground, then although there is 6 feet of Earth now pulling me up, every other point on the Earth is now 6 feet closer, and pulling that much harder. I was taught in Physics that your point of maximum gravity was about 1000 kilometers under the surface, or about 10% of the way to the center. I was thinking along the same lines, and this is indeed a fascinating topic. It seems someone's going to have to bite the bullet and come up with a function for gravitational attraction versus distance from the core for distances less than the radius of the sphere. I'd do it but.. er.. my integrator broke and I haven't bought another one yet. Stimpy?
	__________________ A gun is not a weapon Marge, it's a tool, like a butcher knife or a harpoon or... ah... ah... an alligator. You just need more education on the subject. -- Homer Simpson

18th February 2003, 01:46 AM	#12
Iconoclast Breaker of Icons Join Date: Sep 2001 Posts: 1,797	OK, here's a thread on the subject from the sci.physics.research newsgroup from a few years ago, the thread in question is entitled "Baltimore Sun Says Gravity Inside Sphere is Uncalculable" (the thread title is misleading, that's not what the article was implying). From this discussion it appears that: - For a homogeneous sphere, gravity is always at a maximum at a point on the surface, it falls off above and below that point. Gravity falls off faster as we move outwards from the surface than than it does as we move inwards. - It is believed that the earth has a much higher density at the core than at the surface, thus gravity on earth will increase for quite a few hundred miles as we descend. I didn't know either of those things.
	__________________ A gun is not a weapon Marge, it's a tool, like a butcher knife or a harpoon or... ah... ah... an alligator. You just need more education on the subject. -- Homer Simpson

18th February 2003, 06:57 AM	#13
DrMatt Graduate Poster Join Date: May 2002 Location: USA Posts: 1,422	No matter where you stand on the cube, the whole cube is "below" you... Imagine the Earth--or the whole cube--divided into smaller subparts. The total gravitational force of the earth on you is the vector sum of the gravitational forces supplied by the parts. Divide the thing into a greater number of smaller pieces without bound, and you see that this is an integral. It's fair to assume that, within bounds of measurablility, the Earth has bilateral symmetry both about a polar axis and about a diameter through the equator. Therefore, for a person standing on it in either of those two places, all components of gravitation perpendicular to the diameter have a vector sum of zero, and only the components parallel to the diameter extending through the earth to the person count for anything. In other words, at those locations, the earth only pulls down. From there, it's a matter of grinding through the arithmetic and seeing whether the figures differ...
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18th February 2003, 08:06 AM	#14
boooeee Dart Fener Join Date: Aug 2002 Location: The Lando System Posts: 2,395	Just to clarify a potential misunderstanding, if you dug a hole six feet deep, the mass shell above you would actually exert no force at all. It is one of the happy consequences of the inverse square law that a spherical mass shell exerts no force at all on you if you are anywhere inside the shell, and if you are outside the shell, the force acts as if all the mass were concentrated at a single point in the center. I'm pretty sure that no exact expression exists for the force exerted by an ellipsoid on any arbitrary point, but it is possible that an exact expression might exist for certain special points, like the poles, or along the equator. But, like DrMatt and Iconoclast said, the only way to know is to bite the bullet and start integrating.

18th February 2003, 10:12 AM	#15
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885	Or, of course, start digging. Anyone got a gravitometer they aren't using? Great how the "simple" questions turn out more complicated than you would think. Next question. Does a pendulum clock (well insulated and at standard T & P) run at the same rate at pole and equator. (Ignoring relativistic effects of equatorial rotation velocity)?
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885

18th February 2003, 02:01 PM	#16
CurtC Illuminator Join Date: Aug 2001 Location: Dallas, TX Posts: 4,758	As I was digging into this subject a few years ago, I started to wonder, if you were at a point not at the pole and not at the equator, which direction gravity pulls. For example, let's say you're on the ocean at 45 degrees north latitude, due to the flattened shape of the earth, the direction perpindicular to the surface is not pointing at the center of the earth. My question was whether the force you feel at that point is directed towards the center, or along the surface's normal vector, or somewhere between. After a few non-answers, one smart person finally responded with a simple and inituitive answer. Anyone care to pin it down?

18th February 2003, 03:24 PM	#17
fishbob Seasonally Disaffected Join Date: Jan 2003 Location: Chilly Undieville Posts: 5,666	The direction gravity pulls is "down". Down is toward the center of mass of the earth, no matter where on the earth you are. The center of mass = center of gravity = the point where all the down directions around the planet intersect. Down is also perpendicular to the surface of a body of water. The center of mass of the earth is not necessarily the same point as the geometric center of the earth. I hope this helps.
	__________________ When you believe in things you don't understand, then you suffer . . . " - Stevie Wonder "Stupidity - a callow indifference to facts or data" - Stuart Firestein -neuroscientist. I hate bigots.

18th February 2003, 04:54 PM	#18
garys_2k Muse Join Date: Jan 2003 Posts: 756	Interesting about the effects of the Earth's increasing density causing a weight INCREASE for a while when descending into the hole. I'd assumed a homogeneous sphere which would decrease weight. True, too, that the "shell" of earth above you (where "above" is defined as farther from the center than you are) has no net force on you, all of it cancels out. As you descend into this uniform sphere the effect is that you are "standing on" a smaller sphere. The deeper you go, the smaller the effective planet you are on. In the center you're effectively floating in space. Regarding the local vector of "down," that varies a lot from mountain ranges, mineral deposits and nearby bodies of water. In fact, if you followed a "down arrow" from most spots on earth all the way in it would not follow a straight line, being pulled toward higher density areas at different depths.
garys_2k Muse Join Date: Jan 2003 Posts: 756	__________________ - Gary

18th February 2003, 05:52 PM	#19
Iconoclast Breaker of Icons Join Date: Sep 2001 Posts: 1,797	Quote: Originally posted by boooeee It is one of the happy consequences of the inverse square law that a spherical mass shell exerts no force at all on you if you are anywhere inside the shell. How come everyone except me seemed to already know this? As an electircal engineer I have this sinking feeling that I was supposed to already know this rule, I'm sure it has applications in EE.
	__________________ A gun is not a weapon Marge, it's a tool, like a butcher knife or a harpoon or... ah... ah... an alligator. You just need more education on the subject. -- Homer Simpson

18th February 2003, 05:54 PM	#20
garys_2k Muse Join Date: Jan 2003 Posts: 756	Quote: Originally posted by Iconoclast How come everyone except me seemed to already know this? As an electircal engineer I have this sinking feeling that I was supposed to already know this rule, I'm sure it has applications in EE. You already know about it, in a way. Think Faraday cage.
garys_2k Muse Join Date: Jan 2003 Posts: 756	__________________ - Gary

18th February 2003, 08:03 PM	#21
Ladewig Hipster alien Join Date: Dec 2001 Location: not measurable Posts: 16,798	So the cube thing... Quote: So if I had a uniform cube large enough to easily measure the gravitational force, and I stood at the center of a face, I would weigh more than if I stood at a corner? Yes? And a follow-up. If I were to walk from the center of a face to a corner, then would I be walking uphill? The gravitational vector would not be perpendicular to the surface.
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18th February 2003, 08:04 PM	#22
Iconoclast Breaker of Icons Join Date: Sep 2001 Posts: 1,797	Quote: Originally posted by garys_2k You already know about it, in a way. Think Faraday cage. Well I knew the field inside a faraday cage was not effected by any field outside it, but I was anaware that it was because of this shell rule. I feel so inadequate.
	__________________ A gun is not a weapon Marge, it's a tool, like a butcher knife or a harpoon or... ah... ah... an alligator. You just need more education on the subject. -- Homer Simpson

18th February 2003, 08:17 PM	#23
CurtC Illuminator Join Date: Aug 2001 Location: Dallas, TX Posts: 4,758	Quote: fishbob wrote: The direction gravity pulls is "down". Down is toward the center of mass of the earth, no matter where on the earth you are. The center of mass = center of gravity = the point where all the down directions around the planet intersect. BZZZZZZZZZT! On a perfect ellipsoid, which is extremely close to the shape of the earth, the surface is not normal to the line through the center at all points. Only at the poles and the equator. All all the other points on the surface, the vector that's normal to the surface will not be pointed toward's the ellipsoid's center. So does gravity pull towards the center, or is it perpindicular to the surface?

18th February 2003, 10:48 PM	#24
69dodge Illuminator Join Date: Nov 2002 Posts: 3,607	Quote: Originally posted by CurtC As I was digging into this subject a few years ago, I started to wonder, if you were at a point not at the pole and not at the equator, which direction gravity pulls. For example, let's say you're on the ocean at 45 degrees north latitude, due to the flattened shape of the earth, the direction perpindicular to the surface is not pointing at the center of the earth. My question was whether the force you feel at that point is directed towards the center, or along the surface's normal vector, or somewhere between. After a few non-answers, one smart person finally responded with a simple and inituitive answer. Anyone care to pin it down? Water flows downhill. The water of the ocean moves until its surface becomes perpendicular to the force of gravity. (Well, not "the force of gravity", exactly. Rather, "the net force at that point", due to anything: gravity, centrifugal force, tidal forces from the sun and moon, etc.)
69dodge Illuminator Join Date: Nov 2002 Posts: 3,607

18th February 2003, 11:23 PM	#25
69dodge Illuminator Join Date: Nov 2002 Posts: 3,607	Quote: Originally posted by Iconoclast Well I knew the field inside a faraday cage was not effected by any field outside it, but I was anaware that it was because of this shell rule. I feel so inadequate. Faraday cages don't have to be spherical to work. And spherical mass shells don't shield gravity; they just don't create any gravity inside themselves. So I don't see the connection either. On the other hand, I don't know enough to say there definitely isn't one. Can anyone elaborate? I'm pretty sure that both Faraday cage shielding and the lack of field inside a spherical shell are consequences of the inverse square law, and I think they might even be equivalent to it. So that's some sort of connection. But is there a more direct connection that doesn't go through the inverse square law?
69dodge Illuminator Join Date: Nov 2002 Posts: 3,607

19th February 2003, 12:23 AM	#26
fishbob Seasonally Disaffected Join Date: Jan 2003 Location: Chilly Undieville Posts: 5,666	CurtC sez: Quote: On a perfect ellipsoid, which is extremely close to the shape of the earth, the surface is not normal to the line through the center at all points. Only at the poles and the equator. All all the other points on the surface, the vector that's normal to the surface will not be pointed toward's the ellipsoid's center. You appear to be thinking geometry. The geometric center of the earth is not the same as the center of mass (= center of gravity). As garys_2k said: Quote: In fact, if you followed a "down arrow" from most spots on earth all the way in it would not follow a straight line, being pulled toward higher density areas at different depths.
	__________________ When you believe in things you don't understand, then you suffer . . . " - Stevie Wonder "Stupidity - a callow indifference to facts or data" - Stuart Firestein -neuroscientist. I hate bigots.

19th February 2003, 07:45 AM	#27
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885	Last I heard, Earth is an oblate ellipsoid- slightly pear shaped with the big end in the S. hemisphere. Given convection in the mantle (and possibly parts of the core??), and tidal effects from Sun and Moon, rivers, trucks, plate tectonics and migrating lemmings, can the Earth be said to have one " permanently fixed" centre of mass anyway? (The lemmings may be small, but they are way out on the edge. High angular momentum, lemmings- even more so, migrating continents.) As for the argument that gravity doesn't act upwards inside a sphere- As an idiot, I demand more than just an assertion that this is so. I need an explanation. If I'm a mile under Australia I reckon Oz exerts the same pull upwards as the Australia sized mile thick slab below me does downwards. It's mass can't just vanish. Granted there's a similar slab on the far side of the world, but that's 8000miles away (Inverse squares) and furthermore that antipodean slab was there below me before I went down the shaft as well, whereas Oz wos below me too and now is above me. Something does not ring true here. Upward attraction may well average out to zero at the centre of mass, but I can't believe it does so at every point. After all, where is the surface of the Earth? If I sit under a table, is the upward gravity of the table exactly balanced by the pull of the whole Earth? If so, why is the table not lighter when lifted from beneath? Should it not be weightless? I need a cup of strong tea.
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885

19th February 2003, 08:01 AM	#28
CurtC Illuminator Join Date: Aug 2001 Location: Dallas, TX Posts: 4,758	fishbob, I accept that since the earth is not homogeneous, or even has symmetric density, that the geometric center is not exactly the same as the center of mass. But you stated that all the "down" directions all over the surface point towards that common point, and this is not true. They don't intersect. As 69dodge said, "down" is always the normal to the surface (with water, anyway), and I'm sure you can see that all those normal vectors don't intersect. Soapy Sam, if you're inside a even-density shell, then that shell exerts no net gravity on you. Even though that little piece of land that might be closer to you has a higher gravity per unit volume, there is a lot more volume on the other side doing its part to pull you away, and it just so happens that they exactly cancel. The proof is a classic excercise in freshman college physics. If you're a mile under Oz, then Oz does indeed pull you upwards, but the first one mile in depth of the Atlantic Ocean (on the opposite side of the earth) pulls you the other direction, pretty closely negating the effect of Oz pulling you up. As has been pointed out, there are errors due to the fact that the earth isn't even density, but that's basically how it's supposed to work.

19th February 2003, 08:16 AM	#29
boooeee Dart Fener Join Date: Aug 2002 Location: The Lando System Posts: 2,395	Quote: Originally Posted By Soapy Sam As for the argument that gravity doesn't act upwards inside a sphere- As an idiot, I demand more than just an assertion that this is so. I need an explanation. The best way to prove this is to use Gauss' law, but short of proving Gauss' law, this just replaces one mystery with another. The best way I can think of to explain it is as follows: Let's say you're at some arbitrary point within a sphere, say 5 miles below the surface of a 50 mile radius sphere. You could reason as follows: Obviously, the portion of the sphere "above" me is going to pull up, and the portion of the sphere "below" me is going to pull down. So, you could say: "The portion of ther sphere 'below' me is going to exert a greater force because there is more mass in the 'below' portion. But then you might say: "But the portion 'above' me, although less massive, is closer to me on average, so the 'above' portion should exert a greater force". It turns out that these two effects (mass and distance) completely balance, leaving no net force on you no matter where you are within the sphere. Hopefully I haven't just muddied the waters further. As far as CurtC's question goes, it was running through my head all throughout my evening commute yesterday, and I'm still not sure. I am pretty sure that the force does not necessarily have to point towards the center of mass, but it very well may be the case on the surface of an ellipsoid.

19th February 2003, 01:45 PM	#30
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885	CurtC , Boooeee. Thanks for the attempt gents. If I can't quite credit your argument, I'm sure it's my fault, not yours. I UNDERSTAND the principle but find it hard to believe in a real case. In a sphere of uniform density gradients I can happily accept the averaging argument, but not in a real planet. After all, if the inner average is uniform at every point, then surely that must hold for every point on a polar axis- including the pole itself, where we have already agreed g is higher than at the equator. My confusion is increased by the fact that I have been involved in enough gravimetric surveys to know that there are few things less constant on the Earth's surface than gravity. And Earth is far from isotropic in density. (We know most about the crust, which I grant is pretty insignificant in terms of the mass distribution of the whole planet. But lacking data, we have no proof that mantle or core densities are any less anisotropic locally than those of the crust. ) However, putting that aside for a minute, what about Ladewig's cubic world? ARE the corners downhill from the face centres / edge middles??
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885

19th February 2003, 03:05 PM	#31
Skeptoid Graduate Poster Join Date: Sep 2002 Posts: 1,532	Think of the corners as tall mountains on a sphere where "sea level" is the center of a face of the cube.

20th February 2003, 09:14 AM	#32
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885	What a very interesting image! Thanks for that. Immediately makes me think of Larry Niven's Mount Lookitthat.
Soapy Sam NLH Join Date: Oct 2002 Posts: 25,885

20th February 2003, 10:09 AM	#33
Skeptical Greg Agave Wine Connoisseur Join Date: Jul 2002 Location: Just past 'Resume Speed' Posts: 12,873	Quote: Originally posted by Soapy Sam What a very interesting image! Thanks for that. Immediately makes me think of Larry Niven's Mount Lookitthat. Interesting!!! My rough calculations tell me, such a mountain on earth (assuming a sphere with diameter of approx 8,000 miles), would be about 1,200 miles high!!!!
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20th February 2003, 06:23 PM	#34
garys_2k Muse Join Date: Jan 2003 Posts: 756	Quote: Originally posted by 69dodge Faraday cages don't have to be spherical to work. And spherical mass shells don't shield gravity; they just don't create any gravity inside themselves. So I don't see the connection either. On the other hand, I don't know enough to say there definitely isn't one. Can anyone elaborate? I'm pretty sure that both Faraday cage shielding and the lack of field inside a spherical shell are consequences of the inverse square law, and I think they might even be equivalent to it. So that's some sort of connection. But is there a more direct connection that doesn't go through the inverse square law? Well, to be honest, a Faraday cage can work from two ways. First, all the charges will tend to the outside surface because of mutual repulsion, leaving the inside neutral (assuming there's nothing inside to attract them back). But, if a body with an opposite charge was placed inside a cage, then the inverse-square law would negate any net force in one particular direction, even though the inside would show a charge. I haven't thought about a cubic cage, or one of an arbitrary shape. I suspect the net force would still be zero, but I definitely could be wrong on that.
garys_2k Muse Join Date: Jan 2003 Posts: 756	__________________ - Gary

23rd February 2003, 09:48 AM	#35
pupdog Muse Join Date: Dec 2002 Location: Pennsylvania Posts: 627	It might be worth mentioning the GEOID--this is an equipotential surface of the Earth's gravity field (taking into account both mass and centrifugal force) that corresponds (approximately) to mean sea level. That is, it's a surface of equal gravimetric force. It has some topography (relief of almost 200 meters), so it deviates from a sphere, and can be described by vertical distances from a reference spheroid. At any point on the geoid, the pull of gravity is perpendicular.
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