Wiploc

This doesn't work. You are never going to get heavier by digging down. Let's divide the earth into five segments, with you on the top surface:

You
1st segment
2nd segment
3rd segment
4th segment
5th segment

Now you dig down one segment:

1st segment
You
2nd segment
3rd segment
4th segment
5th segment

So what did you get closer to by digging down? You could have accomplished the same thing by taking dirt from the far side of the planet and piling it above you ...

5th segment
You
1st segment
2nd segment
3rd segment
4th segment

... and you certainly wouldn't expect that to make you heavier. In fact, the segment above you cancels out the segment immediately below you, so the effect of digging down is the same as this:

Nothing
You
Nothing
3rd segment
4th segment
5th segment

Clearly that won't make you heavier than your initial position:

You
1st segment
2nd segment
3rd segment
4th segment
5th segment

crc

Jabu Khan

The difference in weight has to be on an extremely negligible scale because the mass of the earth is so much greater.

It seems like you should be able to measure the angle between two positions of the sun at different times and nowing the lapse of time and the percieved distance that it traveled figure out where you are. But its driving me nutty thinkin about it. You would probably also need to know if the sun was 23 degrees N/S or crossing. If you stayed there long enough you'd at least no you were close if the sun was 23 deg N in the summer and 23 deg S in the winter.

SULPHUR

It is a small amount but the fact that it does occur vindicates the idea of an oblate earth

Yggdrasill

wiploc, you fail to take into account the fact that the 1st segment has less mass than the segments in the middle. The 2nd segment has greater mass than the 1st segment, so digging downwards is not the same as taking mass from the opposite side and piling it on top of yourself. Here is a improved version of your model (look at the attachment):

1st segment has approximately 1 unit of mass
2nd segment has approximately 2 units of mass
3rd segment has approximately 2.2 units of mass
4th segment has approximately 2 units of mass
5th segment has approximately 1 unit of mass

By digging downwards by one segment you are getting closer to the larger segments in the middle. And by being closer to them, they exert a greater gravity upon you. This increase in gravity is greater than the gravity that the segment above you detracts. Do you understand?

Wiploc

I do understand your reasoning; you are perfectly lucid. But I still don't agree with your conclusion.

I love your illustration. What program did you make it with? I assume if I read the FAQ I'd learn how to attach an illustration myself?

Let me ask this: If there were a BB and a telephone pole alone in space, would you grant me that the BB's weight would be greater if it was on the end of the telephone pole than if it was on the middle of the side?

And will you grant me that you would weigh less if the top segment in your illustration were turned upside down?

As I look out my window and see how far away the horizon is, I have to think that there is already plenty of wasted earth off to the side of me that must have a negligible effect on my weight. If we shoveled dirt from segment five around to add it on to the sides of segment one, it would definitely bring the dirt closer to me, but it would also move it off to the side where it would be pulling much more to the side than down.

You were perfectly clear. I hope I'm getting my point across too, but I don't know.
crc

Wiploc

Ha, I think I can do this now.

Yggdrasill, let’s say you are standing on top of segment 1 of your diagram.

Particle 1 is a molecule directly below you, at the bottom of segment 1.

Particle 1 pulls you down with a force of 1gu (gravitational unit).

Particle 2 is also straight down, but twice as far away, at the bottom of segment 2.

Particle 2 pulls you down with a force of ¼ gu because gravitational attraction is proportional to the inverse square of the distance.

You can guess where particle 3 is, and it pulls with 1/9 gu.

Now we’re going to put particle 4 off to one side of particle 3. It will be 4/3rds as far from particle 3 as particle 3 is from you. In other words, we have a 3,4,5 triangle, with you as one corner, particle 3 as the right angle, and particle 4 as the third angle.

Since this is a 3,4,5 triangle, we know the distance from you to particle 4 is 5. That is, it pulls on you with a strength of 1/25th gu.

We also know that the vector represented by that hypotenuse can be broken into vectors represented by the other sides of the triangle. We can dismiss the sideways vector, because we are looking at weight. (And the sideways vector will be offset by the sideways vector of another particle on the other side of Particle 3 anyway.) The vertical vector has a length of three, so the downward pull of particle 4 is 3/5ths of 1/25th gu, which is 3/125ths gu, or .024 gu.

My question: If you dug down thru two segments so that you stood at the top of segment 3, would particle 4 give you more weight or less weight than it did before? We know you are going to be closer to it, but we don’t know how much of the increased pull will be weight.

So, we now have a 1,4,x triangle. The Pythagorean theorem gives the length of x as the square root of 17, which is to say 4.123106.

To find out how hard it pulls, we need only square that, deriving the number --- now this is embarrassing --- 17. So, Particle four used to pull with only 1/25th gu; but now that you’ve dug down two segments, it pulls with 1/17th gu. The attraction is more than half again greater than it was when you were back at the surface.

How much of that is weight? That is, the downward vector comes to what part of that 1/17th gu? We take 1/4.123106 times 1/17th gu, which gives us .014267 gu.

When you were on the surface, Particle 4 provided .024 gu of weight. Now that you have dug down closer to it, it provides only about .014 gu of weight.

How am I doing? Is this totally opaque, or am I making my point?
crc

(Feel free to tell me I did the math wrong. I hope I corrected all my errors, but I did find some in there.)

Wiploc

Point a wide-angle lense straight up, and take a time lapse photograph at night. The stars will make white streaks. At the poles, the streaks will look like circles. As you move toward the equator, the streaks will be less and less curved. If you are right on the equator, the stars directly overhead will move in a straight line. So if your picture shows a straight line down the middle with curved lines on both sides, you are on the equator.

The more wide-angle the lense is, the more obvious the curved lines will be.
crc

Lobstrosity

I have no idea what you and wiploc are doing! Why are you using segments? Just use Gauss' Law--it's a two-line calculation! How can you be using cartesian coordinates on a spherical problem???

Assume earth is a sphere of radius R and has unform density rho. Rho can easily be calculated as the total mass of the earth over its total volume.

Gauss' Law: the closed surface integral of g dotted with the differential surface vector will be proportional to the mass enclosed by the surface. Inside the Earth (r < R), if we choose a spherical surface whose center coincides with the center of earth, the mass enclosed by the surface is 4/3 pi r³ rho. By symmetry we know that g is unform everywhere on this surface and pointed radially (i.e. normal to the surface), so we can break g out of the integral and simply replace the integral with the surface area of our surface. This is just 4 pi r². When you divide the mass enclosed by the surface area, you get a function for g that's proportional to r (that's right, r, not 1/r²).

Gravity never increases as you go down into the hole. Inside the earth, the gravitational force goes as a function of r. Outside it goes as a function of 1/r². The two functions are obviously continuous at the earth's surface. Thus the force of gravity is monotonically decreasing as you go down. The force is maximized exactly on the surface.

Lobstrosity

As an aside, it would be interesting to note that if the mass distribution with in a planet could be given by a density function rho(r) proportional to 1/r, gravity would constant everywhere within the planet. So if you had an elevator from the surface down to the core, your weight would not change at all as you descended. Of course such a planet would have infinite mass density at its very center, so perhaps this isn't the most physically-realistic scenario (though it's still interesting!).

For a uniformly dense planet (as derived in my previous post) your weight would decrease linearly with r, so halfway to the center you would way half as much.

Another thing to note: If the density is a function of r^-n where n > 1, gravity will increase as you move from the surface towards the center, approaching infinity as r approaches zero. The rate at which you approach infinity would be dictated by the value for n.

Yeah, this has nothing to do with the equator, but it's still cool (well, cool in that nerdy way, you know?)

SULPHUR

Lets take the force of attraction.Using F=Gmm/d^2.
G=gravitatiomal constant
m and m=the two masses
d = distance
IS d taken as the the distance between the two centres of gravity or the outer surfaces of the masses, say the earth and moon.
It would alter some of the ideas and maths of the above posts.