Suppose you drilled a hole all the way through the exact center of the earth (land to land so as not drain the seas) one mile in diameter. Then suspended a ball about a thousand miles down and 100 feet from the edge of the hole. Would the gravity of the earth pull the ball to the side or would the earth’s center gravity keep it straight down?
You’d be about 1/4 the way to the centre. My guess is, (Assuming you could stop the hole filling with molten rock and collapsing), the ball would hang directly towards the centre. There would be roughly equal mass on either side of the hole and far greater mass towards the core.
Depends how wide the hole is. The Earth is 12,742 km across (7900 miles) give or take a few.
If the hole you drill is 10,000 miles wide then the ring-like remnant of the Earth would exert a greater sideways pull on your suspended ball than if the hole you drill is only 1km wide, or 200 metres. A ball suspended inside a narrow hole would experience a negligible sideways pull because of the equal distribution of mass surrounding it.
So…given the diameter of the hole as in the OP?
Negligible deflection. The exact size of the deflection is left as an exercise for someone with way too much time on their hands. 
I agree with this. Just to extend the thought experiment, wouldn’t the ball develop a circular motion in its hang due to the earth’s rotation, 'a la a Foucault pendulum?
If we have a ball hanging in a perfect , non-rotating sphere with a big hole drilled into it, and it’s 100 feet from the edge, a thousand miles down, then, yes, it will be drawn toward the edge. Here’s an easy way to show this:
1.) Consider the case of a ball inside a spherical shell. One of the problems presented to undergbrads is to calculate the gravitational force inside a spherical shell. The answer – it’s zero. The forces cancel out in all directions.
2.) therefore, no matter how thick the layers above you are, you can deconstruct them into a collection of concentric spherical shells, and the influence of each is zero. so there’s no attraction if you’re inside a complete (no hole) hollow ball.
3.) Similarly, another undergrad exercise is to calculate the effect of standing on a uniform sphere. The gravitational attraction works out to be the same as if it’s all due to a ball of that mass concentrated at the center.
4.) So if you’re inside a sphere, the mass in radii greater than your location has no effect, and you’re pulled towards the center by a force equal to that of the mass contained within the radius you are at.
5.) To lowest order, that’s the solution. A one mile hole doesn’t make much of a difference to the calculation, and you’re mostly pulled straight towards the center with a force that varies , getting smaller as you go towards the center.
6.) But there is a much smaller effect due to the fact that you’ve excavated a hole a mile in diameter out of that sphere. If you were smack in the center of that hole, symmetry alolne would tell you that the force is all downward toward the center. But you’re not – you’re off to one side.
7.) You can visualize the effect of not having a complete sphere by imagining the sphere disassembled into two parts – the sphere with the hole in it, plus the missing “shaft” (a cylinder with spherically rounded ends) that you removed to make the hole.
8.) the gravitational effect of the sphere with the hole taken out can be visyalized as the effect of gravity of a perfect, hole-less sphere PLUS the “anti-gravity” of that shaft. In other words, you can calculate the effects of your sphere with a hole as the effect of a perfect sphere – you’re pulled toward the center by the force of the mass within your radius – and add to it a “gravitational repulsion” that is the opposite of the attraction you would feel from that “shaft” alone.
The gravitational effect of a cylinder 1 mile in diameter and 7,900 miles can be calculated, but I’m not going to do it. By symmetry, you can tell that it will NOT be directed toward the center of the shaft, but will be somewhere else along its length. (recall, too, that your target is inside the 1 mile diameter). The effect , as I say, is the opposite of this attraction. So your ball, suspended at a depth of 1000 miles and 100 feet from the edge will feel mostly a pull toward the center of the sphere, but there will also be a small force upward, lessening the downward pull by a trifle, and there WILL be an outward component, directed toward the wall of the shaft.
Wow, thanks Cal!
And of course in the real universe the real Earth is not perfectly symmetrical in shape nor perfectly homogenous in makeup. So in addition to everything **CalMeacham **so ably explained, there will be some persistent net pull in one random direction.
And as Oly said upthread, the fact the Earth is rotating will introduce Foucault motion. And the orbiting Moon and Sun will exert tidal forces.
But if you want to stick to textbook spherical cows, then **Cal **hit it out of the park.
ALWAYS assume a spherical cow.
Unless you’re considering planning a farm on a hill. Erecting all those barriers to keep the spherical cows from rolling down to the lower part of of the enclosures is a big waste of money.
Yes, but if you arrange the powered corral gates at the bottom correctly you can play pinball !! 
Note that if you suspend the weight from your unbreakable diamond cable and start lowering it down your hole, it’s very quickly going to hit the side of the shaft, in the direction of the forward spin of the Earth. At the equator an object on the surface is traveling at about 1000 miles per hour, because the Earth’s circumference is about 24,000 miles and it takes 24 hours for the Earth to spin once. But 1000 miles down the circumference is only about 18,000 miles, which means the mantle at that point is only spinning at 785 miles per hour. That means as you lower the weight down on the cable its traveling faster than the shaft and will smack into the leading edge pretty soon.
Or you could assume you built the cable in place and placed the weight on the end of the cable, and the cable and weight are all spinning at the appropriate speed for their depth, in which case nevermind. Or, dig the shaft from the north pole to the south pole. I know you didn’t want to drain the ocean, but if you’re digging a shaft through the center of the Earth a small 2-3 mile wall to keep the ocean from pouring in doesn’t seem so hard.
It’s not that hard. We’re far enough from the ends that the length is irrelevant, and Gauss’s Law can be used, just like it can be for an infinite line of charge. The field will be equal to 2Glambda/r, where lambda is the linear mass density (density of rock times the cross-sectional area), and r is the distance from the center of the cylinder to your point of interest.
Well, we should be able to calculate the angle at which the cable hangs, if we can figure out the repulsion of the hole (which is an interesting way to imagine it).
I shan’t do the exact calculation, because I always get these slightly wrong; but it should be roughly similar to the gravity of Comet 67p, about a ten-thousandth of a gee.
But wait a minute - if the cable that this weight hangs from is 100 feet away from the edge of the hole, and the repulsion of the hole is 10,000 times as small as Earth’s gravity - then the cable will hang at such an angle that it will touch the walls after a million feet - 189 miles. If the repulsive effect of this hole is really 0.0001 gees then the ball will touch the wall about a fifth of the way down. Not negligible after all.
(feel free to substitute the real numbers to prove me wrong).