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Theoretically speaking, if you were to travel to or near the core of the Earth, would you be heavier because the radius would decrease between you and the core; therefore, decreasing the radius in the formula Fg=Gm1m2/r2 and increasing the Fg?
Question Date: 2008-12-18
Answer 1:

Hi, you're halfway there with that reasoning, but you've forgotten that one of the other variables in that equation also changes when you move closer to the center of the Earth: the mass of the object pulling on you, which I'll call M1. More specifically, in the formula
Fgravity = G * M1 * Myou / r2 ,

M1 is the mass enclosed within a sphere that has a surface at radius r. So, when you're on the surface of the Earth, M1 = MEarth. But when you're closer to the center of the Earth, M1 is less than MEarth. And when you're very close to the center, you can approximate the mass density of Earth, rho, as constant, so that the mass contained within a sphere of radius r is just the volume (4/3*pi*r3) times the density:

M1 = rho * 4/3 * pi * r3 .

Plugging that into the force equation gives

F = G * 4/3 * pi * Myou * rho * r .

Since rho is just some number (somewhere between 1 and 10 grams per cubic centimeter), and Myou stays constant, the force goes to 0 as r goes to 0. And that makes sense because at the exact center of the Earth, which direction would gravity pull on you? All directions look the same, so there can't be a force due to gravity! And so the force near the center must be nearly zero.

Physicists call that a "proof by symmetry", which can be very useful while involving no effort, both of which are appealing to physicists! Another example that comes to mind as I drink my coffee: the next time you're holding a glass of liquid, put it down with a slight bump on the table, keeping the bottom as flat as you can. Or think back to the movie "Jurassic Park", when the kid has his water on the car's dashboard and the T. Rex is stomping closer. What do the waves on the surface look like? Better yet, what do they HAVE to look like? Circles, because there aren't any preferred directions in this setup! The waves only "know" about the symmetries of the cup, which are circular.

Symmetries in all aspects of physics can take you a long, long distance, in ways that still blow my mind. Good luck with your studies!

Answer 2:

Actually, the acceleration due to gravity is found from the following formula g= G M(r)/r2 where M(r) is the mass of earth between the center of earth and distance r from the center.

So, given how the mass of the earth between the center and some distance r from center varies, it turns of the g is almost constant from the surface down to a depth of about 3000 km. thereafter g decreases until it would be zero at the center of the earth .

If you google this site:
you can see a graph of gravity (actually the value of little g) vs depth.

Answer 3:

No, actually you would get lighter.
If you are inside of a hollow, massive sphere, you would experience no gravity. This is because the sides of the sphere are pulling on you equally in all directions. Even if you are closer to one edge of the sphere, there is more material on the other side even if it's farther away. You experience no gravity.

Now, the Earth is not a hollow sphere, but what this does mean is that as you burrow into the Earth, the mass of the planet is divided into two parts: (1) the mass that is above your current depth, and (2) the mass still inward toward the center. The mass of (1) doesn't matter, but you do feel gravity from mass (2). However, as you continue going inward, the mass of (2) is progressively getting smaller because the radius out to you is getting smaller. Mass is proportional to volume, which is in turn proportional to the cube of the radius. Gravitational force, alternatively, is inversely proportional to the square of the radius. r3/r2 = r, which means that the amount of gravity you experience is proportional to the radius, and the farther out from the center you are, the more gravity there is, until, of course, you break out again through the surface of the Earth.

Now, in this calculation, I assumed that the Earth has the same density throughout. It doesn't; the core is a lot denser than the mantle or the crust. Because of this, it may be that for a while you would get heavier as you bored in for exactly the reason you suggested in the first place because you're getting closer to the surface of the denser, more massive core. You would have to do the calculation (density of the mantle is about 2.5 thousand kg/m3, density of the core is about 10 thousand kg/m3, the Earth has a radius of 6.4 million meters, and the radius of the core is I think about 3 million meters).

The only time in which you could bore into something and get heavier as you go farther in is if all of the mass is concentrated at the very center - something which only happens in a black hole. There, your weight really does become infinite as you approach the center of the hole.

Answer 4:

You would be lighter because all of earth's mass would be distributed equally all around you. All of it would be pulling on you, but the pull would be equal from every direction, so it would all cancel out. On the other hand, if you squeezed Earth to a smaller radius, so it had the same mass but you were still outside it, then your weight (gravitational force) would go up just like you predict from the Fg equation.

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