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Consider a perfectly spherical earth, and a vertical tunnel perforating it through the center, with an elevator in the tunnel. suppose you are halfway towards t

Consider a perfectly spherical earth, and a vertical tunnel perforating it through the center, with an elevator in the tunnel. suppose you are halfway towards the center, coming from the surface, and you pause the elevator. How strong will the gravitational field be? Can we pretend all of Earth's mass is concentrated in the center? No, let's see why. Consider a perfectly spherical hollow shell of mass, and an observer inside the shell, will she be attracted to the center of mass of the shell? No, let's see why. We know that gravitational field falls of with distance squared ( 1 / r^2 ). Consider a random observer position, and a random direction. Now consider a cone tipped at the observer with the random direction as its axis, and also consider a second cone with the same cone angle but the opposite direction as its axis. So the observer position is where the 2 cone tips meet. Then consider the distance between the observer towards each patch of the spherical shell. The area (and thus mass) of this patch will scale with the distance squared, so if one patch is say 3 times further away than the other, it will be 9 times weaker due to 1 / r^2 gravitational fall off, but also 9 times heavier due to geometric scaling of the patch, so gravity due to the patches will cancel, and since the direction of the axis was arbitrary, the gravity of each part of the shell will be balanced by the gravity of a corresponding opposite part of the shell. So when you are in an elevator halfway down the radius of the hypothetical spherical Earth, all the layers of Earth above you will cancel gravitationally, and the gravitational field will only be due to the mass of the Earth that is contained in the sphere up to your altitute with respect to the center of the Earth. Similarily, when the electron is far enough from the (assumed spherical) proton, we can somewhat pretend the proton is a point particle. but when the electron is inside the proton, then the effective charge of the proton will be lower because of all the charge of the proton that is farther from the proton's center than the electron is invisible to the electron (since the electric field also falls of like 1 / r^2 ). I hope that answers the question?

chasd00 7y ago
forgive my ignorance, but don't you have to assume uniform density in both cases? It seems to me the person in the earth elevator halfway down would feel more gravity toward the center of the earth because there's actually more mass due to the higher density. In a similar way, the proton would have higher mass toward the center because of the higher wave amplitude.
- DoctorOetker 7y ago
  correct, it does assume uniform density, if it didnt, you could say the density was zero throughout the sphere except for a small spherical region, and it would be trivially false then I was illustrating how in principle radii can be derived in a phenomenological sense. I didn't read the actual papers by the experimenters of the proton radius determination, but it would probably require a modification of the Hamiltonian for the hydrogen atom, since usually one treats the nuclei as point particles.
- btown 7y ago
  Yep, due to non-uniform density it's much more complicated. https://en.wikipedia.org/wiki/Gravity_of_Earth#Depth https://en.wikipedia.org/wiki/Gravity_of_Earth#Depth includes a graphic that shows how it varies.

chasd00 7y ago

forgive my ignorance, but don't you have to assume uniform density in both cases? It seems to me the person in the earth elevator halfway down would feel more gravity toward the center of the earth because there's actually more mass due to the higher density. In a similar way, the proton would have higher mass toward the center because of the higher wave amplitude.

DoctorOetker 7y ago
correct, it does assume uniform density, if it didnt, you could say the density was zero throughout the sphere except for a small spherical region, and it would be trivially false then I was illustrating how in principle radii can be derived in a phenomenological sense. I didn't read the actual papers by the experimenters of the proton radius determination, but it would probably require a modification of the Hamiltonian for the hydrogen atom, since usually one treats the nuclei as point particles.
btown 7y ago
Yep, due to non-uniform density it's much more complicated. https://en.wikipedia.org/wiki/Gravity_of_Earth#Depth https://en.wikipedia.org/wiki/Gravity_of_Earth#Depth includes a graphic that shows how it varies.