This is just a quick fun article that may be trivial to some readers but will probably be very interesting to others. Imagine you jump through a hole that goes through the center of the earth like the little man below.

Assuming we work under the assumption that the Earth is just a solid sphere of uniform rock that would allow for such a jump and there is no air resistance, let’s see if we can calculate the type of motion one would take in such a situation.

First, we must make sense of a property of uniform spherical spheres and shells. If one is inside any spherical shell, they experience no force. This is called Newton’s shell theorem and it means if someone was inside a sphere, the only part of the sphere that would have a net gravitational pull on the person would be the sphere whose radius is the distance from the person to the center and whose center coincides with the center of the original sphere. This is made clear in the following image.

Only the dark region acts on the man while the light part has no net gravitational pull. This means that at a distance from the center, the mass that is influencing the man is where is average density. Because the whole sphere has uniform density, where is the total mass and is the total radius of the earth. Combining the two equations, we get . So the gravitational pull at any distance in the sphere is . The value is just a big constant so the equation basically reduces to . This is the exact same as Hooke’s law for springs.

Falling through the Earth would simulate the motion of a spring. One would speed up until they reach the center, then start slowing down until they reach the other end where they would momentarily stop and fly all the way back in. This process would repeat forever in perfect periodic motion.