Earth pulls you with the same amount of force that you pull the Earth. But since Earth is so much larger than any freely moving body, on or above her surface, we treat her gravity as a directional force. This makes the force of gravity relatively simple, so we teach gravity before we teach elasticity. But in reality, elastic cohesion (the drawing together of particles) is the force identified by Newton in his law of universal gravitation, “Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them.”
However, our hot-core model is valid only if we ignore horizontal elastic cohesion.
Elastic cohesion in a solid imparts a constant pull between all its parts. But, no one ever bothered to calculate the strength of that pull in Earth’s shells, because in a schematic of forces diagram the improper use of directional forces (the scientist’s sleight of hand) makes them appear to cancel out; but a gravitational pull cannot cancel out another gravitational pull—only balance. Their pulls are still present.
Now, if we treat Earth’s outer shell as a structurally sound, hollow sphere, subject to horizontal elastic cohesion
Hi Cold Co, welcome to Hypography.
Would you agree that the gravitational potential inside a spherical shell of matter is constant meaning there is no gravitational force inside a spherical shell?
Would you also agree that an object outside a spherical shell is drawn toward the center of the shell with a force equal to a situation where the mass of the shell were replaced with an equal mass at a point in the center of the shell?
I ask because it sounds like you are disagreeing with this kind of vector analysis—believing that an element in the earth is affected gravitationally in a net way by the mass above it and next to it. I would certainly disagree with this, but before showing why I'd like to know if you are indeed saying that.