Question

Question #001d8

Answer 1

This is Newton's Shell Theorem. A nice derivation by integrating circular strips can be found here. Another solution is provided below.

The basic idea is that if you were inside a uniform spherical shell, even if you were closer to one side of the sphere and `F = {GMm}/r^2` was greater for the points on that side of the shell, there are a lot more points behind you. Even though their individual pulls on you are weaker, their vector sum adds up to exactly counteract the sum of the pulls in front of you.

To be a little more rigorous, suppose a point mass is located some distance away from the center of the shell. Refer to the diagram below.

This is a 2D cross-section of a 3D shell, slicing through the point mass. The mass per unit area of the shell is a constant given by `sigma = frac{M}{4piR^2}`.

The `color(red)("red")` and `color(blue)("blue")` areas are on opposite sides of the point mass, with both subtending the same solid angle, `d phi`

In the limit of `d phi -> 0`, we have

`color(blue)("d" A_1) = color(blue)(r_1)^2 d phi`
`color(red)("d" A_2) = color(red)(r_2)^2 d phi`

Notice that the area (proportional to the mass) is proportional to the square of the distance and the gravitational force follows the inverse-square law.

It should be pretty obvious that the 2 gravitational forces cancel each other perfectly. You just need to proof

`abs("d" F_1) = frac{G(sigma color(blue)("d"A_1))m}{color(blue)(r_1)^2} = frac{G(sigma color(red)("d"A_2))m}{color(red)(r_2)^2} = abs("d" F_2)`

This result is true for all choices of `color(blue)("d" A_1)` and `color(red)("d" A_2)`. Hence, the gravitational force is equal to zero anywhere within the sphere .