# Newton's Shell Theorem

Our physics teacher was teaching us about gravitation. He said that the force of gravity in a uniform solid sphere is due to the mass of the smaller sphere inside

(On whose surface it sits)

And that the effect of

Force due to the ring/shell outside is zero

And thus the formulas

$$gd = 4/3 × πρ × (R – d) G.$$

When we asked him why the force due to ring was zero,

He replied

It was because the shell exerts equal force on the object in each direction which get cancelled

I am a bit unconvinced by this explanation

For the ring/shell to exert equal force in all directions on the object [that would cancel each other], the object should be placed at the center of the shell which is also the center of the sphere

However the object is only at a depth, d less than the sphere's radius.

This implies inequal force from all directions that do not cancel each other rrsulting in different values

Where am I wrong? • Is this covered by physics.stackexchange.com/questions/158757/… Nov 21, 2021 at 22:16
• A ring isn't a shell, so the shell theorem does not apply. Nov 21, 2021 at 22:21
• I found Newton's proof enlightening as well.
– rob
Nov 22, 2021 at 3:00

Well, the easy answer is that if you mathematically work it out and do the integral, it's zero. The derivation is something readily available online and you can look it up. Instead, I'll focus on an intuitive explanation.

I'll remind you that you accurately stated that for all the forces to cancel themselves out, the object must be symmetrically located within the shell. That, in fact, is the case. Consider such a shell: The green axis is the $$x$$-axis, and the point $$A$$ is our point mass that lies within the shell on the $$x$$-axis.

Let's take a circular slice of our shell as follows: We can view this slice from the $$xz$$-plane as such (I simply rotated my axes such that the red $$y$$-axis is now sticking out of the page): Notice how the force cancels itself out, because the object is indeed at the geometric center of this circle.

Now, we can rotate our view again, and chop up our shell the same way for all points on the $$x$$-axis. So, we make a bunch of circles that are centered around some point on the $$x$$-axis as shown: Rotating our view again, we stare at it again from the $$xz$$ plane.

In the following image, I drew the force vector in purple generated by the mass due to each circle, and the green vector is the $$z$$-component of that force. We realize that the $$z$$-component of the force of gravity is zero, since the $$z$$-components cancel themselves out. So, by splitting our shell into a bunch of circles around the $$x$$-axis, we were able to show that the symmetry makes the net force in the $$z$$-direction zero.

We do the same for the $$x$$- and $$y$$-forces by splitting our shell into circles along the $$z$$- and $$y$$-axes as follows: If you trust gauss law for gravitation which relates the closed flux integral of the gravitation field g to the mass enclosed by my surface

$$\int g \cdot da = -4\pi GM$$ where M is mass enclosed

For a shell of radius R centre 0,0,0 , of some surface mass density

If I choose my gaussian surface to be a sphere of radius r centre 0,0,0

Where r<R

then the mass enclosed by my gaussian surface is zero

Hence, $$\int g \cdot da = 0$$

due to symetry of my shell , g and da are parralell everywhere on my surface

$$\int g |da| = 0$$

Likewise due to symmetry g has the same magnitude along my gaussian surface so is independent of the integral

g $$\int |da| = 0$$

so g = 0

Which also must mean that the force experienced by an object inside my shell is also zero

Thus it becomes