# Is the shell theorem only an approximation?

I've read the shell theorem during gravitation lectures, i.e. I know it states that the net gravitational field inside a 3D spherical shell or a uniform 2D ring is zero.

Now, assume a thin spherical shell. If I put a particle inside the shell, so that it was infinitesimally close to one of the regions of the shell, shouldn't the particle move towards the shell and touch the portion of the shell it was closest to? (Since as the distance goes to zero, the magnitude of the field between the particle and that portion of the shell should be very high, when compared to the field from other regions.)

But in the same case if I apply the shell theorem, the particle shouldn't move at all! Since it states the net gravitational field inside the shell is zero.

Can anybody explain this difference, or if there isn't any, how am I wrong?

• But how much mass is it infinitesimally close to... ? Jan 11 '15 at 13:54
• This comment makes me think..... Jan 11 '15 at 13:56
• Exactly - as you get closer to the surface of the sphere, the amount of mass that is within some factor of the closest distance to the surface shrinks (as the square of that distance). Jan 11 '15 at 14:00
• You'd have to get close enough to the sphere for local imperfections to matter. If there were no imperfections in the sphere, the force would remain zero. But I don't think a perfect sphere would be physically possible. Jan 11 '15 at 21:19
• Feb 12 '18 at 22:07

If you put a particle very close to the border, the force from matter very close to it will be very strong, as you say. But that is only a small portion of the shell; all the rest is pulling the other way, towards the center. The shell theorem guarantees that these forces cancel exactly.

• I don't think this answers the question. Deal with the case that the distance between the test mass and the sphere tends to zero. Jan 11 '15 at 14:06
• @Rob: The gravitational field has a discontinuity at the shell, but infinitesimally close to the shell is still inside of it, and so the force is zero as long as you're not on the shell. Jan 11 '15 at 14:24
• @Rob: I'm saying that what OP already knows is enough. The question asks what happens when you are very close to the sphere. The theorem says that however small the distance may be the force is zero, which implies that the limit is also zero. Jan 11 '15 at 15:40
• This does answer the question adequately, and in this case I think there's value in having a short, succinct statement of the answer. (Though in general I do prefer answers with proofs.) Jan 11 '15 at 20:15
• @RobJeffries: if answers have to prove the shell theorem, is the question in effect: "I have read the shell theorem in lectures but I don't believe it. What is its proof?" Jan 11 '15 at 23:18

Most of the mathematical formlities are dealt with on the wikipedia page you reference in your question. There, to prove the shell theorem, the shell is taken to have mass per unit area $\sigma$ and split it up into lots of coaxial rings, with the axis running along a diameter that goes through the test mass. $\theta$ is the angle between the diameter through the test mass and a line from the centre of the sphere to one of the rings.

Skipping to the main point, we can find the gravitational field generated by one of the thin rings at a test position that is a distance $r$ from the centre of a sphere of radius $R$ and mass $M$, where $s$ is the distance from the test position to a point on the ring is given by

$$dg = \frac{GM}{4Rr^2}\left( 1 - \frac{R^2-r^2}{s^2}\right)\ ds$$ which integrates to $$g = \left[\frac{GM}{4Rr^2}\left(s + \frac{R^2 - r^2}{s}\right)\right]^{s_{\rm max}}_{s_{\rm min}}.$$

For the full spherical shell the limits are $s_{\rm min} = R-r$ (when $\theta=0$) to $s_{\rm max}=R+r$ (when $\theta=\pi$ and the result is zero - this is the shell theorem and should work for any value of $r\leq R$.

However to answer your specific question, why doesn't the gravitational field due to the piece of the shell closest to the test point blow-up towards infinity as the test point gets very close to the surface of the shell and overwhelm the opposite (but clearly finite) field generated by the mass distributed over the rest of the sphere?

Look at the equation above and how it behaves when $r$ is very close to (but smaller than) $R$. In this case $s_{\rm min} \simeq 0$, and $(R^2-r^2)/s \simeq 2R$ resulting in a finite lower limit.

In words what is happening is that the amount of mass that is "infinitesimally close" to the test mass becomes infinitesimally small, ensuring that the gravitational effect of this mass does not blow up to infinity.

The shell theorem assumes a continuous distribution of matter in the shell.

If you came infinitesimally close to a real, physical shell you would discover that it, too, is made of particles. As you passed through the shell one of two things would happen:

1. You could crash into one of the particles and experience a non-gravitational force.

2. You could pass through a hole in the shell to the outside.

The second case is interesting. For a continuous shell there is a discontinuity between the force inside (zero) and the force outside (equivalent to a point mass at the shell's center). For a particle passing through a small hole in a spherical shell that discontinuity gets smoothed out. If the particle doesn't pass through the center of the hole, the smoothed-out gravitational force will include some component parallel to the surface of the shell; the details depend on exactly how "clumpy" the shell is.

• This is the answer I was coming to write. There is a reason the notion goes by the name "Theorem": it is mathematically exact assuming the preconditions are met. But finding a real system that meets the preconditions of the theorem may be difficult. Jan 12 '15 at 0:30

A few things ...

1) I do not think that the shell theorem applies to a 2-D uniform ring, as stated by the OP. If you go thru the math of the 3D sphere case, you then realize that there is an extra "sin" term in the 3D case. (Or put it another way, a sphere is NOT equal to a collection of 2D uniform rings.)

2) In the pure theoretical world where the 3D sphere is made of 100% uniformly and continuously distributed mass (instead of discrete particles), the discontinuity issue can be resolved as follows:

2A) Imagine that there is small hole on the sphere (with radius x). Suppose a particle travels in a straight line from inside of the sphere to the outside thru the center of the hole. If we plot the net gravitational force on the particle against the distance the particle travels, then it is a continuous function (obviously).

2B) Now consider this plotted function as we make x --> 0. The limit is a discontinuous step function, while for any x > 0, however small, the function itself is always continuous.

2C) Now what happens when there is no hole? (Or you are asking -- what happens, when I am AT the limit?) My understanding is that Newton's Law does not apply with distance between matter is exactly at zero. In the physical world, zero distance does not happen. Yet in the pure theoretical mathematical world, Newton's Law also asks you not to let that happen.

1. The shell theorem is derived mainly from Newton's laws of motion, in particular, the Newton's second law, which only apply to particles.

2. The condition that Newton's second law applies comes from the assumption that the the thickness of the shell is negligible compared to the radius of the shell so that the small part of the shell is treated as the particles and Newton's law applies.

3. Another important reason of the above assumption is that there exists a contradiction in applying math because integral calculus starts from small quantity and you can't treat the shell as "small" when the object particle is very close to the shell. This can be imagined as infinitely zooming in.

4. Shell theorem is only an ideal model in physics like any other laws, there's often some assumptions we may ignore and physics laws are often approximations to real world.

This explains that the shell theorem is not applicable when the particle object is very close to the shell. Since the shell cannot be treated as compositions of particles using calculus when the particle object is very close to the shell.

When I was deriving the theorem using calculus, I found it hard to know what the force will be on the surface of the shell and it turns out that they are underneath the same principle. This question is on Gravitational field on surface of spherical shell.

Much of my thoughts were from the book An Introduction to Mechanics (Kleppner, Kolenkow) (1973)