CONTRARIANTHINKER

Shell Theorem Questioned

By dB

The 'Clangers'

Sir Isaac Newton's shell theorem looks suspect...

A solid shell

Going through the equations - my comments in italics

Consider a spherically symmetric shell of finite thickness, with inner radius $R a$ and outer radius $R b$ ..... what is the force felt by an observer somewhere within the shell (i.e. $R a < r < R b$ )?

Again, this thick shell body may be considered as many concentric thin shells. Writing the force from one shell as $dF R$ , the total force from the body is:

$F_r = \int_{R_a}^{r} dF_R \;dR + \int_{r}^{R_b} dF_R \;dR$

Since all of the shells with $R > r$ have no effect on the observer*, the second term drops out:

*Do we know this to be true??? What if the shell is self-supporting???

$F_r = \frac{Gm}{r^2} \int_{R_a}^{r} dM_R\;dR$

The mass of a shell with radius $R$ is given by:

$dM_R = 4\pi R^2 \rho(R) \;dR$

Therefore,

$F_r = \frac{4\pi Gm}{r^2} \int_{R_a}^{r} R^2\rho(R) \;dR$

If the density is constant throughout the body, $ρ(R) = ρ$ and

$F_r = \frac{4\pi\rho Gm}{r^2} \int_{R_a}^{r} R^2 \;dR$

$F_r = \frac{4\pi\rho Gm}{3r^2} \left(r^3 - {R_a}^3\right)$

In general, for constant $ρ$ :

$F_r = \begin{cases}\frac{4\pi\rho Gm}{3r^2} \left({R_b}^3 - {R_a}^3\right), & r > R_b \\ \frac{4\pi\rho Gm}{3r^2} \left(r^3 - {R_a}^3\right), & R_a < r \le R_b \\ 0, & r \le R_a\end{cases}$

'Fr = 0??? Right under a massive shell? Imagine a shell of almost infinitely large diameter. To all intents and purposes a flat wall. Fr is going to be zero on one side of that wall?' (or words to that effect) was my sceptical reaction to the solution for r<Ra.

A spherically symmetric body can be considered as an infinite number of concentric, infinitesimally thin spherical shells. Consider one such shell:

The force due to the shaded band is

$dF_r = \frac{Gm \;dM}{s^2} \cos\phi$

>>No it's not. S should be drawn from the intersection of the orthogonal centre of gravity plane with the surface of the d(theta) segment. Nitpicking I know.....

The surface density of the entire shell is

$\sigma = \frac{M}{4\pi R^2}$

Yup

and the area of the band is

$dA = 2\pi R^2\sin\theta \;d\theta$

I guess so, polar maths a bit rusty

making the mass of the band

$dM = \sigma \;dA = {\textstyle\frac{1}{2}} M\sin\theta \;d\theta$

>>>simple substitution

The force can then be written

$dF_r = \frac{GMm}{2s^2} \cos\phi \sin\theta \;d\theta$

By the law of cosines,

$\cos\phi = \frac{r^2 + s^2 - R^2}{2rs}$

$\cos\theta = \frac{r^2 + R^2 - s^2}{2rR}$

$\sin\theta \;d\theta = \frac{s}{rR} ds$

thus

$dF_r = \frac{GMm}{4r^2 R} \frac{r^2 + s^2 - R^2}{s^2} ds$

To get the total force, we integrate over $s$ as the shaded band sweeps from the point on the sphere closest to $m$ to the farthest, assuming $r > R$ :

>>>ah! That answers my earlier objection

$F_r = \frac{GMm}{4r^2 R} \int_{r-R}^{r+R} \frac{r^2 + s^2 - R^2}{s^2} ds = \frac{GMm}{r^2}$
So far so good

An interesting result occurs when we consider the case in which $r < R$ , i.e. the observer is within the shell. The lower constant of integration is reversed in this case, giving:

$F_r = \frac{GMm}{4r^2 R} \int_{R-r}^{R+r} \frac{r^2 + s^2 - R^2}{s^2} ds = 0$

Suddenly it is not making sense to me

Therefore, the shell exerts no net force on particles anywhere within its volume. In general, we write:

$F_r = \begin{cases}\frac{GMm}{r^2}, & r > R \\ 0, & r < R\end{cases}$

OK. Here's my problem. Fr = real above the surface, increasing as the surface is reached, then rapidly dropping during transit of the shell, then right after the shell has been passed becoming zero.. Consider a dense hollow vanishingly thin shell and a vanishingly small dense object. This theory predicts a shocking and IMO fantasy discontinuity in the mutual force of attraction between the shell and the object as soon as the object crosses the membrane of the shell. What if the shell is made of depleted uranium wire mesh, and an ununpentium pellet passes through a hole. At what point does it suddenly and instantaneously lose it's accelerating attraction to the shell? How does the referee decide when a goal has been scored? I can see the pellet behaving like a bored child in a department store left standing within the orbit of an automatic door, finding out at what point the sensor triggers opening and relieving his boredom with small un-noticed movements...

(voice of experience....)

Now you feel it, now you don't.

The idea of a force switching off as a shell is crossed is more suitable to quantum mechanical calculations than classical physics. Even although the shell theorem seems to work from the outside looking in, it cannot reflect the real view from the inside looking out. Since we have never actually 'been there' we can but speculate of course.

More likely, the mutual attraction between the shell and the object persists, but becomes progressively weaker as the object travels towards the centre. At the centre the object has no net force upon it from gravity. Anywhere off-centre it will be attracted back to the underside of the shell where it will adhere.

When the planet condensed from stellar material it did so because atomic nuclei have a weak affinity for each other that is vanishingly weak, obeys the inverse square law, and is proportional to mass. How this action-at-a-distance happens is still subject for debate. The condensation seemingly contradicted the law of entropy, since matter became more organised, and the pressure created form gratitational attraction generated huge amounts of heat. Hence the Earth was a gravity-fractionated molten ball of elements and simple stable compounds (e.g. oxides). The ball cooled, outside first and perhaps quite quickly - see evidence of early water from zircons.

'the density function $ρ(R)$ is generally not constant, but tends to be inversely related to $R$ . This can cause some counterintuitive behaviors. For example, one might expect gravity to decrease when descending into a deep mineshaft on Earth. However, since density increases with depth, the gravity initially increases slightly.'

>>>Or maybe gravity increases until the gravitational mean of the shell is passed???

Here are a couple of thought problems that I think prove that it is VERY UNLIKELY Newton's Shell Theorem is valid

1) Gravity Well Function of Ultra Thin Shell according to Shell Theorem

This kind of discontinuity does not jive with the gradualism of gravity fields.

2) The Hemispherical 'Object'...

Newton's Shell Theorem states that any object inside a hollow sphere experiences zero gravity. In this illustration every point of the hemisphere floating inside the shell is closer to the left half than the right. F(g) varies as inverse square of distance. What, pray, physical mechanism is going to persuade the hemisphere to move one iota to the right???

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