Shell theorem

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In classical mechanics, the shell theorem gives gravitational simplifications which can be applied to objects inside or outside a spherically symmetrical body. This theorem has particular application to astronomy.

Isaac Newton used Shell Theorem to show that:

  1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center.
  2. If the body is a spherically symmetric shell (i.e. a hollow ball), no gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.
  3. Inside a solid sphere of constant density the gravitational force varies linearly with distance from the center, becoming zero at the center of mass.

These results were important to Newton's analysis of planetary motion; they are not immediately obvious, but they can be proven with Calculus. (Alternatively, Gauss's law for gravity offers a much simpler way to prove the same results.)

Contents

[edit] Outside the Shell

[edit] General Concepts

A solid, spherically symmetric body can be modeled as an infinite number of concentric, infinitesimally thin spherical shells. If one of these shells can be treated as a point mass, then a system of shells (i.e. the sphere) can also be treated as a point mass. Consider one such shell:

Note: dθ in the diagram refers to the small angle, not the arclength. The arclength is Rdθ.

Applying Newton's Universal Law of Gravitation, the magnitude of the force due to the shaded band is

dF = \frac{Gm \;dM}{s^2}

However, since there is partial cancellation due to the vector nature of the force, the leftover component (in the direction pointing toward m) is given by

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

The total force on m, then, is simply the sum of the force exerted by all the bands. By shrinking the width of each band, and increasing the number of bands, the sum becomes an integral expression:

F_r = \int (dF_r)


Since G and m are constants, they may be taken out of the integral:


F_r = Gm \int \frac{dM \cos\phi} {s^2}


This expression is not easily integrated, because for each thin band that makes up the shell, all three quantities dM, cosφ, and s vary (see animation below). To easily evaluate this integral, a change of variables can be made, using the Law of Cosines and density and surface area considerations.

[edit] Surface Area and Density

If the shell has mass M and radius R, the surface density of the entire shell is

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

The infinitesimal area of the band, dA, is its circumference (2πRsinθ) multiplied by its width (Rdθ):

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

So the mass of the band is the density of the shell multiplied by the area of the band:

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

The force can then be written

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

Now, dM has been replaced by a second quantity that also varies from band to band, θ. This term, as well as the cosφ term, can be rewritten using the law of cosines.

[edit] Applying the Law of Cosines

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}

Performing an implicit differentiation of the previous expression yields

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

[edit] Final Form

By substituting the expressions derived above into the original expression for the force, we obtain

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 (i.e. as θ goes from 0 to π). See the animation below.

The integration limits can be explained by the fact that the rightmost band is at a distance (r - R) from the point mass m, and the leftmost band is a distance (r + R) from the point mass.

Assuming r > R:

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


Evaluating the integral expression yields simply 4R, so the force reduces to:

F_r = \frac{GMm}{r^2}

Therefore, a thin shell can be treated as a point mass, provided the second object is outside the shell.

[edit] Inside a Shell

An interesting result occurs when we consider the case in which r < R, i.e. the point mass m is within the shell. By using symmetry arguments, it is easy to see that if the point mass is at the exact center of the shell the force will be zero, but is this true for all locations 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

Therefore, the shell exerts no net force on particles anywhere within its volume. Reasoning intuitively using inverse-square law, the few pieces of the shell that are close to m exert a large force, but there aren't many of them. By contrast, there are many pieces of the shell far from m but their force contribution is smaller.

In general, we write:

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

[edit] Thick shells

Now consider a spherically symmetric shell of finite thickness, with inner radius Ra and outer radius Rb. The behavior entirely inside or outside the shell is no different than for a thin shell, but what is the force felt by an observer somewhere within the shell (i.e. Ra < r < Rb)?

Again, this thick shell body may be considered as many concentric thin shells. The force contribution from each thin shell is:

dF_r = \frac{G m}{r^2} dM_R

Again, by applying density considerations (this time for a volume rather than a surface area), the mass of a thin shell with radius R and thickness dR is:

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

Therefore,

F_r = \frac{4\pi Gm}{r^2} \left[\int_{R_a}^{r} R^2\rho(R) \;dR + \int_{r}^{R_b} R^2\rho(R) \;dR\right]

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

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}


The factors \frac{4\pi\rho}{3}({R_b}^3 - {R_a}^3) and \frac{4\pi\rho}{3}(r^3 - {R_a}^3) are simply the mass M of each thick shell. Thus the first two cases reduce to Newton's law of universal gravitation.

[edit] Solid spheres

A solid sphere may be treated as a special case of a thick shell where Ra = 0:

Therefore, for r < Rb:

F_r = \frac{4\pi Gm}{r^2} \int_{0}^{r} R^2\rho(R) \;dR

And for constant ρ (renaming Rb to R):

F_r = \begin{cases}\frac{4\pi\rho GmR^3}{3r^2}, & r > R \\ \frac{4\pi\rho Gmr}{3}, & 0 < r \le R \\ 0, & r = 0\end{cases}

Many sufficiently large celestial bodies are a good approximation of spherically symmetrical solids. However, 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. This effect would be even more pronounced on a gas giant planet such as Jupiter.

[edit] See also

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