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 showed that a spherically symmetric body affects other objects gravitationally as though all of its mass were concentrated at a point at its center. 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. Finally, inside a solid sphere the gravitational force varies linearly with distance from the center, becoming zero at the center of mass.

These results are not immediately obvious, but they can be proven with calculus.

1 Derivation
2 Thick shells
3 Solid spheres
4 See also

[edit] Derivation

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

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

The surface density of the entire shell is

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

and the area of the band is

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

making the mass of the band

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

The force can then be written

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

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 (i.e. as $θ$ goes from $0$ to $π$ ). Assuming $r > R$ :

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

An interesting result occurs when we consider the case in which $r < R$ , i.e. the test mass is within the shell:

The lower constant of integration is reversed in this case, giving:

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

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

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

[edit] Thick shells

Now consider a spherically symmetric shell of finite thickness, with inner radius $R a$ and outer radius $R b$ . 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. $R a < r < R b$ )?

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

$dF_r = \frac{4 G m}{r^2} dM_R$

The mass of a thin shell with radius $R$ and thickness $d R$ 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 $R a = 0$ :

Therefore, for $r < R b$ :

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

And for constant $ρ$ (renaming $R b$ 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.

Category: Gravity

Shell theorem

From Wikipedia, the free encyclopedia

Contents

[edit] Derivation

[edit] Thick shells

[edit] Solid spheres

[edit] See also

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