# Shell theorem

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

Isaac Newton proved the shell theorem[1] and said that:

1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre.
2. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

A corollary is that inside a solid sphere of constant density the gravitational force varies linearly with distance from the centre, becoming zero by symmetry at the centre of mass.

This is easy to see: take a point within such a sphere, at a distance $r$ from the centre of the sphere, then you can ignore all the shells of greater radius by the shell theorem. So, the remaining mass $m$ is proportional to $r^3$, and the gravitational force exerted on it is proportional to $m/r^2$, so to $r^3/r^2 =r$, so is linear in $r$.

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.)

In addition to gravity, the shell theorem can also be used to describe the electric field generated by a static spherically symmetric charge density, or similarly for any other phenomenon that follows an inverse square law. The derivations below focus on gravity, but the results can easily be generalized to the electrostatic force.

## Outside a shell

A solid, spherically symmetric body can be modelled 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 R dθ.

Applying Newton's Universal Law of Gravitation, the sum of the forces due to mass elements in 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:

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

To evaluate this integral one must first express dM as a function of

The total surface of a spherical shell is

$4\pi R^2 \,$

while the surface of the thin slice between θ and θ +  is

$2\pi R^2\sin\theta \,d\theta$

If the mass of the shell is M one therefore has that

$dM = \frac {2\pi R^2\sin\theta }{4\pi R^2} M\,d\theta = \textstyle\frac{1}{2} M\sin\theta \,d\theta$

and

$F_r = \frac{GMm}{2} \int \frac{\sin\theta \cos\phi} {s^2}\,d\theta$

By the law of cosines,

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

These two relations link the 3 parameters θ, s and φ that appear in the integral together. When θ increases from 0 to π radians φ varies from the initial value 0 to a maximal value to finally return to zero for θ = π.

s on the other hand increases from the initial value r − R to the final value r + R when θ increases from 0 to π radians.

This is illustrated in the following animation

To find a primitive function to the integrand one has to make s the independent integration variable instead of θ

Performing an implicit differentiation of the second of the "cosine law" expressions above yields

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

and one gets that

$F_r = \frac{GMm}{2rR} \int \frac{\cos\phi} {s}\,ds$

where the new integration variable s increases from r − R to r + R.

Inserting the expression for cos(φ) using the first of the "cosine law" expressions above one finally gets that

$F_r = \frac{GMm}{4r^2 R} \int \left( 1 + \frac{r^2 - R^2}{s^2} \right) \, ds.$

A primitive function to the integrand is

$s - \frac{r^2 - R^2}{s}$

and inserting the bounds r − R, r + R for the integration variable s in this primitive function one gets that

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

saying that the gravitational force is the same as that of a point mass in the centre of the shell with the same mass.

Finally, integrate all infinitesimally thin spherical shell with mass of dM, and we can obtain the total gravity contribution of a solid ball to the object outside the ball

$F_{total} = \int dF_r = \frac{Gm}{r^2} \int dM.$

Between the radius of x to x + dx, dM can be expressed as a function of x, i.e.,

$dM = \frac{4 \pi x^2 dx}{\frac{4}{3} \pi R^3} M = \frac{3Mx^2 dx}{R^3}$

Therefore, the total gravity is

$F_{total} = \frac{3GMm}{r^2 R^3} \int_{0}^{R} x^2 dx = \frac{GMm}{r^2}$

which suggests that the gravity of a solid spherical ball to outer object can be simplified as that of a point mass in the centre of the ball with the same mass.

## Inside a shell

For a point inside the shell the difference is that for θ equal to zero φ takes the value π radians and s the value R - r. When θ increases from 0 to π radians φ decreases from the initial value π radians to zero and s increases from the initial value R - r to the value R + r.

This can all be seen in the following figure

Inserting these bounds in the primitive function

$s - \frac{r^2 - R^2}{s}$

one gets that in this case

$F_r = 0 \,$

saying that the net gravitational forces acting on the point mass from the mass elements of the shell, outside the measurement point, cancel out.

Generalization: If $f=k / r^p$ the resultant force inside the shell is:

$F(r) = \frac{GMm}{4r^2 R} \int_{R-r}^{R+r} \left( \frac{1}{s^{p-2}} + \frac{r^2 - R^2}{s^p} \right) \, ds$

The above results into $F(r)$ being identically zero if and only if $p=2$

Outside the shell (i.e. r>R or r<-R) :

$F(r) = \frac{GMm}{4r^2 R} \int_{r-R}^{r+R} \left( \frac{1}{s^{p-2}} + \frac{r^2 - R^2}{s^p} \right) \, ds$

## Derivation using Gauss's law

The shell theorem is an immediate consequence of Gauss's law for gravity saying that

$\int_S {\mathbf g}\cdot \,d{\mathbf {S}} = -4 \pi GM$

where M is the mass of the part of the spherically symmetric mass distribution that is inside the sphere with radius r and

$\int_S {\mathbf g}\cdot \,d{\mathbf {S}} = \int_S {\mathbf g}\cdot {\mathbf{\hat n}}\,dS$

is the surface integral of the gravitational field g over any closed surface inside which the total mass is M, the unit vector $\mathbf{ \hat n}\,$ being the outward normal to the surface

The gravitational field of a spherically symmetric mass distribution like a mass point, a spherical shell or a homogenous sphere must also be spherically symmetric. If $\mathbf{ \hat n}$ is a unit vector in the direction from the point of symmetry to another point the gravitational field at this other point must therefore be

$\mathbf g = g(r)\mathbf{ \hat n }$

where g(r) only depends on the distance r to the point of symmetry

Selecting the closed surface as a sphere with radius r with center at the point of symmetry the outward normal to a point on the surface, $\mathbf{ \hat n }$, is precisely the direction pointing away from the point of symmetry of the mass distribution.

One therefore has that

$\mathbf {g} = g(r)\mathbf{ \hat n }$

and

$\int_S \mathbf g \cdot \,d{\mathbf {S}} = g(r) \int_S \,d{S} = g(r) 4\pi r^2$

as the area of the sphere is 4πr2.

From Gauss's law it then follows that

$g(r) 4\pi r^2 = -4 \pi GM \,$

i.e. that

$g(r) = -\frac {GM}{r^2}.$

## Converses and generalizations

It is natural to ask whether the converse of the shell theorem is true, namely whether the result of the theorem implies the law of universal gravitation, or if there is some more general force law for which the theorem holds. More specifically one may ask the question:

Suppose there is a force $F$ between masses M and m, separated by a distance r of the form $F = M m f(r)$ such that any spherically symmetric body affects external bodies as if its mass were concentrated at its centre. Then what form can the function $f$ take?

In fact, this allows exactly one more class of force than the (Newtonian) inverse square.[2] The most general force is:

$F = -\frac{G M m}{r^2} - \frac{\Lambda M m r}{3}$

where G and $\Lambda$ can be constants taking any value. The first term is the familiar law of universal gravitation; the second is an additional force, analogous to the cosmological constant term in general relativity.

If we further constrain the force by requiring that the second part of the theorem also hold, namely that there is no force inside a hollow ball, we exclude the possibility of the additional term, and the inverse square law is indeed the unique force law satisfying the theorem.

On the other hand, if we relax the conditions, and require only that the field everywhere outside a spherically symmetric body is the same as the field from some point mass at the centre (of any mass), we allow a new class of solutions given by the Yukawa potential, of which the inverse square law is a special case.

Another generalization can be made for a disc by observing that $dM=\frac{R^2/2 d\theta \sin^2{\theta}}{\pi R^2}M=\frac{ \sin^2{\theta}}{2 \pi}M d\theta$

so:

$F_r = \frac{GMm}{2 \pi} \int \frac{ \sin^2{\theta} \cos\phi} {s^2}d\theta.$

where $M=\pi R^2 \rho$

Doing all the intermediate calculations we get:

$F(r) = \frac{Gm \rho}{8r^3} \int_{R-r}^{R+r} { \frac{(r^2+s^2-R^2)\sqrt{2(r^2R^2+r^2s^2+R^2s^2)-s^4-r^4-R^4}}{s^2} } \, ds$

Note that $\rho$ in this example is expressed in $\frac{kg}{m^2}$.