Shell theorem

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 appearing 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

d F = G m d M 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

d F r = G m d M s 2 cos ⁡ ϕ .

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 = ∫ d F r

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

F r = G m ∫ d M cos ⁡ ϕ s 2 .

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

The total surface of a spherical shell is

4 π R 2

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

2 π R 2 sin ⁡ θ d θ

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

d M = 2 π R 2 sin ⁡ θ 4 π R 2 M d θ = 1 2 M sin ⁡ θ d θ

and

F r = G M m 2 ∫ sin ⁡ θ cos ⁡ ϕ s 2 d θ

By the law of cosines,

cos ⁡ ϕ = r 2 + s 2 − R 2 2 r s cos ⁡ θ = r 2 + R 2 − s 2 2 r R .

These two relations link the three 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:

Note: As viewed from m, the shaded blue band appears as a thin annulus whose inner and outer diameters converge to R sin θ as dθ vanishes.

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 ⁡ θ d θ = s r R d s .

and one gets that

F r = G M m 2 r R ∫ cos ⁡ ϕ s d s

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 = G M m 4 r 2 R ∫ ( 1 + r 2 − R 2 s 2 ) d s .

A primitive function to the integrand is

s − 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 = G M m 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 t o t a l = ∫ d F r = G m r 2 ∫ d M .

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

d M = 4 π x 2 d x 4 3 π R 3 M = 3 M x 2 d x R 3

Therefore, the total gravity is

F t o t a l = 3 G M m r 2 R 3 ∫ 0 R x 2 d x = G M m r 2

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

Another suggestion is that Isaac Newton's first statement can't be applied to bodies with non-homogeneous density like Earth. In that case mass M can't be extracted from integral, how it was done above!

Inside a shell

For a point inside the shell, the difference is that when θ is 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 into the primitive function

s − 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 ) = G M m 4 r 2 R ∫ R − r R + r ( 1 s p − 2 + r 2 − R 2 s p ) d s

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 ) = G M m 4 r 2 R ∫ r − R r + R ( 1 s p − 2 + r 2 − R 2 s p ) d s

Derivation using Gauss's law

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

∫ S g ⋅ d S = − 4 π G M

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

∫ S g ⋅ d S = ∫ S g ⋅ n ^ d S

is the surface integral of the gravitational field g over any closed surface inside which the total mass is M, the unit vector 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 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

g = g ( r ) 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, n ^ , is precisely the direction pointing away from the point of symmetry of the mass distribution.

One therefore has that

g = g ( r ) n ^

and

∫ S g ⋅ d S = g ( r ) ∫ S d S = g ( r ) 4 π r 2

as the area of the sphere is 4πr².

From Gauss's law it then follows that

g ( r ) 4 π r 2 = − 4 π G M

i.e. that

g ( r ) = − G M 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. The most general force as derived in is:

F = − G M m r 2 − Λ M m r 3

where G and Λ 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 holds, 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 d M = R 2 / 2 d θ sin 2 ⁡ θ π R 2 M = sin 2 ⁡ θ 2 π M d θ

so:

F r = G M m 2 π ∫ sin 2 ⁡ θ cos ⁡ ϕ s 2 d θ .

where M = π R 2 ρ

Doing all the intermediate calculations we get:

F ( r ) = G m ρ 8 r 3 ∫ R − r R + r ( r 2 + s 2 − R 2 ) 2 ( r 2 R 2 + r 2 s 2 + R 2 s 2 ) − s 4 − r 4 − R 4 s 2 d s

Note that ρ in this example is expressed in k g m 2 .