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 stated that:

A corollary is that inside a solid sphere of constant density, the gravitational force within the object varies linearly with distance from the center, becoming zero by symmetry at the center of mass. This can be seen as follows: take a point within such a sphere, at a distance $r$ from the center of the sphere. Then you can ignore all of the shells of greater radius, according to the shell theorem (2). But the point can be considered to be external to the remaining sphere of radius r, and according to (1) all of the mass of this sphere can be considered to be concentrated at its centre. The remaining mass $m$ is proportional to $r^{3}$ (because it is based on volume). The gravitational force exerted on a body at radius r will be proportional to $m/r^{2}$ (the inverse square law), so the overall gravitational effect is proportional 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. (Gauss's law for gravity offers an alternative way to state the theorem.)

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.

Newton's proofs[edit]

Introduction[edit]

Propositions 70 and 71 consider the force acting on a particle from a hollow sphere with an infinitesimally thin surface, whose mass density is constant over the surface. The force on the particle from a small area of the surface of the sphere is proportional to the mass of the area and inversely as the square of its distance from the particle. The first proposition considers the case when the particle is inside the sphere, the second when it is outside. The use of infinitesimals and limiting processes in geometrical constructions are simple and elegant and avoid the need for any integrations. They well illustrate Newton's method of proving many of the propositions in the Principia.

His proof of Propositions 70 is trivial. In the following, it is considered in slightly greater detail than Newton provides.

The proof of Proposition 71 is more historically significant. It forms the first part of his proof that the gravitational force of a solid sphere acting on a particle outside it is inversely proportional to the square of its distance from the center of the sphere, provided the density at any point inside the sphere is a function only of its distance from the center of the sphere.

Although the following are completely faithful to Newton's proofs, very minor changes have been made to attempt to make them clearer.

Force on a point inside a hollow sphere[edit]

Fig. 2 is a cross-section of the hollow sphere through the center, S and an arbitrary point, P, inside the sphere. Through P draw two lines IL and HK such that the angle KPL is very small. JM is the line through P that bisects that angle. From the inscribed angle theorem, the triangles IPH and KPL are similar. The lines KH and IL are rotated about the axis JM to form two cones that intersect the sphere in two closed curves. In Fig. 1 the sphere is seen from a distance along the line PE and is assumed transparent so both curves can be seen.

The surface of the sphere that the cones intersect can be considered to be flat, and $\angle PJI=\angle PMK$ .

Since the intersection of a cone with a plane is an ellipse, in this case the intersections form two ellipses with major axes IH and KL, where ${\frac {IH}{KL}}={\frac {PJ}{PM}}$ .

By a similar argument, the minor axes are in the same ratio. This is clear if the sphere is viewed from above. Therefore the two ellipses are similar, so their areas are as the squares of their major axes. As the mass of any section of the surface is proportional to the area of that section, for the two elliptical areas the ratios of their masses $\propto {\frac {PJ^{2}}{PM^{2}}}$ .

Since the force of attraction on P in the direction JM from either of the elliptic areas, is direct as the mass of the area and inversely as the square of its distance from P, it is independent of the distance of P from the sphere. Hence, the forces on P from the two infinitesimal elliptical areas are equal and opposite and there is no net force in the direction JM.

As the position of P and the direction of JM are both arbitrary, it follows that any particle inside a hollow sphere experiences no net force from the mass of the sphere.

Note: Newton simply describes the arcs IH and KL as 'minimally small' and the areas traced out by the lines IL and HK can be any shape, not necessarily elliptic, but they will always be similar.

Force on a point outside a hollow sphere[edit]

Fig. 1 is a cross-section of the hollow sphere through the center, S with an arbitrary point, P, outside the sphere. PT is the tangent to the circle at T which passes through P. HI is a small arc on the surface such that PH is less than PT. Extend PI to intersect the sphere at L and draw SF to the point F that bisects IL. Extend PH to intersect the sphere at K and draw SE to the point E that bisects HK, and extend SF to intersect HK at D. Drop a perpendicular IQ on to the line PS joining P to the center S. Let the radius of the sphere be a and the distance PS be D.

Let arc IH be extended perpendicularly out of the plane of the diagram, by a small distance ζ. The area of the figure generated is $IH\cdot \zeta$ , and its mass is proportional to this product.

The force due to this mass on the particle at P $\propto {\frac {IH\cdot \zeta }{PI^{2}}}$ and is along the line PI.

The component of this force towards the center $\propto {\frac {IH\cdot PQ\cdot \zeta }{PI^{3}}}$ .

If now the arc HI is rotated completely about the line PS to form a ring of width HI and radius IQ, the length of the ring is 2 $π$ ·IQ and its area is 2 $π$ ·IQ·IH. The component of the force due to this ring on the particle at P in the direction PS becomes $\propto {\frac {IH\cdot IQ\cdot PQ}{PI^{3}}}$ .

The perpendicular components of the force directed towards PS cancel out since the mass in the ring is distributed symmetrically about PS. Therefore, the component in the direction PS is the total force on P due to the ring formed by rotating arc HI about PS.

From similar triangles: ${\frac {IQ}{PI}}={\frac {FS}{D}}$ ; ${\frac {PQ}{PI}}={\frac {PF}{D}}$ , and ${\frac {RI}{PI}}={\frac {DF}{PF}}$ .

If HI is sufficiently small that it can be taken as a straight line, $\angle SIH$ is a right angle, and $\angle RIH=\angle FIS$ , so that ${\frac {HI}{RI}}={\frac {a}{IF}}$ .

Hence the force on P due to the ring $\propto {\frac {IH\cdot IQ\cdot PQ}{PI^{3}}}={\frac {a\cdot DF\cdot FS\cdot PF}{IF\cdot PF\cdot D\cdot D}}={\frac {a\cdot DF\cdot FS}{IF\cdot D^{2}}}$ .

Assume now in Fig. 2 that another particle is outside the sphere at a point p, a different distance d from the center of the sphere, with corresponding points lettered in lower case. For easy comparison, the construction of P in Fig. 1 is also shown in Fig. 2. As before, ph is less than pt.

Generate a ring with width ih and radius iq by making angle $fiS=FIS$ and the slightly larger angle $dhS=DHS$ , so that the distance PS is subtended by the same angle at I as is pS at i. The same holds for H and h, respectively.

The total force on p due to this ring is

Clearly $fS=FS$ , $if=IF$ , and $eS=ES$ .

Newton claims that DF and df can be taken as equal in the limit as the angles DPF and dpf 'vanish together'. Note that angles DPF and dpf are not equal. Although DS and dS become equal in the limit, this does not imply that the ratio of DF to df becomes equal to unity, when DF and df both approach zero. In the finite case DF depends on D, and df on d, so they are not equal.

Since the ratio of DF to df in the limit is crucial, more detailed analysis is required. From the similar right triangles, ${\textstyle {\frac {DF}{PF}}={\frac {ED}{ES}}}$ and $ED^{2}=(DF+FS)^{2}-ES^{2}$ , giving ${\frac {\left(PF^{2}-ES^{2}\right)DF^{2}}{PF^{2}}}+2\cdot FS\cdot DF+FS^{2}-ES^{2}=0$ . Solving the quadratic for DF, in the limit as ES approaches FS, the smaller root, $DF=ES-FS$ . More simply, as DF approaches zero, in the limit the $DF^{2}$ term can be ignored: $2\cdot FS\cdot DF+FS^{2}-ES^{2}=0$ leading to the same result. Clearly df has the same limit, justifying Newton’s claim.

Comparing the force from the ring HI rotated about PS to the ring hi about pS, the ratio of these 2 forces equals ${\textstyle {\frac {d^{2}}{D^{2}}}}$ .

By dividing up the arcs AT and Bt into corresponding infinitesimal rings, it follows that the ratio of the force due to the arc AT rotated about PS to that of Bt rotated about pS is in the same ratio, and similarly, the ratio of the forces due to arc TB to that of tA both rotated are in the same ratio.

Therefore, the force on a particle any distance D from the center of the hollow sphere is inversely proportional to $D^{2}$ , which proves the proposition.

Shell theorem

Newton's proofs[edit]

Introduction[edit]

Force on a point inside a hollow sphere[edit]

Force on a point outside a hollow sphere[edit]

Chasles' theorem (gravitation)

Scale height