Inside a ring the pull points outward

The inside half of the shell theorem is the stranger half. A hollow spherical shell, of any thickness and any mass, exerts no force at all on anything inside it. A pebble released anywhere in the cavity stays where it was put: it does not drift toward the centre, and it does not drift toward the nearer wall. The proof is a pair of cones. Through any interior point draw a narrow double cone; it cuts the shell in two small patches, one near and one far. The far patch is further away, so each gram of it pulls more weakly, by the square of the distance. But the far patch is also larger, by the square of the distance, because a cone’s cross-section grows as the square of how far it has gone. The two squares cancel, every pair of patches balances, and the net force is zero.

That argument is exact, and it is also fragile in a way the usual statement of it hides. It works because two numbers happen to match: the power at which a patch’s area grows with distance, and the power at which gravity falls. Change either and the cancellation fails. Change the shape — keep the same mass, the same radius, but put it all in a thin ring rather than spread over a sphere — and the power of the area drops from two to one while gravity’s stays at two. The near side is then under-weighted by less than it is over-pulled, and it wins.

So inside a ring the pull points outward. A body at the exact centre feels nothing, by symmetry, and a body displaced from the centre is pulled further from it. The centre of a ring is an equilibrium, and it is an unstable one.

The wedge that does not balance #

The cone argument, redrawn for a ring, is a wedge. Through an interior point draw two straight lines a few degrees apart. They cut the ring in two short arcs, one on each side of the point. A chord of a circle meets the circle at the same angle at both of its ends — the two ends are mirror images across the chord’s perpendicular bisector — so no obliquity factor distinguishes the two arcs. Each arc’s length is simply its distance from the point times the wedge’s angle.

The arithmetic is short. The near arc holds mass proportional to its distance $d_1$ and pulls each gram with $1/d_1^2$, so its pull is proportional to $1/d_1$. The far arc’s pull is proportional to $1/d_2$. The ratio of the two pulls is $d_2/d_1$, which is greater than one whenever the point is off-centre, so every wedge through the point pulls toward its nearer end. Summed over all wedges, the unbalanced parts do not cancel; they add into a single outward force, toward the part of the ring the point is closest to.

Near the centre that force is proportional to the displacement, as it has to be for any smooth field that vanishes at a point. Expanding the ring’s potential for small offsets gives an outward force of exactly half of $GM r / a^3$ for a ring of mass $M$ and radius $a$, at a distance $r$ from the centre. The shell has the same expansion with a coefficient of zero, which is the whole of the difference between them.

The failure is confined to the plane. A body placed on the ring’s axis, above or below the centre, is pulled back toward the plane, because every element of the ring lies below or above it and their vertical pulls add. So the centre of a ring is a saddle — stable to displacement along the axis, unstable to displacement in the plane — rather than a hilltop. It has the same shape as the three collinear points of the restricted three-body problem, which a spacecraft can hold only by paying for it.

One force law for each shape #

The cancellation inside a shell is sometimes presented as a lucky property of gravity, and sometimes as a deep one. The ring makes it possible to say precisely which. Imagine a force that falls as one over distance to some power $p$, not necessarily two, and compute the net interior force for a ring and for a shell as $p$ varies.

Each shape is empty of force for exactly one law, and the law is set by the shape’s dimension. A shell is a two-dimensional surface, whose patches grow as the square of distance, and its interior is empty under an inverse square. A ring is a one-dimensional curve, whose arcs grow as the first power, and its interior is empty under an inverse first power — which is the force law gravity would have in a flat world of two dimensions, where field lines spreading from a mass cross a circle rather than a sphere and dilute only as the circumference grows.

That is Gauss’s law stated as geometry. In $n$ dimensions the field of a point mass must fall as $1/r^{\,n-1}$ for the flux through every surrounding surface to be the same, and a hollow $(n-1)$-dimensional shell is then empty inside, for the same reason the interior of the Earth has a field that depends only on the mass beneath. The shell theorem is not a property of gravity alone and not a property of spheres alone. It is a statement that a shape and a force law are matched. A ring in three-dimensional space is mismatched by exactly one power, and the near side wins by exactly one power of the distance ratio.

There is a second place where the exponent two is singled out for a reason that has nothing to do with cones, and the two are worth holding together. Among all power laws, only the inverse square and the linear spring produce orbits that close on themselves — and those are precisely the two force laws the shell theorem produces, one outside a sphere and one inside a uniform one. The point-like behaviour of a planet and the closure of its orbit are separate facts that happen to require the same exponent.

A planet inside a rigid ring #

Put a planet at the centre of a rigid ring, the ring much lighter than the planet, and nudge the planet slightly off-centre. The ring pulls the planet toward its nearer side, and by Newton’s third law the planet pulls the ring the other way, so the offset between them grows. Near the centre the relative acceleration is proportional to the offset, with a coefficient $G(M_{\rm planet} + M_{\rm ring})/2a^3$. For a light ring that is half the square of the orbital angular frequency $n$ at the ring’s radius. The offset grows as $e^{nt/\sqrt{2}}$.

Two features of that result are what give it its reach. The first is that the ring’s mass drops out. A heavier ring pushes harder, but a heavier ring is also harder for the planet to push back, and for a ring much lighter than its planet the relative acceleration depends only on the planet’s mass. The growth rate is a property of the orbit, $n/\sqrt{2}$, and not of the ring. The second is that the time to collision depends on the starting offset only logarithmically. A perfectly engineered ring, centred to a billionth of its radius, survives about five orbital periods; a carelessly assembled one survives one. No precision of construction buys an order of magnitude.

The number is fast. For a ring at the distance of Saturn’s bright B ring, a hundred thousand kilometres from the planet’s centre, the orbital period is about nine hours, the e-folding time about two hours, and a rigid ring displaced by one metre would reach the planet within two days. A solid Saturnian ring is not a structure that could have existed briefly and then broken. It could not have existed for a week.

Why Saturn’s rings are not solid #

That conclusion is older than any spacecraft. Laplace showed in the 1780s that a uniform rigid ring around Saturn would be unstable, and suggested that the rings were many narrow rings rather than one. In 1856 the University of Cambridge set the stability of Saturn’s rings as the subject of its Adams Prize, and James Clerk Maxwell’s winning essay settled it. He showed that a rigid ring could be stabilised only by loading one point of it with a concentrated mass about four and a half times the rest of the ring — a ring nobody had seen and nothing could make — and that a fluid ring would break into waves. What remained was a ring of separate particles, each on its own orbit.

A ring of particles escapes the instability for a reason the figure makes plain. The instability is a property of a structure that moves as a whole: displace a rigid ring and every part of it moves together, so the near side stays near. A ring of independent particles has no whole to displace. Each particle orbits the planet under the planet’s own field, at its own Keplerian speed, and the collective pull of the others is a small correction that the ring’s own waves and edges are made of. The particle picture made a prediction a solid ring cannot: the inner edge should orbit faster than the outer. In 1895 James Keeler measured the Doppler shift across the rings with a spectrograph and found exactly that, the inner edge faster, at speeds matching Kepler’s third law.

The same arithmetic undid a fictional ring a century later. A novel of 1970 described an engineered ring one astronomical unit in radius, spinning around a star to give its inner surface gravity. Readers pointed out that it was a rigid ring around a massive body and therefore unstable in its own plane, and the novel’s sequel equipped it with attitude jets. The e-folding time, 0.225 of a year, is about eighty days; the ring’s mass, whatever it is, does not enter. The fictional engineers had the same problem as Maxwell’s imaginary ring, on a clock slower by the ratio of a year to nine hours.

A complete shell escapes it differently. A rigid sphere around a star exerts no force on the star wherever it sits, so it is neither stable nor unstable but neutral: displaced, it stays displaced, and drifts at whatever velocity it was given until it touches. The ring’s instability is the price of the missing dimension, the shell’s indifference is the reward of having it, and neither shape is held in place by gravity.

A flat galaxy is a stack of rings #

A spiral galaxy’s stars and gas lie mostly in a thin disc, and a thin disc is a stack of concentric rings. Every ring outside a given radius pulls a star at that radius outward; every ring inside pulls it inward, more strongly than the same mass at the centre would. So the circular speed at radius $R$ depends on the mass outside $R$ as well as inside it, and the familiar formula $v^2 = GM(<R)/R$ — which is the shell theorem applied to a sphere — is not the disc’s rotation curve.

For the commonest case the correct answer has a closed form. A disc whose surface density falls off exponentially with radius, with scale length $R_d$, has a circular speed that Kenneth Freeman worked out in 1970 in terms of modified Bessel functions. It is drawn here beside what the same enclosed mass would give if it were arranged in spheres.

The 15 per cent is not an obscure correction. The peak of the disc’s curve, at 2.15 scale lengths, is where rotation curves are most often quoted, and it is the speed a maximum-disc decomposition fits to the inner rise. A disc is simply a more efficient arrangement of mass for making things go round in its own plane, because much of the mass is closer to a star in the plane than the centre is, just as the ring outside its own radius pulled a quarter harder than a point.

Turned around, the error is in the mass. A circular speed is measured, not a mass: the 21-centimetre line of hydrogen is Doppler-shifted by the gas’s orbital motion, and a line width stands in for that speed. The mass is then inferred as $v^2R/G$, and that inference is exact only for a sphere.

The error changes sign, which is the point worth keeping. There is no single correction factor for flatness. Inside a scale length the outer rings’ outward pull makes the formula under-weigh the disc; outside, the inner rings’ in-plane concentration makes it over-weigh. A rotation curve measured to a per cent tells the enclosed mass of a flat galaxy only to about a third unless the flatness is modelled, and every published disc–halo decomposition models it. What would go wrong is a quick estimate of a galaxy’s mass from one velocity at one radius, which reads the disc as a sphere and then attributes the discrepancy to something else.

What the pictures cannot show #

Every figure here uses an infinitely thin ring or disc, and nothing real is. A ring of finite thickness has a force that stays finite at its rim rather than diverging, and a disc with the thickness of the Milky Way’s, about a tenth of its scale length, rotates a few per cent more slowly at its peak than the thin disc drawn here. The direction of every effect survives thickening; only the singular rim goes away.

The rigid-ring integration assumes a ring that cannot bend, and no ring of astronomical size is that. A real engineered ring would flex, and its flexing would add modes of its own, most of them also unstable. The integration answers the question Laplace and Maxwell asked about a rigid body, and it is the right answer to that question; it says nothing about whether some cleverly jointed structure could be stabilised by feedback, which is the question the fictional engineers answered with jets.

The disc calculation leaves out the dark halo that dominates the outer parts of every real spiral galaxy. A halo is roughly spherical, and for it the enclosed-mass formula is nearly right, so the misreading drawn above applies to the disc’s share of a rotation curve and not to the whole of it. That is why it matters most in the inner few scale lengths, where the disc dominates, and least where the curve refuses to fall.

And the figures do not show the observation that motivates the whole subject. Nobody has measured the field inside a ring directly. What is measured is its consequences: Keeler’s Doppler shifts across Saturn’s rings, the 21-centimetre rotation curves of galaxies, and the photometry that gives a disc’s scale length. The force inside a ring is inferred from Newton’s law and a shape, as every force in this subject is.

The connection the ring makes #

The same theorem gives the Moon a point-like pull on the Earth, empties the interior of a hollow sphere, and says that a planet inside a ring falls into it; the only thing that changes between the three is the number of dimensions the mass is spread over. That turns a result usually presented as a convenient accident of gravity into an accounting identity — area against pull — that holds or fails according to whether the powers match.

It also explains why flattened systems are where the subject gets hard. Spheres are the one case where the field outside is a point’s and the field inside depends only on what lies beneath. Every other shape pulls with its whole distribution, and flat shapes, which are what rotation makes of everything from ring systems to spiral galaxies, are the furthest from spheres of all.

Still open: which flattenings keep the inside empty #

The ring loses the cancellation completely, but a sphere does not have to become a ring all at once. Newton himself found the intermediate case. A thin shell bounded by two similar, concentric ellipsoids — a homoeoid, thick at its equator and thin at its poles in just the proportion that keeps the two surfaces similar — also exerts no force anywhere inside it, however flattened it is. The cone argument survives, because a cone through an interior point cuts the two ellipsoids in segments of equal length at both ends. Outside, the homoeoid pulls nothing like a point.

That result is what makes the Earth’s own flattening and the flattened haloes of galaxies calculable at all, since any ellipsoidal body is a nest of homoeoids and a star inside one feels only the layers beneath it. How flattened the dark haloes of galaxies actually are, and therefore how far their interior fields depart from the spherical case every argument above has leaned on, is measured from the shapes of stellar streams and the tilts of polar rings, and the answers still disagree with each other by more than their stated errors.