# The gravity of a solid ball

In the simple Newtonian theory, one of the standard examples is a ball of some solid material with uniform density, ρ. The mass inwards from any radius r is (4.π.ρ/3).r3; this times G/r2 gives us the gravitational field strength at radius r as 4.π.G.ρ.r/3, pulling inwards, which we can integrate up to get a gravitational potential of K +2.π.G.ρ.r2/3 for some constant K. This holds true out to the radius R of the ball, for r < R; beyond which the gravitational field strength varies with r > R as (4.π.ρ/3).G.R3/r2, inwards, giving potential H −(4.π.ρ/3).G.R3/r for some constant H. The usual normalisation of potential tending to zero at large radius sets H = 0; at R we then need −(4.π.ρ/3).G.R3/R equal to K +2.π.G.ρ.R2/3, whence −K is 2.π.ρ.G.R2. Our body's mass is M = (4.π.ρ/3).R3, so we can write −K = 3.M.G/R/2 and our potential varies as

• for r > R: −M.G/r
• for r < R: M.G.((r/R)2 −3)/R/2

That was easy enough. So now let's explore what happens when we remove some of the over-simplifications there; the density being uniform (despite the weight of the mass further out pressing in on, so compressing, the mass nearer the middle, raising its density) and the use of Newton's simple theory rather than Einstein's general relativity. But first let's pause to prove some little details I quietly brushed under the carpet there: for a point inside the ball, why can we ignore the mass further from the ball's centre and treat all the mass further in as if it were at the centre ?

## Spherical shells

I'm going to answer that question by decomposing our ball into concentric shells about its centre and looking at the gravitational fields of these shells; when we add those up, we get the field of the ball. To work out the details, though, we're going to need some geometry.

The intersecting chords theorem is really just a proof that the two triangles fromed by two chords, when you connect up an end of one to an end of another on one side, and their opposite ends on the other side of the place where they cross, are similar. When the two chords are close to parallel to one another, and we join the end of each to the nearby end of the other, the length of the short connecting side between their ends on one side of the crossing is a good approximation to the length of the arc of the circle between those same ends; and the parts of the two chords on that side of their crossing are a good approximation to the typical distance of that arc from the crossing point. So we can read the intersecting chords theorem as saying that the length of the short arc, between ends of our nearly-parallel chords, on either side is proportional to how far that arc is from the crossing-point of the two nearly-parllel chords.

So now consider a point inside a sphere and a sharp cone whose apex is that point; let's look at where the two sides of that cone cut the sphere, in two little patches on opposite sides of our point. Each of those patches is roughly elliptical and its area shall be π/4 times the product of its longest and shortest diameters; each of which is an arc of a great circle; and each end of either arcs is connected by a ray along the cone's surface, via the apex, to a relevant end of the corresponding diameter of the other blob, so the lengths of corresponding diameters of the blobs are proportional to the blobs' distances from the apex. When we then multiply two diameters of each blob together, we find that the area in which the cone cuts the sphere, on either side of the apex, is proportional to the square of the distance of the relevant blob from the apex.

Since gravitational attraction drops off as the inverse square of the distance from a mass, if the masses of the two patches in which the cone cuts the sphere are proportional to their areas, then the gravitational attraction felt at the cone's apex from the parts of the sphere opposite each other will be opposite (because on opposite sides) and of equal magnitude (because proportional to mass divided by square of distance, with the mass being proportional to the square of distance). So the gravitational force felt at a point inside a shell of uniform density, due to that shell, is zero: the contribution from any small part of the shell is exactly cancelled by the contribution from the portion directly opposite; and when we integrate that up over the whole shell, we're left with nothing.

So now let's consider a point outside a spherical shell. The shell is, of course, symmetric under rotations about the line through this external point and the shell's centre, so the net force must be along that line – any component off the line from one part of the shell must be ballanced by an equal and opposite component from the part of the shell that the given one maps to under a half turn about this axis. Use that axis as z-axis for cylindrical polar co-ordinates, with radius r outwards from it, and take the point on the sphere closest to the external point as origin, with the external point at z = −h and the sphere's centre at z = r. At each z from 0 to 2.r there's a plane perpendicular to the z-axis that cuts the sphere sphere in a ring whose squared radius is r.r −power(2, r −z) = z.(2.r −z); the centre of that ring, where the plane cuts the axis, is h+z from the external point, so that each point on the ring has squared distance from the external point power(2, h +z) +z.(2.r −z) = h.h +2.z.(h +r). Mass in the spherical shell close to this ring then exterts its force at an angle to the axis; as a proportion of that force, its component parallel to the axis is h+z divided by the distance of the ring from the external point; so a mass close to the ring contributes a force proportional to (h +z) / power(3/2, h.h +2.(h +r).z) times that mass along the axis. Helpfully, by Archimedes's reasoning, the area of the sphere between two planes perpendicular to the axis is simply proportional to the distance between the planes (resulting from two factors – one due to its radius being less than r, the other due to the area being sloped – cancelling) so the total force is proportional to integral(: dz.(h +z)/power(3/2, h.h +2.(h +r).z) ←z; 0≤ z ≤ 2.r :). Substituting u.u = h.h +2.(h +r).z, which ranges from h.h at z = 0 to h.h +4.r.(h +r) = power(2, h +2.r) at z = h +2.r, with dz = u.du / (h +r) and h +z = h +(u.u −h.h)/(h+r)/2 = (h.(h +2.r) +u.u)/(h+r)/2, this becomes

integral(: dz.(h +z)/power(3/2, h.h +2.(h +r).z) ←z; 0≤ z ≤ 2.r :)
= integral(: u.du.(h.(h +2.r) +u.u)/power(3, u) ←u; h ≤ u ≤ h +2.r :) / (h +r) / (h+r) / 2
= integral(: du.(h.(h +2.r)/u/u +1) ←u; h ≤ u ≤ h +2.r :) / (h +r) / (h+r) / 2
= (h.integral(: −power(-1)'(u).du ←u; h ≤ u ≤ h +2.r :).(h +2.r) +2.r) / (h +r) / (h+r) / 2
= (h.(−1/(h +2.r) +1/h).(h +2.r) +2.r) / (h +r) / (h+r) / 2
= (−h +h +2.r +2.r) / (h +r) / (h+r) / 2
= 2.r / (h+r) / (h+r)

## Uniform material, varying density

So let's allow that the density of our material increases with pressure; at pressure P, the density is Q(P) for some function Q; Q(1 Atm) is the usually-quoted density of the material and reasonably close to Q(0 Pa) for most solids; we can expect Q(P) to increase as P increases. We'll still have spherical symmetry, now with the pressure P(r) and ρ(r) varying with radius r, linked by ρ(r) = Q(P(r)). We'll have a gravitational field strength F(r), pulling inwards, and a potential f(r) tending to zero at large radius, with F(r) = −f'(r) and F(r) varying with r in response to the mass inwards from r, controlled by ρ.

So consider a shell bounded by radii r and r+h, with h small and 0 ≤ r < r+h < R. The material in this shell has average density ρ(r) +h.ρ'(r +k) for some k between 0 and h; its volume is (4.π/3).((r +h)3 −r3) = 4.π.(h.r.r +h.h.r +h.h.h/3) and its mass is the result of multiplying this by the density. Then F(r+h) −F(r) is just G times this mass and divided by the square of some radius r+i for some i between 0 and h; but it's also h.F'(r +j) for some j between 0 and h; whence

• F'(r +j) = 4.π.G.(r.r +h.r +h.h/3).(ρ(r) +h.ρ'(r +k))/(r+i)/(r+i)

with i, j and k between 0 and h. As we consider values of h very close to zero, the terms in h, i, j and k become negligible with respect to those they are added to and this reduces to

• F'(r) = 4.π.G.ρ(r)

making F simply the integral of ρ times 4.π.G. Looking once more at such a shell, to see how pressure increases in response to weight, we have P(r+h) −P(r) = h.P'(r +i) for some i between 0 and i; and minus this difference is just the weight of the mass between r and r+h, divided by the surface area over which it's spread; that surface area, times h, is roughly the volume of this mass, with the approximation getting arbitrarily good for small enough h, so the weight over area is just the field strength times density times thickness, h, of the shell. This gives us

• −P'(r) = F(r).ρ(r)

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