### A Hole through the Earth

“I wonder if I shall fall right through the earth”, thought Alice as she fell down the rabbit hole, “and come out in the antipathies”. In addition to the author of the “Alice” books, Lewis Carroll – in real life the mathematician Charles L. Dodgson – many famous thinkers have asked what would happen if one fell down a hole right through the earth’s centre.

Galileo gave the answer to this question: an object dropped down a hole piercing the earth diametrically would fall with increasing speed until the centre, where it would be moving at about 8 km per second, after which it would slow down until reaching the other end, where it would fall back again, oscillating repeatedly between the two ends.

Of course, drastic simplifications are being made here, with no account taken of air resistance or the effects of the earth’s rotation, not to mention the engineering challenge of constructing the hole through molten magma under extreme pressures and temperatures. But let us proceed with this fantasy or, as Einstein might have called it, gedankenexperiment.

Simple Harmonic Motion

The force due to gravity on an object of mass m at the earth’s surface is

F = (GMm/a2) = mg

where M and a are the mass and radius of the earth and G is the universal gravitational constant (G = 6.67×10^-11 m^3kg^-1s^-2), and the acceleration due to gravity is g = (GM/a^2) = 9.8 m s^-2. When the object is beneath the surface, at a distance r from the centre, there are two contributions one from the sphere of radius r beneath the object and one from the spherical shell of thickness (a – r) outside it. For uniform density, the attraction of the inner sphere is as if all its mass were concentrated at the centre. The attraction of the outer shell vanishes; this follows from a simple symmetry argument showing that the attractions of opposite sides of the shell cancel. So, the force at radius r is

F(r) = (GM(r)m/r2) = (GM(a)m/r2)(r/a)3 = (mg/a) r

where M(r) is the mass of the inner sphere of radius r. The equation of motion for the object is now

d2r/dt2 + (g/a) r = 0

which is the equation for simple harmonic motion with frequency ω = √(g/a). The half-period, which is the time to travel through the tunnel, is

τ/2 = π √(a/g)

which comes to 42.2 minutes. This is a very surprising result. It also follows that the maximum speed, reached at the earth’s centre, is = √(ag) with a value of 7900 m/s or about 8 km per second. If you thought it was hot there, it’s just got hotter!

Suppose now that the tunnel runs straight between two points but not through the earth’s centre. The component of gravity along the route is now weaker, making for a longer trip-time. But the distance is less, making the time shorter. These two effects cancel exactly, giving us the amazing result that the time to traverse any straight tunnel through the earth is 42.2 minutes, regardless of the tunnel’s length.

Two curiosities about the motion of an object oscillating in a hole through the earth:

• the period depends only on the density of the planet, not on its radius or total mass.
• the period is precisely that of a satellite orbiting the earth at ground level.

Rapid Rail Transport

In another of Lewis Carroll’s books, Sylvie and Bruno Concluded, the author describes an ingenious method of running trains, based on the above ideas, with gravity as the driving force. The track would be laid in a straight tunnel between two cities. Trains would run downhill to the midpoint, which is closer than the ends to the earth’s centre, gaining enough momentum to carry them uphill to the other end. It is a simple but surprising mechanical result that the journey time is independent of the distance. Two trains leaving Dublin at midday, one bound for London and one for Sydney, would both arrive at 12:42, less than three-quarters of an hour later. Image from The Strand Magazine, 1909: Camille Flammarion: A Hole through the Earth.