No, not the hole (beloved of conspiracy theorists everywhere) where our alien overlords keep their UFO’s, but rather the equally imaginary hole from Pole to Pole (beloved of physics teachers everywhere) where we can drop in objects and watch them travel to the opposite end of the hole and back again, as an example of bodies moving with simple harmonic motion. In reality it’s not so simple, and in future posts I will develop spreadsheet tools to allow us to analyse the motion of a mass dropped into this hole in more detail to see what they would really do, but for now we will look at why, with suitable simplifying assumptions, the mass will move with simple harmonic motion, and what it’s period of oscillation will be.

The simplifying assumptions are:

- The hole passes exactly along the axis of rotation of the Earth, and this line is moving through space with constant velocity (so we can ignore tidal effects)
- The Earth is a perfect sphere of uniform density
- The hole is a perfect vacuum

We want to show firstly that an object dropped into this hole will move with simple harmonic motion, and secondly to calculate how long it will take to pass through the hole, and back again to exactly the point from which it was dropped.

Simple harmonic motion is the motion of a body attached to an ideal spring, with no other forces acting, so that the force on the body is proportional to the extension or compression of the spring (according to Hooke’s Law) and is directed towards the point at which the spring has zero extension.

How does the gravitational force vary along the length of the hole through the middle of the Earth? This can be simply determined by dividing the sphere of the Earth into two concentric parts at any depth down the hole; an outer shell and an inner ball. It can be shown that the gravitational field inside a spherical shell is zero for any thickness of shell, so that the gravitational field at any point down the hole is equal to that due to the sphere below that level. If this sphere has a radius R the mass of the sphere is proportional to R^{3}, and gravitational force at the surface of the sphere is proportional to mass / R^{2}, the gravitational force at distance R from the centre is therefore proportional to R^{3}/ R^{2} = R, and the resulting motion will be simple harmonic.

To determine the period of oscillation of a body dropped down the hole, consider a second body moving at orbital velocity at the surface of the Earth, that is with a radial acceleration, a_{r}, equal to the acceleration due to gravity at the surface, g. As this body moves around the Earth, if the radial line to the body forms an angle θ with the axis of the hole (i.e. the polar axis) then:

- The component of its radial acceleration parallel to the polar axis will be gCosθ
- Its perpendicular distance from the equatorial plane will be RCosθ
- Its acceleration towards the equatorial plane is therefore proportional to its distance from this plane
- Since its initial acceleration was equal to that of an object dropped down the Polar hole, and its component of acceleration in the direction of the hole is proportional to its distance from the plane through the centre of the hole, it follows that the path of the orbiting body projected onto the Polar Axis follows exactly the same path as an object dropped down the hole

Radial acceleration of a body moving with velocity V in a circle of radius R = V^{2}/R

For orbital velocity, radial acceleration = g, therefore V = (gR)^{0.5}

Circumference of a sphere = 2πR, therefore orbital period = 2πR / (gR)^{0.5} = 2π(R/g)^{0.5} = Period of oscillation of a body through the Polar hole.

Taking g = 9.81 m/s^{2} and R = 6,371,000 m, the period of oscillation of the mass dropped down the hole through the middle of the Earth is therefore:

2π(6,371,000/9.81)^{0.5} = 5063 seconds, or 84 minutes and 23 seconds.

The maximum speed of the object falling down the hole will be when it reaches the centre, at which point it will have zero acceleration and will be travelling parallel to the orbiting body, and will therefore have a speed equal to orbital velocity, i.e. (9.81*6,371,000)^{0.5} = 7906 m/s.

I seem to recall working out (a long time ago – a homework problem?) that the period is the same for any two points on the surface. So, assuming the existence of a bunch of underground evacuated tubes everywhere, getting to the other side of the world would take as long as getting across town…

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John – yes that’s right. It’s a feature of simple harmonic motion that the period is the same regardless of the amplitude – that’s why a tuning fork or a plucked string emitts the same frequency note no matter how hard you hit (or pluck) it.

If you had a perfectly evacuated, perfectly straight and horizontal, perfectly frictionless tube with a weight inside on your desk, and the weight was realeased from one end, then it would slide backwards and forwards along the tube, with a period of 84 minutes and 23 seconds, for ever!

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