A Circular Argument

This is a question oft asked by non-physicists:

"If a hole was drilled right through the Earth from one side to the other what would happen if a man jumped into it?"

Ok, we know that the Earth had a liquid core, but let's assume for the sake of argument, that it is solid all the way through, and there is hole right through it.

Incredibly, the answer lies in something I covered very recently, Simple Harmonic Motion.

Ignoring air resistance for now, the guy jumps in *(let's make the hole about 2.5 meters in diameter - the reason will become clear later on) and gravity accelerates him towards the centre, at the rate of "g", the acceleration due to gravity at the surface. The key to understanding this is to realise that the guy's acceleration depends on how much of the Earth is beneath him.

So as he falls, his acceleration gets less, until he reaches the centre. He overshoots the centre, and now starts to slow down, as gravity now acts against him.

Now what? Well, he keeps getting slower and slower, and because of the symmetry, emerges at the other end at exactly the same height at which he jumped in.

He stops for a microsecond, then falls back again, and the same thing happens, but in the other direction. presently, he emerges where he first jumped in. He falls back again, and it starts once more (remember we are ignoring air resistance).

he keeps going back and forth forever.

I guess you can figure out what happens if we include air resistance. he does the same, but each time he changes direction, he will do that at successively lower points inside the hole.

He goes back and forth, each time being shorter than the last, until finally air resistance stops him at the centre of the Earth.

He is suspended at the centre of the Earth, because all parts of the Earth attract him equally. Furthermore, he cannot get out by himself, as he he is too short to grasp both sides of the tunnel to give him traction! A ladder has to be lowered from either end of the hole for him to climb up (a long climb!)

Anyway, in the original scenario, the guy oscillates across the Earth's centre with SHM.

What is even more interesting is this - if you could sit in a stationary orbit (where you are above the same point on the Earth's surface all the time) and watch what happens, of course, with a telescope, you would see the guy disappear, then after some time he would emerge from the other end. NOW, if you hurled an object (again ignoring air resistance) perpendicularly directly away from you so that it entered into orbit, say , one meter above the ground, it would travel around the Earth in such a way (at constant speed, no acceleration) that when the guy emerged at the other end, it would hit him!

The travel time for the guy to travel through the earth at an acceleration, and then a deceleration, is exactly the same as an object (of the same mass) going around the Earth at constant speed.

How do we know all this? The simple equations of SHM tell us. The simplest way of describing it is to say the SHM exhibited by the guy inside the Earth is the "projection" (shadow) of the circular motion of the object in orbit outside the Earth.

Simple Harmonic Motion is the projection onto a plane of circular motion.