Sunday, February 13, 2011

A friend returned from a fascinating visit to the SNO lab, the slow neutron observatory in Sudbury, Ontario, Canada. He described the long ride down into the depths of the mine on the rickety, open elevator, and the remarkable equipment he observed. He made one statement, though, that bears examining, because it's a popular misconception.

He said, "You could feel yourself getting heavier."

Because he was progressing toward the centre of the Earth, he expected to weigh more.

This is a fallacy, though. He would weigh less. In fact, although weight follows the familiar inverse-square law outside, weight is proportional to the radius inside. It drops linearly.

Here's why. (It's a standard university calculus question, but you don't need calculus to see.)

First, the quick, intuitive answer. How much would you weigh at the centre? Zero, of course, because the mass is pulling you in all directions, each molecule on one side being cancelled by one on the other.

Since you weigh more than zero at the surface, and zero at the middle, either it decreases as you go down, or increases for a while before decreasing. Where is the logical turnaround point? (The surface, of course.) Also, it doesn't make sense to say that the force of gravity keeps increasing all the way to the centre, then instantly becomes zero when you travel that last nanometre to the absolute centre.

OK, so now that we expect a decreasing graph of weight versus distance from centre (radius), why is it linear.

The answer is easiest to understand if you consider the force inside a shell. Pretend the Earth is a perfectly spherical egg. You're inside. How much do you weigh? Then answer: zero everywhere. Pretend you are a two-thirds down to the centre. The mass below your feet is twice as far away as the mass above your head, so pulls a quarter strength (inverse-square law). But there is four times as much of the shell below you than above. Four times as much pulling at one-quarter the force…cancels out. Same with side-to-side, of course.

(This is analogous to the electric field inside a charged sphere--zero everywhere.)

So, now you're down a mineshaft. All the shells above you are cancelled out by mass below you, farther out than your radius. What's left is only the ground beneath your feet, acting as if it's all at the centre. Mass goes as volume, radius cubed. So as you go down, the mass of the sphere beneath your feet is decreasing by radius cubed, but acting stronger on you by radius squared.

Result, the force of gravity is decreasing by the radius to the first power. You feel lighter down a mine shaft.

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