MatheMUSEments

Articles for kids about math in everyday life, written by Ivars Peterson for Muse magazine.

June 7, 2007

Earth Hole

Forget about the heat and the pressure and the iron core. Imagine there's a hole that goes all the way through Earth, passing through its center. If you dropped a stone into this hole, what do you think would happen?

Normally when you drop a stone, it falls to the ground, pulled by Earth's gravity. For a stone falling through Earth, the situation is a little different. At the surface, it's attracted by all of Earth. As it travels down the hole, the stone is attracted by parts of Earth that are both above and below it. So, the force pulling the stone downward gets smaller as the stone travels toward Earth's center.

At the center of Earth, the stone is weightless. It's attracted equally in all directions, so it experiences no net force. Everything cancels out.

What does this changing force do to the stone's motion? Because it experiences a downward force all the way to Earth's center, the stone goes faster and faster even as this force gets smaller and smaller. It reaches its highest speed at Earth's center.

Because there's no force a acting on it there, the stone overshoots. As it travels away from the center, the force pulling the stone back toward the center gets stronger and stronger. The stone slows down more and more. When it finally reaches the surface on the other side of Earth, it comes to a stop for an instant. Then, unless someone snatches it out of the air, it begins to fall again, repeating its trip through Earth.

Galileo Galilei argued in 1632 that a cannonball dropped down such a hole would oscillate forever between the drop site on one side of Earth and the exit hole on the other side. Each round trip would take 84.6 minutes. It wouldn't matter whether you dropped a cannonball or a pebble: The trips would take the same amount of time.

If you take air resistance into account, however, a dropped stone wouldn't actually make it all the way to the other side of Earth. Instead, with each trip through the center, it would travel less and less far, eventually getting trapped in the middle. The weightless stone would then simply float at Earth's center.

But Earth also spins. So, as the stone falls, Earth moves around it. For this reason, Isaac Newton argued in 1679 that, unless it was dropped along Earth's spin axis (a line joining the north and south poles), a stone would follow a curved path. The closer to the equator the stone was dropped, the wider the curve would be. A stone dropped at a point on the equator would curve wide enough that it would hit the hole's wall—unless the hole was more than 300 kilometers wide.

What a way to go!


Muse, November/December 2004, p. 17.

2 comments:

guapo011 said...

What if one were to build a scale model of the earth. Lets say the scale model earth was the size of a water tower, or a large building. And lets say that we put a hole through this object. Obviously we would have to enclose it in a vacuum to remove the forces of the earth. I am assuming this phenomenon could be reproduced by man by passing a particle small enough through the center of the hole. My question is this. Would a magnetic field be generated or will the sphere develop a polar charge on each side? If so, would it theoretically be possible to harness this power?

Anonymous said...

Well this wouldn't work on earth, because our planet's gravity would overwhelm that of any model. If you did this out in space with a charged "stone", it would generate a magnetic field, because it's a moving charge.
Unfortunately you couldn't get any power out of it, at least not the perpetual motion kind I assume you're talking about. This is because any mechanism you used to extract energy from this thing, be it magnets, or a paddle wheel, would take energy away from the rock.
Before long you'd just have a rock gravitationally suspend in your sphere and you'd get no more energy out of it.