# Flat Earth model Part 3: Gravity

The flat earth model is based on a highly fanciful model of gravity. In fact I honestly doubt that any advocate of the flat earth model understands gravity at all. So let’s begin with a simple question: What is gravity?

Isaac Newton conjectured that all bodies of matter in the universe exert an attractive force on all other bodies of matter. He summarized this idea in his universal law of gravitation, which says that the attractive force between two bodies of matter is proportional to the product of their masses divided by the square of the distance that separates their centers of mass:

The constant G in the above equation is the constant of proportionality; F is the force; M is the mass of one of the two bodies, m is the mass of the other; and R is the distance that separates them.

This conjecture has two key components. The first is that it applies to all bodies of matter equally– there isn’t one class of bodies that exert a gravitational force and other classes that don’t. This is different from the electric force, for example, in that some types of matter are electrically neutral.

The second key component is that Newton claimed that the gravitational force radiates in all directions uniformly– there isn’t a privileged direction. One often repeated objection to the spherical earth model is that people standing on the “bottom” of the earth would fall off. That’s a perfect example of the failure to understand that Newton’s universal law of gravitation does not identify a privileged direction. In Newton’s universe “down” is the direction that points to the nearest large body of matter. For persons standing on the “bottom” of a spherical earth “down” points toward the center of the earth, just as it does for persons standing on its “top”.

This example illustrates another fundamental principle of our universe, namely its relativity. Every observer’s perception of physical phenomena such as length, mass, time, and down is different from that of any other observer not in the same inertial frame.

But was Newton’s conjecture correct? Is there an attractive force that behaves as he described and can one measure it? In 1797 Henry Cavendish performed the first experiment to observe and measure the force of gravity operating on objects in a laboratory. His experimental design is described in many articles that are available on the Internet.

Cavendish used lead spheres in his testing, but he could just as readily have used spheres of stone, cement, iron, wood– anything. Newton called his law of gravitation universal because he believed it to apply to all material objects, whatever they are made of.

So yes, there is indeed an attractive force between all bodies of matter in the universe, just as Newton had conjectured. But does that force obey an inverse square law as Newton had hypothesized?

This aspect of Newton’s theory of gravitation is due to its second key element, the fact that it is omnidirectional. Think of a body of mass emitting lines of gravitational force. Now surround the body by a concentric sphere. The lines of force intercept the sphere with a density that represents the strength of the gravitational field. Now imagine the sphere increasing in size. As its radius increases its surface area increases in proportion to the square of the radius, and the density of the lines of force intercepting the sphere falls simply because the surface area of the sphere is increasing. The inverse square relationship of Newton’s law is simply a consequence of the fact that the gravitational force radiates in all directions.

In Cavendish’s experiment the attraction between the lead balls was exerted horizontally, not vertically. So we have hard physical evidence that the gravitational force doesn’t only pull vertically.

How would this attractive force work on a disk shaped earth? The standard answer one hears from the advocates of the flat earth model is that it would pull everything “down”. But is that claim really true?

Let’s consider the gravitational force exerted by a disk shaped earth. We’ll construct our disk in such a way that it has the same surface area as our spherical earth, and the same volume. A little mathematics shows that such a disk would have a radius of about 8,000 miles (rather than the spherical earth’s 4,000 miles) and a thickness of about 1500 miles.

Now imagine a building constructed at the very center of a disk shaped earth. For the sake of simplicity we will think of this building as consisting of a stack of stone or concrete blocks. Figure 2: A disk shaped earth with a building at its center

Now let’s consider the force acting on one cubic inch of the building due to the gravitational force of the earth. The key point we have to consider is the first part of Newton’s universal law of gravitation– namely that all bodies of matter in the universe exert an attractive force. So to account for the gravitational force exerted by a disk shaped earth we have to consider the force exerted by every cubic inch of the earth’s mass, and then we have to sum up all of the resulting forces.

To do that let’s think about a cubic inch of the tower, P, and a cubic inch of the earth directly below it, Q some distance below the surface. The direction of the force exerted by Q pulls directly down, as shown in the diagram below: Figure 3: The force exerted on a point P of the building by a point Q of the disk shaped earth

Now let’s consider a cubic inch of earth that is some distance horizontally from the vertical, as shown in the diagram below: Figure 4: The force exerted by a point R offset from the vertical

The diagram shows that the gravitational attraction between the points P and R can be resolved into two components: a vertical component and a horizontal component.

Because a disk is symmetrical around its central axis it is always possible to find another cubic inch of matter in the earth that is at the same depth below the surface, the same distance horizontally from the vertical, and 180 degrees opposite to point R, The diagram below shows the point S that satisfies this condition: Figure 5: Showing a point S with the same horizontal component as point R

The important fact about point S is that its horizontal component is of the same magnitude as that of point R, but it points in the opposite direction Therefore the horizontal components for points R and S will cancel out with the result that we only have to consider the vertical components. Hence the net force operating on the building will be directed vertically down into the planet, just as we would hope.

Now let’s consider another building close to the outer edge of the disk, with a point R offset from the vertical, as shown in the picture below: Figure 6: A building at the edge of the disk shaped earth

As before the diagram shows the gravitational force resolved into a vertical and a horizontal component. But in this case there is no mirror point with which to offset the horizontal force. So the horizontal component of the force exerted by point R would not be offset by that of any other point. And this would be true for every cubic inch of matter of the earth that is horizontally offset from the building.

The net result is that there will be an overwhelming horizontal force that will pull the building horizontally to the left, causing the building to collapse. As a result there would be only one truly safe location on which to build any building on this fantasy earth– at its very center. A building built anywhere else will be ripped down by the planet’s gravitational force.

Incidentally this would give us a very simple way of finding the edge of a disk shaped earth. All one would need is a plumb bob. The outer edge of the disk would be in the direction opposite to that toward which the plumb bob is pulled.

A spherical earth doesn’t have this problem for the simple reason that a sphere doesn’t have an edge. For any location on a spherical earth it is always possible to find a point of matter with which to counterbalance any point that is offset from the vertical, as shown in the diagram below.