Thickness of an infinite Flat Earth plane

If the Flat Earth is infinitely large and has a thickness b and a homogeneous density ρ, all horizontal gravitational force components would cancel each other at every place on earth and only leave the sum of all vertical forces components. There would be no center of mass like on the globe.

But is the sum of the vertical force components of all volume elements of the infinitely large plane not infinite, assuming that Newton's law of universal gravitation is applied like on the Globe Earth?

It turns out: NO

If the earth were an infinite plane of a certain thickness b and homogeneous density ρ, you would get a homogeneous gravity field that acts perpendicular to the surface everywhere.

Such a Flat Earth would not collapse into a sphere, theoretically. But this plane would not be stable. The slightest irregularities in the density would disturb the equilibrium and crunch the earth, except if it is infinitely rigid. An infinite plane would also mean an infinite mass for the earth.

The formula for calculating g for an infinite plane turns out to be really simple. The calculations are shown below:

To achieve g = 9.806 m/s² gravitational acceleration, you can calculate how thick the Flat Earth has to be:

Note: In the equation (5) for the gravitational acceleration g for an infinite plane there is no term for the altitude h like on the Globe Earth! This implies that the gravitaional acceleration is the same at every altitude.

This is due to the homogeneous gravity field that such an infinte plane would create.

How can we explain the Gravitational Gradient?

How can we explain on an infinite flat earth that gravity has a gradient getting weaker as we gain altitude and is dependent on latitude, as we can measure in reality? Some propose that sun, moon and stars have a gravitational influence too.

But this does not work. If the gravitational influence is restricted to a finite region, no matter how big, we have a point gravitational field towards a center of mass in the sky which adds with the homogeneous field from the earth:

• Either the center of mass in the sky is far enough so that its gravity field is approximately homogeneous near the earth. But this does not lead to a gravity gradient that is decreasing with altitude.
• Or it is near enough to disturbe the homogeneity of earth's field. But then net gravity would point to a single source in the sky so net gravity would not act perpendicular on earth anymore.

If the heaven were infinite like the earth, it would produce a homogeneous field like the flat infinite earth and adding both fields together does not produce a gravitational gradient as we observe: homogeneous + homogeneous => homogeneous.

Comparison with Observation

So what do we observe? Is g constant at each altitude h from the surface of the earth or is it changing with 1/(R+h)²? Is it perpendicular to the water surface everywhere? Does gravity depend on latitude?

We know the gravitational acceleration g is changing with 1/(R+h)² from measuring it with a scale and a weight at different altitudes or by using a gravimeter. And it also depends on latitude because the earth rotates and is not a perfect sphere. And it acts perpendicular to the Geoid surface which defines mean sea level.

See Experiments with a Test Mass (Centrifugal and Gravitational Acceleration in an Aircraft)

So Gravity measurements falsify the hypothesis of an infinite Flat Earth plane. Observations of gravity on earth confirm a rotating ellipsoid.

Calculations

For the following calculations we divide the infinite Flat Earth plane into an infinite number of infinitely small volume elements of density ρ and calculate the gravitational acceleration from the sum of all volume elements for any location at an altitude of h from the surface.

I choose a cylindrical coordinate system with the origin at the observer, the horizontal radial distance x from a volume element pointing away from the observer at an azimuthal angle φ and the vertical down is the Z axis.

The distance r of any volume element from the observer is then:

The mass of any volume element in the cylindrical coordinate system is given by:

The magnitude of the gravitational acceleration from a certain volume element i in any direction φ_i is according to Newton:

Because of the symmetry around the Z axis all horizontal components g_x of the gravitational acceleration cancel. We have only to calculate the contributions of all vertical accelerations g_z of each volume element i:

The blue term is derived from the geometry. If we have a right angle triangle with height z and hypothenuse r, the z component of the acceleration is g_z = g · z/r.

From here on I only refer to the vertical z component of the accelerations, so I will not write the index z explicitly anymore.

Lets insert the equation for $\mathrm{d} m$ to get the vertical acceleration component of the i-th volume element:

To calculate the sum of all vertical accelerations, we have to integrate the accelerations of all volume elements. This gives 3 nested integrals: one for the angle from φ = 0 to 2π, one for the horizontal distance from x = 0 to infinity and one for the vertical distance from the observer from z = h to h + b:

Lets solve the integrals one by one, starting with the inner most for the angle φ from 0 to 2π. Because the terms x and z do not depend on φ we can take them out of the integral, which simply results in 2π:

Inserting this result into (11) gives:

Now lets solve the next inner integral for x from 0 to infinity:

This is an amazing simple result. Inserting it into (13) leaves us with the following simple integral:

Now lets solve the last integral for z = h to h + b:

Inserting this result into (15) gives our final equation for the gravitational acceleration on an infinite Flat Earth plane: