📅 Originally Published: May 16, 2016

Author: John Davis

🧠 Common Sense vs Mathematical Reality

Ask someone: “How much gravity would an infinite mass exert?” The gut response? Infinite gravity.

But here’s the twist — mathematics disagrees.

Using Gauss’s Law, a principle most often applied to electric fields, we can demonstrate that an infinite plane of mass can produce a finite gravitational field. This article walks through the logic and math behind that assertion. While some calculus is involved, the core insights are approachable with a bit of curiosity.

🔣 Introduction to Surface Integrals

Before diving into the math, let’s clarify a symbol many are unfamiliar with:

• ∫ – The standard integral, representing an infinite sum over a curve.

• ∮ – The closed surface integral, used to sum over an entire surface area, not just a line.

This integral is crucial to understanding Gauss’s Law.

🌌 Gauss’s Law (for Gravity)

While Gauss’s Law is often seen in electromagnetism, it also applies to gravitational fields due to the inverse square law both phenomena obey.

The gravitational version:

∮ g • n dA = -4πGm

Where:

• g is the gravitational field

• n is a unit vector normal (perpendicular) to the surface

• dA is an infinitesimal surface element

• G is the gravitational constant

• m is the mass enclosed by the surface

🧱 Applying It to an Infinite Plane

Instead of a sphere (used for point masses), we use a “pillbox” shape to evaluate the field above and below an infinite plane of mass.

✖️ Curved Surface Cancels Out

The sides of the pillbox cancel due to symmetry — for every surface point, there’s an opposite one balancing its pull.

This also explains why the plane doesn’t collapse into a sphere. It’s horizontally stable — each point is counteracted by all the surrounding mass.

🟢 The Key Result: A Finite g

After canceling the sides, we’re left with just the top and bottom caps of the pillbox. Here, g and n are directly opposed (parallel but opposite), so:

g • n = -g

Substituting into Gauss’s Law:

-g(2A) = -4πGm

Solving:

g = 2πG * (m / A)

If we express mass as:

m = density × area

We get:

g = 2πG * density

Which is a finite gravitational field, regardless of the plane’s infinite size.

📏 Estimating the Depth of the Flat Earth

Now, consider the plane also has depth, and let:

m = density × area × depth

Then:

g = 2πG × density × depth

Using known values:

• g = 9.81 m/s²

• G = 6.754 × 10⁻¹¹ m³/kg/s²

• density (ρ) = 5.51 g/cm³ = 5,510 kg/m³

We get:

depth = g / (2πGρ)
≈ 4,195.43 kilometers

This is far less than 9,000 km — correcting an earlier rough estimate.

🧾 Conclusion

This exercise shows that an infinite flat Earth can have a stable, finite gravitational field — as long as the math is done right. Using Gauss’s Law, we uncover how structure, symmetry, and surface integrals play a pivotal role in gravitational models — even ones that challenge mainstream views.

📚 Sources & Further Reading

• [Gauss’s Law for Gravity – D.G. Simpson, Ph.D. (2006)]

• [BBC: Do They Really Believe the Earth is Flat?]

• [Wolfram: Bougher Gravity]