Comments on: Flat Earth Calculus: Deriving the Thickness of Flat Earth Using Gauss’s Law for Gravity https://journalofgeocentriccosmology.org/2024/07/24/flat-earth-calculus-deriving-the-thickness-of-flat-earth-using-gausss-law-for-gravity/ Wed, 02 Jul 2025 02:02:27 +0000 hourly 1 https://wordpress.org/?v=6.8.2 By: Dave Thomas https://journalofgeocentriccosmology.org/2024/07/24/flat-earth-calculus-deriving-the-thickness-of-flat-earth-using-gausss-law-for-gravity/#comment-173 Fri, 02 Aug 2024 01:25:45 +0000 https://journalofgeocentriccosmology.org/?p=17114#comment-173 Why are you using the “globe” value for the mass of the earth? That is based on, for example, applying Kepler’s 3rd Law to the earth and one of its satellites (the Moon, the ISS, etc.), along with the universal gravitational constant G. Knowledge of average orbit radius and period, and G, provides mass of the Earth.
I have a derivation of thickness of the infinite flat slab that depends only upon density of soil, thickness of the slab, and G. You just posit a cylindrical pillbox of arbitrary length L, and equal top/bottom areas A. L must be large enough to enclose a cross-sectional volume of the slab, from tip to bottom. Using Gauss’s theorem, the surface integral of field over area = 2gA, where g is the field intensity, and A is the area of the top or bottom of the pillbox. A is arbitrary; because the slab is infinite, no horizontal fields exist, and no part of the surface integral is affected by the sides of the pillbox (which run parallel to the field). By Gauss’s Law, this surface integral must equal a volume integral, which comes to 4 pi G times the mass (M) enclosed within the pillbox. That mass, M, = A*H*rho, where rho is the density of the slab (ie. soil/rock), A is the area of the “caps” of the pillbox, and H is the thickness of the slab itself. I used 1250 kg/m^3 for rho.

Thus, 2gA = 4 pi G M = 4 pi G A H rho, or g = 2 pi G H rho.

Solving for H gives H = g/(2 pi G rho) = 9.8/(2*3.14159…*6.67E-11*1250) = 18,707,000m = 18,707 km, quite close to the value you derived.

Don’t you think this is a simpler approach? No need for your “radius” that describes a fixed height, and is not related to any circle in your development? No need to invoke the heliocentric estimate of Earth’s mass? And so on.

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