Abstract:
This paper explores the application of Gauss’s law for gravity to derive the radius of a finite section of an infinite flat plane, whose mass is equivalent to the Earth’s mass. The mathematical approach employs calculus and classical physics principles to address the problem, elucidating the conceptual use of infinite planes in gravitational field calculations.
Introduction:
The concept of an infinite flat plane is used in theoretical physics to simplify gravitational field calculations and understand the behavior of fields in highly symmetrical environments. Gauss’s law for gravity, analogous to Gauss’s law for electrostatics, provides a powerful tool for calculating the gravitational field of mass distributions exhibiting high symmetry.
Theoretical Background:
Gauss’s Law for Gravity:
The gravitational flux through a closed surface is proportional to the enclosed mass. Mathematically, it is represented as:
∫ g . dA = -4πG M_enc
where g is the gravitational field, dA is the infinitesimal area vector on the closed surface, G is the gravitational constant, and M_enc is the enclosed mass.
For an infinite plane with surface mass density σ, the gravitational field g is uniform and perpendicular to the plane. The symmetry of the problem allows simplifications when applying Gauss’s law, leading to a straightforward calculation of the gravitational field above the plane.
Methodology:
Calculation of Surface Mass Density (σ): To relate the gravitational field at a height above the infinite plane to the surface mass density, we use the formula derived from Gauss’s law: g = 2πGσ Rearranging gives: σ = g / (2πG) where g is the acceleration due to gravity (9.8 m/s^2) and G is the gravitational constant (6.674 x 10^-11 m^3/kg/s^2). Mass of a Circular Section of the Infinite Plane: Assuming a finite section of the plane with a radius R has a mass equivalent to Earth’s mass (5.972 x 10^24 kg), the mass M can be expressed as: M = σπR^2 Solving for R provides: R = sqrt(M / (σπ))
Calculation:
Using the values for g, G, and M (Earth’s mass), the calculations are as follows:
σ = 9.8 / (2π x 6.674 x 10^-11) ≈ 2.34 x 10^10 kg/m^2
R = sqrt(5.972 x 10^24 / (2.34 x 10^10 x π)) ≈ 9019 km
Results and Discussion:
The radius R ≈ 9019 km, derived from the calculations, suggests that if Earth’s mass were spread uniformly over an infinite plane with the calculated surface mass density, a circular section of this plane with a radius of approximately 9019 km would encompass Earth’s mass. This theoretical scenario is useful for understanding the gravitational effects of large, flat surfaces and provides an educational example of applying classical physics concepts to hypothetical geometries.
Conclusion:
This mathematical exploration demonstrates the utility of Gauss’s law for gravity in theoretical physics and offers insights into the gravitational fields produced by large-scale symmetrical mass distributions. The calculated radius serves as a conceptual tool rather than a suggestion of physical reality, emphasizing the importance of symmetry and calculus in gravitational studies.
This paper presents a methodical approach to applying Gauss’s law for gravity to an abstract problem, providing a foundation for further exploration in gravitational physics and theoretical modeling.
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.