Gauss's Law for Gravitation

Introduction

Gauss's Law for gravitation relates the gravitational flux through a closed surface to the mass enclosed. It is the gravitational analog of Gauss's Law in electrostatics. This law provides a powerful method for calculating gravitational fields in situations with high symmetry: spherical, cylindrical, or planar.

Statement of Gauss's Law

The gravitational flux through any closed surface equals -4πG times the total mass enclosed: ∮g·dA = -4πGM_enclosed. The negative sign indicates that gravity is attractive (field points inward toward mass). Flux is defined as ∫g·dA where dA is outward-pointing area element. For positive enclosed mass, flux is negative (net field lines enter surface).

Derivation and Physical Meaning

For a point mass M, field is g = -(GM/r²)\(\hat{r}\). On spherical surface of radius r: g·dA = -(GM/r²)(\(\hat{r}\)·dA) = -(GM/r²)dA since \(\hat{r}\) and dA are parallel. Integrating: ∮g·dA = -(GM/r²)(4πr²) = -4πGM. By superposition, this extends to arbitrary mass distributions. Physical meaning: enclosed mass is source of net flux; external masses contribute zero net flux.

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Introduction

Gauss's Law for gravitation relates the gravitational flux through a closed surface to the mass enclosed. It is the gravitational analog of Gauss's Law in electrostatics. This law provides a powerful method for calculating gravitational fields in situations with high symmetry: spherical, cylindrical, or planar.

Statement of Gauss's Law

The gravitational flux through any closed surface equals -4πG times the total mass enclosed: ∮g·dA = -4πGM_enclosed. The negative sign indicates that gravity is attractive (field points inward toward mass). Flux is defined as ∫g·dA where dA is outward-pointing area element. For positive enclosed mass, flux is negative (net field lines enter surface).

Derivation and Physical Meaning

For a point mass M, field is g = -(GM/r²)\(\hat{r}\). On spherical surface of radius r: g·dA = -(GM/r²)(\(\hat{r}\)·dA) = -(GM/r²)dA since \(\hat{r}\) and dA are parallel. Integrating: ∮g·dA = -(GM/r²)(4πr²) = -4πGM. By superposition, this extends to arbitrary mass distributions. Physical meaning: enclosed mass is source of net flux; external masses contribute zero net flux.

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Application: Spherical Shell

For uniform spherical shell of radius R and mass M: Outside (r > R): ∮g·dA = g(4πr²) = -4πGM, so g = -GM/r² (toward center) - same as point mass. Inside (r < R): No enclosed mass, so ∮g·dA = g(4πr²) = 0, giving g = 0. This confirms Shell Theorem 2: no force inside uniform spherical shell. Field is discontinuous at r = R.

Application: Uniform Solid Sphere

For uniform sphere of radius R and total mass M: Outside (r > R): g = GM/r² (point mass result). Inside (r < R): Enclosed mass M_enclosed = M(r³/R³). Gauss's law: g(4πr²) = -4πG[M(r³/R³)], giving g = -(GM/R³)r. Field increases linearly with r inside, then decreases as 1/r² outside. Maximum at surface (r = R).

Other Symmetric Cases

Infinite line mass (linear mass density λ): Cylindrical Gaussian surface gives g(2πrL) = -4πG(λL), so g = -2Gλ/r (radial, toward line). Infinite plane (surface density σ): Pillbox Gaussian surface gives 2gA = -4πG(σA), so g = -2πGσ (constant, toward plane). These results are much easier via Gauss's law than direct integration.

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Solved Example: Field of Line Mass

An infinite line mass has linear mass density λ = 0.01 kg/m. Find gravitational field at distance r = 5 m from line. Compare with field from point mass 0.5 kg at same distance. Solution: For line mass using Gauss's law: g = 2Gλ/r = 2 × 6.67×10⻹¹ × 0.01 / 5 = 2.67×10⻹³ N/kg. For point mass m = 0.5 kg at same distance: g_point = Gm/r² = 6.67×10⻹¹ × 0.5 / 25 = 1.33×10⻹² N/kg. The point mass produces 5× stronger field despite having 50× less total mass (in any finite length), because the line mass's field falls as 1/r while point mass falls as 1/r². At r = 0.1 m: g_line = 2Gλ/0.1 = 1.33×10⻹¹ N/kg, g_point = Gm/0.01 = 3.33×10^-9 N/kg. At close distances, point mass dominates; at large distances, infinite line would dominate (but infinite line is unphysical). This demonstrates importance of geometry in gravitational field calculations.

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