§ 13.1 — Newton's Law of Universal Gravitation

Fundamental Principle

Every object in the universe attracts every other object with a force that depends only on their masses and the distance between them. This is the essence of Newton's Law of Universal Gravitation, one of the most elegant and powerful principles in physics. Unlike the gravitational effects described by general relativity (which treats gravity as a curvature of spacetime), Newton's formulation provides an excellent approximation for everyday scales and remains the foundation of classical mechanics.

The Mathematical Form

The gravitational force between two point masses \( m_1 \) and \( m_2 \) separated by distance \( r \) is given by:

\[ F = G\frac{m_1 m_2}{r^2} \]

where \( G = 6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 \) is the universal gravitational constant. This constant, determined experimentally through Cavendish's famous torsion balance experiment in 1798, is remarkably small—reflecting why we don't notice the gravitational attraction between everyday objects.

The force follows an inverse-square law: doubling the separation reduces the force by a factor of four. The force is always attractive, and by Newton's third law, both objects experience equal magnitude forces directed along the line connecting their centers.

Vector Form and Superposition

For a more complete description, we express the gravitational force as a vector. If object 1 exerts force on object 2:

\[ \vec{F}_{12} = -G\frac{m_1 m_2}{r^2}\hat{r} \]

where \( \hat{r} \) is the unit vector pointing from mass 1 toward mass 2. The negative sign ensures the force is attractive.

When multiple masses are present, the net gravitational force on any object is the vector sum of forces from all other masses (principle of superposition). This allows us to handle complex systems: planets orbiting a star, galaxies in a cluster, or a small test mass in the field of a massive body.

A Parallel Problem: The Three-Body Configuration

Consider three identical spheres, each with mass \( M = 2.0 \times 10^6 \, \text{kg} \), arranged at the vertices of an equilateral triangle with side length \( a = 500 \, \text{m} \). Find the net gravitational force on one sphere due to the other two.

Solution:

The gravitational force from each of the other two spheres has magnitude:

\[ F_1 = G\frac{M^2}{a^2} = (6.674 \times 10^{-11})\frac{(2.0 \times 10^6)^2}{(500)^2} \]

\[ F_1 = (6.674 \times 10^{-11})\frac{4.0 \times 10^{12}}{2.5 \times 10^5} = 1.07 \times 10^{-3} \, \text{N} \]

Since the triangle is equilateral, the two force vectors make an angle of 60° with each other (considering the geometry from the target sphere's perspective). Setting up coordinates with one force along the positive x-axis:

• Force from sphere 1: \( \vec{F}_1 = 1.07 \times 10^{-3} \hat{x} \, \text{N} \)
• Force from sphere 2: \( \vec{F}_2 = (1.07 \times 10^{-3})\cos(60°)\hat{x} + (1.07 \times 10^{-3})\sin(60°)\hat{y} \)

\[ \vec{F}_2 = 0.535 \times 10^{-3}\hat{x} + 0.927 \times 10^{-3}\hat{y} \, \text{N} \]

The net force has components:

\[ F_x = 1.07 \times 10^{-3} + 0.535 \times 10^{-3} = 1.605 \times 10^{-3} \, \text{N} \]

\[ F_y = 0.927 \times 10^{-3} \, \text{N} \]

\[ F_{\text{net}} = \sqrt{F_x^2 + F_y^2} = \sqrt{(1.605)^2 + (0.927)^2} \times 10^{-3} = 1.85 \times 10^{-3} \, \text{N} \]

The net force points at an angle \( \theta = \tan^{-1}(0.927/1.605) = 30° \) from the line connecting the target sphere to one of the others.

Real-World Applications

Newton's Law explains the orbital motions of planets around the Sun, the Moon's orbit around Earth, and the dynamics of binary star systems. Satellites rely on this principle for their orbital mechanics. In astrophysics, gravitational lensing—where massive objects bend light around them—is explained through general relativity but calculated approximately using Newtonian concepts.

Understanding this law also helps us predict asteroid impacts, design space missions, and comprehend the large-scale structure of the universe.

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