Classical Mechanics: Newton's Laws and Universal Gravitation

Newton’s Laws of Motion - Introduction and Framework

• Historical and Academic Context:     * The lecture notes are dated Wednesday, 25th April 2006.     * Newton’s laws of motion are celebrated for their extraordinary success, both in the breadth and depth of their applications.     * These laws have survived essentially intact for more than three centuries.     * The same longevity applies to Newton’s universal law of gravitation, which specifies the force all masses exert upon each other.
• Scope and Foundation:     * Taken together, Newton’s laws represent virtually the entire foundation of classical mechanics.     * They provide an accurate explanation for a vast range of phenomena, from the behavior of large molecules to the movement of entire galaxies.     * These are laws of physics founded upon experimental evidence. Their validity stands or falls according to the accuracy of their predictions.
• Dynamics vs. Kinematics:     * Kinematics: Concerned purely with the geometry of motion.     * Dynamics: Seeks to answer questions regarding what motion occurs when specified forces act upon a body.     * Newton’s laws of motion provide the rules that make the connection between force and the resulting motion.

The First Law of Motion (The Law of Inertia)

• Conceptual Definition:     * The essence of Newton’s First Law is the concept of inertia.     * The law states that a body remains in a state of rest or continues to move with a constant velocity unless acted upon by an external force.
• Inertial Reference Frames:     * In Newtonian mechanics, the stage of mechanics consists of a framework known as a reference frame.     * A reference frame in which the First Law of Newton is true is called an Inertial Frame.     * A frame moving with a constant velocity relative to an inertial frame is also an inertial frame.     * Consequently, there are infinitely many inertial frames.     * In practice, exact inertial frames are not available. Historically, frames were often defined with respect to the "fixed stars."
• Practical Examples and Exercises:     * Question: Which of the following are inertial frames?         1. A frame fixed to a motor vehicle moving with constant speed around a flat circular track.         2. A frame fixed to a car traveling along a straight road while accelerating with constant acceleration.         3. A frame fixed to a car moving with constant speed up a road with a constant gradient.     * Solution: Only the third scenario (constant speed up a constant gradient) constitutes an inertial frame, as there is no acceleration involved (moving with constant velocity).

The Second Law of Motion (The Law of Motion)

• Central Postulate:     * In an inertial frame, if a particle of mass $m$ is acted on by a force $\mathbf{F}$, then:         $\mathbf{F} = \frac{d\mathbf{p}}{dt}$
• Linear Momentum:     * The term $\mathbf{p} = m\mathbf{v}$ is defined as the linear momentum of the particle, measured relative to the given inertial frame.
• Significance:     * This equation provides the means of determining the trajectory of a particle and is known as the "Law of Motion."

The Third Law of Motion

• Postulate of Interaction:     * The theory of motion is completed by postulating the nature of interaction between any two particles.     * If $\mathbf{F}_{12}$ is the force that particle one ($P_1$) exerts on particle two ($P_2$), and $\mathbf{F}_{21}$ is the force that $P_2$ exerts on $P_1$, then:         $\mathbf{F}_{12} = -\mathbf{F}_{21}$
• Core Principle:     * The mutual actions between particles are always equal in magnitude and opposite in direction.

Multi-Particle Systems and Center of Mass

• Extension of the Laws:     * The standard formulation is extended to a system of $N$ particles.     * Let $m_i$ denote the mass and $\mathbf{r}_i$ denote the position vector of the i$th particle (where i = 1, 2, \dots, N).\n    * The momentum of the i$th particle relative to an inertial frame is $\mathbf{p}_i = m_i \frac{d\mathbf{r}_i}{dt}$.     * The equations of motion are given by:         $\mathbf{F}_i = \frac{d\mathbf{p}_i}{dt}$
• Force Categorization:     * $F_i$ is the total force on the particle, which is the vector sum of inter-particle forces and external forces.     * Inter-particle forces ($\mathbf{F}_{ji}$): Forces exerted on particle $i$ by other particles $j$ within the system.     * External forces ($\mathbf{F}_i^{(e)}$): Forces associated with sources outside the system.     * Total force equation:         $\mathbf{F}_i = \sum_{j \neq i} \mathbf{F}_{ji} + \mathbf{F}_i^{(e)}$
• N-Particle Equations of Motion:     * If masses $m_i$ are constant:         $m_i \frac{d^2\mathbf{r}_i}{dt^2} = \sum_{j \neq i} \mathbf{F}_{ji} + \mathbf{F}_i^{(e)}$     * This represents a set of $N$-coupled differential equations.
• Center of Mass Definition:     * The center of mass of a system of particles (with masses $m_1, m_2, \dots, m_N$ and position vectors $\mathbf{r}_1, \mathbf{r}_2, \dots, \mathbf{r}_N$) is a single point in space with position vector $\mathbf{R}$, defined as:         $\mathbf{R} = \frac{1}{M} \sum_{i=1}^N m_i \mathbf{r}_i$     * Where $M = \sum_{i=1}^N m_i$ is the total mass of the system.     * The center of mass is the weighted mean of the position vectors where the weights are the particle masses.     * Examples:         1. Pair of Particles ($P_1, P_2$): The center of mass is $\mathbf{R} = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2}{m_1 + m_2}$. It lies on the line joining $P_1$ and $P_2$ and divides the line in the ratio $m_2:m_1$.         2. Three Identical Particles ($P_1, P_2, P_3$): The center of mass is $\mathbf{R} = \frac{\mathbf{r}_1 + \mathbf{r}_2 + \mathbf{r}_3}{3}$. This corresponds to the centroid of the triangle formed by the three particles.

Newton’s Universal Law of Gravitation

• Basic Observation:     * Any object with mass attracts any other object with mass with a force called gravitation.
• Formal Definition:     * The gravitational force that two particles exert on each other has a magnitude determined by:         $F = \frac{Gm_1 m_2}{R^2}$     * Where $m_1$ and $m_2$ are the particle masses, and $R$ is the distance between them.
• The Gravitational Constant ($G$):     * $G$ is a universal constant of gravitation.     * It is not dimensionless; its numerical value depends on the units used for mass, length, and force.     * In SI units:         $G = 6.67 \times 10^{-11} \, N \cdot m^2 \cdot kg^{-2}$
• Interaction Properties:     * Per Newton's Third Law, interaction forces are equal in magnitude, opposite in direction, and parallel to the straight line joining the particles.

Gravitational Force by Mass Distributions

• Principle of Multiple Interaction:     * The effective force exerted on a particle is the resultant (vector sum) of the individual forces of interaction exerted on that particle by all other mass elements.
• Example: Particle Attracted by Two Masses (A and B):     * Consider a particle $C$ of mass $m$ attracted by two particles $A$ and $B$, each of mass $M$, separated by a distance $2a$. Let $C$ be at a distance $x$ from the midpoint $D$ of line $AB$.     * The distance from $A$ or $B$ to $C$ is $R = \sqrt{a^2 + x^2}$.     * The force magnitude from one particle is $F' = \frac{GmM}{R^2}$.     * By symmetry, the resultant force points toward $D$ along the $-x$ direction.     * The total magnitude is $F = 2F' \cos(\alpha)$, where $\cos(\alpha) = \frac{x}{R}$.     * Resultant force:         $F = \frac{2GmMx}{(a^2 + x^2)^{3/2}}$     * Asymptotic Behavior: When $x$ is very large ($x \gg a$), the force is approximately:         $F \approx \frac{G m (2M)}{x^2}$     * This shows that at a great distance, the system of masses acts like a single particle of mass $2M$ situated at the center of mass.
• Example: Uniform Rod of Length $2a$ and Mass $M$:     * A particle of mass $m$ is at distance $b$ from the center of the rod along the perpendicular bisector.     * Considering an element of the rod of length $dx$ and mass $\frac{M}{2a} dx$, the force magnitude is calculated via integration:         $F = \int_{-a}^a \frac{Gm(\frac{M}{2a}dx)}{x^2 + b^2} \cos(\theta)$     * Using trigonometric substitution $x = b \tan(\theta)$, the result is:         $F = \frac{GmM}{b \sqrt{a^2 + b^2}}$     * As $b \rightarrow \infty$, $F \approx \frac{GmM}{b^2}$, confirming the asymptotic point-mass behavior.
• Example: Uniform Disk of Radius $a$ and Mass $M$:     * A particle $P$ of mass $m$ is on the axis of the disk at distance $b$ from the center.     * Integration is performed over the area using polar coordinates $(r, \theta)$, where the element of area is $dA = r \, dr \, d\theta$.     * Resultant force formula:         $F = \frac{2GmM}{a^2} \left[ 1 - \frac{b}{\sqrt{a^2 + b^2}} \right]$

Gravitation by a Symmetric Sphere

• Importance in Astronomy:     * The most important case for celestial mechanics and space travel is the spherical mass distribution.     * A "symmetric sphere" is defined as a body where the mass density $\rho$ depends only on the distance $r$ from the center: $\rho = \rho(r)$.     * Newton proved that for a symmetric sphere of total mass $M$, the gravitational force exerted on an external particle is exactly the same as if the entire mass $M$ were replaced by a point particle at the center of the sphere.
• Practical Application (Earth):     * Earth's density is not uniform; it is approximately $3000 \, kg \cdot m^{-3}$ near the surface but reaches $16500 \, kg \cdot m^{-3}$ at the center.     * Because Newton's theorem accounts for varying density $\rho(r)$, it remains valid for the Earth and Sun.
• Proof Sketch (Spherical Polar Coordinates):     * The element of volume in spherical coordinates is $dv = (dr)(r \, d\theta)(r \sin \theta \, d\phi) = r^2 \sin \theta \, dr \, d\theta \, d\phi$.     * The total mass is calculated as $M = \int \rho \, dv = 4\pi \int_0^a \rho(r) r^2 \, dr$.     * The total force magnitude is given by:         $F = G m \int \frac{\rho(r) \cos(\alpha)}{R^2} dv$     * After complex integration using a change of variables from $\theta$ to $R$ (where $R^2 = r^2 + b^2 - 2rb \cos \theta$), the result simplifies to:         $F = \frac{GmM}{b^2}$