Analytical Mechanics: Gravitation and Central Forces

Newton's Law of Gravitation

• Fundamental Definition: Every particle in the universe attracts every other particle with a force whose magnitude is proportional to the product of the masses of the two particles and inversely proportional to the square of the distance between them.
• Mathematical Representation:     * Magnitude: $F = G \frac{m_i m_j}{r_{ij}^2}$     * Vector Form: $\mathbf{F} = -G \frac{m_i m_j}{r_{ij}^2} \mathbf{\hat{r}}_{ij}$
• Universal Gravitational Constant ($G$):     * Value: $G = 6.67 \times 10^{-11}\,\text{Nm}^2/\text{kg}^2$
• Nature of Gravitational Force:     * The force between two particles acts along the line joining them.     * These forces form an action-reaction pair.

Gravitational Force and Extended Objects

• Extended Objects vs. Point Particles:     * In real life, we rarely deal with point particles. Instead, we consider extended objects where dimensions (length, height, etc.) are significant.     * A point particle approximation is only used when dimensions are small enough to be unimportant.     * Extended objects can be a stick of length $L$, a disc of radius $r$, or a solid of some shape.     * The center of mass is one point among an infinite number of points inside the object.
• Evaluating Force for Extended Objects:     * The central question is: "How to evaluate the gravitational force between the extended objects?"     * This is answered by the Shell Theorem.
• Shell Theorem Definition: A uniform shell of matter attracts an external particle as if all the shell's mass were concentrated at its center.

Gravitational Force Between a Uniform Sphere Shell and a Particle

• System Setup:     * Consider a uniform spherical shell of mass ($m_1$) and radius ($R$).     * Let $r$ be the distance from the center $O$ of the shell to a particle $P$ of mass ($m_2$).
• Elemental Force Derivation:     * $df_{ij} = G \frac{m_i \Delta m_j}{r_{ij}^2}$     * $dm_1 = \sigma (R d\theta \times 2 \pi R \sin(\theta)) = \sigma 2 \pi R^2 \sin(\theta) d\theta$     * The gravitational force on $P$ has components $df_x = df_i \cos(\phi)$ and $df_y = df_i \sin(\phi)$.     * Due to symmetry, all vertical ($y$) components exerted on $P$ add up to zero.
• Total Force Integral:     * $df = \frac{G m_2 dm_1}{PQ^2} \cos(\phi)$     * $F_{\text{outside shell}} = \int df = \int \frac{G m_2 \sigma 2 \pi R^2 \sin(\theta) \cos(\phi) d\theta}{s^2}$     * Using the Law of Cosines for $\triangle OPQ$: $s^2 = r^2 + R^2 - 2 r R \cos(\theta)$     * Differentiating leads to: $2 s ds = 2 r R \sin(\theta) d\theta \Rightarrow s ds = r R \sin(\theta) d\theta$     * From $\triangle OPQ$: $R^2 = r^2 + s^2 - 2 r s \cos(\phi) \Rightarrow \cos(\phi) = \frac{r^2 + s^2 - R^2}{2 r s}$
• Changing Limits of Integration:     * At $\theta = 0$, $s^2 = r^2 + R^2 - 2 r R = (r - R)^2 \Rightarrow s = (r - R)$.     * At $\theta = \pi$, $s^2 = r^2 + R^2 + 2 r R = (r + R)^2 \Rightarrow s = (r + R)$.
• Final Force Calculation:     * $F = G m_2 \sigma 2 \pi R^2 \int_{r-R}^{r+R} \frac{s ds}{r R} \frac{r^2 + s^2 - R^2}{2 r s} \frac{1}{s^2}$     * $F = \frac{G m_2 \sigma 2 \pi R^2}{2 r^2 R} \int_{r-R}^{r+R} \frac{s^2 + r^2 - R^2}{s^2} ds$     * Using $M_1 = 4 \pi R^2 \sigma$:     * $F = \frac{G m_2 M_1}{4 R r^2} [2 R + (r^2 - R^2) \frac{2 R}{(r+R)(r-R)}] = \frac{G m_2 M_1}{4 R r^2} [2 R + 2 R] = \frac{G m_2 M_1}{r^2}$

Kepler's Laws of Planetary Motion

• Law of Plane Motion: Suppose a planet of mass ($m$) is in a circular orbit around the sun of mass ($M$) with radius ($r$).     * Gravitational force: $|F| = G \frac{M m}{r^2}$     * Centripetal force: $|F_c| = \frac{m v^2}{r}$     * Equating the two: $G \frac{M m}{r^2} = \frac{m v^2}{r}$     * Using orbital period $T$: $v = \frac{2 \pi r}{T}$     * Substitution results in: $T^2 = \frac{4 \pi^2 r^3}{GM}$     * This shows $T^2 \propto r^3$. While planets generally have elliptical orbits, the relationship holds.
• Law 1: Law of Ellipses: The orbit of each planet is an ellipse, with the sun located at one of its foci.
• Law 2: Law of Equal Areas: A line drawn between the sun and the planet sweeps out equal areas in equal times as the planet orbits the sun.
• Law 3: Harmonic Law: The square of the sidereal period of a planet (the time it takes a planet to complete one revolution about the sun relative to the stars) is directly proportional to the cube of the semi-major axis of the planet's orbit.

Derivation of Kepler's First Law

• Coordinate System: Using polar coordinates, the acceleration is $\mathbf{a} = (\ddot{r} - r\dot{\theta}^2)\mathbf{\hat{r}} + (2\dot{r}\dot{\theta} + r\ddot{\theta})\mathbf{\hat{\theta}}$.
• Equation of Motion:     * The direction of force is toward the sun, so force in $\mathbf{\hat{\theta}}$ is zero.     * $\mathbf{F} = -G \frac{M m}{r^2} \mathbf{\hat{r}} = m(\ddot{r} - r\dot{\theta}^2) \mathbf{\hat{r}}$     * $\ddot{r} - r\dot{\theta}^2 = -\frac{GM}{r^2}$
• Angular Momentum Conservation:     * $\mathbf{L} = |\mathbf{r} \times m\mathbf{v}| = |\mathbf{r} \times m(\dot{r}\mathbf{\hat{r}} + r\dot{\theta}\mathbf{\hat{\theta}})| = mr^2\dot{\theta}$     * Since force is central, $L$ is constant.
• The Orbit Equation:     * The general equation of an ellipse is $r = \frac{a(1 - \epsilon^2)}{1 + \epsilon \cos(\theta)}$.     * $a = \text{semi-major axis}$     * $\epsilon = \text{eccentricity}$     * In polar coordinates, the equation of motion to solve is $\ddot{r} - \frac{L^2}{m^2 r^3} = -\frac{GM}{r^2}$.

Central Force Motion

• Definition: When a particle moves under the influence of a force directed toward a fixed center of attraction, the motion is called central-force motion.
• Examples: Orbital movement of planets and satellites.
• Differential Equation of the Orbit:     * Accelerations: $a_r = \ddot{r} - r\dot{\theta}^2$ and $a_{\theta} = 2\dot{r}\dot{\theta} + r\ddot{\theta}$.     * For central force, $a_{\theta} = 0$, implying $r^2\dot{\theta} = l$ (angular momentum per unit mass) is constant.     * Using the variable transformation $u = 1/r$:         * $\dot{r} = -l \frac{du}{d\theta}$         * $\ddot{r} = -l^2 u^2 \frac{d^2u}{d\theta^2}$     * Substitution into the radial equation $m(\ddot{r} - r\dot{\theta}^2) = F(r)$ yields:         * $\frac{d^2u}{d\theta^2} + u = -\frac{F(u^{-1})}{ml^2 u^2}$     * This is the differential equation for the orbit.

Specific Examples of Force and Motion

• Example 1: Spiral Orbit Force Function     * Given: Spiral orbit $r = c \theta^2$     * Task: Determine force function $f(u^{-1})$.     * Solution: $u = \frac{1}{c \theta^2} \Rightarrow \theta = \frac{1}{\sqrt{cu}}$.     * $\frac{du}{d\theta} = -2 c \theta^3$     * From definition: $\frac{d^2u}{d\theta^2} + u = -\frac{f(u^{-1})}{ml^2 u^2}$     * $f(r) = -\frac{ml^2}{r^2} [\frac{6c}{r} + \frac{1}{r}]$ (Derived from differentiation of $u$ with respect to $\theta$).
• Example 2: Time Variation of Angle     * Given: Spiral orbit $r = c \theta^2$     * Task: Determine how angle $\theta$ varies with time.     * Solution: $l = r^2 \dot{\theta} \Rightarrow \dot{\theta} = \frac{l}{r^2} = \frac{l}{c^2 \theta^4}$.     * $\int \theta^4 d\theta = \int \frac{l}{c^2} dt \Rightarrow \frac{\theta^5}{5} = \frac{l}{c^2} t + D$     * At $t = 0$, $\theta = 0$, so $D=0$.     * $\theta = (\frac{5l}{c^2} t)^{1/5}$.

Inverse Square Law and Conics

• If $f(r) = -\frac{k}{r^2}$, where $k = GMm$, the orbit equation becomes:     * $\frac{d^2u}{d\theta^2} + u = \frac{k}{ml^2}$
• General solution: $u = A \cos(\theta - \theta_0) + \frac{k}{ml^2}$.
• This results in: $r = \frac{1}{A \cos(\theta - \theta_0) + k/ml^2}$.
• Setting $\theta_0 = 0$: $r = \frac{ml^2/k}{1 + (Aml^2/k) \cos(\theta)}$.
• This is the equation of a conic section (ellipse if eccentricity \epsilon < 1).
• Ellipse Parameters:     * $r = \frac{a(1 - \epsilon^2)}{1 + \epsilon \cos(\theta)}$     * Perihelion (closest approach, $\theta = 0$): $r_{\min} = a(1 - \epsilon)$.     * Aphelion (farthest distance, $\theta = \pi$): $r_{\max} = a(1 + \epsilon)$.     * Sum of distances: $r_{\max} + r_{\min} = 2 a$.     * Eccentricity: $\epsilon = \frac{A m l^2}{k}$.

Elliptical Orbit Applications

• Example 3: Halley's Comet     * Data: $\epsilon = 0.967$, Perihelion distance $r_{\min} = 0.59\,\text{AU}$.     * Task: Find aphelion distance $r_{\max}$.     * Formula: $r_{\max} = \frac{r_{\min}(1 + \epsilon)}{(1 - \epsilon)}$     * Calculation: $r_{\max} = \frac{0.59 (1 + 0.967)}{(1 - 0.967)} = \frac{0.59(1.967)}{0.033} \approx 35\,\text{AU}$.     * Conclusion: Halley's comet travels outside the orbit of Neptune.

Derivation of Kepler's Second and Third Laws

• Second Law (Equal Areas):     * Area of triangle segment: $dA = \frac{1}{2} |\mathbf{r} \times d\mathbf{r}| = \frac{1}{2} r^2 d\theta$.     * Areal velocity: $\frac{dA}{dt} = \frac{1}{2} r^2 \dot{\theta} = \frac{l}{2}$.     * Since $l = L/m$ is constant, the areal velocity $\dot{A}$ is constant: $\dot{A} = \frac{L}{2m}$.
• Example 4: Circular Orbits and Areal Velocity     * For circular orbits: $r = \text{constant}$.     * $|f(r)| = m r \dot{\theta}^2$.     * If areal velocity $A$ is identical for all orbits: $A = \frac{1}{2} r^2 \dot{\theta}$.     * $\dot{\theta} = \frac{2A}{r^2}$.     * Substitution: $|f(r)| = m r (\frac{2A}{r^2})^2 = \frac{4 m A^2}{r^3}$.
• Third Law (Harmonic Law):     * Total area of ellipse $A = \pi a b$, where $b = a \sqrt{1 - \epsilon^2}$.     * Integrating areal velocity over one period: $A = \int_0^T \dot{A} dt = \frac{l T}{2}$.     * $T = \frac{2 \pi a b}{l} = \frac{2 \pi a^2 \sqrt{1 - \epsilon^2}}{l}$.     * Using orbital parameter relationship $l^2 = G M a (1 - \epsilon^2)$.     * $T^2 = \frac{4 \pi^2 a^4 (1 - \epsilon^2)}{G M a (1 - \epsilon^2)} = \frac{4 \pi^2 a^3}{G M}$.
• Example 5: Satellite Near Earth Surface     * Task: Estimate period of a satellite close to Earth surface.     * Formula: $T^2 = \frac{4 \pi^2 R_e^3}{G M_e}$.     * Using $g = \frac{G M_e}{R_e^2}$, then $T = 2 \pi \sqrt{\frac{R_e}{g}}$.     * Constants: $R_e = 6.38 \times 10^6\,\text{m}$, $g = 9.8\,\text{m/s}^2$.     * Calculation: $T = 2 \pi \sqrt{\frac{6.38 \times 10^6}{9.8}} \approx 5070\,\text{s} \approx 85\,\text{min}$.