Gravitation

Historical Foundations and Kepler's Laws of Planetary Motion

• Early Planetary Models and Historical Timeline:

  • Nicolaus Copernicus (1473–1543): Proposed a definitive heliocentric model where planets move in circular orbits around a fixed central Sun. The Copernican model was discredited by the church. Galileo Galilei was a notable supporter of the Copernican view and was prosecuted by the state for his beliefs.

  • Tycho Brahe (1546–1601): A Danish nobleman who recorded extensive, precise observations of planetary positions using only the naked eye over his lifetime.

  • Johannes Kepler (1571–1640): Assistant to Tycho Brahe who analyzed Brahe's compiled observational data over sixteen years. Kepler formulated three empirical laws governing planetary motion, deviating from circular trajectories to elliptical paths. Kepler is also recognized as the founder of geometrical optics, being the first to describe what happens to light after it enters a telescope.

• Kepler's First Law (Law of Orbits):

  • Definition: All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse.

  • Geometric Properties of an Ellipse:

  • An ellipse is a closed plane curve where the sum of the distances from any point $T$ on the curve to two fixed points ($F_1$ and $F_2$, called the foci) is constant: $F_1T + F_2T = \text{constant}$.

  • A circle represents a limiting, special case of an ellipse where the two foci merge into a single central point, and the semi-major axis becomes the radius.

  • The line joining the foci intersects the ellipse at two apsides: the closest distance to the focus is the perihelion ($P$), and the farthest distance is the aphelion ($A$).

  • The center $O$ is the midpoint of the major axis line segment $PA$. The semi-major axis length is $a = PO = AO = \frac{1}{2}PA$. The semi-minor axis is denoted by $b$.

  • Mechanical Drawing of an Ellipse:

  • Select two points $F_1$ and $F_2$ on a flat sheet.

  • Secure the two ends of a piece of flexible string at points $F_1$ and $F_2$ using pins.

  • Stretch the string taut with the tip of a pencil and sweep the pencil around the pins while keeping the string taut throughout.

• Kepler's Second Law (Law of Areas):

  • Definition: The line that joins any planet to the Sun sweeps out equal areas in equal intervals of time; that is, the areal velocity $\frac{\Delta \mathbf{A}}{\Delta t}$ is strictly constant.

  • Orbital Consequence: Planets travel faster when they are nearer the Sun (perihelion) and slower when they are farther away (aphelion).

• Kepler's Third Law (Law of Periods):

  • Definition: The square of the time period of revolution ($T$) of a planet is directly proportional to the cube of the semi-major axis ($a$) of the elliptical orbit traced out by the planet:     $T^2 \propto a^3$

  • Planetary Verification Data:

  • Planetary survey measurements confirm the constancy of the quotient $Q = \frac{T^2}{a^3}$ (in units of $10^{-34}\,\text{y}^2\,\text{m}^{-3}$, with $a$ in units of $10^{10}\,\text{m}$ and $T$ in Earth years $\text{y}$):

    • Mercury: $a = 5.79$, $T = 0.24$, $Q = 2.95$

    • Venus: $a = 10.8$, $T = 0.615$, $Q = 3.00$

    • Earth: $a = 15.0$, $T = 1$, $Q = 2.96$

    • Mars: $a = 22.8$, $T = 1.88$, $Q = 2.98$

    • Jupiter: $a = 77.8$, $T = 11.9$, $Q = 3.01$

    • Saturn: $a = 143$, $T = 29.5$, $Q = 2.98$

    • Uranus: $a = 287$, $T = 84$, $Q = 2.98$

    • Neptune: $a = 450$, $T = 165$, $Q = 2.99$

    • Pluto: $a = 590$, $T = 248$, $Q = 2.99$

  • Circularity of Planetary Orbits: Except for Mercury, Mars, and Pluto, planetary orbits are very close to being circular. For Earth, the ratio of the semi-minor axis to the semi-major axis is $\frac{b}{a} = 0.99986$.

Mathematical Analysis and Derivation of Kepler's Laws

• Derivation of the Law of Areas from Angular Momentum Conservation:

  • The Law of Areas is a direct dynamical consequence of conservation of angular momentum under a central force.

  • Definition of Central Force: A central force acts strictly along the position vector $\mathbf{r}$ connecting the source to the moving body. Because the force vector is parallel or anti-parallel to the position vector, the net torque is identically zero:     $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} = 0$

  • Areal Velocity Kinematics:

  • Let the Sun occupy the origin. The position vector of the planet is $\mathbf{r}$, its momentum is $\mathbf{p} = m\mathbf{v}$, and its displacement over a small time increment $\Delta t$ is $\Delta \mathbf{r} = \mathbf{v}\,\Delta t$.

  • The area of the planar triangular segment $\Delta \mathbf{A}$ swept out by the radius vector is:       $\Delta \mathbf{A} = \frac{1}{2}(\mathbf{r} \times \mathbf{v}\,\Delta t)$

  • Dividing both sides by $\Delta t$ and substituting $\mathbf{v} = \frac{\mathbf{p}}{m}$, where $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ is the orbital angular momentum vector:       $\frac{\Delta \mathbf{A}}{\Delta t} = \frac{1}{2}\frac{\mathbf{r} \times \mathbf{p}}{m} = \frac{\mathbf{L}}{2m}$

  • Since $\mathbf{L}$ is a constant of motion for any central force, $\frac{\Delta \mathbf{A}}{\Delta t}$ is constant in both magnitude and direction.

• Orbital Speed and Traversal Time (Worked Example):

  • Setup: Let the speed of a planet at perihelion ($P$) be $v_P$ and the Sun-planet distance be $r_P$. At aphelion ($A$), let the corresponding speed and distance be $v_A$ and $r_A$.

  • Velocity Relation: At both perihelion and aphelion, the position vectors and velocity vectors are mutually perpendicular. The magnitudes of angular momentum are:     $L_P = m_p r_P v_P$     $L_A = m_p r_A v_A$

  • Conservation of angular momentum yields:       $m_p r_P v_P = m_p r_A v_A \implies \frac{v_P}{v_A} = \frac{r_A}{r_P}$

  • Because $r_A > r_P$, it follows that $v_P > v_A$.

  • Traversal Time Comparison: To compare the traversal times along elliptical arcs $BAC$ and $CPB$:

  • The area $SBAC$ bounded by the ellipse and radius vectors $SB$ and $SC$ is strictly greater than the sector area $SBPC$.

  • Because equal areas are swept in equal intervals of time, the planet takes a longer time to traverse arc $BAC$ than arc $CPB$.

Newton's Universal Law of Gravitation and Vector Formulation

• Physical Formulation:

  • Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

  • Formulated by Isaac Newton and published in 1687 in the Philosophiae Naturalis Principia Mathematica.

• Deduction from Terrestrial and Lunar Kinematics:

  • Newton examined the circular centripetal acceleration of the Moon orbiting at radius $R_m$ with orbital period $T$:     $a_m = \frac{V^2}{R_m} = \frac{4\pi^2 R_m}{T^2}$

  • Using the known lunar parameters $R_m \approx 3.84 \times 10^8\,\text{m}$ and $T \approx 27.3\,\text{days} = 27.3 \times 24 \times 3600\,\text{s} = 2.3587 \times 10^6\,\text{s}$:     $a_m \approx 0.00272\,\text{m\,s}^{-2}$

  • Comparing this acceleration to terrestrial surface gravitational acceleration $g \simeq 9.8\,\text{m\,s}^{-2}$:     $\frac{g}{a_m} \approx \frac{9.8}{0.00272} \approx 3600$

  • Assuming an inverse-square force law ($a \propto \frac{1}{r^2}$):     $\frac{g}{a_m} = \frac{R_m^2}{R_E^2}$

  • Given $R_m \approx 60 R_E$, the ratio of the squared distances is:     $\left(\frac{R_m}{R_E}\right)^2 = (60)^2 = 3600$

  • This exact match verified that the force governing falling terrestrial bodies is the identical force maintaining the Moon in its orbit.

• Mathematical and Vector Expression:

  • The scalar magnitude of the gravitational attraction between two point masses $m_1$ and $m_2$ separated by distance $r$ is:     $|\mathbf{F}| = G\frac{m_1 m_2}{r^2}$

  • Vector Formulation: Let $\mathbf{r}_1$ and $\mathbf{r}_2$ be position vectors of masses $m_1$ and $m_2$ relative to an origin. The vector pointing from $m_1$ to $m_2$ is $\mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1$, with unit vector \hat{\mathbf{r}} = \frac{\mathbf{r}}{|\mathbf{r}|}$.\n * The force \mathbf{F}$on point mass$m_2$directed toward point mass$m_1$is attractive (pointing along$-\hat{\mathbf{r}}):\n    \mathbf{F} = G\frac{m_1 m_2}{r^2}(-\hat{\mathbf{r}}) = -G\frac{m_1 m_2}{r^2}\hat{\mathbf{r}} = -G\frac{m_1 m_2}{|\mathbf{r}|^3}\mathbf{r}\n * **Third Law Symmetry:** The mutual forces form an action-reaction pair:\n    \mathbf{F}{12} = -\mathbf{F}{21}\n\n\n\n# Gravitational Superposition and Extended Bodies (Shell Theorems)\n\n* **Principle of Superposition for Point Masses:**\n * The resultant gravitational force acting on a target point mass m_1$due to an assembly of discrete masses$m_2, m_3, \dots, m_n equals the vector sum of the isolated two-body forces:\n    \mathbf{F}1 = \sum{i=2}^n \mathbf{F}{1i} = \frac{G m_2 m_1}{r{21}^2}\hat{\mathbf{r}}{21} + \frac{G m_3 m_1}{r{31}^2}\hat{\mathbf{r}}{31} + \dots + \frac{G m_n m_1}{r{n1}^2}\hat{\mathbf{r}}{n1}\n * Each gravitational interaction operates independently and is uninfluenced by the presence of surrounding bodies.\n\n* **Equilateral Triangle Superposition (Worked Example):**\n * **Configuration:** Three equal point masses m$sit at the vertices of an equilateral triangle$ABC$. A particle of mass$2m$is positioned at the centroid$G$. Let$AG = BG = CG = 1\,\text{m}.\n * **Geometry:** The line segment GC$makes an angle of$30^\circ$with the positive$x$-axis. Segment$GB$makes an angle of$30^\circ$with the negative$x$-axis. Segment$GA$is aligned with the positive$y-axis.\n * **Individual Force Vectors on Mass 2m:**\n    \mathbf{F}{GA} = \frac{G m (2m)}{(1)^2}\hat{\mathbf{j}} = 2Gm^2\hat{\mathbf{j}}\n    \mathbf{F}{GB} = 2Gm^2(-\hat{\mathbf{i}}\cos(30^\circ) - \hat{\mathbf{j}}\sin(30^\circ))\n    \mathbf{F}{GC} = 2Gm^2(\hat{\mathbf{i}}\cos(30^\circ) - \hat{\mathbf{j}}\sin(30^\circ))\n * **Resultant Force Calculation:**\n    \mathbf{F}R = \mathbf{F}{GA} + \mathbf{F}{GB} + \mathbf{F}{GC}\n    \mathbf{F}R = 2Gm^2\hat{\mathbf{j}} + 2Gm^2\left[\hat{\mathbf{i}}(\cos(30^\circ) - \cos(30^\circ)) - 2\hat{\mathbf{j}}\sin(30^\circ)\right]\n    \mathbf{F}_R = 2Gm^2\hat{\mathbf{j}} + 2Gm^2\left[0\hat{\mathbf{i}} - 2\left(\frac{1}{2}\right)\hat{\mathbf{j}}\right] = 2Gm^2\hat{\mathbf{j}} - 2Gm^2\hat{\mathbf{j}} = 0\n * By spatial symmetry, the net vector force vanishes.\n * **Effect of Doubling Vertex Mass:** If the mass at vertex A$is doubled from$m$to$2m:\n    \mathbf{F}'{GA} = \frac{G(2m)(2m)}{(1)^2}\hat{\mathbf{j}} = 4Gm^2\hat{\mathbf{j}}\n    \mathbf{F}'R = \mathbf{F}'{GA} + \mathbf{F}{GB} + \mathbf{F}{GC} = 4Gm^2\hat{\mathbf{j}} - 2Gm^2\hat{\mathbf{j}} = 2Gm^2\hat{\mathbf{j}}\n\n\n\n* **The Shell Theorems for Spherically Symmetric Extended Bodies:**\n * Extended masses cannot be treated as single point masses without integration. Calculus reveals two fundamental theorems for a thin, hollow spherical shell of uniform mass density:\n * **Shell Theorem 1 (External Field):** The force of attraction exerted by a hollow spherical shell of uniform surface density on an external point mass is identical to the force obtained if the entire mass of the shell were concentrated at its geometric center.\n * *Qualitative proof:* Across all surface rings of the shell, force components perpendicular to the line connecting the external point mass to the shell center cancel by azimuthal symmetry, leaving only a net attractive force directed along the line toward the center.\n * **Shell Theorem 2 (Internal Field):** The gravitational force exerted by a uniform hollow spherical shell on any point mass situated anywhere inside its interior cavity is identically zero.\n * *Qualitative proof:* The forces exerted by opposing surface segments on opposite sides of the interior point mass cancel completely due to geometric and inverse-square balance.\n\n# Determination of the Gravitational Constant and Earth's Mass\n\n* **The Cavendish Torsion Balance Experiment (1798):**\n * The universal gravitational constant G was first measured experimentally by English scientist Henry Cavendish.\n * **Apparatus:** A light horizontal bar of length L$supports two small lead spheres (each of mass$m$) attached at opposite ends. The bar is suspended horizontally from a rigid overhead support by a fine, elastic torsion wire. Two large lead spheres (each of mass$M$) are brought into close proximity on opposite sides of the small spheres at an identical center-to-center distance$d.\n * **Torque and Equilibrium Relations:**\n * The gravitational attraction between each big sphere and the nearby small sphere acts as if masses are concentrated at their centers:\n      F = G\frac{Mm}{d^2}\n * The two equal and opposite forces create a gravitational couple, exerting a net torque on the bar:\n      \tau_{\text{grav}} = F \times L = G\frac{Mm}{d^2}L\n * As the suspended wire twists through an angle \theta, an elastic restoring torque is developed:\n      \tau_{\text{restoring}} = \tau_0 \theta\n      where \tau_0 is the restoring couple per unit angle of twist (measured independently by timing torsional oscillations or applying a known calibration torque).\n * At static equilibrium:\n      G\frac{Mm}{d^2}L = \tau_0 \theta \implies G = \frac{\tau_0 \theta d^2}{M m L}\n * Reversing the positions of the large spheres to the opposite sides reverses the direction of the torque, doubling the deflection to increase experimental precision.\n * **Accepted Value of G:**\n * Cavendish's initial measurements have been refined to modern standard values:\n      G = 6.67 \times 10^{-11}\,\text{N\,m}^2/\text{kg}^2\n      G = 6.672 \times 10^{-11}\,\text{N\,m}^2\,\text{kg}^{-2}$$

• **Surface Gravity and