Comprehensive Study Guide to Circular Motion, Gravitation, and Orbital Dynamics

Fundamentals of Force, Velocity, and Circular Motion

• Vector Orientations and Kinematic Outcomes:

  • Parallel Vectors: When the net force vector and velocity vector act in the exact same direction, the object accelerates tangentially. The velocity continuously increases.

  • Anti-parallel Vectors: When the net force vector and velocity vector act in opposite directions, the object decelerates. The linear speed decreases continuously.

  • Perpendicular Vectors: When the net force vector acts at a $90^\circ$ angle to the velocity vector, the force produces no work along the trajectory, leaving the magnitude of velocity (speed) unchanged. Instead, the force continuously alters the direction of motion, compelling the object to trace a circular path.

• Definition of Centripetal Force:

  • Centripetal force designates any center-directed or center-seeking force that acts perpendicular to an object's instantaneous velocity, turning its velocity vector inward toward the center of curvature.

  • Directional sequence during circular motion: When moving northward, an eastward-directed force causes an eastward deflection; as velocity points east, a southward-directed force causes a southward deflection; as velocity points south, a westward force causes a westward deflection; as velocity points west, a northward force completes the continuous cycle of rotation.

• Mechanisms Supplying Centripetal Force Across Physical Systems:

  • String Tension ($F_T$): When swinging an attached ball in a horizontal plane, the mechanical tension acting through the string supplies the inward centripetal force.

  • Gravitational Attraction ($F_g$): Gravity provides the centripetal force governing orbital systems. The mutual gravitational attraction between Earth and the Moon maintains the Moon in orbit, preventing it from escaping along its tangential velocity vector into deep space. Similarly, solar gravity maintains planetary orbits.

  • Static Friction ($f_s$): For an automobile negotiating a turn on a level roadway, the static friction between the tire rubber and the pavement provides the requisite centripetal force to prevent sliding or skidding.

  • Normal Force ($F_N$): When an object traverses a vertical dip or follows the curvature of a track, the surface exerts a normal force. At the bottom of a curve, the normal force must simultaneously balance the object's weight and supply the centripetal acceleration, making $F_N > m g$.

  • Magnetic Force ($F_B$): In electromagnetism, a magnetic field exerts a force on a moving charge given by perpendicular vectors. A positive charge moving rightward through a magnetic field directed into the page experiences an upward perpendicular force, driving it into a circular path.

Mathematical Formulation and Scaling Proportionalities

• Governing Centripetal Force Equation:

  • Centripetal force is calculated via the formula:     $F_c = \frac{m v^2}{r}$     where $m$ is mass, $v$ is tangential speed, and $r$ is the radius of curvature.

• Proportionalities and Factor Scaling Rules:

  • Direct Proportionality to Mass ($F_c \propto m$):

  • Doubling the mass ($2m$) doubles the required centripetal force ($2\times$).

  • Tripling the mass ($3m$) triples the required centripetal force ($3\times$).

  • Quadratic Proportionality to Speed ($F_c \propto v^2$):

  • Doubling the speed ($2v$) quadruples the required centripetal force ($2^2 = 4\times$).

  • Tripling the speed ($3v$) increases the required centripetal force nine-fold ($3^2 = 9\times$).

  • Quadrupling the speed ($4v$) increases the required centripetal force sixteen-fold ($4^2 = 16\times$).

  • Inverse Proportionality to Radius ($F_c \propto \frac{1}{r}$):

  • Doubling the radius ($2r$) halves the required centripetal force ($\frac{1}{2}\times$).

  • Tripling the radius ($3r$) decreases the required centripetal force by a factor of 3 ($\frac{1}{3}\times$).

  • Halving the radius ($\frac{1}{2}r$) doubles the required centripetal force ($\frac{1}{1/2} = 2\times$).

  • Multi-Variable Compound Scaling Scenario:

  • If mass is doubled ($m \rightarrow 2$), speed is tripled ($v \rightarrow 3$), and the radius is reduced to one-fourth ($r \rightarrow \frac{1}{4}$):       $\text{Factor} = \frac{2 \times 3^2}{1/4} = \frac{2 \times 9}{1/4} = \frac{18}{1/4} = 18 \times 4 = 72$

  • The required centripetal force increases by a factor of $72$.

Centripetal Acceleration, Period, and Frequency

• Derivation from Newton's Second Law:

  • Net force in circular motion is expressed as $F_c = m a_c$.

  • Dividing both sides by $m$ yields centripetal acceleration:     $a_c = \frac{v^2}{r}$

  • The centripetal acceleration vector ($\mathbf{a}_c$) always points radially inward toward the center of the circle.

• Frequency and Period Definitions:

  • Frequency ($f$): The total count of complete circular revolutions executed per unit time (measured in seconds):     $f = \frac{\text{revolutions}}{\text{time}}$

  • Period ($T$): The time required to execute one full revolution:     $T = \frac{\text{time}}{\text{revolutions}} = \frac{1}{f}$

• Tangential Speed as a Function of Period:

  • The distance traversed in one complete cycle equals the circumference ($2\pi r$). Dividing by the period ($T$) gives:     $v = \frac{2\pi r}{T}$

• Centripetal Acceleration Expressed in Terms of Period:

  • Substituting linear velocity into the acceleration formula:     $a_c = \frac{v^2}{r} = \frac{1}{r} \left(\frac{2\pi r}{T}\right)^2 = \frac{4\pi^2 r^2}{r T^2} = \frac{4\pi^2 r}{T^2}$

  • Units for centripetal acceleration are $\text{m/s}^2$, which requires $r$ in meters ($\text{m}$) and $T$ in seconds ($\text{s}$).

• Calculation of Kinematic and Dynamic Variables of a Whirled Ball:

  • A ball of mass $m = 2\,\text{kg}$ rotates in a circle of radius $r = 10\,\text{m}$, completing $20\,\text{revolutions}$ in $4\,\text{s}$.

  • Frequency:     $f = \frac{20\,\text{revolutions}}{4\,\text{s}} = 5\,\text{rev/s}$

  • Period:     $T = \frac{4\,\text{s}}{20\,\text{revolutions}} = \frac{1}{5\,\text{rev/s}} = 0.2\,\text{s}$

  • Centripetal acceleration:     $a_c = \frac{4\pi^2 (10)}{(0.2)^2} = \frac{39.478 \times 10}{0.04} = \frac{394.78}{0.04} = 9869.6\,\text{m/s}^2$

  • Centripetal force:     $F_c = m a_c = 2\,\text{kg} \times 9869.6\,\text{m/s}^2 = 19739\,\text{N}$

  • Tangential speed:     $v = \frac{2\pi (10)}{0.2} = 100\pi \approx 314.16\,\text{m/s}$

• Centripetal Acceleration of an Aircraft in Gravitational Units ($g$'s):

  • An airplane moves at $v = 400\,\text{m/s}$ through a turn of radius $r = 3\,\text{km} = 3000\,\text{m}$.

  • Centripetal acceleration:     $a_c = \frac{v^2}{r} = \frac{400^2}{3000} = \frac{160000}{3000} \approx 53.3\,\text{m/s}^2$

  • Expressed in units of standard gravity ($1\,g = 9.8\,\text{m/s}^2$):     $\text{Acceleration in } g\text{'s} = \frac{53.3\,\text{m/s}^2}{9.8\,\text{m/s}^2/g} \approx 5.44\,g$

Tension Analysis in Horizontal and Conical Systems

• Vector Components of Tension in Horizontal Swings:

  • In a true physical system, a ball swung by a string cannot orbit purely horizontally because gravitational force ($m g$) acts downward.

  • The tension vector ($\mathbf{F}_T$) tilts upward at an angle, resolving into orthogonal components:

  • Vertical component: $T_y = m g$ (supports the gravitational weight).

  • Horizontal component: $T_x = F_c = \frac{m v^2}{r}$ (provides inward centripetal acceleration).

  • Net tension via the Pythagorean theorem:       $T = \sqrt{T_x^2 + T_y^2}$

• Speed-Dependent Tension Regimes ($m = 2\,\text{kg}$, $r = 1\,\text{m}$):

  • High-Speed Case ($v = 20\,\text{m/s}$):     $T_x = \frac{2 \times 20^2}{1} = 800\,\text{N}$     $T_y = 2 \times 9.8 = 19.6\,\text{N}$     $T = \sqrt{800^2 + 19.6^2} = \sqrt{640000 + 384.16} = 800.24\,\text{N} \approx 800\,\text{N}$     At high speeds, $T_x \gg T_y$, so the rope appears almost perfectly horizontal, and $T \approx F_c$.

  • Low-Speed Case ($v = 5\,\text{m/s}$):     $T_x = \frac{2 \times 5^2}{1} = 50\,\text{N}$     $T_y = 19.6\,\text{N}$     $T = \sqrt{50^2 + 19.6^2} = \sqrt{2500 + 384.16} \approx 53.7\,\text{N}$     At low speeds, $T_y$ represents a significant portion of the total force; the string dips significantly below the horizontal plane.

• Conical Pendulum (Tetherball Model):

  • A tetherball of mass $m = 0.25\,\text{kg}$ rotates on a string of length $L = 0.75\,\text{m}$, completing $10\,\text{revolutions}$ in $5\,\text{s}$.

  • Period of revolution:     $T_{period} = \frac{5\,\text{s}}{10\,\text{revolutions}} = 0.5\,\text{s}$

  • Angle $\beta$ defined between the string and the vertical pole yields the geometric relation:     $\sin(\beta) = \frac{r}{L} \implies r = L \sin(\beta)$

  • The horizontal component of tension provides the inward centripetal force:     $F_{Tx} = F_T \sin(\beta) = F_c = m a_c = m \left(\frac{4\pi^2 r}{T_{period}^2}\right)$

  • Substituting $r = L \sin(\beta)$:     $F_T \sin(\beta) = m \left(\frac{4\pi^2 L \sin(\beta)}{T_{period}^2}\right)$

  • Dividing both sides by $\sin(\beta)$ isolates the string tension:     $F_T = \frac{4\pi^2 m L}{T_{period}^2}$

  • Numerical calculation:     $F_T = \frac{4\pi^2 (0.25) (0.75)}{(0.5)^2} = \frac{7.4022}{0.25} = 29.6\,\text{N}$

Dynamics of Vertical Circular Motion

• Point-by-Point Force Balance along a Vertical Track:

  • Bottom Position (Point A):

  • Both tension and the normal force point upward toward the center, while gravity pulls downward.

  • The upward force must balance the downward weight and supply centripetal acceleration:       $T_A - m g = F_c \implies T_A = F_c + m g = \frac{m v^2}{r} + m g$

  • The object experiences maximum tension and feels heaviest at the bottom.

  • Top Position (Point C):

  • Both string tension and gravitational force point downward toward the circle's center.

  • Gravity assists the centripetal acceleration:       $T_C + m g = F_c \implies T_C = F_c - m g = \frac{m v^2}{r} - m g$

  • The object experiences minimum tension and feels lightest at the apex.

  • Side Positions (Points B and D):

  • Gravity acts perpendicular to the radial direction and does not contribute to the radial centripetal acceleration.

  • Tension alone supplies the centripetal force:       $T_{side} \approx F_c = \frac{m v^2}{r}$

• Vertical Circle Numerical Example:

  • A ball of mass $m = 2\,\text{kg}$ orbits vertically on a string with radius $r = 50\,\text{cm} = 0.5\,\text{m}$ at speed $v = 20\,\text{m/s}$.

  • Centripetal force component:     $F_c = \frac{m v^2}{r} = \frac{2 \times 20^2}{0.5} = \frac{800}{0.5} = 1600\,\text{N}$

  • Weight force component:     $W = m g = 2 \times 9.8 = 19.6\,\text{N}$

  • Tension at the apex:     $T_{top} = 1600 - 19.6 = 1580.4\,\text{N}$

  • Tension at the bottom:     $T_{bottom} = 1600 + 19.6 = 1619.6\,\text{N}$

  • Tension at mid-height (sides):     $T_{side} = 1600\,\text{N}$

• Threshold Velocity to Maintain Vertical Circular Motion:

  • To prevent a string from going slack or an object from falling out of a circular path at the apex, tension (or normal force) must remain strictly non-negative ($T \ge 0$).

  • The critical boundary condition occurs at $T = 0$:     $0 = \frac{m v^2}{r} - m g \implies m g = \frac{m v^2}{r}$

  • Canceling mass $m$ and isolating velocity yields the minimum required speed:     $v_{min} = \sqrt{r g}$

  • For a ball on a $1.5\,\text{m}$ string ($m = 0.2\,\text{kg}$):     $v_{min} = \sqrt{1.5 \times 9.8} = \sqrt{14.7} \approx 3.83\,\text{m/s}$

Road Curvature, Banked Tracks, and Loop Mechanics

• Unbanked (Flat) Curves with Static Friction:

  • When a car executes a turn on a level roadway, static friction between the tires and the asphalt provides the centripetal force ($f_s = F_c$).

  • The normal force equals the automobile's weight: $F_N = m g$.

  • The maximum allowable static friction before skidding is:     $f_{s,max} = \mu_s F_N = \mu_s m g$

  • Equating maximum friction to centripetal force:     $\mu_s m g = \frac{m v^2}{r} \implies \mu_s g = \frac{v^2}{r} \implies v_{max} = \sqrt{\mu_s g r}$

  • Calculation for $r = 20\,\text{m}$ and $\mu_s = 0.60$:     $v_{max} = \sqrt{0.60 \times 9.8 \times 20} = \sqrt{117.6} \approx 10.84\,\text{m/s}$

  • Exceeding this velocity causes the vehicle to skid outward.

• Frictionless Banked Curves:

  • On a banked turn inclined at an angle $\theta$, the normal force ($F_N$) tilts toward the center of curvature:

  • Vertical equilibrium: $F_N \cos(\theta) = m g \implies F_N = \frac{m g}{\cos(\theta)}$

  • Horizontal radial acceleration: $F_N \sin(\theta) = F_c = \frac{m v^2}{r}$

  • Dividing the horizontal equation by the vertical equation cancels mass and normal force:     $\frac{F_N \sin(\theta)}{F_N \cos(\theta)} = \frac{m v^2 / r}{m g} \implies \tan(\theta) = \frac{v^2}{r g}$

  • The banking angle required to negotiate the turn with zero friction is:     $\theta = \arctan\left(\frac{v^2}{r g}\right)$

  • Numerical calculation for $r = 50\,\text{m}$ at $v = 40\,\text{m/s}$:     $\theta = \arctan\left(\frac{40^2}{50 \times 9.8}\right) = \arctan\left(\frac{1600}{490}\right) = \arctan(3.2653) \approx 72.97^\circ$

• Traversal of Convex Crests and Concave Dips:

  • Concave Dip (Bottom / Point A):

  • The road surface curves upward. The normal force must overcome gravity and accelerate the car upward:       $F_N - m g = \frac{m v^2}{r} \implies F_N = m g + \frac{m v^2}{r}$

  • Convex Hill Crest (Top / Point B):

  • The road surface curves downward. Gravity acts inward toward the center, while the normal force pushes outward:       $m g - F_N = \frac{m v^2}{r} \implies F_N = m g - \frac{m v^2}{r}$

  • Traversal Example ($m = 1000\,\text{kg}$, $v = 20\,\text{m/s}$, $r = 100\,\text{m}$):

  • Gravitational weight: $m g = 1000 \times 9.8 = 9800\,\text{N}$

  • Centripetal force: $F_c = \frac{1000 \times 20^2}{100} = \frac{400000}{100} = 4000\,\text{N}$

  • Normal force at bottom (Point A): $F_N = 9800 + 4000 = 13800\,\text{N}$

  • Normal force at crest (Point B): $F_N = 9800 - 4000 = 5800\,\text{N}$

  • Maximum Speed Over a Crest Without Becoming Airborne:

  • Loss of physical road contact occurs when $F_N = 0$:       $m g - 0 = \frac{m v^2}{r} \implies v_{max} = \sqrt{r g}$

  • Calculation for $r = 100\,\text{m}$:       $v_{max} = \sqrt{100 \times 9.8} = \sqrt{980} \approx 31.3\,\text{m/s}$

• Inverted Roller Coaster Loop Mechanics:

  • For a coaster car at the inverted top of a loop of radius $r$, both track normal force and gravity act downward toward the center of curvature:     $F_N + m g = \frac{m v^2}{r} \implies F_N = \frac{m v^2}{r} - m g$

  • The minimum speed required to maintain track contact ($F_N = 0$) is:     $v_{min} = \sqrt{r g}$

  • Numerical calculation for $r = 15\,\text{m}$:     $v_{min} = \sqrt{15 \times 9.8} = \sqrt{147} \approx 12.1\,\text{m/s}$

Non-Uniform Circular Motion and Vector Kinematics

• Orthogonal Acceleration Components:

  • Radial (Centripetal) Acceleration ($a_r$): Accounts for changes in the direction of the velocity vector, pointing perpendicular to the path toward the center of curvature:     $a_r = \frac{v^2}{r}$

  • Tangential Acceleration ($a_t$): Accounts for changes in the linear speed of the object along the tangent to the curve:     $a_t = \frac{v_f - v_i}{t}$

  • Net Vector Acceleration ($a_{net}$):     $a_{net} = \sqrt{a_t^2 + a_r^2}$

• Vehicle Accelerating on a Circular Curve:

  • A car uniformly accelerates from rest ($v_i = 0$) to $v_f = 40\,\text{m/s}$ in $t = 8\,\text{s}$ along a circular track of radius $r = 600\,\text{m}$.

  • Tangential acceleration component:     $a_t = \frac{40 - 0}{8} = 5\,\text{m/s}^2$

  • Radial centripetal acceleration at $v = 40\,\text{m/s}$:     $a_r = \frac{40^2}{600} = \frac{1600}{600} \approx 2.67\,\text{m/s}^2$

  • Magnitude of net acceleration:     $a_{net} = \sqrt{5^2 + 2.67^2} = \sqrt{25 + 7.1289} = \sqrt{32.1289} \approx 5.67\,\text{m/s}^2$

Universal Gravitation and Planetary Gravitational Fields

• Newton's Law of Universal Gravitation:

  • The mutual attractive gravitational force between two point masses ($m_1$ and $m_2$) separated by a center-to-center distance $r$ is:     $F_g = \frac{G m_1 m_2}{r^2}$     where $G = 6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2$ is the universal gravitational constant.

• Gravitational Attraction between the Earth and Sun:

  • Given: $M_{Sun} = 2 \times 10^{30}\,\text{kg}$, $M_{Earth} = 5.98 \times 10^{24}\,\text{kg}$, $r = 1.5 \times 10^{11}\,\text{m}$.

  • Gravitational force:     $F_g = \frac{(6.67 \times 10^{-11}) (2 \times 10^{30}) (5.98 \times 10^{24})}{(1.5 \times 10^{11})^2} = \frac{7.97732 \times 10^{44}}{2.25 \times 10^{22}} \approx 3.55 \times 10^{22}\,\text{N}$

• Gravitational Scaling Proportionalities:

  • Doubling the Earth's mass doubles $F_g$ ($2\times$).

  • Tripling the Sun's mass triples $F_g$ ($3\times$).

  • Doubling the separation distance decreases $F_g$ by a factor of 4 ($\frac{1}{2^2} = \frac{1}{4}\times$).

  • Tripling the separation distance decreases $F_g$ by a factor of 9 ($\frac{1}{3^2} = \frac{1}{9}\times$).

  • Tripling Earth's mass ($3$), quadrupling the Sun's mass ($4$), and halving the separation distance ($\frac{1}{2}$):     $\text{Factor} = \frac{3 \times 4}{(1/2)^2} = \frac{12}{1/4} = 48\times$

  • Reducing distance by one-third ($\frac{1}{3}r$) increases force by a factor of 9 ($\frac{1}{(1/3)^2} = 9\times$).

• Surface Gravitational Field Strength ($g$):

  • Equating the gravitational force on a test mass $m$ to its local weight $m g$:     $m g = \frac{G M m}{r^2} \implies g = \frac{G M}{r^2}$

  • Earth's surface value ($M_E = 5.98 \times 10^{24}\,\text{kg}$, $r_E = 6.38 \times 10^6\,\text{m}$):     $g_E = \frac{(6.67 \times 10^{-11}) (5.98 \times 10^{24})}{(6.38 \times 10^6)^2} = \frac{3.98866 \times 10^{14}}{4.07044 \times 10^{13}} \approx 9.799 \approx 9.8\,\text{m/s}^2$

  • Moon's surface value:     $g_{Moon} \approx 1.62\,\text{m/s}^2$

• Gravitational Acceleration at High Altitude:

  • At an altitude of $h = 4000\,\text{km} = 4 \times 10^6\,\text{m}$ above Earth's surface:     $r = r_E + h = 6.38 \times 10^6\,\text{m} + 4.0 \times 10^6\,\text{m} = 10.38 \times 10^6\,\text{m} = 1.038 \times 10^7\,\text{m}$

  • Local acceleration:     $g = \frac{(6.67 \times 10^{-11}) (5.98 \times 10^{24})}{(1.038 \times 10^7)^2} = \frac{3.98866 \times 10^{14}}{1.07744 \times 10^{14}} \approx 3.7\,\text{m/s}^2$

• Planetary Scaling Scenarios for $g$:

  • Doubling the planet's mass ($2M$) doubles surface gravity:     $g = 2 \times 9.8 = 19.6\,\text{m/s}^2$

  • Tripling the planet's radius ($3r$) reduces surface gravity by a factor of 9:     $g = \frac{9.8}{3^2} = \frac{9.8}{9} \approx 1.09\,\text{m/s}^2$

  • Halving the planet's radius ($\frac{1}{2}r$) increases surface gravity four-fold:     $g = \frac{9.8}{(1/2)^2} = 9.8 \times 4 = 39.2\,\text{m/s}^2$

Satellite Dynamics and Orbital Physics

• Derivation of Orbital Velocity and Period:

  • The inward gravitational pull on a satellite of mass $m_s$ orbiting a central body of mass $M$ provides the centripetal force:     $\frac{G M m_s}{r^2} = \frac{m_s v^2}{r}$

  • Dividing out satellite mass $m_s$ and one factor of radius $r$ gives orbital speed:     $v = \sqrt{\frac{G M}{r}}$

  • Using $v = \frac{2\pi r}{T}$, the orbital period is:     $T = \frac{2\pi r}{v}$

• Orbit Calculations at $3000\,\text{km}$ Altitude:

  • Altitude: $h = 3000\,\text{km} = 3 \times 10^6\,\text{m}$.

  • Radial distance: $r = 6.38 \times 10^6 + 3 \times 10^6 = 9.38 \times 10^6\,\text{m}$.

  • Orbital speed:     $v = \sqrt{\frac{(6.67 \times 10^{-11}) (5.98 \times 10^{24})}{9.38 \times 10^6}} = \sqrt{\frac{3.98866 \times 10^{14}}{9.38 \times 10^6}} = \sqrt{4.2523 \times 10^7} \approx 6521\,\text{m/s}$

  • Orbital period in seconds:     $T = \frac{2\pi (9.38 \times 10^6\,\text{m})}{6521\,\text{m/s}} = \frac{58936279}{6521} \approx 9037.9\,\text{s}$

  • Orbital period in hours:     $T = \frac{9037.9\,\text{s}}{3600\,\text{s/hour}} \approx 2.51\,\text{hours}$

• Calculation of the Sun's Mass from Earth's Orbital Parameters:

  • Earth orbits the Sun at $r = 1.5 \times 10^{11}\,\text{m}$.

  • Setting gravitational force equal to centripetal force using period:     $\frac{G M_{Sun} M_{Earth}}{r^2} = M_{Earth} \left(\frac{4\pi^2 r}{T^2}\right) \implies M_{Sun} = \frac{4\pi^2 r^3}{G T^2}$

  • Period in seconds ($1\,\text{year}$):     $T = 365 \times 24 \times 60 \times 60 = 31536000\,\text{s}$

  • Sun's mass calculation:     $M_{Sun} = \frac{4\pi^2 (1.5 \times 10^{11})^3}{(6.67 \times 10^{-11}) (31536000)^2} = \frac{4\pi^2 (3.375 \times 10^{33})}{(6.67 \times 10^{-11}) (9.9452 \times 10^{14})} = \frac{1.33239 \times 10^{35}}{6.63345 \times 10^4} \approx 2.01 \times 10^{30}\,\text{kg}$

• Analysis of Geosynchronous Satellites:

  • A geosynchronous satellite remains fixed over a single equatorial point, matching Earth's rotational period of one sidereal day:     $T = 1\,\text{day} = 24 \times 3600 = 86400\,\text{s}$

  • Derivation of orbital radius ($r$):     $\frac{G M_E}{r^2} = \frac{4\pi^2 r}{T^2} \implies r^3 = \frac{G M_E T^2}{4\pi^2} \implies r = \sqrt[3]{\frac{G M_E T^2}{4\pi^2}}$

  • Radial distance calculation:     $r = \sqrt[3]{\frac{(6.67 \times 10^{-11}) (5.98 \times 10^{24}) (86400)^2}{4\pi^2}} = \sqrt[3]{\frac{2.9775 \times 10^{24}}{39.4784}} = \sqrt[3]{7.5421 \times 10^{22}} \approx 42250474\,\text{m} \approx 4.225 \times 10^7\,\text{m}$

  • Height ($h$) above Earth's surface:     $h = r - r_E = 4.225 \times 10^7\,\text{m} - 6.38 \times 10^6\,\text{m} = 3.587 \times 10^7\,\text{m} = 35870\,\text{km}$

  • Orbital speed:     $v = \sqrt{\frac{G M_E}{r}} = \sqrt{\frac{(6.67 \times 10^{-11}) (5.98 \times 10^{24})}{4.225 \times 10^7}} = \sqrt{\frac{3.98866 \times 10^{14}}{4.225 \times 10^7}} \approx 3072.6\,\text{m/s}$

Kepler's Third Law of Planetary Motion

• Statement of the Law:

  • The ratio of the squares of the orbital periods ($T$) of any two bodies orbiting the same central mass is directly proportional to the ratio of the cubes of their mean orbital radii ($r$):     $\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{r_1}{r_2}\right)^3$

• Determining the Orbital Period of Mars:

  • Given: $r_E = 1.5 \times 10^{11}\,\text{m}$, $r_M = 2.28 \times 10^{11}\,\text{m}$, and $T_E = 365\,\text{Earth days}$.

  • Setting up the ratio:     $\left(\frac{T_M}{365}\right)^2 = \left(\frac{2.28 \times 10^{11}}{1.5 \times 10^{11}}\right)^3 = (1.52)^3 = 3.51181$

  • Solving for $T_M$:     $T_M^2 = 365^2 \times 3.51181 = 133225 \times 3.51181 = 467860.887$     $T_M = \sqrt{467860.887} \approx 684\,\text{Earth days}$

• Determining the Orbital Radius of Mercury:

  • Given: $T_{Mercury} = 88\,\text{Earth days}$, $T_E = 365\,\text{Earth days}$, and $r_E = 1.5 \times 10^{11}\,\text{m}$.

  • Setting up the ratio:     $\left(\frac{88}{365}\right)^2 = \left(\frac{r_{Mercury}}{1.5 \times 10^{11}}\right)^3$

  • Cross-multiplying:     $r_{Mercury}^3 \times 365^2 = (1.5 \times 10^{11})^3 \times 88^2$     $r_{Mercury}^3 \times 133225 = (3.375 \times 10^{33}) \times 7744 = 2.6136 \times 10^{37}$     $r_{Mercury}^3 = \frac{2.6136 \times 10^{37}}{133225} = 1.9618 \times 10^{32}$

  • Taking the cube root:     $r_{Mercury} = (1.9618 \times 10^{32})^{1/3} \approx 5.8 \times 10^{10}\,\text{m}$

Apparent Weight and Elevator Dynamics

• Operational Definition of Apparent Weight:

  • A weighing scale measures the contact normal force ($F_N$) it exerts against an object, not the actual force of gravity ($m g$).

  • Formulating the net vertical force equation via Newton's Second Law:     $F_{net} = F_N - m g = m a \implies F_N = m (g + a)$

• Dynamic Cases for a Person of Mass $m = 50\,\text{kg}$ ($W = 50 \times 9.8 = 490\,\text{N}$):

  • Elevator at Rest ($a = 0\,\text{m/s}^2$):     $F_N = 50 (9.8 + 0) = 490\,\text{N}$

  • Elevator Ascending at Constant Velocity ($a = 0\,\text{m/s}^2$):     $F_N = 490\,\text{N}$     Because acceleration is zero, constant velocity causes no sensation of weight change.

  • Elevator Descending at Constant Velocity ($a = 0\,\text{m/s}^2$):     $F_N = 490\,\text{N}$     Zero acceleration results in a scale reading equal to the true gravitational weight.

  • Elevator Accelerating Upward ($a = +2\,\text{m/s}^2$):     $F_N = 50 (9.8 + 2) = 50 (11.8) = 590\,\text{N}$     The scale reading exceeds true weight because the floor must support the mass and provide upward acceleration.

  • Elevator Accelerating Downward ($a = -5\,\text{m/s}^2$):     $F_N = 50 (9.8 - 5) = 50 (4.8) = 240\,\text{N}$     The scale reading drops below true weight as the floor drops away, reducing the normal contact force.