Gravitation, Satellite Motion, and Rotational Kinematics Guide

Universal Law of Gravitation and Weight Calculations

• Inverse Square Law of Gravity: The gravitational force ($F_g$) between two objects is inversely proportional to the square of the distance between their centers ($r$). Mathematically, $F = \frac{G m_1 m_2}{r^2}$. This implies that if the distance from the center of Earth doubles ($r = 2R_{earth}$), the weight of an object decreases to one-quarter ($1/4$) of its weight at the surface.
• Acceleration due to Gravity ($g$) on Various Planets:     * The formula for acceleration due to gravity is $g = \frac{G M}{R^2}$, where $G = 6.67 \times 10^{-11}\,N \cdot m^2/kg^2$, $M$ is the planet's mass, and $R$ is its radius.     * Hypothetical Planet Scenarios:         * If a planet has half the mass of Earth and twice the radius ($M_p = 1/2 M_E$, $R_p = 2R_E$), the acceleration due to gravity is $g/8$.         * If a planet has one-third the mass and one-third the radius of Earth ($M_p = 1/3 M_E$, $R_p = 1/3 R_E$), the gravity $g$ is $3g$.         * If two planets have the same surface gravity but Planet B has twice the radius of Planet A, Planet B must have four times the mass ($4m$).         * If Planet A has twice the mass of Planet B but the radii are unknown, the acceleration due to gravity at the surface cannot be determined without the radius values.
• Gravity at Specific Altitudes:     * At an altitude equal to the radius of the Earth ($r = 2R_{earth}$), the acceleration due to gravity is $g/4$.     * To find the height where gravity is a specific fraction of surface gravity (e.g., $0.54g$), use the ratio $\frac{g_{alt}}{g_{surface}} = \left(\frac{R_E}{R_E + h}\right)^2$. For Earth ($R_E = 6.38 \times 10^6\,m$, $M_E = 5.97 \times 10^{24}\,kg$), an acceleration of $0.54g$ occurs at an altitude of approximately $2300\,km$.     * An acceleration of $7.8\,m/s^2$ on Earth occurs at an altitude of approximately $770\,km$.
• Gravitational Force Examples:     * The gravitational force between two $59\,kg$ people standing $2.0\,m$ apart is approximately $5.8 \times 10^{-8}\,N$.     * The force on a $70\,kg$ person due to the Moon ($M = 7.35 \times 10^{22}\,kg$, distance $r = 3.82 \times 10^8\,m$) is $0.0024\,N$.     * Equilibrium Force Point: The point between Earth ($M_E = 5.97 \times 10^{24}\,kg$) and the Moon ($M_m = 7.35 \times 10^{22}\,kg$) where gravitational forces cancel is located approximately $3.84 \times 10^7\,m$ from the center of the Moon.

Satellite Motion and Kepler's Laws

• Orbital Speed and Mass Independence: The speed ($v$) of a satellite in a circular orbit depends only on the mass of the central body ($M$) and the orbital radius ($r$), not the mass of the satellite itself. The formula is $v = \sqrt{\frac{GM}{r}}$. Therefore, if Satellite A has twice the mass of Satellite B but orbits at the same distance, their speeds are equal.
• Orbital Period ($T$): Defined by Kepler's Third Law as $T^2 = \frac{4 \pi^2 r^3}{GM}$. This means $T^2 \propto r^3$. If the Sun's mass quadrupled, the Earth's orbital period would reduce to $1/2$ of its current length (approx. $6$ months).
• Kepler's Three Laws:     1. First Law: Planets move in elliptical orbits with the Sun at one focus.     2. Second Law: A line joining a planet and the Sun sweeps out equal areas during equal intervals of time, meaning orbital speed increases as the planet nears the Sun (perihelion) and decreases as it moves away (aphelion).     3. Third Law: The ratio $\frac{T^2}{r^3}$ is constant for all objects orbiting the same central mass.
• Weightlessness in Space: Astronauts feel weightless not because gravity is absent, but because they and their satellite are in a state of continuous free-fall toward Earth.
• Orbital Examples:     * International Space Station: At an altitude of $370\,km$, the period is approximately $5.5 \times 10^3\,s$ and the speed is $7.7 \times 10^3\,m/s$.     * A satellite revolving $5$ times a day around Earth requires an orbital radius of $1.44 \times 10^7\,m$.     * Jupiter-Earth Comparison: Using Kepler's Third Law $\frac{T_J^2}{R_J^3} = \frac{T_E^2}{R_E^3}$, given Earth is $1.50 \times 10^{11}\,m$ from the Sun and Jupiter is $7.78 \times 10^{11}\,m$, Jupiter's orbital period is approximately $12$ years.

Escape Speed and Potential Energy

• Escape Speed ($v_{esc}$): The minimum speed required for an object to escape the gravitational pull of a celestial body without further propulsion. It is calculated as $v_{esc} = \sqrt{\frac{2GM}{R}}$.     * Escape speed is independent of the mass of the escaping object; a $20\,N$ rocket and a $20,000\,N$ rocket have the same escape speed from the same planet.     * If a planet has the same mass/radius ratio as Earth ($M/R$ is the same, e.g., Planet X-39 with $1/3$ mass and $1/3$ radius), the escape speed is equal to Earth's escape speed.
• Gravitational Potential Energy ($U$): Defined as $U = -\frac{G m_1 m_2}{r}$. For a system of multiple masses, the total potential energy is the sum of the potential energies of every pair of masses.     * Energy Requirements for Orbit Changes: To move an $825\,kg$ vehicle from an orbit of radius $10,300\,km$ to $17,400\,km$, the energy required is the difference in total mechanical energy ($E = -\frac{GMm}{2r}$), resulting in approximately $1.30 \times 10^{10}\,J$.     * Impact Velocity: An asteroid falling from a very great distance hits Earth with a speed related to the conservation of energy. For an asteroid starting with $6500\,m/s$ far away, it hits Earth at approximately $12,900\,m/s$.

Uniform Circular Motion and Banked Curves

• Centripetal Force and Acceleration: In uniform circular motion, acceleration ($a_c = \frac{v^2}{r}$) is always directed toward the center of the circular path. The force causing this acceleration is the centripetal force ($F_c = \frac{mv^2}{r}$).
• Friction on Curves: On an unbanked (horizontal) curve, static friction between the tires and the road provides the centripetal force ($f_s = \frac{mv^2}{r}$). The maximum safe speed is $v_{max} = \sqrt{\mu_s g r}$.     * If a car hits ice (where friction becomes zero), it moves in a straight-line path in its original direction (tangent to the circle) due to inertia.
• Vertical Circular Motion:     * Loop-the-loop: The minimum speed at the top to prevent falling out (assuming no straps) is $v = \sqrt{gr}$. For a $13.2\,m$ radius loop, this is $11.4\,m/s$.     * Vertical Bumps and Depressions: At the top of a bump, the normal force is $N = mg - \frac{mv^2}{r}$. At the bottom of a depression, it is $N = mg + \frac{mv^2}{r}$.
• Banked Curves: Curves are banked at an angle $\theta$ to allow vehicles to navigate them without relying on friction. The relationship is $\tan(\theta) = \frac{v^2}{rg}$.     * For a radius of $110\,m$ and speed of $24.5\,m/s$, the proper angle is $29.1^{\circ}$.     * The safe speed on a banked curve is independent of the vehicle's mass and the quality of its tires (traction).
• Artificial Gravity: Space stations can simulate gravity by rotating. A cylinder of diameter $380\,m$ needs a rotation period of $28\,s$ to provide normal gravity ($9.8\,m/s^2$).

Rotational Kinematics and Angular Quantities

• Relationship between Angular and Linear Variables:     * Angular displacement: $\theta$     * Angular speed: $\omega = \frac{v}{r}$     * Tangential acceleration: $a_t = r \alpha$     * Radial (Centripetal) acceleration: $a_r = r \omega^2$
• Rigid Body Rotation: For a rigid object rotating about a fixed axis, all points on the object have the same angular speed ($\omega$) and angular acceleration ($\alpha$). However, points farther from the axis have greater tangential speeds ($v$) and radial accelerations ($a_r$).
• Rotational Kinematics Equations (Constant $\alpha$):     1. $\omega = \omega_0 + \alpha t$     2. $\Delta \theta = \omega_0 t + \frac{1}{2} \alpha t^2$     3. $\omega^2 = \omega_0^2 + 2 \alpha \Delta \theta$
• Calculations and Unit Conversions:     * To convert RPM to $rad/s$: $\omega = \text{RPM} \times \frac{2 \pi}{60}$. Example: $813.0\,rpm = 85.14\,rad/s$.     * A centrifuge slowing from $3600\,rev/min$ to rest over $50$ turns has an angular acceleration magnitude of $230\,rad/s^2$ and takes $1.7\,s$ to stop.     * A wheel accelerating from rest ($\omega_0 = 0$) at $4.5\,rad/s^2$ for $2.0\,s$ turns through an angle of $11\,rad$ and reaches a speed of $10\,rad/s$.