Circular Motion of Planets and Satellites PDF

CIRCULAR MOTION OF PLANETS AND SATELLITES INTRODUCTION The motion of planets around the Sun and satellites around planets is a fundamental aspect of celestial mechanics. This motion can often be approximated as circular due to the relatively stable distances involved. Understanding this motion invo...

CIRCULAR MOTION OF PLANETS AND SATELLITES INTRODUCTION The motion of planets around the Sun and satellites around planets is a fundamental aspect of celestial mechanics. This motion can often be approximated as circular due to the relatively stable distances involved. Understanding this motion involves concepts from Newtonian physics, particularly gravitation and centripetal force. Circular Motion Circular motion refers to the movement of an object along the circumference of a circle. It can be uniform (constant speed) or non-uniform (changing speed). Centripetal Force For an object moving in a circle, a net inward force is required to keep it in that circular path. This force is known as centripetal force and is directed towards the center of the circle. Gravitational Force Newton’s Law of Universal Gravitation: The gravitational force between two masses and the distance between the two mass is given by: 𝑮𝒎𝟏 𝒎𝟐 𝑭= 𝒓𝟐 Where: F = gravitational force G = gravitational constant (6.674×10-11 Nm2/kg2) M1 and M2= masses of the two objects r = distance between the centers of the two masses Newton’s law of universal gravitation states: The force of attraction between two bodies (masses) is directly proportional to the product of the masses and inversely proportional to the square of the distance between them Example 1: Gravitational Force Between Earth and a Person Calculate the gravitational force of attraction between the Earth and a 70 kg person standing at sea level, at a distance of 6.38 × 106 𝑚 from the Earth's center. Given: Mass of the Earth, m1 = 5.98 × 1024 𝑘𝑔 Mass of the person, m2= 70kg Distance from the center of the Earth, 𝑟 = 6.38 × 106 𝑚 Gravitational constant, 𝐺 = 6.673 × 10−11 𝑁𝑚2 𝑘𝑔−2 Solution: Using the formula for gravitational force: 𝑮𝒎𝟏 𝒎𝟐 𝑭= 𝒓𝟐 Calculating: 6.673 × 10−11 × 5.98 × 1024 × 70 𝑭= (6.38 × 106 )𝟐 𝑭 =? ? ? ? ? Example 2: Gravitational Force Between Earth and Moon Calculate the gravitational force between the Earth and the Moon. Given: - Mass of the Earth, m1 = 5.97×10^24kg - Mass of the Moon, m2= 7.35×1022 kg - Distance between Earth and Moon, r = 3.844×10^8 m Solution: Using the gravitational force formula: 𝑮𝒎𝟏 𝒎𝟐 𝑭= 𝒓𝟐 Substituting values: 6.673 × 10−11 × 5.98 × 1024 × 7.35 × 1022 𝑭= (3.844 × 108 )𝟐 Calculating: 𝑭 =? ? ? ? ? ? ? 𝑵 Example 3: Gravitational Force on a Falling Object What is the gravitational pull that the Earth exerts on a ball of mass 0.3 kg lying on its surface? Given: Mass of the Earth, M1 =5.97×10^24kg Mass of the ball, M2 = 0.3kg Radius of Earth, r = 6378km = 6378000 m Solution: Using the gravitational force formula: 𝑮𝒎𝟏 𝒎𝟐 𝑭= 𝒓𝟐 Substituting values: 6.673×10−11 ×5.97×1024 ×0.3 𝐹= (6378000)2  𝑭 =? ? ? ? ? ? 𝑵 Example 4: Gravitational Force Between Two Small Masses Calculate the gravitational force between two small masses: one mass is 9.1 × 10−31 𝑘𝑔 (electron), and another is 1.67 × 10−27 𝑘𝑔 (proton), at a distance of 1 × 1010 𝑚 Given: - Mass of electron, M1 = 9.1 × 10−31 𝑘𝑔 - Mass of proton, M2 = 1.67 × 10−27 𝑘𝑔 - Distance, r = 1 × 1010 𝑚 Solution: Using the gravitational force formula: 𝑮𝒎𝟏 𝒎𝟐 𝑭= 𝒓𝟐 𝐹 = 1.013 × 10−47 𝑁 Example 5: Effect of Doubling Distance on Gravitational Force Suppose two objects attract each other with a gravitational force of 16 units at a certain distance apart; what will be the new force if the distance is doubled? Given: - Initial force, Fi =16units. - New distance is double. Solution: According to Newton's Law of Universal Gravitation, if you double the distance (d), then: The new force is given by:  𝐹𝑖 𝐹 16 𝐹𝑓 = ⁄(22 ) = 𝑖⁄4 = = 4𝑢𝑛𝑖𝑡𝑠 4 CONCLUSION These examples illustrate how to apply Newton’s Law of Universal Gravitation in various contexts—from calculating forces between celestial bodies to understanding how distance affects gravitational attraction between smaller masses or objects on Earth’s surface. MOTION OF PLANETS Kepler's Laws of Planetary Motion Johannes Kepler formulated three laws that describe planetary motion: 1. Kepler's First Law (Law of Ellipses): The orbit of a planet around the Sun is an ellipse with the Sun at one focus. 2. Kepler's Second Law (Law of Equal Areas): A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time, meaning that planets move faster when they are closer to the Sun.  3. Kepler's Third Law (Harmonic Law): The square of the orbital period T of a planet is directly proportional to the cube of the semi-major axis r of its orbit: 𝑇2 ∝ 𝑟3 Circular Orbits For planets in nearly circular orbits, the radius r can be approximated as constant, and we can use the following relationships: 1. Centripetal Force: The gravitational force provides the necessary centripetal force for circular motion: 𝑭𝒈𝒓𝒂𝒗𝒊𝒕𝒚 = 𝑭𝒄𝒆𝒏𝒕𝒓𝒊𝒑𝒆𝒕𝒂𝒍 Thus, from centripetal acceleration: 𝒗𝟐 𝒂𝒄 = 𝒓 𝒎 𝒗𝟐 𝑭𝒄𝒆𝒏𝒕𝒓𝒊𝒑𝒆𝒕𝒂𝒍 = 𝑴 × 𝒂𝑪 = 𝒓 𝑮 𝑴 𝟏 𝒎𝟐 𝑭𝒈𝒓𝒂𝒗𝒊𝒕𝒚 = 𝒓𝟐 𝑭𝒈𝒓𝒂𝒗𝒊𝒕𝒚 = 𝑭𝒄𝒆𝒏𝒕𝒓𝒊𝒑𝒆𝒕𝒂𝒍  𝑮 𝑴𝟏 𝒎𝟐 𝒎 𝒗𝟐 = 𝒓𝟐 𝒓 M = mass of the Sun (or central body) m = mass of the planet v = orbital speed of the plane MOTION OF SATELLITES The motion of satellites is a fascinating aspect of physics, governed by Newton's laws of motion and universal gravitation. Types of Satellites Natural Satellites: Moons that orbit planets. Artificial Satellites: Man-made objects placed into orbit for various purposes (communication, weather monitoring, etc.). Types of Satellite Orbits Satellite orbits can be categorized based on their altitude and purpose: Low Earth Orbit (LEO): Altitude: Approximately 160 to 2,000 km above Earth's surface. Purpose: Imaging, communications, scientific research. Characteristics: Short orbital periods (about 90-120 minutes). Medium Earth Orbit (MEO): Altitude: Approximately 2,000 to 35,786 km. Purpose: Navigation (e.g., GPS). Characteristics: Longer orbital periods than LEO. Geostationary Orbit (GEO):  Altitude: Approximately 35,786 km above Earth's equator. Purpose: Weather forecasting, telecommunications. Characteristics: Remains stationary relative to a point on Earth's surface. Polar Orbit: Passes over both poles, allowing coverage of the entire Earth as it rotates. Sun-Synchronous Orbit (SSO): A near-polar orbit that maintains consistent solar lighting conditions for imaging. ESCAPE VELOCITY Escape velocity is defined as the minimum speed an object must reach to break free from the gravitational influence of a celestial body, such as a planet or moon, without any further propulsion. This velocity ensures that the object can move away from the body indefinitely, overcoming its gravitational pull. The escape velocity (ve ) can be derived from the principles of energy conservation in a gravitational field. The formula is given by: 2GM ve = r where: G is the gravitational constant, M is the mass of the celestial body, r is the distance from the center of the body to the point of escape (usually taken as the radius when considering surface escape). Key Characteristics Independence from Mass: The escape velocity is independent of the mass of the object attempting to escape. This means that whether a small rocket or a large spacecraft is launched, they must achieve the same escape velocity to leave Earth's gravitational influence. Effect of Altitude: Escape velocity decreases with altitude. As an object moves away from a celestial body, the gravitational pull weakens, resulting in a lower required speed for escape. Comparison with Orbital Velocity: Escape velocity is approximately 1.414 times greater than the orbital velocity required to maintain a stable orbit at the same altitude. This relationship highlights that achieving escape velocity allows an object to not only orbit but also leave the gravitational influence entirely. Examples 1. Earth: The escape velocity at Earth's surface is approximately 11.2km/s (or about 25,000 mp/h). This means any spacecraft must reach this speed to leave Earth without falling back. 2. Moon: For the Moon, which has a weaker gravitational field, the escape velocity is about 2.4 km/s (or roughly 5,400 mp/h) at its surface. 3. Jupiter: For Jupiter, with its massive size and strong gravity, the escape velocity is significantly higher at approximately 60km/s. Practical Implications Understanding escape velocity is crucial for space exploration and satellite deployment. Rockets must be designed to reach these speeds to ensure successful missions beyond Earth's atmosphere. Additionally, celestial bodies with low escape velocities may struggle to retain atmospheres since gas molecules can easily exceed this speed and drift into space. In summary, escape velocity serves as a fundamental concept in astrodynamics and space exploration, determining how and when objects can leave a planet's gravitational influence effectively. Worked Examples of Escape Velocity Here are a few worked examples to illustrate how to calculate escape velocity for different celestial bodies using the formula: Example 1: Escape Velocity from Earth Given: - Mass of Earth (M) = 5.972 10^24 kg - Radius of Earth (r) = 6.371 10^6 m - Gravitational constant (G) = 6.674 10^-11m^3kg^-1s^-2 Calculation: Using the escape velocity formula:  2GM ve = r Substituting the values: Result: The escape velocity from Earth is approximately 11.2 km/s. Example 2: Escape Velocity from the Moon Given: - Mass of Moon (M) = 7.34767309 *10^22 kg - Radius of Moon (r) = 1.7371 *10^6 m Calculation: Using the escape velocity formula: 2GM ve = r Substituting the values: Result: The escape velocity from the Moon is approximately 2.38 km/s. Example 3: Escape Velocity from Mars Given: - Mass of Mars (M) = 6.4171 × 10^23 kg - Radius of Mars (r) = 3.3895 × 10^6 m Calculation: Using the escape velocity formula: 2GM ve = r Substituting the values: The escape velocity from Mars is approximately 5 km/s. TORQUE Torque is a measure of the force that causes an object to rotate around an axis. It can be thought of as the "twisting force" applied to an object. It is the tendency of a force to produce rotational motion about a pivot point or axis Torque, often referred to as the moment of force, is a measure of the rotational force applied to an object. It determines how effectively a force can cause an object to rotate around a pivot point or axis. Mathematically, torque (  ) is defined as:  = F  r sin  Where:  = Torque F = Magnitude of the applied force r= Distance from the pivot point to the point where the force is applied (also known as the moment arm) sin  = Angle between the force vector and the line from the pivot point to the point of application Units of Torque The SI unit of torque is the Newton-meter (N·m), while in some contexts, it may also be expressed in pound-feet (lb-ft). Importance of Torque Torque plays a crucial role in various mechanical systems and applications. It influences rotational motion and is foundational in understanding how forces cause objects to spin or rotate. Applications of Torque Torque has numerous real-life applications across different fields: 1. Automotive Industry: In vehicles, torque is essential for engine performance. It is responsible for acceleration and moving the vehicle forward. The engine produces torque that is transmitted through the drive train to the wheels, enabling movement. 2. Construction and Tools: Tools like torque wrenches apply a specific amount of torque to bolts and screws, ensuring they are tightened to precise specifications without over-tightening, which could lead to failure. 3. Sports: Athletes use torque in sports like golf, baseball, and tennis. For instance, a golfer generates torque through body rotation to hit the ball with more power. 4. Robotics: In robotic systems, motors generate torque for movement in joints and limbs, allowing for precise control and functionality. Torque is a fundamental concept in physics and engineering that describes how forces cause rotational motion around an axis. Its applications range from everyday tasks to complex machinery in various industries. Understanding torque allows for better design and functionality in mechanical systems, enhancing efficiency and performance across numerous fields. ROTATIONAL EQUILIBRIUM Rotational equilibrium occurs when the net torque acting on an object is zero. This means that the sum of all torques acting about a pivot point must balance out, resulting in no angular acceleration. An object in rotational equilibrium can either be at rest or rotating at a constant angular velocity.  DEFINITION OF CENTRIPETAL ACCELERATION Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is directed toward the center of the circle around which the object is moving. This acceleration is necessary for maintaining circular motion, as it changes the direction of the object's velocity without necessarily changing its speed. CENTRIPETAL ACCELERATION The magnitude of centripetal acceleration (ac) can be expressed using two primary equations: 1. In terms of linear velocity: 𝒗𝟐 𝒂𝒄 = 𝒓 where: 𝑎𝑐 = centripetal acceleration (m/s²) 𝑣 2 = linear velocity (m/s) 𝑟 = radius of the circular path (m) 2. In terms of angular velocity: 𝒂𝒄 = 𝒓𝝎𝟐 where: ω = angular velocity (rad/s) Cause of Centripetal Acceleration Centripetal acceleration occurs due to a net inward force acting on the object, which is often referred to as centripetal force. This force can arise from various sources, such as tension in a string, gravitational force, or frictional force, depending on the context of the circular motion. Characteristics of Centripetal Acceleration Direction: Always points toward the center of the circular path. Magnitude: Increases with higher speeds and smaller radii. Relation to Velocity: Even if an object moves at a constant speed in a circular path, its velocity is continuously changing due to its direction change, hence it experiences acceleration. Worked Examples Example 1: Calculating Centripetal Acceleration CHALLENGE: A car is moving around a circular track with a radius of 50 meters at a speed of 20 meters per second. What is centripetal acceleration? Solution: Using the formula: 𝒗𝟐 𝒂𝒄 = 𝒓 Substituting the values: (20𝑚𝑠) × (20𝑚𝑠) 400 𝑎𝑐 = = = 8𝑚𝑠 2 50𝑚 50 Example 2: Finding Radius from Centripetal Acceleration CHALLENGE: An object has a centripetal acceleration of 10 m/s² while moving at a speed of 5 m/s. What is the radius of its circular path? Solution: Using the formula: 𝑣2 𝑎𝑐 = 𝑟 Rearranging to find r: 𝒗𝟐 𝒓= 𝒂𝒄 Substituting the values: (5𝑚𝑠) × (5𝑚𝑠) 𝑟= = 2.5𝑚 10𝑚𝑠 2 Example 3: Angular Velocity and Centripetal Acceleration Challenge: A rotating wheel has an angular velocity of 4 rad/s and a radius of 0.75 m. Calculate its centripetal acceleration. Solution: Using the formula: 𝒂𝒄 = 𝒓𝝎𝟐 Substituting the values: 𝑎𝑐 = 0.75𝑚(4𝑟𝑎𝑑𝑠)(𝑟𝑎𝑑𝑠) = 0.75(16) = 12𝑚𝑠 2 Centripetal acceleration is crucial for understanding motion in circular paths. It highlights how objects maintain their trajectory when moving in curves and emphasizes that even constant-speed motion requires continuous inward acceleration due to changing direction. The formulas provided allow for calculating centripetal acceleration based on either linear or angular velocities, making it applicable in various physical scenarios ranging from vehicles on curved roads to satellites in orbit. Summary Centripetal acceleration always points toward the center of rotation. It is proportional to the square of velocity and inversely proportional to the radius. The net force causing centripetal acceleration must act inward; otherwise, an object will move off in a straight line due to inertia. ANGULAR VELOCITY Angular velocity (𝜔) is defined as the rate of change of angular displacement with respect to time. It describes how fast an object rotates around a fixed axis and is a vector quantity, indicating both the speed of rotation and the direction of the axis of rotation. The angular velocity can be expressed mathematically as: ∆𝜽 𝝎= ∆𝒕 Where: 𝝎 = angular velocity (in radians per second, rad/s) ∆𝜽 = change in angular position (in radians) ∆𝒕 = change in time (in seconds) Relationship between 𝝎 and Linear Velocity (𝒗) Angular velocity is related to linear velocity (𝒗)by the following equation: 𝒗 = 𝒓. 𝝎 where: 𝒗 = linear velocity (in meters per second, m/s) 𝑟 = radius of the circular path (in meters) This equation shows that the linear velocity of a point on a rotating object increases with distance from the axis of rotation. Types of Angular Velocity Average Angular Velocity: Calculated over a time interval. Instantaneous Angular Velocity: The angular velocity at a specific moment in time, which can be found using calculus as: ∆𝜽 𝝎 = 𝐥𝐢𝐦 ∆𝒕→𝟎 ∆𝒕 Worked Example 1: Calculating Angular Velocity CHALLENGE: A wheel rotates through an angle of 30 radians in 5 seconds. What is its angular velocity? Solution Using the formula for angular velocity:  ∆𝜽 𝟑𝟎𝑟𝑎𝑑 𝝎= = = 𝟔𝑟𝑎𝑑/𝑠 ∆𝒕 𝟓𝑠 CHALLENGE: A wheel rotates through an angle of 45 degrees in 22 seconds. What is its angular velocity in radians per second? Solution Angular velocity is typically expressed in radians per second. To convert degrees to radians, we use the conversion factor: 𝜋𝑟𝑎𝑑 𝑅𝑎𝑑𝑖𝑎𝑛 = 𝐷𝑒𝑔𝑟𝑒𝑒𝑠 × ( ) 1800 For 45 degrees 𝜋 𝜋 ∆𝜃 = 450 × ( ) = 𝑟𝑎𝑑 180 4 Step 2: Calculate Angular Velocity ∆𝜽 The formula for angular velocity (ω) is given by: 𝝎 = ∆𝒕 𝜋 𝜋 𝜔 = 4 = 𝑟𝑎𝑑/𝑠 2𝑠 8 𝜔 ≈ 0.393𝑟𝑎𝑑/𝑠 Classwork: 1. A rotating disk has an angular velocity of 10 rad/s and a radius of 0.50m. What is the linear velocity of a point on the edge of the disk? 2. A car is traveling around a banked curve with a radius of 30 m at an angle of 15∘. What is the speed at which the car should travel to avoid sliding?  Angular velocity can be calculated using changes in angular position over time. The relationship between linear and angular velocities is crucial for understanding motion in circular paths. ANGULAR MOMENTUM Angular momentum (L) is a measure of the amount of rotation an object has, taking into account its mass distribution relative to the axis of rotation. The formula for angular momentum is: 𝐿 = 𝐼𝜔 Where: L= Angular momentum (in kg·m²/s) I = Moment of inertia (in kg·m²) 𝜔 = Angular velocity (in radians per second) Conservation of Angular Momentum Angular momentum is conserved in a closed system where no external torques act. This means that if an object's moment of inertia changes, its angular velocity must change in such a way that the product I𝜔 remains constant: 𝐼1 𝜔1 = 𝐼2 𝜔2 Worked Examples Example 1: Ice Skater's Spin Consider an ice skater with a moment of inertia of 3.0 kg·m2 spinning at an angular velocity of 2.0rad/s. If she pulls her arms in and her new moment of inertia becomes1.5 kg·m2, what will be her new angular velocity? Using conservation of angular momentum: 𝐼1 𝜔1 = 𝐼2 𝜔2 Substituting the known values: 2.0𝑟𝑎𝑑 3.0𝑘𝑔. 𝑚2 × = 1.5𝑘𝑔. 𝑚2 × 𝜔2 𝑠 Calculating: 6.0 = 1.5 × 𝜔2 The skater's new angular velocity is 4.0 rad/s.