Circular Motion and Gravitation Dr Walker Circular Motion

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Circular Motion and Gravitation Dr. Walker

Circular Motion and Gravitation Dr. Walker

Circular Motion • Unit will describe objects that move with uniform circular motion –

Circular Motion • Unit will describe objects that move with uniform circular motion – motion of an object in a circle with a constant or uniform speed – What objects? • • Planets Cars Roller Coasters Anything moving in a circle

Circumference • The distance of one complete cycle around the perimeter of a circle

Circumference • The distance of one complete cycle around the perimeter of a circle is known as the circumference – Average speed = distance/time – In circular motion, the distance is the distance around the outside of the circle – For uniform circular motion, Avg. speed = circumference/time

Average Speed and Circles • Some relationships to consider – Average speed = circumference/time

Average Speed and Circles • Some relationships to consider – Average speed = circumference/time – Circumference = 2 pr, where r is the radius – The period is the time required to make one trip around the circle – Combining these concepts , we get

Circles and Velocity • We said previous that objects moving in a circle do

Circles and Velocity • We said previous that objects moving in a circle do not have a constant velocity, because the direction is constantly changing. – We do have an instantaneous velocity, because an object moving in a circle has a velocity (with direction) at any given moment – The direction of the velocity vector at any instant is in the direction of a tangent line drawn to the circle at the object's location (pesky trig…. )

Circles and Acceleration • Since the velocity is constantly changing even at the same

Circles and Acceleration • Since the velocity is constantly changing even at the same speed, there is acceleration in uniform circular motion – Remember, since velocity is a vector, it can still have the same d/t numerical value, but a different direction – so it’s still different!!

Circular Acceleration • As an object accelerates around a circle, its force is directed

Circular Acceleration • As an object accelerates around a circle, its force is directed toward the center – Remember F = ma, if there’s an acceleration, there’s a force • This is known as centripetal force • This is an unbalanced (net) force – Otherwise, the object would go in a straight line instead of going in a circle based on Newton’s first law

Centripetal vs. Centrifugal • Centripetal force – Net force acting towards the center in

Centripetal vs. Centrifugal • Centripetal force – Net force acting towards the center in circular motion • Centrifugal force – Perception of outward force radiating from an object moving in a circle – This is really the result of a object trying to move in a straight line (tangent) based on its inertia – If there was a outward force, you’d go away from the center…and NOT in a circle! • Remember, Newton’s 1 st law says objects will continue at the same velocity unless acted on by a force…. but the force is IN not OUT!!!

Examples of Centripetal Force http: //www. physicsclassroom. com/class/circles/Lesson-1/The-Centripetal-Force-Requirement

Examples of Centripetal Force http: //www. physicsclassroom. com/class/circles/Lesson-1/The-Centripetal-Force-Requirement

Speed vs. Direction • If an unbalanced force is acting on an object in

Speed vs. Direction • If an unbalanced force is acting on an object in circular motion, shouldn’t the speed change? – Speed = magnitude of the velocity (10 m/s, for example) – The work (f x d) is being directed perpendicular to the current velocity (remember, centripetal – inward force) – Short version - All of the work being done on the object is changing the direction instead of changing the magnitude

Back To The Math… • Notice it says speed instead of velocity • All

Back To The Math… • Notice it says speed instead of velocity • All of these equations are derived from equations for velocity, acceleration, and force and substituted with quantities involved with circles (like T, R, and p)

Example Problem 1 • A 900 -kg car moving at 10 m/s takes a

Example Problem 1 • A 900 -kg car moving at 10 m/s takes a turn around a circle with a radius of 25. 0 m. Determine the acceleration and the net force acting upon the car.

Example Problem 1 • A 900 -kg car moving at 10 m/s takes a

Example Problem 1 • A 900 -kg car moving at 10 m/s takes a turn around a circle with a radius of 25. 0 m. Determine the acceleration and the net force acting upon the car. • a = v 2/R = (10 m/s)2 / 25. 0 m • a = 4. 0 m/s 2

Example Problem 1 • A 900 -kg car moving at 10 m/s takes a

Example Problem 1 • A 900 -kg car moving at 10 m/s takes a turn around a circle with a radius of 25. 0 m. Determine the acceleration and the net force acting upon the car. • Fnet = m x v 2/R = 900 kg x (10 m/s)2 / 25. 0 m • Fnet = 3600 N

Example Problem 2 • A 95 -kg halfback makes a turn on the football

Example Problem 2 • A 95 -kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12 -meters. The halfback makes a quarter of a turn around the circle in 2. 1 seconds. Determine the speed, acceleration and net force acting upon the halfback.

Example Problem 2 • A 95 -kg halfback makes a turn on the football

Example Problem 2 • A 95 -kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12 -meters. The halfback makes a quarter of a turn around the circle in 2. 1 seconds. Determine the speed, acceleration and net force acting upon the halfback. • Speed = 2(pi)R/T • Speed = 2 (3. 14)(12 m)(0. 25)/2. 1 s = 8. 97 m/s

Example Problem 2 • A 95 -kg halfback makes a turn on the football

Example Problem 2 • A 95 -kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12 -meters. The halfback makes a quarter of a turn around the circle in 2. 1 seconds. Determine the speed, acceleration and net force acting upon the halfback. • a = v 2/r • a = (8. 97 m/s)2/(12 m) = 6. 7 m/s 2

Example Problem 2 • A 95 -kg halfback makes a turn on the football

Example Problem 2 • A 95 -kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12 -meters. The halfback makes a quarter of a turn around the circle in 2. 1 seconds. Determine the speed, acceleration and net force acting upon the halfback. • Fnet = mv 2/r • Fnet = (95 kg)(8. 97 m/s)2/(12 m) = 637 N

Example Problem 3 • Determine the centripetal force acting upon a 40 -kg child

Example Problem 3 • Determine the centripetal force acting upon a 40 -kg child who makes 10 revolutions around the Cliffhanger in 29. 3 seconds. The radius of the barrel is 2. 90 meters.

Example Problem 3 • Determine the centripetal force acting upon a 40 -kg child

Example Problem 3 • Determine the centripetal force acting upon a 40 -kg child who makes 10 revolutions around the Cliffhanger in 29. 3 seconds. The radius of the barrel is 2. 90 meters. • V = 2(pi)R/T = 2(3. 14)(2. 90 m)/(29. 3 rev/10 s) = 6. 22 m/s • Fnet = mv 2/r = (40 kg)(6. 22 m/s)2/2. 90 m = 533 N

Kepler • Johannes Kepler – 17 th century German mathematician – Developed laws to

Kepler • Johannes Kepler – 17 th century German mathematician – Developed laws to describe the motion of data based on astronomical data – Explained the movements of planets, but not the why • “Here’s how they move, but we don’t know WHY they move”

Isaac Newton • Mathematician known for inventing calculus • Was bothered by the lack

Isaac Newton • Mathematician known for inventing calculus • Was bothered by the lack of explanation for planetary orbits – There must be a WHY!! – Led to the concept of universal gravitation

Kepler’s Laws • The paths of the planets about the sun are elliptical in

Kepler’s Laws • The paths of the planets about the sun are elliptical in shape, with the center of the sun being located at one focus. (The Law of Ellipses) • An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time. (The Law of Equal Areas) • The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies)

Newton’s Mountain Newton compared motion of the moon to a cannonball fired from the

Newton’s Mountain Newton compared motion of the moon to a cannonball fired from the top of a high mountain. • If a cannonball were fired with a small horizontal speed, it would follow a parabolic path and soon hit Earth below. • Fired faster, its path would be less curved and it would hit Earth farther away. • If the cannonball were fired fast enough, its path would become a circle and the cannonball would circle indefinitely.

The Moon Is Falling!! More of Newton’s conclusions: – Both the orbiting cannonball and

The Moon Is Falling!! More of Newton’s conclusions: – Both the orbiting cannonball and the moon have a component of velocity parallel to Earth’s surface. – The sideways or tangential velocity is sufficient to ensure nearly circular motion around Earth rather than into it. – With no resistance to reduce its speed, the moon will continue “falling” around Earth indefinitely. Remember, the moon is moving around the Earth because of its’ tangential velocity. Tangential velocity is shown in the circular motion above

The Curvature Of The Earth • The curvature of Earth, the surface drops a

The Curvature Of The Earth • The curvature of Earth, the surface drops a vertical distance of nearly 5 meters for every 8000 meters tangent to its surface. • Remember, based on our kinematics equations, an object will drop approximately 5 m in the first second.

The Curvature Of The Earth • Newton figured an object would need a velocity

The Curvature Of The Earth • Newton figured an object would need a velocity of 8000 m/s (or 8 km/s) to orbit the Earth. • This is the speed necessary for the satellite to “fall” at the same rate the Earth curves.

The Connection • The same forces that cause an apple to fall to the

The Connection • The same forces that cause an apple to fall to the ground from a tree (the original Newton story) and keep the moon rotating around the Earth also govern the motion of objects in space – Thus the term…. universal gravitation

The Apple And The Moon • Newton performed another “experiment” to determine the effect

The Apple And The Moon • Newton performed another “experiment” to determine the effect of the Earth on the Moon – The moon was known in Newton’s time to be 60 times further from the center of the Earth than an apple at the surface d 60 d

The Apple And The Moon • Newton figured out that in 1 s, an

The Apple And The Moon • Newton figured out that in 1 s, an apple falls 5 m • In that same 1 s, the circle of the moon’s orbit should falls 1. 4 mm below a point along the straight line where the moon would otherwise be one second later • Using this data, he determined that the gravitational force differed by 1/602. This led to the inverse square law. d 60 d

The Apple And The Moon • Since the distance of an apple from the

The Apple And The Moon • Since the distance of an apple from the center of Earth is 60 times closer than the Moon to the center of the Earth, the attractive force differs by a factor of 602. If you notice gapple/gmoon = 3600 = 602

Inverse Square Law • Newton’s law of universal gravitation states that every object attracts

Inverse Square Law • Newton’s law of universal gravitation states that every object attracts every other object with a force that for any two objects is directly proportional to the mass of each object. • Newton figured out that the force decreases as the square of the distance between the centers of mass of the objects increases.

Applications Of The Inverse Square Law • As two bodies are farther apart, the

Applications Of The Inverse Square Law • As two bodies are farther apart, the gravitational effects of one on the other are reduced • Based on this, objects “weigh” less are they get farther from the Earth.

Law of Universal Gravitation • The law of universal gravitation can be expressed as

Law of Universal Gravitation • The law of universal gravitation can be expressed as an exact equation when a proportionality constant is introduced. • The universal gravitational constant, G, in the equation for universal gravitation describes the strength of gravity.

Basic Example • Suppose that two objects attract each other with a gravitational force

Basic Example • Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is reduced by a factor of 5, then what is the new force of attraction between the two objects?

Basic Example • Suppose that two objects attract each other with a gravitational force

Basic Example • Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is reduced by a factor of 5, then what is the new force of attraction between the two objects? • If you assume d was 1, when d is reduced by a factor of 5, F ~ 1/d 2 • This means if the distance is reduced by a factor of 5, the force increases by a factor of 52 or 25. • As a result, the new gravitation force is 16 units x 25 = 400 units.

Calculating G • The gravitational constant, G, was calculated by Henry Cavendish in 1798

Calculating G • The gravitational constant, G, was calculated by Henry Cavendish in 1798 using a torsion balance • When the large spheres were brought next to the small spheres, there was an attractive force and the rod in the middle was twisted a -11 N m 2/kg 2 G = 6. 67259 x 10 measureable amount http: //www. physicsclassroom. com/class/circles/Lesson-3/Cavendish-and-the-Value-of-G

Gravitational Examples • Two balls have their centers 2. 0 m apart. One ball

Gravitational Examples • Two balls have their centers 2. 0 m apart. One ball has a mass of 8. 0 kg. The other has a mass of 6. 0 kg. What is the gravitational force between them?

Gravitational Examples • Two balls have their centers 2. 0 m apart. One ball

Gravitational Examples • Two balls have their centers 2. 0 m apart. One ball has a mass of 8. 0 kg. The other has a mass of 6. 0 kg. What is the gravitational force between them? • F = Gm 1 m 2/d 2 • F = (6. 67259 x 10 -11 N m 2/kg 2)(8. 0 kg)(6. 0 kg)/(2. 0 m)2 • F = 8. 0 x 10 -10 N – Notice that the attractive forces between normal objects is negligible. Only astronomically significant objects will have a significant attractive force.

Gravitational Examples • Determine the force of gravitational attraction between the Earth and the

Gravitational Examples • Determine the force of gravitational attraction between the Earth and the sun. Their masses are 5. 98 x 1024 kg and 1. 99 x 1030 kg, respectively. The average distance separating the Earth and the sun is 1. 50 x 1011 m. Determine the force of gravitational attraction between the Earth and the sun.

Gravitational Examples • Determine the force of gravitational attraction between the earth and the

Gravitational Examples • Determine the force of gravitational attraction between the earth and the sun. Their masses are 5. 98 x 1024 kg and 1. 99 x 1030 kg, respectively. The average distance separating the Earth and the sun is 1. 50 x 1011 m. Determine the force of gravitational attraction between the earth and the sun. • F = Gm 1 m 2/d 2 • F = (6. 67259 x 10 -11 N m 2/kg 2)(5. 98 x 1024 kg )(1. 99 x 1030 kg)/(1. 50 x 1011 m)2 • F = 3. 53 x 1022 N – Notice how much bigger this force is!!!

Kepler’s Laws (Again) • Kepler's first law (The law of ellipses) – Planets are

Kepler’s Laws (Again) • Kepler's first law (The law of ellipses) – Planets are orbiting the sun in a path described as an ellipse. http: //www. physicsclassroom. com/class/circles/Lesson-4/Kepler-s-Three-Laws

Kepler’s Laws • Kepler’s 2 nd Law (The Law of Equal Areas) – An

Kepler’s Laws • Kepler’s 2 nd Law (The Law of Equal Areas) – An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time. • The speed of a planet’s orbit is highest when it is closest to the sun • If an imaginary line were drawn from the center of the planet to the center of the sun, that line would sweep out the same area in equal periods of time. http: //astronomer. wpengine. netdna-cdn. com/wp-content/uploads/2013/06/kepler 2. gif

http: //hildaandtrojanasteroids. net/Kepler. II. jpg

http: //hildaandtrojanasteroids. net/Kepler. II. jpg

Kepler’s 3 rd Law • Kepler’s 3 rd Law – The Law of Harmonies

Kepler’s 3 rd Law • Kepler’s 3 rd Law – The Law of Harmonies – The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. • Period – Time required for a planet to revolve around the sun • It should make sense that the further away from the sun (or any star) an object is, the longer it takes to revolve around it

Kepler’s 3 rd Law • Prove It!!

Kepler’s 3 rd Law • Prove It!!

Kepler’s 3 rd Law • The Proof http: //www. physicsclassroom. com/class/circles/Lesson-4/Kepler-s-Three-Laws

Kepler’s 3 rd Law • The Proof http: //www. physicsclassroom. com/class/circles/Lesson-4/Kepler-s-Three-Laws

Kepler’s 3 rd Law • Kepler’s 3 rd law not only describes planetary motion

Kepler’s 3 rd Law • Kepler’s 3 rd law not only describes planetary motion around a star, but satellite motion around a planet

Kepler’s 3 rd Law Examples • Galileo is often credited with the early discovery

Kepler’s 3 rd Law Examples • Galileo is often credited with the early discovery of four of Jupiter's many moons. The moons orbiting Jupiter follow the same laws of motion as the planets orbiting the sun. One of the moons is called Io - its distance from Jupiter's center is 4. 2 units and it orbits Jupiter in 1. 8 Earth-days. Another moon is called Ganymede; it is 10. 7 units from Jupiter's center. Make a prediction of the period of Ganymede using Kepler's law of harmonies.

Kepler’s 3 rd Law Examples • Galileo is often credited with the early discovery

Kepler’s 3 rd Law Examples • Galileo is often credited with the early discovery of four of Jupiter's many moons. The moons orbiting Jupiter follow the same laws of motion as the planets orbiting the sun. One of the moons is called Io - its distance from Jupiter's center is 4. 2 units and it orbits Jupiter in 1. 8 Earth-days. Another moon is called Ganymede; it is 10. 7 units from Jupiter's center. Make a prediction of the period of Ganymede using Kepler's law of harmonies. – For Io T 2/R 3 = (1. 8 days)2/(4. 2 units)3 = 0. 0437 – Ganymede should have the same relationship, so plug in to find the period of Ganymede – T 2/(10. 7)3 = 0. 0437 T = 7. 32 days

Kepler’s 3 rd Law Examples • The average orbital distance of Mars is 1.

Kepler’s 3 rd Law Examples • The average orbital distance of Mars is 1. 52 times the average orbital distance of the Earth. Knowing that the Earth orbits the sun in approximately 365 days, use Kepler's law of harmonies to predict the time for Mars to orbit the sun.

Kepler’s 3 rd Law Examples • The average orbital distance of Mars is 1.

Kepler’s 3 rd Law Examples • The average orbital distance of Mars is 1. 52 times the average orbital distance of the Earth. Knowing that the Earth orbits the sun in approximately 365 days, use Kepler's law of harmonies to predict the time for Mars to orbit the sun. • • • Assume the distance for Earth = 1 a. u. For Earth, T 2/R 3 = (365 days)2/(1 a. u. )3 = 133225 R = 1. 52 a. u. for Mars and the ratio will be the same For Mars, T 2/R 3 = T 2/(1. 52 a. u. )3 = 133225 For Mars, T = 684 days

Satellite Motion • A satellite is any object orbiting a star, planet, or any

Satellite Motion • A satellite is any object orbiting a star, planet, or any other massive body – They can be natural (planets, moons) or manmade • Once launched into orbit, the only force governing the motion of a satellite is the force of gravity • If launched with sufficient speed, the projectile would fall towards the earth at the same rate that the earth curves – At large enough speeds, the orbit of a satellite around the Earth is an ellipse

Satellite Motion – The Basics • The velocity of the satellite would be directed

Satellite Motion – The Basics • The velocity of the satellite would be directed tangent to the circle at every point along its path. • The acceleration of the satellite would be directed towards the center of the circle - towards the central body that it is orbiting. • The acceleration is caused by a net (centripetal) force that is directed inwards in the same direction as the acceleration. http: //www. physicsclassroom. com/class/circles/Lesson-4/Circular-Motion-Principles-for-Satellites

Satellite Orbit • Without the centripetal force of gravity, the satellite would continue on

Satellite Orbit • Without the centripetal force of gravity, the satellite would continue on a straight line path instead of staying in its circular orbit http: //www. physicsclassroom. com/class/circles/Lesson-4/Circular-Motion-Principles-for-Satellites

Weightlessness in Orbit • Most would believe the feeling of weightlessness is because of

Weightlessness in Orbit • Most would believe the feeling of weightlessness is because of a lack of gravity… – If this is the case, why do you have the feeling of weightlessness on a roller coaster ride? ? ?

Weightlessness in Orbit • Normally, you feel a normal force (you pushing on the

Weightlessness in Orbit • Normally, you feel a normal force (you pushing on the earth) that is opposed to gravity • On a roller coaster, you ONLY feel gravity, as there is no counterforce (you can’t push the ground!) http: //www. physicsclassroom. com/Class/circles/u 6 l 4 d 2. gif

Weightlessness In Orbit • In orbit, the astronaut is in free fall, continuously around

Weightlessness In Orbit • In orbit, the astronaut is in free fall, continuously around the earth, just like a satellite • Since gravity is the only force acting in orbit (you know…keeping you in orbit), the lack of counterforce makes the astronaut feels weightless!!