1. Circular Motion and Gravitation
Dr. Walker

2. 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 

3. 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

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

5. 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….)

6. 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!!

7. 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

8. 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 1st law says objects will continue at the same velocity unless acted on by a force….but the force is IN not OUT!!!

9. Examples of Centripetal Force

10. 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

11. 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)

12. 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.

13. 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 = v2/R = (10 m/s)2 / 25.0 m
a = 4.0 m/s2

14. 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 v2/R = 900 kg x (10 m/s)2 / 25.0 m
Fnet = 3600 N

15. 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.

16. 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

17. 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 = v2/r
a = (8.97 m/s)2/(12 m) = 6.7 m/s2

18. 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 = mv2/r
Fnet = (95 kg)(8.97 m/s)2/(12 m) = 637 N

19. 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.

20. 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/10s) = 6.22 m/s
Fnet = mv2/r = (40 kg)(6.22 m/s)2/2.90 m = 533 N

21. Kepler Johannes Kepler 17th 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”

22. 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

23. 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)

24. 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.

25. 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 and 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

26. 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.

27. 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.

28. 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

29. 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

30. 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

31. 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 If you notice gapple/gmoon = 3600 = 602

32. 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.

33. 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.

34. 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.

35. 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?

36. 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/d2
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.

37. Calculating G G = 6.67259 x 10-11 N m2/kg2
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 measureable amount
G = x 10-11 N m2/kg2

38. Gravitational Examples
Two balls have their centers 2.0m apart. One ball has a mass of 8.0kg. The other has a mass of 6.0kg. What is the gravitational force between them?

39. Gravitational Examples
Two balls have their centers 2.0m apart. One ball has a mass of 8.0kg. The other has a mass of 6.0kg. What is the gravitational force between them?
F = Gm1m2/d2
F = ( x 10-11 N m2/kg2)(8.0 kg)(6.0 kg)/(2.0 m)2
F = 8.0 x 10-10N
Notice that the attractive forces between normal objects is negligible. Only astronomically significant objects will have a significant attractive force.

40. 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.

41. 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 = Gm1m2/d2
F = F = ( x 10-11 N m2/kg2)(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!!!

42. Kepler’s Laws (Again) Kepler's first law (The law of ellipses)
Planets are orbiting the sun in a path described as an ellipse.

43. Kepler’s Laws Kepler’s 2nd 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. 

45. Kepler’s 3rd Law Kepler’s 3rd 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

46. Kepler’s 3rd Law
Prove It!!

47. Kepler’s 3rd Law The Proof

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

49. Kepler’s 3rd 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.

50. Kepler’s 3rd 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 T2/R3 = (1.8 days)2/(4.2 units)3 =
Ganymede should have the same relationship, so plug in to find the period of Ganymede
T2/(10.7)3 = T = 7.32 days

51. Kepler’s 3rd 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.

52. Kepler’s 3rd 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, T2/R3 = (365 days)2/(1 a.u.)3 =
R = 1.52 a.u. for Mars and the ratio will be the same
For Mars, T2/R3 = T2/(1.52 a.u.)3 =
For Mars, T = 684 days

53. 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

54. 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.

55. Satellite Orbit
Without the centripetal force of gravity, the satellite would continue on a straight line path instead of staying in its circular orbit

56. 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???

57. 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!)

58. 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!!