1. Gravitation

2. Forces Causing Centripetal Acceleration Newton’s Second Law says that the centripetal acceleration is accompanied by a force Newton’s Second Law says that the centripetal acceleration is accompanied by a force F = ma C = mv 2 /r F = ma C = mv 2 /r F stands for any force that keeps an object following a circular path F stands for any force that keeps an object following a circular path Tension in a string Tension in a string Gravity Gravity Force of friction Force of friction

3. Problem Solving Strategy Draw a free body diagram, showing and labeling all the forces acting on the object(s) Draw a free body diagram, showing and labeling all the forces acting on the object(s) Choose a coordinate system that has one axis perpendicular to the circular path and the other axis tangent to the circular path Choose a coordinate system that has one axis perpendicular to the circular path and the other axis tangent to the circular path

4. Problem Solving Strategy, cont. Find the net force toward the center of the circular path (this is the force that causes the centripetal acceleration) Find the net force toward the center of the circular path (this is the force that causes the centripetal acceleration) Solve as in Newton’s second law problems: ∑ F = ma c (2 dimensions) Solve as in Newton’s second law problems: ∑ F = ma c (2 dimensions) The directions will be radial and tangential The directions will be radial and tangential The acceleration on the right side will be the centripetal acceleration. The acceleration on the right side will be the centripetal acceleration. Do not put centripetal force on the left side – it is the right side of the equation already. Do not put centripetal force on the left side – it is the right side of the equation already.

5. Newton’s Law of Universal Gravitation - 1687 “Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.” “Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.” r does not go to zero here because the objects have some radius of their own.

6. Law of Gravitation, cont. G is the constant of universal gravitational G is the constant of universal gravitational G = 6.673 x 10 -11 N m² /kg² G = 6.673 x 10 -11 N m² /kg² This is an example of an inverse square law - force is proportional to 1/r 2 This is an example of an inverse square law - force is proportional to 1/r 2 Do the units of G make sense to you? Could you derive them?

7. Features of gravitation 1. The gravitational force is a field force that always exists between two particles regardless of the medium that separates them. 2. The force varies as one over the square of the distance between the particles and therefore decreases rapidly with increasing separation.

8. Features of gravitation 3. The force is proportional to the product of the particles’ masses. 4. The gravitational force is actually very weak, as shown by the size of G (6.67 x 10 -11 Nm 2 /kg 2 )

9. Gravitation Constant Determined experimentally Determined experimentally by Henry Cavendish in 1798 by Henry Cavendish in 1798 using this torsion balance. using this torsion balance. The small spheres m are The small spheres m are attracted by large spheres attracted by large spheres M and the rod rotates. M and the rod rotates. The light beam and mirror The light beam and mirror serve to amplify the motion. serve to amplify the motion.

11. Quick Quiz – which are false? Scenario: A ball falls to the ground Scenario: A ball falls to the ground The force that the ball exerts on the Earth is equal in magnitude to the force that Earth exerts on the ball. The force that the ball exerts on the Earth is equal in magnitude to the force that Earth exerts on the ball. The ball undergoes the same acceleration as Earth. The ball undergoes the same acceleration as Earth. Earth is much more massive than the ball, so Earth pulls much harder on the ball than the ball pulls on the Earth. Therefore the ball falls while Earth remains stationary. Earth is much more massive than the ball, so Earth pulls much harder on the ball than the ball pulls on the Earth. Therefore the ball falls while Earth remains stationary.

12. Quick Quiz Superman circles Earth at a radius of 2R where R is Earth’s radius. He then moves out to a radius of 4R. The gravitational force on him at this second orbit as compared to the first is a) the same, b) 2x more, c) 4x more, d) ½ as great, e) ¼ as great. Superman circles Earth at a radius of 2R where R is Earth’s radius. He then moves out to a radius of 4R. The gravitational force on him at this second orbit as compared to the first is a) the same, b) 2x more, c) 4x more, d) ½ as great, e) ¼ as great.

13. Example 1 – 2D problem Find net force on cue ball 1 Find net force on cue ball 1 F 21 = (6.67E-11)(.3)(.3)/(.4) 2 F 21 = (6.67E-11)(.3)(.3)/(.4) 2 = 3.75E-11 N = 3.75E-11 N F 31 = (6.67E-11)(.3)(.3)/(.3) 2 F 31 = (6.67E-11)(.3)(.3)/(.3) 2 = 6.67E-11 N = 6.67E-11 N F net = (F 21 2 + F 31 2 ) 1/2 F net = (F 21 2 + F 31 2 ) 1/2 =7.65E-11 N =7.65E-11 N Can use ϴ= tan -1 (Fy/Fx) to find angle of F net Mass of all balls= 0.3 kg

14. Example 2– Find total force on B A C B 4 cm 6 cm m A = 10 kg m B = 8 kg m C = 12 kg

15. Applications of Universal Gravitation Find the mass of the earth Find the mass of the earth Use an example of an object close to the surface of the earth Use an example of an object close to the surface of the earth r ~ R E, W = weight r ~ R E, W = weight

16. Example 2 Find the mass of Earth. Use an object like a baseball of mass m, falling to Earth. The magnitude of the gravitational force exerted by the Earth on the ball = the weight of the ball. mg = GM e m /R E 2 Cancel m on both sides, solve for M e M e = g R E 2 / G = (9.8) (6.38E6 )2 / 6.67E-11 = 5.98E24 kg is the mass of the Earth

17. Applications of Universal Gravitation We can rearrange the equation used on the last 2 pages to find the acceleration due to gravity at various radius positions. We can rearrange the equation used on the last 2 pages to find the acceleration due to gravity at various radius positions. g will vary with altitude g will vary with altitude (1000 km is about 600 miles)

18. Gravitational Potential Energy This next section has to do with Energy. We actually have our Energy unit as Unit 6, later in the semester. This next section has to do with Energy. We actually have our Energy unit as Unit 6, later in the semester. That means you will use some equations here that we haven’t introduced earlier, but trust me, the proof will come later in unit 6. That means you will use some equations here that we haven’t introduced earlier, but trust me, the proof will come later in unit 6. No matter which order you do Physics in, there are always some before/after entanglements, and this is one of them. No matter which order you do Physics in, there are always some before/after entanglements, and this is one of them.

19. Energy equations Here’s just a quick intro to some of the energy equations we will be using Here’s just a quick intro to some of the energy equations we will be using Potential energy PE = mgh where m=mass, g=acceleration due to gravity and h = height above the ground (or some reference plane) Potential energy PE = mgh where m=mass, g=acceleration due to gravity and h = height above the ground (or some reference plane) Kinetic energy KE = ½ mv 2 where m = mass, v = velocity Kinetic energy KE = ½ mv 2 where m = mass, v = velocity Work W = Force x distance Work is in Joules, the unit of energy, and it is equal to the net applied force x the distance that the object is moved. Work W = Force x distance Work is in Joules, the unit of energy, and it is equal to the net applied force x the distance that the object is moved. Total energy E = mgh + ½ mv 2 This is PE + KE Total energy E = mgh + ½ mv 2 This is PE + KE

20. Gravitational Potential Energy

21. Here, PE is zero at infinite distance from Earth’s center, because gravity goes to zero at infinite distance from Earth. (Distant planets can’t feel the pull of the Earth.) Here, PE is zero at infinite distance from Earth’s center, because gravity goes to zero at infinite distance from Earth. (Distant planets can’t feel the pull of the Earth.) So if r gets smaller, PE decreases! So if r gets smaller, PE decreases! The negative sign indicates that the work done by an external force in moving an object from infinity to a distance r away from Earth’s center is negative, or the gravitational potential energy decreases in such a process. (it doesn’t gain PE because it gets pulled in w/o an external force) The negative sign indicates that the work done by an external force in moving an object from infinity to a distance r away from Earth’s center is negative, or the gravitational potential energy decreases in such a process. (it doesn’t gain PE because it gets pulled in w/o an external force) Another way to say that is that the work is negative because the external force has to hold the object back against the attractive force of gravity. Another way to say that is that the work is negative because the external force has to hold the object back against the attractive force of gravity.

22. Potential Energy Let’s review: Looking at this graphic, can you see how the potential energy goes to zero when the object is very far away from Earth? Looking at this graphic, can you see how the potential energy goes to zero when the object is very far away from Earth? Do you see how the PE function is always negative? Do you see how the PE function is always negative? Note that the value of PE at the Earth’s surface is Note that the value of PE at the Earth’s surface is PE = - GM E m / R E PE = - GM E m / R E

23. Total Energy in Planetary Systems

24. Total Energy in Planetary Systems (continued)

25. Escape Speed

26. Newton’s Cannon applet and PhET simulation This applet allows you to see the trajectory of a projectile as a function of launch speed to explore escape speed. This applet allows you to see the trajectory of a projectile as a function of launch speed to explore escape speed. http://waowen.screaming.net/revision/force&motion/ncananim.htm http://waowen.screaming.net/revision/force&motion/ncananim.htm http://waowen.screaming.net/revision/force&motion/ncananim.htm This PhET simulation models the sun and earth. You can change the masses and watch what happens to the trajectory. You can look it up oneline at PhET too. This PhET simulation models the sun and earth. You can change the masses and watch what happens to the trajectory. You can look it up oneline at PhET too. gravity-and-orbits_en.jar gravity-and-orbits_en.jar gravity-and-orbits_en.jar

27. Kepler’s Laws – 3 laws 1) All planets move in elliptical orbits with the Sun at one of the focal points. 2) A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals. 3) The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.

28. Kepler’s Laws, cont. Based on observations made by Tyco Brahe Based on observations made by Tyco Brahe Newton later demonstrated that these laws were the consequence of the gravitational force between any two objects together with Newton’s laws of motion Newton later demonstrated that these laws were the consequence of the gravitational force between any two objects together with Newton’s laws of motion

29. Kepler’s First Law All planets move in elliptical orbits with the Sun at one focus. All planets move in elliptical orbits with the Sun at one focus. Any object bound to another by an inverse square law will move in an elliptical path (F ~ 1/r 2 ) Any object bound to another by an inverse square law will move in an elliptical path (F ~ 1/r 2 ) Second focus is empty Second focus is empty

30. Kepler’s Second Law A line drawn from the A line drawn from the Sun to any planet will sweep out equal areas in equal times. Area from A to B and Area from A to B and C to D are the same C to D are the same This is a result of conservation of angular momentum (a topic for AP Physics). This is a result of conservation of angular momentum (a topic for AP Physics).

31. Kepler’s Second Law

32. So did you notice that when Earth gets closer to the Sun, it has to go faster, to sweep out that area? Did you notice that when Earth is farther away from the Sun, it goes slower? Why? If the Earth and the Sun are farther apart, they have more potential energy, because they are attracted to each other. If they are closer together, they have less PE. To make up for losing that PE as Earth approaches the sun, it speeds up, so the total energy will remain the same. Check out the PhET simulation if you want to see that!

33. Kepler’s Second Law Do you see how PE decreases (gets more negative) when you go from r= 100 to r=10?

34. Kepler’s Third Law

35. Kepler’s Third Law: (T=period) Kepler’s Third Law: (T=period) (T A / T B ) 2 = (r A /r B ) 3

36. Example of using Kepler’s Third If the average distance from the sun for Venus is 1.082E8 m and for Saturn is 1.434E9 m, and the period of revolution for Saturn is 29.42 Earth years, what is the period for Venus? If the average distance from the sun for Venus is 1.082E8 m and for Saturn is 1.434E9 m, and the period of revolution for Saturn is 29.42 Earth years, what is the period for Venus? let A=Venus, B=Saturn (T A /T B ) 2 =(r A /r B ) 3 (T A /T B ) 2 =(r A /r B ) 3 T A = ((r A /r B ) 3 x T B 2 ) 1/2 T Venus = ((1.082E8/1.434E9) 3 x 29.42 2 ) 1/2 T Venus = ((1.082E8/1.434E9) 3 x 29.42 2 ) 1/2 T Venus = 0.61 Earth years T Venus = 0.61 Earth years

37. Kepler’s Third Law application M s = Mass such as the Sun or other celestial body that has something orbiting it. M s = Mass such as the Sun or other celestial body that has something orbiting it. M p = Mass of planet M p = Mass of planet Assuming a circular orbit is a good approximation Assuming a circular orbit is a good approximation Eccentricity is low for many planets. Eccentricity is low for many planets.

38. Eccentricity, e Mercury.206 Venus.0068 Earth.0167 Mars.0934 Jupiter.0485 Saturn.0556 Uranus.0472 Neptune.0086

39. Kepler’s Third Law Derivation < Set force due to gravity equal to centripetal force < if orbit is assumed to be circular, then v = distance/time = (2πr/T) where T= period and v = 2πr/T. < collect r’s all on right, T’s on left, items in parenthesis are constants which are grouped to be called “K”

40. Example problem (like HW)

41. Example problem: Geosynchronous Orbit

42. Ex: Geosynchronous Orbit

43. Professor Lewin – MIT http://www.youtube.com/watch?v=MJYQGPl3MNI http://www.youtube.com/watch?v=MJYQGPl3MNI http://www.youtube.com/watch?v=MJYQGPl3MNI 4’ video on why gravitational force falls away as 1/R^2 4’ video on why gravitational force falls away as 1/R^2

44. Gravity Visualized https://www.youtube.com/watch?v=MTY1Kje0yLg https://www.youtube.com/watch?v=MTY1Kje0yLg https://www.youtube.com/watch?v=MTY1Kje0yLg 9 minute video on visualizing gravity 9 minute video on visualizing gravity Ties to PhET simulation “My Solar System” Ties to PhET simulation “My Solar System” Space video on what happens when you try to wring water out of a wet wash cloth (great video even if it’s not exactly on the topic of gravity, rather lack of gravity) Space video on what happens when you try to wring water out of a wet wash cloth (great video even if it’s not exactly on the topic of gravity, rather lack of gravity) http://apod.nasa.gov/apod/ap130424.html http://apod.nasa.gov/apod/ap130424.html http://apod.nasa.gov/apod/ap130424.html