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# Ch8.1 Universal Gravitation All matter is attracted to all other matter.

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1. Physics – Ch8,9 Ch8.1 Universal Gravitation All matter is attracted to all other matter. - More matter = larger force - Distance apart is important (Inverse square law)

2. Physics – Ch8,9 Ch8.1 Universal Gravitation All matter is attracted to all other matter. - More matter = larger force - Distance apart is important (Inverse square law) Newton’s Law of Universal Gravitation Universal gravitational constant Ex1) Compute the gravitational attraction between 2 60kg students sitting 0.5m apart.

3. Physics – Ch8,9 Ch8.1 Universal Gravitation All matter is attracted to all other matter. - More matter = larger force - Distance apart is important (Inverse square law) Newton’s Law of Universal Gravitation Universal gravitational constant Ex1) Compute the gravitational attraction between 2 60kg students sitting 0.5m apart.

4. Ex2) What is the gravitational attraction between a 75 kg person and the earth? Solve 2 ways. ME = 5.97x1024 kg RE= 6.4x106 m

5. Ex2) What is the gravitational attraction between a 75 kg person and the earth? Solve 2 ways. ME = 5.97x1024 kg RE= 6.4x106 m Fg = m∙g = (75kg)(9.8m/s2) = 735 N

6. Ex3) Two masses both of mass m are set a distance d apart, so that the force of attraction between them is F How does the force change if a) Both masses double b) Instead the distance between them doubles c) Instead both masses triple and the distance between them is cut in half Ch8 HW#1 1 –5

7. Ex3) Two masses both of mass m are set a distance d apart, so that the force of attraction between them is F F = Gmm d2 How does the force change if a) Both masses double F1 = G( 2∙m )( 2∙m ) = 4∙Gmm 4X bigger ( d )2 d 2 b) Instead the distance between them doubles F2 = G( m )( m ) = Gmm 4X smaller ( 2∙d )2 4d 2 c) Instead both masses triple and the distance between them is cut in half F3 = G( 3∙m )( 3∙m ) = 9∙Gmm 36X bigger ( ½ d )2 ¼ d 2 Ch8 HW#1 1 –5

8. Ch8 HW#1 1 – 5 1. What is the gravitational attraction between 2 7.25kg bowling balls placed 2m apart? 2. What is the gravitational attraction between you (60kg) and a 0.1kg spoon, placed 0.25m away?

9. Ch8 HW#1 1 – 5 1. What is the gravitational attraction between 2 7.25kg bowling balls placed 2m apart? 2. What is the gravitational attraction between you (60kg) and a 0.1kg spoon, placed 0.25m away?

10. 3. What is the gravitational attraction between the moon and the earth? 4. What is the gravitational attraction between the moon and the sun?

11. 3. What is the gravitational attraction between the moon and the earth? 4. What is the gravitational attraction between the moon and the sun?

12. 5. Two masses both of mass m are set a distance d apart, so that the force of attraction between them is F How does the force change if a) 1 mass doubles b) Both masses double and the distance between them doubles c) 1 mass tripled, 1 mass cut in half, and the distance quadrupled

13. Ch8.2 – Using Big G - There must be something that causes matter to seek smallest volume - Objects appear to be circling each other - Tycho Brahe and Johannes Kepler

14. Ch8.2 – Using Big G - There must be something that causes matter to seek smallest volume - Objects appear to be circling each other - Tycho Brahe and Johannes Kepler Kepler’s Laws 1.) Paths of all planets are ellipses, with Sun at one focus 2.) An imaginary line from the Sun sweeps out equal areas in equal time intervals 3.) The square of the ratio of the periods of 2 planets = the cube of the ratio of radii

15. Satellite Orbits

16. Satellite Orbits

17. Satellite Orbits

18. Satellite Orbits Earth drops off by 4.9 meters every 8000 meters (curvature). Orbital speed for a low satellite: 8000 m/s What type of force deals with circles? What force holds the satellite from flying off? Orbital Speed:

19. Satellite Orbits Earth drops off by 4.9 meters every 8000 meters (curvature). Orbital speed for a low satellite: 8000 m/s What type of force deals with circles? FC What force holds the satellite from flying off? FG FC = FG Orbital Speed: M is the mass of the object being circled. r is the radius of orbit from center. (both equations independent of mass of satellite.)

20. Ex3) A satellite orbits Earth 225km above its surface. x • What speed must it have? h • rE • r = rE + h = 6.38x106m + 225,000m = 6.605x106m r

21. Ex3) A satellite orbits Earth 225km above its surface. x • a. What speed must it have? h • r rE • r = rE + h = 6.38x106m + 225,000m • = 6.605x106m • a.

22. Ex4) Satellite at 150 km above earth. Orbital speed: Ch8 HW#2 6 – 10

23. Ex4) Satellite at 150 km above earth Orbital speed: r = rE + h = 6.38x106m + 150,000m = 6.53x106m Einstein's theory of Gravity -Newton’s Laws are great approximation F = Gmm can get us to the moon r2 -Einstein: ( part of General Theory of Relativity ) Mass distorts space-time, kind of like a bowling ball distorts a bed. Masses are attracted to each other because of this. Black holes (predicted by Einstein ) – If an object is large enough, it’s g-force becomes so large that its escape speed becomes larger than the speed of light. Light gets pulled back in so it can’t be seen. Ch8 HW#2 6 – 10

24. Ch8 HW#2 6 – 10 6. What is the gravitational attraction between the earth and a 2000kg satellite, sitting in a satellite factory, on the earth’s surface? (U can solve this the long or short way) 7. What is the gravitational attraction between a 2000kg satellite and the earth, when the satellite is 100 km above the earth? dE = ( rE + h ) = ( 6.38 x 106 + 100,000 ) = 6.48 x 106 m

25. Ch8 HW#2 6 – 10 6. What is the gravitational attraction between the earth and a 2000kg satellite, sitting in a satellite factory, on the earth’s surface? (U can solve this the long or short way) | F = m.g | = (2000kg)(9.8m/s2) | = 19,600 N | | 7. What is the gravitational attraction between a 2000kg satellite and the earth, when the satellite is 100 km above the earth? dE = ( rE + h ) = ( 6.38 x 106 + 100,000 ) = 6.48 x 106 m

26. Ch8 HW#2 6 – 10 7. What is the gravitational attraction between a 2000 kg satellite and the earth, when the satellite is 100 km above the earth? dE = ( rE + h ) = ( 6.38 x 106 + 100,000 ) = 6.48 x 106 m 8. What speed does that satellite have to travel to maintain a circular orbit?

27. 9. F = Gmm d2 a. Mass 1 doubles, mass 2 is ¼ the size, distance is ⅓ the size b. Both masses double, distance is 10x smaller c. Mass1: 4x bigger, mass2: 3x bigger, distance 8x greater

28. 10. Speed of a satellite 1000 km above earth: r = rE + h = 6.38x106m + 1,000,000m = 7.38x106m

29. 10. Speed of a satellite 1000 km above earth: r = rE + h = 6.38x106m + 1,000,000m = 7.38x106m

30. Ch9.1 – Momentum momentum = mass ∙ velocity Units: p = m∙v Ex1) What is the momentum of a 5 gram bullet traveling at 300m/s ?

31. Ch9.1 – Momentum momentum = mass ∙ velocity Units: p = m∙v Ex1) What is the momentum of a 5 gram bullet traveling at 300m/s ? p = m∙v = ( .005 kg )( 300 m/s ) = 1.5 kg∙m/s Ex2) What is the change in momentum of a 1000 kg car traveling at 50m/s that breaks to a stop?

32. Ch9.1 – Momentum momentum = mass ∙ velocity Units: p = m∙v Ex1) What is the momentum of a 5 gram bullet traveling at 300m/s ? p = m∙v = ( .005 kg )( 300 m/s ) = 1.5 kg∙m/s Ex2) What is the change in momentum of a 1000 kg car traveling at 50m/s that breaks to a stop? ∆p = pf – pi = mvf – mvi = (1000)(0) – (1000)(50) = – 50,000

33. Impulse: force x time (F∙t) Ex3) What kind of impulse is placed on a baseball that is hit with a bat with a force of 2500N over 50 ms (milliseconds)? Ex4) A 47g golf ball is blasted from rest to 70m/s when hit by a driver. If the impact lasts 1 ms, what impulse is acted on the ball? t = .001 sec vi = 0 vf = 70 m/s m = .047 kg

34. Impulse – force x time (F∙t) Ex3) What kind of impulse is placed on a baseball that is hit with a bat with a force of 2500N over 50 ms (milliseconds)? Impulse = F∙t = ( 2,500 N )( .05 sec ) = 125 N∙s Ex4) A 47g golf ball is blasted from rest to 70m/s when hit by a driver. If the impact lasts 1 ms, what impulse is acted on the ball? t = .001 sec vi = 0 vf = 70 m/s m = .047 kg

35. Impulse – force x time (F∙t) Ex3) What kind of impulse is placed on a baseball that is hit with a bat with a force of 2500N over 50 ms (milliseconds)? Impulse = F∙t = ( 2,500 N )( .05 sec ) = 125 N∙s Ex4) A 47g golf ball is blasted from rest to 70m/s when hit by a driver. If the impact lasts 1 ms, what impulse is acted on the ball? t = .001 sec vf = vi + at vi = 0 vf = 70 m/s a = vf – vi m = .047 kg t F = m∙a a = 70,000 m/s2 = (0.047kg)(70,000 m/s2) Alternate method: = 3290N F∙t = ∆p Imp = F∙t = mvf – mvi = 3290N.(0.001s) = 3.29 Ns

36. Impulse causes a change in momentum F∙t = ∆p Ex5) A 60kg person is sitting in a car traveling at 26m/s. In an accident, the person is brought to a stop by the seatbelt in 0.22 seconds. How much force acts on the person? Imp = ∆p F∙t = mvf – mvi Ch9 HW #1 1 – 6

37. Impulse causes a change in momentum F∙t = ∆p Ex5) A 60kg person is sitting in a car traveling at 26m/s. In an accident, the person is brought to a stop by the seatbelt in 0.22 seconds. How much force acts on the person? Imp = ∆p F∙t = mvf – mvi F = mvf – mvi = 0 – ( 60 )( 26 ) = -7,091 N t ( .22 ) Ch9 HW #1 1 – 6

38. Ch9 HW#1 1 – 6 1. Compact car of mass 725 kg moving at 28 m/s east. Find momentum. A second car of mass 2,175 kg has same momentum. Find its velocity. 2. The driver of the 725kg compact car applies the brakes for 2 seconds. There was an average force of 5x103 N. a. What is the impulse? F∙t = b. Final velocity? F∙t = ∆p

39. Ch9 HW#1 1 – 6 1. Compact car of mass 725 kg moving at 28 m/s east. Find momentum. A second car of mass 2,175 kg has same momentum. Find its velocity. 2. The driver of the 725kg compact car applies the brakes for 2 seconds. There was an average force of 5x103 N. a. What is the impulse? F∙t = 5,000N ∙ 2s = 10,000 N∙s b. Final velocity? F∙t = ∆p F∙t = mvf – mvi –10,000 = 725(vf) – 725(28) vf = 14 m/s

40. 3. A driver accelerates a 240 kg snowmobile, speeding it up from 6 m/s to 28 m/s over 60 seconds. a. Find ∆p b. Impulse c. Force

41. 3. A driver accelerates a 240 kg snowmobile, speeding it up from 6 m/s to 28 m/s over 60 seconds. a. Find ∆p b. Impulse F∙t = ∆p = 5,280 N∙s c. Force F = ∆p = 5,280 N∙s = 88 N t 60 s

42. 4. A 75kg running back is running at 5 m/s North on the football field. a. What is his change in momentum if he slows to 2 m/s in 2 seconds? b. What average force does his muscles exert to slow him in the 2 sec?

43. 4. A 75kg running back is running at 5 m/s North on the football field. a. What is his change in momentum if he slows to 2 m/s in 2 seconds? b. What average force do his muscles exert to slow him in the 2 sec? 5. A 75kg running back is running at 5 m/s North on the football field. a. What is his change in momentum if he is hit and stops in 0.2 sec? b. What average force does the opposing player exert on him? 6. A 75kg running back is running at 5 m/s North on the football field. a. What is his change in momentum if he turns and runs south at 5 m/s? b. What force to do this in 2 sec?

44. Ch9.2 – Conservation of Momentum pi = pf The total momentum = the total momentum before a collision after the collision Two types: Elastic collision (objects don’t stick together) Inelastic collision (objects stick together)

45. Ch9.2 – Conservation of Momentum pi = pf The total momentum = the total momentum before a collision after the collision Two types: Elastic collision (objects don’t stick together) m1v1i + m2v2i = m1v1f + m2v2f Inelastic collision (objects stick together) m1v1i + m2v2i = ( m1 + m2 )vf

46. Elastic collision m1v1i + m2v2i = m1v1f + m2v2f(Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf(Stick together) Ex1) A 0.510kg lab cart moving at 1.5m/s collides with and sticks to a 0.605kg lab cart at rest. What speed do they roll away at?  2 2 1 1

47. Elastic collision m1v1i + m2v2i = m1v1f + m2v2f(Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf(Stick together) Ex1) A 0.510kg lab cart moving at 1.5m/s collides with and sticks to a 0.605kg lab cart at rest. What speed do they roll away at?  m1v1i + m2v2i = ( m1 + m2 )vf (.510)(1.5) + 0 = (1.115)vf vf = .686 m/s 2 2 1 1

48. Elastic collision m1v1i + m2v2i = m1v1f + m2v2f(Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf(Stick together) Ex1) A 2,275kg car going 28m/s rear-ends a 875kg compact car going 16m/s in the same direction. The two cars stick together. If friction is negligible, how fast does the wreckage move after the collision? 

49. Elastic collision m1v1i + m2v2i = m1v1f + m2v2f(Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf(Stick together) Ex1) A 2,275kg car going 28m/s rear-ends a 875kg compact car going 16m/s in the same direction. The two cars stick together. If friction is negligible, how fast does the wreckage move after the collision? m1v1i + m2v2i = (m1 + m2)vf (2275)(+28) + (875)(+16) = (3150)vf 63700 + 14000 = 3150 vf 77700 = 3150 vf vf = 24.7 m/s Ch9 HW#2 7 – 11

50. Lab9.1 – Conservation of Momentum in Inelastic Collisions - due tomorrow - Ch9 HW#2 due at beginning of period