1. Universal Gravitation Eleanor Roosevelt High School Chin-Sung Lin Lesson 12

2. Isaac Newton 1643 - 1727

3. Newton & Physics

4. Newton’s Law of Universal Gravitation states that gravity is an attractive force acting between all pairs of massive objects. Gravity depends on:  Masses of the two objects  Distance between the objects Universal Gravitation

5. Universal Gravitation - Apple

6. Universal Gravitation - Moon

8. Newton’s question: Can gravity be the force keeping the Moon in its orbit? Newton’s approximation: Moon is on a circular orbit Even if its orbit were perfectly circular, the Moon would still be accelerated v v v v Universal Gravitation

9. The Moon’s Orbital Speed radius of orbit: r = 3.8 x 10 8 m Circumference: 2  r = ???? m orbital period: T = 27.3 days = ???? sec orbital speed: v = (2  r)/T = ??? m/sec

10. The Moon’s Orbital Speed radius of orbit: r = 3.8 x 10 8 m Circumference: 2  r = 2.4 x 10 9 m orbital period: T = 27.3 days = 2.4 x 10 6 sec orbital speed: v = (2  r)/T = 10 3 m/sec = 1 km/s

11. The Moon’s Centripetal Acceleration The centripetal acceleration of the moon: orbital speed: v = 10 3 m/s orbital radius: r = 3.8 x 10 8 m centripetal acceleration: A c = v 2 / r = ???? m/s 2

12. The Moon’s Centripetal Acceleration The centripetal acceleration of the moon: orbital speed: v = 10 3 m/s orbital radius: r = 3.8 x 10 8 m centripetal acceleration: A c = v 2 / r A c = (10 3 m/s) 2 / (3.8 x 10 8 m) = 0.00272 m/s 2

13. At the surface of Earth (r = radius of Earth) g = 9.8 m/s 2 At the orbit of the Moon (r = 60x radius of Earth) a =0.00272 m/s 2 What’s relation between them? The Moon’s Centripetal Acceleration

14. At the surface of Earth ( r = radius of Earth ) g = 9.8 m/s 2 At the orbit of the Moon ( r = 60x radius of Earth ) a =0.00272 m/s 2 9.8 m/s 2 / 0.00272 m/s 2 = 3600 / 1 = 60 2 / 1 The Moon’s Centripetal Acceleration

15. r 2r 3r 4r 5r 6r 60r g g g g g g g 1 4 9 16 25 36 3600 Bottom Line The Moon’s Centripetal Acceleration

16. Bottom Line If the acceleration due to gravity is inverse proportional to the square of the distance, then it provides the right acceleration to keep the Moon on its orbit (“to keep it falling”) The Moon’s Centripetal Acceleration

17. Bottom Line The Moon’s Centripetal Acceleration

18. Bottom Line The acceleration due to gravity is inverse proportional to the square of the distance g ~ 1/r 2 The gravity is inverse proportional to the square of the distance Fg = mg Fg ~g Fg ~ 1/r 2 Gravity’s Inverse Square Law

19. Bottom Line Gravity is reduced as the inverse square of its distance from its source increased Fg ~ 1/r 2 Gravity’s Inverse Square Law r 2r 3r 4r 5r 6r 60r FgFg FgFg FgFg FgFg FgFg FgFg FgFg 1 4 9 16 25 36 3600

20. Bottom Line Gravity’s Inverse Square Law Fg ~ 1/r 2

21. Bottom Line Gravity’s Inverse Square Law

22. Bottom Line Gravity’s Inverse Square Law Gravity decreases with altitude, since greater altitude means greater distance from the Earth's centre If all other things being equal, on the top of Mount Everest (8,850 meters), weight decreases about 0.28%

23. Bottom Line Gravity’s Inverse Square Law Astronauts in orbit are NOT weightless At an altitude of 400 km, a typical orbit of the Space Shuttle, gravity is still nearly 90% as strong as at the Earth's surface

24. Location Distance from Earth's center (m) Value of g (m/s 2 ) Earth's surface 6.38 x 10 6 m9.8 1000 km above 7.38 x 10 6 m7.33 2000 km above 8.38 x 10 6 m5.68 3000 km above 9.38 x 10 6 m4.53 4000 km above 1.04 x 10 7 m3.70 5000 km above 1.14 x 10 7 m3.08 6000 km above 1.24 x 10 7 m2.60 7000 km above 1.34 x 10 7 m2.23 8000 km above 1.44 x 10 7 m1.93 9000 km above 1.54 x 10 7 m1.69 10000 km above 1.64 x 10 7 m1.49 50000 km above 5.64 x 10 7 m0.13 Bottom Line Gravity’s Inverse Square Law

25. Bottom Line Law of Universal Gravitation Newton’s discovery Newton didn’t discover gravity. In stead, he discovered that the gravity is universal Everything pulls everything in a beautifully simple way that involves only mass and distance

26. Bottom Line Law of Universal Gravitation Universal gravitation formula F g = G m 1 m 2 / d 2 F g : gravitational force between objects G:universal gravitational constant m 1 :mass of one object m 2 :mass of the other object d: distance between their centers of mass

27. Bottom Line Law of Universal Gravitation p.83 m1m1 m2m2 d FgFg FgFg

28. Bottom Line Law of Universal Gravitation F g = G m 1 m 2 / d 2 Gravity is always there Though the gravity decreases rapidly with the distance, it never drop to zero The gravitational influence of every object, however small or far, is exerted through all space

29. Bottom Line Law of Universal Gravitation Example Mass 1Mass 2DistanceRelative Force m1m1 m2m2 dF 2m 1 m2m2 d m1m1 3m 2 d 2m 1 3m 2 d m1m1 m2m2 2d m1m1 m2m2 3d 2m 1 2m 2 2d

30. Law of Universal Gravitation Example Mass 1Mass 2DistanceRelative Force m1m1 m2m2 dF 2m 1 m2m2 d2F m1m1 3m 2 d3F 2m 1 3m 2 d6F m1m1 m2m2 2dF/4 m1m1 m2m2 3dF/9 2m 1 2m 2 2dF

31. Universal Gravitational Constant The Universal Gravitational Constant (G) was first measured by Henry Cavendish 150 years after Newton’s discovery of universal gravitation

32. Henry Cavendish 1731 - 1810

33. Universal Gravitational Constant Cavendish’s experiment  Use Torsion balance (Metal thread, 6-foot wooden rod and 2” diameter lead sphere)  Two 12”, 350 lb lead spheres  The reason why Cavendish measuring the G is to “Weight the Earth”  The measurement is accurate to 1% and his data was lasting for a century

34. Cavendish’s Experiment

35. G = F g d 2 / m 1 m 2 = 6.67 x 10 -11 N·m 2 /kg 2 F g = G m 1 m 2 / d 2 Universal Gravitational Constant

36. G = 6.67 x 10 -11 N·m 2 /kg 2 F g = G M m / r 2 The force (F g ) that Earth exerts on a mass (m) of 1 kg at its surface is 9.8 newtons The distance between the 1-kg mass and the center of Earth is Earth’s radius (r), 6.4 x 10 6 m Calculate the Mass of Earth

37. G = 6.67 x 10 -11 N·m 2 /kg 2 F g = G M m / r 2 9.8 N = 6.67 x 10 -11 N·m 2 /kg 2 x 1 kg x M / (6.4 x 10 6 m) 2 where M is the mass of Earth M = 6 x 10 24 kg Calculate the Mass of Earth

38. Universal Gravitational Force

39. G = 6.67 x 10 -11 N·m 2 /kg 2 Gravity is is the weakest of the presently known four fundamental forces Universal Gravitational Force

40. ForceStrong Electro- magnetic WeakGravity Strength11/13710 -6 6x10 -39 Range10 -15 m ∞ 10 -18 m ∞

41. Universal Gravitation Example Calculate the force of gravity between two students with mass 55 kg and 45kg, and they are 1 meter away from each other

42. Universal Gravitation Example Calculate the force of gravity between two students with mass 55 kg and 45kg, and they are 1 meter away from each other F g = G m 1 m 2 / d 2 F g = (6.67 x 10 -11 N·m 2 /kg 2 )(55 kg)(45 kg)/(1 m) 2 = 1.65 x 10 -7 N

43. Universal Gravitation Example Calculate the force of gravity between Earth (mass = 6.0 x 10 24 kg) and the moon (mass = 7.4 x 10 22 kg). The Earth-moon distance is 3.8 x 10 8 m

44. Universal Gravitation Example Calculate the force of gravity between Earth (mass = 6.0 x 10 24 kg) and the moon (mass = 7.4 x 10 22 kg). The Earth-moon distance is 3.8 x 10 8 m F g = G m 1 m 2 / d 2 F g = (6.67 x 10 -11 N·m 2 /kg 2 )(6.0 x 10 24 kg) (7.4 x 10 22 kg)/(3.8 x 10 8 m) 2 = 2.1 x 10 20 N

45. Acceleration Due to Gravity Law of Universal Gravitation: F g = G m M / r 2 Weight F g = m g Acceleration due to gravity g = G M / r 2 F g : gravitational force / weight G: univ. gravitational constant M: mass of Earth m: mass of the object r: radius of Earth g: acceleration due to gravity

46. Universal Gravitation Example Calculate the acceleration due to gravity of Earth (mass = 6.0 x 10 24 kg, radius = 6.37 × 10 6 m )

47. Universal Gravitation Example Calculate the acceleration due to gravity of Earth (mass = 6.0 x 10 24 kg, radius = 6.37 × 10 6 m ) g = G M / r 2 g = (6.67 x 10 -11 N·m 2 /kg 2 )(6.0 x 10 24 kg)/(6.37 x 10 6 m) 2 = 9.86 m/s 2

48. Universal Gravitation Example In The Little Prince, the Prince visits a small asteroid called B612. If asteroid B612 has a radius of only 20.0 m and a mass of 1.00 x 10 4 kg, what is the acceleration due to gravity on asteroid B612?

49. Universal Gravitation Example In The Little Prince, the Prince visits a small asteroid called B612. If asteroid B612 has a radius of only 20.0 m and a mass of 1.00 x 10 4 kg, what is the acceleration due to gravity on asteroid B612? g = G M / r 2 g = (6.67 x 10 -11 N·m 2 /kg 2 )(1.00 x 10 4 kg)/(20.0 m) 2 = 1.67 x 10 -9 m/s 2

50. Universal Gravitation Example The planet Saturn has a mass that is 95 times as massive as Earth and a radius that is 9.4 times Earth’s radius. If an object is 1000 N on the surface of Earth, what is the weight of the same object on the surface of Saturn?

51. Universal Gravitation Example The planet Saturn has a mass that is 95 times as massive as Earth and a radius that is 9.4 times Earth’s radius. If an object is 1000 N on the surface of Earth, what is the weight of the same object on the surface of Saturn? F g = G m M / r 2 F g ~ M / r 2 F g = 1000 N x 95 / (9.4) 2 = 1075 N

52. Relative Weight on Each Planet

54. Isaac Newton’s Influence People could uncover the workings of the physical universe Moons, planets, stars, and galaxies have such a beautifully simple rule to govern them Phenomena of the world might also be described by equally simple and universal laws

55. Summary Isaac Newton Universal gravitation – Apple and Moon? Moon’s centripetal acceleration Gravity’s inverse square law Law of universal gravitation Universal gravitational constant – Henry Cavendish Calculate the mass of Earth Weak gravitational force Acceleration due to gravity Newton’s influence