Newton's Moon Orbit Mystery Unraveled

1. Newton and the His 3464 Spring 2007

2. Newton at 83 Portrait by Enoch Seeman

3. Sir James Thronhill’s portrait of Newton at 67

4. Newton in 1701 (at 59)

5. Newton at 46

6. The Problem: ? ? ? ? ? Why does the moon orbit the earth?

7. Complicating the Problem Galileo’s explanation: inertial motion circular

9. cosmic string

10. The legend of the apple "After dinner, the weather being warm, we went into the garden and drank tea, under the shade of some apple trees, only he and myself. Amidst other discourse, he told me he was in the same situation as when formerly the notion of gravitation came into his mind. It was occasioned by the fall of an apple as he sat in a contemplative mood." William Stuckley, 1726

11. Given the following statements: p = The same force that affects apples also affects the moon q = The moon falls 16 feet in one minute The structure of Newton’s argument is A. If p then q So he must prove both A and B B. q C. Therefore p

12. Gravity likely weakens, but by how much? Newton figures this out by combining his own insights with those of his predecessors

13. What he already knew a = 32 ft/sec2 (by measuring it) d = 1/2 a t2 (from Galileo) Inertial motion (his own “corrected” version) f of tension in a string The relation between a planet’s period and its distance from the sun

14. If r is the average distance of a planet from the sun And T is the time it takes the planet to circle the sun r1 r2 r T2  r3 Then Kepler’s 3rd Law says:

15. To find the tension in the cosmic string Newton examined the case of a rock on a string cosmic string

17. m v2 /r f 

18. Now Newton applies this result to the moon case: f  mv2/r v = d/t = 2Br/T m(4B2r2/T2) x 1/r f  m(2Br/T)2 /r = m4B2r2 1 m4B2r x = r T2 T2

19. m4B2r So f  T2 But, from Kepler’s Third Law T2 r3 m4B2 m4B2r m4B2r = = r2 r3 T2 2 k Thus f " r2

20. 1 f is  r2 2r r At 1 earth radius: At 2 earth radii: 1 1 1 F  = some amount F1 X F1 = F  F 1 r2 22 (2r)2

21. Newton’s next move: To show that the moon (like apples) can be considered a falling body

22. From Newton’s PrincipiaMathematica

24. Two cases of projectile fall Thrown hard Thrown slowly

26. (89.4) B 1 mi Throw at 4.92 mi/sec for 18.66 sec D (4001) d = 89.4 mi C (4001)2 + (89.4)2 CB = = 4002 - CB BD = CD BD = ½ x 32 ft/sec2 x (18.166)2 = 5280 ft So CD = 4002 mi - 1 mi = 4001 mi

27. Sincef  a, a  f 1 f is also  r2 1 Therefore a is  r2 At 2 earth radii: At 1 earth radius: a= 1/22 x 32 = 8 ft/sec2 a= 32 ft/sec2

28. The moon is 60 earth radii away Therefore at the distance of the moon 1 a = x 32 602 32 or ft/sec2 602

29. In one minute (60 sec) the moon will “fall” 32 x = 602 d = 1/2 a t2 1/2 x = 602 16 feet He has thus proven A (if p then q)

30. To prove B (q: the moon “falls” 16 in 1 min) Newton drew the following construction A B D F C. E ADF AED

31. His task is to find BD, the distance the moon “falls” in a given amount of time. A B F D C. E

32. To find BD Newton thinks like an engineer A B F D C. E

33. AE AD AD2 AE AF SO = = X AD AF AD2 AF = THUS AE A B F D C. E ADF AED

34. A D Earth Arc AD represents 1 minute. The whole circle is 1 month 1 min Therefore arc length AD is x moon’s orbit minutes in a month We now know AD

35. AD2 and since AE is known Since AF = AE A B F D C. ? E Newton calculated AF to be

36. Recall the structure of Newton’s argument : If p then q, If the “apple force” is affecting the moon, then the moon “falls” 16 feet in 1 minute. q: The moon does fall 16 feet in one minute. therefore p Therefore the apple force affects the moon.

37. But, look at the following argument: p: If you have enough money, you go to the movies q: You go to the movies. Therefore you have enough money.

38. Nature was more complicated than Newton ever dreamed.

39. "The most beautiful experience we can have is the mysterious. It is the fundamental emotion that stands at the cradle of true art and true science. Whoever does not know it and can no longer wonder, no longer marvel, is as good as dead.”