2. Universal Gravitation E M Ah, the same force that pulls the apple to the ground pulls moon out of its inertial path into circular motion around the earth! Therefore, the forces must be proportional to each other!

3. E M Now, if the earth pulls the apple at a rate of 9.8 m/s 2, then, the same earth must pull to moon at a proportional rate to that. 60 re If the moon is 60 x further from the apple, and all forms of energy obey the Inverse Square Law, then, the acceleration of the moon should be 1/60 2 of that of the apple, or 9.8 m/s 2 x 1/60 2 = 0.0027 m/s 2 And, in one second it should fall d = 1/2 (0.0027m/s 2 )(1sec) 2 or, 0.0014 m (d = ½ at 2 )

4. Universal Gravitation   FEFE Force of Earth on moon FMFM Force of Moon on Earth F E = F M 3 rd Law

5. Universal Gravitation   FEFE Force of Earth on moon FMFM Force of Moon on Earth F E = F M 3 rd Law Because of the 3 rd Law and the Inverse Square Law : F = Gm 1 m 2 /r 2

6. Universal Gravitation F = Gm 1 m 2 /r 2 If “F” is the weight of an object, “F w ”, then, F w = m 2 g and, m 2 g = Gm 1 m 2 /r 2 or, g = Gm 1 /r 2 (m 2 ’s cancel) or, m 1 = gr 2 /G

7. Universal Gravitation If gravity is the force that causes an object to travel in circular motion, then, F = F c or, Gm 1 m 2 /r 2 = m 2 v 2 /r or, (m 2 ’s cancel)m 1 = v 2 r/G or, r = Gm 1 /v 2 or, v 2 = Gm 1 /r or, m 1 = v 2 r/G v = Gm 1 /r

8. Universal Gravitation If gravity is the force that causes an object to travel in circular motion, then, F = F c or, Gm 1 m 2 /r 2 = m 2 v 2 /r or, Gm 1 m 2 /r 2 = m 2 4  2 r/T 2 transpose extremes T 2 /r 2 = m 2 4  2 r/Gm 1 m 2 divide by “r” and cancel m 2 T 2 /r 3 = 4  2 /Gm 1 or, r 3 /T 2 = Gm 1 /4  2 This equation is called “Newton Variation of Kepler’s 3 rd Law”

9. Universal Gravitation r 3 /T 2 = Gm 1 /4  2 Note that for any object circling a superior object that Gm 1 /4  2 remains constant!!!! (k). Therefore, r 3 /T 2 is also constant for all objects circling that superior object

10. Universal Gravitation r 3 / T 2 = “k” for all circling objects Therefore,for two objects circling the same superior object... r 1 3 / T 1 2 = r 2 3 /T 2 2

11. Kepler’s Laws 1 st Law…all planets circle the Sun in ellipital paths with the Sun at one focus Sun planet F2F2 F2F2

12. Kepler’s Laws 1 st Law…all planets circle the Sun in ellipital paths with the Sun at one focus 2 nd Law…Each planet moves around the sun in equal area sweep in equal periods of time

13. Kepler’s Laws 2 nd Law…Each planet moves around the sun in equal area sweep in equal periods of time 1 2 3 4 b a Area 12a = Area 43b

14. Kepler’s Laws 1 st Law…all planets circle the Sun in ellipital paths with the Sun at one focus 2 nd Law…Each planet moves around the sun in equal area sweep in equal periods of time 3 rd Law…the ratio of the squares of the periods to the cube of their orbital radii is a constant

15. Kepler’s Laws 3 rd Law…the ratio of the squares of the periods to the cube of their orbital radii is a constant r 3 /T 2 = “k” for all circling objects Therefore,for two objects circling the same superior object... r 1 3 / T 1 2 = r 2 3 /T 2 2

16. Sample Problems What is the gravitational attraction between the Sun and Mars? F = ? m s = 1.99 x 10 30 kg m m = 6.42 x 10 23 kg r m = 2.28 x 10 11 m F = Gm s m m /r m 2 F = 6.67 x 10 -11 N m 2 /kg 2 (1.99 x 10 30 kg)(6.42 x 10 23 kg) (2.28 x 10 11 m) 2 F = 1.64 x 10 21 N

17. Sample Problems What velocity does Mars circle the Sun at? v = ? m s = 1.99 x 10 30 kg m m = 6.42 x 10 23 kg r m = 2.28 x 10 11 m F = F c Gm s m m /r m 2 = m m v 2 /r m v 2 = Gm s /r v 2 = 6.67 x 10 -11 Nm 2 /kg 2 (1.99 x 10 30 kg)/2.28 x 10 11 m v = 2.4 x 10 4 m/s

18. Sample Problems What is the period of Mars as it circles the Sun? T = ? m s = 1.99 x 10 30 kg m m = 6.42 x 10 23 kg r m = 2.28 x 10 11 m F = F c Gm s m m /r m 2 = m m 4  2 r/T 2 T 2 = 4  2 r 3 /Gm s T 2 = 4  2 (2.28 x 10 11 m) 3 /6.67 x 10 -11 )1.99 x 10 30 kg T = 5.9 x 10 7 s or,T = 685 days

19. Sample Problems What is the period of Mars? This time use Kepler’s 3 rd Law to find it! T m = ? T e = 365.25 da r e = 1.5 x 10 11 m r m = 2.28 x 10 11 m F = F c Gm s m m /r m 2 = m m 4  2 r/T 2 r 3 /T 2 = Gm s /4  2 = k r m 3 /T m 2 = /r e 3 /T e 2 (2.28 x 10 11 m) 3 / T m 2 = (1.5 x 10 11 m) 3 / (365.25 da) 2 T m = 684 days T m = (2.28 x 10 11 m) 3 x (365.25 da) 2 (1.5 x 10 11 m) 3

20. Sample Problems What is the force of attraction (gravitational attraction) between a 85 kg person and a 55 kg person 0.90 m apart? 0.90 m 55 kg 85 kg F = Gm 1 m 2 /r m 2 F = 6.67 x 10 -11 Nm 2 /kg 2 (85 kg)(55 kg) (0.90 m) 2 F = 3.85 x 10 -7 N