1. PHYS 20 LESSONS Unit 5: Circular Motion Gravitation Lesson 5: Gravitation

2. Reading Segment #1: Early Theories Kepler’s Laws To prepare for this section, please read: Unit 5: pp. 16-17

3. Introduction to Gravitation Throughout history, people have wondered about how and why the stars and planets move.

4. Early Greeks (6th / 7th century B.C.) They believed that the stars revolved around the Earth in perfect circles. This is called the geocentric view of the heavens. "geo" = Earth "centric" = centreSo, Earth-centred

5. Copernicus (1543) Copernicus suggested that the planets revolve about the sun in perfect circles. This is called the heliocentric ("sun-centred") view of the heavens. This was a key advance for modern understanding.

6. D1. Kepler's Laws In the 16th century, Tyco Brahe carefully observed planetary motion for 20 years. He then gave all of this data to his student, Johann Kepler In 1609, Kepler found patterns in the data. He summarized these patterns into 3 laws.

7. Law #1: Elliptical Orbits - the planets revolve about the sun in ellipses - the sun is one of the two foci of the ellipse i.e.  F 1 F 2

8. Law #2: Equal Areas - the planets sweep out equal areas in equal periods of time i.e. A 1 A 2 A 1 = A 2

9. Animation Kepler's 2nd Law: http://www.walter-fendt.de/ph11e/keplerlaw2.htm

10. Law #3: Kepler's Equation The ratio T 2 / R 3 is constant for all planets revolving about the sun (or about any central mass). i.e. k = T 1 2 = T 2 2 R 1 3 R 2 3 where R is the average radius of the oribit (in m) T is the orbital period, or time for 1 lap (in s) k is Kepler's constant

11. Ex. 1An asteroid has a period of 8.1  107 seconds. What is its mean (average) radius around the sun? Note: Kepler's constant for objects around the sun is 2.985  10 -19 s 2 / m 3

12. k = T 2 R 3 k R 3 = T 2 R 3 = T 2 k R = T 2 = (8.1  10 7 s) 2 3 k 3 (2.985  10 -19 s 2 / m 3 ) = 2.8  10 11 m

13. Practice Problems Try these problems in the Physics 20 Workbook: Unit 5 p. 19 #1 - 4

14. Reading Segment #2: Inverse Square Law To prepare for this section, please read: Unit 5: p.18

15. Inverse Square Law (Newton, 1687) Kepler could not explain why the planets moved this way, but Isaac Newton could. When an apple fell on Newton's head due to Earth's gravity, he wondered: Could the Moon be held in orbit due to Earth's gravity as well? Is it not the same kind of force (F g ) ?

16. Consider the Moon going in uniform circular motion around the Earth:F g Moon (m) a c Earth The force of gravity is the centripetal force, holding the Moon in orbit.

17. It follows that F g = F c F g = m 4  2 r T 2

18. It follows that F g = F c F g = m 4  2 r T 2 Now, based on Kepler's 3rd law, k = T 2 R 3 orT 2 = k R 3

19. CombiningF g = m 4  2 r and T 2 = k R 3 T 2 we find F g = m 4  2 R k R 3 or F g = m 4  2 k R 2 where R is the distance from the centre of the planet

20. Newton's Conclusion:Based on F g = m 4  2 k R 2 Fg has an inverse square relationship with R. i.e. F g  1 R 2 Newton was able to use this formula to accurately predict the centripetal acceleration of the Moon (a c = 2.7  10 -3 m/s 2 )

21. Ex. 2An object experiences a F g of 18 N at a distance D from the centre of the Earth. What would this F g be if the distance was 6D ?

22. Based on Newton's inverse square law, F g  1 R 2 So, if R  6, then F g  6 2 Thus, the new F g would be 18 N = 0.50 N 6 2