2. 9.4.1State that gravitation provides the centripetal force for circular orbital motion. 9.4.2Derive Kepler’s third law. The third law states that the period squared of a planet in orbit is proportional to the cube of the radius of its orbit. Topic 9: Motion in fields 9.4 Orbital motion

3. State that gravitation provides the centripetal force for circular orbital motion.  Consider a baseball in circular orbit about Earth.  Clearly the only force that is causing the ball to move in a circle is the gravitational force.  Thus the gravitational force is the centripetal force for circular orbital motion. EXAMPLE: A centripetal force causes a centripetal acceleration a c. What are the two forms for a c ? SOLUTION: Recall from Topic 2 that a c = v 2 /r.  Then from the relationship v = 2  r/T we see that a c = v 2 /r = (2  r/T) 2 /r = 4  2 r 2 /(T 2 r) = 4  2 r/T 2. Topic 9: Motion in fields 9.4 Orbital motion

4. EXAMPLE: Suppose a 0.500-kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). Given that the earth has a radius of R E = 6400000 m, find the speed of the ball. SOLUTION: The ball is traveling in a circle of radius r = 6408850 m.  F c is caused by the weight of the ball so that F c = mg = (0.5)(9.8) = 4.9 n.  Since F c = mv 2 /r we have 4.9 = (0.5)v 2 /6408850 v = 7925 m s -1 ! State that gravitation provides the centripetal force for circular orbital motion. Topic 9: Motion in fields 9.4 Orbital motion FYI  We assumed that g = 10 ms -2 at the top of Everest.

5. PRACTICE: Find the period T of one complete orbit of the ball. SOLUTION:  r = 6408850 m.  F c = 4.9 n.  F c = ma c = 0.5a c so that a c = 9.8.  But a c = 4  2 r/T 2 so that T 2 = 4  2 r/a c T 2 = 4  2 (6408850)/9.8 T = 5081 s = 84.7 min = 1.4 h. State that gravitation provides the centripetal force for circular orbital motion. Topic 9: Motion in fields 9.4 Orbital motion a c = v 2 /r centripetal acceleration a c = 4  2 r/T 2

6. Derive Kepler’s third law. The third law states that the period squared of a planet in orbit is proportional to the cube of the radius of its orbit. Topic 9: Motion in fields 9.4 Orbital motion EXAMPLE: Show that for an object in a circular orbit about a body of mass M that T 2 = (4  2 /GM)r 3. SOLUTION:  In circular orbit F c = ma c and F c = GMm/r 2.  But a c = 4  2 r/T 2. Then ma c = GMm/r 2 4  2 r/T 2 = GM/r 2 4  2 r 3 = GMT 2 T 2 = [4  2 /(GM)]r 3 FYI  The IBO expects you to be able to derive this relationship.

7. PRACTICE: Using Kepler’s third law find the period T of one complete orbit of the baseball from the previous example. SOLUTION:  r = 6408850 m.  G = 6.67×10 −11 N m 2 kg −2.  M = 5.98×10 24 kg.  Then T 2 = (4  2 /GM)r 3 so that T 2 = [4  2 /(6.67×10 −11 ×5.98×10 24 )](6408850) 3 T = 5104 s = 85.0 min = 1.4 h. State that gravitation provides the centripetal force for circular orbital motion. Topic 9: Motion in fields 9.4 Orbital motion T 2 = [4  2 /(GM)]r 3 Kepler’s third law FYI  Note the slight discrepancy in the period (it was 5081 s before). How do you account for it?

8. 9.4.3Derive expressions for the kinetic energy, potential energy and total energy of an orbiting satellite. 9.4.4Sketch graphs showing the variation with orbital radius of the kinetic energy, gravitational potential energy and total energy of a satellite. 9.4.5Discuss the concept of weightlessness in orbital motion, in freefall and in deep space. 9.4.6Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion

9. Derive expressions for the kinetic energy, potential energy and total energy of an orbiting satellite.  An orbiting satellite has both kinetic energy and potential energy.  As we learned in Topic 9.2, the gravitational potential energy of an object of mass m in the gravitational field of Earth is E P = -GMm/r.  As we learned in Topic 2.3, the kinetic energy of an object of mass m moving at speed v is E K = (1/2)mv 2.  Thus the total mechanical energy of an orbiting satellite of mass m is Where M is the mass of the earth. Topic 9: Motion in fields 9.4 Orbital motion E = E K + E P total energy of an orbiting satellite E = (1/2)mv 2 - GMm/r

10. Derive expressions for the kinetic energy, potential energy and total energy of an orbiting satellite. Topic 9: Motion in fields 9.4 Orbital motion EXAMPLE: Show that the kinetic energy of an orbiting satellite at a distance r from the center of Earth is E K = GMm/(2r). SOLUTION:  In circular orbit F c = ma c and F c = GMm/r 2.  But a c = v 2 /r. Then ma c = GMm/r 2 mv 2 /r = GMm/r 2 mv 2 = GMm/r (1/2)mv 2 = GMm/(2r) kinetic energy of an orbiting satellite E K = (1/2)mv 2 = GMm/(2r)

11. Derive expressions for the kinetic energy, potential energy and total energy of an orbiting satellite. Topic 9: Motion in fields 9.4 Orbital motion EXAMPLE: Show that the total energy of an orbiting satellite at a distance r from the center of Earth is E = -GMm/(2r). SOLUTION: From E = E K + E P and the expressions for E K and E P we have E = E K + E P E = GMm/(2r) - GMm/r E = GMm/(2r) - 2GMm/(2r) E = -GMm/(2r) FYI  IBO expects you to derive these relationships. total energy of an orbiting satellite E = -GMm/(2r) E K = GMm/(2r) E P = -GMm/r

12. Sketch graphs showing the variation with orbital radius of the kinetic energy, gravitational potential energy and total energy of a satellite. Topic 9: Motion in fields 9.4 Orbital motion total energy of an orbiting satellite E = -GMm/(2r) E K = GMm/(2r) E P = -GMm/r EXAMPLE: Graph the kinetic energy vs. the radius of orbit for a satellite of mass m about a planet of mass M and radius R. SOLUTION: Use E K = GMm/(2r). Note that E K decreases with radius. It has a maximum value of E K = GMm/(2R). EKEK r R2R2R 3R3R 4R4R5R5R GMm 2R

13. Sketch graphs showing the variation with orbital radius of the kinetic energy, gravitational potential energy and total energy of a satellite. Topic 9: Motion in fields 9.4 Orbital motion total energy of an orbiting satellite E = -GMm/(2r) E K = GMm/(2r) E P = -GMm/r EXAMPLE: Graph the potential energy vs. the radius of orbit for a satellite of mass m about a planet of mass M and radius R. SOLUTION: Use E P = -GMm/r. Note that E P increases with radius. It becomes less negative. EPEP r R2R2R 3R3R 4R4R5R5R GMm R -

14. Sketch graphs showing the variation with orbital radius of the kinetic energy, gravitational potential energy and total energy of a satellite. Topic 9: Motion in fields 9.4 Orbital motion total energy of an orbiting satellite E = -GMm/(2r) E K = GMm/(2r) E P = -GMm/r EXAMPLE: Graph the total energy E vs. the radius of orbit and include both E K and E P. SOLUTION: GMm R - E r R2R2R 3R3R 4R4R5R5R GMm 2R - GMm 2R + EKEK EPEP

15. PRACTICE: If the elevator were to accelerate upward at 2 ms -2, what would Dobson observe the dropped ball’s acceleration to be? SOLUTION:  Since the elevator is accelerating upward at 2 ms -2 to meet the ball which is accelerating downward at 10 ms -2, Dobson would observe an acceleration if 12 ms -2.  If the elevator were accelerating downward at 2, he would observe an acceleration of 8 ms -2. Discuss the concept of weightlessness in orbital motion, in freefall and in deep space.  Consider Dobson inside an elevator which is not moving…  If he drops a ball, it will accelerate downward at 10 ms -2 as expected. Topic 9: Motion in fields 9.4 Orbital motion

16. PRACTICE: If the elevator were to accelerate downward at 10 ms -2, what would Dobson observe the dropped ball’s acceleration to be? SOLUTION:  He would observe the acceleration of the ball to be zero!  He would think that the ball was “weightless!” FYI  The ball is NOT weightless, obviously. It is merely accelerating at the same rate as Dobson!  How could you get Dobson to accelerate downward at 10 ms -2 ? Cut the cable! Discuss the concept of weightlessness in orbital motion, in freefall and in deep space. Topic 9: Motion in fields 9.4 Orbital motion

17. The “Vomit Comet”

18. PRACTICE: We have all seen astronauts experiencing “weightlessness.” Explain why it only appears that they are weightless. SOLUTION: The astronaut, the spacecraft, and the tomatoes, are all accelerating at a c = g.  They all fall together and appear to be weightless. Discuss the concept of weightlessness in orbital motion, in freefall and in deep space. Topic 9: Motion in fields 9.4 Orbital motion International Space Station

19.  In deep space, the r in F = GMm/r 2 is so large for every m that F, the force of gravity, is for all intents and purposes, zero. Discuss the concept of weightlessness in orbital motion, in freefall and in deep space.  Only in deep space – which is defined to be far, far away from all masses – will a mass be truly weightless.

20. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion  KE is POSITIVE and decreasing.  GPE is NEGATIVE and increasing (becoming less negative).

21. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion T 2 = (4  2 /GM)r 3 Kepler’s third law  T 2 = [4  2 /(GM)]r 3  r 3 = [GM/(4  2 )]T 2.

22. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion  g = GM/r 2 (from Topic 9.2).  E K = GMm/(2r) = (1/2)mv 2. GM/r = v 2. GM/r 2 = v 2 /r g = v 2 /r

23. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion R x  g = GM/x 2 (from Topic 9.2).  a c = GM/x 2 (since a c = g in circular orbits).  v 2 /x = GM/x 2 (since a c = v 2 /r).  v 2 = GM/x so that v = GM/x.

24. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion R x  But E K = (1/2)mv 2.  Thus E K = (1/2)mv 2 = (1/2)m(GM/x) = GMm/(2x).  Use E P = mV and V = -GM/x.  Then E P = m( - GM/x) = -GMm/x.  From (a) v 2 = GM/x.

25. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion R x E = E K + E P E = GMm/(2x) + - GMm/x [ from (b)(i) ] E = 1GMm/(2x) + - 2GMm/(2x) E = -GMm/(2x)

26. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion R x The satellite will begin to lose some of its mechanical energy in the form of heat.

27. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion R x  Refer to E = -GMm/(2x) [ from (b)(ii) ]:  If E gets smaller x must also get smaller (since E is negative).  If r gets smaller the atmosphere must get thicker and more resistive.  Clearly the orbit will continue to decay (shrink).

28. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion T 2 = (4  2 /GM)r 3 Kepler’s third law T 2 = (4  2 /GM)r 3 T = [(4  2 /GM)r 3 ] 1/2 T = (4  2 /GM) 1/2 r 3/2 T  r 3/2

29. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion M2M2 M1M1 R1R1 R2R2 P  It is the gravitational force.

30. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion M2M2 M1M1 R1R1 R2R2 P  Note that F G = GM 1 M 2 /(R 1 +R 2 ) 2.  M 1 experiences F c = M 1 v 1 2 /R 1.  Since v 1 = 2  R 1 /T, v 1 2 = 4  2 R 1 2 /T 2.  Then F c = F G  M 1 v 1 2 /R 1 = GM 1 M 2 /(R 1 +R 2 ) 2. M 1 (4  2 R 1 2 /T 2 )/R 1 = GM 1 M 2 /(R 1 +R 2 ) 2 4  2 R 1 (R 1 +R 2 ) 2 = GM 2 T 2 T 2 = R 1 (R 1 +R 2 ) 2 4  2 GM 2

31. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion M2M2 M1M1 R1R1 R2R2 P  From (b) T 2 = (4  2 /GM 2 )R 1 (R 1 +R 2 ) 2.  From symmetry T 2 = (4  2 /GM 1 )R 2 (R 1 +R 2 ) 2. (4  2 /GM 2 )R 1 (R 1 +R 2 ) 2 = (4  2 /GM 1 )R 2 (R 1 +R 2 ) 2 (1/M 2 )R 1 = (1/M 1 )R 2 M 1 /M 2 = R 2 /R 1  Since R 2 > R 1, M 1 > M 2.

32. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion total energy of an orbiting satellite E = -GMm/(2r) E K = GMm/(2r) E P = -GMm/r  If r decreases E K get bigger.  If r decreases E P get more negative (smaller).

33. Solve problems involving orbital motion. Topic 9: Motion in fields 9.4 Orbital motion T 2 = (4  2 /GM)r 3 Kepler’s third law  T X 2 = (4  2 /GM)r X 3.  T Y 2 = (4  2 /GM)r Y 3.  T X = 8T Y  T X 2 = 64T Y 2. T X 2 /T Y 2 = (4  2 /GM)r X 3 /[(4  2 /GM)r Y 3 ] 64T Y 2 /T Y 2 = r X 3 /r Y 3 64 = (r X /r Y ) 3 r X /r Y = 64 1/3 = 4