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Physics 2053C – Fall 2001

Physics 2053C – Fall 2001. Chapter 5 Circular Motion & Gravitation. Quiz on Wednesday - Topics. Applications of Newton’s Laws Forces in two dimensions Circular Motion Gravity Motion in two dimensions. Quiz on Wednesday. Sample Questions: Chap 4: 11, 22 Chap 5: 4, 9 Sample Problems:

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Physics 2053C – Fall 2001

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  1. Physics 2053C – Fall 2001 Chapter 5 Circular Motion & Gravitation Dr. Larry Dennis, FSU Department of Physics

  2. Quiz on Wednesday - Topics • Applications of Newton’s Laws • Forces in two dimensions • Circular Motion • Gravity • Motion in two dimensions

  3. Quiz on Wednesday • Sample Questions: • Chap 4: 11, 22 • Chap 5: 4, 9 • Sample Problems: • Chap 3: 35 • Chap 4: 13, 39 • Chap 5: 7, 53

  4. mg mg mg mg Vo  Quiz on Wednesday • Chapters 4 and 5 – but don’t forget about kinematics. F = ma So a = g (downward) everywhere.

  5. Forces • Newton’s Laws: • If the force on an object is zero, then it’s velocity is constant. • The acceleration = net force / mass. • Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. F = m a

  6. F = ma No acceleration in the horizontal direction. F = ma So a = g (downward) everywhere. mg mg mg mg Vo  Quiz on Wednesday • How do Newton’s Laws apply?

  7. Newton’s Law of Gravition • Assumed gravity was the same everywhere in the solar system. F = G M1 M2 R2 This is an attractive force. M1 M2 R

  8. Newton’s Law of Gravition • It is a very weak force and perhaps one of the least understood. F = G M1 M2 R2 This is an attractive force. M1 F M2 F

  9. RO RE Gravity • F = G M1 M2/R2 = Force of gravity • Near the Earth we use mg • But note that: M1g = G M1 ME/RE2 F = G M1 ME/RE2 = M1 g  g = G ME/RE2

  10. FO = G M1 M2/RO2 FE G M1 M2/RE2 Circular Motion & Gravity • F = G M1 M2/R2 = Force of gravity RO FO/ FE = RE2/RO2 RE

  11. RO RE Gravity FO/ FE = RE2/RO2 Twice Earth’s radius ie, 12,760 km. RE 2RE 3RE 4RE

  12. FO = G M1 M2/RO2 FE G M1 M2/RE2 Circular Motion & Gravity • F = G M1 M2/R2 = Force of gravity • F =mv2/R for uniform circular motion RO FO/ FE = RE2/RO2 RE

  13. Circular Motion & Gravity F = G m ME /R2 = mV2/R G ME /R2 = V2/R V = 2R/T G ME /R2 = (2R/T)2/R T2 = 42R3 /GME R

  14. Kepler’s Laws • Planets orbit the earth in an ellipse. • An imaginary line between the planet and the sun sweeps out equal areas in equal times. • T2 = constant * R3. Newton’s Laws were able to derive these results from more fundamental concepts.

  15. CAPA #9 • A satellite is in circular orbit around the earth. The period of the satellite is 24.0 hrs. Calculate the radius of the orbit of the satellite. The mass of the earth is 5.98 X 1024 kg. T2 = 42R3 /GME T = 24.0 hr * 3600 s/hr= 86400 s R3 = T2 (GME)/ 42 R3 = 864002(6.67x10-11 * 5.98X1024)/(4*3.141592)

  16. CAPA #9 • A satellite is in circular orbit around the earth. The period of the satellite is 24.0 hrs. Calculate the radius of the orbit of the satellite. The mass of the earth is 5.98 X 1024 kg. R3 = 864002(6.67x10-11* 5.98X1024)/(4*3.141592) R3 = (8.64x104)2(6.67x10-11*5.98X1024)/(4*3.141592) R3 = 75.4X10+(8-11+24) = 75.4X10+(21) m3 R = 4.23x107 m

  17. Other CAPA Problems Worked #6, #8 and #10 in class

  18. Next Time • Quiz. • Chapter 6 – Work and Energy. • Please see me with any questions or comments. See you Wednesday.

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