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PHY 113 C General Physics I 11 AM - 12:15 P M MWF Olin 101 Plan for Lecture 13:

PHY 113 C General Physics I 11 AM - 12:15 P M MWF Olin 101 Plan for Lecture 13: Chapter 13 – Fundamental gravitational force law and planetary motion Gravitational force law; relationship with g near Earth’s surface Orbital motion of planets, satellites, etc.

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PHY 113 C General Physics I 11 AM - 12:15 P M MWF Olin 101 Plan for Lecture 13:

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  1. PHY 113 C General Physics I • 11 AM - 12:15 PM MWF Olin 101 • Plan for Lecture 13: • Chapter 13 – Fundamental gravitational force law and planetary motion • Gravitational force law; relationship with g near Earth’s surface • Orbital motion of planets, satellites, etc. • Energy associated with gravitation and planetary motion PHY 113 C Fall 2013-- Lecture 13

  2. PHY 113 C Fall 2013-- Lecture 13

  3. Question from Webassign Assignment #11 PHY 113 C Fall 2013-- Lecture 13

  4. t3 X t2 t1 PHY 113 C Fall 2013-- Lecture 13

  5. t3 X t2 t1 PHY 113 C Fall 2013-- Lecture 13

  6. Question from Webassign Assignment #11 X PHY 113 C Fall 2013-- Lecture 13

  7. Question from Webassign Assignment #11 iclicker question: A. Ki=Kf ? B. Ki=Kf ? PHY 113 C Fall 2013-- Lecture 13

  8. Universal law of gravitation  Newton (with help from Galileo, Kepler, etc.) 1687 PHY 113 C Fall 2013-- Lecture 13

  9. Newton’s law of gravitation: m2attracts m1 according to: y m1 r2-r1 G=6.67 x 10-11 N m2/kg2 m2 r1 r2 x PHY 113 C Fall 2013-- Lecture 13

  10. Vector nature of Gravitational law: y m3 d x m1 m2 d F12 F13 PHY 113 C Fall 2013-- Lecture 13

  11. Gravitational force of the Earth RE m Note: Earth’s gravity acts as a point mass located at the Earth’s center. PHY 113 C Fall 2013-- Lecture 13

  12. Question: Suppose you are flying in an airplane at an altitude of 35000ft~11km above the Earth’s surface. What is the acceleration due to Earth’s gravity? a/g = 0.997 PHY 113 C Fall 2013-- Lecture 13

  13. Attraction of moon to the Earth: Acceleration of moon toward the Earth: F = MM a  a = 1.99x2020 N/7.36x1022kg =0.0027 m/s2 REM PHY 113 C Fall 2013-- Lecture 13

  14. PHY 113 C Fall 2013-- Lecture 13

  15. Gravity on the surface of the moon Gravity on the surface of mars PHY 113 C Fall 2013-- Lecture 13

  16. iclicker question: • In estimating the gravitational acceleration near the surfaces of the Earth, Moon, or Mars, we used the full mass of the planet or moon, ignoring the shape of its distribution. This is a reasonable approximation because: • The special form of the gravitational force law makes this mathematically correct. • Most of the mass of the planets/moon is actually concentrated near the center of the planet/moon. • It is a very crude approximation. PHY 113 C Fall 2013-- Lecture 13

  17. REM F Stable circular orbit of two gravitationally attracted objects (such as the moon and the Earth) v a PHY 113 C Fall 2013-- Lecture 13

  18. iclicker question: In the previous discussion, we saw how the moon orbits the Earth in a stable circular orbit because of the radial gravitational attraction of the moon and Newton’s second law: F=ma, where a is the centripetal acceleration of the moon in its circular orbit. Is this the same mechanism which stabilizes airplane travel? Assume that a typical cruising altitude of an airplane is 11 km above the Earth’s surface and that the Earth’s radius is 6370 km. (a) Yes (b) No PHY 113 C Fall 2013-- Lecture 13

  19. Stable (??) circular orbit of two gravitationally attracted objects (such as the airplane and the Earth) REa F v a PHY 113 C Fall 2013-- Lecture 13

  20. Summary: RE m REM F v a MM Newton’s law of gravitation: Earth’s gravity: Stable circular orbits of gravitational attracted objects: PHY 113 C Fall 2013-- Lecture 13

  21. m2 R2 R1 m1 More details If we examine the circular orbit more carefully, we find that the correct analysis is that the stable circular orbit of two gravitationally attracted masses is about their center of mass. PHY 113 C Fall 2013-- Lecture 13

  22. m2 R2 R1 m1 Radial forces on m1: T2 ? PHY 113 C Fall 2013-- Lecture 13

  23. m2 R2 R1 m1 iclicker question: What is the relationship between the periods T1 and T2 of the two gravitationally attracted objects rotating about their center of mass? (Assume that m1 < m2.) (A) T1=T2 (B) T1<T2 (C) T1>T2 PHY 113 C Fall 2013-- Lecture 13

  24. m2 R2 R1 m1 PHY 113 C Fall 2013-- Lecture 13

  25. iclicker questions: • How is it possible that all of these relations are equal? • Magic. • Trick. • Algebra. PHY 113 C Fall 2013-- Lecture 13

  26. m2 R2 R1 m1 For m2>>m1 R2 << R1: Larger mass m2 R1 m1 PHY 113 C Fall 2013-- Lecture 13

  27. What is the physical basis for stable circular orbits? • Newton’s second law? • Conservation of angular momentum? L = (const) Note: Gravitational forces exert no torque PHY 113 C Fall 2013-- Lecture 13

  28. m2 R2 R1 m1 v2 L1=m1v1R1 L2=m2v2R2 L = L1 + L2 v1 Question: How are the magnitudes of L1 and L2 related? Note: More generally, stable orbits can be elliptical. PHY 113 C Fall 2013-- Lecture 13

  29. Satellites orbiting earth (approximately circular orbits): RE ~ 6370 km Examples: PHY 113 C Fall 2013-- Lecture 13

  30. Planets in our solar system – orbiting the sun PHY 113 C Fall 2013-- Lecture 13

  31. m2 R2 R1 m1 Review: Circular orbital motion about center of mass v2 CM v1 PHY 113 C Fall 2013-- Lecture 13

  32. Review: Circular orbital motion about center of mass m2 R1 v1 m1 PHY 113 C Fall 2013-- Lecture 13

  33. Example: Satellite in circular Earth orbit PHY 113 C Fall 2013-- Lecture 13

  34. Work of gravity: ri rf PHY 113 C Fall 2013-- Lecture 13

  35. Gravitational potential energy Example: PHY 113 C Fall 2013-- Lecture 13

  36. U=mgh h • iclicker exercise: • We previously have said that the gravitational potential of an object of mass m at a height h is • U=mgh. How is this related to • No relation • They are approximately equal • They are exactly equal PHY 113 C Fall 2013-- Lecture 13

  37. h g • Near the surface of the Earth • U=mgh is a good approximation • to the gravitation potential. PHY 113 C Fall 2013-- Lecture 13

  38. iclicker exercise: • How much energy kinetic energy must be provided to an object of mass m=1000kg, initially on the Earth’s surface to outer space? • This is rocket science and not a fair question. • It is not possible to escape the Earth’s gravitational field. • We can estimate the energy by simple conservation of energy concepts. PHY 113 C Fall 2013-- Lecture 13

  39. Total energy of a satellite in a circular Earth orbit PHY 113 C Fall 2013-- Lecture 13

  40. Total energy of a satellite in a circular Earth orbit • iclicker question: • Compared to the energy needed to escape the Earth’s gravitational field does it take • more • less • energy to launch a satellite to orbit the Earth? PHY 113 C Fall 2013-- Lecture 13

  41. PHY 113 C Fall 2013-- Lecture 13

  42. Energy involved with changing orbits h’ PHY 113 C Fall 2013-- Lecture 13

  43. Stable Elliptical Orbits PHY 113 C Fall 2013-- Lecture 13

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