1 / 15

Announcements

Announcements. Exam 1 is two weeks away. Will be similar to homework problems. Bring a calculator. Homework Set 3: Chapter 3 #27, 29 & 32 + Supplemental Problems. Kepler’s Laws of Planetary Motion. Empirical laws developed by Johannes Kepler based on the observational data of Tycho Brahe.

jaunie
Download Presentation

Announcements

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Announcements • Exam 1 is two weeks away. Will be similar to homework problems. Bring a calculator. • Homework Set 3: Chapter 3 #27, 29 & 32 + Supplemental Problems

  2. Kepler’s Laws of Planetary Motion Empirical laws developed by Johannes Kepler based on the observational data of Tycho Brahe

  3. Kepler’s 1st Law The Law of Ellipses The planets move in elliptical orbits with the Sun located at one focus

  4. Kepler’s 2nd Law Law of Areas A line drawn from a planet to the Sun will sweep out equal areas in equal time periods

  5. Kepler’s 3rd Law The Law of Harmonies The ratio of the square of the orbital period to the cube of the orbital semimajor axis (the radius) is the same for all the planets For objects orbiting the Sun, if the orbital period (P) is measured in years and the semimajor axis (A) is measured in AU then If units other than AU and years are used or the object is orbiting something other than the Sun, then a conversion constant must be added

  6. Using Kepler’s 3rd Law The constant k can be determined by looking at the orbital properties of other bodies orbiting the same parent body. Be sure to use consistent units. For the Sun: Earth orbits the Sun in 1 year with a semimajor axis of 1 AU so Or, using 1 AU =1.496x1011m and 1 year = 3.156x107s

  7. Using Kepler’s 3rd Law A new asteroid is discovered orbiting the Sun in a circular orbit at a distance of 8 AU. What is it’s orbital period? Jupiter’s moon Io orbits Jupiter at an average distance of 421,600 km with an orbital period of 1.769 days. If a new moon is discovered orbiting Jupiter at an average distance of 2,350,000 km, what will it’s orbital period be?

  8. Solution 1 Given: orbital radius (A) = 8AU Also know orbital radius of Earth (1.0AU) and orbital period of Earth (1year) Equation to use: In this case, using orbital radius in AU and period in years, the constant is 1AU/year so

  9. Solution 2 Given: Io orbital radius = 421,600 km Io orbital period = 1.769 days New moon orbital radius = 2,350,000 km Rather than calculating what the constant is, set up the problem as equal ratios

  10. Newton’s 1st Law of Motion The Law of InertiaAn object in straight line uniform motion will continue that motion unchanged unless some external force acts on it This was law based entirely on the work of Galileo

  11. Newton’s Second Law The Force Law: F = maThe acceleration a body experiences is directly proportional to the net force acting on it and inversely proportional to its mass Acceleration and force are actually vectors

  12. Vectors have both magnitude and direction A vector can be broken into horizontal and vertical components using a few simple trigonometric relationships. These relationships come from the Pythagorean Theorem relating the sides and angles of a right triangle.

  13. Newton’s Third Law The Action-Reaction LawFor every force there is an equal and opposite reaction force

  14. Using Newton’s Laws A 215 kg sailboat experiences a force from the wind of 75.0 N at 30° north of east while the current pushes it with a force of 50.0 N due south. What is the magnitude and direction of the acceleration of the boat?

  15. Solution North-South Components Wind: (75.0N)sin30° = 37.5N Current: (50.0N)sin270° = -50.0N Net = 37.5N – 50.0N = -12.5N East-West Components Wind: (75.0N)cos30° = 64.45N Current: (50.0N)cos270° = 0.0N Net = 64.45N + 0.0N = 64.45N The acceleration is 0.31 m/s2 at 10° south of east

More Related