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Planetary Dynamics § 13.4–13.8. Closed Orbits U g + K tr = constant < 0 The closer the satellite is to the main body, the faster it moves Objects do not.

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Presentation on theme: "Planetary Dynamics § 13.4–13.8. Closed Orbits U g + K tr = constant < 0 The closer the satellite is to the main body, the faster it moves Objects do not."— Presentation transcript:

1 Planetary Dynamics § 13.4–13.8

2 Closed Orbits U g + K tr = constant < 0 The closer the satellite is to the main body, the faster it moves Objects do not meet at rest without a non- conservative force acting (crash!) E UgUg 0 E r

3 Escape Speed if an object’s E = U g + K tr ≥ 0, it can “escape” from a gravitational field. The force never disappears, but the object does not return. E UgUg 0 E r

4 Example Problem Calculate the escape speed of Earth, that is, the speed with which a rocket must be launched if it is to completely escape the gravitational pull of Earth. (This means that the kinetic energy is zero when it reaches a distance of infinity.) M earth = 5.976  10 24 kg R earth = 6378 km G = 6.67  10 –11 Nm 2 /kg 2

5 Group Question Would it be possible for a planet to have the same surface gravitational field as earth, but a different escape speed from the surface?

6 Orbits E > 0: Open –hyperbola E = 0: Escape speed –parabola E < 0: Closed –ellipse

7 Group Work Find the tangential speed v of an object in circular orbit a distance r from an attractor of mass M. –Use the fact that the centripetal force is the force of gravity. Find the square of the orbital period T 2 as well. –Use the fact that v = 2  r/T.

8 Kepler’s Laws of planetary motion 1.Planets travel in elliptical paths with one focus at the Sun. 2.At all times, a planet’s path traces out equal areas. 3.The square of a planet’s orbital period is directly proportional to the cube of the semi-major axis of the orbit.

9 Kepler’s Laws 2.Equal-area law 3.T 2  a 3 –For planets, y and AU are easy units

10 Kepler’s laws Work for other systems too (planets and moons, etc.)

11 Example Problem Calculate the orbital distance of the Moon based on the observation that the Moon orbits the Earth once every 27.3 days. The mass of the Moon is 7.35  10 22 kg and the mass of the Earth is 6.0  10 24 kg.

12 Double Systems Both objects orbit system’s center of mass Orbital radii < separation Centripetal force = gravity

13 Weightlessness Actually free fall No force opposes gravity

14 Example Problem The planet Uranus has a radius of 25,362 km and a surface gravity of 8.87 N/kg at its poles. Its moon Miranda is in a circular orbit at a distance of 129,560 km from Uranus’s center. Miranda has a mass of 6.6  10 19 kg and a radius of 235 km. a.Calculate the mass of Uranus from the given data.

15 Example Problem The planet Uranus has a radius of 25,362 km and a surface gravity of 8.87 N/kg at its poles. Its moon Miranda is in a circular orbit at a distance of 129,560 km from Uranus’s center. Miranda has a mass of 6.6  10 19 kg and a radius of 235 km. b.Calculate the magnitude of Miranda’s orbital acceleration due to its orbital motion about Uranus.

16 Example Problem The planet Uranus has a radius of 25,362 km and a surface gravity of 8.87 N/kg at its poles. Its moon Miranda is in a circular orbit at a distance of 129,560 km from Uranus’s center. Miranda has a mass of 6.6  10 19 kg and a radius of 235 km. c.Calculate the acceleration due to Miranda’s gravity at the surface of Miranda.

17 Example Problem The planet Uranus has a radius of 25,362 km and a surface gravity of 8.87 N/kg at its poles. Its moon Miranda is in a circular orbit at a distance of 129,560 km from Uranus’s center. Miranda has a mass of 6.6  10 19 kg and a radius of 235 km. d.Do the answers to parts b and c mean that an object released 1 m above Miranda’s surface on the side toward Uranus will fall up relative to Miranda? Explain what is happening.


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