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# Universal Gravitation

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### Universal Gravitation

1. 1. + Chapter 7 Pg. 232-261
2. 2. + 7.1 Circular Motion Pg. 234- 239
3. 3. + Circular Motion  Any object that revolves about a single axis  The line about which the rotation occurs is called the axis of rotation
4. 4. + Tangential Speed (vt)  Speed of an object in circular motion  Uniform circular motion: vt has a constant value  Onlythe direction changes  Example shown to the right  How would the tangential speed of a horse near the center of a carousel compare to one near the edge? Why?
5. 5. + Centripetal Acceleration (ac)  Acceleration directed toward the center of a circular path  Acceleration is a change in velocity (size or direction).  Direction of velocity changes continuously for uniform circular motion.
6. 6. + Centripetal Acceleration (magnitude)  How do you think the magnitude of the acceleration depends on the speed?  How do you think the magnitude of the acceleration depends on the radius of the circle?
7. 7. + Example  A testcar moves at a constant speed around a circular track, If the car is 48.2m from the center and has the centripetal acceleration of 8.05m/s2, what is the car’s tangential speed?  ac = vt2 r
8. 8. + Example  ac = vt2 r  acr = vt2  vt =√ac r  vt= √(8.05) (48.2)  vt= 19.69 m/s
9. 9. + Tangential Acceleration Occurs if the speed increases Directed tangent to the circle Example: a car traveling in a circle  Centripetal acceleration maintains the circular motion.  directed toward center of circle  Tangential acceleration produces an increase or decrease in the speed of the car.  directed tangent to the circle
10. 10. + Centripetal Acceleration Click below to watch the Visual Concept. Visual Concept
11. 11. + Centripetal Force  Maintains motion in a circle  Can be produced in different ways, such as  Gravity  A string  Friction  Which way will an object move if the centripetal force is removed?  Ina straight line, as shown on the right
12. 12. + Centripetal Force (Fc) Fc m ac 2 vt an d a c r 2 m vt so Fc r
13. 13. + Example  A pilot is flying a small plane at 56.6 m/s in a circular path with a radius of 188.5m. The centripetal force needed to maintain the plane’s circular motion is 1.89 X 104 N. What is the plane’s mass?  Given:  vt= 56.6 m/s r= 188.5 m  Fc= 1.89X 104 N m= ??
14. 14. + Example  m vt 2 rearranged Fc r Fc m 2 r vt m = (1.89 X 104)(188.5m) (56.6)2 m = 1112.09 kg
15. 15. + Describing a Rotating System  Imagine yourself as a passenger in a car turning quickly to the left, and assume you are free to move without the constraint of a seat belt.  How does it “feel” to you during the turn?  How would you describe the forces acting on you during this turn?  Thereis not a force “away from the center” or “throwing you toward the door.”  Sometimes called “centrifugal force”  Instead, your inertia causes you to continue in a straight line until the door, which is turning left, hits you.
16. 16. + 7.2 Newton’s Law of Universal Gravitation Pg. 240-247
17. 17. + Gravitational Force  The mutual force of attraction between particles of matter.  Gravitational force depends on the masses and the distance of an object.
18. 18. + Simpson’s Video  http://www.lghs.net/ourpages/users/dburn s/ScienceOnSimpsons/Clips_files/3D- Homer.m4v
19. 19. + Family Guy Video
20. 20. + Newton’s Thought Experiment  What happens if you fire a cannonball horizontally at greater and greater speeds?  Conclusion: If the speed is just right, the cannonball will go into orbit like the moon, because it falls at the same rate as Earth’s surface curves.  Therefore, Earth’s gravitational pull extends to the moon.
21. 21. + Law of Universal Gravitation  Fg is proportional to the product of the masses (m1m2).  Fgis inversely proportional to the distance squared (r2).  Distance is measured center to center. G converts units on the right (kg2/m2) into force units (N). G = 6.673 x 10-11 N•m2/kg2
22. 22. + Law of Universal Gravitation
23. 23. + The Cavendish Experiment  Cavendish found the value for G.  He used an apparatus similar to that shown above.  He measured the masses of the spheres (m1 and m2), the distance between the spheres (r), and the force of attraction (Fg).  Hesolved Newton’s equation for G and substituted his experimental values.
24. 24. + Gravitational Force  Ifgravity is universal and exists between all masses, why isn’t this force easily observed in everyday life? For example, why don’t we feel a force pulling us toward large buildings?  The value for G is so small that, unless at least one of the masses is very large, the force of gravity is negligible.
25. 25. + Ocean Tides  What causes the tides?  How often do they occur?  Why do they occur at certain times?  Are they at the same time each day?
26. 26. +
27. 27. + Ocean Tides  Newton’s law of universal gravitation is used to explain the tides.  Since the water directly below the moon is closer than Earth as a whole, it accelerates more rapidly toward the moon than Earth, and the water rises.  Similarly, Earth accelerates more rapidly toward the moon than the water on the far side. Earth moves away from the water, leaving a bulge there as well.  As Earth rotates, each location on Earth passes through the two bulges each day.
28. 28. + Gravity is a Field Force  Earth,or any other mass, creates a force field.  Forces are caused by an interaction between the field and the mass of the object in the field.  The gravitational field (g) points in the direction of the force, as shown.
29. 29. + Calculating the value of g Since g is the force acting on a 1 kg object, it has a value of 9.81 N/m (on Earth).  The same value as ag (9.81 m/s2) The value for g (on Earth) can be calculated as shown below. Fg GmmE GmE g 2 2 m mr r
30. 30. + Classroom Practice Problems  Find the gravitational force that Earth (mE= 5.97 1024 kg) exerts on the moon (mm= 7.35 1022 kg) when the distance between them is 3.84 x 108 m.  Answer: 1.99 x 1020 N  Findthe strength of the gravitational field at a point 3.84 x 108 m from the center of Earth.  Answer: 0.00270 N/m or 0.00270 m/s2
31. 31. + 7.3 Motion in Space Pg. 248- 253
32. 32. + Kepler’s Laws  Johannes Kepler built his ideas on planetary motion using the work of others before him.  Nicolaus Copernicus and Tycho Brahe
33. 33. + Kepler’s Laws  Kepler’s first law  Orbits are elliptical, not circular.  Some orbits are only slightly elliptical.  Kepler’s second law  Equal areas are swept out in equal time intervals.  Basically things travel faster when closer to the sun
34. 34. + Kepler’s Laws Kepler’s third law  Relatesorbital period (T) to distance from the sun (r)  Period is the time required for one revolution.  Asdistance increases, the period increases.  Not a direct proportion  T2/r3 has the same value for any object orbiting the sun
35. 35. + Equations for Planetary Motion  UsingSI units, prove that the units are consistent for each equation shown below.
36. 36. + Classroom Practice Problems  A large planet orbiting a distant star is discovered. The planet’s orbit is nearly circular and close to the star. The orbital distance is 7.50 1010 m and its period is 105.5 days. Calculate the mass of the star.  Answer: 3.00 1030 kg  What is the velocity of this planet as it orbits the star?  Answer: 5.17 104 m/s
37. 37. + Weight and Weightlessness Bathroom scale  A scale measures the downward force exerted on it.  Readings change if someone pushes down or lifts up on you.  Your scale reads the normal force acting on you.
38. 38. + Apparent Weightlessness  Elevator at rest: the scale reads the weight (600 N).  Elevatoraccelerates downward: the scale reads less.  Elevator in free fall: the scale reads zero because it no longer needs to support the weight.
39. 39. + Apparent Weightlessness  Youare falling at the same rate as your surroundings.  No support force from the floor is needed.  Astronautsare in orbit, so they fall at the same rate as their capsule.  Trueweightlessness only occurs at great distances from any masses.  Even then, there is a weak gravitational force.

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