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# Chapter 11 GRAVITATION

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### Chapter 11 GRAVITATION

1. 1. Newton's Universal Law of Gravitation
2. 2. Newton's Universal Law of Gravitation ๏ฎ The gravitational force that exists between two masses ๐1 and ๐2 is given by ๐ญ ๐ = ๐ฎ ๐ ๐ ๐ ๐ ๐ ๐ where ๐ - is the distance of separation between their centers. ๐บ = 6.67 ๐ฅ 10โ11 Nm2/kg2 - universal gravitation constant ๐ ๐1 ๐2๐น๐ ๐น๐
3. 3. Relationship between Force, mass and distance.
4. 4. Relationship between Force, mass and distance.
5. 5. ๐1 ๐2 ๐3 ๐1 ๐2 ๐น31 ๐น21 NET GRAVITATIONAL FORCE For three or more objects, finding the net gravitational force is by component! REVIEW YOUR COMPONENT METHOD!!! ๐น31 ๐น21 ๐น1 Cartesian Plane:
6. 6. Gravitational Field Strength & Gravitational Acceleration A gravitational field (popularly known as acceleration due to gravity) is created by an object causing masses inside it to experience the gravitational force. e๐ฅ๐๐๐๐๐: ๐ ๐ธ = 9.8 m/๐ 2
7. 7. Gravity Near The Earthโs Surface At the Earthโs surface: ๐น๐ = ๐ค ๐บ ๐๐ ๐ธ ๐๐ธ 2 = ๐๐ ๐ธ ๐ ๐ธ = ๐บ ๐ ๐ธ ๐๐ธ 2 = 9.8 m/s2 In general: ๐ ๐ = ๐บ ๐ ๐ ๐๐ 2 Gravitational field at the objectโs surface where ๐ ๐ = mass of the object having the field ๐๐ = radius of that object ๐ ๐ธ = mass of the earth ๐๐ธ = radius of the earth
8. 8. ๐ The effective g, gโ As you go far from the Earthโs surface, the gravitational field decreases. So, the effective g (gโ): where ๐ฆ = distance above objectโฒ s surface ๐ = ๐๐ + ๐ฆ (๐ > ๐๐) ๐โฒ = ๐บ ๐ ๐ ๐2 Since ๐ ๐ = ๐บ ๐ ๐ ๐ ๐ 2 ๐โฒ = ๐ ๐ ๐๐ ๐ 2
9. 9. Sample Problems: 1. Two objects attract each other with a gravitational force of magnitude 1.00 ๐ฅ 10โ8 N when separated by 20.0 cm. If the total mass of the objects is 5.00 kg, what is the mass of each? 2. Calculate the effective value of g, at 3200 m and 3200 km above the earthโs surface. 3. Calculate the velocity of a satellite moving in a stable circular orbit about the Earth at a height of 3600 km.
10. 10. Satellite Motion and Weightlessness without gravity With gravity Artificial satellite is put into orbit by accelerating it to a sufficiently tangential speed with the use of the rocket. If the speed is too high, the satellite will escape. If the speed is too low, it will fall back to earth. Fg
11. 11. ๐น๐ = ๐ ๐ฃ2 ๐ ๐บ ๐๐ ๐2 = ๐ ๐ฃ2 ๐ ๐ฃ = ๐บ ๐ ๐ Speed of satellite at orbit radius r where ๐ = mass of the object/planet that the satellite ๐ is orbiting
12. 12. Satellite Motion and Weightlessness The โweightlessnessโ experienced by a person in a satellite orbit close to Earth is the same apparent weightlessness experienced in a freely falling elevator.
13. 13. Keplerโs Laws of Planetary Motion Keplerโs First Law: The path of each planet about the Sun is an ellipse with the Sun at one focus An Ellipse is a closed curve such that the sum of the distances from any point P on the curve to two fixed points (called the foci, F1 and F2) remains constant.
14. 14. Keplerโs Laws of Planetary Motion Keplerโs Second Law: Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal periods of time.
15. 15. Sun 4 3
16. 16. Keplerโs Laws of Planetary Motion Keplerโs Third Law: The ratio of the squares of the periods of any two planets revolving around the Sun is equal to the ratio of the cubes of their mean distances from the Sun.
17. 17. Sample Problems 1. Four 7.5-kg spheres are located at the corners of a square of side 0.60 m. Calculate the net gravitational force on one sphere due to the other three. 2. Calculate the effective value of g, at 3200 m and 3200 km above the earthโs surface. 3. Calculate the velocity of a satellite moving in a stable circular orbit about the Earth at a height of 3600 km. 4. Neptune is an average distance of 4.5 x 109 km from the Sun. Estimate the length of the Neptunian year given that the Earth is 1.50 x 108 km from the Sun on the average.
18. 18. TERMINAL VELOCITY
19. 19. Fg =mg 1. Object about to start falling. V=0 W=mg 2. Object is falling. V>0 Friction a 9.8 m/s2 Friction=Fg a= m/s2 TERMINAL VELOCITY 3. The object now moves with TERMINAL VELOCITY. An object is dropped from REST. V = max
20. 20. Fg =mg 1. Object about to start falling. V=0 =mg 2. Object is falling. V>0 Friction a=10m/s2 The object accelerates towards the earth. a<10 m/s2 Acceleration decreased! Friction= a= 0 m/s2 TERMINAL VELOCITY 3. The object now moves with TERMINAL VELOCITY. An object is dropped from REST. V = maxFres < Fg