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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 2/21 Do Now A particle of mass m rotates with a uniform speed on the inside.

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Presentation on theme: "Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 2/21 Do Now A particle of mass m rotates with a uniform speed on the inside."— Presentation transcript:

1 Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 2/21 Do Now A particle of mass m rotates with a uniform speed on the inside of a bowl’s parabolic frictionless surface in a horizontal circle of radius R = 0.4 meters as shown below. At the position of the particle the surface makes an angle θ = 17 o with the vertical. What is the angular velocity of the particle?

2 Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley PowerPoint ® Lectures for University Physics, Twelfth Edition – Hugh D. Young and Roger A. Freedman Lectures by James Pazun Chapter 12 Gravitation

3 Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Goals for Chapter 12 To study Newton’s Law of Gravitation To consider gravitational force, weight, and gravitational energy To compare and understand the orbits of satellites and celestial objects

4 Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. 12.1 Newton’s Law of Gravitation

5 Lowercase g is the acceleration due to gravity, which relates the weight w of a body to its mass m: w = mg. The value of g is different at different locations on the earth's surface and on the surfaces of different planets. By contrast, capital G relates the gravitational force between any two bodies to their masses and the distance between them. Henry Cavendish measured the value of G in 1798. We call G a universal constant because it has the same value for any tow bodies, no matter where in space they are located. CAUTIONDon’t confuse g and G

6 Example 12.1 Calculating gravitational force The mass m 1 of one of the small spheres of a Canvendish balance is 0.0100 kg, the mass m 2 of one of the large spheres is 0.500 kg. And the center-to-center distance between each large sphere and the nearer small one is 0.0500 m. Find the gravitational force F g on each sphere due to the nearest other sphere.

7 Suppose one large sphere and one small sphere are detached from the apparatus in Example 12.1 and placed 0.0500 m from each other at a point in space far removed from all other bodies. What is the magnitude of the acceleration of each? Example 12.2 Acceleration due to gravitational attraction

8 Example 12.3 superposition of gravitational forces Many stars in the sky are actually systems of two or more stars held together by their mutual gravitational attraction. The figure shows a three-star system at an instant when the stars are at the vertices of a 45 o right triangle. Find the magnitude and direction of the total gravitational force exerted on the small star by the two large ones.

9 We defined the weight of a body as the attractive gravitational force exerted on it by the earth. The broaden definition of weight is: The weight of a body is the total gravitational force exerted on the body by all other bodies in the universe. When the body is near the surface of the earth, we can neglect all other gravitational forces and consider the weight as just the earth’s gravitational attraction. At the surface of the moon we consider a body’s weight to be the gravitational attraction of the moon, so on. 12.2 Weight

10 If we model the earth as a spherically symmetric body with radius R E and mass m E, the weight w of a small body of mass m at the earth’s surface is We also know that weight w of a body is the force that causes the acceleration g of free fall, w = mg. Equating this with the above equation, we find The acceleration due to gravity g is independent of the mass m of the body. This equation allows us to calculate the mass of the earth:

11 An unmanned Lander is sent to the surface of the planet Mars, which has radius R M = 3.40 x 10 6 m and mass m M = 6.42 x 10 23 kg. The earth weight of the Mars Lander is 3290 N. Calculate its weight F g and the acceleration g M due to the gravity of Mars: a. 6.0 x 10 6 m above the surface of Mars b. At the surface of Mars. Example 12.4 Gravity on Mars

12 “Journey to the center of the earth” Suppose we drill a hole through the earth (radius R E, mass m E ) along a diameter and drop a mail pouch (mass m) down the hole. Derive an expression for the gravitational force on the pouch as a function of its distance r from the center. Assume that the density of the earth is uniform.

13 Weight Gravity (and hence, weight) decreases as altitude rises outside Earth and also decrease as approaching the center of Earth and becomes zero at the center of Earth.

14 The apparent weight of a body on earth differs slightly from the earth’s gravitational force because the earth rotates and is therefore not precisely an inertial frame of reference. We have used the fact that the earth is an approximately spherically symmetric distribution of mass. But this does not mean that the earth is uniform. In fact, the density of the earth decreases with increasing distance from its center.

15 Rank the following hypothetical planets in order from highest to lowest surface gravity: 1.Mass = 2 times the mass of earth radius = 2 times the radius of the earth; 2.Mass = 4 times the mass of the earth, radius = 4 times the radius of the earth 3.Mass = 4 times the mass of the earth, radius = 2 times the radius of the earth; 4.Mass = 2 times the mass of the earth, radius = 4 times the radius of the earth. 3, 1, 2, 4 Test Your Understanding 12.2

16 Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 2/24 Do Now Ball B (moment of inertial about its center ⅔ MR 2 ), rolls down the distance x along the inclined plane without slipping. In terms of acceleration due to earth’s gravity g, what is the acceleration of ball B along the inclined plane? Extra credit is due Test is on Tuesday, 3/4

17 Brief review MC packet: #21, 23, 24, 25, 32,36

18 12.3 Gravitational Potential Energy To determine the gravitational potential energy of an object at a height that is way above the earth’s surface, we need to calculate work done by the gravitational force on the object and use equation: W grav = - ∆U = - (U 2 – U 1 ) = U 1 – U 2

19 Gravitational potential energy depends on the distance r between the body of mass m and the center of the earth. When the body moves away from the earth, r increases, the gravitational force does negative work, and U increases (becomes less negative). When the body “falls” toward earth, r decreases, the gravitational work is positive, energy decreases (becomes more negative). Note that U is relative. Only ∆U is significant.

20 In Jules Verne’s 1865 story with this title, three men were sent to the moon in a shell fired from a giant cannon sunk in the earth in Florida. a.Find the muzzle speed needed to shoot the shell straight up to a height above the earth equal to the earth’s radius. b.Find the escape speed – that is, the muzzle speed that would allow the shell to escape from the earth completely. Neglect air resistance, the earth’s rotation, and the gravitational pull of the moon. (R E = 6380 km, M E = 5.97 x 10 24 kg) Example 12.5 “From the earth to the moon”

21 Extra credit on test – use the above equation to prove that near Earth’s surface the potential energy is More on Gravitational Potential Energy

22 Is it possible for a planet to have the same surface gravity as the earth (that is the same value of g at the surface) and yet have a greater escape speed? Test Your Understanding 12.3

23 Example The planet Saturn has about 100 times the mass of the earth and is about 10 times farther from the sun than the earth is. Compared to the acceleration of the earth caused by the sun’s gravitational pull, how great is the acceleration of Saturn due to the sun’s gravitation? [show work] 1.100 times greater 2.10 times greater 3.The same 4.1/10 as great 5.1/100 as great.

24 12.4 Satellite motion Trajectories 1 through 7 show the effect of increasing the initial speed. Trajectories 1 through 5 close on themselves and are called closed orbits. All closed orbits are ellipses; Trajectory 4 is a circle, a special case of an ellipse. Trajectories 6 and 7 are open orbits. For these paths the projectile never returns to its starting point but travels ever farther away from the earth.

25 A circular orbit is the simplest and also an important case because artificial satellites as well as planets around the sun have nearly circular orbits. Since the only force acing on a satellite in circular orbit around the earth is the earth’s gravitational attraction, which is directed toward the center of the earth and hence toward the center of the orbit, the satellite is in uniform circular motion with constant speed. Satellites: Circular Orbits

26 To find the constant speed v of a satellite in a circular orbit with radius r measured from the center of the earth, we need to use Newton’s 2 nd law: F net = ma This relationship shows that we can’t chose the orbit radius r and the speed v independently; for the given radius r, the speed v for a circular orbit is determined. The larger the orbit, the slower the speed. The equation also shows that the motion of a satellite does not depend on its mass. Any satellite launched in the same orbit will have the same speed.

27 An astronaut on board a space shuttle is a satellite of the earth in the same orbit, has the same velocity and acceleration as the shuttle. The only force acting on the astronaut is gravity, so the astronaut is experiencing apparent weightlessness as in a freely falling elevator; Both the international space station and the moon are satellites of the earth. The moon orbits much farther from the center of the earth than does the space station, so its has a slower orbital speed and a longer orbital period. International space station R = 6800 km V = 7.7 km/s T = 93 min Moon Distance from center of earth = 384,000 km V = 1.0 km/s T = 27.3 days

28 In circular motion: The relationship between r and T The equation shows that larger orbits longer periods.

29 Comparing circular orbit speed with escaping speed Circular orbit speed: escaping speed: The escape speed from a spherical body with radius R is times greater than the speed of a satellite in a circular orbit at that radius. If our spacecraft is in circular orbit around any planet, we have to multiply our speed by a factor of to escape to infinity, regardless of the planet’s mass.

30 The total mechanical energy in a circular orbit is negative and equal to ½ the potential energy. Increasing the orbit radius r means increasing the mechanical energy (making E less negative). If the satellite is in a relatively low orbit that encounters the outer fringes of earth’s atmosphere, mechanical energy decreases due to negative work done by the force of air resistance; as a result, the orbit radius decreases until the satellite hits the ground or burns up in the atmosphere. The total mechanical energy E = K + U Circular orbit speed:

31 Example 12.6 A satellite orbit Suppose you want to place a 1000 kg weather satellite into a circular orbit 300 km above the earth’s surface. a.What speed, period and radial acceleration must it have? b.How much work has to be done to place this satellite into orbit? c.How much additional work would have to be done to make this satellite escape the earth? (R E =6380 km, m E =5.97 x 10 24 kg)

32 Your personal spacecraft is in a low-altitude circular orbit around the earth. Air resistance from the outer regions of the atmosphere does negative work on the spacecraft, causing the orbital radius to decrease slightly. Does the speed of the spacecraft 1.remain the same 2.increase 3.decrease Test Your Understanding 12.4

33 12.5 Kepler’s laws for planetary motion 1 st Law: Each planet moves in an elliptical orbit with the sun at one focus. 2 nd Law: A line connecting the sun to a given planet sweeps out equal areas in equal times.

34 Perihelion is the point closest to the sun; Aphelion is the point furthest from the sun. Semi-Major axis is the length of a ea = e∙a: the distance from each focus to the center of the ellipse. Eccentricity (e): a dimensionless number between 0 and 1. If e=0, ea=0, the ellipse is a circle.

35 Where m s is the mass of the sun. a is the semi major axis. Note that the period does not depend on the eccentricity e. An asteroid in an elongated elliptical orbit with semi-major axis a will have the same orbital period as a planet in a circular orbit of radius a. The key difference is that the asteroid moves at different speeds at different points in its elliptical orbit, while the planet’s speed is constant around its circular orbit. The periods of the planets are proportional to the 3 / 2 powers of the semi-major axis lengths in their orbits Kepler’s Third Law

36 Example 12.7 Orbital Speeds The speed v is maximum at perihelion. At what point in an elliptical orbit does a planet have the greatest speed?

37 Example 12.8 Kepler’s third Law The asteroid Pallas has an orbital period of 4.62 years and an orbital eccentricity of 0.233. Find the semi-major axis of its orbit. (m s =1.99x10 30 kg)

38 Example 12.9 Comet Halley Comet Halley moves in an elongated elliptical orbit around the sun. at perihelion, the comet is 8.75 x 10 7 km from the sun; at aphelion, it is 5.26 x 10 9 km from the sun. Find the semi-major axis, eccentricity, and period of the orbit.

39 Planetary Motions and the Center of Mass

40 The orbit of Comet X has a semi-major axis that is four times larger than the semi-major axis of Comet Y. What is the ratio of the orbital period of X to the orbital period of Y? T 1 / T 2 = (a 1 /a 2 ) 3/2 = (4) 3/2 = 8 Test Your Understanding 12.5

41 Example The planet Saturn has about 100 times the mass of the earth and is about 10 times farther from the sun than the earth is. Compared to the acceleration of the earth caused by the sun’s gravitational pull, how great is the acceleration of Saturn due to the sun’s gravitation? [show work] 1.100 times greater 2.10 times greater 3.The same 4.1/10 as great 5.1/100 as great. F g = Gm E m s /r 2 a E = Gm s /r E 2 a Saturn = Gm s /r Saturn 2 a Saturn = Gm s /(10r E ) 2 a Saturn = (1/100) a E

42 Example How far from a very small 150 kg ball would a particle have to be placed so that the ball pulled on the particle just as hard as the earth does?

43 Example What is the escape speed from an asteroid of diameter 255 km with a density of 2720 kg/m 3 ? 88.7 m/s

44 Example a.How far from a very small 120 kg ball would a particle have to be placed so that the ball pulled on the particle just as hard as the earth does? b.Is it reasonable that you could actually set up this as an experiment? Why? 2.86 x10 -5 m

45 On Earth’s surface, r 1 ≈ r 2 = R E ; (radius of the earth) and r 2 – r 1 = h (height above Earth’s surface) More on Gravitational Potential Energy


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