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PHY 151: Lecture 6A 6.1 Newton’s Law of Universal Gravity 6.2 Gravitational Formula 6.3 Extending Particle in Uniform Circular Motion Model.

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Presentation on theme: "PHY 151: Lecture 6A 6.1 Newton’s Law of Universal Gravity 6.2 Gravitational Formula 6.3 Extending Particle in Uniform Circular Motion Model."— Presentation transcript:

1 PHY 151: Lecture 6A 6.1 Newton’s Law of Universal Gravity 6.2 Gravitational Formula 6.3 Extending Particle in Uniform Circular Motion Model

2 PHY 151: Lecture 6A Circular Motion Other Applications of Newton’s Laws 6.1 Newton’s Law of Universal Gravity

3 Newton’s Law of Universal Gravity The Story - 1 Newton observes apple falling from a tree Apple is accelerated, since its velocity changes from zero as it moves toward the ground By Newton’s 2 nd Law there must be a force acting on the apple to cause acceleration Call the force “gravity” Imagine tree is twice as high The apple is still accelerated towards the ground Suggests that this force reaches to top of the tallest tree

4 Newton’s Law of Universal Gravity The Story - 2 Newton has a brilliant insight If the force of gravity reaches the top of the highest tree, might it not reach even further Might it not reach all the way to the orbit of the Moon The orbit of the Moon about the Earth could be a consequence of the gravitational force Acceleration due to gravity could change the velocity of the Moon in such a way that it follows an orbit around the earth

5 Newton’s Law of Universal Gravity The Story - 3 Suppose we fire a cannon ball horizontally from a high mountain The projectile will eventually fall to Earth As we increase the muzzle velocity, the projectile will travel further and further before returning to Earth If the cannon fired the cannon ball with the right velocity, the projectile would travel completely around the Earth

6 Newton’s Law of Universal Gravity The Story - 4 The projectile is always falling because of gravity but never reaching the Earth, which is curving away at the same rate that the projectile falls That is, the cannon ball would have been put into orbit around the Earth Newton concluded that the orbit of the Moon was of exactly the same nature The moon continuously “fell” in its path around the Earth due to gravity, thus producing an orbit

7 Newton’s Law of Universal Gravity The Story - 5 Newton came to the conclusion that any two objects in the Universe exert gravitational attraction on each other

8 PHY 151: Lecture 6 Circular Motion Other Applications of Newton’s Laws 6.2 Gravitational Formula

9 Dependence on Distance - 1 Acceleration on earth’s surface is 9.81 m/s 2 Center of moon is 3.84 x 10 8 meters from center of earth Moon’s orbital period is 27.3 days = 2.36 x 10 6 seconds

10 Dependence on Distance - 2 Centripetal acceleration of moon due to earth’s gravitational attraction is a c = 4  2 r/T 2 (we will soon learn formula)  T is orbital period of moon 4  2 r/T 2 4  2 (3.84x10 8 )/(2.36x10 6 ) 2 0.00272 m/s 2

11 Dependence on Distance - 3 ratio of earth’s acceleration to moon’s acceleration 9.81/0.0272 = 3607 ratio of earth’s radius to distance of moon’s center from earth’s center is 60 3607 is approximately 60 2 bigger the radius the smaller the acceleration therefore, dependence on distance is 1/r 2

12 Mass Dependence – 1 Galileo showed that different masses fall with the same acceleration Let m 1 be mass of falling object and m 2 mass of earth Therefore, the formula for the force of gravity must be like m 1 h(m 2 )/r 2 = m 1 a h is some formula containing only m 2 Cancellation of m 1 from both sides of the equation means that a does not depend on the mass m 1

13 Mass Dependence - 2 If Newton’s Third Law applies to gravity Then earth attracts moon and moon attracts earth with forces of the same magnitude It should not matter which mass is called m 1 and which is called m 2 Dependence on mass must be m 1 h(m 2 ) h is a function

14 Mass Dependence - 3 h is a function functions For example:  h is m 2, (m 2 ) 2, sin 2 (m 2 ), e 2(m2), … If the interchange of m 1 and m 2 produces the same force Then h must be m 2 Force of gravity is proportional to m 1 m 2 /r 2

15 Constant of Proportionality meter and kilogram were defined without any concern for the force of gravity The constant of proportionality, G, adjusts for the fact the meter and kilogram are not appropriate to gravity

16 Newton’s Law of Universal Gravity Every particle in the universe exerts an attractive force on every other particle A particle is a piece of matter, small enough in size to be regarded as a mathematical point For two particles that have mass m 1 and m 2 and are separated by a distance r, the force that each exerts on the other is directed along the line joining the particles and has a magnitude given by G is the universal gravitational constant, whose value is found experimentally G = 6.673 x 10 -11 Nm 2 /kg 2

17 Newton’s Law of Universal Gravity Special Requirements Newton’s law of gravitation applies only to particles Newton proved that an object of finite size can be considered a particle for purposes of the gravitation law, provided the mass of the object is distributed with spherical symmetry about its center In this case, r is the distance between the centers of the spheres and not between the outer surfaces The gravitational forces that the spheres exert on each other are the same as if the entire mass of each were concentrated at its center

18 Newton’s Law of Universal Gravity Example Calculate the gravitational force between the Earth and the Moon  F = GM e M m /r em 2  F = 6.673x10 -11 x 5.98 x 10 24 x 7.35 x 10 22 /(3.85 x 10 8 ) 2  F = 2.0 x 10 20 N

19 Newton’s Law of Universal Gravity Weight - 1 Weight of an object on or above the earth is the gravitational force that the earth exerts on the object Weight always acts downward, toward the center of the earth On or above another astronomical body, the weight is the gravitational force exerted on the object by the body SI Unit of Weight: newton (N) Symbol for Weight: W

20 Newton’s Law of Universal Gravity Weight - 2 m is the mass of the object M E is the mass of the Earth r is the distance from the center of the earth to the object r must be equal to or greater than the radius of the Earth

21 Newton’s Law of Universal Gravity Relation between Mass and Weight Mass  A quantitative measure of inertia  An intrinsic property of matter  Does not change as an object is moved from one location to another Weight  Gravitational force acting on the object  Can vary, depending on how far the object is above the earth’s surface or whether it is near another body such as the moon

22 Newton’s Law of Universal Gravity Weight at Earth’s Surface m is the mass of the object M E is the mass of the Earth R e is radius of the Earth

23 Newton’s Law of Universal Gravity Mass / Weight - Example What is the mass of a person weighing 740 N on the Earth?  W = mg  m = W/g  m = 740 / 9.8 = 75.5 kg

24 PHY 151: Lecture 6A Circular Motion Other Applications of Newton’s Laws 6.3 Extending Particle in Uniform Circular Motion Model

25 Circular Motion Two problems using Newton’s Laws of Motion have been developed The models have been applied to linear motion Newton’s Laws can be applied to other situations: –Objects traveling in circular paths –Motion observed from an accelerating frame of reference –Motion of an object through a viscous medium Many examples will be used to illustrate the application of Newton’s Laws to a variety of new circumstances

26 Uniform Circular Motion, Acceleration A particle moves with a constant speed in a circular path of radius r with an acceleration The magnitude of the acceleration is given by The centripetal acceleration,, is directed toward the center of the circle The centripetal acceleration is always perpendicular to the velocity

27 Uniform Circular Motion, Force A force,, is associated with the centripetal acceleration The force is also directed toward the center of the circle Applying Newton’s Second Law along the radial direction gives

28 Uniform Circular Motion, cont. A force causing a centripetal acceleration acts toward the center of the circle It causes a change in the direction of the velocity vector If the force vanishes, the object would move in a straight-line path tangent to the circle –See various release points in the active figure


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