Chapter_13

# Chapter_13 - Chapter 13 Gravitation In this chapter we will...

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Chapter 13 Gravitation In this chapter we will explore the following topics: - Newton’s law of gravitation that describes the attractive force between two point masses and its application to extended objects -The acceleration of gravity on the surface of the earth, above it, as well as below it. - Gravitational potential energy - Kepler’s three laws on planetary motion (13-1)

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m m Newton realized that the force which holds the moon it its orbit around the earth is of the same nature as the force that makes an apple drop near the surface of the earth. N Newton's Law of Gravitation ewton infered that every object in the universe attracts every other object via gravitation. Newton formulated a force law known as Every particle any other particle with a gravitational force that has the following characteristics: attracts 1. Newton's law of gravitation. 1 2 1 2 2 The force acts along the line that connects the two particles Its magnitude is given by the equation: Here and are the masses of the two particles, is their separation an i d m m r m F G m G r = 2. 11 2 2 s the gravitational constant. Its value is: 6.67 10 N.m / G kg - = × (13-2)
m m F r 12 1 2 21 2 1 12 21 The gravitational force exerted on by is equal in magnitude to the force exerted on by but opposite in direction. The two forces obey Newton's third law: 0 F m m F m m F F + = r r 1 2 1 2 Newton proved that a uniform shell attarcts a particle that is outside the shell as if the shell's mass were concentrated at the shell center If the particle is inside the shell, the m m F G r = Note : net force is zero Consider the force the earth (radius , mass ) exerts on an apple of mass . The earth can be thought of as consisting of concentric shells. Thus from the apple's point of view the earth behaves l F R M m ike a point mass at the earth center. The magnitude of the force is given by the equation: (13-3) m m r F 1 2 2 m m F G r = 2 mM F G R =

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(12-16) 2 2 24 How to measure the mass of the earth? Gravitational force on apple = = Solve for M: = 5.96 10 kg = × mMG mg R gR M G
The net gravitational force exerted by a group of particles is equal to the vector sum of the contribution from each particle. Gravitation and the Principle of Superposition 1 1 2 3 12 13 1 2 3 1 1 12 13 For example the net force exerted on by and is Here and are the forces exerted on by and , respectively. In general the force exerted on by n pa rt = + F m m m F F m m F m m F F r r r r r r 1 12 13 14 1 1 2 icles is ... = = + + + + = n n i i F F F F F F r r r r r r 1 The gravitation force exerted by a continuous extended object on a particle of mass can be calculated using the principle of superposition. The object is divided into elements of mass . the net m dm 1 1 1 1 2 force on is the vector sum of the forces exerted by each element. The sum takes the form of an integral Here is the force exerted on by = = m F dF dF m dm Gm dm dF r r r r (13-4) dm r dF r object

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2 If we assume that the earth is a sphere of mass , the magnitude of the gravitational force on an object of mass by earth ( : distance from the ce = Gravitation near the earth's Surface M F
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Chapter_13 - Chapter 13 Gravitation In this chapter we will...

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