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Chapter09

# Chapter09 - Chapter 9 Gravitation 9.1 9.2 9.3 9.4 9.5 9.6...

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1 m 2 m r F F Chapter 9 Gravitation 9.1 Newton’s Law of Universal Gravitation 9.2 Gravitational Field 9.3 Gravitational Potential Energy 9.4 Satellite Orbits 9.5 Launching a Satellite 9.6 Energy in Satellite Motion 9.7 Kepler’s Laws 9.1 Newton’s Law of Universal Gravitation Newton’s law of universal gravitation states that the gravitational force F between any two bodies of mass m 1 and m 2 separated by a distance r is described by 2 2 1 r m m F . It is a center-to-center attraction between all forms of matter. The force of gravity between any two bodies is directly proportional to the product of their masses, and inversely proportional to the square of their separation. The proportionality constant is a universal constant called the gravitational constant G whose value is equal to 6.67 × 10 -11 N m 2 kg -2 . So we can write: 2 2 1 r m m G F = 1 r F

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Example 9-1 Twin asteroids X and Y , having the same mass ( M = 3.5 × 10 18 kg) are located 3.00 km apart. Find the net gravitational force of the spaceship when it is located at positions A and B as shown in the figure. Given that the mass of the spaceship is m = 2.50 × 10 7 kg. Answer: (1) When the spaceship is at A : The distance AX = AY = m 3350 1500 3000 2 2 = + . The angle o 1 6 . 26 3000 1500 tan = = - θ . The attractive force between the spaceship and asteroid X is 2 ) ( AX mM G F = along AX . The attractive force between the spaceship and asteroid Y is 2 ) ( AY mM G F = along AY . But the vertical components of the two attractive forces are of opposite direction and equal in magnitudes, they counterbalance each other. Hence the net force is equal to the summation of the two horizontal forces. cos ) ( cos ) ( 2 2 net AY mM G AX mM G F + = cos ) ( 2 2 AX mM G = Thus, N 10 30 . 9 6 . 26 cos ) m 3350 ( ) kg 10 50 . 3 )( kg 10 50 . 2 ( ) kg m N 10 67 . 6 ( 2 8 o 2 18 7 2 - 2 11 net × = × × × = - F (2) When the spaceship is at B : The attractive force between the spaceship and asteroid X is 2 ) ( BX mM G F = along BX . The attractive force between the spaceship and asteroid Y is 2 ) ( BY mM G F = along BY . Note that they are of the same magnitude but in opposite direction. Hence the net force acting on the spaceship is zero. 2 X Y 3.00 km B m M M A 1.50 km 1.50 km
9.2 Gravitational Field Strength If we place a test body of small mass m 0 near the Earth, it will experience a gravitational force F which has a definite direction and magnitude at each point in space. F always points toward the center of the Earth and its magnitude is m 0 g where g is the acceleration that a body would experience if it were released. We define the gravitational field strength of an object (which tells us how large the gravitational effect of that object is) as the gravitational force that it exerts on a test body per unit mass: 0 m F g = . (Note that the gravitational field strength of an object may have different values at different

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Chapter09 - Chapter 9 Gravitation 9.1 9.2 9.3 9.4 9.5 9.6...

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