381_PartUniversity Physics Solution

# 381_PartUniversity Physics Solution - 12-2 12.5. Chapter 12...

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12-2 Chapter 12 12.5. IDENTIFY: Use Eq.(12.1) to calculate g F exerted by the earth and by the sun and add these forces as vectors. (a) SET UP: The forces and distances are shown in Figure 12.5. Let E F ! and S F ! be the gravitational forces exerted on the spaceship by the earth and by the sun. Figure 12.5 EXECUTE: The distance from the earth to the sun is 11 1.50 10 m. r Let the ship be a distance x from the earth; it is then a distance rx from the sun. ES FF = says that 22 // ( ) Gmm x Gmm r x =− () 2 2 E mx m sr x and SE ( / ) rx xmm −= / rxxmm and (1 / ) mm =+ 11 8 30 24 1.50 10 m 2.59 10 m 1/ 1+ 1.99 10 kg/5.97 10 kg r x × == = × + ×× (from center of earth) (b) EVALUATE: At the instant when the spaceship passes through this point its acceleration is zero. Since " this equal-force point is much closer to the earth than to the sun. 12.6. IDENTIFY: Apply Eq.(12.1) to calculate the magnitude of the gravitational force exerted by each sphere. Each force is attractive. The net force is the vector sum of the individual forces. SET UP: Let + x be to the right. EXECUTE: (a) 11 2 2 11 g 5.00 kg 10.0 kg 6.673 10 N m /kg 0.100 kg 2.32 10 0.400 m 0.600 m x F ⎡⎤ = ×⋅ + = × Ν ⎢⎥ ⎣⎦ , with the minus sign indicating a net force to the left. (b) No, the force found in part (a) is the net force due to the other two spheres. EVALUATE: The force from the 5.00 kg sphere is greater than for the 10.0 kg sphere even though its mass is less, because r is smaller for this mass. 12.7. IDENTIFY: The force exerted by the moon is the gravitational force, M g 2 Gm m F r = . The force exerted on the person by the earth is wm g = . SET UP: The mass of the moon is 22 M 7.35 10 kg m . 11 2 2 6.67 10 N m /kg G . EXECUTE: (a) 22 11 2 2 3 moon g 82 (7.35 10 kg)(70 kg) (6.67 10 N m /kg ) 2.4 10 N (3.78 10 m) × × = × × . (b) 2 earth (70 kg)(9.80 m/s ) 690 N Fw = . 6 moon earth /3 . 5 1 0 . EVALUATE: The force exerted by the earth is much greater than the force exerted by the moon. The mass of the moon is less than the mass of the earth and the center of the earth is much closer to the person than is the center of the moon. 12.8. IDENTIFY: Use Eq.(12.2) to find the force each point mass exerts on the particle, find the net force, and use Newton&s second law to calculate the acceleration. SET UP: Each force is attractive. The particle (mass m ) is a distance 1 0.200 m r = from 1 8.00 kg m = and therefore a distance 2 0.300 m r = from 2 15.0 kg m = . Let + x be toward the 15.0 kg mass. EXECUTE: 11 2 2 8 1 1 1 (8.00 kg) (6.67 10 N m /kg ) (1.334 10 N/kg) (0.200 m) Gm m m Fm r −− × , in the x -direction. 11 2 2 8 2 2 2 (15.0 kg) (6.67 10 N m /kg ) (1.112 10 N/kg) (0.300 m) Gm m m r = × , in the x + -direction. The net force is 88 9 12 ( 1.334 10 N/kg 1.112 10 N/kg) ( 2.2 10 N/kg) xxx FF F m m =+= × + × = × . 92 2.2 10 m/s x x F a m −× . The acceleration is × , toward the 8.00 kg mass. EVALUATE: The smaller mass exerts the greater force, because the particle is closer to the smaller mass.

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Gravitation 12-3 12.9. IDENTIFY: Apply Eq.(12.1) to calculate the magnitude of each gravitational force. Each force is attractive.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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381_PartUniversity Physics Solution - 12-2 12.5. Chapter 12...

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