YF_ISM_12 - GRAVITATION 12 Use the law of gravitation,...

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12-1 G RAVITATION 12.1. IDENTIFY and SET UP: Use the law of gravitation, Eq.(12.1), to determine g . F EXECUTE: SM S on M 2 SM (S sun, M moon); mm FG r == EM E on M 2 EM (E earth) r 2 2 EM S on M S M S EM 2 E on M SM E M E SM r Fm m m r G Fr G m m m r ⎛⎞ ⎜⎟ ⎝⎠ EM , r the radius of the moon’s orbit around the earth is given in Appendix F as 8 3.84 10 m. × The moon is much closer to the earth than it is to the sun, so take the distance SM r of the moon from the sun to be SE , r the radius of the earth’s orbit around the sun. 2 30 8 S on M 24 11 E on M 1.99 10 kg 3.84 10 m 2.18. 5.98 10 kg 1.50 10 m F F ×× EVALUATE: The force exerted by the sun is larger than the force exerted by the earth. The moon’s motion is a combination of orbiting the sun and orbiting the earth. 12.2. IDENTIFY: The gravity force between spherically symmetric spheres is 12 g 2 Gm m F r = , where r is the separation between their centers. SET UP: 11 2 2 6.67 10 N m /kg G . The moment arm for the torque due to each force is 0.150 m. EXECUTE: (a) For each pair of spheres, 11 2 2 7 g 2 (6.67 10 N m /kg )(1.10 kg)(25.0 kg) 1.27 10 N (0.120 m) F ×⋅ × . From Figure 12.4 in the textbook we see that the forces for each pair are in opposite directions, so net 0 F = . (b) The net torque is 78 net g 2 2(1.27 10 N)(0.150 m) 3.81 10 N m Fl τ −− × . (c) The torque is very small and the apparatus must be very sensitive. The torque could be increased by increasing the mass of the spheres or by decreasing their separation. EVALUATE: The quartz fiber must twist through a measurable angle when a small torque is applied to it. 12.3. IDENTIFY: The force exerted on the particle by the earth is wm g = , where m is the mass of the particle. The force exerted by the 100 kg ball is g 2 Gm m F r = , where r is the distance of the particle from the center of the ball. SET UP: 11 2 2 6.67 10 N m /kg G , 2 9.80 m/s g = . EXECUTE: g Fw = gives ball 2 Gmm mg r = and 11 2 2 5 ball 2 (6.67 10 N m /kg )(100 kg) 2.61 10 m 0.0261 mm 9.80 m/s Gm r g = × = . It is not feasible to do this; a 100 kg ball would have a radius much larger than 0.0261 mm. EVALUATE: The gravitational force between ordinary objects is very small. The gravitational force exerted by the earth on objects near its surface is large enough to be important because the mass of the earth is very large. 12.4. IDENTIFY: Apply Eq.(12.2), generalized to any pair of spherically symmetric objects. SET UP: The separation of the centers of the spheres is 2 R . EXECUTE: The magnitude of the gravitational attraction is 22 2 2 /(2 ) /4 . GM R GM R = EVALUATE: Eq.(12.2) applies to any pair of spherically symmetric objects; one of the objects doesn't have to be the earth. 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.
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This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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YF_ISM_12 - GRAVITATION 12 Use the law of gravitation,...

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