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386_PartUniversity Physics Solution

386_PartUniversity Physics Solution - Gravitation EXECUTE...

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Gravitation 12-7 E XECUTE : 11 2 2 22 2 2 6 2 (6.67 10 N m /kg )(4.8 10 kg) 1.30 m/s (1.569 10 m) Gm g R × × = = = × . rad g a = gives 2 1.30 m/s 60 s 1 rev (0.553 rad/s) 5.28 rpm 4.25 m 1 min 2 rad g r ω π ⎞⎛ = = = = ⎟⎜ ⎠⎝ . E VALUATE : The radius of Europa is about one-fourth that of the earth and its mass is about one-hundredth that of earth, so g on Europa is much less than g on earth. The lander would have some spatial extent so different points on it would be different distances from the rotation axis and rad a would have different values. For the ω we calculated, rad a g = at a point that is precisely 4.25 m from the rotation axis. 12.23. I DENTIFY and S ET U P : Example 12.5 gives the escape speed as 1 2 / , v GM R = where M and R are the mass and radius of the astronomical object. E XECUTE : 11 2 2 12 1 2(6.673 10 N m /kg )(3.6 10 kg)/700 m 0.83 m/s. v = × × = E VALUATE : At this speed a person can walk 100 m in 120 s; easily achieved for the average person. We can write the escape speed as 2 4 1 3 , v GR πρ = where ρ is the average density of Dactyl. Its radius is much smaller than earth°s and its density is about the same, so the escape speed is much less on Dactyl than on earth. 12.24. I DENTIFY : In part (a) use the expression for the escape speed that is derived in Example 12.5. In part (b) apply conservation of energy. S ET U P : 3 4.5 10 m R = × . In part (b) let point 1 be at the surface of the comet. E XECUTE : (a) The escape speed is 2 GM v R = so 2 3 2 13 11 2 2 (4.5 10 m)(1.0 m/s) 3.37 10 kg 2 2(6.67 10 N m /kg ) Rv M G × = = = × × . (b) (i) 2 1 1 1 2 K mv = . 2 1 0.100 K K = . 1 GMm U R = − ; 2 GMm U r = − . 1 1 2 2 K U K U + = + gives 2 2 1 1 1 1 2 2 (0.100)( ) GMm GMm mv mv R r = . Solving for r gives 2 2 1 3 11 2 2 13 1 1 0.450 1 0.450(1.0 m/s) 4.5 10 m (6.67 10 N m /kg )(3.37 10 kg) v r R GM = = × × × and 45 km r = . (ii) The debris never loses all of its initial kinetic energy, but 2 0 K as r → ∞ . The farther the debris are from the comet°s center, the smaller is their kinetic energy. E VALUATE : The debris will have lost 90.0% of their initial kinetic energy when they are at a distance from the comet°s center of about ten times the radius of the comet. 12.25. I DENTIFY : The escape speed, from the results of Example 12.5, is 2 / . GM R S ET U P : For Mars, 23 6.42 10 kg M = × and 6 3.40 10 m R = × . For Jupiter, 27 1.90 10 kg M = × and 7 6.91 10 m R = × . E XECUTE : (a) 11 2 2 23 6 3 2(6.673 10 N m /kg )(6.42 10 kg)/(3.40 10 m) 5.02 10 m/s. v = × × × = × (b) 11 2 2 27 7 4 2(6.673 10 N m /kg (1.90 10 kg)/(6.91 10 m) 6.06 10 m/s. v = × × × = × (c) Both the kinetic energy and the gravitational potential energy are proportional to the mass of the spacecraft. E VALUATE : Example 12.5 calculates the escape speed for earth to be 4 1.12 10 m/s × . This is larger than our result for Mars and less than our result for Jupiter. 12.26. I DENTIFY : The kinetic energy is 2 1 2 K mv = and the potential energy is GMm U r = − S ET U P : The mass of the earth is 24 E 5.97 10 kg M = × .
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