405_Physics ProblemsTechnical Physics

405_Physics ProblemsTechnical Physics - Chapter 13 P13.66...

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Chapter 13 407 P13.66 (a) From the data about perigee, the energy of the satellite-Earth system is Em v GM m r p E p =− = × ×× × 1 2 1 2 1 60 8 23 10 6 67 10 5 98 10 1 60 702 10 23 2 11 24 6 .. . . af ej e j or E × 367 10 7 . J (b) Lm v r m v r pp == ° = × × sin sin . . . . . θ 90 0 1 60 8 23 10 7 02 10 924 10 36 10 kg m s m kg m s 2 bg e j (c) Since both the energy of the satellite-Earth system and the angular momentum of the Earth are conserved, at apogee we must have 1 2 2 mv GMm r E a a −= and mv r L aa sin . 90 0 °= . Thus, 1 2 160 6 67 10 5 98 10 1 60 2 11 24 7 . . . e j a f v r a s × J and 1 60 9 24 10 10 kg kg m s 2 vr . Solving simultaneously, 1 2 6 67 10 5 98 10 1 60 1 60 2 11 24 10 7 . . . . . a f e j v v a a × × which reduces to 0 800 11 046 3 672 3 10 0 27 vv −+ × = so v a = ±− × 11 046 11 046 4 0 800 3 672 3 10 20800 2 7 a f . . This gives v a = 8 230 m s or 5580 ms . The smaller answer refers to the velocity at the
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