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402_Physics ProblemsTechnical Physics

# 402_Physics ProblemsTechnical Physics - 404*P13.60...

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404 Universal Gravitation *P13.60 For both circular orbits, F ma = : GM m r mv r E 2 2 = v GM r E = FIG. P13.60 (a) The original speed is v i = × × × + × = × 6 67 10 5 98 10 6 37 10 2 10 7 79 10 11 24 6 5 3 . . . . N m kg kg m m m s 2 2 e je j e j . (b) The final speed is v i = × × × = × 6 67 10 5 98 10 6 47 10 7 85 10 11 24 6 3 . . . . N m kg kg m m s 2 2 e je j e j . The energy of the satellite-Earth system is K U mv GM m r m GM r GM r GM m r g E E E E + = = = − 1 2 1 2 2 2 (c) Originally E i = − × × × = × 6 67 10 5 98 10 100 2 6 57 10 3 04 10 11 24 6 9 . . . . N m kg kg kg m J 2 2 e je j b g e j . (d) Finally E f = − × × × = × 6 67 10 5 98 10 100 2 6 47 10 3 08 10 11 24 6 9 . . . . N m kg kg kg m J 2 2 e je j b g e j . (e) Thus the object speeds up as it spirals down to the planet. The loss of gravitational energy is so large that the total energy decreases by
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