402_Physics ProblemsTechnical Physics

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

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404 Universal Gravitation *P13.60 For both circular orbits, Fm a = : 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 10 637 10 2 10 779 10 11 24 65 3 .. . . Nm kg kg m m ms 22 ej e j . (b) The final speed is v i = × × 6 67 10 5 98 10 647 10 785 10 11 24 6 3 . . m e j . The energy of the satellite-Earth system is KU m v GM m r m GM r GM r GM m r g EE E E += = = 1 2 1 2 (c) Originally E i =− × × × 6 67 10 5 98 10 100 2 6 57 10 304 10 11 24 6 9 . . N m kg kg kg m J e j bg . (d) Finally E f × × × 6 67 10 5 98 10 100 2 6 47 10 308 10 11 24 6 9 . . N m kg kg kg m J 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 if −= × ×
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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