393_Physics ProblemsTechnical Physics

393_Physics ProblemsTechnical Physics - Chapter 13 P13.38...

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Chapter 13 395 P13.38 The gravitational force supplies the needed centripetal acceleration. Thus, GM m Rh mv E E E + = + bg 2 2 or v GM E E 2 = + (a) T r v E GM E E == + + 2 2 π T GM E E = + 2 3 (b) v GM E E = + (c) Minimum energy input is EK U K U fg f ig i min =+ ej e j . It is simplest to launch the satellite from a location on the equator, and launch it toward the east. This choice has the object starting with energy Km v ii = 1 2 2 with v RR i EE 2 100 2 86 400 ππ . d a y s and U GM m R gi E E =− . Thus, Em GM GM m m RG M m R E E E E E min = + F H G I K J + L N M M O Q P P + 1 2 1 2 4 86 400 22 2 s or EG M m RR h Rm E E E min = + + L N M M O Q P P 2 2 2 86 400 2 s P13.39 E GMm r tot 2 E GMm rr E if F H G I K J = ×× + + F H G I K J = 2 11 6 67 10 5 98 10 2 10 1 6370 100 1 6370 200 469 10 469 11 24 3 8 .. . e j kg 10 m J M J 3 P13.40 g Gm r E E E = 2 g Gm r U U U = 2 (a)
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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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