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Unformatted text preview: Ph1a Solutions 9 Chefung Chan (cchan@caltech.edu) 27 November 2004 Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself. Use these instead. 9.1 FP4 (5 points) 9.1.a (1 point) The total energy E is given by: E = K + U = 1 2 mv 2 GMm r = 1 2 m parenleftbigg GM 5 R parenrightbigg GMm 5 R E = GMm 10 R Because E < 0, the orbit will be elliptical. 9.1.b (1 point) The magnitude of the angular momentum vector L is  vector L  =  vector r vector p  = (5 R )( mv ) sin 135  vector L  = m radicalBig 5 2 GMR 9.1.c (1.5 points) The equations for conservation of energy and angular momentum are respectively: E = GMm 10 R = 1 2 mv 2 p GMm r p  vector L  = m radicalBig 5 2 GMR = mv p r p Solving the second equation for r p gives r p = radicalBig 5 2 GMR/v 2 p Substituting into the energy equation: GMm 10 R = 1 2 mv 2 p GMm v p radicalBig 5 2 GMR v 2 p 2 parenleftBigg radicalbigg 2 GM 5 R parenrightBigg v p + GM 5 R = 0...
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This note was uploaded on 03/30/2010 for the course PH 1a taught by Professor Goodstein during the Fall '07 term at Caltech.
 Fall '07
 Goodstein
 Energy, Work

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