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Unformatted text preview: Ph1a Solutions 9 Chefung Chan ([email protected]), Fall 2007 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 v p = radicalbigg 2 GM 5 R ± radicalbigg 2 GM 5 R GM 5 R = radicalbigg 2 GM 5 R ± radicalbigg GM 5 R 1 At closest approach the speed will be maximal, so v p = (1 + √ 2) radicalbigg GM 5 R 9.1.d (1.5 points) All that is required is that we calculate the total energy with the new mass M ′ = M/ 2: E = K + U = 1 2 mv 2 GM ′ m r = 1 2 m parenleftbigg GM 5 R parenrightbigg GMm 10 R = 0 The special case E = 0 describes a parabolic orbit. 9.2 FP6 (5 points) 9.2.a (2 points) For an object (of finite mass, m ) to escape, it’s total energy must be nonnegative ( E ≥ 0). For the event horizon (distance R ), this implies the limiting condition E = 0. E = K + U = 1 2 mv 2 GMm/R = 0 1 2 mv 2 = GMm/R R = 2 GM/v 2 Since the speed of light is v = c , the event horizon is at a distance R = 2 GM/c 2 from the center of the black hole. For a solarmass black hole: R = 2(6 . 7 × 10 − 11 N m 2 kg − 2 )(2 . × 10 30 kg) (3 . × 10 8 m / s) 2 = 3 . 0 km 9.2.b (1.5 points) The difference in acceleration a is given by δa = ∂a ∂r δr = ∂ ∂r parenleftbigg GM r 2 parenrightbigg δr = 2...
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 Fall '07
 Goodstein
 Energy, Kinetic Energy, Mass, Potential Energy, Work

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