Ph1a-2007-HW9-soln

Ph1a-2007-HW9-soln - Ph1a Solutions 9 Chefung Chan...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ph1a Solutions 9 Chefung Chan (cchan@caltech.edu), 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, its total energy must be non-negative ( 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 solar-mass 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 a = a r r = r parenleftbigg- GM r 2 parenrightbigg r = 2...
View Full Document

Page1 / 6

Ph1a-2007-HW9-soln - Ph1a Solutions 9 Chefung Chan...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online