set5-06-sol

set5-06-sol - Place two such boxes next to each other and...

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Astro 346, Spring Semester 2006 Homework, 5th set, solutions 11.15 In class we derived v = v 0 1 + 4 π r 3 3 M 0 ρ 0 3 M 0 v 0 4 π r 3 ρ 0 for M À M 0 which with the numbers listed in 11.14 gives r = 8 · 10 19 cm τ = 2 · 10 13 s The values of Q and f are then Q 0 . 01 f 0 . 01 Obviously Q needs to increase approximately by a factor one hundred. This could arise from an expansion of the overpressurized interior regions of supernova remnants. 12.3 The condition of hydrostatic equilibrium sets the pressure equal to the weight per unit area. Insert the expression for the turbulent pressure and obtain P = ρ K g z H P = ρ K v 2 z g z = v 2 z H Gauss’ law relates a volume integral to an integral over the closed surface of that volume. Applied to the gravitational acceleration we have I d ~ A ~g = Z dV div ~g = Z dV 4 π G ρ The direction of the surface element is inward. Choose a rectangular box that extends far above the Galactic plane on either side.
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Unformatted text preview: Place two such boxes next to each other and apply Gauss law for the two boxes individually and combined. This shows that the contributions from the side walls of the box vanish and we only have to consider the top and bottom, giving 2 A g z = 4 G M V = g z 2 G = M V A The gravitational acceleration is zero in the galactic mid-plane. If the acceleration increases linearly with distance from the mid-plane, then the average acceleration is half that on top or bottom of the box, g z = 0 . 5 g z . The surface mass density implied by teh numbers in the textbook is then . 017 g/cm 2 , which is about twice that observed in the form of stars and gas, thus indicating the existence of dark matter in the local environment....
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This note was uploaded on 02/29/2012 for the course PHYS 227 taught by Professor Rabe during the Fall '08 term at Rutgers.

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