Homework 6 Solutions

# Now substitute the expressions for r and hr 0 rr0

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Unformatted text preview: f material integrated perpendicular to the disk, R0 is an arbitrary reference radius, and α, β, ρ0, Σ0 and H0 are all constants. Note that surface density is related to the total density by: +∞ ρ(r, z )dz Σ( r ) = −∞ i.e., it is the integrated mass perpendicular to the disk (units of mass/area) (a) [10 pts] Show that Σ0 ρ(r, 0) = √ π H0 r R0 α−β Start with the expression for surface density: +∞ +∞ − ρ(r, z )dz = ρ(r, 0) Σ(r) = e z2 H (r ) 2 dz −∞ −∞ In the integral, H(r) is effectively a constant; change variables to x = z/H(r), dx = dz/H(r) +∞ Σ(r) = ρ(r, 0)H (r) √ 2 e−x dx = ρ(r, 0)H (r) π −∞ where we have used the well ­known solution to the exponential integral. Now substitute the expressions for Σ(r) and H(r): √ Σ0 (r/R0 )α = ρ(r, 0)H0 (r/R0 )β π Σ0 (r/R0 )α−β ⇒ ρ(r, 0) = √ π H0 (b) [10 pts] If the disk has a total mass MD, show that, for α ≠  ­2: MD (α + 2) Σ0 = 2 2π Rout R0 Rout α 1− Rin Rout α+2 −1 Where Rin and Rout are the inner and outer radii of the disk. The total mass of the disk is just an integral of the density over volume. The vertical integral (z) has already been done; that’s the surface density, so now we just have to integrate the surface density over the disk annulus: 2π Rmax Σ0 (r/R0 )α rdrdθ MD = 0 Rmin Rmax = (r/R0 )α (r/R0 )(dr/R0 ) 2 2π Σ 0 R 0 Rmin th...
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