Homework 6 Solutions

Now substitute the expressions for r and hr 0 rr0

Info icon This preview shows page 1. Sign up to view the full content.

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

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...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern