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Unformatted text preview: Physics 341: Problem Set #8 Solutions 1. The vertical motion of stars in spiral galaxies depends on the gravity exerted by the disk, so it allows us to “weigh” the disk. (a) Use dimensional analysis to derive an estimate of the mass density ρ of a spiral galaxy disk, in terms of its scale height h z , its vertical velocity dispersion σ z , and a relevant physical constant. The things we have to work with are: scale height h z = [ L ] vertical velocity dispersion σ z = [ LT 1 ] gravity G = [ M 1 L 3 T 2 ] We are looking for a mass density, which has dimensions ρ = [ M L 3 ] To get this, we clearly need G 1 to get [ M ]. Then we need σ 2 z to get rid of [ T 2 ]. So far we have σ 2 z G = [ M L 1 ] We then need h 2 z to get two more factors of [ L ] in the denominator. Thus, our dimensional analysis estimate is ρ ∼ σ 2 z Gh 2 z (b) In the neighborhood of the Sun, the Milky Way has h z ≈ 350 pc and σ z ≈ 16 km s 1 for the thin disk, and h z ≈ 1 kpc and σ z ≈ 35 km s 1 for the thick disk. Use these values and your result from part (a) to estimate the mass density of the Milky Way’s disk, in M pc 3 . Do the thin and thick disks give a consis tent density estimate to the level of precision we might expect from dimensional analysis? Using the thin disk we estimate ρ ∼ (16 × 10 5 cm s 1 ) 2 (6 . 67 × 10 8 cm 3 g 1 s 2 ) × (350 × 3 . 086 × 10 18 cm) 2 ∼ 3 . 3 × 10 23 g cm 3 × 1 M 1 . 99 × 10 33 g × 3 . 086 × 10 18 cm 1 pc 3 ∼ . 49 M pc 3 1 Using the thick disk we estimate ρ ∼ (35 × 10 5 cm s 1 ) 2 (6 . 67 × 10 8 cm 3 g 1 s 2 ) × (3 . 086 × 10 21 cm) 2 ∼ 1 . 9 × 10 23 g cm 3 ∼ . 29 M pc 3 The answers differ, but by less than a factor of 2 and in dimensional analysis we cannot be too concerned about such factors....
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This note was uploaded on 10/18/2011 for the course PHYSICS 341 taught by Professor Keeton during the Fall '08 term at Rutgers.
 Fall '08
 Keeton
 Gravity, Mass

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