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Unformatted text preview: lenoir (wml297) – homework 02 – Turner – (58220) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points The radius of the moon is 1 . 74 × 10 6 m , the radius of the sun is 6 . 96 × 10 8 m, the average moonearth distance is 3 . 84 × 10 8 m , and the average sunearth distance is 1 . 496 × 10 11 m. What is the apparent angle the diameter of the moon subtends, as seen from the earth? Correct answer: 0 . 0090625 rad. Explanation: If you have an angle θ that is very small ( ≈ 1 ◦ ) in measure, then θ ≈ tan θ, for θ measured in radians. r earth − moon r earth − moon d moon θ Since the distance from the earth to the moon or the sun ≫ than the diameters in volved, the angle can thus be approximated by its tangent, and θ ≈ tan θ = diameter distance Alternate Solution: The definition of an gular displacement is the arc length divided by the radius. The angular displacement us ing this definition is labeled using rad (actu ally a pure number). For this problem, the “arc length” is ap proximately the diameter 2 × R moon = D moon = 2 (1 . 74 × 10 6 m) = 3 . 48 × 10 6 m of the moon and the radius is the “distance” R earth moon = 3 . 84 × 10 8 m from the earth to the moon. Thus θ ≡ arc length radius = 2 R moon d EM = 2 (1 . 74 × 10 6 m) 3 . 84 × 10 8 m = . 0090625 rad ....
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This note was uploaded on 04/15/2010 for the course PHY 12343 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Turner

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