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Unformatted text preview: Astronomy 210 Spring 2011 Homework Set #7: Solutions 1. A Transiting Exoplanet . (a) [5 points] Since a 3 = GMP 2 / (4 π 2 ), we have a = parenleftbigg GM 4 π 2 parenrightbigg 1 / 3 P 2 / 3 (1) It is fine to just use this formula, being careful with units. A shortcut is to divide this result by the value for the earth a ⊕ = parenleftbigg GM ⊙ 4 π 2 parenrightbigg 1 / 3 P 2 / 3 ⊕ (2) where of course P ⊕ = 1 year and a ⊕ = 1 AU. Dividing, we get a 1AU = parenleftbigg M M ⊙ parenrightbigg 1 / 3 parenleftbigg P 1yr parenrightbigg 2 / 3 (3) and thus for M = 1 . 35 M ⊙ and P = 4 . 89 day = 0 . 0134 yr, we get a = 0 . 062 AU, in good agreement with the professional value. (b) [5 points] We want the star’s speed v ⋆ about the center of mass. Since both the star and the planet move about the center of mass with period P and angular speed ω = 2 π/P , the star’s speed is v ⋆ = ωr ⋆ (4) where r s tar is the star’s distance from the center of mass, which we need to find. We have r ⋆ + r p = a (5) and m ⋆ r ⋆ = m p r p (6) Since we want r ⋆ , we eliminate r p : m ⋆ r ⋆ = m p ( a − r ⋆ ) (7) ( m ⋆ + m p ) r ⋆ = m p a (8) r ⋆ = m p m ⋆ + m p a (9) and thus we have v ⋆ = m p m ⋆ + m p 2 πa P = m p m ⋆ + m p v (10) where v is the speed of the planet relative to the star ....
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This note was uploaded on 10/06/2011 for the course ASTRO 210 taught by Professor Fields during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Fields
 Astronomy

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