solutions-ps05

solutions-ps05 - Physics 341 Problem Set#5 Solutions 1 In...

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Physics 341: Problem Set #5 Solutions 1. In 2003 I helped discover a transiting planet orbiting OGLE-TR-56 (a faint star in the constellation Sagittarius). We used the Keck telescope on Mauna Kea, Hawaii to measure the star’s Doppler shift. The radial velocity curve is sinusoidal, with a radial velocity amplitude of 212 m s - 1 and a period of just 1.212 days. During the transits, the planet blocks 1.1% of the star’s light. Our best estimate of the star’s mass and radius are M ? = 1 . 17 M ± and R ? = 1 . 15 R ± . (a) What is the orbital separation between the star and the planet (in AU)? Because the radial velocity curve is sinusoidal, we know that the orbits are circular, and the orbital separation is constant. So then we can use Kepler’s Third Law, and use the approximation that the mass of the planet is negligible compared to he mass of the star: a 3 = G ( m ? + m planet ) P 2 4 π 2 Gm ? P 2 4 π 2 = a = ± Gm ? P 2 4 π 2 ² 1 / 3 a = ³ (6 . 67 × 10 - 8 cm 3 g - 1 s - 2 )(1 . 17 × 1 . 99 × 10 33 g)(1 . 212 × 24 × 3600 s) 2 4 π 2 ´ 1 / 3 = 3 . 51 × 10 11 cm = 0 . 023 AU (b) What are the mass and radius of the planet (in Jupiter units, M J and R J )? Because the planet is much less massive than the star, we derived that m planet sin i = ± m 2 ? P 2 πG ² 1 / 3 K ? and because we know the planet transits the star from our point of view, the inclination must be close to edge-on so that sin i 1. So we have m planet = ± m 2 ? P 2 πG ² 1 / 3 K ? = ³ (1 . 17 × 1 . 99 × 10 33 g) 2 (1 . 212 × 24 × 3600 s) 2 π (6 . 67 × 10 - 8 cm 3 g - 1 s - 2 ) ´ 1 / 3 (2 . 12 × 10 4 cm s - 1 ) = 2 . 35 × 10 30 g = 1 . 24 M J The fraction of light that the planet blocks is just the ratio of the projected area of the planet to the projected area of the star, so πR 2 planet πR 2 ? = 0 . 011 = R planet = 0 . 011 R ? = 0 . 105 R ? = R planet = 0 . 105(1 . 15 × 6 . 96 × 10 10 cm) = 8 . 40 × 10 9 cm = 1 . 18 R J 1
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(c) What is the average density of the planet? Is it rocky or gaseous? The average density is just ρ = M V = 3 M 4 πR 3 = 3(2 . 35 × 10 30 g) 4 π (8 . 40 × 10 9 cm) 3 = 0 . 95 g cm - 3 This is below the density of a few g cm - 3 for rocks and rocky planets, so the planet must be mostly gaseous (or liquid!). 2. The Kepler space mission was launched by NASA in March 2009 to look for transiting Earth-mass planets. (a)
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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.

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solutions-ps05 - Physics 341 Problem Set#5 Solutions 1 In...

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