PS06_exoplanets_tidal_forces_solutions

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Dec 6, 2023

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Physics 341: Problem Set #6 due November 1 by 2:00 PM in PDF format on Canvas [40 pts] You are encouraged to work in groups on these problems, but each student must write up the solutions individually. You must also list your collaborators on your solutions, and cite any external sources you used (other than the course notes or textbook). Using online solutions to these or similar problems is not allowed. Show your work and explain your reasoning. The number of points for each problem is listed; partial credit will be given. Be careful with units. Questions asking about concepts require at most a few sentences to answer. Please make your answers clear and concise . Figure 1: Radial velocities of the star TrES-3, from Sozzetti et al. (2009, ApJ, 691, 1145). The period is 1.31 days. 1. TrES-3, shown in Figure 3, is a transiting extrasolar planetary system, with an edge-on circular orbit. At what orbital phase does the planet go in front of the star? [5 pts] A. 0.25 B. 0.50 C. 0.75 D. 1.00 E. either 0.50 or 1.00; can’t tell which with just the radial velocity curve Answer: B 1
2. Imagine that the Kepler space telescope has discovered an “Earth-like” planet outside of our solar system that is twice the radius of our own Earth, but the same mass. The newly discovered planet happens to have a moon the same size, mass, and distance as our own moon. Which of the following is true? [5 pts] A. The magnitude of the gravitational force on the new Earth from its moon is less than the magnitude of the gravitational force in our own Earth-Moon system. B. The tidal force on the new Earth from its moon is larger than the tidal force in our own Earth-Moon system. C. The tidal force on the new Earth from its moon is exactly the same as in our own Earth-Moon system. D. The tidal force on the new Earth from its moon is less than the tidal force in our own Earth-Moon system. Answer: B 3. Shown is the light-curve for a planet transiting in front of a star. The time difference t B t A is 32 minutes and the time difference t C t A is 227 minutes. [10 pts] (a) The star has a radius of 1 . 5 R . What is the radius of the planet in units of the radius of Jupiter? 2
The fraction of the light blocked by the planet is f = 0 . 02 . This is related to the radii of the star and the planet as f = R 2 planet R 2 star R planet = p fR star = 0 . 02(1 . 5 R ) = (0 . 21 R ) 6 . 955 × 10 8 m 1 R 1 R J 7 . 149 × 10 7 m = 2 . 06 R J Alternatively, one can see that the radius of the planet and the star must be given by: 2 R star = ( v star + v planet )( t C t A ) 2 R planet = ( v star + v planet )( t B t A ) . Therefore, we can solve for the planet’s radius R planet R star = t B t A t C t A R planet = t B t A t C t A R star = 32 227 (1 . 5 R ) = (0 . 141)(1 . 5 R ) 6 . 955 × 10 8 m 1 R 1 R J 7 . 149 × 10 7 m = 2 . 06 R J Finally, one could solve directly for v planet + v star in this step, and then use the previous equations to solve for R planet , solving for (b) as in intermediate step. (b) The star is measured to have a maximum radial velocity of 0 . 5 km/s, how fast is the planet moving? We know that the time difference t B t A is related to the velocities of the planet and the star by 2 R planet = ( v star + v planet )( t B t A ) . Therefore, v planet = 2 R planet t B t A v star 3
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