PS06_exoplanets_tidal_forces_solutions

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

= 2 × 2 . 06 × 7 . 149 × 10 7 m 32 min × 60 s/min − 500 m/s = (153406 − 500) m/s = 152906 m/s = 153 km/s . Alternatively, you can use the time difference t C − t A , which is related to the radius of the star and the velocities by: v planet = 2 R star t C − t A − v star = 2 × 1 . 5 × 6 . 955 × 10 8 m 227 min × 60 s/min − 500 m/s = (153194 − 500) m/s = 152694 m/s = 153 km/s . 4. Here is the Doppler velocity curve for the star HD 192263, which has a mass of 0 . 82 M ⊙ and radius of 0 . 78 R ⊙ . What can you say about the companion’s mass, size, and/or distance from the star? Be quantitative when possible (you cannot be super precise when reading numbers from a graph, but do your best). If some properties cannot be determined from these data, explain why. [10 pts] 0 20 40 60 time (days) - 50 - 25 0 25 50 velocity (m/s) The period is P = 24 . 4 days and the velocity amplitude is K 1 = 52 . 7 m/s. The curve appears to be sinusoidal, so we take the eccentricity to be 0. Then assuming the planet’s mass is small compared with the star’s mass, we can use Kepler III to find the semimajor axis: a = Gm 1 P 2 4 π 2 1 / 3 4

= (6 . 67 × 10 − 8 cm 3 g − 1 s − 2 ) × (0 . 82 × 1 . 99 × 10 33 g) × (24 . 4 × 86 , 400 s) 2 4 π 2 1 / 3 = 2 . 30 × 10 12 cm = 0 . 15 AU We can use the equation for Doppler planets to find m 2 sin i = m 2 1 P 2 πG 1 / 3 K 1 = (0 . 82 × 1 . 99 × 10 33 g) 2 × (24 . 4 × 86 , 400 s) 2 π (6 . 67 × 10 − 8 cm 3 g − 1 s − 2 ) 1 / 3 × 5270 cm s − 1 = 1 . 25 × 10 30 g = 0 . 66 M J We cannot determine the planet’s mass itself without knowing the inclination, but this gives a lower limit. We cannot say anything about the planet’s size unless we have transit information, which is not given here. 5. Mars has a mass of 6 . 4 × 10 26 g (about one tenth M ⊕ ) and a radius of 3400 km (about half R ⊕ ). Its small moon Phobos has a mass of 1 . 1 × 10 19 g and a radius of just 11 km . Phobos orbits Mars with a semimajor axis of 9380 km . [10 pts] (a) What are the mean densities of Mars and Phobos, in g cm − 3 ? The average densities are just ρ Mars = 3 M Mars 4 πR 3 Mars = 3(6 . 4 × 10 26 g) 4 π (3 . 4 × 10 8 cm) 3 = 3 . 9 g cm − 3 ρ Phobos = 3(1 . 1 × 10 19 g) 4 π (1 . 1 × 10 6 cm) 3 = 2 . 0 g cm − 3 (b) What is the Roche limit for the Mars/Phobos system? Is Phobos inside it? The Roche limit is r T = f ρ planet ρ moon 1 / 3 R planet where f ≈ 2 . 5. Plugging in the numbers from part (a) we find: r T = 2 . 5 3 . 9 g cm − 3 2 . 0 g cm − 3 1 / 3 R Mars = 3 . 1 R Mars = 10 , 500 km That means that Phobos’ orbit ( a = 9380 km) is inside the Roche limit today! 5