Jovian Exoplanets: Formation and Detection Techniques Explained

L3310-7169 Homework 6 Donald Gaeta TOTAL POINTS 95 / 100 QUESTION 1 1 (14.19) Present theory suggests that giant planets cannot form without condensation of water ice, which becomes vapor at the high temperatures close to a star. So how can we explain the presence of jovian- sized exoplanets closer to their star than Mercury is to our Sun? 10 / 10 ✓ - 0 pts Correct - 10 pts Astronomers now believe that large planets may form far from their star, where low temperatures allow for condensation of ice, and then migrate inward closer to the star. This idea would also explain the higher eccentricity of some exoplanet orbits, which could occur during migration of a planet through a planetary system. - 10 pts This answer is nearly word-for-word the same as an answer found on line and cannot be accepted for credit. QUESTION 2 2 (21.12) Which types of planets are most easily detected by Doppler measurements? By transits? 10 / 10 ✓ - 0 pts Correct - 10 pts The Doppler technique measures the motion of the star caused by the pull of one or more planets around it. The gravitational force is proportional to the mass of the planet and inversely proportional to the square of the separation. So the easiest planets to detect with this method are massive and close to the star. That’s why the hot Jupiters were found first. The transit probability (the chance that the orbit will bring the planet in front of the star for a transit) is greatest for close-in planets. The size of the planet must be big enough to give a measureable decrease in the brightness of the star. Since astronomers need to wait for three transits before they feel comfortable confirming their observation, that means that the longer the planet takes to orbit its star, the longer it will take to confirm the existence of a transiting planet. So this method works best for planets of larger size, orbiting close to their stars. - 10 pts This answer is nearly word-for-word the same as an answer found on line and cannot be accepted for credit. - 5 pts The Doppler technique measures the motion of the star caused by the pull of one or more planets around it. The gravitational force is proportional to the mass of the planet and inversely proportional to the square of the

separation. So the easiest planets to detect with this method are massive and close to the star. That’s why the hot Jupiters were found first. - 5 pts The transit probability (the chance that the orbit will bring the planet in front of the star for a transit) is greatest for close-in planets. The size of the planet must be big enough to give a measureable decrease in the brightness of the star. Since astronomers need to wait for three transits before they feel comfortable confirming their observation, that means that the longer the planet takes to orbit its star, the longer it will take to confirm the existence of a transiting planet. So this method works best for planets of larger size, orbiting close to their stars. QUESTION 3 3 (21.23) An exoplanetary system has two known planets. Planet X orbits in 290 days and Planet Y orbits in 145 days. Which planet is closest to its host star? If the star has the same mass as the Sun, what is the semi-major axis of the orbits for Planets X and Y? 5 / 10 - 0 pts Correct - 5 pts Planet Y is in the shorter orbital period and therefore resides closest to its star. ✓ - 2.5 pts For Planet X, the period is 0.79 years, so (using Kepler’s laws) the semi-major axis is $$p^2 = a^3$$. So $$p = 290 d × (1 y/365.25 d) = 0.79 y$$, $$p^2 = (0.79 y)^2 = 0.624 y^2$$, $$1 y^2 = 1 AU^3$$, $$0.624 AU^3 = a^3$$, $$a = 0.85 AU$$. ✓ - 2.5 pts For Planet Y, the period is 0.40 years so the semi-major axis is $$p^2 = a^3$$. So $$p = 145 d × (1 y/365.25 d) = 0.40 y$$, $$p^2 = (0.40 y)^2 = 0.16 y^2$$, $$1 y^2 = 1 AU^3$$, $$0.16 AU^3 = a^3, a = 0.54 AU$$. - 1 pts If using the formula: $$a = (\frac{GMp^2}{4\pi^2})^{\frac{1}{3}}$$, p needs to be converted to seconds to cancel with the units of $$G$$ - 1 pts If using $$P^2 = a^3$$ then $$P$$ needs to be in years - 2 pts unkown calculator error QUESTION 4 4 (21.26) If a transit depth of 0.00001 can be detected with the Kepler spacecraft, what is the smallest planet that could be detected around a 0.3 Rsun M dwarf star? 10 / 10 ✓ - 0 pts Correct - 10 pts $$Depth = \left( \frac{R_{planet}}{R_{star}} \right)^2$$ - 5 pts $$R_{planet} = R_{star} \sqrt{Depth}$$ - 2 pts $$R_{planet} = 0.3 R_{sun} \sqrt{0.00001}$$ - 2 pts $$R_{planet} \sim 0.001 R_{sun} $$ - 10 pts No Work QUESTION 5 Using a web site such as the exoplanet encyclopedia (http://exoplanet.eu) or the exoplanet archive

(https://exoplanetarchive.ipac.caltec h.edu) make the following plots: 30 pts 5.1 Orbital period vs planet mass for all known planets 10 / 10 ✓ + 10 pts Correct + 9 pts Orbital period vs planet mass means orbital period on the y and mass on the x + 5 pts Axis limits make this data un- interpretable + 0 pts I don't see this plot 5.2 Orbital period vs planet mass for transiting planets 10 / 10 ✓ + 10 pts Correct + 9 pts Orbital period vs. planet mass means planet mass on the x + 5 pts Axis limits renders this data un- interpretable + 0 pts I don't see this plot 5.3 Orbital period vs planet mass for planets detected via radial velocity 10 / 10 ✓ + 10 pts Correct + 9 pts Period vs. mass means mass should be on the x + 5 pts Axis Limits make this data un- interpretable + 0 pts I don't see this plot QUESTION 6 6 Homework submitted 30 / 30 ✓ + 30 pts Correct Page 3

Your preview ends here