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PH709 UNIVERSITY OF KENT SCHOOL OF PHYSICAL SCIENCES EXOPLANETS Assignment 1 (Smith) 1. State the physical property which differentiates planets from brown dwarfs and the estimated critical mass which separates them. [40 marks] Deuterium burning is absent. 2. By analysing the superposition of frequencies and amplitudes in the radial velocity data for the star 55 Cancri A, the first three exoplanets in the system were deduced. The star was found to have the following observed parameters: 55 Cancri b corresponds to Vobs = 71.8 m s-1, and P = 14.6 days, 55 Cancri c corresponds to Vobs = 10.0 m s-1, and P = 43.9 days and 55 Cancri d corresponds to Vobs = 47.2 m s-1, and P = 5218 days. Assuming the orbits are coplanar and circular, determine (i) which planet is furthest from the star, (ii) which planet has the lowest mass and (iii) which planet has the highest mass. [20 marks each = 60 marks] The planet corresponding to the longest period is furthest from the star. This is 55 Cancri d. 1/ 3 PM s 2 The planet masses are M p 2G Vobs sin i 1/ 3 In this case, planet masses are proportional to: P Vobs For the data: 55 Cancri b Vobs = 71.8 m s-1, and P = 14.6 days 55 Cancri c Vobs = 10.0 m s-1, and P = 43.9 days 55 Cancri d Vobs = 47.2 m s-1, and P = 5218 days This has a value using the arbitrary units: b = 175.3 c = 35.2 d = 816.3 turn over 2 Therefore 55 Cancri d is the highest mass (iii) From (ii), 55 Cancri c is the lowest mass Assignment 2 1. The planet transitting a mildly metal-poor G0V star is named CoRoTExo_1b. It has a period P = 1.5 days. The estimated mass and radius of the star are 0.95 solar masses and 1.11 solar radii. The planet has a radius of 1.49 Jupiter radii. If the planet has the reported particularly low mean density of 0.38 g cm-3, calculate its mass and, hence, the maximum radial velocity (of the star) that was measured. (i) Mass = average density x volume = average density x (4/3) R3 = 380 x 1.33 x3.14 x (1.49 x 7.1 x 107)3 kg = 1.9 x 1027 / 1.9 x` 1027 Mj = 1 Mj [50 marks] (ii) The planet orbital speed Vp is given by Vp2 = GM*/a the star-planet distance. where a is Distance a is given by Kepler’s law: a = 3.7 x 109 m Note that we require the star’s speed, not the planet’s speed: V* = (Mp/M*) Vp =180 m/s approx. [50 marks] Assignment 3 2. The planet transitting the yellowish-white F0V dwarf main-sequence star CoRoT-4 is named CoRoT-Exo-4b. It has an orbital period P = 9.2 days, an inclination angle of 90o and an eccentricity of 0. The estimated mass and radius of the star are 1.1 solar masses and 1.2 solar radii. The planet has a radius of 1.2 Jupiter radii. 3 (i) (ii) (iii) (iv) If the planet has the reported low mean density of 530 kg m-3, calculate its mass in Jupiter masses (RJ = 7.1 × 107 m , MJ = 1.9 × 1027 kg). Estimate the maximum radial velocity of the star (G = 6.7 10-11 m3 kg-1 s-2, Msun = 2.0 × 1030 kg) due to the orbiting planet. Determine how close the planet is to the star in AU (1 AU = 1.5 x 1011 m). Estimate the change in magnitude during a transit (Rsun = 7.0 × 108 m). ANSWERS (i) Mass = average density x volume = average density x (4/3) R3 = 530 x 1.33 x3.14 x (1.2x 7.1 x 107)3 kg = 530 x 1.33 x3.14 x (1.2x 7.1 x 107)3 / 1.9 x` 1027 Mj = 0.7 Mj [10 marks] (ii) The planet orbital speed Vp is given by Vp2 = GM*/a where a is the star-planet distance. The star is about one solar mass and the period is about 9/360 = 1/40 of a year. Since in the solar system P2 = a3 in years and AU, a3 = 1/1600 approximately Hence a = 1/12 roughly i.e. a = 0.08 AU roughly. So Vp2 = 6.7 x 10-11 x 2 x 1030 / (0.08 x 1.5 x 1011) = 1.4 x 1010 m2/s2 So Vp = 1.2 x 105 m/s . V* = (Mp/M*) Vp = 0.7 x 0.001 x 1.2 x 105 m/s = 800 m/s approx. [10 marks] (iii) Rearrange and take mB = 0 for the planet to yield a = 1.2 1010 m = 0.08 AU. [10 marks] (iv) turn over 4 Relative change in flux change dF/F = (Rp/R*)2 = 0.01 [10 marks] Magnitude change dM = -2.5 log ( 1 + dF/F) = 0.01 [10 marks] Assignment 4 : SMITH 1. State the technique through which the majority of the population of extrasolar planets have so far been detected. State why this is an indirect method. Sketch diagrams to show how the eccentricity of the exoplanet orbit is determined. Describe the location relative to the star of exoplanets detected by this method with (i) low eccentricity and (ii) high eccentricity. Technique: Radial velocity It is the most successful method so far . Observable is the line of sight velocity of the star orbiting centre of mass of star - planet binary system. Direct imaging of the planets is difficult because of the enormous difference in brightness between the star and the planet, and the small angular separation between them.. If the planet has low eccentricity, then the star’s orbit (same eccentricity) traces out a radial velocity curve which is close to sinusoidal as a function of time (i.e. orbital phase). High eccentricity: there is a short period in which the radial speed differs by a large amount: Low e case: 5 High e case: Low eccentricity planets are observed to be located preferentially close to the star. High eccentricity exoplanets tend to be located far from the star. Assignment 5 : SMITH 1. Suppose that two exoplanets are observed to transit the same star. They are both in circular orbits with an inclination of 90 degrees. One produces periodic dips with a period of 4 days and the other produces periodic dips with a period of 108 days. The decrease in luminosity caused by both exoplanets is 1%. Determine (i) how much further away the outer planet is from the central star and (ii) how much larger the outer exoplanet is than the inner planet. Assuming they have the same density, calculate which exoplanet will cause (i) the highest variation in radial velocity and (ii) the largest wobble in location of the central star. Kepler’s law as developed by Newton to include the star’s mass is where omega is the angular velocity. However, use Kepler’s law in any form except the one valid for just the solar system, since in this case we don’t know the star’s mass. 2 Hence P = b a3 Where b is a constant. For the given two planets orbiting the same star, this applies to both. turn over 6 Since the ratio of the periods is 104/4 = 27, the outer planet is 27 from the star than the inner planet. 2/3 = 9 times further away The planets must have the same size in order to produce the same decrease in flux during the transit. Remember the formula: This implies that they would roughly have the same mass. The inner planet would thus produce the greater gravitational force and the highest variation in radial velocity. The outer planet would nevertheless produce the greatest wobble since the centre of mass of the star/outer planet system would be fixed.: æ mp ö ÷÷ ´ a a1 = çç è M* ø Assignment 6 : SMITH 1. Describe how planetesimals form and grow from protoplanetary discs in the core-accretion model. Explain why observed exoplanets are unlikely to have formed from protostellar accretion discs or from debris discs. Planetesimals form as dust/grains separate/diffuse from the gas, forming a denser dust disc near the mid-plane. Inelastic ollisions between grains are more common in this disc and sticking/coalescence occurs. Turbulence stirs up the gas increasing the collision rate OR Gravity becomes more important as planetesimals grow, providing further growth Exoplanets cannot form easily from protostellar discs due to the gas-grain friction which causes the planetesimals to lose energy and spiral in. (rapid ‘migration, also accepted) The material observed as debris discs are believed to form well after the planet formation epoch where impacts generate some dust and small debris. The mass is not sufficient to allow planets to form either through coalescence or gravity. 7 Assignment 7 Describe the observed distribution of eccentricities of exoplanet orbits as presently derived from known exoplanetary systems. Explain how the eccentricity distribution is observed to depend on thesemi-major axis of the exoplanet Explain the physical property that the parameter [Fe/H] is used to indicate and why its measurement is providing support for the core accretion hypothesis for planet formation (a) Planets orbiting within 0.1 AU of their host star all have close to circular orbits. All detected planets with semi-major axis < 0.07 AU have low e. 50% of exoplanets have higher e than Pluto's e = 0.244. Beyond 0.1 AU, the distribution of eccentricities appears essentially uniform between 0 and 0.8. Other correct relevant facts concerning eccentricity will get 2 marks, up to the maximum of 6. The parameter [Fe/H] is the star’s metallicity. The more massive planets are preferentially found around stars with enhanced metal abundance. Enhanced metals would provide the dusty material needed to efficiently form the planetesimals which then form the accretion cores. Assignment 8 A rocky planet of radius 0.001 R orbits a main-sequence G star of radius 1.0 R and mass 1.0 M at an orbital radius of a = 0.2 AU. (a) Photometric properties: calculate the orbital period in days, the (approximate) transit duration in hours, and the depth (fractional change in flux) of an observed transit in one of these systems if the inclination is i = 90°. (b) Spectroscopic properties: determine the amplitude of the radial velocity oscillation of the star in this system if the planet has a mass of 10 -6 M . turn over 8 (c) Astrometric properties: determine the amplitude of the stellar motion in the plane of the sky in arcseconds if the star is a distance of 10 parsecs from the Earth. (a) The period P is given by 2 a / vp where vp2 = GMstar /a is the planet orbital speed. = 6.67 x10-11 2 x1030 / 3.0 x1010 = 4.4 x109 m2 s-2 vp = 6.5 x104 m/s P = 2 x 3.14 x 3.0 x1010 / 6.5 x104 s = 3 x 106 s = 40 days Duration of transit = 2 Rstar /Vp = 2x 7x108 m / 6.5 x104 m/s = 21,000s = 6 hours Depth of transit = (Rp/Rstar)2 = 10-6 (b) The radial velocity of the star is given by the motions relative to the centroid of the system: vstar = vp (Mp/Mstar) = 6.5 x104 x 10-6 m/s = 0.065 m/s (c) The same centroid argument gives the semi-major axis: astar = a (Mp/Mstar) and = astar/d radians where d = 10pc Hence æ m p öæ a ö ÷÷ç ÷ Dq = çç M è * øè d ø radians = 10-6 (3x 1010 /3 x1017 ) radians = 10-13 radians 9 = 10-13 x 60x60x60 arcsec = 2 10-8 arcsec turn over