ASTB01 Exoplanets Lab Author: Anders Johansen <[email protected]> Revision date: $Date: 2015/08/28 14:55:59 $ Planets orbiting stars other than the Sun are called exoplanets. Stellar light reflected off exoplanets is extremely hard to detect, mainly because the photons from the host star swamp the planetary signal, even for very large telescopes. Therefore mostly indirect methods are used for finding exoplanets. In indirect methods the light from the host star is analysed in order to infer the presence of an unseen planet. The gravitational pull of the exoplanet on the host star gives a periodic redshift and blueshift in the planetary light. This shift is proportional to the radial velocity variations of the star and can be detected with very precise spectrographs. From the measured radial velocity one can infer the semi major axis of the planet and its minimum mass. If our line-of-sight goes approximately along the orbital plane of the planet, a periodic reduction in the stellar luminosity can be seen as the planet transits in front of the star. The transit method allows the determination of the planet's radius and orbital inclination, which combined with the minimum mass obtained from radial velocity measurements yields the mass density, an important characteristic of the planet. Today several thousand exoplanets have been detected, mainly by the radial velocity method and the transit method. The goal of this lab is to get acquainted with the methods for detecting exoplanets and to get a hands-on experience with exoplanet data through the NASA Kepler website and the interactive catalogue of the Extrasolar Planets Encyclopedia. Reflected light We first consider the stellar light reflected off a planet and estimate how hard it is to detect this light from Earth. L p L * r The figure shows the star with luminosity L? shining on a planet at a distance r away. The planet of radius Rp reflects the luminosity Lp. The stellar flux received by the planet is L F = ? : (1) ? 4πr2 The planet reflects the fraction A (the albedo) of the incoming light on its dayside, 2 Lref = AπRpF? : (2) 1 The ratio of planet light to stellar light received at Earth at a distance d from the system is 2 Fref (d) Lref =(2πd ) 2Lref = 2 = F?(d) L?=(4πd ) L? 2 2 2 2AπRpL?=(4πr ) A Rp = = 2 L? 2 r R !2 r −2 ≈ 4 × 10−9A p : (3) RJup 5 AU Here the final equation is normalised to a planet of Jupiter radius RJup at 5 AU distance from the host star. The typical contrast of 2:5 × 108 corresponds to 21 in magnitude difference between planet and star. Considering a Sun-like star of absolute magnitude M = 5 at a distance of 10 pc gives a magnitude of m = 26 for a Jupiter-size planet at 5 AU from the host star, in principle observable with a good telescope (such as the Hubble Space Telescope). The problem is that the star, effectively a point source, seen by a telescope of diameter D will be imaged as an Airy pattern due to diffraction: The angular diameter of central Airy disc is λ λ ! D −1 θ ≈ 1:22 = 0:126" : (4) D 500 nm 1 m Recall that 1 AU at 1 pc distance extends 1" in angle, so 5 AU (the distance between the Sun and Jupiter) extends 0.5" in angle at 10 pc. This means that planets within 5 AU from the host star will be lost in the Airy disc or in one the secondary peaks of the Airy pattern. Planets in wider orbits, on the other hand, have very low luminosities in their reflected light. That is the reason why astronomers resort to indirect methods to detect exoplanets. Radial velocity method The radial velocity method exploits that two gravitating bodies in fact orbit around their common centre of mass. The orbital motion of the star around the centre of mass can be detected from the periodic redshift and blueshift of its spectral lines. The amplitude of the radial velocity yields the semi-major axis of the planet as well as its minimum mass. Let us assume that the star and the planet interact only via gravity: 2 F* Fp r* rp We place the origin of the coordinate frame at an arbitrary position and denote the position vector of the star r? and the position vector of the planet rp. The force of gravity felt by the star is denoted F ? and the force of gravity felt by the planet is denoted F p. These two forces are given by Gm m r ¨ ? p ?p F ? = m?r? = 2 ; (5) r?p r?p Gm m r ¨ ? p p? F p = mprp = 2 : (6) rp? rp? Here we introduced the mass of the star m?, the mass of the planet mp, the gravity constant G, the vector pointing from star to planet r∗p = rp − r?, as well as the vector pointing from planet to star rp∗ = r? −rp. We immediately see that F p = −F ?. Gravity clearly should obey Newton's third law! Equations (5) and (6) can be subtracted to yield a single equation for r ≡ r∗p, G(m + m ) r r¨ = − ? p : (7) r2 r This equation corresponds to the simpler situation of a mass-less (non-gravitating) parti- cle orbiting around a central object with mass m? +mp. The solution is that the vector r follows the familiar circular or eccentric orbit. We were able to transform the complicated two-body problem into a much simpler one-body problem by choosing a reference frame fixed on the central star. Note that r is measured relative to the position of the star, but the star is of course moving as well. Going instead to the (inertial) centre-of-mass frame will illuminate the actual orbit of the star. In the centre-of-mass frame the relative positions must obey m?r? + mprp = 0 : (8) Here we redefined the coordinate system so that r? and rp are measured relative to the centre of mass. Using r ≡ r?p = rp − r? (which follows a Keplerian orbit) we get m?r? + mp(r? + r) = 0 : (9) 3 The positions of star and planet follow as mp mp r? = − r ≈ − r for mp m? ; (10) m? + mp m? m? rp = + r ≈ r for mp m? : (11) m? + mp We see here how both the planet and the star follow the Keplerian orbit vector r, but with different coefficients in front, which depend on their relative masses. The star is always on the opposite side of the planet, in order to conserve the centre of mass at the origin of the coordinate frame. In the limit where the star is much more massive than the planet, the planet's orbit will be approximately rp ≈ r and the star's orbit will be the much smaller r? ≈ −(mp=m?)r. The centre of mass between Sun and the planets in the Solar System is shown in the following table: Planet CM with Sun [km] CM with Sun [R ] Mercury 9.6 km 1:4 × 10−5 Venus 265 km 3:8 × 10−4 Earth 449 km 6:4 × 10−4 Mars 74 km 1:1 × 10−4 Jupiter 743,000 km 1:07 Saturn 408,000 km 0:59 Uranus 125,000 km 0:18 Neptune 232,000 km 0:33 Jupiter has by far the largest effect on the Sun. In the presence of only Jupiter the Sun would orbit around a point just outside the surface of the Sun. The orbital frequency of the planet Ωp ≡ Ω must equal that of the star Ω? in order to maintain the centre of mass. Therefore we will know the orbital frequency of the planet by measuring the orbital frequency of the star from the periodic redshift and blueshift of the spectral lines. The period of the planet, P = 2π/Ω, then directly yields the semi-major axis of the planet a, provided that the star's mass is known, from v u 3 s 3 2π u a a P = = 2πt ≈ 2π : (12) Ω G(m? + mp) Gm? In order to get the mass of the planet we need to measure the speed of the star around the centre of mass. The star's speed is given by 2πr v = ? : (13) ? P The centre of mass of the system is located at a distance r? = (mp=m?)a from the centre of the star, so knowing v? from the spectrum we can infer r? and hence mp (provided that the star's mass is known). The speed of the Sun around the centre of mass with the planets of the Solar System is shown in the following table. 4 Planet RV of Sun Mercury 0.008 m/s Venus 0.09 m/s Earth 0.09 m/s Mars 0.008 m/s Jupiter 12.5 m/s Saturn 2.8 m/s Uranus 0.3 m/s Neptune 0.3 m/s Jupiter again gives the largest effect, of approximately 12:5 m/s. This is nevertheless an extremely low speed in astronomical terms. The Earth gives a velocity signal of only 10 cm/s. The problem is that only the component of the star's velocity that is perpendicular to the plane of the sky can be measured. This radial velocity is obtained directly from the spectrum. The result is that we measure RV? = v? sin i / mp sin i. Thus we see only the 0 effect of a planet with mass mp = mp sin i.