Keplerian Orbits and Dynamics of Exoplanets Carl D. Murray Queen Mary University of London Alexandre C.M. Correia University of Aveiro Understanding the consequences of the gravitational interaction between a star and a planet is fundamental to the study of exoplanets. The solution of the two-body problem shows that the planet moves in an elliptical path around the star and that each body moves in an ellipse about the common center of mass. The basic properties of such a system are derived from first principles and described in the context of detecting exoplanets. 1. INTRODUCTION planet F2 Themotionof a planetarounda star can beunderstoodin r m the context of the two-body problem, where two bodies ex- star F1 2 ert a mutual gravitational effect on each other. The solution to the problem was first presented by Isaac Newton (1687) m1 in his Principia. He was able to show that the observed el- r2 liptical path of a planet and the empirical laws of planetary r1 motion derived by Kepler (1609, 1619) were a natural con- sequence of an inverse square law of force acting between O a planet and the Sun. According to Newton’s universal law of gravitation, the magnitude of the force between any two Fig. 1.— The forces acting on a star of mass m1 and a planet of masses m1 and m2 separated by a distance r is given by mass m2 with position vectors r1 and r2. m m F = G 1 2 (1) r2 The relative motion of the planet with respect to the star − is given by the vector r = r r . The gravitational forces where G = 6.67260 10−11Nm2kg 2 is the universal 2 1 acting on the star and the planet− are gravitational constant.× The law is applicable in a wide va- riety of circumstances. For example, the two bodies could m1m2 F1 = m1¨r1 =+G r , (2) be a moon orbiting a planet or a planet orbiting a star. New- r3 ton’s achievement was to show that motion in an ellipse is m1m2 F2 = m2¨r2 = G r (3) the natural consequence of such a law. A more difficult − r3 task is to find the position and velocity of an object in the respectively. Now consider the motion of the planet m two-body problem; this is commonly referred to as Kepler 2 with respect to the star m1. If we write ¨r = ¨r2 ¨r1 we can problem. In this chapter we derive the basic equations of use Eq. (1) to obtain − the two-body problem and solve them to show how ellip- arXiv:1009.1738v2 [astro-ph.EP] 25 Feb 2011 tical motion arises. We then proceed to solve the Kepler r ¨r + G(m1 + m2) =0 . (4) problem showing how motion around the common center r3 of mass of the two-body system can be used to infer the If we take the vector product of r with Eq. (4) we have presence of planetary companions to a star. Finally we give r ¨r =0 which can be integrated directly to give a few representative examples among extra-solar planets al- × ready detected. For the most part we follow the approach of r r˙ = h (5) Murray & Dermott (1999). × where h is a constant vector which is simultaneously per- pendicular to both r and r˙. Therefore the motion of the 2. BASIC EQUATIONS planet about the star lies in a plane (the orbit plane) perpen- dicular to the direction defined by h. Another consequence Consider a star and a planet of mass m1 and m2, respec- of this result is that the position and velocity vectors will tively, with position vectors r1 and r2 referred to an origin O fixed in inertial space (Fig. 1). always lie in the same plane (see Fig. 2). Equation (5) is 1 r h planet r θ = f + ̟ y r planet apoapse f star ae periapse star ̟ x b reference a = θ 0 direction Fig. 2.— The motion of m2 with respect to m1 defines an orbital plane (shaded region), because r × r is a constant vector, h, the ˙ Fig. 3.— The geometry of the ellipse of semi-major axis a, semi- angular momentum vector and this is always perpendicular to the minor axis b, eccentricity e and longitude of periapse ̟. orbit plane. In order to find r as a function of θ we need to make the often referred to as the angular momentum integral and h substitution u = 1/r. By differentiating r with respect to represents a constant of the two-body motion. time and making use of Eq. (9) we can eliminate time in the In order to solve Eq. (4) we transform to a polar coor- differential equation. We obtain dinate system (r, θ) referred to an origin centered on the star with an arbitrary reference line corresponding to θ =0. d2u d2u r¨ = h θ˙ = h2u2 (13) In polar coordinates the position, velocity and acceleration − dθ2 − dθ2 vectors can be written as and hence Eq. (12) can be written r = r ˆr (6) d2u G(m + m ) + u = 1 2 . (14) r˙ =r ˙ ˆr + rθ˙ θˆ (7) dθ2 h2 1 d This is a second order, linear differential equation, often ¨r = (¨r rθ˙2)ˆr + r2θ˙ θˆ . (8) − r dt referred as Binet's equation, with a general solution ˆ G(m1 + m2) where ˆr and θ denote unit vectors along and perpendicular u = [1 + e cos(θ ̟)] , (15) to the radius vector respectively. Substituting Eq. (7) into h2 − Eq. (5) gives h = r2θ˙ zˆ, where ˆz is a unit vector perpendic- where e (an amplitude) and ̟ (a phase) are two constants ular to the plane of the orbit forming a right-handed triad of integration. Substituting back for r gives ˆ with ˆr and θ. The magnitude of this vector gives us p r = , (16) 2 ˙ 1+ e cos(θ ̟) h = r θ . (9) − where p = h2/G(m + m ). This is the general equation Therefore, although r and θ vary as the planet movesaround 1 2 in polar coordinates of a set of curves known as conic sec- the star, the quantity r2θ˙ remains constant. The area ele- tions where e is the eccentricity and p is a constant called ment dA swept out by the star-planet radius vector in the the semilatus rectum. For a given system the initial con- time interval dt is given in polar coordinate by ditions will determine the particular conic section (circle, r 1 ellipse, parabola, or hyperbola) the planet follows. We con- dA = r dr dθ = r2dθ , (10) 0 2 sider only elliptical motion for which and thus p = a(1 e2) , (17) − 1 2 1 where a, a constant, is the semi-major axis of the ellipse. A˙ = r θ˙ = h = constant . (11) 2 2 The quantities a and e are related by This is equivalent to Kepler’s second law of planetary mo- b2 = a2(1 e2) , (18) tion which states that the star-planet line sweeps out equal − areas in equal times. where b is the semi-minor axis of the ellipse (see Fig. 3). Using Eq. (6) and comparing the ˆr components of Therefore, for any given value of θ the radius is calcu- Eqs. (4) and (8) gives the scalar differential equation lated using the equation 2 G(m + m ) a(1 e ) r¨ rθ˙2 = 1 2 . (12) r = − . (19) 2 1+ e cos(θ ̟) − − r − 2 Hence the path of the planet around the star is an ellipse which can be integrated to give with the star at one focus; this is Kepler’s first law of plan- 1 G(m + m ) etary motion. Note that in the special case where e = 0 (a v2 1 2 = C, (28) circular orbit), r = a and the angle ̟ is undefined. 2 − r The angle θ is called the true longitude. Equation (19) where v2 = r˙ r˙ is the square of the velocity and C is a shows that the minimum and maximum values of r are constant of the· motion. Equation (28) is called the vis viva a(1 e) (at θ = ̟) and a(1 + e) (at θ = ̟ + π), re- integral. It shows that the orbital energy per unit mass of spectively.− These points are referred to as the periapse and the system is conserved. the apoapse, respectively, although for motion around a star Because ̟ is a constant, θ˙ = f˙ and Eq. (7) gives they can also be referred to as the periastron and apastron. The angle ̟ (pronounced “curly pi”) is called the lon- v2 = r˙ r˙ =r ˙2 + r2f˙2 . (29) gitude of periapse or longitude of periastron of the planet’s · orbit and gives the angular location of the closest approach By differentiating Eq. (20) we obtain with respect to the reference direction. If we define the true anomaly to be the angle f = θ ̟ (see Fig. 3) then f is r fe˙ sin f − r˙ = (30) measured with respect to the periapse direction and Eq. (19) 1+ e cos f can be written and hence, using Eqs. (9) and (16), we have a(1 e2) r = − . (20) na 1+ e cos f r˙ = e sin f (31) √1 e2 In this case if we define a cartesian coordinate system cen- − tered on the star with the x-axis pointing towards the peri- and apse (see Fig. 3), then the position vector has components na rf˙ = (1 + e cos f) . (32) √1 e2 x = r cos f (21) − y = r sin f .