Lecture 1: The Two-Body Problem

Lecture 1 The Two Body Problem Newton’s Two-Body Equations of Motion Force = Mass × Acceleration Gm1 m2 (r2 − r1 ) d r = m1 21 2 r r dt Gm2 m1 (r1 − r2 ) d2 r2 = m 2 r2 r dt2 2 =⇒ 1687 d2 (m r + m2 r2 ) = 0 dt2 1 1 G(m1 + m2 ) (r2 − r1 ) d2 − = (r − r1 ) r2 r dt2 2 Conservation of Total Linear Momentum d2 rcm =0 dt2 =⇒ rcm = c1 t + c2 Page 96 or def where Two-Body Equation of Relative Motion µ d2 r + 3r = 0 2 dt r #3.1, #3.3 rcm = m1 r1 + m2 r2 m1 + m2 Page 108 dv µ = − 3r dt r r = r2 − r1 r = |r| = |r2 − r1 | µ = G(m1 + m2 ) where Vector Notation • Position Vectors r1 = x1 ix + y1 iy + z1 iz r2 = x2 ix + y2 iy + z2 iz r = r2 − r1 = x ix + y iy + z iz   x1 r1 =  y1  z1   x2 r2 =  y2  z2   x  r = r 2 − r1 = y  z • Two-Body Equations of Motion in Rectangular Coordinates µ d2 x + 3x = 0 2 dt r • Velocity Vectors d2 y µ + 3y = 0 2 dt r d2 z µ + 3z = 0 2 dt r   dx/dt dr dx dy dz v= = ix + iy + i =  dy/dt  dt dt dt dt z dz/dt • Polar Coordinates r = r ir v= ir = cos θ ix + sin θ iy iθ = − sin θ ix + cos θ iy = dr dr di dθ dθ dr = ir + r r = ir + r iθ = vr ir + vθ iθ dt dt dθ dt dt dt 16.346 Astrodynamics Lecture 1 dir dθ Kepler’s Second Law 1609 Equal Areas Swept Out in Equal Times Assume z = 0 so that the motion is confined to the x-y plane d2 x d � dy dx � dy dx d2 y 0=x 2 −y 2 = x −y =⇒ x −y = Constant dt dt dt dt dt dt dt Using polar coordinates x = r cos θ y = r sin θ =⇒ x dx d dθ dy −y = r2 = Constant ≡ h = 2 × Area dt dt dt dt Josiah Willard Gibbs (1839–1908) Vector Analysis for the Engineer Appendix B–1 Image removed due to copyright restrictions. r1 · r2 = x1 x2 + y1 y2 + z1 z2 = r1 r2 cos
� � �i � � x iy iz � r1 × r2 = �� x1 y1 z1 �� = r1 r2 sin
in �x y z � 2 2 2 � � �x y z � 1 1� � 1 r1 × r2 · r3 = �� x2 y2 z2 �� �x y z � 3 3 3 (r1 × r2 ) × r3 = (r1 · r3 )r2 − (r2 · r3 )r1 r1 × (r2 × r3 ) = (r1 · r3 )r2 − (r1 · r2 )r3 Kepler’s Second Law 1609 Conservation of Angular Momentum d dv = (r × v) = 0 =⇒ h = r × v = Constant dt dt Motion takes place in a plane and angular momentum is conserved r× In polar coordinates dr dθ dr =v= ir + r i = vr i r + vθ i θ dt dt dt θ so that the angular momentum of m2 with respect to m1 is r = r ir m2 r vθ = m2 r2 • Rectilinear Motion: 16.346 Astrodynamics dθ def = m2 h = Constant dt For r  v, then h = 0 . Lecture 1 The quantity h is called the angular momentum but is actually the massless angular momentum. In vector form h = h i z so that h = r×v and is a constant in both magnitude and direction. This is called Kepler’s second law even though it was really his first major result. As Kepler expressed it, the radius vector sweeps out equal areas in equal time since 1 dθ h dA = Constant = r2 = dt 2 dt 2 Kepler’s Law is a direct consequence of radial acceleration! Eccentricity Vector d dv µ µh µh dθ di (v × h) = × h = − 3 r × h = − 2 i r × ih = 2 i θ = µ i θ = µ r dt dt r r r dt dt Hence µe = v × h − µ r = Constant r The vector quantity µe is often referred to as the Laplace Vector. We will call the vector e the eccentricity vector because its magnitude e is the eccentricity of the orbit. Kepler’s First Law 1609 The Equation of Orbit If we take the scalar product of the Laplace vector and the position vector, we have µe · r = v × h · r − Also µe · r = µre cos f r= µ r · r = r × v · h − µr = h · h − µr = h2 − µr r where f is the angle between r and e p 1 + e cos f or r = p − ex where so that def p = h2 µ is the Equation of Orbit in polar coordinates. (Note that r cos f = x .) The angle f is the true anomaly and p , called the parameter, is the value of the radius r for f = ± 90 ◦ . The pericenter ( f = 0) and apocenter ( f = π ) radii are rp = p 1+e and ra = p 1−e If 2a is the length of the major axis, then rp + ra = 2a 16.346 Astrodynamics Lecture 1 =⇒ p = a(1 − e2 ) Kepler’s Third Law 1619 The Harmony of the World Archimedes was the first to discover that the area of an ellipse is πab where a and b are the semimajor and semiminor axes of the ellipse. Since the radius vector sweeps out equal areas in equal times, then the entire area will be swept out in the time interval called the period P . Therefore, from Kepler’s Second Law πab h = = P 2 � √ µp µa(1 − e2 ) = 2 2 √ Also, from the elementary properties of an ellipse, we have b = a 1 − e2 so that the Period of the ellipse is � P = 2π a3 µ Other expressions and terminology are used Mean Motion 2π = n= P � µ a3 or µ = n2 a3 or a3 = Constant P2 The last of these is known as Kepler’s third law. • Kepler made the false assumption that µ is the same for all planets. Units for Numerical Calculations A convenient choice of units is Length Time Mass The astronomical unit (Mean distance from Earth to the Sun) The year (the Earth’s period) The Sun’s mass (Ignore other masses compared to Sun’s mass) Then µ = G(m1 + m2 ) = G(msun + mplanet ) = G(msun ) = G so that, from Kepler’s Third Law, we have µ = G = 4π 2 or k= where G is the Universal Gravitation Constant. 16.346 Astrodynamics Lecture 1 √ G = 2π Josiah Willard Gibbs (1839–1908) was a professor of mathematical physics at Yale College where he inaugurated the new subject — three dimensional vector analysis. He had printed for private distribution to his students a small pamphlet on the “Elements of Vector Analysis” in 1881 and 1884. Gibbs’ pamphlet became widely known and was finally incorporated in the book “Vector Analysis” by J. W. Gibbs and E. B. Wilson and published in 1901. Gibb’s Method of Orbit Determination Pages 131–133 • Given r1 , r2 , r3 with r1 × r2 · r3 = 0 • To determine the eccentricity vector e and the parameter p r ×r ·n r ×r ·n r2 = αr1 + βr3 with n = r1 × r3 =⇒ α = 2 23 and β = 1 22 n n αr − r2 + βr3 p= 1 0 = e · (r2 − αr1 − βr3 ) = p − r2 − α(p − r1 ) − β(p − r3 ) =⇒ α−1+β • To determine the eccentricity vector, we observe that: n × e = (r1 × r3 ) × e = (e · r1 )r3 − (e · r3 )r1 = (p − r1 )r3 − (p − r3 )r1 Then, since (n × e) × n = n2 e e= it follows that 1 [(p − r1 )r3 × n − (p − r3 )r1 × n] n2