MIT OpenCourseWare http://ocw.mit.edu 16.346 Astrodynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Lecture 1 The Two Body Problem Newton’s Two-Body Equations of Motion 1687 #3.1, #3.3 Force = Mass × Acceleration d2 Gm1m2 (r2 − r1) d2r1 (m1r1 + m2r2)= 0 = m1 dt2 r2 r dt2 = ⇒ G(m1 + m2) (r2 − r1) d2 Gm2m1 (r1 − r2)= m2 d2r2 − r2 r = dt2 (r2 − r1) r2 r dt2 Conservation of Total Linear Momentum Page 96 d2r def m1r1 + m2r2cm = 0 = r cm = c1t + c2 where r cm = dt2 ⇒ m1 + m2 Two-Body Equation of Relative Motion Page 108 r = r2 − r1 d2r µ or ddt v = − rµ 3 r where r = |r| = |r2 − r1|+ r = 0 dt2 r3 µ = G(m1 + m2) Vector Notation Position Vectors • r1 = x1 i x + y1 i y + z1 i z  x1   x2    r2 = x2 i x + y2 i y + z2 i z r1 =  y1  r2 =  y2  r = r2 − r1 =  y  r = r2 − r1 = x i + y i y + z i z1 z2 z x z Two-Body Equations of Motion in Rectangular Coordinates • d2xµ d2yµ d2zµ+ x =0 + y =0 + z =0 dt2 r3 dt2 r3 dt2 r3 Velocity Vectors •   dx/dt v = dr = dx i x + dy i y + dz i z =  dy/dt  dt dt dt dt dz/dt Polar Coordinates • di rr = r ii = cos θ i + sin θ i iθ = − sin θ i + cos θ i = r r x y x y dθ dr dr di dθdr dθ v == i r + r r = i r + r iθ = v r i r + vθ iθdt dt dθ dt dt dt 16.346 Astrodynamics Lecture 1 x= r 2 = Constant ≡ h =2 Area θ ⇒ xdt − y = dt dt × dt ibbs (1839–1908) Vector Analysis for the Engineer Appendix B–1 r1 · r2 = x1x2 + y1y2 + z1z2 = r 1r2 cos i x i y i z r1 × r2 = �� �� ���x y z r1r 1 1 1 � ����= 2 sin in x2 y2 z2 x1 y1 z1 r1 × r2 · r3 = ����x2 y2 z2 x y3 z�3 3 ���(r1 × r2) × r3 =(�� r1 � �· r3)r2 − (r2 · r3)r1 r1 × (r2 × r3)=(r1 · r3)r2 − (r1 · r2)r3 Law 1609 Conservation of Angular Momentum � � 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 d2yd2xd dy dx dy dx 0= x = Constant = = x x dt2 − y dt2 dt dt − y dt ⇒ dt − y dt Using polar coordinates x = r cos θ dy dx dθ d y = r sinJosiah Willard GKepler’s Seconddv d r × dt = dt(r × v)= 0 = ⇒ = Constant Motion takes place in a plane and angular momentum is conserved In polar coordinates h = r × v dr dr dθ r = r ir dt = v = dt i r + rdt iθ = v r i r + vθ iθ so that the angular momentum of m2 with respect to m1 is 2 dθ def m2 rvθ = m2 r = m2 h = Constant dt • Rectilinear Motion: For r v, then .h =0 16.346 Astrodynamics Lecture 1 Image removed due to copyright restrictions.The quantity h is called the angular momentum but is actually the massless angular momentum. In vector form h = h i so that h = r×v and is a constant in both magnitude z 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 dA 1 2 dθ h = Constant = r = dt 2 dt 2 Kepler’s Law is a direct consequence of radial acceleration! Eccentricity Vector ddv µ µh µh dθ di dt(v × h)= dt × h = −r3 r × h = − r2 i r × ih = r2 iθ = µdt iθ = µ dt r Hence µ µe = v × h − 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 − µr r = r × v h − µr = h h − µr = h2 − µr· · r · · · Also µe r = µre cos f where f is the angle between r and e so that · p h2 defr = or r = p − ex where p = 1+ e cos f 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 r = p and r = p p a1+ e 1 − e If 2a is the length of the major axis, then r + r =2a = p = a(1 − e 2)p a ⇒ 16.346 Astrodynamics Lecture 1� � 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 √µp � µa(1 − e2)= = = P 2 2 2 2Also, from the elementary properties of an ellipse, we have b = a√1 − e so that the Period of the ellipse is 3aP =2π µ Other expressions and terminology are used 2πµMean Motion orn == P a3 2 3 µ = n a or 3a= 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 The astronomical unit (Mean distance from Earth to the Sun) Time The year (the Earth’s period) Mass The Sun’s mass (Ignore other masses compared to Sun’s mass) Then µ = G(m1 + m2)= G(m sun + mplanet)= G(m sun)= …