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.� Exercises 01 1. Given a =2i x − i + i y z b = i +2i y − i x z c = i + i y − 2i x z find a vector parallel to the plane of b and c and perpendicular to a. 2. Find the angle between the vectors a = i + i +2i x y z b =2i x − i + i y z 3. The vectors from the origin to the points A, B , C are a = i + i y − i x z b =3i +3i +2i x y z c =3i x − i y − 2i z Find the distance from the origin to the plane ABC . 4. By means of products express the condition that three vectors a, b, c be parallel to a plane. 5. Prob G–4 Consider a triangle with sides a, b and c.If a, b and c are vectors representing the sides of the triangle, use Vector Algebra to derive the Law of Cosines c 2 = a 2 + b2 − 2ab cos θ 6. Prob 3–13 From the pericenter and apocenter radii r and r , the semimajor and p a semiminor axes and the parameter of an orbit can be conveniently obtained. Show that 11� 11 � 2r r p a=+b = r r a = 12 (r + r ) or p = p ap ap 2 r r r + rp a p a7. Prob 3–1 To derive the equations of motion in polar coordinates we differentiate the vector representing the velocity in polar coordinates. We have dv � d2r �dθ �2 � � drdθ d2θ � µa = dt = dt2 − r dt i r + 2 dt dt + rdt2 iθ = − r2 i r so that d2r �dθ �2 µ drdθ d2θ dt2 − r dt + r2 = 0 and 2 dt dt + rdt2 =0 8. Prob. 3–6 The second of these equations of motion can be integrated to produce Kepler’s second law 2 dθ r = h dt which, in this form, also provides a transformation of independent variable from t to θ as given by dr drdθ hdr = = dt dθ dt r2 dθ Similarly, d2r hd � h dr � h2 d2rh2 �dr �2 = = 2 2 4 5dt2 r dθ r dθ r dθ2 − 2 r dθ Substituting in the first of the equations of motion gives 1 d2r 2 �dr �2 1 µ 1 r2 dθ2 − r3 dθ − r = − h2 = − p Next, we replace the dependent variable r by its reciprocal 1/r to obtain d �1� 1 dr = −dθ r r2 dθ d2 �1� 1 d2r 2 �dr �2 11 dθ2 r = − r2 dθ2 + r3 dθ = p − r so that d2 �1� 1 1 + = dθ2 r r p is obtained as a linear, constant-coefficient, second-order differential equation for 1/r . The solution provides an independent derivation of the equation of orbit and is readily obtained as 11 = + c1 cos θ + c2 sin θ r p First θ = 2 1 π = c2 = 0 Then θ =0= c1p = e Hence⇒ ⇒ 11 e p= + cos θ or r = r p p 1+ e cos θ��� ��� ��� ��� ��� ��� ��� � � � � ��� � � � � ��� ��� ��� ��� ��� ��� 9. Prob. 3–14 Suppose the force of attraction is proportional to the distance separating m1 and m2 rather than inversely proportional to the square of the distance. The equations of motion would be d2r1 = Gm1m2)m1 dt2 (r2 − r1dv = + G(m1 + m2)r = 0 m2 d2r2 = Gm2m1(r1 − r2) ⇒ dt dt2 where r = r2 − r1 and µ = G(m1 + m2). Then from dv d r × dt = dt(r × v)= −µr × r = 0 we have h = r × v so that angular momentum is preserved. The equations of motion are linear with constant coefficients and can be solved directly. We have r = cos √µ t c1 + sin √µ t c2 where c1 and c2 are constant vectors. The motion is planar so, for convenience, we can assume that the orbit is confined to the x, y plane. Hence, the equations of motion are x = cos √µtc11 + sin √µtc21 y = cos √µtc12 + sin √µtc22 Next, solve for cos √µ t and sin √µ t xc21 c11 x cos √µ t = yc22 sin √µ t c12 y= c11 c21 c11 c21 c12 c22 c12 c22 and then eliminate the trigonometric functions by squaring and adding � � + � � = � � yc22 c12 y c12 c22 The result is an equation for the ellipse (c22x − c21y)2 +(c11y − c12x)2 =(c11c22 − c12c21)2 or (c 2 + c 2 )x 2 )xy +(c 2 + c 2 )y )2 22 12 − 2(c22c21 + c11c12 21 112 =(c11c22 − c12c21 with the mass m1 at the center of the ellipse. 2 2 2 xc21 c11 x c11