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by: Hugh Wilkinson


Hugh Wilkinson
GPA 3.56

Kurt Anderson

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About this Document

Kurt Anderson
Class Notes
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This 18 page Class Notes was uploaded by Hugh Wilkinson on Monday October 19, 2015. The Class Notes belongs to MANE 4100 at Rensselaer Polytechnic Institute taught by Kurt Anderson in Fall. Since its upload, it has received 127 views. For similar materials see /class/224913/mane-4100-rensselaer-polytechnic-institute in Mechanical and Aerospace Engineering at Rensselaer Polytechnic Institute.

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Date Created: 10/19/15
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wzzyx Vi WHERE DO swquotX AND lkil m23 F0 ELLIPTIL rRwsrm H36 swhquot7 AND JUN Fan Hypguvm rmwsrsns Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 1 Problem 112 The absolute position velocity and acceleration of O are r0 3001 200i 100R m v0 101 30j 50Rn1s a0 251 40j 1512 msz The angular velocity and acceleration of the moving frame are 12 061 04j 101 rads Q 041 03j 1012 rads2 The unit vectors of the moving frame are 1 0577351 057735j 0577351 5 07242961 066475j 007820612 12 0338641 047410j 08127412 The absolute position of P is r 1591 200i 30012 In The velocity and acceleration of P relative to the moving frame are vrel 201 253 7012 ms arel 751 85 6012 ms Calculate the absolute velocity VP and acceleration ap of P Wlnertial frame Y Solution Velocity analysis From Equation 166 VP ZVO DXI39re1 Vre1 From the given information we have Howard D Curtis 15 Copyright 2010 Elsevier Inc Solutions Manual Orbital Mechanicsfor Engineering Students Second Edition Chapter 1 v0 101 301 5012 2 rrel r r0 1501 2001 30012 3001 2001 10012 1501 4001 20012 3 1 1 12 n x rml 06 04 10 3201 2701 30012 4 150 400 200 vrel 201 251 7012 20 0577351 0577351 05773512 25 0742961 0664751 007820612 70 0338641 0474101 08127412 so that vrel 538261 281151 4730012ms 5 Substituting 2 3 4 and 5 into 1 yields VP 101 301 5012 3201 2701 30012 538261 281151 4730012 VP 256171 268121 302712 vp 47868 0535161 0560111 063236K ms Acceleration mm lysis From Equation 170 aaoQerelQXQere1ZQere1arel 6 Using the given data together with 4 and 5 we obtain a0 251 401 1512 7 1 1 12 Q X gel 04 03 10 3401 2301 20512 8 150 400 200 1 1 12 QXQgtltrre1 06 04 10 39015001 3412 9 320 270 300 1 1 12 29 x vre1 2 06 04 10 2 91511 822061 3839912 10 53826 28115 47300 Howard D Curtis Copyright 2010 Elsevier Inc Solutions Manual Orbital Mechanicsfor Engineering Students Second Edition Chapter 1 arel 753 85 6012 75 057735i 057735j 05773512 85 074296i 066475j 007820612 60 033864f 047410j 08127412 arel 86134l 41649j85418flt 11 Substituting 7 8 9 10 and 11 into 6 yields a 25f 40f 1512 340i 230i 20512 390i 500i 3412 29151i 82206j 38399K86134i 41649j 85418flt a 102i 60142j 8774312 3 61629 016551f 097588j 01432712 msz Howard D Curtis 17 Copyright 2010 Elsevier Inc Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 1 Problem 115 At 29 north latitude what is the deviation 1 from the vertical of a plumb bob at the end of a 30 In string due to the earth39s rotation Solution From Equation 190b with z 0 a QZRE sin cos i QZRE cos2 112 From ZFy 2 may we get T sin 6 InQZRE sin 1 cos or T InQZRE sin cos sin 6 From 2132 111512 we obtain TcosB mg 2 1n22RE cos2 1 Substituting 1 52212 m Esm COS c080 mg 2 1n22RE cos2 1 Sin 6 from which we find QZRE sin 1 cos 1 tan6 g QZRE cos2 1 Since 51 LtanQ we deduce d L QZRE sin cos g QZRE cos2 1 Setting L 30 m RE 2 6378 X 1000 6378 X 106 m 1 29 g 981 ms2 27 Q 72921 x 10 5 rads 239344 x 3600 yields Howard D Curtis 23 1 2 Copyright 2010 ElseVier lnc Solutions Manual Orbital Mechanicsfor Engineering Students Second Edition Chapter 1 d 441 mm to the south Howard D Curtis 24 Copyright 2010 Elsevier Inc Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 2 Problem 22 Three particles of identical mass In are acted on only by their mutual gravitational attraction They are located at the vertices of an equilateral triangle with sides of length 1 Consider the motion of any one of the particles about the system center of mass G and using G as the origin of the inertial frame employ Newton39s second law to determine the angular velocity u required for d to remain constant Solution The center of mass of the three equal masses lies at the centroid of the equilateral triangle whose altitude h is given by h 2 do sin 60 The distance r of each mass from the center of mass is therefore 2 2 r h d0sin60 Relative to an inertial frame with the center of mass as its origin the acceleration of each particle is 2 a wozr 3600de sin 60 and this acceleration is directed toward the center of mass The net force on each particle is the vector sum of the gravitational force of attraction of its two neighbors This net force is directed towards the center of mass so that its magnitude focusing on particle 1 in the figure is 2 G n 1 cos30 Gquot 1 cos30 2 dmz 0 0 0 Fmt Fgl2 cos30 Fgl3 cos30 G cos30 Setting Fmt 2 ma we get Gin2 2 2 d0 cos 30 111 600de sin 60 2 3G1n cos 30 3G1n we 3 0 3 do s1n 60 do 3G1n Howard D Curtis 42 Copyright 2010 Elsevier Inc Solutions Manual Orbital Mechanicsfor Engineering Students Second Edition Chapter 2 Problem 211 Relative to a nonrotating earth centered Cartesian coordinate system the position and velocity vectors of a spacecraft are r 8900i 1690 52101 km and v 6i 45 151 kms Calculate the orbit s a eccentricity vector and b the true anomaly Solution a i j k h r x v 8900 1690 5210 25 980i 44 610 29 91012 kas 6 415 15 From Equation 240 v X h r e 7 7 u r i 12 6 45 15 e 25 980 44 610 29910 8900 1690 521012 398600 J 89002 16902 52102 201 510 140 490 384 57012 8900 1690 521012 c 398 600 10 450 e 050554 035246 09648012 085164 016172 04985512 e 03461005141804662512 b re cos9i re 8900 1690 521012 034610 051418 04662512 cos9 89002 16902 52102 0346102 0514182 0466252 1520l cos9 10450077560 cos9 018754 Howard D Curtis 58 Copyright 2010 Elsevier Inc Solutions Manual Orbital Mechanicsfor Engineering Students Second Edition Chapter 2 Problem 217 A spacecraft is in a circular orbit of the moon at an altitude of 80 km Calculate its speed and its period Solution The mass of the moon is 7348 X 1022 kg Therefore for a satellite orbiting the moon 3 3 p Gmm00n 667259 X10720 km 52 7348 x1022 kg 4903 k The radius of the moon as 1738 km Hence p 4903 7 7 1642 k v if jlm go Q Q Ter2 2 1738802 6956 sec 1 hr 56 min 4 14903 Howard D Curtis 64 Copyright 2010 ElseVier lnc


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