MANUAL SOLUTIONS FOR Orbital Mechanics for Engineering Students: Revised Reprint (Aerospace Engineering) 4th Edition by Howard D. Curtis ISBN:978-0128240250 COMPLETE GUIDE ALL CHAPTERS COVERED 100% VERIFIED A+ GRADE ASSURED!!!!! NEW LATEST UPDATE!!!!!

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ORBITAL MECHANICS FOR ENGINEERING STUDENTS
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Howard D. Curtis
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Embry-
Riddle Aeronautical University Daytona Be
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ach, Florida
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,Solutions Manual Orbital Mechanics for Engineering Students Chapter 1


Problem 1.1 mf


(a)
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A A = A x iˆ + Ay ˆj + A zkˆ  A x iˆ + Ay ˆj + A zkˆ
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= Ax iˆ  Ax iˆ + Ay ˆj + A zkˆ + A y ˆ j  Ax iˆ + Ay ˆj + A zkˆ + A zkˆ  A x iˆ + A y ˆ j + A zkˆ
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= Ax 2 (iˆ  iˆ ) + Ax Ay i ˆ  ˆ j + Ax Az (i ˆ  kˆ ) +  Ay Ax ˆ j  i ˆ + Ay 2 ˆj  ˆ j + Ay Az ˆ j  k ˆ  ( ) ( ) ( ) ( )
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   
+ AzAx (kˆ  iˆ)+ AzAy kˆ  ˆj + Az2 (kˆ kˆ ) ( )
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= Ax 2 (1 ) + Ax Ay (0 ) + Ax Az (0 ) + Ay Ax (0 ) + Ay 2 (1 ) + Ay Az (0 ) + Az Ax (0 ) + Az Ay (0 ) + Az2 (1 )
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= Ax 2 + Ay 2 + Az2
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But, according to the Pythagorean Theorem, xA 2 y+ A 2z + A 2 = A2 , where A = A
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, the magnitude of
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the vector A . Thus A  A = A2 .
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(b)
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Cx Cy Cz

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or

A  (B  C) = Ax By Cz + Ay BzCx + AzBx Cy − Ax BzCy − Ay Bx Cz − AzBy Cx
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(1)

Note that (A  B )  C = C  (A  B ) , and according to (1)
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C  (A  B ) = Cx Ay Bz + Cy AzBx + Cz Ax By − Cx AzBy − Cy Ax Bz − Cz Ay Bx
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(2)

The right hand sides of (1) and (2) are identical. Hence A  ( B  C) = (A  B )  C .
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(c)
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A  (B  C) = Ax iˆ + Ay ˆj + Azkˆ  Bx By
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= Ay Bx Cy − By Cx − Az (BzCx − Bx Cz ) iˆ + Az By Cz − BzCy − Ax Bx Cy − By Cx  ˆj
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+ A ( B C − B C ) − A B C − B C  k ˆ
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= Bx (Ay Cy + AzCz )− Cx (Ay By + AzBz ) iˆ + By (Ax Cx + AzCz )− Cy (Ax Bx + AzBz ) ˆj
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1

, Solutions Manual Orbital Mechanics for Engineering Students Chapter 1



Add a nd s ubtract the underlined terms to get
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2