TEST BANK FOR SOLUTION MANUAL for Orbital Mechanics for Engineering Students 4th Edition By Howard Curtis ISBN 978-0081021330 COMPLETE GUIDE | A+ GRADE ASSURED!!! LATEST UPDATE!!!! GUARANTEED SUCCESS!!!!

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to accompany
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ORBITAL MECHANICS FOR ENGINEERING STUDENTS
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Howard D. CurtisK K




Embry-
Riddle Aeronautical University Daytona Beac
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h, Florida
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,Solutions Manual K Orbital Mechanics for Engineering Students K K K K Chapter 1
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Problem 1.1 K




(a)
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A A = Axiˆ + Ayˆj + Azkˆ  Axiˆ + Ayˆj + Azkˆ
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= Ax2 (iˆ  iˆ)+ AxAy iˆ  ˆj + AxAz (iˆ kˆ ) + AyAx ˆj iˆ + Ay2 ˆj  ˆj + AyAz ˆjkˆ  ( ) ( ) ( ) ( )
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   
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+ AzAx (kˆ iˆ)+ AzAy kˆ ˆj +Az2 (kˆ kˆ) ( )
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= Ax2 (1)+ AxAy (0)+ AxAz (0) + AyAx (0)+ Ay2 (1)+ AyAz (0) + AzAx (0)+ AzAy (0)+ Az2 (1)
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= Ax + Ay + Az2
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But, according to the Pythagorean Theorem, A 2 x+ A 2 +y A 2 =z A2 , where A = A , the magnitude of
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the vector A . Thus A  A = A .
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(b)
iˆ ˆj kˆ
A (B  C) = A  Bx
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Cx Cy Cz
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A  (B  C) = AxByCz + AyBzCx + AzBxCy − AxBzCy − AyBxCz − AzByCx
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(1)

Note that (A  B) C = C  (A  B) , and according to (1)
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C  (A  B) = CxAyBz + Cy AzBx + Cz AxBy − CxAzBy − Cy AxBz − Cz AyBx
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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)
iˆ ˆj kˆ iˆ ˆj kˆ


A (BC)= Axiˆ + Ayˆj+ Azkˆ  Bx
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+ A (B C − B C )− A B C − B C kˆ
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( x z x y z y x x z y y z)
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= Bx (AyCy + AzCz )− Cx (AyBy + AzBz )iˆ + By (AxCx + AzCz )− Cy (AxBx + AzBz )ˆj
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z( x x y y) z( x x y y)
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 
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1

, Solutions Manual
K Orbital Mechanics for Engineering Students
K K K K Chapter 1
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Add and subtract the underlined terms to get
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2