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Orbital Mechanics for Engineering Students 4th Edition By Howard Curtis (Solution Manual) $16.49
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Orbital Mechanics for Engineering Students 4th Edition By Howard Curtis (Solution Manual)

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Orbital Mechanics for Engineering Students, 4e Howard Curtis (Solution Manual) Orbital Mechanics for Engineering Students, 4e Howard Curtis (Solution Manual)

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  • August 12, 2023
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  • 2023/2024
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SOLUTIONS MANUAL for ORBITAL MECHANICS FOR ENGINEERI NG STUDENTS Fourth Edition Howard D. Curtis NOTE: For Complete File, Download link at the end of this File Solutions Manual Orbital Mechanics for Engineering Students Fourth Edition Chapter 1 Howard D. Curtis 1–1 Problem 1.1 Given the three vectors A=Axˆi+Ayˆj+Azˆk, B=Bxˆi+Byˆj+Bzˆk and C=Cxˆi+Cyˆj+Czˆk, show analytically that (a) A⋅A=A2 (b) A⋅B×C( ) =A×B ( ) ⋅C (c) A×B×C( ) =B A ⋅C( ) −C A ⋅B( ) Solution A⋅A=Axˆi+Ayˆj+Azˆk ( ) ⋅Axˆi+Ayˆj+Azˆk ( ) =Axˆi⋅Axˆi+Ayˆj+Azˆk ( ) +Ayˆj⋅Axˆi+Ayˆj+Azˆk ( ) +Azˆk⋅Axˆi+Ayˆj+Azˆk ( ) =Ax2ˆi⋅ˆi( ) +AxAyˆi⋅ˆj( ) +AxAzˆi⋅ˆk( ) ⎡⎣⎤⎦+AyAxˆj⋅ˆi( ) +Ay2ˆj⋅ˆj( ) +AyAzˆj⋅ˆk( ) ⎡⎣⎤⎦ +AzAxˆk⋅ˆi( ) +AzAyˆk⋅ˆj( ) +Az2ˆk⋅ˆk( ) ⎡⎣⎤⎦ =Ax21( )+AxAy0( )+AxAz0( ) ⎡⎣⎤⎦+AyAx0( )+Ay21( )+AyAz0( ) ⎡⎣⎤⎦+AzAx0( )+AzAy0( )+Az21( ) ⎡⎣⎤⎦ =Ax2+Ay2+Az2 But, according to the Pythagorean Theorem, Ax2+Ay2+Az2=A2, where A=A, the magnitude of the vector A. Thus A⋅A=A2. (b) A⋅B×C( ) =A⋅ˆi ˆj ˆk
BxByBz
CxCyCz =Axˆi+Ayˆj+Azˆk ( ) ⋅ˆiByCz−BzCy ( ) −ˆjBxCz−BzCxA ( ) +ˆkBxCy−ByCx ( )⎡
⎣⎤
⎦ =AxByCz−BzCy ( ) −AyBxCz−BzCx ( ) +AzBxCy−ByCx ( ) or A⋅B×C( ) =AxByCz+AyBzCx+AzBxCy−AxBzCy−AyBxCz−AzByCx (1) Note that A×B ( ) ⋅C=C⋅A×B ( ), and according to (1) C⋅A×B ( ) =CxAyBz+CyAzBx+CzAxBy−CxAzBy−CyAxBz−CzAyBx (2) The right hand sides of (1) and (2) are identical. Hence A⋅B×C( ) =A×B ( ) ⋅C. (c) Solutions Manual Orbital Mechanics for Engineering Students Fourth Edition Chapter 1 Howard D. Curtis 1–2 A×B×C( ) =Axˆi+Ayˆj+Azˆk ( ) ׈i ˆj ˆk
BxByBz
CxCyCz=ˆi ˆj ˆk
AxAyAz
ByCz−BzCyBzCx−BxCyBxCy−ByCx =AyBxCy−ByCx ( ) −AzBzCx−BxCz ( ) ⎡⎣⎤⎦ˆi+AzByCz−BzCy ( ) −AxBxCy−ByCx ( )⎡⎣⎤⎦ˆj +AxBzCx−BxCz ( ) −AyByCz−BzCy ( )⎡⎣⎤⎦ˆk =AyBxCy+AzBxCz−AyByCx−AzBzCx ( )ˆi+AxByCx+AzByCz−AxBxCy−AzBzCy ( )ˆj
+AxBzCx+AyBzCy−AxBxCz−AyByCz ( )ˆk =BxAyCy+AzCz ( ) −CxAyBy+AzBz ( )⎡⎣⎤⎦ˆi+ByAxCx+AzCz ( ) −CyAxBx+AzBz ( ) ⎡⎣⎤⎦ˆj
+BzAxCx+AyCy ( ) −CzAxBx+AyBy ( )⎡⎣⎤⎦ˆk Add and subtract the underlined terms to get A×B×C( ) =BxAyCy+AzCz+AxCx ( ) −CxAyBy+AzBz+AxBx ( )⎡⎣⎤⎦ˆi
+ByAxCx+AzCz+AyCy ( )−CyAxBx+AzBz+AyBy ( )⎡
⎣⎤
⎦ˆj
+BzAxCx+AyCy+AzCz ( ) −CzAxBx+AyBy+AzBz ( )⎡⎣⎤⎦ˆk =Bxˆi+Byˆj+Bzˆk ( ) AxCx+AyCy+AzCz ( ) −Cxˆi+Cyˆj+Czˆk ( ) AxBx+AyBy+AzBz ( ) =Bxˆi+Byˆj+Bzˆk ( ) A⋅C( ) −Cxˆi+Cyˆj+Czˆk ( ) A⋅B( ) Or, A×B×C( ) =B A ⋅C( ) −C A ⋅B( ) Solutions Manual Orbital Mechanics for Engineering Students Fourth Edition Chapter 1 Howard D. Curtis 1–3 Problem 1.2 Use just the vector identities in Problem 1.1 to show that A×B( ) ⋅C×D( ) =A⋅C( ) B⋅D( ) −A⋅D( ) B⋅C( ) Solution From Problem 1.1(b) A×B ( ) ⋅C×D( ) = A×B ( ) ×C [ ] ⋅D (1) But A×B ( ) ×C [ ] ⋅D=− C×A×B ( ) [ ] ⋅D Using Problem 1.1(c) on the right yields A×B ( ) ×C [ ] ⋅D=−A C ⋅B( ) −B C ⋅A( ) [ ] ⋅D or A×B ( ) ×C [ ] ⋅D=−A⋅D( ) C⋅B( ) +B⋅D( ) C⋅A( ) (2) Substituting (2) into (1) we get A×B( ) ⋅C×D( ) =A⋅C( ) B⋅D( ) −A⋅D( ) B⋅C( )

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