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