Solucion al problema 89 del capitulo 12
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Transcript of Solucion al problema 89 del capitulo 12
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 12, Solution 89.
For earth, 63960 mi 20.909 10 ftR = = × ( ) ( )( )22 6 15 3 2earth 32.2 20.909 10 14.077 10 ft /sGM gR= = × = ×
For sun, ( ) ( )( )3 15 21 3 2sun 332.8 10 14.077 10 4.6849 10 ft /sGM = × × = ×
For circular orbit of earth, 6 992.96 10 mi 490.8 10 ftEr = × = ×
( ) 213sun
94.6849 10 97.70 10 ft/s490.8 10E
E
GMv
r×
= = = ××
For transfer orbit AB, 6 9, 141.5 10 mi 747.12 10 ftA E B Mr r r r= = = × = ×
( ) ( )( )397.70 10 1.83 5280 107.36 ft/sA E Av v v= + ∆ = × + =
A A B Bmr v mr v=
( )( )9 3
3
490.8 10 107.36 1070.527 ft/s
747.12 10A A
BB
r vvr
× ×= = =
×
For circular orbit of Mars, 6 9141.5 10 mi = 747.12 10 ftMr = × ×
( ) 213sun
94.6849 10 79.187 10 ft/s747.12 10M
M
GMv
r×
= = = ××
Speed increase at B.
( ) 3 3 379.187 10 70.527 10 8.660 10 ft/sM BBv v v∆ = − = × − × = ×
( ) 1.640 mi/sBv∆ =