()221sincoscos02ztgt vtvtαωαλ′′=−++=DDor 2sin2sin2coscosvvtgvgωα=≈′−DWe have ignored the second term in the denominator—since v′Dwould have to be impossibly large for the value of that term to approach the magnitude g For example, for 4145ooandλα==oscosgvωαλ′′−≈−or 144gks′≈≈Dmv! Substituting t into equation 5.4.15b to find the lateral deflection gives 324cossinsinsincosvytvtg′DD±5.18Let…aDG= acceleration of object relative to Earth GD= ˆkDAD= its angular speed RG0RGrGyxG= acceleration of satellite ˆk=G= its angular speed ( )2aavr Aωω=+×+××+GGGGGG(Equation 5.2.14) GGGGGG2aa Avr=−− ×−××As in prob em 5.7 Evaluate the termGGl… ( )aArRRRR= − −××= × × −××−×× −DDDDGGGGGG G G( )aR∆=−GGa∆… given the condition that 23RR=3231
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Velocity, Cos, Physical quantities, Ω, 144