AA311.Homework3.Solutions

# AA311.Homework3.Solutions - AA311 Homework 3 Due Solution...

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AA311 Homework 3. Due October 17, 2011. Solution. Give all your answers in SI units. Problem 1. Recall that, when developing the equations for the ISA and the notion of “geopotential height,” we assumed that g ( h G ) g 0 . The geopotential height h was then deﬁned such that ΔPE( h ) = ΔPE( h G ) h = ± r earth r earth + h G ² h G This is a “zeroth order” approximation of the true acceleration due to gravity g ( h G ) = g 0 ± 1 1 + ( h G /r earth ) ² 2 A “ﬁrst order” approximation is g ( h G ) g 0 ± 1 - 2 h G r earth ² 1. Verify the correctness of this ﬁrst order approximation by expanding f ( ± ) = ± 1 1 + ± ² 2 in a Taylor series about ± = 0. Show the ﬁrst three terms (i.e., to quadratic order in ± ). 2. Develop an expression for a new altitude ˜ h ( h G ) deﬁned such that ΔPE( ˜ h ) = ΔPE( h G ) where ΔPE( ˜ h ) = Z ˜ h 0 m g 0 1 - 2 ˜ h r earth ! d ˜ h Solution. Recall that the Taylor series expansion of a function f ( ± ) about ± = 0 is f ( ± ) = X n =0 1 n ! d n f n ³ ³ ³ ³ ± =0 ± n = f (0) + df ³ ³ ³ ³ ± =0 ± + 1 2 d 2 f 2 ³ ³ ³ ³ ± =0 ± 2 + ... Thus, f ( ± ) = 1 + ± - 2 ± 1 1 + ± ²² ± =0 ± + 1 2 ± 2 ± 1 1 + ± ²² ± =0 ± 2 + ... = 1 - 2 ± + ± 2 + ... Integrating the potential energy function given for ˜ h gives ΔPE( ˜ h ) = Z ˜ h 0 m g 0 1 - 2 ˜ h r earth ! d ˜ h = m g 0 ˜ h 1 - ˜ h r earth ! Setting this equal to ΔPE( h G ) = m g 0 h G ± r earth r earth + h G ² 1

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0 200 400 600 800 1000 6.5 7 7.5 8 8.5 9 9.5 10 Geometric Altitude (km) Acceleration Due To Gravity (m/s 2 ) True Constant Model
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AA311.Homework3.Solutions - AA311 Homework 3 Due Solution...

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