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
d±
n
³
³
³
³
±
=0
±
n
=
f
(0) +
df
d±
³
³
³
³
±
=0
±
+
1
2
d
2
f
d±
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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Geometric Altitude (km)
Acceleration Due To Gravity (m/s
2
)
True
Constant Model
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 Fall '09
 Aerodynamics, hG hG

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