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Problem GPS 4.21 (2
nd
ed. 4.22)
If we neglect the “centrifugal force” term in GPS Eq.(4.91), the equation of
motion for a particle of mass
m
in the rotating frame of the earth is
m
d
v
dt
=
m
g
+2
m
v
×
ω
,
(1)
where
v
is the particle velocity,
ω
is the earth’s angular velocity vector, and
g
is
the acceleration of gravity at the earth’s surface. All the vectors and derivatives
aremeasuredinabody
f
xed frame in which the
z
axis is normal to the earth’s
surface and the
x

y
plane is tangent to the earth’s surface and is located at
z
=0
. Assume that we are in the Northern hemisphere, and let the
y
axis
point north and the
x
axis point east. As we saw in class, the solution to Eq.(1)
to
f
rst order in
ω
is
r
=
r
0
+
v
0
t
+
1
2
g
t
2
+(
v
0
×
ω
)
t
2
+
1
3
(
g
×
ω
)
t
3
,
(2)
where the subscript
0
denotes an initial value. Since the initial velocity is upward
(vertical), we can write
v
0
=
v
0
e
z
,
(3)
and gravity acts downward, so
g
=
−
g
e
z
.
(4)
With
r
0
=0
,the
f
rst three terms of Eq.(2) are seen to determine the height
h
and total time
t
tot
of the particle’s trajectory as
t
tot
=2
t
h
=2
v
0
/g ,
(5)
h
=
1
2
gt
2
h
,
(6)
where
t
h
is the time required for the particle to fall to the earth when it is
released at the height
h
above the surface.
The fourth and
f
fth terms in Eq.(2) are responsible for the de
F
ection of the
particle from the vertical. Because
ω
lies, generally, in the body
y

z
plane, the
cross products generate vectors along the body
x
axis, i.e.,
e
z
×
ω
=
−
ω
y
e
x
=
−
ω
sin
θ
e
x
,
(7)
where
θ
is the polar angle, or colatitude, measured from the earth’s rotation
axis. (The equator is at
θ
=
π/
2
.) Thus
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 Fall '10
 Wilemski
 mechanics, Centrifugal Force, Force, Mass

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