Problem GPS 3.13
If the center of force is located at the origin of an
xy
coordinate system, then the orbit is a circle
that passes through the origin as, for example, drawn below.
This circle has radius
a
, and its
origin is at (
a
,
0
). The equation that describes it is (
x
±
a
)
2
+
y
2
=
a
2
.
In terms of polar coordinates
measured from the center of force, we have
x
=
r
cos
θ
and
y
=
r
sin
θ
.
After substituting for
x
and
y
in the equation of the circle, we obtain the representation of the orbit in polar coordinates
as
r
= 2
a
cos
θ
.
Note that a full orbit is traversed as
θ
varies from
±
π
/2 to
π
/2 instead of the
usual 0 to 2
π
.
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View Full DocumentProblem Goldstein 36
(a)
Start with the di
f
erential equation of the orbit in terms of the variable
u
=1
/r
,
l
2
m
u
2
μ
d
2
u
d
θ
2
+
u
¶
=
−
f
(
1
u
)
,
(1)
where
l
is the angular momentum,
m
is the mass of the particle, and
u
=(2
a
cos
θ
)
−
1
.
(2)
The required derivatives are
du
d
θ
=
tan
θ
2
a
cos
θ
,
(3)
and
d
2
u
d
θ
2
=
1
2
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 Fall '10
 Wilemski
 mechanics, Polar Coordinates, Angular Momentum, Force, Coordinate system, Polar coordinate system, Conic section, Eqs.

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