Physics 105B Problem Set 1
April 8, 2008
Je
ff
Schonert:
schonert (at) physics.ucsb.edu
Taylor 8.7
a)
Mass
m
1
is orbiting mass
m
2
, and the radially directed force between the two masses is
given by Newton’s law of gravitation. If the distance of separation is
r
, and mass
m
1
rotates
with angular velocity
ω
, then by Newton’s second law we have
Gm
1
m
2
r
2
=
m
1
ω
2
r.
(1)
But
ω
= 2
π
/T
, where
T
is the period of one revolution, so plugging this in and rearranging
gives
T
2
=
4
π
2
r
3
Gm
2
.
(2)
b)
Now we have the same masses
m
1
and
m
2
, but the mass
m
2
is allowed to move freely.
The only di
ff
erence in this problem is that in the expression for the centripetal force,
m
1
→
μ
,
where
μ
is the reduced mass. Then we have
Gm
1
m
2
r
2
=
μ
ω
2
r,
(3)
which implies that
T
2
=
4
π
2
r
3
GM
,
(4)
where
M
=
m
1
+
m
2
.
If
m
2
is much greater than
m
1
,
M
=
m
1
+
m
2
≈
m
2
, so that the result in part b) reduces
to the result in part a).
c)
Since
m
1
=
m
2
,
M
= 2
m
2
= 2
M
sun
. Therefore,
T
=
2
π
r
3
/
2
√
2
GM
sun
=
1
√
2
(1 year) =
.
71 years
.
(5)
Taylor 8.9
a)
For two particles of equal mass
m
, the kinetic energy is
m
2
(˙
r
2
1
+ ˙
r
2
2
)
.
Using equation (8.10), this is equal to
M
2
(
˙
R
)
2
+
μ
2
(˙
r
)
2
,
1
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where
M
= 2
m
and
μ
=
m/
2. Now we are told to express ˙
r
2
in terms of twodimensional
polar coordinates (
r,
φ
). This means that the kinetic energy is
T
=
M
2
(
˙
R
)
2
+
μ
2
˙
r
2
+
r
2
˙
φ
2
.
(6)
The displacement from the equilibrium length
L
is
r

L
, so the potential energy is
V
=
k/
2(
r

L
)
2
. The Lagrangian is
L
=
T

V
.
b)
We are asked to express the Lagrangian in terms of the cartesian centerofmass
coordinates (
X, Y
). Since
R
2
=
X
2
+
Y
2
,
L
=
M
2
˙
X
2
+
˙
Y
2
+
μ
2
˙
r
2
+
r
2
˙
φ
2

k
2
(
r

L
)
2
.
(7)
The Lagrangian is independent of
X
and
Y
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 Spring '08
 MARTINIS
 mechanics, Force, Mass, C2 C2

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