3
35
3
30
2
21
1
3
2
12
(2.7 10 m/s) (1.70 days)(86400 s/day)
6.90 10 kg
()
2
2
(
6
.
6
7
1
0
m
/
k
g
s
)
3.467
,
s
mv
T
mm
G
M
ππ
−
×
==
=
×
+×
⋅
=
where
30
1.99 10 kg
s
M
=×
is the mass of the sun. With
1
6
s
mM
=
, we write
2
s
α
=
and solve the following cubic equation for
:
3
2
3.467
0
(6
)
−=
+
.
The equation has one real solution:
9.3
=
, which implies
2
/9
s
≈
.
56. The two stars are in circular orbits, not about each other, but about the twostar
system’s center of mass (denoted as
O
), which lies along the line connecting the centers
of the two stars. The gravitational force between the stars provides the centripetal force
necessary to keep their orbits circular. Thus, for the visible, Newton’s second law gives
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Center Of Mass, Force, Mass, Orbits

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