Phys 341: Homework #4 Solutions
1.
(a) Recall that the center of mass is defined such that
m
A
a
A
=
m
B
a
B
, so we have
m
A
m
B
=
a
B
a
A
=
1
0
.
466
= 2
.
15
Since
m
A
/m
B
>
1, star A is more massive than star B.
(b) Recall that the distance can be determined from the parallax angle as
d
= 1 pc
×
p
1

1
So for
p
= 0
.
377 , we have
d
= 2
.
65 pc.
(c) The phrase “the semimajor axis of the reduced mass” simply refers to the sum of the
semimajor axes of the two star,
a
A
+
a
B
. This subtends an angle of 7
.
62 ; to get the
corresponding length we multiply by the distance, finding
a
A
+
a
B
=
2
.
65 pc
×
7
.
62
×
π
180
×
3600
=
9
.
79
×
10

5
pc
=
3
.
03
×
10
14
cm
The last factor on the first line converts from arcseconds to radians. Kepler’s Third Law
now tells us the sum of the masses,
m
A
+
m
B
=
4
π
2
(
a
A
+
a
B
)
3
GP
2
=
4
π
2
(3
.
03
×
10
14
cm)
3
(6
.
67
×
10

8
g

1
cm
3
s

2
)
×
(49
.
94
×
365
×
86400 s)
2
=
6
.
64
×
10
33
g
=
3
.
34
M
In the second line, the factor of 365
×
86400 in the denominator converts from years
to seconds (86400 seconds per day, 365 days per year). Combining this with the result
m
B
= 0
.
466
m
A
from part (a), we get
m
A
(1 + 0
.
466) = 3
.
34
M
⇒
m
A
= 2
.
28
M
which then yields
m
B
= 1
.
06
M
2.
(a) For circular orbits, recall that the velocities are
v
A
=
2
πa
A
P
=
2
π
P
m
B
m
A
+
m
B
a
(1)
v
B
=
2
πa
B
P
=
2
π
P
m
A
m
A
+
m
B
a
(2)
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where
a
=
a
A
+
a
B
. Taking the ratio, we reproduce the result from class that the mass
ratio can be determined from the velocity ratio,
m
A
m
B
=
v
B
v
A
=
22
.
4 km/s
5
.
4 km/s
= 4
.
15
(b) From the expressions for the velocity above, we have
a
A
+
a
B
=
P
2
π
(
v
A
+
v
B
)
So we can write Kepler’s Third Law as
m
A
+
m
B
=
4
π
2
GP
2
P
2
π
3
(
v
A
+
v
B
)
3
=
P
2
πG
(
v
A
+
v
B
)
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 Spring '08
 Keeton
 Physics, Center Of Mass, Mass, Work, Ratio, Interferometry, Southern African Large Telescope

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