Chapter 19: Star Formation
1
Chapter 18 Problems
2.
The distance to Alpha Centauri is 1.33 pc
×
3.1
×
10
16
m/pc = 4.1
×
10
16
m. The volume of space is
4.1
×
10
16
m
3
.
Multiplying this by the mass density from the previous question gives a total mass of
7.0
×
10
8
kg.
This is equal to 0.070 mg.
4.
Assume the shape is a cylinder of length L and crosssectional area
π
r
2
, the product of which is the
volume.
From the scale of the diagram, the diameter is about 10
7
m and a length of about three
times this.
Using this to calculate the volume and multiplying by the density will give the mass.
3
×
10
7
×
π
×
(5
×
10
8
)
2
×
3000 = 7.1
×
10
18
kg
5.
Every 5 pc it gets fainter by a factor of 2.
For example, in 10 pc it gets 2
×
2 = 4 times fainter.
It
gets reduced by 2 a total of 60/5 = 12 times as it passes through the cloud.
This is 2
×
2
×
2
×
.....
, or
more easily 2
12
= 4096.
The light is seen 4096 times fainter than it was before it entered the cloud.
On the magnitude scale, the light is reduced in intensity by 2.5Log(4096) = 9 magnitudes.
7.
M
= 6 and
m
= 14, 14  (6) = 5Log(d)  5,
d
= 10
5
pc.
Solving for
m
, from the known distance of 5,000 pc, gives
m
 (6) = 5Log(5000)  5,
m
= 7.5
It has
been dimmed by 6.5 magnitudes over 5 kpc or 1.3 mag/kpc.
9.
M
= 5 and
m
= 10.
10  (5) = 5Log(
d
)  5 + (
d
/1000)
×
2.
Reduce this equation to 4 = Log(
d
) +
d
/2500.
You can not solve directly for
d
, so try various values for
d
, solve the right hand side and if
the result is greater than 4, choose a smaller
d
; if smaller than 4, choose a larger
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 Fall '09
 MehmetYali
 Star Formation, Atom, Mass, Interstellar medium

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