Download How fast do stars form out of the ISM?

How fast do stars form out of the ISM?
Let's consider the gravitational collapse of a cloud into globules and further into a proto-star. The
molecular clouds are very dark because they are cold (~10-40 K) and dense (~1000 grains per centimeter cubed). Gravity is an attractive force between any two masses (the same way as the electromagnetic
force makes two oppositely charged magnets attract each other). So all the grains in the cloud attract each
other causing the entire cloud to collapse under its own gravity.
Normally, when gases get denser they heat up (analogy: when you pump the tire of your bicycle, the
tire will bet warm). However, during the initial stages of the collapse, the dust grains are shielding almost all
the heat that is produced in the centers of those clouds (analogy: if the clouds in the sky are dense enough,
we do not see the sun through them). Thus the outer parts of the cloud fall towards the center of the cloud
the same way as an apple would fall off a tree. Thus the outer parts of the clouds collapse in a time scale
that is equivalent to the "free-fall" time scale, which is given by:
t≈
1
Gρ
where G is the gravitational constant and ρ the density of the cloud.
If you are interested read through the following argument, otherwise turn the page. Consider the apple
is falling towards the ground. It falls faster and faster in the same time interval. In fact it is accelerated
towards the earth. From high school you may remember a relation between the distance, d, the
acceleration, a, and the time interval, t:
d=
1 2
2d
at or t 2 =
2
a
And we need to know (maybe from high school) that the gravitational acceleration is
a=
GM
R2
Assuming that the apple falls towards the center of the earth (i.e. d = R), we get:
2d
2R
2R3
t =
=
=
GM GM
a
R2
2
The density, ρ, is defined as mass per volume, so assuming the cloud is roughly spherical, its density is:
ρ=
Mass
M
4
or M = πR 3 ρ
=4
Volume
3
πR 3
3
and the time-scale in terms of the density is:
2 R3
t =
=
GM
2
2 R3
1
1
=
≈
4 3
2
Gρ
G πR ρ
πGρ
3
3
The real derivation involves calculus, but the general result is the same:
t≈
1
Gρ
Thus the collapse time of a cloud only depends on its density.
How fast does a Bok Globule (a very dense molecular cloud) collapse to form a star? If the cloud has
a mass of 100M and a size of 1 light year, its density is:
ρ=
100 × 2 × 10 30 kg
100 × 2 × 10 30 kg
kg
=
= 5.6 × 10 −17 3
4
4
3
3
m
π × (1ly )
π × (9.5 × 1015 m)
3
3
The gravitational constant, G, is
 kg × m 2

m
 sec 2 
N ×m
m3
−11
−11
= 6.67 × 10
= 6.67 × 10
kg 2
kg 2
kg × sec 2
2
G = 6.67 × 10 −11
(Note in the back of your book there is a mistake in the value of G. It is to the power of 1011 but it should
have been 10-11! This is the reason why so many of you got absolutely ridiculous numbers. Also, in most
other books, the units of the gravitational constant are given in m3/kg×sec2 rather than in Nm2/kg2. This
would have made the assignment much easier and a lot more reasonable! And I am very sorry I did not
notice this earlier, particularly the minus-sign error.) The collapse time then is:
t≈
1
=
Gρ
1
3
m
kg
−17
6.67 × 10 −11
2 × 5.6 × 10
kg × sec
m3
= 16
. × 1013 sec
Since one year has roughly 3 ×107 seconds (1yr = 3.156 ×107sec), the age in years is
t = 1.6 × 10 13 sec =
16
. × 1013 sec
≈ 500,000 years
sec
7
3 × 10
yr
How long is this collapse relative to the main-sequence lifetime of the sun? 500 thousand to 10 billion years
is only a small fraction of that, i.e., 0.0005. So the collapse time into a proto-star is very fast. In real life it is
a little longer, because in the last stages of the collapse, the cloud will get heated, and because heat makes
gases expand, it will slow down the final stages of the collapse. At this stage a proto-star forms…
And if the sun was to collapse like one of the clouds in space, it would take:
t≈
1
=
Gρ
1
G×
M
4 3
πR
3
1
=
6.67 × 10
−11
m
2 × 1030 kg
×
3
kg × sec 2 4
π (7 × 10 8 m)
3
3
which is less than one hour. Well, I am glad this is not happening…
= 3,300 sec = 54 min