Download 1117 Discussion Notes

Wednesday, November 27, 2004
Please come to the front of the room and take back your papers.
If you absolutely cannot attend discussion on Nov. 24, then you may
make it up by writing a one to two page paper (double spaced) on the
Cassini Mission that has recently arrived at Saturn
(http://saturn.jpl.nasa.gov). This should be handed in to me before the
24th. Late submissions will not be accepted. For details, see last week’s
discussion notes online.
Black Hole film
You should be able to reproduce and label the pictures of the Milky Way
Galaxy from different angles (ex. figure 14.1).
Disk Stars
Found in the disk
Spend all of time in disk
Orbit in same direction, bobbing up and down
Halo Stars
Found in or near the halo
Spend most of time in halo
Orbit in random directions
Bulge Stars
Found in bulge
Spend all of time in bulge
Orbit in random directions
Measuring masses – finding dark matter!
Remember how we could measure the mass of a binary system using
Newton’s version of Kepler’s third law? To do so, we needed to know
the period of the binary system and the distance between the two stars.
In a similar way, we can determine the mass of the galaxy interior to the
orbit of a star if we know the period of the star’s orbit about the galactic
center and the distance between the star and the galactic center.
For instance, we can carry out this calculation for our Sun. Recall that
Newton’s version of Kepler’s third law is:
4 2
2
P 
a3
G( M 1  M 2 )
where P is the orbital period (in seconds), a is the average distance (in
meters), M1 and M2 are the masses of the things being considered (in
kilograms), and G is the gravitational constant. In this case, the things
being considered are the entire Milky Way galaxy and the Sun.
Therefore, a = the average distance between the sun and the galactic
center, and P = the orbital period of the Sun about the galactic center.
 9.46 1012 km 
  2.6488 1017 km  2.6488 1020 m
a  28,000 light - years 
 1 light - year 
In the text, we are not given the Sun’s period, but we are given it’s
velocity. To find the period from a and the velocity of 220 km/s:


distance traveled 2πa 2π 2.6488 1017 km
P


 7.56496 1015 s
velocity
v
220 km/s
Now plug into Newton’s version of Kepler’s third law and solve for mass:
7.56496  10 s
15
6.67  10
11
2
3
4 2


2.6488  10 20 m 
11
3
2
6.67  10 m /kg  s (M 1  M 2 )
m 3 /kg  s 2 7.56496  1015 s 
2
4 2 2.6488  10 20 m 
3

1
(M 1  M 2 )
4 2 2.6488  10 20 m 
3
(M 1  M 2 ) 
6.67  10
11
( M 1  M 2 )  2  10 41 kg
m 3 /kg  s 2 7.56496  1015 s 
2
 1.922  10 41 kg
Why is it that this only tells us the mass of the material within the Sun’s
orbit? Newton’s version of Kepler’s third law is based upon gravity. (If
we had derived this law from basic physical principles rather than
simply presenting it to you, this would be obvious. As it is, just trust me
on this one.) The force of gravity from the material outside of the orbit
is in all different directions, so it practically cancels out. The force of
gravity from the material within the orbit is all in the same direction, so
it adds up.
We can, in theory, do similar calculations for stars that are closer to the
edge of the disk to find the mass of a greater portion of the galaxy.
Since the bulge looks so big and bright, we expect that most of the
mass of the galaxy is located in the center. However, in order for this
to be true the orbital periods of the stars closer to the edge of the disk
would have to be larger (i.e. their velocities would have to be
smaller.) Conversely, if most of the mass is in the halo, the orbital
periods of the stars closer to the edge of the disk would have to be
about the same as the stars near the bulge (i.e. their velocities would
have to be about the same as the stars near the bulge.)
Why, you ask? Well, you don’t REALLY need to know this for the
course… but let’s take another look at Newton’s version of Kepler’s
third law, slightly rewritten.
4 2 a 3
(M1  M 2 ) 

GP2
4 2 a 3
 distance traveled
G
velocity




2

4 2 a 3
 2πa 
G

 v 
2
 av 2
If most of the mass is located in the bulge, and you look at the total
amount of mass within a disk star’s orbit, that total mass will not change
much as you examine stars further and further away from the bulge.
Your a is getting larger and larger, but the mass is not changing by
much, so the only other variable, the period, must be increasing enough
to counterbalance the increasing a. The period is dependent on the
velocity, so if the period is increasing, then the velocity must be
decreasing.
Contrary to our expectations, stars that are far from the galactic center
have velocities that are comparable to stars near the bulge. We explain
this by saying that the bulk of the mass in our galaxy is not in our bulge,
but in our halo. We call this mass “dark matter” because we do not
detect it emitting light.