So, so far I've been telling you stories about what people tell us the galaxy is
made of, I've told you how much the mass of the disc is, how much the mass of the
halo is and most of these, other than the gas discovered by Chandra where estimates
based on basically star counts. We count the stars and therefore there's
some uncertainty, because rememeber. The number of stars in the Milky Way is
uncertain to almost a factor of two, depending on what the prevalence of
brown, brown dwarfs is. But, brown dwarfs don't comprise a
significant fraction of the mass of the stars since they may be numerous but
they're not very massive. So we have these estimates, but it would
be good to make an independent Comparison of a direct measurement of the mass the
galaxy, How do you weight a galaxy? Well, by now we know how to weigh anything.
You just need something to orbit it. If only we had something orbiting the
Milky Way, oh yea, we do. We orbit the Milky Way, so all we need is
the orbital parameters of the sun and you think that's an easy problem.
If fact it's a lot trickier then you think.
the distance of the sun from the center of the galaxy that I quoted as a 8
kiloparsecs is one of the least precisely low numbers in astronomy, in fact it's
somewhere between 7 1/2 and 8 1/2 kiloparsecs, so there's a 12% uncertainty
in or 6% uncertainty in this distance. It is a difficult measurement to make,
likewise the speed with which the sun moves as it orbits the galaxy.
You'd think that's an easy thing to measure, but it's not easy because
remember all around us are stars that are moving in their own peculiar motion as
an, along with that, with some average motion that is essentially ours.
And so, getting correct value for the speed with which the sun moves around
the, milky way is also not trivial. I'm going to use the established values.
There are some uncertainties, but they're not going to affect what we're going to
do significantly, and if you compute 2 pi R divided by, v, you find that the sun
orbits the Milky Way once every 230 million years, which means that in the 5
billion, year history of the sun, we've gone around the Milky Way, about 20 or 25
times. So the sun has, sampled, all kinds of
places, and space, in the course of its history.
And now, we're back to Newton. We know exactly how to do this.
We have an object, that is orbiting. We know its period, we know its, the
radius of its orbit. The sun's orbit is roughly circular And
so we can make a calculation, the easiest way we know how to do that is to do our
favorite scaling. Compared to the Earth's orbit around the
sun the mass that the sun is orbiting is related to the solar mass by a The ratio
of the periods to the -2, radius, ratio of the radii ^ 3, plug in all the numbers
and you find 88 billion solar masses is what the sun is orbiting.
Now, that is a little bit higher than the numbers I quoted for the galaxy, but
remember That, and i am going to make a slight fudge here and i will explain if
the galaxy were a symmetric object then the sun would be orbiting precisely that
fraction of a galaxy that is within its own orbit the the galaxy being
spherically symmetric everything outside the everything outside the suns orbit
would have no impact on our orbit. Everything inside it could, It could
equivalently be placed at the center. We talked about that when we did
Newtonian gravity. Now the galaxy is not spherically
symmetric it's distinctly a disc, this adds a a small change to the calculation.
It is still true that everything outside the sun's orbit has zero impact on our
orbital acceleration. however there's a small geometric factor.
that the factor or order one that is associated to the fact that the mass
inside our orbit is mostly located in a disc, were that the case, and, but our
estimate is still good. And we found is what the sun is orbiting
is slightly more than the total mass that we had for the entire Milky Way galaxy.
Galaxy, that's interesting, typically, so, so we want to figure out what it is
we've missed, are there a lot more brown dwarfs or whatever that we thought? To
study this, the way this investigation is done, is typically instead of using this
nice scaling relation, which I . confess is my favorite way to do it.
The way we usually do it is we write that an object that is orbiting at a radius r
with speed v has a c, centripetal acceleration which we computed in the
second week of v^2 over R and that is given by the gravitational, acceleration
about whatever it's orbiting, so, we would put R over here, where M(R) is the
mass inside, the circle of radius R, which this object orbits and I should say
R^2. And now that I've written the correct
formula, I can cancel this R and we see that v^2 is GM(R)/R.
This is exactly one half the escape velocity, if you remember that
calculation. And so the way this is usually written,
is that v squared gives you, if you measure the speed with which something is
orbiting and the radius from the center of which it is orbiting, you can figure
out the mass enclosed within its orbit. Okay the study of galactic rotation
curves is something extremely interesting.
What is it that we expect and what is it that we find? So, here's what we expect,
in the red here I've drawn 2 red graph, curves the first is the Measured stellar
density, and you see that it's an exponential density in the disc, the
stellar population is densest in the center of the disc, and then smoothly
falls off, you see that by the time you get to the sun, it has fallen to about
1/6th of its density at the center, and it continues to fall off rapidly.
this, if you compute how much mass is with With this, density, you find this
red curve. We're going to make an approximation
because it's, it allows me to do a calculation, the approximation I'm
going to do is I'm going to replace, this crazy disc by a uniform disc with a
uniform mass distribution. So I'm going to imagine that I have a
disc of some thickness, say, T. This is a, again assumed uniform, we can
make it some kiloparsec or something like that.
And, some density, row of, and kilo per meter cubed or.
Billion, 10 billion solar masses per kiloparsec cubed, would be the more
standard units to be using here. And so I have a density, and I have a
thickness, and I'm going to give my disc some radius R.
And the way I've fixed all of these parameters, is I fixed T and row, so that
the stellar density, out where the sum is at 8 kiloparsecs agrees.
With the actual stellar density, and then I fix the radius so that the total mass
of this uniform blue disc, is the same as the total mass of the stellar disc.
So, what is M of R, in this case? Well, in this case, M(R) is pretty easy to
write. M(R) is row times the volume.
Of, all of the stars m of by burning the mass of all of the stars enclosed in a
circle of radius r, those stars enclosed in a circle of radius r, form a cylinder.
I'm great at drawing cylinders. A cylinder of radius r and height t.
So. its volume.
Is pi R^2 times t, and I will sometimes fall into using, lower case r, for the
radius, because that's, the way it's done in the field,
So, this is M or R, our calculation said, that V of R, should be if this is what
you are orbiting, V of R, Should be, let me make a correction to this in a minute.
V of R should be the root of G M of R over R.
Now, what is thing going to give me? Well M(R) is row times the area of a disc of
radius R times T as long as R is less. Then R0, which is the radius of my disc,
as long as your orbiting inside what I call the uniform disc, then the farther
you are, the more you are orbiting. Of course if R is bigger than R0, then
the mass inside your orbit is simply N or row times Pi R0^2 t, you're not orbiting
any more. By orbiting farther outside in the
vacuum, if you're orbiting at distances out here.
And so, what is V of R I can do this calculation, and if R is less than R 0,
there's a whole bunch of constants that I'm not going to worry about, but M goes
like R^2. Right? And, and dividing by R, so V is
going to like,well okay. The root of row pi t G times R.
It' going to grow like the square root of R.
So the farther you are from the middle, the faster your orbiting.
by the square root of R, the reason, remember, in the solar system the farther
you are from the sun, the slower you are, but that's because we all orbit the same
mass. We're all orbiting the sun, here the
farther you're orbiting, the more mass you're orbiting, therefore, you're
orbital speed will increase. But, once you hit R0, then of course, the
result is, M no longer depends on R will call of this M.
And so, now it's just root of GM/R. The only R dependence is in the
denominator. So we expect something that grows like
the square root of R, and then decreases like 1 over the square root of R to be
the plot of V as a function of R, which is what I'm about.
To plot, but I wanted you to show, to show you that this is not some,
misticism, we can actually do the calculation.
And so, here is the prediction, Indeed, the blue line does what we
expected, it grows like the square root of R.
This is the plot of a square root, and then once you, are outside the disc,
you've reached the maximum speed, you're orbiting as much as you're ever In orbit,
farther out you're orbiting the same thing but at a larger radius, so your
orbital velocity decreases. Notice that the mass curve the m of r
curve or the blue curve is not a bad approximation considering the simple
assumptions I made, to the red curve they're different inside the sun's
radius, they certainly agree almost exactly past the sun's radius.
Because most of the mass is within that. So, the red curve differs the prediction
for v of r differs closer in, than the sum but past the sun's radius, they
pretty much agree. Okay, this is not yet a real galaxy.
To have a real galaxy, you have to take into account that there's also a bulge in
the middle. So how do we add a bulge in the middle
well, just to be very, sinful, I have added in this picture, this is the red
graph we had before, I've added a bulge which is a uniform from a genius fear of
radius 3 who's mass is adjusted 3 kiloparsecs.
Who's mass is adjusted to be exactly the mass, I've predicted for the the galactic
bulge and so up to radius 3 there's this extra increase.
Because you're not orbiting more and more in the bulge and then past the outside of
the bulge, the total mass orbited gets a constant contribution for the mass and so
the total mass you're orbiting is now this black graph that I've drawn over
here. And from this I can again reproduce a
graph of v of R. I've broken it up into contributions.
this is, this would be your orbit of velocity.
If you only orbited the bulge if you want the fact that this increases in a
straight line. it has to do with the fact that the mass
you're orbiting inside of radius R. As long as you're inside the bulge the
mass you're orbiting is M(R) for R inside the bulge.
M($) is row of the bulge if we're just talking about the bulge * the volume of a
sphere of radius R. Which is R^3.
And so M of R / R which is what goes into G into V.
Goes like lots of constants * R^2 and v^2 is going like R ^ 2 means v goes like R
in a homogeneous ball of homogeneous density, your radial velocity increases
linearly. Okay, and, not your radial, orbital
velocity increases linearly, the outside has to orbit bit faster.
This is great, because this also means, that, V of R proportional to R, is what
would happen if you just took that ball and rotated it rigidly.
So, a homogeneous ball rotating, orbiting itself under the force of gravity rotates
like a rigid ball, at least in the plane of the orbit.
In the plain of the equator, all a straight line drawn from the origin to
the, equator will continue to stay a straight line, everybody orbits the
period of everybody on the equator is the same.
That's an interesting observation. It turns out that this does describe in, the
real galaxy. In real galaxies, the behavior at very small radii.
And, then we have the red curve, which is the same curve we had before.
Adding them together we get this black curve.
The sharp corner here is an artifact, of the sharp corner here, of my, approximate
model of the bulge, but I wasn't about to insert bars, it gets to complicated.
So, I wanted to see that we could make a reasonable prediction as to what the
velocity curve should look like. We already know that we're falling short
because I inserted the measured masses. So this gives the sun a velocity of about
180 meter, kilometers per second, whereas the measured value is 230 kilometers per
second. But we knew we had a short falling in the
mass there. The big question is going to be what
happens out here? Of outside the solar radius and, we have measurements, here
are the measurements. So remember, 8 kilo parsecs is where the
sun is, so this is, the solar radius, the sun is here.
Are measured at about, 225 kilometers per second, I told you that measurement is
subject to corrections. And here is a function of R, is the
observed rotation, speed and, what it is that's being measured are carbon monoxide
clouds and H molecular H. hydrogen clouds and 21 centimeter
emission from atomic hydrogen, Various things have been used to make this
measurement and what we see is indeed the steep.
linear with R, when R = 0, then they decline, when you leave the dense region,
this structure is associated with the galactic bulge, and modelling it more
precisely, we get out to the sun and we are little underestimating.
And then what goes on here is a disaster, the BFR graph starts, Decreasing, as we
predicted. Remember, it was supposed to start
declining, because you're outside, essentially, the entire mass of the disc,
and it's not declining. It's staying constant.
So not only are we missing some mass inside the sun's orbit, but the farther
and farther out we get, the more and more mass we're missing.
Because what we're finding is that M(R)/R is a constant.
The mass enclosed within radius R is growing linearly with R.
And where it stops growing linearly with R? This graph is not going to show us.
it becomes tricky to measure it here, we'll have more luck actually measuring
this with other galaxies. So the resolution of this problem will
wait until we see whether it's some crazy quirk of the Milky Way and of us being
stuck in a disc and not measuring very well, or actually a more general thing.
But I wanted to bring it up, and also we have to do a calculate.