Using publicly available data from NASA of actual satellite observations of astronomical x-ray sources, we explore some of the mysteries of the cosmos, including neutron stars, black holes, quasars and supernovae. We will analyze energy spectra and time series data to understand how these incredible objects work. We utilize an imaging tool called DS9 to explore the amazing diversity of astronomical observations that have made x-ray astronomy one of the most active and exciting fields of scientific investigation in the past 50 years. Each week we will explore a different facet of x-ray astronomy. Beginning with an introduction to the nature of image formation, we then move on to examples of how our imaging program, DS9, can aid our understanding of real satellite data. You will using the actual data that scientists use when doing their work. Nothing is "canned". You will be able to appreciate the excitement that astronomers felt when they made their important discoveries concerning periodic binary x-ray sources, supernovae and their remnants, and extragalactic sources that have shaped our understanding of cosmology.
Lecture 4 Of GK-Per and White Dwarfs, Part 3
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From the course by Rutgers the State University of New Jersey
Analyzing the Universe
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From the lesson
Stellar Evolution and White Dwarfs
This week is our first in-depth look at an x-ray source, and it involves a white dwarf in a binary system. So sharpen up your detective skills, keep your copy of DS9 at the ready, and let's get down to business. It should be an exciting week.
Meet the Instructors
Dr. Terry A. MatilskyProfessor
Physics and Astronomy
[BLANK_AUDIO]
So we have an incredibly accurate clock in the sky, stable to a spectacularly high
precision over about 20 years at least. What could provide such a mechanism?
Again some clues are given to us by
the optical spectrum of this object shown here.
Looking at this visible light, we see something quite strange,
all the spectral lines that are there, we expect to see coming form a very
cool star, even cooler than our Sun. Except
for one line, at 4686 angstroms.
It is the strongest line in the spectrum, and is the fingerprint
of helium-2, or ionized helium. This is incredible!
Why?
Because in order to ionize helium, you
need an environment that is unbelievably hot.
At least ten times hotter than the Sun.
So here we have an example of a composite spectrum.
One which can only be explained by the existence of double star.
One very cool and one very, very hot.
As we saw earlier, when we studied the HR diagram,
hot stars come in basically two flavors; one group astonishingly
bright, much more massive than the Sun, and the other
the stellar graveyard of objects very much like our Sun.
Cooling down like a dying ember and being very small.
How do we know they are small?
Remember that the luminosity of any star can be given
by its surface area, 4 pi r squared,
times sigma, a constant, times T to the 4th power.
We know that the luminosity is very small for these objects.
The temperature is very high. The only way we can
have a small luminosity is if the size of the star, its radius is very small indeed.
And in fact, the radii of most white
dwarfs are about equal to that of the Earth.
Imagine that!
A mass about equal to the Sun in a size no bigger than a fairly
small planet like the Earth. So back to GK Per.
We have an object with a star similar to the Sun.
A bit cooler being a K star, and a hot white component.
We know it can't be very luminous in the visible part of the spectrum.
Because we know its distance. And at that distance, a hot O or B star
would make GK Per hundreds of thousands of times brighter than we observe it.
So the second component must be a white dwarf.
Is there any other clue that this is the correct picture?
Yes!
Again, using Doppler measurements, we can see the composite spectral
lines moving back and forth over a period of about two days.
You can see that in the data shown here. This cosmic dance, as we shall see,
is indicative of a double star system whose components orbit each other.
With a period of about two days. But what about our 352
second period, what can that be and which star is responsible.
It can't be an orbital period because that's already been spoken
for in the two day clock observed in the spectral lines.
But now, let's look at the next obvious choice.
Rotation, of one of the stars.
Let's start with the K star. We know from theoretical considerations,
that these are objects about 0.7 times the radius of
our own Sun. This will make this star a ball equal
to a radius of about 5 times 10 to the 8th meters.
We can also use the binary periodicity observations.
To show that the mass of the k star is about
one half a solar mass, or about 1 times 10
to the 13th kilograms. Now if
this star is spinning around in 352 seconds, This
means that the speed at the equator must be 2 pi R,
which is the distance traveled, divided
by the amount of time it takes for
it to travel once around 2 pi R over T.
This means it's speed is about 9 times 10 to
the 6 meters per second.
Boy, that's pretty fast, what can possibly prevent
the stuff from flying apart at the equator.
We would need some sort of force to hold it together.
Is gravity upto the task?
Let's see.
You might remember that objects moving in a circle of radius r are accelerated
to the centre of that circle through their centripetal acceleration.
V squared over R.
If you haven't seen this before, don't worry, we will demonstrate it shortly.
But right now let's just use this fact to see what we can find out.
If we are on the surface of a k star, spinning around every 352
seconds, we need an acceleration of v squared over r.
And if you put in the values that we just found, you find that the
acceleration of that material on that, the equator of the star
ought to be about 1.6 times 10 to the 5th
meters per second per second.
Is gravity up to this task? We know from Newton's
Law, which we will also discuss later, that a test
object of mass M, in the vicinity of an object of capital
M will feel a force. F equals m a equals
G big M little m divided by the
distance squared between the
two objects. Thus the acceleration that
gravity can provide towards the center of the K star is.
a, which will indicate a gravitational here, equaling G times capital M,
divided by the radius of the star squared.
You have something sitting on the equator of the star.
It's a distance R from the center of the star.
The mass of the big star is M, and the acceleration that gravity
will provide will be equal to this quantity.
Now, G, the gravitational interaction constant,
is in these units, 6.7 times 10 to the -11.
You can look these numbers up, I mean there really is good
reason to have something like Wikipedia handy so you can check on this.
We know what the mass is.
And we know what the size of the object is.
And if you work this out, rather than doing
it on the blackboard, it would be a very good
exercise for you to do this yourself, you
find that this amount of acceleration due to
gravity on the surface of a k-star is about
2.7 times 10 to the 2 meters per second
squared. Now, if you compare
these two numbers, the number that is provided or can be provided
for by gravity to the number that we
actually would have for a mass sitting on the
surface of a spinning k-star. You can see that
gravity simply can't do the trick. The acceleration by,
provided by gravity is grossly inadequate, to prevent the star
from flying apart at the equator. It needs much more of a pull
than gravity on the surface of a K star can provide.
Well, what about the white dwarf? Again from spectral observations
we find that the mass of this component is about equal to the mass
of the Sun. Which is 2 times 10 to the 30th kilograms.
So why don't we just kind of change this to, mass of white dwarf, make this a 2.
Okay,
and the radius of this object is
about equal to the radius of the Earth. 6 times 10 to
the 6 meters. So the radius of our
white dwarf is about equal
to the radius of the Earth which is 6 times 10 to
the 6 meters oh, not meters right.
Oh yes, no, okay, well it's not kilometers.
That is correct. 6 times 10 to the 6 meters.
Now we can go through the same calculation as we did before.
We could calculate the speed at the surface of the white dwarf.
Calculate the acceleration where now
you use the white dwarf speed and the radius of the white dwarf.
And if you do that, you will find, I'm not going to do it for you, you
do it yourself, you've got an example now.
And you ought to find that the
acceleration at the
surface of a white dwarf is equal to about
1.6 times 10 to the 3 meters per second squared,
1.6 times 10 cubed
meters per second per second. Now, is gravity up to this task?
Again, we calculate this value, and instead
of finding The value that we had at the surface
of the K star, we find that gravity
can provide an acceleration of about 4
times 10 to the 6 meters
per second squared. Yes!
Gravity can do the trick. 4.6, 4 times 10 to the 6
meters per second squared is much more than what we need to
prevent something on the surface of a white dwarf from flying
off into space. So gravity can provide more
than enough pull to keep the star intact and still have it spin around.
Every 352 seconds.
So it appears we have solved the mystery. But what is it that's changing every
352 seconds, that allows us to see the star varying at all?
I mean, just because an object is rotating doesn't
mean that we're going to be able to see it.
There must be something that is tied to the rotation
period as the star spins around on
its axis, and in order to see what
that might be, we will examine our
next X-ray source, Cen X-3.
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