[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.
