Here you can see, the frequency seems to be starting out about here. And going out here it's centred around oh, about 0.2071 cycles per second. So our little spike turns out to be a set of frequencies between 0.2068 cycles per second, and 0.2074 cycles per second. It's not quite constant. And if we follow the values of our new clock, which corresponds to a period, of 1 over f. Now these two are different periods right? This is a two day period, and this is going to correspond to something else within the star. We get for this, periods of about 4.836 seconds to 4.822 seconds. And let's see, this is the larger frequency, so it corresponds to the smaller period. I think I might even have that right, but you can check me on that. It's changing its frequency. And guess what? It does so smoothly, varying back and forth over, what do you think the time period is? Right, 2.09 days. We have apparently uncovered a classic case of an eclipsing binary star. But wait just a cotton picking minute. If every 2.09 days we really see an eclipse that lasts about 40,000 seconds, can we predict anything from this observation? Sure. If the x-ray emitting object is going in and then coming out from behind another star, as it does so, maybe we should see a change in the x-ray energy spectrum, similar to the way the sun and sky changes color at sunrise and sunset, as the sunlight passes through more and more of the Earth's atmosphere. So, the picture is this. We have an object and we have our x-ray source that goes behind the object and then comes out of eclipse. And as it goes through the limb of our companion star, and as it comes out from the other limb of the companion star, it is possible that the energy that we see from the x-rays will change its composition, just as the sky changes to red during sunrise and some, and sunset. And indeed, although the mechanisms are very different for x-rays, the result is that we do see the source change as it goes into and out of eclipse. So we are beginning to put together a satisfyingly coherent picture of our system. A source of cosmic x-rays which somehow, has an internal clock of about 4.8 seconds is orbiting another object every 2.09 days. But what are these objects, and what is the 4.8 second periodicity due to? Well, let's roll up our sleeves and get back to work. Let's examine what we already have and where it can get us. First, it appears that the regularity of our range of different periods is due to the Doppler shift of the x-ray emitting star travelling around its companion. Since delta f over f equals v over c, our change in frequency over the frequency is equal to the speed divided by the velocity of light, and our range in frequencies is 0.2074 To 0.2068 Hertz. We end up with, Delta f over f, equals v over c, equals 0.2074 minus 0.2068, this is the change in frequency, divided by 2, all over the regular frequency, or the average frequency, 0.2071. See if you can understand, why there's a factor of two in there. That means, that v over c, is equal to and we probably should put in approximately equal things, just to show that we're really not exactly 100% accurate here. This is going to be equal to 0.0003 divided by 0.2071. And that equals approximately 0.00145. So when we solve for v, v turns out to be around 430 Kilometers per second, if we have a circular orbit. And in fact, these objects really are mostly in circular orbits because of other factors of their orbital circumstances. But now, we can find the size of the orbit. Since the circumference of the orbit must be equal to 2 pi r, okay, the object goes around at a radius r from the companion, it goes once around and if the speed is constant at 430 kilometers per second, that distance is nothing more than the velocity times the amount of time it takes to go once around, namely 2.09 days. So now we can figure out what r is. R is going to be equal to the velocity times the orbital period divided by 2 pi, and our velocity is 430 kilometres per second times the time, which is 2.09 days times 86,400 seconds in a day divided by about 6. And that's going to be in kilometers. And if you do the math, you find that this radius is equal to about 1.2 times 10 to the 7 kilometers. This is about one quarter the size of Mercury's orbit around the Sun. So, these two objects are really close together. Now we can do something really neat. Since, the x-ray source is eclipsed for about 40,000 seconds every orbit, we can estimate crudely the size of the other object. Just take for its diameter that 40,000 seconds worth of blackout, and multiply by the speed that the neutron star is going as it goes around in its orbit. So 40,000 seconds times 430 kilometers per second is equal to 1.7 times 10 to the 7 kilometers. This would be a crude estimate of the diameter of our companion star. It's approximating the arc of a circle with a straight line. So the companion star's radius is about half that, or 8.5 times 10 to the 6 kilometers. Thus, not only is the orbit small, the x-ray source must be very close to the surface of the other star. Here's the radius of the orbit and the radius of the star, r star, is about 8.5 times 10 to the 6 kilometers. So what we're doing is we're imagining that we have a star and that the object is just moving behind it but in a straight line instead of a circle. But it's actually not a really bad approximation. But what it means is the radius of the star is over 10 times the size of the sun. Now, I do have to say we did make some approximations here. And we talked about one of those approximations, okay. What are some of the other assumptions that we made? And I will leave that to you to figure out. So if our theory about the nature of the low state in Cen X-3's light curve is correct, we have a prediction. It is that the companion star should be quite large and massive. Can we test that theory? Yes! Remember when we talked about gravity? We derived an equation relating the period of an orbit to its size, or radius. Better known as Kepler's Third Law, it states that, the square of the period of an object in an orbit is proportional to the cube of the radius of that orbit or T squared equals 4 pi squared over GM times the radius of the orbit cubed. Well, now we know what T is, it's 2.09 days. We know what r is, it's 1.2 times 10 to the 7 kilometers. We can solve for M. It turns out that our derivation of this equation is a somewhat simplified version of reality. In actuality, when you do a more precise derivation of this equation, the mass in the denominator here is really the sum of the two masses in the system. But, this is really not a big problem at all. If we solve for M, you get, and I urge you to do this, you get about 3 times 10 to the 34 grams. About 15 solar masses. This is an excellent problem with which to practice unit conversion; do it on your own. This result is reassuring. We see that the system is a massive one, which is what we suspected on the basis of the light curve analysis.