Okay, gang, let's roll. Last time, we obtained a bit of insight into that most famous of all equations in classical mechanics, better known as Newton's 2nd law. That says F=ma. This particular equation is surprisingly subtle, and we're going to ignore most of those subtleties, but I should at least tell you that one, this is a vector equation. Force has a particular direction, and the acceleration also has a particular direction. Also, you need to sum all of the forces that might be acting on an object. So we really have a net force, because forces can pull in different directions, and so this particular equation can yield insight into the way astronomical bodies can move. We also derived the acceleration, a, for circular motion, and we found that the magnitude of that acceleration was given by the square of the speed divided by the distance between where the center of the circle was and where the object was rotating around. Now, we are ready to see what the left hand side of Newton's law has in store for us by applying it to simple astronomical systems. Newton found out after countless generations of scientists and would-be scientists over the centuries tried their luck and skill at figuring out the problem, that the gravitational force between two point like objects was F equals some constant, G, capital G, times big M times little m, divided by r squared. Where we usually just use capital M and small m to represent the masses of the two objects involved. Again, we're going to assume that this is a vector, and we're not going to worry about putting arrows on everything, because sometimes that can get a little bit cumbersome. This is simple, but profound. G is a constant that is surprisingly difficult to get accurately because gravity turns out to be a very weak force, but we do know the value of g to about a tenth of one percent. Let's see what the consequences of our understanding of these ideas are. First, imagine that M is the sun and little m is any planet. Then, we have F equals G-M-m over r squared and that's going to be little mv squared over r. Where now we're substituting for the acceleration, our understanding for how things move in a circular orbit. So, if we solve for v squared, we find that v squared is equal to G-M, the mass of the sun, divided by r. Notice that this is independent of our small m. So if we look at the speeds of planets in their orbits, and these are pretty close to being circular, we should see that the velocity or speed will be proportional to 1 over the square root of r. Let's see how that works out in practice. You can see from this diagram that there is exquicite agreement between what is predicted and what is observed. Now, let's instead of considering the Sun, let's consider the Earth, where we know that at the surface, the acceleration of the Earth is given by 9.8 meters per second, per second. So now, we can rewrite our force equation as F equals G-M, Earth now. And this is just the symbol for our planet. Times little m, over r squared. And that's going to be given by m-a. So now if we have something like a piece of chalk, which is our little m, we see that that will cancel from both sides of the equation, and so we can actually solve for the mass of the Earth. Since a is equal to 9.8 meters per second, per second, we find after looking at this equation for just a few seconds, that the mass of the Earth will be given by the number 9.8 times r squared over G, just solving this equation for the mass of the Earth. So if we know G, and the radius of our planet. We have actually weighed the Earth. So sometimes we actually say that the determination of the gravitational constant is the weighing of the Earth. And now you can see that since r, we know from Era, Eratosthenes right, I mean he was the one who did it for the first time thousands of years ago, if r is equal to six times ten to the sixth meters from the, experiment done by Eratosthenes. G, we know, is given by 6.7 times 10 to the minus 11, the units of G are a little strange, so I'm just going to say that were in the mks system where we measure mass in kilograms here. We end up with the mass of the Earth, being equal to, and you can work this out for yourself, 5 times 10 to the 24 kilograms. That's pretty neat! Now, let's proceed to circular orbits. We know that the speed is constant. This allows us to eliminate the speed by considering an entire orbit of a body, say one star orbiting another more massive star. Since the speed is constant we know that the velocity, or speed in this case, if we ignore the direction for a moment, is given by an extremely simple idea. We just take the distance and divide it by the time, and we can choose, any amount of distance, and any amount of time, because the speed is constant, and, if we choose the complete orbit of the object, just going around, the center of the orbit, OK, we get V equals 2 pi r, where r now is the distance from one object to the other object and 2 pi r is nothing more than the circumference of the circle. So it goes one complete revolution about its orbit in the time capital T, where T is the orbital period. Now, since we know once again that v squared over r is equal to G-M over r squared, we can substitute this expression for v here, and we get, of course, v squared is going to be given by 4 pi squared, r squared over T squared. So we end up with T squared equaling 4 pi squared over G-M, times r cubed. Or in words, the square of the period is proportional to the cube of the radius of the orbit. Now this only works for circles, actually at least we've derived it for circles. But in fact, a similar equation can be derived for elliptical orbits. Where instead of r, we end up defining something called the semi-major axis of the orbit. But, it works in any event. And, let's see how this actually turns out for the planets in our solar system. You can see here what that law looks like for every large body in the neighborhood of the Sun. Wow! Everything works! We're done, right? Problem solved! Not quite. After a while we realised that something was not right. Mercury wasn't moving according to specifications even when you included the ellipticity of the orbit, and all the gravitational effects of the other planets. The amount of the discrepancy was quite small. But it was real, since it was over 100 times more than the probable error, associated with the data. It couldn't be ignored. Here is the crux of the situation. When you have an elliptical orbit, there is a point in the motion where you are closest to the sun. That is called the perihelion point. And the position of Mercury's perihelion was changing a bit more than Newton's law predicted. Here, you see the situation. How much was it off by? You see this penny? If you held this penny up to the sky at a distance of about 100 meters, the angle it would subtend, would be equal to the discrepancy in Mercury's orbit, over the period of 100 years. That's pretty amazing. Well, you might say since the discrepancy is so small, the correction to the law must be small as well. Wrong! This tiny problem along with several others, lead to a profound change in our entire understanding of the structure of space and how it affects the motions of all bodies in the universe. The important point here is that sometimes incredible revelations can be the result of measurements that deviate oh so little from what we anticipated. It is the job of the scientist to pay attention to these details, because on occasion, these details can be the key to some overarching principal that would otherwise be overlooked. But let's get back to the business at hand, which in this case is using Newton's idea of mechanics, to yield insight into some cosmic X-ray sources. The summary of our results can be expressed in four very simple equations. The first equation was that the speed of an object in its orbit is given by the circumference of its orbit, 2 pi r, divided by its orbital period. The second equation is nothing more than a restatement of Newton's 2nd law, F=ma, and that's equal to capital G, a constant, times the product of the two masses involved, divided by the square of the distance between them. And we also know that for circular orbits, the acceleration is given by V squared over r, and last but not least is Kepler's Law that says the square of the period is equal to the cube of the size of the orbit, times a constant which is given by 4 pi squared over G times big M. That's it! These four equations allow us to do some incredibly interesting and beautiful things, and we're going to use these results to explore one of the most fascinating x-ray sources in the sky. Centaurus X-3, this object will be the focal point of the next lectures in analyzing the Universe. [BLANK_AUDIO]
