0:11

Okay, gang, let's roll. Last time, we obtained a bit of

Â insight into that most famous of all equations in classical mechanics,

Â better known as Newton's 2nd law. That says

Â F=ma. This particular

Â equation is surprisingly subtle, and we're going to ignore

Â most of those subtleties, but I should at least tell you that

Â one, this is a vector equation. Force has a particular

Â direction, and the acceleration also has a particular direction.

Â Also, you need to sum all of the forces that might be acting on an object.

Â So we really have a net force, because forces can pull in different

Â directions, and so this particular equation can

Â yield insight into the way astronomical bodies can move.

Â 1:19

We also derived the acceleration, a, for

Â circular motion, and we found that the magnitude

Â of that acceleration was given by the square of the speed

Â divided by the distance between where the center of the

Â circle was and where the object was rotating around.

Â Now, we are ready to see what the left hand

Â side of Newton's law has in store for us by applying

Â it to simple astronomical systems. Newton found

Â out after countless generations of scientists and would-be scientists

Â over the centuries tried their luck and skill at figuring out the problem,

Â that the gravitational force between two point

Â like objects was F equals some constant,

Â G, capital G, times big M times little m,

Â divided by r squared. Where we usually just use capital M

Â and small m to represent the masses of the two objects involved.

Â Again, we're going to assume that this is a

Â vector, and we're not going to worry about putting

Â arrows on everything, because sometimes that can get a little bit cumbersome.

Â This is simple, but profound.

Â G is a constant that is surprisingly difficult to get

Â accurately because gravity turns out to be a very weak force,

Â but we do know the value of g to about a tenth of one percent.

Â Let's see what the consequences of our understanding of these

Â ideas are. First, imagine

Â that M is the sun and little m is any

Â planet. Then, we have F equals

Â G-M-m over r squared and

Â that's going to be little mv squared

Â over r. Where now we're substituting for the

Â acceleration, our understanding for how things move in

Â a circular orbit. So, if we solve

Â for v squared, we find that v squared is

Â equal to G-M, the mass of

Â the sun, divided by r. Notice

Â that this is independent of our

Â small m. So if we look at the speeds of planets in

Â their orbits, and these are pretty close to being circular, we should

Â see that the velocity or speed will be proportional to 1 over

Â the square root of r. Let's see how that works out in practice.

Â 4:57

Now, let's instead of considering the Sun, let's consider

Â the Earth, where we know that at the surface, the acceleration

Â of the Earth is given by 9.8 meters per

Â second, per second. So now, we can rewrite our

Â force equation as F equals

Â G-M, Earth now. And this is just the symbol for

Â our planet. Times little m, over r squared.

Â And that's going to be given by m-a. So now if

Â we have something like a piece of chalk, which is our little m, we see that that

Â will cancel from both sides of the equation, and so

Â we can actually solve for the mass of the

Â Earth. Since a is equal to

Â 9.8 meters per second, per second, we find

Â after looking at this equation for just

Â a few seconds, that the mass of the Earth will be given by

Â the number 9.8 times r squared

Â over G, just solving this equation for the mass of the Earth.

Â So if we know G, and the radius of our planet.

Â We have actually

Â weighed the Earth. So sometimes we actually say that the

Â determination of the gravitational constant is the weighing of the Earth.

Â 6:53

And now you can see that since r, we know from Era, Eratosthenes

Â right, I mean he was the one who did it for the first time thousands of years ago,

Â if r is equal to six times ten to

Â the sixth meters from the, experiment

Â done by Eratosthenes. G, we know, is

Â given by 6.7 times 10 to the minus

Â 11, the units of G are a little strange, so I'm just

Â going to say that were in the mks system

Â where we measure mass in kilograms

Â here. We end up with the mass of the Earth,

Â 8:03

Now, let's proceed to circular orbits. We know

Â that the speed is constant. This allows us to eliminate the

Â speed by considering an entire orbit of a

Â body, say one star orbiting another more massive star.

Â Since the speed is constant we know that the

Â velocity, or speed in this case, if we ignore the

Â direction for a moment, is given by an extremely

Â simple idea. We just take the distance

Â 8:47

and divide it by the time, and we can choose,

Â any amount of distance, and any amount of time,

Â because the speed is constant, and, if we choose the

Â complete orbit of the object, just going around, the

Â center of the orbit, OK, we get

Â V equals 2 pi r, where r

Â now is the distance from one object

Â to the other object and 2 pi r is nothing

Â more than the circumference of the circle. So it goes one

Â complete revolution about its orbit in the time

Â capital T, where T is the orbital period.

Â Now, since we know once again that

Â v squared over r is equal to G-M over

Â r squared, we can substitute this

Â expression for v here, and we get,

Â of course, v squared is going to be

Â given by 4 pi squared, r squared

Â over T squared. So we end

Â up with T squared

Â equaling 4 pi squared

Â over G-M, times

Â r cubed. Or in words, the

Â square of the period is proportional to the cube of

Â the radius of the orbit. Now this only works

Â for circles, actually at least we've derived it for circles.

Â But in fact, a similar equation can be derived for

Â elliptical orbits. Where instead of r, we

Â end up defining something called the semi-major

Â axis of the orbit. But, it works in any event.

Â And, let's see how this actually turns out for the planets in our

Â solar system. You can see here what that law looks

Â like for every large body in the neighborhood of the Sun.

Â 11:42

Not quite.

Â After a while we realised that something was not right.

Â Mercury wasn't moving according to specifications

Â even when you included the ellipticity

Â of the orbit, and all the gravitational effects of the other planets.

Â The amount of the discrepancy was quite small.

Â But it was real, since it was over 100

Â times more than the probable error, associated with the data.

Â It couldn't be ignored.

Â Here is the crux of the situation. When you have an elliptical

Â orbit, there is a point in the motion where you are closest to the sun.

Â 12:28

That is called the perihelion point. And the position

Â of Mercury's perihelion was changing a bit more

Â than Newton's law predicted. Here, you see the

Â situation. How much was it off by?

Â You see this penny? If you held this penny up to the sky

Â at a distance of about 100 meters, the

Â angle it would subtend, would be equal to

Â the discrepancy in Mercury's orbit, over

Â the period of 100 years.

Â That's pretty amazing. Well, you might say since

Â the discrepancy is so small, the correction

Â to the law must be small as well.

Â Wrong!

Â This tiny problem along with several others, lead to a profound

Â change in our entire understanding of the structure of space

Â and how it affects the motions of all bodies in the universe.

Â The important point here is

Â that sometimes incredible revelations can be the result of measurements

Â that deviate oh so little from what we anticipated.

Â It is the job of the scientist to pay attention to these details, because on

Â occasion, these details can be the key to

Â some overarching principal that would otherwise be overlooked.

Â 14:09

But let's get back to the business at hand, which in this case is using

Â Newton's idea of mechanics, to yield insight into some cosmic

Â X-ray sources. The summary of our results can be

Â expressed in four very simple equations. The first

Â equation was that the speed of an object in its orbit is

Â given by the circumference of its orbit, 2 pi r,

Â divided by its orbital period. The second equation is

Â nothing more than a restatement of Newton's 2nd law,

Â F=ma, and that's equal to capital G,

Â a constant, times the product of the two masses involved,

Â divided by the square of the distance between them.

Â 15:06

And we also know that for circular orbits, the acceleration is

Â given by V squared over r, and last but

Â not least is Kepler's Law that says the square

Â of the period is equal to the cube of

Â the size of the orbit, times a constant which

Â is given by 4 pi squared over

Â G times big M. That's

Â it! These four equations allow us to do some

Â incredibly interesting and beautiful things, and we're going to

Â use these results to explore one of the most fascinating x-ray sources in the sky.

Â Centaurus X-3, this object will be the focal point

Â of the next lectures in analyzing the Universe.

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Â