0:11

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.

1:19

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.

4:57

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.

6:53

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,

8:03

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

8:47

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.

11:42

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.

12:28

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.

14:09

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.

15:06

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.

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