Selasa, 26 April 2011

Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan
Fisika di tingkat Universitas

» Download this transcript (PDF)

Today, I will talk to you about elliptical orbits and Kepler's famous laws.

I first want to review with you briefly what we know about circular orbits, so I wrote on the blackboard everything we know about circular orbits.

There's an object mass little m going in a circle around capital M.

This could be the Sun; it could be the Earth.

It has radius R, circular.

We know there in equation one how to derive the time that it takes to go around.

The way we found that was by setting the centripetal force onto little m the same as the gravitational force.

Also, the velocity in orbit--

maybe I should say speed in orbit--

also follows through the same kind of reasoning.

Then we have the conservation of mechanical energy--

the sum of kinetic energy and potential energy.

It's a constant; it's not changing.

You see there first the component of the kinetic energy, which is the one-half mv-squared, and then you see the term which is the potential energy.

We have defined potential energy to be zero at infinity, and that is why all bound orbits have negative total energy.

If the total energy is positive, the orbit is not bound.

And when you add these two up, you have an amazing coincidence that we have discussed before.

We get here a very simple answer.

The escape velocity you find by setting this E total to be zero, so this part of the equation is zero.

Out pops that speed with which you can escape the gravitational pull of capital M, which is the square root of two times larger than this V.

And I want to remind you that for near Earth orbits, the period to go around the Earth is about 90 minutes, and the speed-- this velocity, then, that you see in equation two--

is about eight kilometers per second, and the escape velocity from that orbit would be about 11.2 kilometers per second.

And for the Earth going around the Sun, the period would be about 365 days, and the speed of the Earth in orbit is about 30 kilometers per second, just to refresh your memory.

Now, circular orbits are special.

In general, bound orbits are ellipses, even though I must add to it that most orbits of our planets in our solar system--

very close to circular, but not precisely circular.

But the general solutions call for a elliptical orbit.

And I first want to discuss with you the three famous laws by Kepler from the early 17th century.

These were brilliant statements that he made.

The interesting thing is that before he made these brilliant statements, he published more nonsense than anyone else.

But finally he arrived at two... three golden eggs.

And the first golden egg then is that the orbits are ellipses--

he talked always about planets-- and the Sun is at one focus.

That's Kepler's law number one.

These are from around 1618 or so.

The second...

Kepler's second law is--

quite bizarre how he found that out, an amazing accomplishment.

If you take an ellipse, and you put the Sun here at a focus--

this is highly exaggerated because I told you that most orbits look sort of circular--

and the planet goes from here to here in a certain amount of time, and you compare that with the planet going from here to here in a certain amount of time, then Kepler found out that if this area here is the same as that area here, that the time to go from here to here is the same as to go from there to there.

An amazing accomplishment to come up with that idea.

And this is called "equal areas, equal times."

Somehow, it has the smell of some conservation of angular momentum.

And then his third law was that if you take the orbital period of an ellipse, that is proportional to the third power of the mean distance to the Sun.

And he was so pleased with that result that he wrote jubilantly about it.

I will show you here the data that Kepler had available in 1618, largely from the work done by, of course, astronomers, observers like Tycho Brahe and others.

You see here the six planets that were known at the time, and the mean distance to the Sun.

For the Earth, it is one because we work in astronomical units.

Everything is referenced to the distance of the Earth.

This is 150 million kilometers.

And it takes the Earth 365 days to go around the Sun;

Jupiter, about 12 years; and Saturn, about 30 years.

And then when he takes this number to the power of three and this number squared, and he divides the two, then he gets numbers which are amazingly constant.

And that is his third law.

The third law leads immediately to the inverse square dependence of gravity, which he was not aware of, but Newton later put that all together.

But he very jubilantly writes:

And he wrote that in 1619.

So, orbits in general are ellipses.

And now I want to review with you what I have there on the blackboard about ellipses.

You see an ellipse there?

Capital M-- could be the Earth, could be the Sun--

is at location Q.

The ellipse has a semimajor axis A, so the distance P to A-- perigee to apogee-- is 2a.

If M were the Earth, capital M, then we would call the point of closest approach "perigee," and the point farthest away from the Earth, we would call that "apogee." If capital M were the Sun, we would call that "aphelion" and "perihelion." So you see the little m going in orbit;

you see the position vector r of q.

It has a certain velocity v.

And so the total mechanical energy is conserved.

The sum of kinetic energy and potential energy doesn't change.

The first term is the kinetic energy-- one-half mv squared, and the second term is the potential energy--

no different from equation three for circular orbits, except that now capital R, which was a fixed number in a circle, is now a little r, and little r changes, of course, with time.

Also that velocity, v, in that equation number five will also change with time because it's an elliptical orbit.

It will not change in time in equation number two and in equation number three.

Now I give you a result which I didn't prove, and that is that the total mechanical energy, which has these two terms in it which you do fully understand, also equals minus mMG divided by 2a, and a is the semimajor axis.

And compare number five with number three, then you see they are brothers and sisters.

The only change is that what was capital R before is now little a, the semimajor axis.

And if you want to calculate the time to go around the ellipse, then you get an equation for T squared, which is almost identical to number one for the circular orbit, except that now the radius has to be replaced by a, which is the semimajor axis.

And the escape velocity you can calculate in exactly the same way that you calculate the escape velocity under equation number four.

All you do is you make the total energy zero, and then you solve equation three and equation five, and out pops the speed that you need to make it all the way out to infinity.

And that is in the case of the circular orbit, square root of two times the speed in orbit.

So these are the numbers that we are going to use today, the equations.

And there's one thing which is already quite remarkable and very nonintuitive-- very nonintuitive, to say the least.

That if you have various orbits which have the same semimajor axis, that the period is the same and the energy is the same, and that's by no means obvious.

So, this is one orbit-- think of it as being an ellipse--

and this is another one.

This distance is the same as this distance.

I've just done it that way.

That means, according to equation five and six, that both orbits have exactly the same mechanical energy, and both orbits have the same periods.

So to go around this circular orbit will take the same amount of time as to go around this one, and that is by no means obvious.

I now want to start with a very general initial condition of an object, little m, in orbit... in an elliptical orbit.

And I want to see how we can get all the information about the ellipse that we would like to find out.

So I'm only giving you the initial conditions.

So here is an ellipse, here is P and here is A.

If this is an ellipse around the Earth, then this would be perigee and this would be apogee.

The mass is capital M, this is point Q.

Let me get a ruler so that I can draw some nice lines.

So the distance AP equals 2a-- a being the semimajor axis--

and our object happens to be here-- mass little m--

and this distance equals R zero.

Think of it as being time zero.

And at time zero, when it is there, it has a velocity in that ellipse.

Let this be v zero.

And there is an angle between the position vector and v zero; I call that phi zero.

So I'm giving you M, I'm giving you v zero, I'm giving you r zero, I'm giving you phi zero.

And now I'm going to ask you, can we find out from these initial conditions how long it takes for this object to go around? Can we find out what QP is? Can we find out what the semimajor axis is? Can we find out what the velocity is at point P, at closest approach when this angle is 90 degrees? And can we find out what the velocity is when the object, little m, is farthest away-- apogee? Can we find all these things? And the answer is yes.

a is the easiest to find-- the semimajor axis.

I turn to equation number five, which is the conservation of mechanical energy.

And the conservation of mechanical energy says that the total energy is the kinetic energy plus the potential energy equals one-half mv zero squared.

That is when the object is here at location D minus mMG divided by this r zero at location D.

This can never change.

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

Tidak ada komentar: