PROFESSOR: I think we've now reached our last problem.
Let's do a time check.
I need 6 more minutes, and boy, I have 6 minutes.
It's perfect.
1.E.3 deals with a geosynchronous satellite.
Let's first consider a circular motion.
Radius r.
And there's an object going around with a constant speed.
Speed is v. Here the speed is the same.
That is v. But be careful.
The velocity vector here is different from the velocity
vector there, because the direction has changed.
The magnitude, which we call speed, that is only the
magnitude of the vector, but the direction of the vector is
very important, and the direction changes.
And so since the direction changes, there has to be an
acceleration.
Otherwise, the vector v could not possibly change.
In this situation, which I will give you without proof,
the acceleration is always pointed towards the center.
That's why we call it centripetal acceleration.
It's therefore, also changing.
a itself is not constant and v is not constant.
a is pointing in this direction and
a is pointing here.
a is constant.
The magnitude of the acceleration is constant.
And that I will give you as our proof is v
squared divided by r.
And sometimes you see a little c here, which indicates
centripetal.
It's pointing towards the center of the circle.
Now, you can take an apple and swirl an apple above your head
in more or less a horizontal plane.
If you can do this at constant speed, then you have a
situation like this.
A car that's going around in a circle, if the speed of the
car were not changing, you would have a
situation like this.
Take a satellite going around the earth, putting it in a
circular orbit, you will have a situation like this.
Take a planet around the sun, put it in a circular orbit,
you will have a situation like this.
And in your particular case, you're being asked to
calculate the centripetal acceleration for a
geosynchronous orbit.
A geosynchronous orbit is an orbit whereby the satellite
takes 24 hours to go around the earth.
And if you launch that satellite in the plane of the
Equator, then-- and you should think about that--
when you look from any point of earth at that satellite, it
will not move relative to the rotating earth, because the
earth rotates in 24 hours and the satellite
rotates in 24 hours.
So the satellite will appear to stand still in the sky
relative to the earth.
Of course, not relative to the stars.
It has to be launched in the equatorial plane.
You think about that.
That's a must.
Otherwise, it does not stand still in the sky.
Well, the orbital period is 24 hours and the radius of that
orbit was given to you.
There will be a time that you can calculate that.
If you have Newton's laws and Kepler's laws, so the velocity
will be 2 pi r divided by 24 times 3,600.
That would be in kilometers per second.
And so the centripetal acceleration, which is v
squared divided by r--
I think I found something like 0.2 meters per second squared.
But you better check that.