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All right...

long weekend ahead of us.

One more lecture to go.

If I have an object, mass m, in gravitational field, gravitational force is in this direction, if this is my increasing value of y, then this force, vectorially written, equals minus mg y roof.

Since this is a one-dimensional problem, we will often simply write F equals minus mg.

This minus sign is important because that's the increasing value of y.

If this level here is y equals zero, then I could call this gravitational potential energy zero.

And this is y.. .

Then the gravitational potential energy here equals plus mg y.

This is u.

So if I make a plot of the gravitational potential energy as a function of y, then I would get a straight line.

This is zero.

So this equals u.. .

equals mg y, plus sign.

If I'm here at point A and I move that object to point B, I, Walter Lewin, move it, I have to do positive work.

Notice that the gravitational potential energy increases.

If I do positive work, the gravity is doing negative work.

If I go from A to some other point —

call it B prime —

then I do negative work.

Notice the gravitational potential energy goes down.

If I do negative work, then gravity is doing positive work.

I could have chosen my zero point of potential energy anywhere I please.

I could have chosen it right here and nothing would change other than that I offset the zero point of my potential energy.

But again, if I go from A to B, the gravitational potential energy increases by exactly the same amount —

I have to do exactly the same work.

So you are free to choose, when you are near Earth, where you choose your zero.

Now we take the situation whereby we are not so close to the Earth.

Here is the Earth itself.

Of course you can also replace that by the sun if you want to.

And this is increasing value of r.

The distance between here and this object m equals r.

I now know that there is a gravitational force on this object —

Newton's Universal Law of Gravity —

and that gravitational force equals minus m M-Earth G divided by this r squared, r roof so this is a vectorial notation.

Since it is really one-dimensional, we would...

Just like we did there, we would delete the arrow and we would delete the unit vector in the positive r direction and so we would simply write it this way.

The gravitational potential energy we derived last time equals minus m M-Earth G divided by r —

and notice, here is r and here is r squared —

and if you plot that, then the plot goes sort of like this.

This is r, this is increasing potential energy —

all these values here are negative —

and you get a curve which is sort of like this.

This is proportional to one over r.

Now, of course, if the Earth had a radius which is this big, then, of course, this curve does not exist, it stops right here.

If I move from point A to point B, with a mass m in my hand, notice that the gravitational potential energy increases.

I have to do positive work, there is no difference.

If I go from A to another point, B prime, which is closer to the Earth, notice that the gravitational potential energy decreases.

I do negative work.

If I do positive work, gravity is doing negative work.

If I do negative work, gravity is doing positive work.

Right here near Earth, where this one-over-r curve hits the Earth, that is, of course, exactly that line.

long weekend ahead of us.

One more lecture to go.

If I have an object, mass m, in gravitational field, gravitational force is in this direction, if this is my increasing value of y, then this force, vectorially written, equals minus mg y roof.

Since this is a one-dimensional problem, we will often simply write F equals minus mg.

This minus sign is important because that's the increasing value of y.

If this level here is y equals zero, then I could call this gravitational potential energy zero.

And this is y.. .

Then the gravitational potential energy here equals plus mg y.

This is u.

So if I make a plot of the gravitational potential energy as a function of y, then I would get a straight line.

This is zero.

So this equals u.. .

equals mg y, plus sign.

If I'm here at point A and I move that object to point B, I, Walter Lewin, move it, I have to do positive work.

Notice that the gravitational potential energy increases.

If I do positive work, the gravity is doing negative work.

If I go from A to some other point —

call it B prime —

then I do negative work.

Notice the gravitational potential energy goes down.

If I do negative work, then gravity is doing positive work.

I could have chosen my zero point of potential energy anywhere I please.

I could have chosen it right here and nothing would change other than that I offset the zero point of my potential energy.

But again, if I go from A to B, the gravitational potential energy increases by exactly the same amount —

I have to do exactly the same work.

So you are free to choose, when you are near Earth, where you choose your zero.

Now we take the situation whereby we are not so close to the Earth.

Here is the Earth itself.

Of course you can also replace that by the sun if you want to.

And this is increasing value of r.

The distance between here and this object m equals r.

I now know that there is a gravitational force on this object —

Newton's Universal Law of Gravity —

and that gravitational force equals minus m M-Earth G divided by this r squared, r roof so this is a vectorial notation.

Since it is really one-dimensional, we would...

Just like we did there, we would delete the arrow and we would delete the unit vector in the positive r direction and so we would simply write it this way.

The gravitational potential energy we derived last time equals minus m M-Earth G divided by r —

and notice, here is r and here is r squared —

and if you plot that, then the plot goes sort of like this.

This is r, this is increasing potential energy —

all these values here are negative —

and you get a curve which is sort of like this.

This is proportional to one over r.

Now, of course, if the Earth had a radius which is this big, then, of course, this curve does not exist, it stops right here.

If I move from point A to point B, with a mass m in my hand, notice that the gravitational potential energy increases.

I have to do positive work, there is no difference.

If I go from A to another point, B prime, which is closer to the Earth, notice that the gravitational potential energy decreases.

I do negative work.

If I do positive work, gravity is doing negative work.

If I do negative work, gravity is doing positive work.

Right here near Earth, where this one-over-r curve hits the Earth, that is, of course, exactly that line.

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