So, I think we have hit that hard enough.
Let's go to--
my goodness--
4.B.5.
It's Monday morning and it's early, believe it or not.
The gravitational and electric potential energy.
Let's first do gravity and there is no fiction.
We have assumed no friction, no air drag of any kind.
Here is the sun.
It has a mass M sun.
This symbol stands for the sun.
Here's an object of mass little m at position 1 at
separation r1 from the sun.
And here is that same mass at position 2 at separation r2
for the sun.
And I now want to evaluate the forces, gravitational forces,
pulling from the sun.
This is F1.
It's regularly pointing towards the sun, and the
magnitude is m times the mass of the sun times g divided by
r1 squared.
And this force, which is smaller because it's farther
away from the sun, F2, equals m times the mass of the sun
times the gravitational constant
divided by r t squared.
And when I go all the way to infinity, F then at infinity,
the gravitational pull from the sun equals 0.
Because 1 over r squared becomes 0.
Now if I bring the object from position 1 to position 2, I
have to overcome the gravitational force.
So I move my hand to the right, I pull to the right,
and I move it to the right, so I do positive work.
And per definition, U2, that is the gravitational potential
energy here, minus U1, the gravitational
potential energy here.
[INAUDIBLE]
per definition, the work I do in moving that object from
position 1 to position 2.
And since here r2 is larger than r1, U2 minus U1
is larger than 0.
So I do positive work.
If I bring it from here to here, then r2 is
smaller than r1.
r2 minus U1 is smaller than 0, and I do--
no.
U2 minus U1 is always larger than 0.
But if I go from here to here, then I go from a higher
potential energy to a lower potential energy, and so I do
negative work.
So I confused you. r2 is always larger
than r1 in this case.
U2 minus U1 is always larger than 0.
If I go from 1 to 2, I therefore have
to do positive work.
If I go from 2 to 1, I therefore, have
to do negative work.
I go from a higher potential to a lower potential.
Now before we continue, I would like to examine with you
a somewhat simpler case.
Suppose here I have an object with mass m and I have
here a force mg.
And here is my force Walter Lewin.
I'm bringing it from-- here's the floor.
Here's a table top.
I call this table top arbitrarily, y equals 0.
I could've called this something else.
This table top is at location A. This is at location B. And
I bring it up to a position h, so y equals h here.
Now to bring this mass from A to B I have
to do positive work.
And per definition, Ub minus UA is the work that I have to
do to bring it from A to B. And it's immediately obvious
that that is mgh.
This is easy to remember.
mgh stands for Massachusetts General Hospital.
At least, that's the way I always remember it.
So that's the work I do when I bring it from A to B.
Now here, since we are moving it, that distance is very
small compared to the size of the earth, the gravitational
force is constant all the way when I go from A to B. So my
force is constant.
So that's why I can do it in this simple way.
However, if we are going back to the problem with the sun,
then of course, the situation is somewhat different.
If I go back to the sun, it is obvious that this force is not
the same as this force.
So now I cannot simply when I do the integration in going
from 1 to 2, I cannot assume that the force is constant.
So F1 is not the same as F2 because r1 is smaller than r2.
So now what is the work that I do?
Well that is the integral r1 to r2 of m M sun times g
divided by r squared dr.
I go from here and I move it out and I'm at some random
location r.
And I move it out over the distance dr. Over this small
trajectory I assume that the force remains constant.
This is the magnitude of my force when I have to pull it
to an increasing value of r.
dr is positive here and my force itself is also positive.
I have to pull away from the center.
You could call this force negative if you wanted.
Well, clearly, I'm dealing here with a dot product
between my force, Walter Lewin, and dr. But since the
angle is 0 between the two, the cosine angle equals 0.
So I can ignore that for now.
And so I get that U at r2 minus Ur1, which is the work
that I have to do to bring the object from 1 to 2 becomes m M
sun times g times 1 over r1 minus 1 over r2.
And I have assumed that there is no fiction here.
Now if r2 is larger than r1, which is the case here, then
this whole thing is larger than 0.
And I do positive work.
That's immediately obvious because as I move the object
farther away from the sun, clearly I have to overcome the
gravitational pull.
So it is immediately obvious that I do positive work, and
so the potential energy at location 2 is larger than the
potential energy at location 1.
What now is the potential energy at position 1?
See when you're dealing with very close to earth situations
in our laboratory, you can effectively choose the 0 level
any way you want it.
You can choose it on the floor, you can choose it at
the table top, or you can choose it at that location B.
Remember, I was moving it from the table top to that location
B. It doesn't really matter, the 0 has no consequence.
What counts in physics is the difference
between potential energy.
Difference between potential energy determines the force.
The force determines the acceleration.
That means the change of velocity.
So no matter how you decide, where you decide to choose
your 0, gravitational potential energy in here in
the laboratory system, the physics is not different.
Here however, it is meaningful and useful to choose that the
potential energy of gravity at infinity equals 0.
So I prefer that the choice is not yours now.
Now the consequence of that is, if I fill in for r2
infinity, then I get r equals infinity.
Which we defined to be 0.
Minus U or r.
For r1 I put here now a random location r equals m mass of
the sun times g times 1/r because this is 0.
This is the scalar and this is 0, so if I rewrite this a
little bit as Ur, I get a minus sign here.
So we get Ur gravitational potential energy equals minus
m M sun times g divided by r.
And it is a scalar.
It has no direction.
Let's put this in a nice red box because it's a very
important equation.
This negative sign is the price that we have to pay for
the fact that we call U at infinity equals 0.
Put in r equals infinity and you see that U
indeed becomes 0.
If we wanted to draw a graph of the potential energy as a
function of r, here is r.
Here is U potential.
These values would all be positive.
This will be 0.
And these values would all be negative.
Then, the gravitational potential energy is 1 over r.
It's a hyperbola.
So it would go like this.
Goes to 0 at infinity.
But of course, this is proportional
by the way, to 1/r.
But when you get inside the sun, then of course, if this
is r sun, then this has to be reevaluated.
This certainly does not continue as a hyperbola.
And as r decreases, if r decreases, U decreases.
For B, to move it further in, I have to do negative work.
As r increases, U increases.
Notice if it is negative here, but it's still increasing.
And for me to move it further out, I have
to do positive work.
Now if you know how U depends on x, y, or z, or r, for that
matter, you can easily find the vector components of the
forces that are responsible for the potential energy.
F of x--
that is the x component of that force--
equals minus dU dx.
What does that mean?
I move the object only in the x direction.
I keep y and z constant.
And then, I watch how the potential energy is changing.
And this equation then gives you the x
component of that force.
Equally, Fy equals minus dU dy and Fz equals minus dU dz.
And if you have something that only depends on the radius,
like the case of the sun, Fr equals minus dU dr.
Let's take the situation of the table.
On the table we had U as a function of y.
y was increasing in this way.
Let's call it mgy.
I then arbitrarily choose--
at y equals 0 I choose U equals 0.
Feel free to do this, but that makes it in this
case, rather easy.
Well, if this is a scalar, if now I calculate the y
component of the gravitational force, I get minus dU dy, so I
get minus mg.
And this minus sign is simply telling you that it is in the
opposite direction as increasing y.
So this force is in downwards direction, which is exactly
what it should be.
Fx equals 0 and Fz equals 0 because there's no dependents
of the potential energy on either x or z for that matter.
Now the minus sign is also easy to see.
In the case of the sun where we have only a radial
dependence, the gravitational force would then be minus dU
dr. I have to do positive work to increase U. If dU is
positive, it means that dr is positive.
I go outwards.
And we all know that the gravitational force is
directed inwards. that's the meaning now
of this minus sign.
Just in a similar way as this minus sign was telling you
that it is in the direction opposite to an increasing
value of y.
This is telling you that this minus sign is pointing in an
opposite direction than in the direction of increasing r.
So the force of Walter Lewin, if I bring it out, would be
plus dU dr. Because my force is exactly in the opposite
direction than the force from gravity.
So U in the case of gravity, it was the sun equals m M sun
times g divided by r.
Notice that when r increases that U increases.
Because of this minus sign.
Now dU dr equals plus m M sun times g divided by r squared.
So F in the radial direction of gravity is minus dU dr. So
F in the radial direction equals minus m M sun g divided
by r squared.
And the minus sign then indicates that at this point
in the opposite direction as increasing r I prefer a
different notation.
If here is the sun and here is r, I would like to introduce--
but I think Professor Guth is not doing that.
I would like to introduce a unit vector r roof.
It has lengths 1 and it is radially pointing out.
And I would like to put here an r roof and here a vector.
Then there is simply no confusion whatsoever.
You see immediately that the gravitational force is
opposing the unit vector r roof, which is pointing
outwards to increasing value of r But this is a matter of
taste, and I'll leave it up to you what you prefer.
Now we turn to electric potential energy.
Now we have a charge plus q and plus capital Q. And unlike
gravity, these repel.
So already you feel in your stomach that everything is
going to be very, very similar, except that minus
signs may change to plus signs and plus signs may change to
minus signs.
Yeah, that's true.
But remember, in electricity the charges can also have
opposite signs.
And if that's the case, they attract each other.
So then, you would expect to see something extremely
similar to what we have seen with gravity.
Because with gravity, masses cannot be
negative and positive.
So with gravity we always have attracting forces.
All right, here we have plus q and here we have plus capital
Q. They are at a distance r and they repel, so the
electric force is in this direction outwards.
And the electric force here equaling strength in the
opposite direction.
And the magnitude of this electric force equals q times
Q divided by 4 pi epsilon 0.
Something you will see later in 802.
It's a constant.
Divided by r square, inverse r square.
If I increase the distance between these two objects, I
have to do negative work.
If I have to bring them closer together, I have to overcome
this repelling force.
Then I have to do positive work.
And if again, I assume that the potential energy at
infinity is 0, I now get that the
electrostatic potential energy--
not gravitational, but electrostatic equals plus q
times capital Q divided by 4 pi epsilon 0 r.
This is nearly identical to gravity except that with
gravity we had a minus and here we have a plus.
Now, the nice thing about this equation is that if you change
the signs of q and capital Q-- little q and capital Q-- if
you make them both negative, the plus sign
still remains plus.
Because minus times minus is still plus.
So this still holds.
If however, they have opposite signs, either this plus and
this minus or this plus and this minus, then the upstairs
here becomes negative and you get a situation, which is
completely analogous to gravity.
Which is exactly what you predict because now you have
attracting forces.
And when you have attracting forces, the upstairs here,
once you make this multiplication, the upstairs
becomes negative.
So it's very nice and very cute that you see here the
beautiful symmetry with gravity in case that the
electric charges attract each other.