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We would now like to calculate
the potential function

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for a universal law of gravity.

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So let's set up a
coordinate system first.

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And in this case, we're going
to make a very special case--

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although, the general result
doesn't depend on that.

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So suppose we have a
planet of radius rp

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and we have a small object.

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And initially, our
object is a distance r

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away from the planet.

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This is our initial state.

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And the planet has
mass to the planet.

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And our small object has m.

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And this object is moving.

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Of course, orbits
really aren't like that,

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so we should be a
little bit more careful.

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But our orbit might be some type
of hyperbolic section-- conic

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section.

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And the orbit over here
is at some final position.

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In our eyes, the distance
from the object to the center

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of the planet as our final is.

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Now because there's a central
point to this problem,

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the natural coordinate
system to use

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is polar coordinates
in the plane.

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And let's just imagine
that at this moment

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we have the universal
force of gravity.

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Here's our object.

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And it's being
displaced a distance s.

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So our coordinate
system is polar.

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Let's call r hat this way
and theta hat that way.

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And let's blow up our
little displacement

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in terms of our
coordinate system,

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just so that we see
what's happening.

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So let's imagine
that our little ds

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vector is pointing like this.

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And our center point here,
we have tge r hat pointing

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radially outward
and the theta hat

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pointing towards the center.

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Now to exaggerate this
picture, if this is r,

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then the arc length
here is rd theta.

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And the difference as
the object moves from one

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point here to a closer point
towards the center, will be dr.

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And so our ds vector, we can
write as a radial piece r hat.

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Now, I don't put
a sign in there,

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because only the end
points of my integral

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will tell me whether I was
going towards the planet

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or away from the planet.

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And rd theta, theta hat.

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And our gravitational
force is minus

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g mass mass of the planet over
the distance squared r hat.

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So now what we see
is that when we

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take the dot product
of these two forces,

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r hat dot r hat is 1
in polar coordinates.

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Why is that the case,
r hat dot r hat is 1?

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Because these vectors are
in the same direction,

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the angle between them is 0.

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And remember, that any two
vectors that are perpendicular

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have dot product 0.

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r hat and theta hat
are perpendicular.

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And so r hat dot r hat is 1.

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r hat dot theta hat is 0.

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And so we only get
minus GMP r squared dr.

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And that's the first
step in calculating

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our potential difference because
U of r final minus U of r

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initial by definition is
minus the work done in going

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from the initial state
to the final state, which

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are described by these
parameters r initial

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and r final of F
gravitation dot ds.

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And now we have,
actually, minus sign

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in the definition, another
minus sign coming from the dot

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product.

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And lets make this our
integration variable,

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r prime, r initial, r
prime equals r final.

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And there's going
to be a third minus

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sign because the integral of
dr r prime squared is minus 1

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over r prime.

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So there would be
3 minus signs-- one

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from the definition, one
from the scalar product,

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and one from the integration.

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And so we get r final minus r
initial equals 3 minus signs

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given overall minus
sign planet and planet

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1 over r final minus
1 over r initial.

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So this is the change in
gravitational potential energy

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as my small object goes
from some initial state

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to some final state.

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What about the
potential function?

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Where should we choose
our reference point?

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So our reference state here
is a little bit unusual.

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And it will be at infinity.

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And we'll choose as a potential
for our reference potential

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to be 0 at infinity.

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Imagine initial state very,
very far away from the planet.

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And we'll take us a final
state, an arbitrary state,

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just where the object is some
distance r from the planet.

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Now, we have to be a
little bit careful,

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because in this calculation
our final state must always

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be outside the planet.

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And the reasons
for that is subtle,

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but the gravitational
force inside the planet

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is no longer minus g
and 1 M2 over r-squared.

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So our analysis only applies
for r bigger than r planet.

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And if we put these values
for r and infinity in here,

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we get U of r minus the
potential at our reference

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point, which will be
0, equals minus GMNP.

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And here's the
reason why we choose

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that reference point, 1 over r
arbitrary state is just minus 1

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over r.

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But 1 over infinity, if
we use the initial state

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as the reference point,
1 over infinity is 0.

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So that's minus 0.

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And so what we get is that
the potential function

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is equal to our
reference potential--

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actually, minus GM mass
of the planet over r for r

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bigger than our reference.

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And that is the
potential function

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for the gravitational problem.

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But now remember, we're choosing
that to be our reference point.

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And so the real conclusion
is U of r equals minus GMN

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planet over r with U at
infinity equals to 0.

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Now, let's just finish
this by saying in words

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precisely what this means.

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So if I start my
system at infinity

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and I end my system a distance
r away from the planet,

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and I calculate the work done
by the gravitational force

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in going from infinity
to a distance r

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away from the planet-- I
take a minus that sign.

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So the negative of the
gravitational work--

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then that's what that
number corresponds to.