We would now like to calculate the potential function
for a universal law of gravity.
So let's set up a coordinate system first.
And in this case, we're going to make a very special case--
although, the general result doesn't depend on that.
So suppose we have a planet of radius rp
and we have a small object.
And initially, our object is a distance r
away from the planet.
This is our initial state.
And the planet has mass to the planet.
And our small object has m.
And this object is moving.
Of course, orbits really aren't like that,
so we should be a little bit more careful.
But our orbit might be some type of hyperbolic section-- conic
section.
And the orbit over here is at some final position.
In our eyes, the distance from the object to the center
of the planet as our final is.
Now because there's a central point to this problem,
the natural coordinate system to use
is polar coordinates in the plane.
And let's just imagine that at this moment
we have the universal force of gravity.
Here's our object.
And it's being displaced a distance s.
So our coordinate system is polar.
Let's call r hat this way and theta hat that way.
And let's blow up our little displacement
in terms of our coordinate system,
just so that we see what's happening.
So let's imagine that our little ds
vector is pointing like this.
And our center point here, we have tge r hat pointing
radially outward and the theta hat
pointing towards the center.
Now to exaggerate this picture, if this is r,
then the arc length here is rd theta.
And the difference as the object moves from one
point here to a closer point towards the center, will be dr.
And so our ds vector, we can write as a radial piece r hat.
Now, I don't put a sign in there,
because only the end points of my integral
will tell me whether I was going towards the planet
or away from the planet.
And rd theta, theta hat.
And our gravitational force is minus
g mass mass of the planet over the distance squared r hat.
So now what we see is that when we
take the dot product of these two forces,
r hat dot r hat is 1 in polar coordinates.
Why is that the case, r hat dot r hat is 1?
Because these vectors are in the same direction,
the angle between them is 0.
And remember, that any two vectors that are perpendicular
have dot product 0.
r hat and theta hat are perpendicular.
And so r hat dot r hat is 1.
r hat dot theta hat is 0.
And so we only get minus GMP r squared dr.
And that's the first step in calculating
our potential difference because U of r final minus U of r
initial by definition is minus the work done in going
from the initial state to the final state, which
are described by these parameters r initial
and r final of F gravitation dot ds.
And now we have, actually, minus sign
in the definition, another minus sign coming from the dot
product.
And lets make this our integration variable,
r prime, r initial, r prime equals r final.
And there's going to be a third minus
sign because the integral of dr r prime squared is minus 1
over r prime.
So there would be 3 minus signs-- one
from the definition, one from the scalar product,
and one from the integration.
And so we get r final minus r initial equals 3 minus signs
given overall minus sign planet and planet
1 over r final minus 1 over r initial.
So this is the change in gravitational potential energy
as my small object goes from some initial state
to some final state.
What about the potential function?
Where should we choose our reference point?
So our reference state here is a little bit unusual.
And it will be at infinity.
And we'll choose as a potential for our reference potential
to be 0 at infinity.
Imagine initial state very, very far away from the planet.
And we'll take us a final state, an arbitrary state,
just where the object is some distance r from the planet.
Now, we have to be a little bit careful,
because in this calculation our final state must always
be outside the planet.
And the reasons for that is subtle,
but the gravitational force inside the planet
is no longer minus g and 1 M2 over r-squared.
So our analysis only applies for r bigger than r planet.
And if we put these values for r and infinity in here,
we get U of r minus the potential at our reference
point, which will be 0, equals minus GMNP.
And here's the reason why we choose
that reference point, 1 over r arbitrary state is just minus 1
over r.
But 1 over infinity, if we use the initial state
as the reference point, 1 over infinity is 0.
So that's minus 0.
And so what we get is that the potential function
is equal to our reference potential--
actually, minus GM mass of the planet over r for r
bigger than our reference.
And that is the potential function
for the gravitational problem.
But now remember, we're choosing that to be our reference point.
And so the real conclusion is U of r equals minus GMN
planet over r with U at infinity equals to 0.
Now, let's just finish this by saying in words
precisely what this means.
So if I start my system at infinity
and I end my system a distance r away from the planet,
and I calculate the work done by the gravitational force
in going from infinity to a distance r
away from the planet-- I take a minus that sign.
So the negative of the gravitational work--
then that's what that number corresponds to.