In this part, we will begin
to look closely at the geometry of satellite orbits as a preparation for the
discussion in the next Chapter: Curves in Art and Nature of the geometry
of plane curves. Newton's analysis is based on the very specific geometry that
is implied by his gravitational law. That analysis will be the Art by
means of which the Calculus will imitate Nature.
Geosynchronous Satellite Orbit
We have actually already solved
the problem: "What is the altitude of a geosynchronous satellite?"
We did this from energy considerations. We solved it when we studied the orbit
of the space station in Docking with a Satellite. The satellite moves
in a circle, so its distance from the center of the Earth is constant. The space
station also moved uniformly in a circle. We saw:
The radius of the Earth is
roughly 
.
And according to Newton's law, there is a universal constant G
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with 
such that if 
represents the distance of our geosynchronous satellite from the center of the
Earth, then the acceleration due to gravity is:
measured in 
Thus, the acceleration is
radial and must be due entirely to gravitation,
so
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Now for the geosynchronous
satellite's circular orbit, 
are constant, so actually,
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The constant speed of the satellite is thus
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so we can conclude that
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This establishes a relationship
between 
and 
for such orbits. In particular, for each radius 
,
there is only one speed 
that gives a circular orbit. Now, if the satellite is to make one revolution
each 24 hours, that is, to be in geocentric orbit, there is another relationship
between 
and 
. It must travel a distance 
in 24 hours.
Since there are 86400 seconds in 24 hours, the new equation says that
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Let's put these equations
together.
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So, measured in Kilometers,
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This determines the unique
radius for a geosynchronous orbit. Now the altitude above the surface of the
Earth is just
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Considering that the shuttle
orbits at approximately 300 km, this satellite has a long way to climb. This
satellite must be roughly 
the distance to the Moon!
Now the constructions of this
Chapter will lead us directly to the idea of plane curvature in the next Chapter:
Curves in Art and Nature. We will develop the machinery and the language
for describing the curvature of curves in that chapter, but will give a brief
preview here.
Suppose you swing a weight
of mass 1 kg tethered to a rope in a circle. If the weight moves in a circular
path, say of radius R, at a uniform (constant) speed V, then the
rope must exert a force that causes the weight to accelerate towards the center
of the circle at each point of its path. The direction of the acceleration is
always toward the center of the circle. We have shown that this acceleration
towards the center must be
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It is called the centripetal
acceleration for uniform circular motion. Let's use Cartesian coordinates
to see why this is true.
The path that this object
will follow is a curve in the plane:
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The uniform angular velocity 
guarantees that the actual speed of the weight will be V because the
velocity V is given by
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where 
is the constant rate at which the angle (measured in radians) changes with time.
The larger the radius, the more slowly you must turn the rope to achieve the same speed V.
Now given a curve
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the velocity V is defined to be the curve
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and the acceleration is defined to be the curve
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It is easy to see that the acceleration for
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is a curve is also a circle whose radius is 
since 
In fact, this acceleration
is the curve
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The radius of the acceleration curve is the measure of the centripetal (inward) acceleration.
So much
for curves that are circles traversed uniformly (equal angles in equal times).
Let us return to the general Cartesian curve |
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with the single assumption:
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It happens that at each point
along such a curve there is a unique circle 
called the "osculating circle" that is tangent to the curve at the
point 
, and that has the following interesting property. We will introduce and discuss
"velocity moving frames" in detail in the next Chapter, but for now,
the following description should suffice.
Let us construct a "velocity
moving frame" at 
,
with unit vectors 
,
where 
is a vector of length 1 tangent to c at 
and 
is
obtained by rotating 
by 90 degrees. Let 
be the velocity vector
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and let 
be its length
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and let 
be the acceleration vector
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and let 
be its length
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then if we calculate the 
coordinate of the vector
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that is if,
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in that "velocity moving frame" then this component is 
.
If 
then 
is the radius of the osculating circle . The remarkable fact is that it does
not matter how the curve c is parametrized (as long as 
never vanishes). A different parametrization of c, say
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would yield the same circle at the point 
where 
.
The number 
will be called the plane curvature of the curve c at 
in the next Chapter. If 
is 0 then of course there is no osculating circle, and we say the curve is flat
at 
.
As you can see from the picture
below, the osculating circle "kisses" the curve at 
in the sense that it gives a very good approximation to it.
The red vector is 
and
its projection on 
is 
for which 
is the curvature and 
is the radius of the osculating circle 
.
Now imagine that a particle
was moving with uniform speed 
around this osculating circle at the instant 
. It would feel the centripetal acceleration towards the center of the circle
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This is the projection of
the actual acceleration 
on 
. The osculating circle, and in fact, the curvature of c at 
give a way to visualize the motion along the curve instantaneously as
though it was uniform circular motion. It gives an approximation that is "better"
than the tangent line approximation.
Exploration: Plane curvature and the osculating circle
Now the experiment on this
page may serve as preparation for our discussion of plane curves in the next
Chapter.
Begin by defining your curve 
by giving its component functions
Specify the smallest and largest values for t in the boxes
You will then be able to select
a point on your curve by clicking in the window at the bottom of the screen
and dragging the pointer to the desired position. As you do, a sprite will move
along the curve.
At the same time, the current values for the point will be continually displayed at the top of the window.
Finally, press the 
button to draw the
osculating circle. You will see something like:
The projection of the light
red vector on the dark red one gives the curvature (here, roughly 0.5631). This
projection when multiplied by 
is the acceleration a particle would feel if it moved uniformly about
the circle with speed 
.
The red vector when multiplied by 
is the actual acceleration at that point:
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