Osculating Circles and Trajectories

1
Osculating Circles and Trajectories
Just Kidding
2
Osculating Circles and Trajectories
M
q
r
m
v
Mechanical Energy
3
Circular Orbit
How do we get the mass, M, of the gravitating
object? Apply Newtons Second Law to the mass,
mltltM
C
r
4
General Orbits, Conic Sections
parabola, e1
hyperbola, egt1
ellipse, 0ltelt1
r
circle, e0
q
In polar coordinates
5
General Orbits, Conic Sections, polar unit vector
notation
In terms of unit vector notation we have
Differentiating the position vector with respect
to time we have the velocity
or,
and,
In polar coordinates
so,
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General Orbits, Conic Sections, polar unit vector
notation
Differentiating the velocity vector with respect
to time we have the acceleration
or,
or,
or,
In polar coordinates
so,
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General Orbits, Conic Sections, polar unit vector
notation
Recall that
And note that
Then we get for the acceleration,
In polar coordinates
Combining terms in unit vectors,
8
General Orbits, Conic Sections, normal and
tangential unit vector notation
In polar coordinates
In terms of unit vector notation we have
Differentiating the velocity vector with respect
to time we have the acceleration
Note that
or,
and,
C
so,
and
so,
9
General Orbits, Conic Sections
We now have the acceleration in normal and
tangential unit vector notation
In polar coordinates
Compare this with the acceleration in polar
coordinate unit vector notation
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General Orbits, Conic Sections, continued
parabola, e1
hyperbola, egt1
When q 0, we have
ellipse, 0ltelt1
circle, e0
v
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General Orbits, Conic Sections, continued part 2
At the minimum distance from the massive body the
acceleration is purely radial or normal depending
upon your point of view.
When q 0, we have
Lets write the acceleration at q 0 for each
basis.
becomes
v
and
becomes
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General Orbits, Conic Sections, continued part 3
In the r,q basis we have
In the t,n basis we have
we get
Because
13
The Osculating Circle at closest approach
We have
and because
C
we get
v
Lets obtain
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Polar coordinates
Lets re-write this as
And take two successive derivatives with respect
to time to obtain
Now let q 0 to get
which now gives
or
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The Osculating Circle at closest approach
We now know
so
C
and
v
or
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The Osculating Circle at closest approach
continued
Compare this with the polar form of a conic
section
Clearly
C
The semi-latus rectum for conic sections is
nothing more than the radius of the osculating
circle at the distance of closest approach.
v
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Polar Coordinates
The equation for a conic section could now be
written as
Apply Newtons second law to m
M
C
m
Solving for M we get
v
or
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General Orbits, Conic Sections
parabola, e1
hyperbola, egt1
When q 0, we have
ellipse, 0ltelt1
circle, e0
M
Cp
Ch
Ce
Cc
v