Math 32A, Lecture 3 (Rogawski): Notes on Kepler’s Laws
These notes contain some challenging mathematics. Read them care-
fully and try to put all the pieces together. Here are the headings:
1. Newton’s Second Law
2. A Simple but Important Result
3. Angular Momentum
4. Area Swept Out by the Radial Vector
5. The Planetary Differential Equation
6. Law of Equal Areas in Equal Times
7. The Velocity Circle
8. Grand Finale: The Law of Ellipses
9. Summary
1.
Newton’s Second Law
Let
r
(
t
) denote the path of a particle of mass
m
in three-dimensional
space. The velocity and acceleration vectors are
v
(
t
) =
r
′
(
t
)
,
a
(
t
) =
r
′′
(
t
)
When convenient, we drop the reference to
t
and write
r
,
v
,
a
, but it
is understood these are vector-valued functions of time.
Newton’s Second Law of Motion is often stated in scalar form:
F
=
ma
(force equals mass times acceleration).
However, both force and
acceleration are vectors, and the Law of Motion is true as a vector
equation:
F
=
m
a
2.
A simple but important result
Theorem.
For any vector-valued function
h
(
t
)
,
d
dt
h
(
t
)
×
h
′
(
t
) =
h
(
t
)
×
h
′′
(
t
)
Proof.
By the Product Rule for cross products,
d
dt
h
(
t
)
×
h
′
(
t
) =
h
′
(
t
)
×
h
′
(
t
)
bracehtipupleft
bracehtipdownrightbracehtipdownleft
bracehtipupright
this is zero
+
h
(
t
)
×
h
′′
(
t
) =
h
(
t
)
×
h
′′
(
t
)
Note that
h
′
(
t
)
×
h
′
(
t
) = 0 because
v
×
v
=
0
for any vector
v
.
square
1

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2
Actually, this result is not quite as simple as it seems. Suppose that
h
(
t
) =
(
x
(
t
)
, y
(
t
)
,
0
)
. Then
h
(
t
)
×
h
′
(
t
) =
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
i
j
k
x
y
0
x
′
y
′
0
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
= (
xy
′
−
x
′
y
)
k
and the derivative simplifies due to cancellation:
d
dt
h
(
t
)
×
h
′
(
t
) =
d
dt
(
xy
′
−
x
′
y
)
k
= (
xy
′′
+
x
′
y
′
−
x
′
y
′
−
xy
′′
)
k
= (
xy
′′
−
xy
′′
)
k
This is equal to
h
(
t
)
×
h
′′
(
t
) as claimed since
h
(
t
)
×
h
′′
(
t
) =
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
i
j
k
x
y
0
x
′′
y
′′
0
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
= (
xy
′′
−
x
′′
y
)
k
3.
Angular Momentum
The
angular momentum
of a moving particle is the vector
m
J
where
J
(
t
) =
r
(
t
)
×
r
′
(
t
)
Since
m
is fixed in our case, we focus on
J

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