This preview shows pages 1–3. Sign up to view the full content.
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 DiFerential Equation
6. Law of Equal Areas in Equal Times
7. The Velocity Circle
8. Grand ±inale: The Law of Ellipses
9. Summary
1.
Newton’s Second Law
Let
r
(
t
) denote the path of a particle of mass
m
in threedimensional
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 vectorvalued 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
(
r
) =
m
a
When we write the law in this form, we are implicitly assuming that
the force depends only the particle’s position
r
and not on time. ±or
example, gravitational force depends on position but not time. If the
force did depend on time (for example, the force on an electron due to
a varying electric ²eld), it would be necessary to write
F
(
t,
r
) in place
of
F
(
r
).
2.
A simple but important result
Theorem 1.
For any vectorvalued function
h
(
t
)
,
d
dt
h
(
t
)
×
h
′
(
t
) =
h
(
t
)
×
h
′′
(
t
)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document 2
Proof.
By the Product Rule for cross products,
d
dt
h
(
t
)
×
h
′
(
t
) =
h
′
(
t
)
×
h
′
(
t
)
b
B±
²
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
.
s
Actually, this result is not quite as simple as it seems. Suppose that
h
(
t
) =
a
x
(
t
)
, y
(
t
)
,
0
A
. Then
h
(
t
)
×
h
′
(
t
) =
v
v
v
v
v
v
i
j k
x
y
0
x
′
y
′
0
v
v
v
v
v
v
= (
xy
′
−
x
′
y
)
k
and the derivative simpliFes 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
) =
v
v
v
v
v
v
i
j
k
x
y
0
x
′′
y
′′
0
v
v
v
v
v
v
= (
xy
′′
−
x
′′
y
)
k
3.
Angular Momentum
The
angular momentum
of a moving particle is the vector
m
J
where
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 08/19/2009 for the course MATH 32A taught by Professor Gangliu during the Spring '08 term at UCLA.
 Spring '08
 GANGliu
 Math, Multivariable Calculus

Click to edit the document details