572
C H A P T E R
14
CALCULUS OF VECTORVALUED FUNCTIONS
(ET CHAPTER 13)
SOLUTION
Kepler’s Third Law states that the period
T
of the orbit is given by:
T
2
=
4
π
2
GM
a
3
or
T
=
2
π
√
GM
a
3
/
2
If
a
is increased fourfold the period becomes:
2
π
√
GM
(
4
a
)
3
/
2
=
8
·
2
π
√
GM
a
3
/
2
That is, the period is increased eightfold.
Exercises
1.
Kepler’s Third Law states that
T
2
/
a
3
has the same value for each planetary orbit. Do the data in the following table
support this conclusion? Estimate the length of Jupiter’s period, assuming that
a
=
77
.
8
×
10
10
m.
Planet
Mercury
Venus
Earth
Mars
a
(10
10
m)
5.79
10.8
15.0
22.8
T
(years)
0.241
0.615
1.00
1.88
SOLUTION
Using the given data we obtain the following values of
T
2
/
a
3
, where
a
, as always, is measured not in
meters but in 10
10
m:
Planet
Mercury
Venus
Earth
Mars
T
2
/
a
3
2
.
99
·
10
−
4
3
·
10
−
4
2
.
96
·
10
−
4
2
.
98
·
10
−
4
The data on the planets supports Kepler’s prediction. We estimate Jupiter’s period (using the given
a
) as
T
≈
√
a
3
·
3
·
10
−
4
≈
11
.
9 years.
2.
A satellite has initial position
r
=
1
,
000
,
2
,
000
,
0
and initial velocity
r
=
1
,
2
,
2
(units of kilometers and
seconds). Find the equation of the plane containing the satellite’s orbit.
Hint:
This plane is orthogonal to
J
.
SOLUTION
The vectors
r
(
t
)
and
r
(
t
)
lie in the plane containing the satellite’s orbit, in particular the initial position
r
=
1000
,
2000
,
0 and the initial velocity
r
=
1
,
2
,
2 . Therefore, the cross product
J
=
r
×
r
is perpendicular to the
plane. We compute
J
:
J
=
r
×
r
=
i
j
k
1
,
000
2
,
000
0
1
2
2
=
2
,
000
0
2
2
i
−
1
,
000
0
1
2
j
+
1
,
000
2
,
000
1
2
k
=
4
,
000
i
−
2
,
000
j
=
4000
,
−
2000
,
0
We now use the vector form of the equation of the plane with
n
=
J
=
4000
,
−
2000
,
0
and
x
0
,
y
0
,
z
0
=
r
=
1000
,
2000
,
0 , to obtain the following equation:
4000
,
−
2000
,
0
·
x
,
y
,
z
=
4000
,
−
2000
,
0
·
1000
,
2000
,
0
2000 2
,
−
1
,
0
·
x
,
y
,
z
=
2000 2
,
−
1
,
0
·
1000
,
2000
,
0
2
x
−
y
=
2
,
000
−
2
,
000
+
0
=
0
2
x
−
y
=
0
The plane containing the satellite’s orbit is, thus:
P
= {
(
x
,
y
,
z
)
:
2
x
−
y
=
0
}
3.
The earth’s orbit is nearly circular with radius
R
=
93
×
10
6
miles (the eccentricity is
e
=
0
.
017). Find the rate at
which the earth’s radial vector sweeps out area in units of ft
2
/
s. What is the magnitude of the vector
J
=
r
×
r
for the
earth (in units of squared feet per second)?
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
S E C T I O N
14.6
Planetary Motion According to Kepler and Newton
(ET Section 13.6)
573
SOLUTION
The rate at which the earth’s radial vector sweeps out area is
d A
dt
=
1
2
J
;
J
=
r
(
t
)
×
r
(
t
)
(1)
Since
J
is a constant vector, its length is constant. Moreover, if we assume that the orbit is circular then
r
(
t
)
lies on a
circle, and therefore
r
(
t
)
and
r
(
t
)
are orthogonal. Using properties of the cross product we get:
J
=
r
(
t
)
×
r
(
t
)
=
r
(
t
)
r
(
t
)
=
R
r
(
t
)
=
const
We conclude that the speed
v
=
r
(
t
)
is constant. We find the speed using the following equality:
2
π
R
=
v
T
⇒
v
=
2
π
R
T
.
Therefore,
J
=
R
·
2
π
R
T
=
2
π
R
2
T
.
Substituting in (1) we get:
d A
dt
=
1
2
·
2
π
R
2
T
=
π
R
2
T
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '09
 Park
 Multivariable Calculus, Planet, Kepler's laws of planetary motion, Celestial mechanics

Click to edit the document details