572
CHAPTER 14
CALCULUS OF VECTOR-VALUED FUNCTIONS
(ET CHAPTER 13)
SOLUTION
Kepler’s Third Law states that the period
T
of the orbit is given by:
T
2
=
Ã
4
π
2
GM
!
a
3
or
T
=
2
√
a
3
/
2
If
a
is increased four-fold the period becomes:
2
√
(
4
a
)
3
/
2
=
8
·
2
√
a
3
/
2
That is, the period is increased eight-fold.
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
Using the given data we obtain the following values of
T
2
/
a
3
,whe
re
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
)a
s
T
≈
√
a
3
·
3
·
10
−
4
≈
11
.
9 years.
2.
A satellite has initial position
r
=
h
1
,
000
,
2
,
000
,
0
i
and initial velocity
r
0
=
h
1
,
2
,
2
i
(units of kilometers and
seconds). Find the equation of the plane containing the satellite’s orbit.
Hint:
This plane is orthogonal to
J
.
The vectors
r
(
t
)
and
r
0
(
t
)
lie in the plane containing the satellite’s orbit, in particular the initial position
r
= h
1000
,
2000
,
0
i
and the initial velocity
r
0
= h
1
,
2
,
2
i
. Therefore, the cross product
J
=
r
×
r
0
is perpendicular to the
plane. We compute
J
:
J
=
r
×
r
0
=
¯
¯
¯
¯
¯
¯
ij
k
1
,
000
2
,
000
0
12
2
¯
¯
¯
¯
¯
¯
=
¯
¯
¯
¯
2
,
000
0
22
¯
¯
¯
¯
i
−
¯
¯
¯
¯
1
,
000
0
¯
¯
¯
¯
j
+
¯
¯
¯
¯
1
,
000
2
,
000
¯
¯
¯
¯
k
=
4
,
000
i
−
2
,
000
j
= h
4000
,
−
2000
,
0
i
We now use the vector form of the equation of the plane with
n
=
J
=
h
4000
,
−
2000
,
0
i
and
h
x
0
,
y
0
,
z
0
i
=
r
=
h
1000
,
2000
,
0
i
, to obtain the following equation:
h
4000
,
−
2000
,
0
i · h
x
,
y
,
z
i = h
4000
,
−
2000
,
0
i · h
1000
,
2000
,
0
i
2000
h
2
,
−
1
,
0
i · h
x
,
y
,
z
i =
2000
h
2
,
−
1
,
0
i · h
1000
,
2000
,
0
i
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
0
for the
earth (in units of squared feet per second)?