576
CHAPTER 14
CALCULUS OF VECTORVALUED FUNCTIONS
(ET CHAPTER 13)
4
π
2
R
2
v
2
=
4
2
R
3
GM
1
v
2
=
R
⇒
M
=
R
v
2
G
13. Mass of the Milky Way
The sun revolves around the center of mass of the Milky Way galaxy in an orbit that
is approximately circular, of radius
a
≈
2
.
8
×
10
17
km and velocity
v
≈
250 km
/
s. Use the result of Exercise 12 to
estimate the mass of the portion of the Milky Way inside the sun’s orbit (place all of this mass at the center of the orbit).
SOLUTION
Let
M
be the mass of the portion of the Milky Way inside the sun’s orbit, assuming that all this mass is at
the center of the sun’s orbit. By Exercise 12, the following equality holds:
M
=
a
v
2
G
.
We substitute the values
a
=
2
.
8
×
10
20
m,
v
=
250
×
10
3
m
/
sand
G
=
6
.
673
×
10
−
11
m
3
kg
−
1
s
−
1
and compute the
mass
M
.Thisgives:
M
=
2
.
8
·
10
20
·
(
250
·
10
3
)
2
6
.
673
·
10
−
11
=
2
.
6225
×
10
41
kg
.
The mass of the sun is 1
.
989
×
10
30
kg, hence
M
is 1
.
32
×
10
11
times the mass of the sun (132 billions times the mass
of the sun).
14. Conservation of Energy
The total mechanical energy (kinetic plus potential) of a planet of mass
m
orbiting a sun
of mass
M
with position
r
and speed
v
=k
r
0
k
is
E
=
1
2
m
v
2
−
GMm
k
r
k
11
Use (2) and (9) to show that
E
is conserved, that is,
dE
dt
=
0.
We differentiate the total mechanical energy
E
with respect to the time
t
, obtaining:
=
1
2
m
·
2
v
d
v
−
d
±
1
k
r
k
¶
=
m
k
r
0
k
d
k
r
0
k+
1
k
r
k
2
d
k
r
k
=
nm
k
r
0
k
d
k
r
0
k
r
k
2
d
k
r
k
(1)
By formula (9),
d
k
r
k=
r
·
r
0
k
r
k
. Substitution into (1) gives (for
k
=
):
=
m
k
r
0
k
d
k
r
0
k
r
k
2
r
·
r
0
k
r
k
=
m
k
r
0
k
d
k
r
0
k
r
k
3
r
·
r
0
=
m
k
r
0
k
d
k
r
0
±
m
k
k
r
k
3
r
¶
·
r
0
(2)
By formula (2),
r
00
=−
k
k
r
k
3
r
, hence:
=
m
k
r
0
k
d
k
r
0
k−
m
r
00
·
r
0
(3)
Using formula (9) for
r
0
in place of
r
gives:
d
k
r
0
r
0
·
r
00
k
r
0
k
(4)
We substitute (4) into (3) to obtain:
=
m
k
r
0
k
r
0
·
r
00
k
r
0
k
−
m
r
00
·
r
0
=
m
r
0
·
r
00
−
m
r
00
·
r
0
=
0
15.
Show that the total energy (11) of a planet in a circular orbit of radius
R
is
E
/(
2
R
)
.
Hint:
Use Exercise
8.
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View Full DocumentSECTION
14.6
Planetary Motion According to Kepler and Newton
(ET Section 13.6)
577
SOLUTION
The total energy of a planet in a circular orbit of radius
R
is
E
=
1
2
m
v
2
−
GMm
k
r
k
=
1
2
m
v
2
−
R
(1)
In Exercise 8 we showed that
v
2
=
GM
R
(2)
Substituting (2) in (1) we obtain:
E
=
1
2
m
R
−
R
=−
1
2
R
2
R
.
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 Winter '08
 GANGliu
 Kepler's laws of planetary motion, rper

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