homework 13 – FIERRO, JEFFREY – Due: Mar 3 2008, 11:00 pm
1
Question 1, chap 9, sect 4.
part 1 of 2
10 points
The period of the earth around the sun
is 1 year and it is at a distance 150 million
kilometers from the sun.
An asteroid in a
circular orbit around the sun is at a distance,
181 million kilometers from the sun.
What is the period,
T
a
, of the asteroid
orbit?
Correct answer: 1
.
3255
year (tolerance
±
1
%).
Explanation:
From Kepler’s laws,
T
2
e
r
3
e
=
T
2
a
r
3
a
T
a
=
parenleftbigg
r
a
r
e
parenrightbigg
3
2
T
e
=
parenleftbigg
181 million
150 million
parenrightbigg
3
2
1 year
= 1
.
3255 year
Question 2, chap 9, sect 4.
part 2 of 2
10 points
What is the orbital velocity of the asteroid?
Assume there are 365 days in one year.
Correct answer: 27206
.
4 m
/
s (tolerance
±
1
%).
Explanation:
v
a
=
2
π r
a
T
a
=
2
π
(1
.
81
×
10
11
m )
(1
.
3255 year)
=
2
π
(1
.
81
×
10
11
m )
(4
.
18011
×
10
7
s)
=27206
.
4 m
/
s
.
Question 3, chap 9, sect 5.
part 1 of 1
10 points
Given:
G
= 6
.
67259
×
10

11
N m
2
/
kg
2
Calculate the work required to move a
planet’s satellite of mass 1490 kg from a cir
cular orbit of radius 2
R
to one of radius 3
R
,
where 6
.
54
×
10
6
m is the radius of the planet.
The mass of the planet is 4
.
52
×
10
24
kg.
Correct answer: 5
.
72612
×
10
9
J (tolerance
±
1 %).
Explanation:
Applying the equation for the total energy,
E
, of a satellite in a circular orbit,
E
=

G M m
2
r
,
where
r
is the radius of the orbit, we obtain
for the total initial and final energies
E
i
=

G M m
4
R
,
E
f
=

G M m
6
R
.
Therefore, the work required to increase the
energy of the system is
W
=
E
f

E
i
=
G M m
12
R
= 6
.
67259
×
10

11
N m
2
/
kg
2
×
4
.
52
×
10
24
kg
×
1490 kg
12
×
6
.
54
×
10
6
m
= 5
.
72612
×
10
9
J
.
If we wish to determine how the energy is dis
tributed after the work is done on the system,
we find from the equation
1
2
m v
2
=
G M m
2
r
,
that the change in kinetic energy is
Δ
K
=

G M m
12
R
(it decreases), while the corresponding change
in potential energy is
Δ
U
=
G M m
6
R
(it increases).
Thus, the work done on the
system is
W
= Δ
K
+ Δ
U
=
G M m
12
R
,
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homework 13 – FIERRO, JEFFREY – Due: Mar 3 2008, 11:00 pm
2
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 Spring '08
 Turner
 Mass, Potential Energy, Work, Celestial mechanics

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