homework 07 – VENNES, ROSS – Due: Oct 19 2007, 1:00 am
1
Question 1, chap 9, sect 1.
part 1 of 1
10 points
According to some nineteenthcentury geo
logical theories (now largely discredited), the
Earth has been shrinking as it gradually cools.
If so, how would
g
have changed over geo
logical time?
1.
It would decrease; the Earth’s radius is
decreasing.
2.
It would not change;
the mass of the
Earth remained the same.
3.
It would increase;
g
is inversely propor
tional to the square of the radius of the Earth.
correct
Explanation:
The acceleration
g
of gravity would increase
if the Earth shrank, but its mass stayed the
same, since
g
∝
1
r
2
.
Question 2, chap 9, sect 1.
part 1 of 2
10 points
Given:
G
= 6
.
67259
×
10
−
11
N m
2
/
kg
2
In Larry Niven’s science fiction novel
Ring
world
, a ring of material rotates about a star.
n
Star
F
g
The rotational speed of the ring is 1
.
38
×
10
6
m
/
s, and its radius is 1
.
3
×
10
11
m. The
inhabitants of this ring world experience a
normal contact force
vector
n
.
Acting alone, this
normal force would produce an inward accel
eration of 9
.
27 m
/
s
2
.
Additionally, the star
at the center of the ring exerts a gravitational
force on the ring and its inhabitants.
What is the total centripetal acceleration
of the inhabitants?
Correct answer: 14
.
6492 m
/
s
2
(tolerance
±
1
%).
Explanation:
The centripetal acceleration of the inhabi
tants is
a
c
=
v
2
r
=
(1
.
38
×
10
6
m
/
s)
2
1
.
3
×
10
11
m
= 14
.
6492 m
/
s
2
.
Question 3, chap 9, sect 1.
part 2 of 2
10 points
The difference between the total acceler
ation and the acceleration provided by the
normal force is due to the gravitational at
traction of the central star.
Calculate the approximate mass of the star.
Correct answer: 1
.
36242
×
10
33
kg (tolerance
±
1 %).
Explanation:
The difference Δ
a
between the total accel
eration and the acceleration provided by the
normal force is due to the gravitational at
traction of the central star and is equal to
Δ
a
=
a
c
−
a
i
.
On the other hand, this difference can be
expressed from Newton’s law of gravity
Δ
a
=
G M
r
2
,
where
M
is the mass of the central star. Using
the values for
a
c
and
a
i
, the mass of the star
is
M
=
r
2
Δ
a
G
=
r
2
(
a
c
−
a
i
)
G
= (1
.
3
×
10
11
m)
2
×
(14
.
6492 m
/
s
2
)
−
(9
.
27 m
/
s
2
)
6
.
67259
×
10
−
11
N m
2
/
kg
2
= 1
.
36242
×
10
33
kg
.
Question 4, chap 9, sect 99.
part 1 of 1
10 points
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homework 07 – VENNES, ROSS – Due: Oct 19 2007, 1:00 am
2
Given:
The radius of the earth is
R
Earth
=
6
.
37
×
10
6
m .
How high does a rocket have to go above the
Earth’s surface until its weight is 0
.
62 times
its weight on the Earth’s surface?
Correct answer: 1719
.
91
km (tolerance
±
1
%).
Explanation:
Basic concepts
By Newton’s Universal Law of Gravitation,
W
∝
1
r
2
Let :
n
0
.
62
.
Solution:
The radius of the Earth is
R
Earth
= 6
.
37
×
10
6
m
.
The rocket will be at a distance of
h
+
R
Earth
from the center of the Earth when it weighs
n W
. Thus
n W
W
=
1
(
h
+
R
Earth
)
2
1
R
2
Earth
=
R
2
Earth
(
h
+
R
Earth
)
2
,
so
R
2
Earth
=
n
(
h
+
R
Earth
)
2
R
2
Earth
n
= (
h
+
R
Earth
)
2
R
Earth
√
n
=
h
+
R
Earth
,
so
h
=
R
Earth
√
n
−
R
Earth
=
R
Earth
parenleftbigg
1
√
n
−
1
.
0
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 Spring '08
 Turner
 Energy, Force, Mass, Potential Energy, Work, General Relativity

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