Morby, Grant – Homework 27 – Due: Apr 14 2006, noon – Inst: Drummond
1
This
printout
should
have
11
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
A newly discovered planet has twice the mass
of the Earth, but the acceleration due to grav
ity on the new planet’s surface is exactly the
same as the acceleration due to gravity on the
Earth’s surface.
The radius
R
p
of the new planet in terms of
the radius
R
of Earth is
1.
R
p
=
√
2
R
.
correct
2.
R
p
= 2
R
.
3.
R
p
= 4
R
.
4.
R
p
=
1
2
R
.
5.
R
p
=
√
2
2
R
.
Explanation:
From Newton’s second law and the law of
universal gravitation, the gravitational force
near the surface is
F
g
=
m g
=
G
M m
r
2
g
=
G M
r
2
.
Now
M
p
= 2
M
e
and
g
p
=
g
e
, so
G M
e
R
2
=
G M
p
R
2
p
=
2
G M
e
R
2
p
1
R
2
=
2
R
2
p
R
p
=
√
2
R .
002
(part 1 of 1) 10 points
Two identical stars, a fixed distance
D
apart,
revolve in a circle about their mutual center
of mass, as shown below. Each star has mass
M
and speed
v
.
G
is the universal gravitational constant.
D
v
v
M
M
Which of the following is a correct relation
ship among these quantities?
1.
v
2
=
G M
2
D
correct
2.
v
2
=
G M
D
2
3.
v
2
=
4
G M
2
D
4.
v
2
=
G M
D
5.
v
2
=
M G D
6.
v
2
=
2
G M
D
7.
v
2
=
2
G M
2
D
8.
v
2
=
4
G M
D
Explanation:
From circular orbital movement, the cen
tripetal acceleration is
a
=
v
2
D
2
.
Using the Newton’s second law of motion,
we know the acceleration is
a
=
F
M
, where
F
is the force between two stars and is totally
supplied by the universal force.
So we obtain
2
v
2
D
=
a
=
F
M
=
G M
D
2
=
⇒
v
2
=
G M
2
D
.
003
(part 1 of 1) 10 points
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Morby, Grant – Homework 27 – Due: Apr 14 2006, noon – Inst: Drummond
2
How much less would you weigh on the top
of Mount Everest (elevation 8850 m) than at
sea level?
1.
0.5%
2.
5%
3.
0.3%
correct
4.
2%
5.
1.5%
6.
3%
7.
0.2%
8.
0.4%
9.
0.1%
Explanation:
Let :
M
E
= 6
×
10
24
kg
,
R
E
= 6
.
4
×
10
6
m
,
h
= 8850 m
,
and
G
= 6
.
67
×
10

11
N
·
m
2
/
kg
2
.
At sea level,
g
=
G M
E
R
2
E
=
(6
.
67
×
10

11
N
·
m
2
/
kg
2
) (6
×
10
24
kg)
(6
.
4
×
10
6
m)
2
= 9
.
77051 m
/
s
2
.
At the top of Mount Everest,
g
0
=
G M
E
(
R
E
+
h
)
2
=
(6
.
67
×
10

11
N
·
m
2
/
kg
2
) (6
×
10
24
kg)
(6
.
4
×
10
6
m + 8850 m)
2
= 9
.
74354 m
/
s
2
.
Since
W
=
m g ,
then
Δ
W
W
=
g

g
0
g
= 1

g
0
g
=
1

9
.
74354 m
/
s
2
9
.
77051 m
/
s
2
¶
×
100%
=
0
.
27599%
.
004
(part 1 of 1) 10 points
A
m
= 78
.
7 kg object is released from rest at
a distance
h
= 0
.
624272 R above the Earth’s
surface, where
R
is the radius of the earth.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Find the speed of the object when it strikes
the Earth’s surface. (Neglect any atmospheric
friction.
G
= 9
.
8 m
/
s
2
.
For the Earth,
R
= 6
.
38
×
10
6
m,
M
= 5
.
98
×
10
24
kg.
It
implies that the gravitational acceleration at
the surface of the earth is
g
= 9
.
8
m/s
2
.
)
Caution:
You must take into account that
the gravitational acceleration depends on dis
tance between the object and the center of the
earth.
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 Spring '09
 KLEINMAN
 Physics, Mass, Work, General Relativity, Standard gravity

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