oldmidterm 04 – FRENNEA, KYLE – Due: Apr 28 2007, 4:00 am
1
Question 1
part 1 of 1
10 points
Given:
G
= 6
.
6726
×
10
−
11
N m
2
/
kg
2
Three 3 kg masses are located at points in
the xy plane as shown in the figure.
36 cm
55 cm
What is the magnitude of the resultant
force (caused by the other two masses) on
the mass at the origin?
Correct answer: 5
.
04111
×
10
−
9
N (tolerance
±
1 %).
Explanation:
Let :
m
= 3 kg
x
= 36 cm
,
and
y
= 55 cm
.
Basic Concepts:
Newton’s Law of Grav
itation:
F
g
=
G
m
1
m
2
r
2
We calculate the forces one by one, and
then add them using the superposition prin
ciple.
The force from the mass on the right
is pointing in the
x
direction, and has magni
tude
f
1
=
G
m m
x
2
=
G m
2
x
2
=
(6
.
6726
×
10
−
11
N m
2
/
kg
2
)(3 kg)
2
(0
.
36 m)
2
= 4
.
63375
×
10
−
9
N
.
The other force is pointing in the
y
direction
and has magnitude
f
2
=
G
m m
y
2
=
G m
2
y
2
=
(6
.
6726
×
10
−
11
N m
2
/
kg
2
)(3 kg)
2
(0
.
55 m)
2
= 1
.
98524
×
10
−
9
N
.
f
2
f
1
F
θ
Now we simply add the two forces, using
vector addition. Since they are at right angles
to each other, however, we can use Pythago
ras’ theorem as well:
F
=
radicalBig
f
2
1
+
f
2
2
=
bracketleftBig
(4
.
63375
×
10
−
9
N)
2
+ (1
.
98524
×
10
−
9
N)
2
bracketrightBig
1
2
=
5
.
04111
×
10
−
9
N
.
As you see, this force is very small. If the
masses in the picture are standing on a table
(the view being from above), typically the
force of static friction will not be overcome.
In agreement with common sense, then, the
masses will not move.
Note:
The angle
θ
shown in the figure is
θ
= arctan
bracketleftbigg
f
2
f
1
bracketrightbigg
= arctan
bracketleftbigg
(1
.
98524
×
10
−
9
N)
(4
.
63375
×
10
−
9
N)
bracketrightbigg
= 23
.
1917
◦
.
Question 2
part 1 of 1
10 points
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oldmidterm 04 – FRENNEA, KYLE – Due: Apr 28 2007, 4:00 am
2
Planet X has five times the diameter and
three times the mass of the earth.
What is the ratio of gravitational accelera
tion at the surface of planet X to the gravita
tional acceleration at the surface of the Earth,
g
x
g
e
?
1.
g
x
g
e
=
2
9
2.
g
x
g
e
=
3
25
correct
3.
g
x
g
e
=
5
36
4.
g
x
g
e
=
4
25
5.
g
x
g
e
=
5
9
6.
g
x
g
e
=
6
49
7.
g
x
g
e
=
9
49
8.
g
x
g
e
=
1
7
9.
g
x
g
e
=
2
49
10.
g
x
g
e
=
3
16
Explanation:
Let :
M
X
= 3
M
e
,
R
X
= 5
R
e
.
Because
m g
e
=
G M
e
m
R
2
e
,
where
M
e
is the mass of the earth and
R
e
is the radius.
So that
g
e
=
G M
e
R
2
e
.
Similar
equation applies to the planet X
g
x
=
G M
x
R
2
x
.
The ratio between the surface gravitational
accelerations is
g
x
g
e
=
M
x
R
2
e
M
e
R
2
x
=
3
M
e
R
2
e
M
e
(5
R
e
)
2
=
3
25
.
Question 3
part 1 of 1
10 points
An astronaut weighs 147 N on the Moon’s
surface.
He is in a circular orbit about the
Moon at an altitude of twoninth the moon’s
radius.
What gravitational force does the Moon
exert on him?
Correct answer: 98
.
405 N (tolerance
±
1 %).
Explanation:
Let :
h
=
2
9
R
moon
,
R
astronaut
=
parenleftbigg
1 +
2
9
parenrightbigg
R
moon
,
W
=
m g
= 147 N
.
The astronaut’s weight on the Moon is
w
=
m g
moon
= 147 N
,
where the gravitational acceleration
g
moon
on
the Moon’s surface is
g
moon
=
G
M
moon
R
2
moon
.
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 Spring '08
 Turner
 Physics, Mass, General Relativity, Wavelength, Correct Answer, Standing wave

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