PHYSICS 171
AQ 2009
Solutions to Homework
#6
1. Giancoli Chapter 6, Problem 10
Assume that the two objects can be treated as point masses, with m1  m and
m2  4.00 kg  m. The gravitational force between the two masses is given by the
following.
2
m  4.00  m 
mm
11
2
2 4.00m  m
F  G 12 2  G

6.67

10
N

m
kg
 2.5  10 10 N


2
2
r
r
 0.25 m 
This can be rearranged into a quadratic form of m 2  4.00m  0.234  0 . Use the quadratic
formula to solve for m, resulting in two values which are the two masses.
m1  3.94 kg , m2  0.06 kg
2. Giancoli Chapter 6, Problem 12
With the assumption that the density of Europa is the same as Earth’s, the radius of Europa
can be calculated.
 Europa   Earth 
g Europa 
GM Europa

2
rEuropa
M Europa
4
3
3
 rEuropa

M Earth
4
3
GM Europa
1/ 3

 M Europa  
 rEarth 
 

M Earth  



3
 rEarth

2
1/ 3
 rEuropa
1/ 3
2/3
GM Europa
M Earth

2
rEarth
 M Europa 
 rEarth 

 M Earth 
1/ 3
GM Earth M Europa
2
rEarth
1/ 3
M Earth
1/ 3
 M Europa 

 M Earth 
 g Earth 
1/ 3
 4.9  1022 kg 
 5.98  1024 kg 



 9.80 m s 2 
 1.98 m s 2  2.0 m s 2
3. Giancoli Chapter 6, Problem 17
Each mass M will exert a gravitational force on mass m. The vertical components of the two
forces will sum to be 0, and so the net force on m is directed horizontally. That net force
will be twice the horizontal component of either force.
GMm
FMm  2

x 2  R2
R
x  R2
x

FMm x 
GMm
x
2
 R2
Fnet x  2 FMm x 


cos  
GMm
x
2
2GMmx
x
2
 R2

3/ 2
 R2

x
x2  R2

R
GMmx
x
2
 R2

3/ 2
x 2  R2


r
4. Giancoli Chapter 6, Problem 20
We can find the actual g by taking g due to the uniform Earth,
subtracting away g due to the bubble as if it contained uniform Earth
matter, and adding in g due to the oil-filled bubble. In the diagram, r
= 1000 m (the diameter of the bubble, and the distance from the
surface to the center of the bubble). The mass of matter in the bubble
is found by taking the density of the matter times the volume of the
bubble.
g oil  g uniform  g bubble  g bubble 
present
Earth
g  g oil
(Earth
matter)
rE  r
rE
(oil)
 g uniform  g bubble  g bubble
present
Earth

r2
(Earth
matter)
GM bubble
GM bubble
 oil 
(oil)


G

3
 2 M bubble  M bubble   2   oil   Earth  43  rbubble
r   oil 
r 
(Earth 
matter 
matter) 

(Earth
matter)

G
r2
The density of oil is given, but we must calculate the density of a uniform Earth.
m
5.98  1024 kg
 Earth  4 E 3 
 5.50  103 kg m 3
3
6
4
 rE 3   6.38  10 m 
matter
3
g 
G

3
 oil   Earth  43  rbubble

r 
matter 
2
 6.67  10 N m kg  8.0  10

1.00  10 m 
11

2
3
2
2
2
kg m3  5.50  103 kg m 3
  5.0  10 m 
4
3
2
3
 1.6414  104 m s2  1.6  104 m s2
Finally we calculate the percentage difference.
g
1.6414  104 m s2
 100  1.7  103%
% 
g
9.80 m s2
The negative sign means that the value of g would decrease from the uniform Earth
value.
5. Giancoli Chapter 6, Problem 30
The speed of an object in an orbit of radius r around the Earth is given in Example 6-6 by
v  G M Earth r , and is also given by v  2 r T , where T is the period of the object in
orbit. Equate the two expressions for the speed and solve for T. Also, for a “near-Earth”
orbit, r  REarth .
G
M Earth

r
T  2
2 r
 T  2
T
r3
GM Earth
 6.38 10 m 
 5070 s  84.5 min
N m kg  5.98  10 m 
3
6
3
REarth
 2
GM Earth
 6.67 10
11
2
2
24
No , the result does not depend on the mass of the satellite
6. Giancoli Chapter 6, Problem 38
Knowing the period of the Moon and the distance to the Moon, we can calculate the speed of
the Moon by v  2 r T . But the speed can also be calculated for any Earth satellite by
v  G M Earth r , as derived in Example 6-6. Equate the two expressions for the speed, and
solve for the mass of the Earth.
G M Earth r  2 r T
M Earth 
4 2 r 3
GT
2



4 2 3.84  108 m
 6.67 10
11
2
N m kg
2

3
  27.4 d 86, 400 s d 
2
 5.98  1024 kg
7. Giancoli Chapter 6, Problem 46
d
dN
F
(a) In a short time  t , the planet will travel a
v

t
N
distance vt along its orbit. That distance is
Sun
essentially a straight line segment for a short
time duration. The time (and distance moved)
during  t have been greatly exaggerated on the
diagram. Kepler’s second law states that the area
swept out by a line from the Sun to the planet during the planet’s motion for the  t is
the same anywhere on the orbit. Take the areas swept out at the near and far points, as
shown on the diagram, and approximate them as triangles (which will be reasonable for
short  t ).
 Area  N   Area  F

1
2
 vN t  d N  12  vF t  d F

vN vF  d F d N
(b) Since the orbit is almost circular, an average velocity can be found by assuming a
circular orbit with a radius equal to the average distance.
vavg 
2 r

2
1
2
dN  dF 

2
1
2
1.47  10
11
m  1.52  1011 m
  2.973  10
4
ms
T
T
3.16  10 s
From part (a) we find the ratio of near and far velocities.
vN vF  d F d N  1.52 1.47  1.034
For this small change in velocities (3.4% increase from smallest to largest), we assume
that the minimum velocity is 1.7% lower than the average velocity and the maximum
velocity is 1.7% higher than the average velocity.
7
vF t
vN  vavg 1  0.017   2.973  104 m s 1.017   3.02  10 4 m s
vF  vavg 1  0.017   2.973  104 m s  0.983  2.92  10 4 m s
8. Giancoli Chapter 7, Problem 16
Use Eq. 7.4 to calculate the dot product.


A B  Ax Bx  Ay By  Az Bz  2.0 x 2 11.0    4.0 x  2.5 x    5.0  0   22 x 2  10 x 2
 12 x 2
9. Giancoli Chapter 7, Problem 22
Both vectors are in the first quadrant, so to find the angle between them, we can simply subtract
the angles of each of them.


 2.0 N  2   4.0 N 2


1.0 m 2   5.0 m  2
F  2.0ˆi  4.0ˆj N  F 
d  1.0ˆi  5.0ˆj m  d 








20 N ; F  tan 1
 tan 1 2.0

W  Fd cos   
(b)
W  Fx d x  Fy d y   2.0 N 1.0 m    4.0 N  5.0 m   22 J

4.0
2.0
5.0
26 m ; d  tan 1
 tan 1 5.0
1.0
(a)

20 N  

26 m cos  tan 1 5.0  tan 1 2.0  22 J

10. Giancoli Chapter 7, Problem 32
To be perpendicular to the given vector means that the dot product will be 0. Let the
unknown vector be given as uˆ  ux ˆi  u y ˆj.


uˆ 3.0ˆi  4.0ˆj  3.0u x  4.0u y  u y =  0.75u x ; unit length  u x2  u 2y  1 
ux2  u 2y  ux2   0.75ux   1.5625ux2  1  ux  
2
1
1.5625
 0.8 , u y  0.6
So the two possible vectors are uˆ  0.8ˆi  0.6ˆj and uˆ  0.8ˆi  0.6ˆj .
Note that it is very easy to get a non-unit vector perpendicular to another vector in two
dimensions, simply by interchanging the coordinates and negating one of them. So a nonunit vector perpendicular to  3.0ˆi  4.0ˆj could be either  4.0ˆi  3.0ˆj or  4.0ˆi  3.0ˆj .
Then divide each of those vectors by its magnitude (5.0) to get the possible unit vectors.
11. Giancoli Chapter 7, Problem 54
We assume the train is moving 20 m/s (which is about 45 miles per hour), and that the
distance of “a few city blocks” is perhaps a half-mile, which is about 800 meters. First find
the kinetic energy of the train, and then find out how much work the web must do to stop the
train. Note that the web does negative work, since the force is in the OPPOSITE direction of
the displacement.
2
Wto stop  K  12 mv22  12 mv12  0  12 104 kg   20 m s   2  106 J
train
Wweb   kx  2  10 J  k 
1
2
2
6


800 m 
2 2  106 J
6N m
2
Note that this is not a very stiff “spring,” but it does stretch a long distance.
12. Giancoli Chapter 7, Problem 62
FT
(a) From the free-body diagram for the load being lifted, write Newton’s second law for
the vertical direction, with up being positive.
 F  FT  mg  ma  0.150mg 

FT  1.150mg  1.150  265 kg  9.80 m s
2
  2.99 10 N
(b) The net work done on the load is found from the net force.

Wnet  Fnet d cos 0o   0.150mg  d  0.150  265 kg  9.80 m s 2
 8.96  103 J
(c) The work done by the cable on the load is as follows.

Wcable  FT d cos 0o  1.150mg  d  1.15  265 kg  9.80 m s 2
(d) The work done by gravity on the load is as follows.

WG  mgd cos180o  mgd    265 kg  9.80 m s 2
(e)
  23.0 m 
  23.0 m   6.87 10 J
  23.0 m  
4
5.97  104 J
Use the work-energy theorem to find the final speed, with an initial speed of 0.
Wnet  K 2  K1  mv  mv
1
2
2
2
1
2
2
1
 v2 
2Wnet
m
v 
2
1
mg
3

2 8.96  103 J
265kg
  0  8.22 m s