Download chapter 12 blm answer key

CHAPTER 12 BLM ANSWER KEY
BLM 12-2: Newton’s Law of Universal
Gravitation/Reinforcement
BLM 12-4: The Orbital Speed of
Planets/Reinforcement
Answers
Answers
m1m2
, where Fg is the force of gravity, G is
r2
the universal gravitational constant, m1 is the mass of the
Sun, m2 is the mass of the planet, and r is the orbital radius
of the planet, gives the following answers.
Equating the centripetal force equation, Fc 
Using Fg  G
Sun-Mercury: 1.30  1022 N
Sun-Saturn:
3.68  1022 N
Sun-Venus:
5.50  1022 N
Sun-Uranus:
1.42  1021 N
Sun-Earth:
3.58  1022 N
Sun-Neptune: 6.75  1020 N
Sun-Mars:
1.64  1021 N
Sun-Pluto:
Sun-Jupiter:
4.17  1023 N
mv 2
, and
r
mm
Newton’s law of universal gravitation, Fg  G 1 2 2 ,
r
where m1 is the mass of the Sun, m2 is the mass of the
planet, and r is the orbital radius of the planet, gives
m
v  G 1 . Solving this expression for each planet gives
r
the orbital speed of that planet.
4.96  1016 N
Mercury:
4.79 104 m/s
Saturn:
9.64  103 m/s
Venus:
3.51  104 m/s
Uranus:
6.80  103 m/s
BLM12-3: The Acceleration Due to
Gravity/Reinforcement
Earth:
2.99  104 m/s
Neptune:
5.43  103 m/s
Mars:
2.41  104 m/s
Pluto:
4.7  103 m/s
Answers
Jupiter:
1.31  104 m/s
m1m2
, where m1 is the mass of the planet,
r2
m
m2 is the mass of an object, and Fg  mg , gives g  G 21 .
r
Solving this equation for each planet gives the acceleration
due to gravity close to the surface of that planet.
Equating Fg  G
Mercury: 3.31 m/s2
Saturn:
10.4 m/s2
Venus:
8.09 m/s2
Uranus:
8.24 m/s2
Earth:
9.80 m/s2*
Neptune: 11.2 m/s2
Mars:
Jupiter:
3.75
m/s2
24.6
m/s2
Pluto:
0.72
BLM 12-5: Chapter 12 Test/Assessment
Answers
1.
(a)
7.
Brahe believed in an Earth-centred universe in which
all planets other than Earth orbited the Sun. In
formulating his laws of planetary motion, Kepler
applied Brahe’s data to the motions of planets in a
Sun-centred system.
8.
During the period between July and December, Earth
is moving closer to the Sun, and as any planet moves
closer to the Sun, its orbital speed increases.
m/s2
*The value of 9.81 m/s2 that is traditionally used for the acceleration
due to gravity on Earth is obtained when the value is calculated from
data that have more significant figures than those given in the Table
of Planetary Data in BLM 12-1.
2.
(b)
3.
(b)
4.
(a)
5.
(b)
6.
(b)
9. (a)
If the mass of one object doubled, the gravitational
force between the two bodies would be doubled.
(b)
If the distance between two bodies was doubled,
the gravitational force between them would be one
fourth as great.
10.
To find a planet’s period using Kepler’s third law, you
need to know the period of another planet and the
orbital radii of both planets.
11.
The surface gravity of each planet varies inversely
with the square of the radius of the planet. Jupiter’s
radius is much greater (almost 11 times) than
Earth’s radius.
CHAPTER 12 BLM ANSWER KEY
12.
The satellite moves with uniform circular motion, so
the curvature of Earth’s surface exactly matches the
curvature of the trajectory of the satellite. Earth’s
surface “falls away” from the satellite at the same rate
that the satellite falls toward Earth.
13.
The force of attraction will be greatest between the
spheres in (c) set A and C.
16.
The spheres would have to be 11.3 m apart to main
the gravitational force between them.
Gm1m2
r
Fg
(6.67  1011 Nkgm2 )(40.0 kg)(20.0 kg)
2
B
mm
Fg  1 2 2
r
(32 kg)(25 kg)
Fg 
(20 cm) 2
mm
Fg  1 2 2
r
(80 kg)(20 kg)

(40 cm) 2
2.0 kg 2
cm 2

4.17  1010 N
r  11.3 m
A
Fg 
r
17.
The distance between the Sun and Mars is 1.52 AU.
2
 TMars   rMars 

 

 TEarth   rEarth 
2 
 
T 
rMars  rEarth  3  Mars  
  TEarth  


1.0 kg 2
cm 2
C
mm
Fg  1 2 2
r
(90 kg)(10 kg)

(15 cm) 2
4.0 kg 2

cm 2
14.
rMars
18.
Fg  1.67  10
9
N
The satellite is 1.84 108 m from Earth.
2
 Tsatellite   rsatellite 

 

 TMoon   rMoon 
The gravitational force between spheres is
4.17 1010 N.
mm
Fg  G 1 2 2
r

N  m 2  (20 kg)(20 kg)
Fg   6.67  1011

kg 2  (8.00 m) 2

The force between the spheres will be 1.67 109 N.
mm
Fg  G 1 2 2
r

N  m 2  (20 kg)(20 kg)
Fg   6.67  1011

kg 2  (4.00 m) 2

  686 d 2 

 1 AU  3 
  365 d  


rMars  1.52 AU
rsatellite
3
2
 
 
T
satellite

 rMoon 3 
  TMoon  


  9.1d  2 

rsatellite  3.84  108 m  3 
  27.3d  


8
rsatellite  1.84  10 m
Fg  4.17  1010 N
15.
3
19.
The mass of the larger sphere is 50 kg.
Fg r 2
m2 
Gm1
m2
 2.50  10

 6.67 10
8
N  1.05 m 
11 N m 2
kg 2
m2  5.0  101 kg
2
 8.2 kg 
CHAPTER 12 BLM ANSWER KEY
20. (a)
The period of the object is 32 years.
2
3
 Tobject   robject 

 

 TEarth   rEarth 
 robject 


 rEarth 
3
 10 AU 
1 a 

 1 AU 
 32 a
3
Tobject  TEarth
Tobject
Tobject
(b)
The speed of the object is 9.41103 m/s.
GmSun
v
r
v
(6.67  1011
N m2
kg 2
)(1.99  1030 kg)
1.49  1012 m
m
v  9.44  103
s