Download Conceptual Physics Review Chapter 12, 13, 32

Conceptual Physics Review Chapter 12, 13, 32, & 33
Solutions
Chapter 12 – Universal Gravitation
• Explain Newton’s idea of why the apple falls to Earth.
• Explain why the moon does not fall to Earth.
• Explain how Earth is falling.
• State Newton’s law of universal gravitation.
• Explain the significance of an inverse-square law..
• Be able to define all vocabulary words in bold from the chapter sections 12.1-12.5.
Chapter 13 – Gravitational Interactions
• Describe the gravitational field outside Earth.
• Describe the gravitational field inside Earth.
• Explain why an astronaut in Earth orbit seems weightless even through there is a gravitational
force.
• Explain ocean tides.
• Give examples of tides other than those in water.
• Be able to identify high or low tide based on your location on earth relative to the moon.
• Be able to describe what a spring and neap tide are.
• Be able to identify spring and neap tide based on the relative locations of the earth moon and sun.
• Describe black holes.
• Be able to define all vocabulary words in bold from the chapter sections 13.1, 13.3-13.6.
Chapter 32 – Electrostatics
• Describe Electrostatics.
• Be able to explain the subatomic particles relative charge, relative mass, and location.
• Be able to describe how like and opposite charges interact.
• Describe the three methods of charging.
• Be able to define and use Coulomb’s Law. (listed below)
• Be able to list the three methods of giving something a static charge.
• Be able to define all vocabulary words in bold from the chapter sections 32.1-32.2
Chapter 33 – Electric Fields and Potential
• Be able to define all vocabulary words in bold from the chapter sections 33.1-33.2
• Be able to define an electric field.
• Know what the lines represent on a drawing of an electric field. (direction, distance from one
another)
Review all homework assignments and worksheets from Chapters 12, 13, 32, & 33, especially for
examples of explanations you will be expected to know.
Formulas:
Fg = G
2
m1m2
-11 N ⋅ m
,
where
G=6.672
×10
r2
kg 2
Fe = k
2
q1q2
9 N ⋅m
,
where
k=9.0
×10
r2
C2
Sample Calculations and Questions (Remember to write a formula first)
1. Describe Newton’s Law of Universal Gravitation, both in words and using a formula.
Newton’s Law of Universal Gravitation states that the force of gravitational attraction between
any two masses is proportional to the magnitude of the masses, and inversely proportional to the
square of the distance between the centers of the masses.
mm
N ⋅ m2
Fg = G 1 2 2 , where G=6.672 ×10 -11
r
kg 2
2. Describe the relationship between the moon’s velocity, the moon’s acceleration, and the force of
gravitational attraction between the earth and the moon. How are these three quantities related?
The moon orbits the earth with a tangential velocity—without that velocity, it would fall into the
earth. The moon has a centripetal acceleration as it orbits the earth, due to the changing direction
of its velocity. This acceleration is caused by a centripetal Force of gravitational attraction, due to
Newton’s Law of Universal Gravitation.
3. Two objects, when placed 1 meter apart, are attracted to each other by a 4 ×10 −9 N force of gravity.
What is the gravitational attraction between the two objects when:
a. they are 2 meters apart?
Objects are twice as far apart, so Force between them is (1/2)2 times as big, or 1/4 of the
1
original value, so ( 4 ×10 −9 N ) = 1×10 −9 N
4
b. they are 1/3 of a meter apart?
Now that they are three times closer together, the force of gravity is 32 times stronger, or 9
times greater than the original value at 1 meter. 9 ( 4 ×10 −9 N ) = 36 ×10 −9 N = 3.6 ×10 −8 N
c. the mass of each object is doubled
Assuming the original situation, if the mass of each object is doubled, than the Force will
be increased by a factor of 4.
mm
Fg = G 1 2 2 = 4 ×10 −9 N
r
# mm &
2m 2m
Fg = G 1 2 2 = 4 % G 1 2 2 ( = 36 ×10 −9 N = 3.6 ×10 −8 N
$
r
r '
d. the mass of each object is tripled, and they are 1/3 of a meter apart?
If the mass of each object is tripled, the Force of gravity is increased by a factor of 9.
Moving the objects to 1/3 the distance means the Force is increased by another factor of 32,
or 9. Thus, the new force is 9 time 9, or 81 times greater than the original force!
4. We say that astronauts in space are weightless. Explain what is meant by this statement.
Astronauts in space, like objects in free-fall, are said to be “weightless” because they don’t have a
supporting Normal Force pushing up on them from below. It’s the sensation of the ground pushing
up on us that gives us our sense of weight. Obviously the objects still have a weigh—Earth’s
gravity is still pulling on these objects—but we say that their apparent weight has changed, due to
their acceleration. Objects with an acceleration downward have a smaller apparent weight, and
objects with an acceleration upwards have a greater apparent weight.
5. Is there actually a force of gravity between two students standing one meter apart? Explain.
Of course there’s a force of gravity between two students. There is a force of gravity between all
objects that have mass. The force will not be something that the students can sense or feel,
although instruments (including the Cavendish balance, used to establish the value of the
gravitational constant G) can measure it.
6. When a space shuttle orbits the earth at constant speed, is it accelerating? Is it weightless?
Yes, it is accelerating towards the center of the earth at the same time that it has a tangential
velocity. It is weightless in the sense that there is no Normal force supporting it, but it does
experience a gravitational Force due to the earth. It’s that Force that keeps the space shuttle in
orbit!
7. There is a gravity field that surrounds the earth. What quantities does this gravity field depend
upon?
The gravity field around any object, including the earth, is related to the mass of that object and
the distance at which one is located relative to that object. The gravity field g can be calculated as
M
g=G 2
r
24
8. The earth has a mass of 5.98 ×10 kg , and a radius of 6.38 ×10 6 m . Determine the strength of the
gravity field g at a point 12000 meters above the surface of the earth.
The gravity field at the surface of the earth is 9.8 m/s2, so we expect that the gravity field some
distance above the surface of the earth will be a bit less than this.
M
g=G 2
r
#
(5.98 ×1024 kg) = 9.77m / s2
m2 &
g = % 6.672 ×10 −11 N 2 (
kg ' ( 6.38 ×10 6 m +12000m )2
$
9. Draw a picture of the earth, the moon, and the moon’s gravitational influence on the earth’s oceans.
Why are there two high tides each day when the earth only spins on its axis once per day?
In this exaggerated diagram, the moon can be seen to be
exerting a stronger influence on the near oceans, and a
weaker influence on the far oceans, producing a “doublebulge” effect on the earth’s oceans. We experience two high
tides each day (as well as two low tides each day) as the
earth rotates once every 24 hours, with its surface passing
beneath the bulging oceans.
10. Explain what happens during a spring tide and neap tide.
During spring tides, the sun, earth, and moon are aligned in a
straight line so that the gravitational effects are compounded,
producing high tides that “spring” to an even higher level than
before.
During neap tides, the sun and moon are oriented at a right angle
relative to the earth, so their gravitational influences produce
smaller-than-usual tides.
11. Describe Coulomb’s Law, both in words and using a formula.
Coulomb’s Law states that the force of electrical attraction or repulsion between any two charges
is proportional to the magnitude of the charge, and inversely proportional to the square of the
distance between the centers of the charges. Like charges repel, and opposite charges attract.
qq
N ⋅ m2
Fe = k 1 2 2 , where k=8.99 ×10 9
r
C2
12. What is the relationship between the masses of a proton and an electron? What is the relationship
between the charges of a proton and an electron?
The proton has a mass approximately 1800 times larger than that of an electron: mproton =
1.67 ×10 −27 kg , and melectron = 9.11×10 −31 kg .
The charge of a proton and an electron are the same magnitude but opposite polarities: qproton =
+1.6 ×10 −19 Coulombs and qelectron = −1.6 ×10 −19 Coulombs .
13. How are the Law of Universal Gravitation and Coulomb’s Law similar? How are they different?
Both laws describe the forces acting between two bodies, and both laws are inverse-square laws:
the magnitude of the force varies in proportion to the inverse-square of the distance between them.
The laws are different in that depends on mass, the other depends on charge. Also, gravity forces
only attract, whereas electric forces may attract or repel, depending on the charges involved.
14. There is an electric field that surrounds positive and negative charges. What quantities does this
electric field depend upon?
The electric field around any charge is related to the magntude of the charge, its polarity (positive
or negative) and the distance at which one is located relative to that charge. The electric field E
can be calculated as
Q
E=k 2
r
15. Determine the force of repulsion between a -2.00 Coulomb charge and a -3.00 Coulomb charge,
separated by a distance of 2.00 meters.
qq
Fe = k 1 2 2
r
#
N ⋅ m 2 & (−2.00C)(−3.00C)
Fe = % 9 ×10 9
= +1.35 ×1010 N
(
2
2
C '
(2.00m)
$
a. How does this force change if the charges are only 1.00 meter apart?
Charges are half as far apart, so force increases by a factor of 22, or 4.
4 (1.35 ×1010 N ) = 5.4 ×1010 N
b. How does this force change if the charges are 4.00 meters apart, and each charge is
doubled?
If both charges are doubled, that increases the force by 2 × 2 = 4 . But moving the objects
twice as far apart, from 2 meters to 4 meters, decreases the force by a factor of 22 = 4. So
the net force of repulsion remains the same, 1.35 ×1010 N .
16. The diameter of an atom is about 1 Ångstrom, or 1×10 −10 m , or 0.1 nm. Calculate the electric force
between a proton in the nucleus and an electron in its shell, and indicate whether this is an
attraction or a repulsion.
This is an attraction because the proton and electron have opposite charges. The magnitude of the
force depends on the charges and the distance between them, which is half the diameter of the
atom:
qq
Fe = k 1 2 2
r
#
N ⋅ m 2 & (+1.602 ×10 −19 C)(−1.602 ×10 −19 C)
Fe = % 9 ×10 9
= −4.62 ×10 −18 N
(
2
2
C
#
&
$
'
1
−10
% (1×10 m ) (
$2
'