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Chapter 11 and 13 Homework
Chapter 11 and 13 Homework
Due: 10:00pm on Wednesday, April 23, 2014
You will receive no credit for items you complete after the assignment is due. Grading Policy
Exercise 11.1
A 0.100-kg, 40.2-cm-long uniform bar has a small 0.080-kg mass glued to its left end and a small 0.150-kg mass glued
to the other end. You want to balance this system horizontally on a fulcrum placed just under its center of gravity.
Part A
How far from the left end should the fulcrum be placed?
ANSWER:
d
= 24.4
cm
Correct
Introduction to Static Equilibrium
Learning Goal:
To understand the conditions necessary for static equilibrium.
Look around you, and you see a world at rest. The monitor, desk, and chair—and the building that contains them—are
in a state described as static equilibrium. Indeed, it is the fundamental objective of many branches of engineering to
maintain this state in spite of the presence of obvious forces imposed by gravity and static loads or the more
unpredictable forces from wind and earthquakes.
The condition of static equilibrium is equivalent to the statement that the bodies involved have neither linear nor angular
acceleration. Hence static mechanical equilibrium (as opposed to thermal or electrical equilibrium) requires that the
forces acting on a body simultaneously satisfy two conditions:
⃗
∑F = 0
and ∑ τ ⃗ = 0;
that is, both external forces and torques sum to zero. You have the freedom to choose any point as the origin about
which to take torques.
Each of these equations is a vector equation, so each represents three independent equations for a total of six. Thus to
keep a table static requires not only that it neither slides across the floor nor lifts off from it, but also that it doesn't tilt
about either the x or y axis, nor can it rotate about its vertical axis.
Part A
Frequently, attention in an equilibrium situation is confined to a plane. An example would be a ladder leaning
against a wall, which is in danger of slipping only in the plane perpendicular to the ground and wall. By orienting a
Cartesian coordinate system so that the x and y axes are in this plane, choose which of the following sets of
quantities must be zero to maintain static equilibrium in this plane.
Hint 1. Simplifying the equations
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The motion (or possible motion) is confined to a plane, the xy plane in this case, when there are no forces
acting out of that plane (e.g., all Fz = 0 or all z-component forces occur in pairs that are applied at the
same points). Recalling that torque is defined as a cross product, you can eliminate the need for two of the
three equations for the components of torque since they will equal zero.
ANSWER:
∑ Fx
and ∑ τz and ∑ Fy
∑ Fz
and ∑ τx and ∑ τy
∑ τx
and ∑ Fx and ∑ τy and ∑ Fy
∑ τx
and ∑ Fx and ∑ τy and ∑ Fy and ∑ τz
Correct
Part B
As an example, consider the case of a board of length L and negligible mass. Take the x axis to be the horizontal
axis along the board and the y axis to be the vertical axis perpendicular to the board. A mass of weight W is
strapped to the board a distance x from the left-hand end. This is a static equilibrium problem, and a good first step
is to write down the equation for the sum of all the forces
in the y direction since the only nonzero forces of
⃗
∑F = 0
What is
that exist are in the y direction.
∑ Fy
? Your equation for the net force in the y
direction on the board should contain all the forces acting
vertically on the board.
Express your answer in terms of the weight W and
the tensions in the two vertical ropes at the left and
right ends TL and TR . Recall that positive forces
point upward.
ANSWER:
∑ Fy = 0 =
=
TL + TR − W
Correct
The only relevant component of the torques is the z component; however, you must choose your pivot point before
writing the equations. This point could be anywhere; in fact, the pivot point does not even have to be at a point on the
body. You should choose this point to your advantage. Generally, the best place to locate the pivot point is where some
unknown force acts; this will eliminate that force from the resulting torque equation.
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Part C
What is the equation that results from choosing the pivot point to be the point from which the mass hangs (where
W acts)?
Express your answer in terms of the unknown quantities TL and
TR
and the known lengths x and
L
.
Recall that counterclockwise torque is positive.
ANSWER:
∑ τW ,z = 0 =
=
TR (L − x) − TL x
Correct
This gives us one equation involving two unknowns,
TL
and TR . We can use this result and ∑ Fy
= 0
to
solve for TL and TR .
Part D
What is the equation that results from choosing the pivot point to be the left end of the plank (where TL acts)?
Express your answer in terms of TL ,
TR
,
W
, and the dimensions L and
. Not all of these variables may
x
show up in the solution.
ANSWER:
∑ τL,z = 0 =
=
TR L − W x
Correct
Part E
What is the equation that results from choosing the pivot point to be the right end of the plank (where TR acts)?
Express your answer in terms of TL ,
TR
,
W
, and the dimensions L and
. Not all of these variables may
x
show up in the solution.
ANSWER:
∑ τR,z = 0 =
=
W (L − x) − TL L
Correct
Part F
Solve for TR , the tension in the right rope.
W
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Express your answer in terms of W and the dimensions L and
in the solution.
. Not all of these variables may show up
x
Hint 1. Choose the correct equation
Which single equation of the ones you've derived can be used to solve for TR in terms of known quantities?
ANSWER:
∑ Fy = 0
∑ τW = 0
∑ τL = 0
∑ τR = 0
ANSWER:
TR
=
W
x
L
Correct
Part G
Solve for TL , the tension in the left rope.
Express your answer in terms of W and the dimensions L and
in the solution.
. Not all of these variables may show up
x
Hint 1. Choose the correct equation
Which single equation of the ones you've derived can be used to solve for TL in terms of known quantities?
ANSWER:
∑ Fy = 0
∑ τW = 0
∑ τL = 0
∑ τR = 0
ANSWER:
TL
=
W
L−x
L
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Correct
Part H
Solve for the tension in the left rope,
TL
, in the special case that
. Be sure the result checks with your
x = 0
intuition.
Express your answer in terms of W and the dimensions L and
in the solution.
. Not all of these variables may show up
x
ANSWER:
TL
=
W
Correct
Only one set of forces, exactly balanced, produces static equilibrium. From this perspective it might seem
puzzling that so much of the world is static. One must realize, however, that many forces—like those of the
tensions in the ropes here or those between the floor and an object resting on it—increase very quickly as the
object moves. If there is a slight imbalance of the forces, the object accelerates so that its position changes
until the object has adjusted itself to restore the force balance. It then oscillates about this point until friction or
some other dissipative mechanism causes it to become stationary at the exact equilibrium point.
Precarious Lunch
A uniform steel beam of length L and mass m1 is attached via a hinge to the side of a building. The beam is supported
by a steel cable attached to the end of the beam at an angle θ , as shown. Through the hinge, the wall exerts an
unknown force, F , on the beam. A workman of mass m2 sits eating lunch a distance d from the building.
Part A
T
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Find T , the tension in the cable. Remember to account for all the forces in the problem.
Express your answer in terms of m1 ,
m2
,
L
, d, θ , and g, the magnitude of the acceleration due to
gravity.
Hint 1. Pick the best origin
This is a statics problem so the sum of torques about any axis a will be zero. In order to solve for T , you
want to pick the axis such that T will give a torque, but as few as possible other unk nown forces will enter
the equations. So where should you place the origin for the purpose of calculating torques?
ANSWER:
At the center of the bar
At the hinge
At the connection of the cable and the bar
Where the man is eating lunch
Hint 2. Calculate the sum torques
Now find the sum of the torques about center of the hinge. Remember that a positive torque will tend to
rotate objects counterclockwise around the origin.
Answer in terms of T ,
L
, d,
m1
,
m2
, θ , and g.
ANSWER:
Σ τa
=0=
(
m1 L
2
+ m 2 d)g − T Lsin ( θ)
ANSWER:
T
=
g( m1
L
2
+ m2 d)
Lsin(θ)
Correct
Part B
Find Fx , the x-component of the force exerted by the wall on the beam ( F ), using the axis shown. Remember to
pay attention to the direction that the wall exerts the force.
Express your answer in terms of T and other given quantities.
Hint 1. Find the sign of the force
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The beam is not accelerating in the x-direction, so the sum of the forces in the x-direction is zero. Using the
given coordinate system, is Fx going to have to be positive or negative?
ANSWER:
Fx
=
−T cos(θ)
Correct
Part C
Find Fy , the y-component of force that the wall exerts on the beam ( F ), using the axis shown. Remember to pay
attention to the direction that the wall exerts the force.
Express your answer in terms of T , θ ,
m1
,
m2
, and g.
ANSWER:
Fy
=
g(m1 + m2 ) − T sin(θ)
Correct
If you use your result from part (A) in your expression for part (C), you'll notice that the result simplifies
somewhat. The simplified result should show that the further the luncher moves out on the beam, the lower the
magnitude of the upward force the wall exerts on the beam. Does this agree with your intuition?
Exercise 11.5
A ladder of length 20.0m is carried by a fire truck. The ladder has a weight of 3000N and its center of gravity is at its
center. The ladder is pivoted at one end (A) about a pin ; you
can ignore the friction torque at the pin. The ladder is raised
into position by a force applied by a hydraulic piston at C.
Point C is a distance 8.0m from A, and the force F ⃗ exerted
by the piston makes an angle of θ = 40∘ with the ladder.
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Part A
What magnitude must
F
⃗
have to just lift the ladder off the support bracket at B?
Express your answer using two significant figures.
ANSWER:
= 5800
F
N
Correct
Tidal Forces
Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularly
visible on the earth's surface in the form of tides. To understand the origin of tidal forces, consider the earth-moon
system to consist of two spherical bodies, each with a spherical mass distribution. Let re be the radius of the earth, m
be the mass of the moon, and G be the gravitational constant.
Part A
Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the
acceleration ae of the earth due to the gravitational pull of the moon?
Express your answer in terms of G ,
, and r.
m
Hint 1. How to approach the problem
Apply the law of gravitation to find the magnitude of the force exerted on the earth by the moon and use
Newton's 2nd law to determine the magnitude of the acceleration of the earth. Recall that the gravitational
interaction of two spherically symmetric bodies is calculated as if the mass of each body were concentrated
in its center.
Hint 2. Find the gravitational force exerted on the earth by the moon
If me is the mass of the earth, what is the gravitational force Fg exerted by the moon on the earth?
Express your answer in terms of G ,
me
,
, and r.
m
ANSWER:
Fg
=
Gme m
r2
ANSWER:
ae
=
Gm
r2
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Correct
Just as the earth accelerates toward the moon, the moon accelerates towards the earth. These are the
centripetal accelerations that cause the earth and moon to follow circular orbits around their mutual center of
rotation.
Part B
Since the gravitational force between two bodies decreases with distance, the acceleration anear experienced by a
unit mass located at the point on the earth's surface closest to the moon is slightly different from the acceleration
afar experienced by a unit mass located at the point on the earth's surface farthest from the moon. Give a general
expression for the quantity
anear − afar
Express your answer in terms of G ,
.
, , and
m r
re
.
Hint 1. Find the acceleration at the point on earth nearest the moon
What is
anear
, the magnitude of the acceleration due to the gravitational pull of the moon of a unit mass of
water located at the point on the earth's surface nearest the moon?
Express your answer in terms of G , r,
re
, and
.
m
Hint 1. How to approach the problem
Repeat the same calculations as in Part A, taking into account, however, that a body located at the
point on the earth's surface nearest the moon is closer to the moon than the center of the earth.
Hint 2. Find the distance between the center of the moon and the point on the earth's
surface closest to the moon
If you draw a line joining the center of the earth and the center of the moon, you will see that the point
on the earth's surface nearest the moon is shifted toward the moon with respect to the center of the
earth a distance equal to the earth's radius. How far is that point, then, from the center of the moon?
Express your answer in terms of r and
re
.
ANSWER:
r − re
ANSWER:
anear
=
Gm
(r−re )
2
Hint 2. Find the acceleration at the point on earth farthest from the moon
What is
afar
, the magnitude of the acceleration of the same amount of water at the point on the earth's
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Chapter 11 and 13 Homework
surface farthest from the moon?
Express your answer in terms of G ,
, ,
m r re
.
Hint 1. Find the distance between the center of the moon and the point on the earth's
surface farthest from the moon
If you draw a line joining the center of the earth and the center of the moon, you will see that the point
on the earth's surface farthest from the moon is shifted away from the moon with respect to the
center of the earth a distance equal to the earth's radius. How far is that point, then, from the center
of the moon?
Express your answer in terms of r and
re
.
ANSWER:
r + re
ANSWER:
afar
=
Gm
2
(r+re )
ANSWER:
anear − afar
=
Gm
2
(r−re )
−
Gm
2
(r+re )
Correct
If you simplified your answer, you found that
anear − af ar = Gm
Note that, since r ≫
re
4re
r3
(1 − (
re
r
2
) )
−2
.
, the difference is very well approximated by the expression 4Gmre /r3 .
On the side of the earth nearest the moon (near side), water has a 7% greater acceleration than on the farthest
side (far side) and bulges out, causing a high tide. Water on the far side is less strongly attracted toward the
moon and thus another tidal bulge occurs. In total then, the earth experiences two high tides. Note that these
tidal bulges of water do not appear from nowhere; instead they are formed by water flowing away from other
areas of the planet, where low tides are observed.
Part C
The earth is subject not only to the gravitational force of the moon but also to the gravitational pull of the sun.
However, the earth is much farther away from the sun than it is from the moon. In fact, the center of the earth is at
11
an average distance of 1.5 × 10 m from the center of the sun. Given that the mass of the sun is
30
1.99 ×
kg
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30
1.99 × 10
kg
, which of the following statements is correct?
Hint 1. How to approach the problem
Since the force exerted on a body is proportional to its acceleration, to compare the gravitational effect of the
sun with that of the moon simply compare the acceleration of the earth due to the moon's gravitational pull
(which you calculated in Part A) with the same acceleration due to the sun's pull.
Hint 2. Find the acceleration of the earth due to the sun's gravitational pull
What is the magnitude of the acceleration ae of the earth due to the gravitational force of the sun?
Express your answer numerically in meters per second squared.
ANSWER:
ae
= 5.90×10−3
2
m/s
ANSWER:
The force exerted on the earth by the sun is weaker than the corresponding force exerted by the moon.
The force exerted on the earth by the sun is stronger than the corresponding force exerted by the moon.
The force exerted on the earth by the sun is of the same order of magnitude of the corresponding force
exerted by the moon.
Correct
Although the sun is much farther away from the earth than the moon, it is much more massive. As a result, the
gravitational force exerted on the earth by the sun is about 180 times stronger than the corresponding pull from
the moon!
Part D
The occurrence of tidal forces on the earth's surface is not limited to the gravitational effects of the moon. Tidal
forces are produced every time different parts of a body are subject to different gravitational forces exerted by a
second body. Therefore, tidal forces due to the gravitational effects of the sun are also present on the earth's
surface. What can you conclude about the relative effects of these two tidal forces on the earth's surface?
Hint 1. How to approach the problem
The strength of the tides is related to the difference between the acceleration of water on opposite sides of
the earth due to the sun. Compute the difference and compare this to the result you obtained for the moon.
Hint 2. Find the difference in acceleration
Find the difference between the acceleration on opposite sides of the earth due to the sun. Use the
approximate formula for anear − afar from the follow-up comment to Part B (but of course plug in the sun's
mass rather than the moon's).
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Express your answer numerically in meters per second squared.
ANSWER:
anear − afar
= 1.00×10−6
2
m/s
ANSWER:
The moon exerts a stronger tidal force on the earth than the sun does.
The sun exerts a stronger tidal force on the earth than the moon does.
The moon and the sun cause tidal forces of equal magnitude.
Correct
Even though the gravitational force exerted on the earth by the sun is about 180 times stronger than the
corresponding pull from the moon, the differential pull is smaller. In fact, the difference in the sun's pull on
opposite sides of the earth is about half the difference in the moon's pull. The effects of the tidal forces caused
by the sun, however, become particularly evident when the sun, the moon, and the earth are aligned. On this
occasion, the gravitational effects of the sun are added to the gravitational effects of the moon and the highest
tides are observed (called spring tides, although they have nothing to do with the spring season). When the
sun is at an angle of 90∘ with respect to the line joining the moon and the earth, instead, the gravitational
effects of the sun partially cancel the effects of the moon and the least difference between high and low tide is
observed (called neap tides).
Exercise 13.5
Two uniform spheres, each of mass 0.260 kg, are fixed at points
A
and B (the figure ).
Part A
Find the magnitude of the initial acceleration of a uniform sphere with mass 0.010 kg if released from rest at point
P
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A
B
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Chapter 11 and 13 Homework
and acted on only by forces of gravitational attraction of the spheres at
A
and B.
Express your answer using two significant figures.
ANSWER:
a
= 2.1×10−9
2
m/s
Correct
Part B
Find the direction of the initial acceleration of a uniform sphere with mass 0.010 kg.
ANSWER:
upward
to the right
downward
to the left
Correct
Gravitational Acceleration inside a Planet
Consider a spherical planet of uniform density ρ. The distance from the planet's center to its surface (i.e., the planet's
radius) is Rp . An object is located a distance R from the center of the planet, where R < Rp . (The object is located
inside of the planet.)
Part A
Find an expression for the magnitude of the acceleration due to gravity,
Express the acceleration due to gravity in terms of ρ,
,
, and
R π
, inside the planet.
g(R)
G
, the universal gravitational constant.
Hint 1. Force due to planet's mass outside radius R
From Newton's Principia, Proposition LXX, Theorem XXX:
If to every point of a spherical surface there tend equal centripetal forces decreasing as the square of the
distances from those points, I say, that a corpuscle placed within that surface will not be attracted by those
forces any way.
In other words, you don't have to worry about the portion of the planet's mass that is located outside of the
radius R. The net gravitational force from this "outer shell" of mass will equal zero. You only have to worry
about that portion of the planet's mass that is located within a radius R.
R
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Hint 2. Find the force on an object at distance
R
Suppose the object has a mass m. Find the magnitude of the gravitational force acting on this object when
it is located a distance R from the center of the planet.
Express the force in terms of m, ρ,
,
, and
π R
G
, the universal gravitational constant.
Hint 1. Find the mass within a radius R
Find the net mass of the planet located within the radius
3
(4/3)πR
. Remember, the volume of a sphere is
R
.
Express your answer in terms of ρ,
, and
π
.
R
ANSWER:
M (R)
=
4
3
3
πR ρ
Hint 2. Magnitude of gravitational force
The general equation for the magnitude of the gravitational force is
2
Fgrav = GM m/R
.
ANSWER:
F (R)
=
4
3
πGmρR
Hint 3. Finding
g(R)
from F (R)
According to Newton's 2nd law, the net force acting on an object is given by
a = g(R)
and Fnet
g(R) = F (R)/m
= F (r)
Fnet = ma
. In this problem,
since the only force acting on the object is the gravitational force. Therefore,
, where F (R) is the force you found in the previous hint.
Note that in this usage, both F (R) and a are magnitudes and hence are positive. By convention,
g(R)
g
(or
in this case) is the magnitude of the gravitational field. This gravitational field is a vector, with direction
downward.
ANSWER:
g(R)
=
4
3
πGρR
Correct
Part B
g(R)
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Rewrite your result for g(R) in terms of gp , the gravitational acceleration at the surface of the planet, times a
function of R.
Express your answer in terms of gp ,
, and
R
Rp
.
Hint 1. Acceleration at the surface
Note that the acceleration at the surface should be equal to the value of the function g(R) from Part A
evaluated at the radius of the planet:
g
p
= g(Rp )
.
ANSWER:
g(R)
=
g (
p
R
Rp
)
Correct
Notice that
g
increases linearly with R, rather than being proportional to 1/R2 . This assures that it is zero at
the center of the planet, as required by symmetry.
Part C
Find a numerical value for ρearth , the average density of the earth in kilograms per cubic meter. Use 6378
the radius of the earth,
−11
G = 6.67 × 10
3
2
m /(kg ⋅ s )
, and a value of g at the surface of 9.80
2
m/s
km
for
.
Express your answer to three significant figures.
Hint 1. How to approach the problem
You already derived the relation needed to solve this problem in Part A:
g(R) = (4/3)πGρR.
At what distance R is
g(R)
known so that you could use this relation to find ρ?
ANSWER:
ρearth
= 5500
3
kg/m
Correct
Exercise 13.11
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Part A
At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.620m/s2 if the
acceleration due to gravity at the surface has magnitude 9.80 m/ s2 ?
ANSWER:
1.90×107
m
Correct
Energy of a Spacecraft
Very far from earth (at R = ∞), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational
force of the earth were to act on it (i.e., neglect the forces from the sun and other solar system objects), the spacecraft
would eventually crash into the earth. The mass of the earth is M e and its radius is Re . Neglect air resistance
throughout this problem, since the spacecraft is primarily moving through the near vacuum of space.
Part A
Find the speed s e of the spacecraft when it crashes into the earth.
Express the speed in terms of M e ,
Re
, and the universal gravitational constant G .
Hint 1. How to approach the problem
Use a conservation-law approach. Specifically, consider the mechanical energy of the spacecraft when it is
(a) very far from the earth and (b) at the surface of the earth.
Hint 2. Total energy
What is the total mechanical energy of the spacecraft when it is far from earth, at a distance R = ∞?
ANSWER:
E
= 0
Hint 3. Potential energy
If the spacecraft has mass M , what is its potential energy U at the surface of the earth? Note that, at the
surface of the earth, the spacecraft is a distance Re from the center of the earth.
Express your answer in terms of M ,
Me
,
Re
, and the universal gravitational constant G .
Hint 1. Formula for the potential energy
The gravitational potential energy
Ug
of a system of 2 masses
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r
is
Ug = −
Gm1 m2
r
.
ANSWER:
U (Re )
−(GM e M )
=
Re
ANSWER:
−
−−
−−
se
=
√
2GM e
Re
Correct
Part B
Now find the spacecraft's speed when its distance from the center of the earth is
R = αRe
, where the coefficient
.
α ≥ 1
Express the speed in terms of s e and
.
α
Hint 1. General approach
This problem is very similar to the problem that you've just done. Note that the potential energy of the
spacecraft at a distance αRe is different from its potential energy at the earth's surface.
Hint 2. First step in finding the speed
Find the spacecraft's speed at
R = αRe
in terms of M e ,
G
,
Re
, and α.
ANSWER:
−
−
−−
−
sα
=
√
2M e G
αRe
ANSWER:
sα
=
se
√α
Correct
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Chapter 11 and 13 Homework
Properties of Circular Orbits
Learning Goal:
To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the
earth.
The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were
observed by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the
speed of the space shuttle and the height of geostationary satellites) is the simplest orbit--a circular one. This problem
concerns the properties of circular orbits for a satellite orbiting a planet of mass M .
For all parts of this problem, where appropriate, use G for the universal gravitational constant.
Part A
Find the orbital speed v for a satellite in a circular orbit of radius
Express the orbital speed in terms of G ,
M
, and
.
R
.
R
Hint 1. Find the force
Find the radial force F on the satellite of mass
. (Note that
m
m
will cancel out of your final answer for v.)
Express your answer in terms of m, M , G , and R. Indicate outward radial direction with a
positive sign and inward radial direction with a negative sign.
ANSWER:
F
−GM m
=
R
2
Hint 2. A basic kinematic relation
Find an expression for the radial acceleration ar for the satellite in its circular orbit.
Express your answer in terms of v and R. Indicate outward radial direction with a positive sign
and inward radial direction with a negative sign.
ANSWER:
ar
2
−v
=
R
Hint 3. Newton's 2nd law
Apply
⃗
F = ma⃗
to the radial coordinate.
ANSWER:
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Chapter 11 and 13 Homework
−
−
−
−
v
=
√
GM
R
Correct
Part B
Find the kinetic energy
K
of a satellite with mass
Express your answer in terms of m,
M
,
G
m
, and
in a circular orbit with radius
.
R
.
R
ANSWER:
K
=
GM m
2R
Correct
Part C
Express the kinetic energy
K
in terms of the potential energy
U
.
Hint 1. Potential energy
What is the potential energy
U
of the satellite in this orbit?
Express your answer in terms of m,
M
,
G
, and
.
R
ANSWER:
U
−GM m
=
R
ANSWER:
K
=
−
U
2
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Chapter 11 and 13 Homework
Correct
This is an example of a powerful theorem, known as the Virial Theorem. For any system whose motion is
periodic or remains forever bounded, and whose potential behaves as
n
U ∝ R
,
Rudolf Clausius proved that
⟨K ⟩ =
n
2
⟨U ⟩,
where the brackets denote the temporal (time) average.
Part D
Find the orbital period T .
Express your answer in terms of G ,
M
,
, and
R
.
π
Hint 1. How to get started
Use the fact that the period is the time to make one orbit. Then time = distance/velocity.
ANSWER:
3
T
=
2πR
2
√GM
Correct
Part E
Find an expression for the square of the orbital period.
Express your answer in terms of G ,
M
,
, and
R
.
π
ANSWER:
T
2
=
4π
2
R
3
GM
Correct
This shows that the square of the period is proportional to the cube of the semi-major axis. This is Kepler's
Third Law, in the case of a circular orbit where the semi-major axis is equal to the radius, R.
Part F
L
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Chapter 11 and 13 Homework
Find L, the magnitude of the angular momentum of the satellite with respect to the center of the planet.
Express your answer in terms of m,
M
,
G
, and
.
R
Hint 1. Definition of angular momentum
Recall that
⃗
⃗
L = R × p⃗
, where p ⃗ is the momentum of the object and R⃗ is the vector from the pivot point.
Here the pivot point is the center of the planet, and since the object is moving in a circular orbit,
p⃗
is
perpendicular to R⃗ .
ANSWER:
L
=
−
−−
−−
m√GM R
Correct
Part G
The quantities v, K , U , and L all represent physical quantities characterizing the orbit that depend on radius
Indicate the exponent (power) of the radial dependence of the absolute value of each.
Express your answer as a comma-separated list of exponents corresponding to v,
order. For example, -1,-1/2,-0.5,-3/2 would mean
−1
v ∝ R
,
−1/2
K ∝ R
K
,
U
, and
L
.
R
, in that
, and so forth.
Hint 1. Example of a power law
The potential energy behaves as
power for this is
−1
(i.e.,
U = GM m/R
−1
U ∝ R
, so U depends inversely on R. Therefore, the appropriate
).
ANSWER:
-0.500,-1,-1,0.500
Correct
Geosynchronous Satellite
A satellite that goes around the earth once every 24 hours is called a geosynchronous satellite. If a geosynchronous
satellite is in an equatorial orbit, its position appears stationary with respect to a ground station, and it is known as a
geostationary satellite.
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Chapter 11 and 13 Homework
Part A
Find the radius R of the orbit of a geosynchronous satellite that circles the earth. (Note that
the center of the earth, not the surface.) You may use the following constants:
The universal gravitational constant
The mass of the earth is
The radius of the earth is
G
24
5.98 × 10
6
6.38 × 10
is
−11
6.67 × 10
kg
m
2
2
N m /kg
R
is measured from
.
.
.
Give the orbital radius in meters to three significant digits.
Hint 1. Find the force on the satellite
If we just consider the earth-satellite system, then there is only one force acting on the satellite. Suppose
the mass of the satellite is m, the mass of the earth is M , and the radius of the satellite's orbit is R. What
is the magnitude of the force that acts on the satellite?
Answer in terms of m,
numerical values.)
M
,
, and the universal gravitational constant G . (Use variables, not
R
ANSWER:
F
=
GmM
R
2
Hint 2. Angular frequency of satellite
The gravitational force on the satellite provides a centripetal acceleration that pulls the satellite inward,
holding it in a circular orbit. A generic formula for the magnitude of the centripetal acceleration is a = Rω2 ,
where ω is the angular frequency of the satellite's orbit.
What is the numerical value for ω in radians per second for a geosynchronous satellite?
Hint 1. How to calculate
ω
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Chapter 11 and 13 Homework
Calculate ω using the definition of a geosynchronous orbit; the angular velocity should be such that
the satellite makes one orbit per day. The equation relating the angular velocity ω and the time period
T is
ω=
2π
T
.
Hint 2. What is T ?
Here the time period T is one day, but you are asked for the angular velocity in radians per second.
ANSWER:
= 7.27×10−5 radians/s
ω
Hint 3. Find an expression for
Type an expression for the radius
R
R
of the circular orbit of a satellite orbiting the earth.
Express your answer in terms of G ,
satellite.
M
(the mass of the earth), and
ω
, the angular velocity of the
Hint 1. Putting it all together
Using Newton's 2nd law,
, gives us
F = ma
GM m
R
2
2
= mω R.
Find R from this equation.
ANSWER:
−
−−
R
=
3
√
GM
2
ω
ANSWER:
R
= 4.23×107 m
Correct
A Satellite in a Circular Orbit
Consider a satellite of mass m1 that orbits a planet of mass m2 in a circle a distance r from the center of the planet.
The satellite's mass is negligible compared with that of the planet. Indicate whether each of the statements in this
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Chapter 11 and 13 Homework
problem is true or false.
Part A
The information given is sufficient to uniquely specify the speed, potential energy, and angular momentum of the
satellite.
Hint 1. What constitutes sufficient initial conditions?
An initial position and velocity plus a knowledge of the forces at all points will suffice to specify the
subsequent motion in Newtonian mechanics. Are you able to determine the satellite's velocity from the
information given?
ANSWER:
true
false
Correct
Part B
The total mechanical energy of the satellite is conserved.
Hint 1. When is mechanical energy conserved?
A system's total mechanical energy is conserved if no nonconservative forces act on the system. Is gravity a
conservative force?
ANSWER:
true
false
Correct
Part C
The linear momentum vector of the satellite is conserved.
Hint 1. When is linear momentum conserved?
The linear momentum vector of a system is conserved if the net external force acting on it is zero.
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ANSWER:
true
false
Correct
Part D
The angular momentum of the satellite about the center of the planet is conserved.
Hint 1. When is angular momentum conserved?
Angular momentum about a particular axis is conserved if there is no net torque about that axis.
ANSWER:
true
false
Correct
Part E
The equations that express the conservation laws of total mechanical energy and linear momentum are sufficient to
solve for the speed necessary to maintain a circular orbit at R without using F ⃗ = ma⃗ .
Hint 1. How are conservation laws used?
Conservation laws are generally applied to a system that has some "initial" and "final" conditions that are
different. Motion in a circular orbit, however, possesses no obvious initial and final points that are different.
ANSWER:
true
false
Correct
Kepler's 3rd Law
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Chapter 11 and 13 Homework
A planet moves in an elliptical orbit around the sun. The mass of the sun is
of the planet from the sun are R1 and R2 , respectively.
Ms
. The minimum and maximum distances
Part A
Using Kepler's 3rd law and Newton's law of universal gravitation, find the period of revolution P of the planet as it
moves around the sun. Assume that the mass of the planet is much smaller than the mass of the sun.
Use G for the gravitational constant.
Express the period in terms of G ,
Ms
,
R1
, and
R2
.
Hint 1. Kepler's 3rd law
Kepler's 3rd law states that the square of the period of revolution of a planet around the sun is proportional to
the cube of the semi-major axis of its orbit. Try finding the period of a circular orbit and then using Kepler's
3rd law (which applies equally to circular and elliptical orbits) to extend your result to an elliptical orbit.
Hint 2. Find the semi-major axis
Find the semi-major axis a.
Express the semi-major axis in terms of R1 and
R2
.
Hint 1. Definition of semi-major axis
The semi-major axis of an ellipse is half of its major axis. The sun is at the focus of the elliptical orbit
and the focus lies on the major axis.
ANSWER:
a
=
R 1 +R 2
2
Hint 3. Find the period of a circular orbit
Find the period P of a planet in a circular orbit of semi-major axis a.
Express the period in terms of a,
Ms
, and
G
.
Hint 1. Formula for the period
The period is
2πr/v
, where r is the radius of the orbit and v is the speed of the object. Note that this
is the distance traveled in one orbit divided by the speed.
Hint 2. Find the velocity
Find the velocity
v
acceleration acent
of an object in an orbit of radius
2
= v /r
r
by setting the magnitude of the centripetal
equal to the magnitude of the acceleration due to gravity.
Ms
G
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Chapter 11 and 13 Homework
Express your answer in terms of r,
Ms
, and
G
.
ANSWER:
=
v
−−−−
√
MsG
r
Hint 3. Radius of the orbit
For a circle, the semi-major axis is just the radius.
ANSWER:
−
−−−
P
=
2π√
a
3
GM s
ANSWER:
−−
−−−−−
−−
P
=
3
2
√
π (R 1 + R 2 )
2GM s
Correct
Exercise 13.26
In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at
64,000 km. Pluto already was known to have a large satellite Charon, orbiting at 19,600 km with an orbital period of
6.39 days.
Part A
Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using
the mass of Pluto.
Enter your answers numerically separated by a comma.
ANSWER:
T1 , T2
= 24.5,37.7 days
Correct
Exercise 11.10
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Chapter 11 and 13 Homework
A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder
weighs 160 N. The coefficient of static friction between the foot of the ladder and the ground is 0.40. A man weighing
740 N climbs slowly up the ladder.
Part A
What is the maximum frictional force that the ground can exert on the ladder at its lower end?
Express your answer using two significant figures.
ANSWER:
Ffr
=
N
Part B
What is the actual frictional force when the man has climbed 1.0 m along the ladder?
Express your answer using two significant figures.
ANSWER:
Ffr
=
N
Part C
How far along the ladder can the man climb before the ladder starts to slip?
Express your answer using two significant figures.
ANSWER:
s
=
m
Exercise 13.4
Two uniform spheres, each with mass
M
and radius
, touch one another.
R
Part A
What is the magnitude of their gravitational force of attraction?
Express your answer in terms of the variables M ,
, and appropriate constants.
R
ANSWER:
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Chapter 11 and 13 Homework
Orbiting Satellite
A satellite of mass m is in a circular orbit of radius R2
around a spherical planet of radius R1 made of a material
with density ρ. ( R2 is measured from the center of the
planet, not its surface.) Use G for the universal gravitational
constant.
Part A
Find the kinetic energy of this satellite,
K
.
Express the satellite's kinetic energy in terms of G ,
,
,
m π R1
,
R2
, and ρ.
Hint 1. Kinematics of circular motion
The circular trajectory of the satellite implies that there is a certain inward radial acceleration
2
aradial = −R2 ω , which must be due to the gravitational force Fg .
Hint 2. Mass of the planet
Find the mass of the planet in terms of its size and density.
Express m in terms of ρ and
R1
.
ANSWER:
m
=
4πR1
3
ρ
3
Hint 3. Gravitational force
Fg
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Chapter 11 and 13 Homework
Find Fg , the magnitude of the gravitational force on the satellite.
Express your answer in terms of G ,
,
,
,
R2
, and ρ.
,
,
,
R2
, and ρ.
m π R1
ANSWER:
3
Gm⋅4πR1
=
Fg
R2
2
ρ
⋅3
Hint 4. Speed of the satellite
Find the speed v of the satellite.
Express your answer in terms of G ,
m π R1
ANSWER:
−
−−−−−−
v
=
√
G⋅4πR1
3
ρ
3R2
ANSWER:
K
=
Gρ
2
3
3
πR1 m
R2
Correct
Part B
Find U , the gravitational potential energy of the satellite. Take the gravitational potential energy to be zero for an
object infinitely far away from the planet.
Express the satellite's gravitational potential energy in terms of G ,
,
,
m π R1
,
R2
, and ρ.
Hint 1. What physical principle to use
The gravitational potential energy associated with a
1
R
2
force is best remembered. If you have to work it out,
remember that the potential energy of an object found at height h = B, relative to height h = A, is equal to
the negative of the work done by the gravitational force when the object is brought from A to B:
U (B) = −WA→B
,
where
W
In this case,
Fg
A→B
= ∫
B
A
Fg dh .
is a function of the radius of the satellite trajectory,
R = ∞
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, and the zero point of the energy is
R
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Chapter 11 and 13 Homework
conventionally taken at
, so you want to use
R = ∞
W∞→R
2
= ∫
R2
∞
Fg (R) dR .
Hint 2. How to handle the math
You will need to solve an integral of the form
−C/x
∫ (C /x ) dx ,
2
where C is a constant. This evaluates to
, and you will need to evaluate this between the limits of integration.
ANSWER:
U
=
−Gρ
4
3
3
πR1 m
R2
Correct
Part C
What is the ratio of the kinetic energy of this satellite to its potential energy?
Express
K
U
in terms of parameters given in the introduction.
ANSWER:
K
U
= -0.500
Correct
The result of this problem may be expressed as
(i.e.
n
F (R) = −G ⋅ M ⋅ m ⋅ R
K
U
=
n+1
2
where n
= −2
is the exponent of the force law
). This is a specical case of a general and powerful theroem of advanced
classical mechanics known as the Virial Theorem. The theorem applies to the average of the kinetic and
potential energies of of any one or multiple objects moving over any closed (or almost closed) path that returns
very close to itself provided that all objects interact via potentials with the same power law dependence on their
separation. Thus it applies to stars in a galaxy, or masses tied together with springs (where
the force law is
1
F (R) = −k ⋅ R
⟨K ⟩
⟨U ⟩
=1
since
).
Score Summary:
Your score on this assignment is 108%.
You received 14.19 out of a possible total of 14 points, plus 0.99 points of extra credit.
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