Chapter 11 and 13 Homework

4/24/2014 Chapter 11 and 13 Homework

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Chapter 11 and 13 HomeworkDue: 10:00pm on Wednesday, April 23, 2014

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Exercise 11.1

A 0.100- , 40.2- -long uniform bar has a small 0.080- mass glued to its left end and a small 0.150- mass gluedto 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:

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—arein a state described as static equilibrium. Indeed, it is the fundamental objective of many branches of engineering tomaintain this state in spite of the presence of obvious forces imposed by gravity and static loads or the moreunpredictable forces from wind and earthquakes.

The condition of static equilibrium is equivalent to the statement that the bodies involved have neither linear nor angularacceleration. Hence static mechanical equilibrium (as opposed to thermal or electrical equilibrium) requires that theforces acting on a body simultaneously satisfy two conditions:

and ;that is, both external forces and torques sum to zero. You have the freedom to choose any point as the origin aboutwhich to take torques.

Each of these equations is a vector equation, so each represents three independent equations for a total of six. Thus tokeep a table static requires not only that it neither slides across the floor nor lifts off from it, but also that it doesn't tiltabout 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 leaningagainst a wall, which is in danger of slipping only in the plane perpendicular to the ground and wall. By orienting aCartesian coordinate system so that the x and y axes are in this plane, choose which of the following sets ofquantities must be zero to maintain static equilibrium in this plane.

Hint 1. Simplifying the equations

kg cm kg kg

= 24.4 d cm

∑ = 0F ⃗ ∑ = 0τ ⃗



The motion (or possible motion) is confined to a plane, the xy plane in this case, when there are no forcesacting out of that plane (e.g., all 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 thethree equations for the components of torque since they will equal zero.

ANSWER:

Correct

Part B

As an example, consider the case of a board of length and negligible mass. Take the x axis to be the horizontalaxis along the board and the y axis to be the vertical axis perpendicular to the board. A mass of weight isstrapped to the board a distance from the left-hand end. This is a static equilibrium problem, and a good first stepis to write down the equation for the sum of all the forcesin the y direction since the only nonzero forces of

that exist are in the y direction.

What is ? Your equation for the net force in the y

direction on the board should contain all the forces actingvertically on the board.

Express your answer in terms of the weight andthe tensions in the two vertical ropes at the left andright ends and . Recall that positive forcespoint upward.

ANSWER:

Correct

The only relevant component of the torques is the z component; however, you must choose your pivot point beforewriting the equations. This point could be anywhere; in fact, the pivot point does not even have to be at a point on thebody. You should choose this point to your advantage. Generally, the best place to locate the pivot point is where someunknown force acts; this will eliminate that force from the resulting torque equation.

= 0Fz

and and

and and

and and and

and and and and

∑ Fx ∑ τz ∑ Fy

∑ Fz ∑ τx ∑ τy

∑ τx ∑ Fx ∑ τy ∑ Fy

∑ τx ∑ Fx ∑ τy ∑ Fy ∑ τz

L

W

x

∑ = 0F ⃗

∑ Fy

W

TL TR

= ∑ = 0 =Fy + − WTL TR



Part C

What is the equation that results from choosing the pivot point to be the point from which the mass hangs (where acts)?

Express your answer in terms of the unknown quantities and and the known lengths and .Recall that counterclockwise torque is positive.

ANSWER:

Correct

This gives us one equation involving two unknowns, and . We can use this result and to

solve for and .

Part D

What is the equation that results from choosing the pivot point to be the left end of the plank (where acts)?

Express your answer in terms of , , , and the dimensions and . Not all of these variables mayshow up in the solution.

ANSWER:

Correct

Part E

What is the equation that results from choosing the pivot point to be the right end of the plank (where acts)?

Express your answer in terms of , , , and the dimensions and . Not all of these variables mayshow up in the solution.

ANSWER:

Correct

Part F

Solve for , the tension in the right rope.

W

TL TR x L

= ∑ = 0 =τW,z (L − x) − xTR TL

TL TR ∑ = 0Fy

TL TR

TL

TL TR W L x

= ∑ = 0 =τL,z L − WxTR

TR

TL TR W L x

= ∑ = 0 =τR,z W(L − x) − LTL

TR

W L



Express your answer in terms of and the dimensions and . Not all of these variables may show upin the solution.

Hint 1. Choose the correct equation

Which single equation of the ones you've derived can be used to solve for in terms of known quantities?

ANSWER:

ANSWER:

Correct

Part G

Solve for , the tension in the left rope.


Hint 1. Choose the correct equation

Which single equation of the ones you've derived can be used to solve for in terms of known quantities?

ANSWER:

ANSWER:

W L x

TR

∑ = 0Fy

∑ = 0τW

∑ = 0τL

∑ = 0τR

= TR W xL

TL

W L x

TL

∑ = 0Fy

∑ = 0τW

∑ = 0τL

∑ = 0τR

= TL W L−xL



Correct

Part H

Solve for the tension in the left rope, , in the special case that . Be sure the result checks with your

intuition.


ANSWER:

Correct

Only one set of forces, exactly balanced, produces static equilibrium. From this perspective it might seempuzzling that so much of the world is static. One must realize, however, that many forces—like those of thetensions in the ropes here or those between the floor and an object resting on it—increase very quickly as theobject moves. If there is a slight imbalance of the forces, the object accelerates so that its position changesuntil the object has adjusted itself to restore the force balance. It then oscillates about this point until friction orsome other dissipative mechanism causes it to become stationary at the exact equilibrium point.

Precarious Lunch

A uniform steel beam of length and mass is attached via a hinge to the side of a building. The beam is supportedby a steel cable attached to the end of the beam at an angle , as shown. Through the hinge, the wall exerts anunknown force, , on the beam. A workman of mass sits eating lunch a distance from the building.

Part A

TL x = 0

W L x

= TL W

L m1θ

F m2 d

T



Find , the tension in the cable. Remember to account for all the forces in the problem.

Express your answer in terms of , , , , , and , the magnitude of the acceleration due togravity.

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 , youwant to pick the axis such that will give a torque, but as few as possible other unknown forces will enterthe equations. So where should you place the origin for the purpose of calculating torques?

ANSWER:

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 torotate objects counterclockwise around the origin.

Answer in terms of , , , , , , and .

ANSWER:

ANSWER:

Correct

Part B

Find , the -component of the force exerted by the wall on the beam ( ), using the axis shown. Remember to

pay attention to the direction that the wall exerts the force.

Express your answer in terms of and other given quantities.

Hint 1. Find the sign of the force

T

m1 m2 L d θ g

T

T

At the center of the bar

At the hinge

At the connection of the cable and the bar

Where the man is eating lunch

T L d m1 m2 θ g

= 0 = Σ τa ( + d)g−TLsin(θ)Lm12

m2

= Tg( + d)m1

L

2m2

Lsin(θ)

Fx x F

T



The beam is not accelerating in the -direction, so the sum of the forces in the -direction is zero. Using thegiven coordinate system, is going to have to be positive or negative?

ANSWER:

Correct

Part C

Find , the y-component of force that the wall exerts on the beam ( ), using the axis shown. Remember to pay

attention to the direction that the wall exerts the force.

Express your answer in terms of , , , , and .

ANSWER:

Correct

If you use your result from part (A) in your expression for part (C), you'll notice that the result simplifiessomewhat. The simplified result should show that the further the luncher moves out on the beam, the lower themagnitude of the upward force the wall exerts on the beam. Does this agree with your intuition?

Exercise 11.5

A ladder of length 20.0 is carried by a fire truck. The ladder has a weight of 3000 and its center of gravity is at itscenter. The ladder is pivoted at one end (A) about a pin ; youcan ignore the friction torque at the pin. The ladder is raisedinto position by a force applied by a hydraulic piston at C.Point C is a distance 8.0 from A, and the force exertedby the piston makes an angle of = 40 with the ladder.

x x

Fx

= Fx −Tcos(θ)

Fy F

T θ m1 m2 g

= Fy g( + ) − Tsin(θ)m1 m2

m N

m F ⃗ θ ∘



Part A

What magnitude must have to just lift the ladder off the support bracket at B?

Express your answer using two significant figures.

ANSWER:

Correct

Tidal Forces

Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularlyvisible on the earth's surface in the form of tides. To understand the origin of tidal forces, consider the earth-moonsystem to consist of two spherical bodies, each with a spherical mass distribution. Let be the radius of the earth, be the mass of the moon, and be the gravitational constant.

Part A

Let denote the distance between the center of the earth and the center of the moon. What is the magnitude of theacceleration of the earth due to the gravitational pull of the moon?

Express your answer in terms of , , and .

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 useNewton's 2nd law to determine the magnitude of the acceleration of the earth. Recall that the gravitationalinteraction of two spherically symmetric bodies is calculated as if the mass of each body were concentratedin its center.

Hint 2. Find the gravitational force exerted on the earth by the moon

If is the mass of the earth, what is the gravitational force exerted by the moon on the earth?

Express your answer in terms of , , , and .

ANSWER:

ANSWER:

F ⃗

= 5800 F N

re mG

r

ae

G m r

me Fg

G me m r

= FgG mme

r2

= aeGmr2



Correct

Just as the earth accelerates toward the moon, the moon accelerates towards the earth. These are thecentripetal accelerations that cause the earth and moon to follow circular orbits around their mutual center ofrotation.

Part B

Since the gravitational force between two bodies decreases with distance, the acceleration experienced by a

unit mass located at the point on the earth's surface closest to the moon is slightly different from the acceleration 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 .


Hint 1. Find the acceleration at the point on earth nearest the moon

What is , 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?



Repeat the same calculations as in Part A, taking into account, however, that a body located at thepoint 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'ssurface 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 pointon the earth's surface nearest the moon is shifted toward the moon with respect to the center of theearth 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 and .

ANSWER:

ANSWER:

Hint 2. Find the acceleration at the point on earth farthest from the moon

What is , the magnitude of the acceleration of the same amount of water at the point on the earth's

anear

afar

−anear afar

G m r re

anear

G r re m

r re

r − re

= anearGm

(r− )re2

afar



surface farthest from the moon?

Express your answer in terms of , , , .

Hint 1. Find the distance between the center of the moon and the point on the earth'ssurface 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 pointon the earth's surface farthest from the moon is shifted away from the moon with respect to thecenter of the earth a distance equal to the earth's radius. How far is that point, then, from the centerof the moon?

Express your answer in terms of and .

ANSWER:

ANSWER:

ANSWER:

Correct

If you simplified your answer, you found that

.

Note that, since , the difference is very well approximated by the expression .

On the side of the earth nearest the moon (near side), water has a 7% greater acceleration than on the farthestside (far side) and bulges out, causing a high tide. Water on the far side is less strongly attracted toward themoon and thus another tidal bulge occurs. In total then, the earth experiences two high tides. Note that thesetidal bulges of water do not appear from nowhere; instead they are formed by water flowing away from otherareas 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 atan average distance of from the center of the sun. Given that the mass of the sun is

G m r re

r re

r + re

= afarGm

(r+ )re2

= −anear afar−Gm

(r− )re2

Gm

(r+ )re2

− = Gmanear afar4re

r3 (1 − )( )re

r

2 −2

r ≫ re 4Gm /re r3

1.5 × m1011

1.99 × kg30



, which of the following statements is correct?


Since the force exerted on a body is proportional to its acceleration, to compare the gravitational effect of thesun 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 of the earth due to the gravitational force of the sun?

Express your answer numerically in meters per second squared.

ANSWER:

ANSWER:

Correct

Although the sun is much farther away from the earth than the moon, it is much more massive. As a result, thegravitational force exerted on the earth by the sun is about 180 times stronger than the corresponding pull fromthe moon!

Part D

The occurrence of tidal forces on the earth's surface is not limited to the gravitational effects of the moon. Tidalforces are produced every time different parts of a body are subject to different gravitational forces exerted by asecond body. Therefore, tidal forces due to the gravitational effects of the sun are also present on the earth'ssurface. What can you conclude about the relative effects of these two tidal forces on the earth's surface?


The strength of the tides is related to the difference between the acceleration of water on opposite sides ofthe 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 theapproximate formula for from the follow-up comment to Part B (but of course plug in the sun's

mass rather than the moon's).

1.99 × kg1030

ae

= 5.90×10−3 ae m/s2

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 forceexerted by the moon.

−anear afar



Express your answer numerically in meters per second squared.

ANSWER:

ANSWER:

Correct

Even though the gravitational force exerted on the earth by the sun is about 180 times stronger than thecorresponding pull from the moon, the differential pull is smaller. In fact, the difference in the sun's pull onopposite sides of the earth is about half the difference in the moon's pull. The effects of the tidal forces causedby the sun, however, become particularly evident when the sun, the moon, and the earth are aligned. On thisoccasion, the gravitational effects of the sun are added to the gravitational effects of the moon and the highesttides are observed (called spring tides, although they have nothing to do with the spring season). When thesun is at an angle of with respect to the line joining the moon and the earth, instead, the gravitationaleffects of the sun partially cancel the effects of the moon and the least difference between high and low tide isobserved (called neap tides).

Exercise 13.5

Two uniform spheres, each of mass 0.260 , are fixed at points and (the figure ).

Part A

Find the magnitude of the initial acceleration of a uniform sphere with mass 0.010 if released from rest at point

= 1.00×10−6 −anear afar m/s2

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.

90∘

kg A B

kgP A B



and acted on only by forces of gravitational attraction of the spheres at and .


ANSWER:

Correct

Part B

Find the direction of the initial acceleration of a uniform sphere with mass 0.010 .

ANSWER:

Correct

Gravitational Acceleration inside a Planet

Consider a spherical planet of uniform density . The distance from the planet's center to its surface (i.e., the planet'sradius) is . An object is located a distance from the center of the planet, where . (The object is locatedinside of the planet.)

Part A

Find an expression for the magnitude of the acceleration due to gravity, , inside the planet.

Express the acceleration due to gravity in terms of , , , and , the universal gravitational constant.

Hint 1. Force due to planet's mass outside radius

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 thedistances from those points, I say, that a corpuscle placed within that surface will not be attracted by thoseforces any way.In other words, you don't have to worry about the portion of the planet's mass that is located outside of theradius . The net gravitational force from this "outer shell" of mass will equal zero. You only have to worryabout that portion of the planet's mass that is located within a radius .

P A B

= 2.1×10−9 a m/s2

kg

upward

to the right

downward

to the left

ρRp R R < Rp

g(R)

ρ R π G

R

R

R

R



Hint 2. Find the force on an object at distance

Suppose the object has a mass . Find the magnitude of the gravitational force acting on this object whenit is located a distance from the center of the planet.

Express the force in terms of , , , , and , the universal gravitational constant.

Hint 1. Find the mass within a radius

Find the net mass of the planet located within the radius . Remember, the volume of a sphere is .


ANSWER:

Hint 2. Magnitude of gravitational force

The general equation for the magnitude of the gravitational force is .

ANSWER:

Hint 3. Finding from

According to Newton's 2nd law, the net force acting on an object is given by . In this problem,

and since the only force acting on the object is the gravitational force. Therefore,

, where is the force you found in the previous hint.

Note that in this usage, both and are magnitudes and hence are positive. By convention, (or

in this case) is the magnitude of the gravitational field. This gravitational field is a vector, with direction

downward.

ANSWER:

Correct

Part B

R

m

R

m ρ π R G

R

R

(4/3)πR3

ρ π R

= M(R) π ρ43

R3

= GMm/Fgrav R2

= F(R) πGmρR43

g(R) F(R)

= maFnet

a = g(R) = F(r)Fnet

g(R) = F(R)/m F(R)

F(R) a g

g(R)

= g(R) πGρR43

g(R)



Rewrite your result for in terms of , the gravitational acceleration at the surface of the planet, times a

function of R.


Hint 1. Acceleration at the surface

Note that the acceleration at the surface should be equal to the value of the function from Part A

evaluated at the radius of the planet: .

ANSWER:

Correct

Notice that increases linearly with , rather than being proportional to . This assures that it is zero at

the center of the planet, as required by symmetry.

Part C

Find a numerical value for , the average density of the earth in kilograms per cubic meter. Use for

the radius of the earth, , and a value of at the surface of .

Express your answer to three significant figures.


You already derived the relation needed to solve this problem in Part A:.

At what distance is known so that you could use this relation to find ?

ANSWER:

Correct

Exercise 13.11

g(R) gp

gp R Rp

g(R)= g( )gp Rp

= g(R) ( )gpRRp

g R 1/R2

ρearth 6378 km

G = 6.67 × /(kg ⋅ )10−11m3 s2 g 9.80 m/s2

g(R) = (4/3)πGρR

R g(R) ρ

= 5500 ρearth kg/m3



Part A

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.620 if the

acceleration due to gravity at the surface has magnitude

ANSWER:

Correct

Energy of a Spacecraft

Very far from earth (at ), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitationalforce of the earth were to act on it (i.e., neglect the forces from the sun and other solar system objects), the spacecraftwould eventually crash into the earth. The mass of the earth is and its radius is . Neglect air resistancethroughout this problem, since the spacecraft is primarily moving through the near vacuum of space.

Part A

Find the speed of the spacecraft when it crashes into the earth.

Express the speed in terms of , , and the universal gravitational constant .


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 ?

ANSWER:

Hint 3. Potential energy

If the spacecraft has mass , what is its potential energy at the surface of the earth? Note that, at thesurface of the earth, the spacecraft is a distance from the center of the earth.

Express your answer in terms of , , , and the universal gravitational constant .

Hint 1. Formula for the potential energy

The gravitational potential energy of a system of 2 masses and separated by a distance

m/s2

9.80 m/ ?s2

1.90×107 m

R = ∞

M e Re

se

M e Re G

R = ∞

= 0E

M U

Re

M Me Re G

Ug m1 m2



is

.

ANSWER:

ANSWER:

Correct

Part B

Now find the spacecraft's speed when its distance from the center of the earth is , where the coefficient

.

Express the speed in terms of and .

Hint 1. General approach

This problem is very similar to the problem that you've just done. Note that the potential energy of thespacecraft at a distance 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 in terms of , , , and .

ANSWER:

ANSWER:

Correct

r

= −UgGm1m2

r

= U( )Re−(G M)Me

Re

= se2GMe

Re

− −−−−√

R = αRe

α ≥ 1

se α

αRe

R = αRe Me G Re α

= sα2 GMe

αRe

− −−−−√

= sα

se

α√



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 theearth.

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that wereobserved by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as thespeed of the space shuttle and the height of geostationary satellites) is the simplest orbit--a circular one. This problemconcerns the properties of circular orbits for a satellite orbiting a planet of mass .

For all parts of this problem, where appropriate, use for the universal gravitational constant.

Part A

Find the orbital speed for a satellite in a circular orbit of radius .

Express the orbital speed in terms of , , and .

Hint 1. Find the force

Find the radial force on the satellite of mass . (Note that will cancel out of your final answer for .)

Express your answer in terms of , , , and . Indicate outward radial direction with apositive sign and inward radial direction with a negative sign.

ANSWER:

Hint 2. A basic kinematic relation

Find an expression for the radial acceleration for the satellite in its circular orbit.

Express your answer in terms of and . Indicate outward radial direction with a positive signand inward radial direction with a negative sign.

ANSWER:

Hint 3. Newton's 2nd law

Apply to the radial coordinate.

ANSWER:

M

G

v R

G M R

F m m v

m M G R

= F−GMm

R2

ar

v R

= ar−v2

R

= mF ⃗ a⃗



Correct

Part B

Find the kinetic energy of a satellite with mass in a circular orbit with radius .


ANSWER:

Correct

Part C

Express the kinetic energy in terms of the potential energy .

Hint 1. Potential energy

What is the potential energy of the satellite in this orbit?


ANSWER:

ANSWER:

= vGM

R

− −−−√

K m R

m M G R

= KGMm

2R

K U

U

m M G R

= U−GMm

R

= K −U2



Correct

This is an example of a powerful theorem, known as the Virial Theorem. For any system whose motion isperiodic or remains forever bounded, and whose potential behaves as

,Rudolf Clausius proved that

,

where the brackets denote the temporal (time) average.

Part D

Find the orbital period .


Hint 1. How to get started

Use the fact that the period is the time to make one orbit. Then .

ANSWER:

Correct

Part E

Find an expression for the square of the orbital period.


ANSWER:

Correct

This shows that the square of the period is proportional to the cube of the semi-major axis. This is Kepler'sThird Law, in the case of a circular orbit where the semi-major axis is equal to the radius, .

Part F

U ∝ Rn

⟨K⟩= ⟨U⟩n2

T

G M R π

time = distance/velocity

= T 2πR32

GM√

G M R π

= T 2 4π2R3

GM

R

L



Find , the magnitude of the angular momentum of the satellite with respect to the center of the planet.


Hint 1. Definition of angular momentum

Recall that , where is the momentum of the object and 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, is

perpendicular to .

ANSWER:

Correct

Part G

The quantities , , , and 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 , , , and , in thatorder. For example, -1,-1/2,-0.5,-3/2 would mean , , and so forth.

Hint 1. Example of a power law

The potential energy behaves as , so depends inversely on . Therefore, the appropriate

power for this is (i.e., ).

ANSWER:

Correct

Geosynchronous Satellite

A satellite that goes around the earth once every 24 hours is called a geosynchronous satellite. If a geosynchronoussatellite is in an equatorial orbit, its position appears stationary with respect to a ground station, and it is known as ageostationary satellite.

L

m M G R

= ×L⃗ R⃗ p ⃗ p ⃗ R⃗

p ⃗

R⃗

= L m GMR− −−−−√

v K U L R

v K U L

v ∝ R−1 K ∝ R−1/2

U = GMm/R U R

−1 U ∝ R−1

-0.500,-1,-1,0.500



Part A

Find the radius of the orbit of a geosynchronous satellite that circles the earth. (Note that is measured fromthe center of the earth, not the surface.) You may use the following constants:

The universal gravitational constant is .

The mass of the earth is .

The radius of the earth is .

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. Supposethe mass of the satellite is , the mass of the earth is , and the radius of the satellite's orbit is . Whatis the magnitude of the force that acts on the satellite?

Answer in terms of , , , and the universal gravitational constant . (Use variables, notnumerical values.)

ANSWER:

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 ,where is the angular frequency of the satellite's orbit.

What is the numerical value for in radians per second for a geosynchronous satellite?

Hint 1. How to calculate

R R

G 6.67 × N /k10−11 m2 g2

5.98 × kg1024

6.38 × m106

m M R

m M R G

= FGmM

R2

a = Rω2

ω

ω

ω



Calculate using the definition of a geosynchronous orbit; the angular velocity should be such thatthe satellite makes one orbit per day. The equation relating the angular velocity and the time period

is

.

Hint 2. What is ?

Here the time period is one day, but you are asked for the angular velocity in radians per second.

ANSWER:

Hint 3. Find an expression for

Type an expression for the radius of the circular orbit of a satellite orbiting the earth.

Express your answer in terms of , (the mass of the earth), and , the angular velocity of thesatellite.

Hint 1. Putting it all together

Using Newton's 2nd law, , gives us

.

Find from this equation.

ANSWER:

ANSWER:

Correct

A Satellite in a Circular Orbit

Consider a satellite of mass that orbits a planet of mass in a circle a distance 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

ω

ω

T

ω = 2πT

T

T

= 7.27×10−5 radians/s ω

R

R

G M ω

F = ma

= m RGMm

R2 ω2

R

= RGM

ω2

− −−√3

= 4.23×107 m R

m1 m2 r



problem is true or false.

Part A

The information given is sufficient to uniquely specify the speed, potential energy, and angular momentum of thesatellite.

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 thesubsequent motion in Newtonian mechanics. Are you able to determine the satellite's velocity from theinformation given?

ANSWER:

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 aconservative force?

ANSWER:

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.

true

false

true

false



ANSWER:

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:

Correct

Part E

The equations that express the conservation laws of total mechanical energy and linear momentum are sufficient tosolve for the speed necessary to maintain a circular orbit at without using .

Hint 1. How are conservation laws used?

Conservation laws are generally applied to a system that has some "initial" and "final" conditions that aredifferent. Motion in a circular orbit, however, possesses no obvious initial and final points that are different.

ANSWER:

Correct

Kepler's 3rd Law

true

false

true

false

R = mF ⃗ a ⃗

true

false



A planet moves in an elliptical orbit around the sun. The mass of the sun is . The minimum and maximum distancesof the planet from the sun are and , respectively.

Part A

Using Kepler's 3rd law and Newton's law of universal gravitation, find the period of revolution of the planet as itmoves around the sun. Assume that the mass of the planet is much smaller than the mass of the sun.

Use for the gravitational constant.

Express the period in terms of , , , and .

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 tothe cube of the semi-major axis of its orbit. Try finding the period of a circular orbit and then using Kepler's3rd 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 .

Express the semi-major axis in terms of and .

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 orbitand the focus lies on the major axis.

ANSWER:

Hint 3. Find the period of a circular orbit

Find the period of a planet in a circular orbit of semi-major axis .

Express the period in terms of , , and .

Hint 1. Formula for the period

The period is , where is the radius of the orbit and 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 of an object in an orbit of radius by setting the magnitude of the centripetalacceleration equal to the magnitude of the acceleration due to gravity.

M sR1 R2

P

G

G M s R1 R2

a

R1 R2

= a+R1 R2

2

P a

a Ms G

2πr/v r v

v r

= /racent v2

M s G




ANSWER:

Hint 3. Radius of the orbit

For a circle, the semi-major axis is just the radius.

ANSWER:

ANSWER:

Correct

Exercise 13.26

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 and the other at64,000 . Pluto already was known to have a large satellite Charon, orbiting at 19,600 with an orbital period of6.39 days.

Part A

Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without usingthe mass of Pluto.

Enter your answers numerically separated by a comma.

ANSWER:

Correct

Exercise 11.10

r M s G

= v GMs

r

− −−−√

= P 2π a3

GMs

− −−−√

= P π2( + )R1 R23

2GMs

− −−−−−−−−√

kmkm km

= 24.5,37.7 days ,T1 T2



A uniform ladder 5.0 long rests against a frictionless, vertical wall with its lower end 3.0 from the wall. The ladderweighs 160 . The coefficient of static friction between the foot of the ladder and the ground is 0.40. A man weighing740 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?


ANSWER:

Part B

What is the actual frictional force when the man has climbed 1.0 m along the ladder?


ANSWER:

Part C

How far along the ladder can the man climb before the ladder starts to slip?


ANSWER:

Exercise 13.4

Two uniform spheres, each with mass and radius , touch one another.

Part A

What is the magnitude of their gravitational force of attraction?

Express your answer in terms of the variables , , and appropriate constants.

ANSWER:

m mN

N

= Ffr N

= Ffr N

= s m

M R

M R



Orbiting Satellite

A satellite of mass is in a circular orbit of radius around a spherical planet of radius made of a materialwith density . ( is measured from the center of theplanet, not its surface.) Use for the universal gravitationalconstant.

Part A

Find the kinetic energy of this satellite, .

Express the satellite's kinetic energy in terms of , , , , , and .

Hint 1. Kinematics of circular motion

The circular trajectory of the satellite implies that there is a certain inward radial acceleration , which must be due to the gravitational force .

Hint 2. Mass of the planet

Find the mass of the planet in terms of its size and density.

Express in terms of and .

ANSWER:

Hint 3. Gravitational force

m R2R1

ρ R2G

K

G m π R1 R2 ρ

= −aradial R2ω2 Fg

m ρ R1

= m4π ρR1

3

3

Fg



Find , the magnitude of the gravitational force on the satellite.

Express your answer in terms of , , , , , and .

ANSWER:

Hint 4. Speed of the satellite

Find the speed of the satellite.

Express your answer in terms of , , , , , and .

ANSWER:

ANSWER:

Correct

Part B

Find , the gravitational potential energy of the satellite. Take the gravitational potential energy to be zero for anobject infinitely far away from the planet.

Express the satellite's gravitational potential energy in terms of , , , , , and .

Hint 1. What physical principle to use

The gravitational potential energy associated with a force is best remembered. If you have to work it out,

remember that the potential energy of an object found at height , relative to height , is equal tothe negative of the work done by the gravitational force when the object is brought from to :

,

where

.

In this case, is a function of the radius of the satellite trajectory, , and the zero point of the energy is

Fg

G m π R1 R2 ρ

= FgGm⋅4π ρR1

3

⋅3R22

v

G m π R1 R2 ρ

= vG⋅4π ρR1

3

3R2

− −−−−−−√

= KGρ π m2

3R1

3

R2

U

G m π R1 R2 ρ

1R2

h = B h = A

A B

U(B) = −WA→B

= dhWA→B ∫ B

AFg

Fg R

R = ∞



conventionally taken at , so you want to use

.

Hint 2. How to handle the math

You will need to solve an integral of the form , where is a constant. This evaluates to

, and you will need to evaluate this between the limits of integration.

ANSWER:

Correct

Part C

What is the ratio of the kinetic energy of this satellite to its potential energy?

Express in terms of parameters given in the introduction.

ANSWER:

Correct

The result of this problem may be expressed as where is the exponent of the force law

(i.e. ). 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 andpotential energies of of any one or multiple objects moving over any closed (or almost closed) path that returnsvery 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 since

the force law is ).

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.

R = ∞

= (R) dRW∞→R2∫ R2

∞ Fg

∫ (C/ ) dxx2 C

−C/x

= U−Gρ π m4

3R1

3

R2

KU

= -0.500KU

=KU

n+12

n = −2

F(R) = −G ⋅ M ⋅ m ⋅ Rn

= 1⟨K⟩

⟨U⟩

F(R) = −k ⋅ R1

Chapter 11 and 13 Homework

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