Physics 111physics.valpo.edu/courses/p111/lectures/lecture20.pdf · where m1 and m2 are the masses...

Physics 111

Title page

Tuesday,November 9, 2004

Physics 111 Lecture 20

• Ch 12: GravityUniversal LawPotential Energy

Kepler’s Laws

• Ch 15: Fluidsdensityhydrostatic equilibriumPascal’s Principle

• Wednesday, 8 - 9 pm in NSC 118/119• Sunday, 6:30 - 8 pm in CCLIR 468

Help sessions

AnnouncementsTuesNov

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This week’s lab will be another physicsworkshop - on fluids this time. No quizthis week.

labs

AnnouncementsTuesNov

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Gravity

We extend our set of forces to includeNewton’s Universal Law of Gravity. We alsolook at how this new conservative force resultsin a new Gravitational Potential energy. Finally,we look briefly at Kepler’s Laws.

Ch 12: Gravity

Ch 12: GravityTuesNov

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We’ve mentioned last week that any twobodies that have gravitational mass exertan attractive force on one another.

So far, we’ve only looked at the form thisforce takes near the surface of the Earth,namely

Fg =

W = mg

Newton’s Universal Gravity

Ch 5: Applying Newton’s LawsTuesNov

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where m1 and m2 are the masses of the twoobjects, r is the distance between them, and is a unit vector pointing from m2 to m1. r2→1

But this form has been derived from a more generalform known as Newton’s Universal Law of Gravity:

F

g 2→1= −

Gm1m

2

r 2r2→1

the equation

G = 6.67 ×10−11 Nm2

kg2

Universal GravitationalConstant


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Well, we’ve seen unit vectors for the directionsin a Cartesian coordinate system. Remember

x The unit vector in the x-direction

y The unit vector in the y-direction

z The unit vector in the z-direction

R - hat


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What’s this r ?

y

z

x

r r

It’s the unit vectorthat points radiallyout from the originof the coordinatesystem to the pointof interest in space.

The quantity r (the radial distancefrom the origin to the point ofinterest) is one of three coordinatesin the spherical coordinate system.

con’t


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What’s this r ?

m1 m2

• Notice, these two forces are equal inmagnitude and opposite in direction.

• In fact, they ARE a 3rd Law Pair!

• We also note that these forces act at adistance: that is, the two objects have nodirect physical contact with one another.

F

g 2→1 F

g1→2

…and 3rd Law Pairs


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Worksheet #1

CQ1 universal gravity


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Worksheet #1

CQ1 – universal gravity ?

Two satellites A and B have the same mass and go aroundplanet Earth in concentric orbits. The distance of satelliteB from Earth’s center is twice that of satellite A. What isthe ratio of the gravitational force of the Earth on satelliteB to the gravitational force of the Earth on satellite A?

PI, Mazur (1997)

≥≥

1. 1/82. 1/43. 1/24.5. 1


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12

Such action-at-a-distance forces are termedfield forces. This means that we can definea new physical quantity known in this caseas the gravitational field.

G

2= −

Gm2

r 2r→2

Every mass has anassociated gravitational

field around it.

Gravitational Fields


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F

g 2→1= m

1

G

2

If we put a mass m1 in thefield G2 created by thepresence of mass m2, itfeels a force given by

Fields & forces

G 2 = −

Gm2

r2r→2


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reconciliation

How do we reconcile these two forms?

F

g 2→1= m

1

G

2 W = mg

True universally! True only forexperimentsnear the surfaceof planet Earth.

Let’s see what value G2 hasnear the surface of Earth.

G 2 = −

Gm2

r2r→2


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bingo

m⊕ = 5.98 ×1024 kg

r⊕ = 6.37 ×106 m

m2 will be the mass of the Earth and r will be theradius of the Earth. Since our experiments are atthe surface of the Earth, the distance from ourobjects to the center of the Earth applies.

G⊕ =

(6.67 ×10−11 Nm2

kg2 )(5.98 ×1024 kg)

(6.37 ×106m)2

= 9.83 m

s2

Let’s see what value G2 hasnear the surface of Earth.

G 2 = −

Gm2

r2r→2


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Worksheet #2

CQ2 orbital speed


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Worksheet #2

CQ2 – orbital speed

Two satellites A and B have the same mass and goaround planet Earth in concentric orbits. Thedistance of satellite B from Earth’s center is twicethat of satellite A. What is the ratio of thetangential speed of B to that of A?

PI, Mazur (1997)

≥≥

1. 1/22.3. 14.5. 2


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12

2

We’ve seen that near theEarth’s surface, the functionfor gravitational potentialenergy takes the form

U m g yg = | |

But we noted that this form is correctonly for problems that take place at ornear the Earth’s surface.

For problems in outer-space…we need to usea more general form of this function...

Gravitational Pot’l Energy


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This form is derived using calculus and therelationship between force and potential energy(also from calculus) with the assumption that

U = 0 when r = ∞

UGmm

rg = − 1 2

Formula for gravitational pot’lenergy


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If we have a system that involves several masses,we can compute the total potential energy of thesystem as the sum of the potential energiesbetween each pair of masses in the system.

So, for a 3-masssystem...

U

tot=U

12+U

13+U

23= −G

m1m

2

r12

+m

1m

3

r13

+m

2m

3

r23

⎛

⎝⎜⎞

⎠⎟

m1 m3

m2

r13

r12 r23

3 body system


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m1 m3

m2

r12 r23

r13Notice that this resultis simply the sum of theenergy changes thatresult when each massis brought from infinityto its final location.

Mass 1 is “free.”

Bringing up mass 2 inthe presence of mass 1results in the 1st term.

con’t


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U

tot=U

12+U

13+U

23= −G

m1m

2

r12

+m

1m

3

r13

+m

2m

3

r23

⎛

⎝⎜⎞

⎠⎟

The last two termsresult from bringingup mass 3 in thepresence of masses2 and 1.

con’t


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U

tot=U

12+U

13+U

23= −G

m1m

2

r12

+m

1m

3

r13

+m

2m

3

r23

⎛

⎝⎜⎞

⎠⎟

m1 m3

m2

r12 r23

r13

Bringing up mass 2 inthe presence of mass 1results in the 1st term.Mass 1 is “free.”

How much energy is required to move a 1000-kgmass from the Earth’s surface to a distance thatis twice the Earth’s radius away from the centerof the Earth? Use for the Earth a mass of6 X 1024 kg and a radius of 6400 km.

? Earth launch… (W3)

Worksheet #3

ΔU

g=U

f−U

i= −

GmEm

sat

rsat

−−Gm

Em

sat

rE

ΔU = 3.13 × 1010 JEnough energy to power a100-W light bulb for 10 years!


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The planets in our solarsystem move aroundthe Sun in roughlycircular orbits.

This means that someforce must be acting onthe planets causing acentripetal acceleration.

What force is responsible?

Planetary orbits


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Given that the Earth (m = 6 X 1024 kg) orbits theSun (m = 2 X 1030 kg) in roughly a circular orbit(r = 1.5 X 1011 m) once per year, calculated themean orbital speed of the Earth.

Fr= −F

G

mEv2

rES

=Gm

Em

S

rES2

v2 =

GmS

rES

vGm

rS

ES

=

? V earth (W4)

Worksheet #4


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v =

GmS

rES

=(6.67 ×10−11 Nm2

kg2)(2 ×1030 kg)

1.5×1011m

v = 29.8 km/s

Spaceship Earth travels through the Cosmosat a surprisingly high speed!

? V earth (W4)


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Given that the Earth (m = 6 X 1024 kg) orbits theSun (m = 2 X 1030 kg) in roughly a circular orbit(r = 1.5 X 1011 m) once per year, calculated themean orbital speed of the Earth.

The planets do not move around the Sunin perfectly circular orbits. The first personto figure out the correct shape of the orbitswas Johannes Kepler.

Kepler’s First Law says that the planets movearound the Sun in elliptical orbits, with theSun at one focus.

Kepler’s First Law


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foci

Major axis

center

ellipses

The sum of the distancefrom any point on theellipse to each of the twofoci is constant


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Kepler’s Second Law says a line drawnconnecting the Sun to a planet will sweep out anequal area in the ellipse in equal time intervals.

The law tells us how fast a planet moves at variouspoints in its orbit: close to the Sun the planet willhave a greater speed than far from the Sun.

Makes sense since gravity goes as 1/distance2

Kepler’s Second Law


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A

B

If it takesone monthfor the planetto go from A to B...

C

D

It will also take one month to gofrom C to D, if the areas of theblue & red triangles are the same.

cartoon


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Kepler’s Third Law says the square ofthe period of the orbit of a planet isproportional to the cube of the lengthof the semi-major axis of the orbit.

For a circular orbit, the semi-major axis issimply the radius of the circle (the diameterbeing the major axis).

Kepler’s Third Law


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Skip derivation

Let’s demonstrate this law for the case ofa circular orbit. Simply use Newton’s 2ndLaw where the Radial Force is theGravitational Force.

−FG= F

r

GmSun

mpl .

rSun− pl2

=m

plv

pl2

rSun− pl

period of circular orbit


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The planet will complete one orbit in one period.

The circumference of the orbit is the distancethe planet will travel in one period. So...

v

pl= C

T=

2πrSun− pl

T

Now plug this value ofvelocity into theabove equation...

con’t

GmSun

mpl .

rSun− pl2

=m

plv

pl2

rSun− pl


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GmSun

mpl .

rSun− pl2

=m

pl(2πr / T )2

rSun− pl

GmSun

rSun− pl

= 4π 2r 2

T 2T 2 =

4π 2

GmSun

⎛⎝⎜

⎞⎠⎟

r 3

T^2 and r^3

GmSun

mpl .

rSun− pl2

=m

plv

pl2

rSun− pl


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T K rSun2 3=

We can simplify thisexpression by calculatingthe constants...

where...

K

Sun= 4π 2

GmSun

= 2.97 ×10−19 s2

m3

simpler form


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T K a2 3=

Kepler’s Laws apply to the orbits of planets about the Sun moons about a planet satellites about a planet comets about a star

….You name it! If it’s in orbit, consult Kepler!

Other bodies?

Naturally, the constant K depends upon the bodybeing orbited!

a = semi-major axis


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? Pluto and Saturn (W5)

The semi-major axis of the orbit of Pluto is about4 times as great as that of the orbit of Saturn. IfSaturn orbits the Sun in about 30 years, how longdoes it take Pluto to orbit the Sun once?

1) 30 years2) 120 years3) 165 years4) 225 years5) 240 years

Worksheet #5


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Fluid Mechanics

We’ve spent a lot of time looking at systemsof solid objects. But matter also comes inliquid and gaseous states. We can describethe motions of such substances using theextension of Newtonian mechanics knownas fluid dynamics.

Ch 15: Fluid Mechanics

Ch 15: Fluid MechanicsTuesNov

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Let’s start by characterizinga solid mass of uniformcomposition.

What quantities can we directly measure?

If our block was made of copper, and we doubledits size, what would happen to its mass?

Mass & volume


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Remember our scale arguments (way back inthe first week of the course)!!!

L 2L

V0 = L3

V = (2L)3

V = 8L3 = 8V0

M

8M

scaling


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In fact, if we made a plot of the mass of ourcopper block versus its volume, we’d find

m

V

This line has a slopethat characterizes thetype of material fromwhich the block is made.We define the slope ofthis line as the density(ρ) of the material.

slope

= ρ

How would the slope of the line for a block ofstyrofoam compare to that for a block of lead?

density


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ρ ≡ MV

The unit of mass, the gram, was chosen to bethe mass of 1 cm3 of liquid water. So water hasa density of 1 g/cm3 or 1 kg/L or 1000 kg/m3.

The term specific gravity refers to theratio of the density of a given substanceto that of water. Objects with a specificgravity less than 1 will float in water;those greater than 1 will sink.

definition


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ρ ≡ M

V [ρ] = [ M ]

[V ]

[ρ] = kg

m3

Note: density is oftenwritten in grams percubic centimeter or g/cc.There is a factor of 1000difference between thetwo sets of units.

units


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PF

A≡

When a force is applied over an area, wesay that the object feels pressure.

Usually, we talk about pressure of a fluidor a gas (like the atmosphere).

[P] = [F]

[ A]= N

m2= Pa

“pascal”

pressure


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Our green fish is completely submerged. Thewater exerts a pressure on the fish. From whichdirection does the fish feel the pressure?

It’s all around you!

In fact, the pressure is exertedin the direction normal to thebody of the fish all over the fish.

Where is pressure?


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Think about your own experience walkingaround outside (when there’s NOT a hardwind blowing). You don’t notice any differencein pressure over the surface of your body.

What would happenif there was apressure differenceacross your hand?

P1

P2

P1> P

2


Pressure gradient


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Your hand would feel a net force acting to theRIGHT in this case. Therefore, your hand wouldstart to accelerate to the right, or you would haveto exert a force through your arm to counteractthis force known as a “pressure gradient” force.

Fnet

P1

P2


and force


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12PP>

So, now let’s look at how pressure changeswith altitude. We know it’s a lot harder tobreathe at the top of a mountain than atsea level--there are fewer oxygen moleculesand fewer molecules in general up there.

Climbing the mountain


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At the Earth’s surface, we sit at the bottom of anentire column of molecules in the atmosphere.These molecules exert a pressure on us at thesurface of about 101.325 kPa. We define thispressure to be 1 atmosphere (atm).

As we go up through theatmosphere, what happensto the pressure we feel?

WHY?

Atmospheric pressure


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Let’s look at the force balance for a little layerof atmosphere as we head up the mountain.

P1

P2W

The pressure ofall the moleculesabove our layer.

The pressure fromall the moleculesbelow our layer.

The gravitationalforce acting on thelayer itself. P1

< P2

A = surface area

Force on layer of atmosphere


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P1

P2W

A = surface area

The pressure P1 on the top surface of area Aresults in a force downward of F1 = P1 A.

The pressure P2 on the bottomsurface of area A results ina force upward of F2 = P2 A.

Net force


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If our system is in equilibrium,the net force must be 0. So...

F2= F

1+W P2

A = P1A+ mg

But what is the mass of ourlittle layer of atmosphere?Δh

m = ρV = ρ( AΔh)ρ is the density of air.

weight


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P1

P2W

A = surface area

P2A = P

1A+ mg

P2A = P

1A+ ρ( AΔh)g

P2= P

1+ ρg(Δh) Δ ΔP g h= ρ ( )

equilibrium


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If our system is in equilibrium,the net force must be 0. So...

Physics 111physics.valpo.edu/courses/p111/lectures/lecture20.pdf · where m1 and m2 are the masses...

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