Lecture Outlines Chapter 12 Physics, 3

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Lecture Outlines Chapter 12

Physics, 3

Edition

James S. Walker

Chapter 12

Gravity

Units of Chapter 12

• Newton’s Law of Universal Gravitation

• Gravitational Attraction of Spherical Bodies

• Kepler’s Laws of Orbital Motion

• Gravitational Potential Energy

• Energy Conservation

• Tides*

12-1 Newton’s Law of Universal Gravitation

Newton’s insight:

The force accelerating an apple downward is the same force that keeps the Moon in its orbit.

Hence, Universal Gravitation.

Figure 12-2

Dependence of the gravitational force on separation distance, r

12-1 Newton’s Law of Universal Gravitation

The gravitational force is always attractive, and points along the line connecting the two

masses:

The two forces shown are an action-reaction

Exercise 12-1

G är ett mycket litet tal (G = 6,67 10

Nm

/kg

).

Beräkna gravitationskraften mellan en husse (m

=105 kg) och hans hund (m

=11,2 kg) när de är på avståndet a) 1,0 m b) 10,0 m från varandra.

F(r = 1,00 m) = Gm

m

/r

= 7,84•10

N

F(r = 10,0 m) = Gm

m

/r

= 7,84•10

N

12-1 Newton’s Law of Universal Gravitation

G is a very small number; this means that the force of gravity is negligible unless there is a very large mass involved (such as the Earth).

If an object is being acted upon by several

different gravitational forces, the net force on it is the vector sum of the individual forces.

This is called the principle of superposition.

Example 12-1

How Much Force Is With You?

Exemple 12-1 How much force is with you?

Massan av rymdskeppet, Millenium Eagle, är 25,0 Mkg och vardera asteroidmassan är 350 Gkg.

Beräkna gravitationskraften på rymdskeppet i läge A och i läge B. Betrakta rymdskepp och asteroider

som punktobjekt.

r

= [(3,00 km)

+ (1,50 km)

]

= 3,35 km θ

= θ

= arctan(0,5) = 26,7°

F(r

) = Gm

m

/r

= 52,0 N

Totala kraften (i x-led) = 2•52,0 N•cos(26,7°) = 93,0 N

r = 1,50 km

12-2 Gravitational Attraction of Spherical Bodies

Gravitational force between a point mass and a

sphere: the force is the same as if all the mass

of the sphere were concentrated at its center.

12-2 Gravitational Attraction of Spherical Bodies

What about the gravitational force on objects at

the surface of the Earth? The center of the Earth is one Earth radius away, so this is the distance we use:

Therefore,

Photo 12-2 Global model of the Earth’s gravitational strength Gravitationen är starkast vid brun färg, svagast för blå.

Example 12-2

The Dependence of Gravity on Altitude

Exemple 12-2 The Dependence of Gravity on Altitude

Vad är accelerationen på grund av gravitationen på toppen av Mount Everest (h = 8850 m)?

Vid h = 0 gäller g = 9,81 m/s

dvs F = GMm/r

= gm

Vid h = 8850 m gäller

g

= GM/(r

+h)

= g/(1+h/r

)

= g/(1+0,00139)

=

= g • 99,7% = 9,78 m/s

12-2 Gravitational Attraction of Spherical Bodies

The acceleration of gravity decreases “slowly”

with altitude:

12-2 Gravitational Attraction of Spherical Bodies

Once the altitude becomes comparable to the radius of the Earth, the decrease in the

acceleration of gravity is much larger:

Exercise 12-2 The Dependence of Gravity on Altitude

Vad är accelerationen på månen? (M

= 7,35 • 10

kg och radie R

= 1,74 • 10

m)

F = mGM

/R

= g

m g

= GM

/R

=

= (6,67 • 10

) • (7,35 • 10

)/(1,74 • 10

)

= 1,62 m/s

Månlandaren har massan 225 kg. Vad vägde den a) på jorden? 225 kg • 9,81 m/s

= 2210 N

b) på månen? 225 kg • 1,62 m/s

= 364 N

12-2 Gravitational Attraction of Spherical Bodies

The Cavendish experiment allows us to measure

the universal gravitation constant:

12-2 Gravitational Attraction of Spherical Bodies

Even though the gravitational force is very small, the mirror allows measurement of tiny deflections.

Measuring G also allowed the mass of the Earth to

be calculated, as the local acceleration of gravity

and the radius of the Earth were known.

Exercise 12-3 Hur stor är jordens massa?

Vid havets nivå gäller g = 9,81 m/s

dvs F = GMm/r

= gm

M = g • r

/G = 9,81•(6,37•10

m)

/(6,67•10

) =

= 5,97•10

kg

Example 12-3 Mars Attracts!

Exemple 12-3 Mars attracts!

Hur stor är Mars massa?

Dess radie är 3,39•10

m och vid Marsytan gäller

g

= 3,73 m/s

dvs

M = g

• R

/G = 3,73•(3,39•10

m)

/(6,67•10

) =

= 6,43 • 10

kg

12-3 Kepler’s Laws of Orbital Motion

Johannes Kepler made detailed studies of the

apparent motions of the planets over many years, and was able to formulate three empirical laws:

1. Planets follow elliptical orbits, with the Sun at

one focus (brännpunkt) of the ellipse.

12-3 Kepler’s Laws of Orbital Motion

2. As a planet moves in its orbit, it sweeps out an

equal amount of area in an equal amount of time.

Conceptual Checkpoint 12-1 Compare speeds

Jordens bana runt solen är lätt elliptisk, så jorden är något närmare solen under vintern (!). Är då jordens hastighet < = > än hastigheten under

sommaren?

12-3 Kepler’s Laws of Orbital Motion

3. The period, T, of a planet increases as its

mean distance from the Sun, r, raised to the 3/2 power.

This can be shown to be a consequence of the

inverse square form of the gravitational force.

Figure 12-10

Kepler’s third law and some near misses

Härledning av Keplers tredje lag a

= v

/r

F = ma

= m v

/r = m (2πr/T)

/r = mr 4π

/T

men kraften är ju också

F = GM

m/r

sätter man dessa uttryck lika får man

T

= r

(4π

/GM

)

12-3 Kepler’s Laws of Orbital Motion

A geosynchronous satellite is one whose orbital period is equal to one day. If such a satellite is orbiting above the equator, it will be in a fixed position with respect to the ground.

These satellites are used for communications and

and weather forecasting.

Example 12-4

The Sun and Mercury

Example 12-4 The Sun and Mercury

Jorden kretsar kring solen på ett medelavstånd till solen av 1,50•10

m. Beräkna solens massa.

T

= r

(4π

/GM

) M

= r

4π

/GT

=

= (1,5•10

)

•4π

/{(6,67•10

)•(π•10

)

}

= 2,02•10

kg

Beräkna periodtiden för Merkurius vars

medelavstånd till solen är 5,79•10

m.

Photo 12-5

Geosynchronous orbit

Active Example 12-1 Find the altitude of a Geosynchronous Satellite

Satelliten kretsar kring jorden med en periodtid av 24h. (M

= 5,97 • 10

kg, R

= 6370 km)

T

= r

(4π

/GM

)

r

= M

GT

/4π

= (5,97•10

)(6,67•10

)(86400)

/4π

r = (7,52953•10

)

= 42226909 m

h = r - R

= 42,23 Mm – 6,37 Mm ≈ 35,8 Mm

12-3 Kepler’s Laws of Orbital Motion

GPS satellites are not in geosynchronous orbits;

their orbit period is 12 hours. Triangulation of

signals from several satellites allows precise

location of objects on Earth.

12-3 Kepler’s Laws of Orbital Motion

Kepler’s laws also give us an insight into possible orbital maneuvers.

[Conceptualcheckpoint 12-2 Which rockets to use?]

12-4 Gravitational Potential Energy[+ Exercise 12-4]

Gravitational potential energy of an object of

mass m a distance r from the Earth’s center:

12-4 Gravitational Potential Energy

Very close to the Earth’s surface, the

gravitational potential increases linearly with altitude:

Gravitational potential energy, just like all other forms of energy, is a scalar. It

therefore has no components; just a sign.

Potentiell Energi p.376

U = - mGM

/(R

+h)

bilda skillnaden och serieutveckla

ΔU = U

– U

= {R

>>h} ≈ - mGM

/R

•(1 – h/R

) - 1} =

= m(GM

/R

)h = mgh

Example 12-5 Simple Addition

Example 12-5 Simple Addition p.377

Beräkna systemets potentiella energi för de tre massorna i exempel 12-5 med m

= 2,5 kg, m

= 0,75 kg och m

= 0,75 kg.

U

= - Gm

m

/r

U

= - G • 2,5 kg • 0,75 kg/1,25 m = - 1,0 • 10

J

U

= - G • 2,5 kg • 0,75 kg/1,25√2 m = - 0,71 • 10

J U

= - G • 0,75 kg • 0,75 kg/1,25 m = - 0,30 • 10

J

U

= - 2,0 • 10

J

12-5 Energy Conservation

Total mechanical energy of an object of mass m a distance r from the center of the Earth:

This confirms what we already know – as an

object approaches the Earth, it moves faster

and faster.

12-5 Energy Conservation

12-5 Energy Conservation

Another way of visualizing the gravitational

potential well:

Example 12-6a

Armageddon Rendezvous

Example 12-6a Armageddon Rendez-vous

Anta att asteroiden startat i vila på oändligt avstånd från jorden. Beräkna dess hastighet då den är på

”månavstånd” från jorden.

E

= E

(konservativt kraftfält!) 0 = mv

/2 - GmM

/60R

v

= (2GM

/60R

)

=

= [2•(6,67•10

)•(5,97•10

)/60•6,37•10

]

=

= 1440 m/s

Example 12-6b

Armageddon Rendezvous

12-5 Energy Conservation: Escape speed (p.381) Anta att man vill skjuta iväg en raket med massan m med en begynnelsehastighet så att den kan

undfly jordens dragningskraft.

E

= mv

/2 - GmM

/R

När raketen är på oändligt avstånd från jorden så är avtar dess kinetiska och potentiella energi till noll.

Då kan den (lägsta) flykthastigheten beräknas v

= (2GM

/R

)

=

= [2•(6,67•10

)•(5,97•10

)/6,37•10

]

=

Exercise 12-5 Calculate the escape speed from the Moon (p.382) + Conceptual checkpoint 12-3 (m>< ?)

Anta att man vill skjuta iväg en raket med massan m med en begynnelsehastighet så att den kan

undfly månens dragningskraft.

E

= mv

/2 - GmM

/R

När raketen är på oändligt avstånd från månen så är avtar dess kinetiska och potentiella energi till noll.

Då kan den (lägsta) flykthastigheten beräknas v

= (2GM

/R

)

=

= [2•(6,67•10

)•(7,35•10

)/1,74•10

]

=

≈ 2370 m/s

Example 12-7 Half Escape

Exemple 12-7 Half Escape

Anta att man skjuter iväg en raket med en

begynnelsehastighet som är halva flykthastigheten.

Hur högt når raketen när dess hastighet = 0 ? E

= mv

/2 - GmM

/R

v

= (2GM

/R

)

så att begynnelseenergin nu blir E

= 75% (- GmM

/R

)

E

= - GmM

/r

Sätts energierna lika fås

r = 4 R

/3 (dvs den når bara höjden R

/3 ≈ 2000 km)

12-5 Energy Conservation

Speed of a projectile as it leaves the Earth,

for various launch speeds

12-5 Energy Conservation

Black holes:

If an object is sufficiently massive and sufficiently small, the escape speed will equal or exceed the speed of light –

light itself will not be able to escape the surface.

This is a black hole.

12-5 Energy Conservation

Light will be bent by any gravitational field; this can be seen when we view a

distant galaxy beyond a

closer galaxy cluster. This is

called gravitational lensing,

and many examples have

been found.

12-6* Tides

Usually we can treat planets, moons, and stars as though they were point objects, but in fact they are not.

When two large objects exert gravitational

forces on each other, the force on the near side is larger than the force on the far side, because the near side is closer to the other object.

This difference in gravitational force across an

object due to its size is called a tidal force.

12-6* Tides

This figure illustrates a general tidal force on

the left, and the result of lunar tidal forces on

the Earth on the right.

12-6* Tides

Tidal forces can result in orbital locking, where the moon always has the same face towards the planet – as does Earth’s Moon.

If a moon gets too close to a large planet, the tidal forces can be strong enough to tear the moon apart. This occurs inside the Roche

limit; closer to the planet we have rings, not

moons.

Summary of Chapter 12

• Force of gravity between two point masses:

• G is the universal gravitational constant:

• In calculating gravitational forces,

spherically symmetric bodies can be replaced

by point masses.

Summary of Chapter 12

• Acceleration of gravity:

• Mass of the Earth:

• Kepler’s laws:

1. Planetary orbits are ellipses, Sun at one focus

2. Planets sweep out equal area in equal time 3. Square of orbital period is proportional to

cube of distance from Sun

Summary of Chapter 12

• Orbital period:

• Gravitational potential energy:

• U is a scalar, and goes to zero as the

masses become infinitely far apart

Summary of Chapter 12

• Total mechanical energy:

• Escape speed:

• (Tidal forces are due to the variations in

gravitational force across an extended body)