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Lecture Outlines
Chapter 12
Physics, 3rd Edition
James S. Walker
© 2007 Pearson Prentice Hall
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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
pair.
Exercise 12-1
G är ett mycket litet tal (G = 6,67 10-11 Nm2/kg2).
Beräkna gravitationskraften mellan en husse
(m1=105 kg) och hans hund (m2=11,2 kg) när de
är på avståndet a) 1,0 m b) 10,0 m från varandra.
F(r = 1,00 m) = Gm1m2/r2 = 7,84•10-8 N
F(r = 10,0 m) = Gm1m2/r2 = 7,84•10-12 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.
rA = [(3,00 km)2 + (1,50 km)2]1/2 = 3,35 km
θ1 = θ2 = arctan(0,5) = 26,7°
F(rA) = Gm1m2/rA2 = 52,0 N
Totala kraften (i x-led) = 2•52,0 N•cos(26,7°) = 93,0 N
rB = 1,50 km
Totala kraften (riktad i y-led) blir F(rB) - F(rB) = 0
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/s2 dvs
F = GMm/rE2 = gm
Vid h = 8850 m gäller
gME = GM/(rE+h)2 = g/(1+h/rE)2 = g/(1+0,00139)2 =
= g • 99,7% = 9,78 m/s2
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? (Mm= 7,35 • 1022 kg och
radie Rm = 1,74 • 106 m)
F = mGMm/Rm2 = gmm
gm = GMm/Rm2 =
= (6,67 • 10-11) • (7,35 • 1022)/(1,74 • 106)2 = 1,62 m/s2
Månlandaren har massan 225 kg. Vad vägde den
a) på jorden? 225 kg • 9,81 m/s2 = 2210 N
b) på månen? 225 kg • 1,62 m/s2 = 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/s2 dvs
F = GMm/rE2 = gm
M = g • rE2/G = 9,81•(6,37•106 m)2/(6,67•10-11) =
= 5,97•1024 kg
Example 12-3
Mars Attracts!
Exemple 12-3 Mars attracts!
Hur stor är Mars massa?
Dess radie är 3,39•106 m och vid Marsytan gäller
gM = 3,73 m/s2 dvs
M = gM • RM2/G = 3,73•(3,39•106 m)2/(6,67•10-11) =
= 6,43 • 1023 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
acp = v2/r
F = macp = m v2/r = m (2πr/T)2/r = mr 4π2/T2
men kraften är ju också
F = GMs m/r2
sätter man dessa uttryck lika får man
T2 = r3(4π2/GMs)
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•1011 m. Beräkna solens massa.
T2 = r3(4π2/GMs)
Ms = r3 4π2/GT2 =
= (1,5•1011)3•4π2/{(6,67•10-11)•(π•107)2}
= 2,02•1030 kg
Beräkna periodtiden för Merkurius vars
medelavstånd till solen är 5,79•1010 m.
T = r3/2 •2π/(GMs)1/2 = 7579138 s = 0,241 y
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. (ME = 5,97 • 1024 kg, RE = 6370 km)
T2 = r3(4π2/GME)
r3 = MEGT2/4π2 = (5,97•1024)(6,67•10-11)(86400)2/4π2
r = (7,52953•1022)1/3 = 42226909 m
h = r - RE = 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 = - mGME/(RE +h)
bilda skillnaden och serieutveckla
ΔU = Uh – U0 = {RE>>h} ≈ - mGME/RE •(1 – h/RE) - 1} =
= m(GME/RE2)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 m1 = 2,5 kg, m2 =
0,75 kg och m3 = 0,75 kg.
Uab = - Gmamb/rab
U12 = - G • 2,5 kg • 0,75 kg/1,25 m = - 1,0 • 10-10 J
U13 = - G • 2,5 kg • 0,75 kg/1,25√2 m = - 0,71 • 10-10 J
U23 = - G • 0,75 kg • 0,75 kg/1,25 m = - 0,30 • 10-10 J
Utotal = - 2,0 • 10-10 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.
Ei = Ef (konservativt kraftfält!)
0 = mvf2/2 - GmME/60RE
vf = (2GME /60RE )1/2 =
= [2•(6,67•10-11)•(5,97•1024)/60•6,37•106]1/2 =
= 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.
Ei = mvi2/2 - GmME/RE
När raketen är på oändligt avstånd från jorden så är
avtar dess kinetiska och potentiella energi till noll.
Då kan den (lägsta) flykthastigheten beräknas
vflykt = (2GME /RE )1/2 =
= [2•(6,67•10-11)•(5,97•1024)/6,37•106]1/2 =
≈ 11180 m/s
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.
Ei = mvi2/2 - GmMM/RM
När raketen är på oändligt avstånd från månen så är
avtar dess kinetiska och potentiella energi till noll.
Då kan den (lägsta) flykthastigheten beräknas
vflykt,månen = (2GMM/RM )1/2 =
= [2•(6,67•10-11)•(7,35•1022)/1,74•106]1/2 =
≈ 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 ?
Ei = mvi2/2 - GmME/RE
vflykt = (2GME/RE)1/2
så att begynnelseenergin nu blir
Ei = 75% (- GmME/RE)
Ef = - GmME/r
Sätts energierna lika fås
r = 4 RE/3 (dvs den når bara höjden RE/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)