chapter 13 gravitation
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Chapter 13 Gravitation. Newton’s law of gravitation Any two (or more) massive bodies attract each other Gravitational force (Newton's law of gravitation) Gravitational constant G = 6.67*10 –11 N*m 2 /kg 2 = 6.67*10 –11 m 3 /(kg*s 2 ) – universal constant. - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 13
Gravitation
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Newton’s law of gravitation
• Any two (or more) massive bodies attract each other
• Gravitational force (Newton's law of gravitation)
• Gravitational constant G = 6.67*10 –11 N*m2/kg2 = 6.67*10 –11 m3/(kg*s2) – universal constant
rr
mmGF ˆ
221
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Gravitation and the superposition principle
• For a group of interacting particles, the net gravitational force on one of the particles is
• For a particle interacting with a continuous arrangement of masses (a massive finite object) the sum is replaced with an integral
n
iinet FF
21,1
body
body FdF
,1
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Chapter 13Problem 5
Three uniform spheres of mass 2.00 kg, 4.00 kg and 6.00 kg are placed at the corners of a right triangle. Calculate the resultant gravitational force on the 4.00-kg object, assuming the spheres are isolated from the rest of the Universe.
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Shell theorem
• For a particle interacting with a uniform spherical shell of matter
• Result of integration: a uniform spherical shell of matter attracts a particle that is outside the shell as if all the shell's mass were concentrated at its center
shell
shell FdF
,1
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Gravity force near the surface of Earth
• Earth can be though of as a nest of shells, one within another and each attracting a particle outside the Earth’s surface
• Thus Earth behaves like a particle located at the center of Earth with a mass equal to that of Earth
g = 9.8 m/s2
• This formula is derived for stationary Earth of ideal spherical shape and uniform density
jR
mmGF
Earth
EarthEarth
ˆ2
1,1
jmgjm
R
Gm
Earth
Earth ˆˆ112
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Gravity force near the surface of Earth
In reality g is not a constant because:
Earth is rotating, Earth is approximately an ellipsoid with a non-uniform density
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Gravitational field
• A gravitational field exists at every point in space
• When a particle is placed at a point where there is gravitational field, the particle experiences a force
• The field exerts a force on the particle
• The gravitational field is defined as:
• The gravitational field is the gravitational force experienced by a test particle placed at that point divided by the mass of the test particle
m
Fg g
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Gravitational field
• The presence of the test particle is not necessary for the field to exist
• The source particle creates the field
• The gravitational field vectors point in the direction of the acceleration a particle would experience if placed in that field
• The magnitude is that of the freefall acceleration at that location
gR
Gm
Earth
Earth 2
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Gravitational potential energy
• Gravitation is a conservative force (work done by it is path-independent)
• For conservative forces (Ch. 8):
f
i
r
r
rdFU
f
i
r
r
Earth drr
mGm2
1
fiEarth rr
mGm11
1
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Gravitational potential energy
• To remove a particle from initial position to infinity
• Assuming U∞ = 0
fiEarthif rr
mGmUUU11
1
i
Earth
iEarthi r
mGm
rmGmUU 11
11
i
Earthii r
mGmrU 1)(
r
mGmrU 21)(
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Gravitational potential energy
r
mGmrU 21)(
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Escape speed
• Accounting for the shape of Earth, projectile motion (Ch. 4) has to be modified:
gRvgR
vac
2
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Escape speed
• Escape speed: speed required for a particle to escape from the planet into infinity (and stop there)
002
12
1 planet
planet
R
mGmvm
ffii UKUK
planet
planetescape R
Gmv
2
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Escape speed
• If for some astronomical object
• Nothing (even light) can escape from the surface of this object – a black hole
csmR
Gmv
object
objectescape /103
2 8
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Chapter 13Problem 30
(a) What is the minimum speed, relative to the Sun, necessary for a spacecraft to escape the solar system, if it starts at the Earth’s orbit? (b) Voyager 1 achieved a maximum speed of 125 000 km/h on its way to photographJupiter. Beyond what distance from the Sun is this speed sufficient to escape the solar system?
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Kepler’s laws
Three Kepler’s laws• 1. The law of orbits: All planets move in elliptical orbits, with the Sun at one focus• 2. The law of areas: A line that connects the planet to the Sun sweeps out equal areas in the plane of the planet’s orbit in equal time intervals• 3. The law of periods: The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit
Johannes Kepler(1571-1630)
Tycho Brahe/Tyge Ottesen
Brahe de Knudstrup(1546-1601)
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First Kepler’s law
• Elliptical orbits of planets are described by a
semimajor axis a and an eccentricity e
• For most planets, the eccentricities are very small
(Earth's e is 0.00167)
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Second Kepler’s law
• For a star-planet system, the total angular momentum is constant (no external torques)
• For the elementary area swept by vector
rpL
))((2
1 rdrdA dt
dr
dt
dA 2
2
m
L
dt
dA
2
))(( mvr ))(( rmr 2mr const
2
2r
r
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Third Kepler’s law
• For a circular orbit and the Newton’s Second law
• From the definition of a period
• For elliptic orbits
))(( 22
rmr
GMmmaF
2
22 42
TT
32
r
GM
32
2 4r
GMT
32
2 4a
GMT
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Satellites
• For a circular orbit and the Newton’s Second law
• Kinetic energy of a satellite
• Total mechanical energy of a satellite
r
vm
r
GMm 2
2)(maF
2
U
2
2mvK
UKE r
GMm
r
GMm
2 r
GMm
2 K
r
GMm
2
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Satellites
• For an elliptic orbit it can be shown
• Orbits with different e but the same a have the same total mechanical energy
a
GMmE
2
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Chapter 13Problem 26
At the Earth’s surface a projectile is launched straight up at a speed of 10.0 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth.
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Questions?
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Answers to the even-numbered problems
Chapter 13
Problem 22.67 × 10−7 m/s2
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Answers to the even-numbered problems
Chapter 13
Problem 43.00 kg and 2.00 kg
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Answers to the even-numbered problems
Chapter 13
Problem 10(a) 7.61 cm/s2
(b) 363 s(c) 3.08 km(d) 28.9 m/s at 72.9° below the horizontal
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Answers to the even-numbered problems
Chapter 13
Problem 24(a) −4.77 × 109 J(b) 569 N down(c) 569 N up