msu physics 231 fall 2015 1 physics 231 topic 9: gravitation alex brown october 30, 2015
TRANSCRIPT
MSU Physics 231 Fall 2015 1
Physics 231Topic 9: Gravitation
Alex BrownOctober 30, 2015
MSU Physics 231 Fall 2015 2
What’s up? (Friday Oct 30)
1) The correction exam is now open and is due at 10 pm Tuesday Nov 3th.
The exam grades will available on Wednesday Nov 4..
2) Homework 07 is due Tuesday Nov 10th and covers Chapters 9 and 10. It is
a little longer that usual so you may want to start early.
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Key Concepts: Gravitation
Newton’s Law of Gravitation
Gravitational Acceleration
Planetary MotionKepler’s Laws
Gravitational Potential Energy
Conservation of ME
Artificial Satellites
Covers chapter 9 in Rex & Wolfson
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The gravitational force
221
r
mmGF
G=6.673·10-11 N m2/kg2
N m2/kg2 = m3/(kg s2)
Newton:
The gravitational force works between every two massesin the universe.
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Gravitation between two objects
AB
The gravitational force exerted by the sphericalobject A on B can be calculated as if all of A’s mass would is concentrated in its center andlikewise for object B.
Conditions: B must be outside of AA and B must be ‘homogeneous’
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The force of the earth on the moon is equal and opposite to the force of the moon on the earth!
Gravitation between two objects
221
r
mmGF
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Clicker Quiz! Earth and Moon
a) one quarter
b) one half
c) the same
d) two times
e) four times
If the distance to the Moon were
doubled, then the force of attraction
between Earth and the Moon would be:
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The gravitational force depends inversely on the
distance squared. So if you increase the distance by
a factor of 2, the force will decrease by a factor of 4.
Clicker Quiz! Earth and Moon
a) one quarter
b) one half
c) the same
d) two times
e) four times
If the distance to the Moon were
doubled, then the force of attraction
between Earth and the Moon would be:
2r
MmGF
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Gravitational acceleration at the surface
of planet with mass M
2R
mMGF F=m
g
g = GM/R2
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Gravitational acceleration at the surface of the earth
g=GMe/Re2
G = 6.67x10-11 Me=5.97x1024 kg
Radius from Earth’s Center (km)
Gravitational Acceleration (m/s2)
Earth’s Surface 6366 9.81
Mount Everest 6366 + 8.85 9.78
Mariana Trench 6366 - 11.03 9.85
Polar Orbit Satellite 6366 + 1600 6.27
Geosynchronous Satellite
6366 + 36000 0.22
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Gravitational potential energySo far, we used: PEgravity=mgh Only valid for h near
earth’s surface.
More general: PEgravity=-GMm/r
R
Earlier we noted that we could define the zero of PEgravity anywhere we wanted. So the surface of the earth is as good as anywhere!
22
1
R
MmGPE
R
MmGPE
R < R2
Thus…
PE < PE2
R2=R+h
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Gravitational potential energySo far, we used: PEgravity=mgh Only valid for h near
earth’s surface.
More general: PEgravity=-GMm/r
R
Earlier we noted that we could define the zero of PEgravity anywhere we wanted. So the surface of the earth is as good as anywhere!
mghhR
MGm
hRRmGM
R
MmG
hR
MmG
R
MmG
R
MmGPEPE
2
212
)11
(
)()(
)(
R2=R+h
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Gravitational potential energyPEgravity=mgh only valid for h near earth’s surface.
More general: PEgravity=-GMm/r
PE = 0 at infinite distance from the center of the earth (r = ∞)Application: what should the minimum initial velocity of a rocket be if we want to make sure it will not fall back to earth?
KEi + PEi = ½mv2 - GMm/R
KEf + PEf = 0 v = (2GM/R) = (2gR) = 11.2 km/s
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Second cosmic speed
Second cosmic speed: speed needed to break free from a planet of mass Mp and radius Rp (gp = GMp/Rp
2)
v2 =(2GMp/Rp) = (2gpRp)
For earth: g = 9.81 m/s2 R = 6.37x106 m v2 = 11.2 km/s
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Orbital Velocities
What does the word orbit mean?
An orbit is the gravitationally curved path of an object around a point in space.
To orbit the object, you need to satisfy the kinematic conditions of that type of orbit (more on this shortly…)
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4 km/s 6 km/s 8 km/slaunchspeed
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First cosmic speed
First cosmic speed: speed of a satellite of mass m on a low-lying circular around a planet with orbit of Mp and radius Rp (gp = GMp/Rp
2)
rsatellite ≈ Rp
F = mac mgp = m v2/Rp so v1 =(gpRp)
For earth: g = 9.81 m/s2 R = 6.37x106 m v1=7.91 km/s
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Period for all orbitsConsider an object in circular motion around a larger one
32
2
332
2
22
2
2
4
4
2
KrT
GMK
KrrGM
T
rT
mrmr
mv
r
GMm
maF c
M
m
r
v
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/10297sun)(
1099.1sun)(3221
30
msK
kgM
Two common cases
Planets and other objects orbiting the sun
Moon and satellites orbiting the earth
/1099 earth)(
1097.5earth)(3215
24
msK
kgM
GM
K24
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e) The Moon does not crash into Earth because of its high
speed. If it stopped moving, it would fall directly into
Earth. With its high speed, the Moon would fly off into
space if it weren’t for gravity providing the centripetal
force.
Clicker Quiz! Averting Disaster
The Moon does
not crash into
Earth because:
a) it’s in Earth’s gravitational field
b) the net force on it is zero
c) it is beyond the main pull of Earth’s gravity
d) it’s being pulled by the Sun as well as by Earth
e) its velocity is large enough to stay in orbit
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The Moon does not crash into Earth because of its high
speed. If it stopped moving, it would fall directly into
Earth. With its high speed, the Moon would fly off into
space if it weren’t for gravity providing the centripetal
force.
Clicker Quiz! Averting Disaster
The Moon does
not crash into
Earth because:
a) it’s in Earth’s gravitational field
b) the net force on it is zero
c) it is beyond the main pull of Earth’s gravity
d) it’s being pulled by the Sun as well as by Earth
e) its velocity is large enough to stay in orbit
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Synchronous orbit
Synchronous orbit of a satellite: rotation period of satellite of mass m is the same as rotation period of the planet
For earth: period T = 24 hours = 86 x 103 s
r3 = T2/K = 75 x 1021
r = 42 x 106 m
Re = 6.4 x 106 m
(r/Re) = 6.6
/1099 earth)( 3215 msK
GMK
KrT2
32
4
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Total mechanical energy for Orbits
Consider a planet in circular motion around the sun:
M
m
r
v
r
GMmTE
r
GMm
r
GMm
r
GMmKEPETE
r
GMmmvKE
r
GMv
r
mv
r
GMm
maF c
2
22
22
2
2
2
2
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launch speed = 10 km/s
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Kepler’s laws
Johannes Kepler(1571-1630)
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Kepler’s First law
An object A bound to another object B by a force that goes with 1/r2 moves in an elliptical orbit around B, with B beingin one of the focus point of the ellipse.
A B
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area abA
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2
1
a
b
a
ce
Eccentricity: e
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2
1
a
b
a
ce
Eccentricity: e
circle when e = 0
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Kepler’s First law
2
1
a
b
a
ce
An object A bound to another object B by a force that goes with 1/r2 moves in an elliptical orbit around B, with B beingin one of the focus point of the ellipse.
Eccentricity: e
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Kepler’s second lawA line drawn from the sun to the elliptical orbit of a planetsweeps out equal areas in equal time intervals.
Area(D-C-SUN) = Area(B-A-SUN)
)area(A 2
revolution onefor 2
momentumangular L
constant 2
abT
mAL
m
L
T
A
m
L
t
A
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Kepler’s second law
rmin speed and kinetic energy are largest
rmax
speed and kinetic energy are smallest
PEgravity=-GMEarthm/r
rmaxrmin
constant vrmL
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Kepler’s third lawPEgravity=-GMEarthm/r
3221
32
/10297 msK
aKT
same as for a circular orbit except r is replaced by
the semi-major axis (red line)
a = ½(rmin+rmax)
a
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An Example
A
Bstar
Two planets are orbiting a star. The orbit of A has a radius of 1x108 km.
The distance of closest approach of B to the star is 5x107 km and its maximum distance from the star is 1x109 km.
If A has a rotational period of 1 year, what is the rotational period of B?
Need to use Kepler’s 3rd Law
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An Example
A
Bstar
RA = 1x108 kmRB = ½(Rmin+Rmax) = ½(5x107 + 1x109) = 5.25x108 km
Rmin = distance of closest approach = 5x10710(perihelion)Rmax = maximum distance = 1x109 (aphelion)
R3/T2 = constant RA3/TA
2= RB3/TB
2 so TB2=(RB
3/RA3)TA
2
So TB=(5.253 x (1 yr)2) = 12 years
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Quantities that are constant for a given orbit
ellipse area
2 momentumangular
abAT
AmL
a
b
32 period aKT
a
GMmTE
2energy total
2/12/32/1)(
)(22
a
b
aK
abm
T
mAL
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Quantities that depend on distance r
a
b
2
2
)2/1(energy kinetic
locity angular ve
velocity
energy potential
mvKE
mr
Lmr
Lv
r
MmGPE
r
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