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Physics 231
Topic 9: Gravitation
Alex Brown
October
30,
MSU Physics 231
Fall 2015
2015
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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 Motion
Kepler’s Laws
Gravitational Potential Energy
Conservation of ME
Artificial Satellites
Covers chapter 9 in Rex & Wolfson
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The gravitational force
Newton:
m1m2
F G 2
r
G=6.673·10-11 N m2/kg2
N m2/kg2 = m3/(kg s2)
The gravitational force works between every two masses
in the universe.
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Gravitation between two objects
A
B
The gravitational force exerted by the spherical
object A on B can be calculated as if all
of A’s mass would is concentrated in its center and
likewise for object B.
Conditions: B must be outside of A
A and B must be ‘homogeneous’
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Gravitation between two objects
m1m2
F G 2
r
The force of the earth on the moon is equal and
opposite to the force of the moon on the earth!
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Clicker Quiz! Earth and Moon
a) one quarter
If the distance to the Moon were
b) one half
doubled, then the force of attraction
c) the same
between Earth and the Moon would be:
d) two times
e) four times
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Clicker Quiz! Earth and Moon
a) one quarter
If the distance to the Moon were
b) one half
doubled, then the force of attraction
c) the same
between Earth and the Moon would be:
d) two times
e) four times
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.
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Mm
F G 2
r
10
Gravitational acceleration at the surface
of planet with mass M
mM
F G 2
R
F=mg
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 energy
So far, we used: PEgravity=mgh Only valid for h near
earth’s surface.
More general: PEgravity=-GMm/r
R2=R+h
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!
R
Mm
R < R2
R
Mm
PE2  G
R2
Thus…
PE1  G
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PE < PE2
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Gravitational potential energy
So far, we used: PEgravity=mgh Only valid for h near
earth’s surface.
More general: PEgravity=-GMm/r
R2=R+h
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!
Mm
Mm
PE2  PE1  G
 (G
)
R2
R
R
 G
Mm
Mm
 (G
)
( R  h)
R
 M
1
1
 mGM ( 
)  m G 2
 R
R
R

h

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
h  mgh


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Gravitational potential energy
PEgravity=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/Rp2)
v2 =(2GMp/Rp) = (2gpRp)
For earth: g = 9.81 m/s2
R = 6.37x106 m
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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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launch
speed
4 km/s
6 km/s
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8 km/s
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First cosmic speed
First cosmic speed: speed of a satellite of mass m on a lowlying circular around a planet with orbit of Mp and radius Rp
(gp = GMp/Rp2)
rsatellite ≈ Rp
F = mac
mgp = m v2/Rp
For earth: g = 9.81 m/s2
so
v1 =(gpRp)
R = 6.37x106 m
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v1=7.91 km/s
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Period for all orbits
Consider an object in circular motion around a larger one
F  mac
2
v
GMm mv 2
2



2


m

r

m

 r
2
r
r
m
 T 
2

 3
4

2
r
 r  Kr 3
T  
GM 

M
 4 2 

K  
 GM 
T 2  Kr 3
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 4 2 
Two common cases

K  
 GM 
Planets and other objects orbiting the sun
M (sun)  1.99 1030 kg
K (sun)  297 10  21 s 2 / m 3
Moon and satellites orbiting the earth
M (earth)  5.97 10 24 kg
K (earth)  99 10 15 s 2 / m 3
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Clicker Quiz! Averting Disaster
a) it’s in Earth’s gravitational field
The Moon does
b) the net force on it is zero
not crash into
c) it is beyond the main pull of Earth’s gravity
Earth because:
d) it’s being pulled by the Sun as well as by
Earth
e) its velocity is large enough to stay in orbit
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.
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Clicker Quiz! Averting Disaster
a) it’s in Earth’s gravitational field
The Moon does
b) the net force on it is zero
not crash into
c) it is beyond the main pull of Earth’s gravity
Earth because:
d) it’s being pulled by the Sun as well as by
Earth
e) its velocity is large enough to stay in orbit
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.
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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
T 2  Kr 3
 4 2
K  
 GM



K (earth)  99 10 15 s 2 / m 3
(r/Re) = 6.6
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Total mechanical energy for Orbits
Consider a planet in circular motion around the sun:
F  mac
v
m
r
M
GMm mv 2

2
r
r
GM
2
v 
r
mv 2 GMm
KE 

2
2r
GMm GMm
GMm
TE  PE KE  


r
2r
2r
GMm
TE  
2r
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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
B
A
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 being
in one of the focus point of the ellipse.
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area A   ab
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Eccentricity: e
c
b
e   1  
a
a
2
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Eccentricity: e
c
b
e   1  
a
a
2
circle when e = 0
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Kepler’s First law
Eccentricity: e
c
b
e   1  
a
a
2
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 being
in one of the focus point of the ellipse.
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Kepler’s second law
A line drawn from the sun to the elliptical orbit of a planet
sweeps out equal areas in equal time intervals.
Area(D-C-SUN) = Area(B-A-SUN)
A
L

 constant
t 2m
L  angular momentum
A
L

for one revolution
T 2m
2mA
L
(A  area   ab)
T
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PEgravity=-GMEarthm/r
Kepler’s second law
L  m r v  constant
rmax
rmin
rmin
rmax
speed and kinetic energy are
smallest
speed and kinetic energy are largest
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PEgravity=-GMEarthm/r
Kepler’s third law
T 2  K a3
K  297 10 21 s 2 / m 3
a
same as for a circular orbit
except r is replaced by
a = ½(rmin+rmax)
the semi-major axis (red line)
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An Example
star
A
Two planets are orbiting a star. The
B 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
star
B
A
Rmin = distance of closest approach = 5x10710(perihelion)
Rmax = maximum distance = 1x109 (aphelion)
RA = 1x108 km
RB = ½(Rmin+Rmax) = ½(5x107 + 1x109) = 5.25x108 km
R3/T2 = constant  RA3/TA2= RB3/TB2 so TB2=(RB3/RA3)TA2
So TB=(5.253 x (1 yr)2) = 12 years
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Quantities that are constant for a given orbit
period T 2  K a 3
A
angular momentum L  2m
T
area A   ab ellipse
b
a
2mA 2m (ab)
b
L

 1/ 2
1/ 2 3 / 2
T
(K ) a
a
GMm
total energy TE  
2a
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Quantities that depend on distance r
b
a
r
Mm
potential energy PE  G
r
L
velocity v 
mr
L
angular velocity  
mr 2
kinetic energy KE  (1 / 2)mv 2
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