Download Physics 231 Topic 8: Gravitation Wade Fisher October 24-26 2012

Physics 231
Topic 8: Gravitation
Wade Fisher
October
2012
MSU Physics24-26
231 Fall 2012
1
Clicker Quiz!
3 children are sitting on a rotating disc in a playground.
The disc starts to spin faster and faster.
Which of the three is most likely to start sliding first?
top view
A
B
C
a)
b)
c)
d)
child A
child B
child C
the same for all three
ac= 2r
A= B= c
rA<rB<rc
acA<acB<acC
The centripetal force needed to stay in place is largest for C
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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, revisited
Newton:
m1 m2
F G 2
r
G=6.673·10-11 Nm2/kg2
The gravitational force works between every two massive
particles in the universe, yet is the least well understood
force known.
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Gravitation between two objects
A
B
The gravitational force exerted by the spherical
object A on B can be calculated by assuming that all
of A’s mass would be 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 gravitational force exerted by the spherical
object A on B can be calculated by assuming that all
of A’s mass would be concentrated in its center and
likewise for object B.
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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Gravitation between two objects
m1m2
F G 2
r
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Gravitational acceleration
m1mEARTH
F G
2
F=mg
r
g=GmEARTH/r2
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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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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Gravitational potential energy
So far, we used: PEgravity=mgh Only valid for h near
earth’s surface.
More general: PEgravity=-GMEarthm/r
PE=0 at infinite distance from the center of the earth (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!
PE1  G
R2
R1
mEARTH
m
R1
mEARTH
PE2  G
m
R2
MSU Physics 231 Fall 2012
R1<R2
Thus…
PE1 < 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=-GMEarthm/r
PE=0 at infinite distance from the center of the earth (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!
R2=R1+h
R2
R1
PE2  PE1  G
mEARTH
m
m  (G EARTH m)
R2
R1
mEARTH
mEARTH
 G
m  (G
m)
( R1  h)
R1
 mEARTH
1
1
 mGmEARTH ( 
)  m G
2
R1 R1  h
R
1

MSU Physics 231 Fall 2012

h  mgh

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Gravitational potential energy
So far, we used: PEgravity=mgh Only valid for h near
earth’s surface.
More general: PEgravity=-GMEarthm/r
PE=0 at infinite distance from the center of the earth (r=∞)
Example: escape speed (a.k.a. second cosmic speed )
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 - GMEarthm/Rearth
KEf + PEf = 0
v = (2GMearth/REarth) = 11.2 km/s
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Some definitions
First cosmic speed: speed of a satellite on a low-lying circular orbit
(rsatellite ≈ rplanet):
Msg = Msv2/r so v1 =(gr) where g = Gmplanet/rp2
For earth:
g=9.81 m/s2 r=6.366x106 m
v1=7.91x103 m/s
Second cosmic speed: speed needed to break free from a planet:
v1 =(2GMp/Rp) = (2gr)
For earth:
g=9.81 m/s2 r=6.366x106 m
v2=1.12x104 m/s
Synchronous orbit of a satellite: rotation period of satellite is the same
as rotation period of the planet
For earth: period T = 24 hours
Angular speed
 = 2/(243600) = 7.27x10-5 rad/s
For earth what is its radius?
Use Fg=mac so Gmearthmsat/r2 = msat2r (note rrearth !)
So r3=Gme/2
r=(Gme/2)1/3 = 3.54x107 = 35400 km
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launch speed = 10 km/s
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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.
Mm
F G 2
R
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Kepler’s laws
Johannes Kepler
(1571-1630)
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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 First law
p+q = constant
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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PEgravity=-GMEarthm/r
Kepler’s second law
Area(D-C-SUN)=Area(B-A-SUN)
Potential energy is maximal,
Kinetic energy is minimal
Potential energy is close to
0 here, but still negative
Potential energy is minimal, kinetic energy is maximal
(potential energy is very negative here)
A line drawn from the sun to the elliptical orbit of a planet
sweeps out equal areas in equal time intervals.
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Kepler’s third law
Consider a planet in circular motion around the sun:
G
M sun M planet
r
2

M planet v 2planet
r
s 2r
v

t
T
 4 2  3
2
r  K s r 3
T  
 GM sun 
T2
 Ks
3
r
K s  2.97 10 19 s 2 / m 3
r3
C
2
T
C  3.36 1018 m 3 / s 2
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Kepler’s third law
Consider a planet in circular motion
around the sun:
r3
r3
C
2
T
C  3.36 1018 m3 / s 2
T2
r3=C T2
r3=constantT2
T: period-time it takes to make
one revolution
If the orbit is an ellipse, replace r with a, the semi-major axis.
a=(half the sum of the smallest and greatest distance from the sun)
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An Example
Two planets are orbiting a star.
B The orbit of A has a radius of 108 km.
The distance of closest approach of B
to the star is 5x107 km and its
maximum distance from the star is 109
km. If A has a rotational period of 1
year, what is the rotational period of
B?
star
A
Need to use Kepler’s 3rd
Law, but Planet B orbits in
an ellipse
Recall: replace R with the
average radius on the
semi-major axis.
R=½(a+b)
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An Example
A
Two planets are orbiting a star. The
B orbit of A has a radius of 108 km. The
star
distance of closest approach of B to
the star is 5x107 km and its maximum
distance from the star is 109 km. If A
has a rotational period of 1 year, what
is the rotational period of B?
Rmin = distance of closest approach.
= perihelion
Rmax = maximum distance
= aphelion
RA = 108 km
RB = ½(rmin+rmax) = ½(5x107+109) = 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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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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For Next Week
Chapter 10:
Solids and Fluids
Homework Set 8
Due Wednesday Nov 2!
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