Transcript Angular Momentum

Chapter 11 – Gravity
Lecture 1
April 6, 2010
Kepler’s Laws
Newton’s Universal Law of
gravitation
Gravitational and inertia mass
Gravitational potential energy
Kepler’s (empirical) Laws
• Kepler’s First Law:
– All planets move in elliptical orbits
with the sun at one focus
• Kepler’s Second Law:
– The radius vector drawn from the Sun to a planet
sweeps out equal areas in equal times,
dA/dt = const.
(the law of equal areas)
• Kepler’s Third Law:
T 2  Cr 3
– The square of the orbital period of any planet is proportional to the
cube of the semimajor axis of the elliptical orbit.
Today we can understand the physical reasons for these
laws …
Let’s remind us first of the geometry of the ellipse and then discuss the three
laws.
Ellipses
e = c/a
Planetary Orbits
Kepler found that the orbit of Mars was
an ellipse, not a circle.
Kepler’s Third Law
Kepler had access to very
good data from the
astronomer Tycho Brahe in
Prague. See table for today’s
data.
After many years of work
Kepler found an intriguing
correlation between the orbital
periods and the length of the
semimajor axis of orbits.
T  Cr
2
3
Of the satellites shown revolving around Earth, the
one with the greatest speed is
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1
2
3
4
5
Of the satellites shown revolving around Earth, the
one with the greatest speed is
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1
2
3
4
5
The constant-area law.
The orbits of two planets orbiting a star are shown. The
semimajor axis of planet A is twice that of planet B. If the
period of planet B is TB, the period of planet A is
A.
2TB
B.
2 2 TB
C. 3TB
D.
E.
3 TB
4TB
AÕ
rA
rB
B
A
The orbits of two planets orbiting a star are shown. The
semimajor axis of planet A is twice that of planet B. If the
period of planet B is TB, the period of planet A is
A.
2TB
B.
2 2 TB
C. 3TB
D.
E.
3 TB
4TB
AÕ
rA
rB
B
A
Newton’s Universal Law of Gravity
m
M
F12
F21
r
Newton’s Law of Gravity
• Newton’s law of gravity will provide a physical theory of Kepler’s laws.
m
M
F12
F21
r
mM
F 12  G 2 rˆ12
R
Magnitude of force
mM
Fg  G 2
r
G  6.67 10
11
Nm 2
kg
The Cavendish experiment
Measuring G
• G was first measured by
Henry Cavendish in 1798
• The apparatus shown here
allowed the attractive force
between two spheres to
cause the rod to rotate
• The mirror amplifies the
motion
• It was repeated for various
masses
mM
Magnitude  G 2
r
7/20/2015
G  6.67  10 11 N m 2 / kg 2
Physics 201, UW-Madison
16
Gravitational and inertial mass
• Gravitation is a force that acts on the gravitational mass
Fg  G
mg M g
r2
(the masses are the source)
• Newton’s Law of motion acts on the inertial mass
F  mi a
• In principle, they are not necessarily related, that the
gravitational mass mg is not the same as mi
(but they are, up to the current experimental accuracy)
Kepler’s Third Law derived from Newton’s Law
Easily derived for a circular orbit
Centripetal force = gravitational force
Fcent
v2
 m  mR 2
R
mM
Fg  G 2
R
mM
R2
4 2
M
R 2 G 2
T
R
mR 2  G

2
T
R 3 GM

 const.
2
2
T
4
Kepler’s Third Law
2
4

T2 
R3
GM S
Kepler’s Third Law derived from Newton’s Law
4 2 3
T 
R
GM S
Extension for elliptical orbits,
(Without proof R  a)
Kepler’s Third Law
2
2
4

T2 
a3
GM S
Where a is the semimajor axis
A woman whose weight on Earth is 500 N is lifted to a height
of two Earth radii above the surface of Earth. Her weight
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decreases to one-half of the original amount.
decreases to one-quarter of the original amount.
does not change.
decreases to one-third of the original amount.
decreases to one-ninth of the original amount.
A woman whose weight on Earth is 500 N is lifted to a
height of two Earth radii above the surface of Earth. Her
weight
A.
B.
C.
D.
E.
decreases to one-half of the original amount.
decreases to one-quarter of the original amount.
does not change.
decreases to one-third of the original amount.
decreases to one-ninth of the original amount.
From work to
gravitational
potential energy.
In the example before,
it does not matter on
what path the person is
elevated to 2 Earth radii
above.
Only the final height (or
distance) matters for the
total amount of work
performed.
Potential Energy
m
2
r
r
r12
M1
Force (1,2)   G
mM
rˆ1,2
2
r1,2
G  6.67  1011 N m2 / kg2
Work done to bring mass m from initial to final position.
f
f
mM 

PE  W    F dr     G 2  dr 

r 
i
i
 1 1
 GmM  r dr  GmM   
 rf ri 
i
f
2
Zero point is arbitrary. Choose zero at infinity.
Mm
r
PE(r  )  0
PE  G
Mm
U(r)  G
r
Demo