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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 - PowerPoint PPT Presentation
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Kepler’s Laws
Newton’s Universal Law of gravitation
Gravitational and inertia mass
Gravitational potential energy
Chapter 11 – Gravity Lecture 1 April 6, 2010
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: – 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.
T 2 =Cr3
Ellipses
e = c/a
Planetary Orbits
Kepler found that the orbit of Mars was an ellipse, not a circle.
Kepler had access to very good data from the astronomer Tycho Brahe in Prague. See table for today’s data.
Kepler’s Third Law
After many years of work Kepler found an intriguing correlation between the orbital periods and the length of the semimajor axis of orbits.
T 2 =Cr3
Of the satellites shown revolving around Earth, the one with the greatest speed is
• 1 • 2 • 3 • 4 • 5
Of the satellites shown revolving around Earth, the one with the greatest speed is
• 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
B
B
B
B
B
TT
TT
T
4.E3.D
3.C22.B
2.A
rA A
B
rB
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
B
B
B
B
B
TT
TT
T
4.E3.D
3.C22.B
2.A
rA A
B
rB
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.
Fur
12 =GmMR 2 r̂12
m M
F12 F21
r
G =6.67 ⋅10−11 Nm2
kgFg =G
mMr2
Magnitude of force
The Cavendish experiment
04/22/23 Physics 201, UW-Madison 16
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
Magnitude = Gm Mr2
G =6.67 ×10−11 Nm 2 / kg2
Gravitational and inertial mass
• Gravitation is a force that acts on the gravitational mass (the masses are the source)
• Newton’s Law of motion acts on the inertial mass
• 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)
Fg =Gm gMg
r2
F =m ia
Kepler’s Third Law derived from Newton’s Law
Centripetal force = gravitational force
Easily derived for a circular orbit
Fcent =mv2
R=mRω 2 Fg =G
mMR 2
mRω 2 =GmMR 2
R4π 2
T 2 =GMR 2
ω =2πT
R3
T 2 =GM4π 2 =const.
T 2 =4π 2
GMS
R 3Kepler’s Third Law
Kepler’s Third Law derived from Newton’s Law
Extension for elliptical orbits, (Without proof R a)
T 2 =4π 2
GMS
a3Kepler’s Third Law
T 2 =4π 2
GMS
R 3
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
• 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. decreases to one-half of the original amount. B. decreases to one-quarter of the original amount. C. does not change. D. decreases to one-third of the original amount. E. 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 Energym
M
r
Force (1,2) = −Gm Mr1,22
r̂1,2 G =6.67 ×10−11 Nm 2 / kg2
Work done to bring mass m from initial to final position.
PE =−W =−rFgdrr
i
f
∫ =− −GmMr2
⎛⎝⎜
⎞⎠⎟
i
f
∫ dr⇓
=GmM r−2i
f
∫ dr=−GmM1rf
−1ri
⎛⎝⎜
⎞⎠⎟
PE =−GMmr
PE(r=∞)=0
U(r)=−GMmrZero point is arbitrary. Choose zero at infinity.
rr
12
1
2
Demo
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