Announcements
• CAPA Set #7 due Friday at 10 pm
• This week in Section 
Assignment 4: Circular Motion & Gravity
• Finish reading all sections of Chapter 5, start Chapter 6 in
Friday’s lecture this week
• Reminder  Exam #2 on Tuesday, October 11
See course web link (“exam info”) for details
• Reminder about office hours …
Nagle (Monday 2-3 in office, Wednesday 1:45-3:45 pm help room)
Kinney (Thursday 4-5 pm help room)
Uzdensky (Tuesday 11am-noon help room)
Lab Make-Up Procedure
• You must complete at least 5 of the 6 labs to receive a passing grade
in this course.
•
If you missed a lab, you can make it up during one of two
Review/Lab make-up weeks:
– October 24-28 (labs 1-3)
– December 5-9 (labs 4-6)
• Even if you don’t need to make up a lab, you still must attend your
section those weeks for the Review Recitation.
• To make up a lab, contact your TA ahead of time.
You will need to arrange attending twice: (1) for lab make-up
and (2) for the review recitation. You can attend any other section
(in addition to your regular section), if you have that section’s TA
permission in advance.
Contact: Professor Uzdensky if there are any questions.
Earth at the Center of Everything
Geocentrism
- Epicyles
- Equant
- Deferent
Claudius Ptolemaeus
(Ptolemy)
90 – 168 AD
Heliocentrism (Sun at the Center)
Nicolaus Copernicus
1473-1543
De revolutionibus orbium coelestium
(On the Revolutions of the Celestial Spheres)
Heliocentrism
Tycho Brahe
1546-1601
astronomical observations
Johannes Kepler
1571-1630
Three Laws
Epitome astronomia Copernicanae
(Epitome of Copernican Astronomy)
Kepler’s Laws of Planetary Motion
Law #1: The orbit of each planet is an ellipse,
with the Sun at one focus.
Elliptical Geometry
The sum of the distances from any point P on the ellipse to
those two foci is constant and equal to the major diameter
( PF1 + PF2 = 2a ).
Kepler’s Laws of Planetary Motion
Law#2: An imaginary line drawn from each
planet to the Sun sweeps out equal areas in
equal times.
Clicker Question
Room Frequency BA
Notice how a planet with an elliptical orbit moves closer to and
further away from the Sun. The point of closest approach to the Sun
is called perihelion. The furthest point is called aphelion.
At which position is the asteroid moving
at the largest magnitude velocity?
A) Far left (aphelion)
B) Far right (perihelion)
C) Top
D) Bottom
E) Same velocity magnitude everywhere
Kepler’s Laws of Planetary Motion
Law #3: The ratio of the square of a planet’s orbital period is
proportional to the cube of its mean distance from the Sun.
T R
2
R
3
T
2
3
 3.35 10 km / yr
24
3
2
What is the relationship between velocity and orbit Radius?
Consider F=ma in the radial direction: +inward
Fnet  ma R  G
mM
Rs
2
Force of Gravity Only
Rs
Condition for
Circular Motion
Re
Rs 
h
GM
v
2
elevation
Rs = Re + h
m
v2
Rs
GM
v
Rs
Room Frequency BA
Clicker Question
Satellite 1 is in circular orbit about
the Earth and is moving with
tangential speed v1.
Rs
Satellite 2 is also in orbit but with
speed v2 > v1.
Re
h
Satellite 2’s orbit radius is:
elevation
Rs = Re + h
GM
v
Rs
GM
Rs  2
v
A) Bigger than Satellite 1’s
B) Smaller than Satellite 1’s
C) Unclear from the information
given.
Room Frequency BA
Clicker Question
Satellite 1 is in circular orbit about
the Earth with radius R1.
Rs
Satellite 2 is also in circular orbit
with a radius R2 and a speed twice
that of Satellite 1.
Re
What is true about the radii?
h
elevation
Rs = Re + h
GM
v
Rs
GM
Rs  2
v
A)
B)
C)
D)
E)
R1 = 2 x R2
R1 = ½ x R2
R1 = 4 x R2
R1 = ¼ x R2
Unclear from the information
given.
Room Frequency BA
Clicker Question
GM
v
Rs
GM
Rs  2
v
Period T
Rs
2Rs
v
T
2 2
4 Rs
GM
2
v 

2
Rs
T
Which of the following correctly presents
the relationship between Rs and T?
Re
h
elevation
Rs = Re + h
Eliminate velocity in place of period (T)
 4 2 
A)   
Rs

 GM 
2

 2
4

2
B)   
Rs

 GM 
2

 3
4

2
C)   
Rs

 GM 
 GM  3
D)    2  Rs
 4 
2
GM 4

Rs
T
3
Rs
2
2
2
Rs
2
T

GM
4
2
Recall Kepler’s Empirical Value
Rs3
T2

GM
4 2

6.67 10

R
3
T
2
11
 3.35 10 km / yr
24

Nm 2 / kg 2 1.98 10 30 kg
4 2
3
  3.310
18
2
m3 / s 2
 3.3 1024 km3 / yr 2
Knowledge of a planet’s period (easy to measure)
allows us
to determine it orbital radius.
Where is Mars?
Rs3
T
2

GM
4
3
Rsun
 mars
Tmars
2
3
Rsun
 mars
3
Rsun
 earth
Rsun mars
Rsunearth
2


 constant
3
Rsun
earth
Tearth2
Tmars
 constant
2
Mars is 1.52 times as far
away as the earth is
from the sun.
Tearth2
 Tmars

 Tearth





2/3
 687 days 

 
 365 days 
2/3
 1.52
Exoplanets
Upsilon Andromedae is a binary star located about 44 light-years
away from the Earth. The primary star is a yellow-white dwarf star
that is younger than the Sun. There is a second star that is a red
dwarf in a wide orbit.
As of 2010, four confirmed extrasolar planets have been discovered.
Rs3
2
T

(b)  (0.0595)3/(4.61)2=9.9x10-6 AU3/days2
(c)  (0.832)3/(241.3)2=9.9x10-6 AU3/days2
(d)  (2.53)3/(1278.1)2=9.9x10-6 AU3/days2
(e)  (5.24)3/(3848.8)2=9.7x10-6 AU3/days2
Discovery of accelerating universe!
Need a large scale repulsion force (not gravity).
No one knows what it is – just called Dark Energy!