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PHYS 100: Lecture 11
UNIVERSAL GRAVITATION
and
SPRINGS
x0
FES
FSE
R
x
FSE = G
MSME
R2
F
x
F
F = -k(x – x0)
PHYS 100 Lecture 11, Slide 1
Music
Who are the Artists?
BB
A)
B)
C)
D)
E)
Whitney Houston and Tina Turner
Nina Simone and Patti LaBelle
Etta James and Bonnie Raitt
The Dixie Cups
Marcia Ball, Irma Thomas, Tracy Nelson
Why?
Three Weeks Until Jazzfest !!!!! But who’s counting?
New Orleans Jazzfest Poster from year’s past:
The Sweet Soul Queen of New Orleans: Irma Thomas
PHYS 100 Lecture 11, Slide 2
WHAT DID YOU FIND DIFFICULT?
I find all of this confusing... please help me decipher physics :(
OK We will touch on everything today !!
“I would like to talk more about the "heavenly bodies" and what we can use the more
general gravity equations for.”
“The new Equations looks are a lot tougher to interpret especially Kepler's Third law
Equation in the context of gravitational force.“
The first job is to understand
1)
2)
What is really new
What are the consequences of the really new
PHYS 100 Lecture 11, Slide 3
THE BIG IDEAS
NOTE: THE BIG IDEAS ARE ALWAYS GIVEN IN THE LAST SLIDE
We’ll now try to create some structure here by starting with what’s new
PHYS 100 Lecture 11, Slide 4
What’s New?
There are TWO new PHYSICS things in this Prelecture
The Universal Gravitational Force Law
in mks units
G = 6.67 X 10-11 Nm2/kg2
The Spring Force Law
PHYS 100 Lecture 11, Slide 5
What’s New?
There is ONE new MATH thing in this Prelecture
The Gravitational force at the surface of the Earth
To find the gravitational force on an object at the surface of the Earth:
Add up gravitational forces from every “piece of Earth”,
r
r
F = ∫ dF
Result: Same as if all mass of Earth were concentrated at the center of Earth
M Em
F =G 2
RE
Everything else in the prelecture is a consequence of using this new
information in the context of Newton’s Laws
PHYS 100 Lecture 11, Slide 6
(Some of) The Consequences
Newton’s Leap to the Stars
THE BIG IDEA:
1. The moon orbits the Earth. Therefore, some real force on the moon must be responsible
for its acceleration. That force is the gravitational force exerted by the Earth.
2. The weight of any object on Earth is really the gravitational force exerted by the
Earth on that object.
3. Therefore, we can relate the acceleration due to gravity on Earth to the acceleration
of the moon in its orbit.
Moon:
Object on Earth:
F =G
M Em
RE2
F = mg
M
g = G 2E
RE
amoon
M
=G 2 E
REmoon
F =G
M E M moon
2
REmoon
2
REmoon
g = amoon 2
RE
PHYS 100 Lecture 11, Slide 7
(Some of) The Consequences
Kepler’s Third Law
THE BIG IDEA:
1. All planets orbit the Sun. Therefore, some real force on each planet must be responsible
for its acceleration. That force is the gravitational force exerted by the Sun.
2. Newton’s Second Law can be used to determine a single relationship among the orbit
parameters that must hold for all planets.
FSplanet = G
aSplanet
M S m planet
2
RSplanet
v 2planet
=
RSplanet
PHYSICS
To get usual form of Kepler’s
Third Law, we just need to
do some algebra:
A word on the constant
Numerical value
depends on units:
GM S
:
2
4π
R in AU
P in years
2πRSplanet
Period = P =
v
GM S
=1
2
4π
v 2planet = G
MS
RSplanet
This equation relates
the speed of the planet
to its orbital radius.
3
RSplanet
GM S
=
P2
4π 2
PHYS 100 Lecture 11, Slide 8
CheckPoint 1
Which is a greater gravitational force, the force of gravity from the Sun on
the Earth or the force of gravity from the Earth on the Sun?
2
(A) Force of gravity from the Sun on the Earth
(B) Force of gravity from the Earth on the Sun
(C) They are equal
100
80
60
40
20
0
A
B
C
Great Job!!
“They are equal because of Newtons Third Law.
The forces are equal and opposite.”
PHYS 100 Lecture 11, Slide 9
Acceleration
An apple of mass m falls from rest from a tree. It hits the ground with velocity v.
BB
Just before it hits the ground, what is its acceleration?
M
(A) a = G 2Earth
REarth
v2
(B) a =
REarth
(C)
a=G
M Earth m
2
REarth
(D)
a=g
(E)
a = mg
TWO CORRECT ANSWERS !!
(D)
a=g
M Earth
(A) a = G 2
REarth
We know all objects fall with acceleration g
Acceleration is due to gravitational
force that Earth exerts on mass m
F = W = mg
F =G
M Earth m
2
REarth
Therefore we know g, the acceleration due to gravity on Earth, in terms
of G, the universal gravitation constant, and Earth parameters (ME, RE)
g =G
M Earth
2
REarth
PHYS 100 Lecture 11, Slide 10
Weight
Suppose the radius of the Earth were half of what it is now, but its mass was the
same.
BB
How would your weight change?
(B) Decrease by factor of 2
(C) Stay the Same
(A) Decrease by factor of 4
(D) Increase by factor of 2
(E) Increase by factor of 4
We know g, the acceleration due to gravity on Earth, in terms of G, the
universal gravitation constant, and Earth parameters (ME, RE)
g =G
M Earth
2
REarth
We know your weight is determined by your mass and g:
W = mg
If mass of Earth (and your mass) do not change:
W∝
1
2
Earth
R
Wnew
2
Rold
= Wold 2 = 4Wold
Rnew
PHYS 100 Lecture 11, Slide 11
CheckPoint 2
On Earth your weight is WE. Suppose you are now on Planet X
whose mass is twice the mass of the Earth’s but whose
diameter is also twice the Earth’s diameter.
4
BB
Compare WX, your weight on Planet X to your weight on Earth.
(A) WX = ¼ WE
(B) WX = 1/2 WE
(C) WX = WE
(D) WX = 2 WE
(E) WX = 4 WE
You said:
• When you double the mass of the planet, you increase
the numerator by a factor of 2. Doubling the radius will
increase the denominator by a factor of 4, thus WX =
1/2W
• When you double the mass of the planet, you increase
the numerator by a factor of 2. Doubling the radius will
increase the denominator by a factor of 4, thus WX =
1/2W
50
40
30
20
• The diameter of Earth is two times that of Planet X.
Thus, one's weight is 2 times the weight of one's
weight on Earth.
10
0
A
B
C
D
E
PHYS 100 Lecture 11, Slide 12
CheckPoint 2
On Earth your weight is WE. Suppose you are now on Planet X
whose mass is twice the mass of the Earth’s but whose
diameter is also twice the Earth’s diameter.
Compare WX, your weight on Planet X to your weight on Earth.
(A) WX = ¼ WE
(B) WX = 1/2 WE
(C) WX = WE
(D) WX = 2 WE
(E) WX = 4 WE
We know your weight is determined by your mass and g:
W = mg
We know g, the acceleration due to gravity on a planet, in terms of G,
the universal gravitation constant, and planet parameters (MP, RP)
g planet = G
M planet
2
R planet
Therefore your weight is proportional to gplanet
Wplanet
M planet
∝ 2
R planet
2
M X REarth
WX = WEarth 2
RX M Earth
2
2 M Earth REarth
WX = WEarth
(2 REarth ) 2 M Earth
WX = WEarth
2
1
=
WEarth
2
( 2) 2
PHYS 100 Lecture 11, Slide 13
Speed
Two satellites are in circular orbits about the Earth. Satellite 2 is twice as far from
the Earth as is Satellite 1.
BB
How is the speed of Satellite 2 (v2) related to the speed of Satellite 1 (v1) ?
(A) v2 < v1
(B) v2 = v1
(D) Cannot determine: depends on
their masses
(C) v2 > v1
(D) Is NOT CORRECT: that is the point of Kepler’s Third Law !!
WHY??
1) Both accelerations (a = v2/R) are caused by the same force !!
2) That force (gravitation) is proportional to the satellite’s mass
M m
F1 = G E 2 1
R1
PHYSICS
a1 =
BUT
F1
M
= G 2E
m1
R1
v12
a1 =
R1
Satellites in larger orbits have smaller speeds.
v12 = G
v2 ∝
ME
R1
1
R
PHYS 100 Lecture 11, Slide 14
CheckPoint 3
Case I: a satellite orbits the Earth once every 2 days at an orbital distance of Ro.
Case II: a satellite travels around the Earth at an orbital distance of 4Ro.
BB
3
How long would take the satellite in Case II to make one complete revolution around the Earth?
32
(A) 1 day
(B) 2 days
(C) 4 days
(D) 8 days
(E) 16 days
(F) 32 days
You said:
• Period is directly proportional to radius. Therefore, if
the radius increases by a factor of 4, so does the
period.
• the period is in the denominator, and is a squared
term. for simplicity we can say that the radius cubed /
period squared for case one is 1/4. so, the second term
having a radius 4 times larger will need a period 16 days.
• If R0 takes two days, 4 squared is 12, and multiply by
the original days, 2 to get 32 days.
60
50
40
30
20
10
0
A
B
C
D
E
F
PHYS 100 Lecture 11, Slide 15
CheckPoint 3
Case I: a satellite orbits the Earth once every 2 days at an orbital distance of Ro.
Case II: a satellite travels around the Earth at an orbital distance of 4Ro.
How long would take the satellite in Case II to make one complete revolution around the Earth?
32
(A) 1 day
(B) 2 days
(C) 4 days
(D) 8 days
(E) 16 days
Recall the “speed” question: larger orbits correspond to slower speeds
Using definition of period (P = 2πR/v), we get Kepler’s Third Law:
Case I
Case II
R03
(4 R0 )3
=
2
P1
P22
43 R03
P = P1
= 64 P12
3
R0
2
2
2
(F) 32 days
v12 = G
ME
R1
R3
= constant
2
P
P2 = 8P1
PHYS 100 Lecture 11, Slide 16
Directions
BB
The springs have equal spring constants (k) and unstretched lengths (x0) in the two cases
Compare the directions of the forces that the spring exerts on the mass in the two cases
I II
(D)
(C)
(B)
(A)
I II
I II
I II
Springs ALWAYS exert a RESTORING FORCE
Force direction ALWAYS towards UNSTRETCHED POSITION
PHYS 100 Lecture 11, Slide 17
CheckPoint 4
The springs have equal spring constants (k) and unstretched lengths (x0) in the two
cases
BB
2
Compare XI and XII, the distances the srings are stretched/compressed in the two cases
(A) XI < XII
(B) XI = XII
(C) XI > XII
You said:
• Draw a freebody diagram. The forces are the same. F
spring=mg sinX equals k abs(x - x0) = mg sinx
• X1 is greater because the string is at the top of the
incline and the box is moving down so the distance
between the base of the spring and the box are
increasing so it is stretching. X2 is smaller because the
spring is placed at the base of the incline and the box is
going to slide that way naturally due to gravity so it is
compressing the spring and making it smaller.
70
60
50
40
30
20
10
0
A
B
C
PHYS 100 Lecture 11, Slide 18
CheckPoint 4
The springs have equal spring constants (k) and unstretched lengths (x0) in the two cases
Compare XI and XII, the distances the srings are stretched/compressed in the two cases
(A) XI < XII
(B) XI = XII
THE KEY:
(C) XI > XII
FREE BODY DIAGRAMS
They are the SAME !!
Fspring = mg sin θ
k x − x0 = mg sin θ
x − x0 =
m
g sin θ
k
PHYS 100 Lecture 11, Slide 19
REMINDERS
• Homework on Universal Gravitation & Springs Due
Tuesday
• Final
•
•
•
Exam
Monday May 7 at 1:30pm
112 Transportation Building
Contact me if you have a conflict
PHYS 100 Lecture 11, Slide 20