Transcript Physics 131: Lecture 14 Notes

Physics 151: Lecture 27
Today’s Agenda

Today’s Topic
Gravity
Planetary motion
Physics 151: Lecture 27, Pg 1
See text: 14

New Topic - Gravity
Sir Isaac developed his laws of motion largely to
explain observations that had already been made
of planetary motion.
Earth
Sun
Moon
Note : Not to scale
Physics 151: Lecture 27, Pg 2
See text: 14.1
Gravitation
(Courtesy of Newton)
Things Newton Knew,
1. The moon rotated about the earth with a
period of ~28 days.
2. Uniform circular motion says, a = w2R
4. Acceleration due to gravity at the surface of
the earth is
g ~ 10 m/s2
5. RE = 6.37 x 106
6. REM = 3.8 x 108 m
Physics 151: Lecture 27, Pg 3
See text: 14.1
Gravitation
(Courtesy of Newton)
Things Newton Figured out,
1. The same thing that causes an apple to
fall from a tree to the ground is what causes
the moon to circle around the earth rather than
fly off into space. (i.e. the force accelerating
the apple provides centripetal force for the
moon)
2. Second Law, F = ma
So, acceleration of the apple (g) should have
some relation to the centripetal acceleration of
the moon (v2/REM).
Physics 151: Lecture 27, Pg 4
Moon rotating about the Earth :

Calculate angular velocity :
w = v / REM = 2  REM / T REM = 2  / T
w=
1 rot
1 day
rad
x
x 2
 2 .66 x10 6 s -1
27 .3 day 86400 s
rot

So w = 2.66 x 10-6 s-1.

Now calculate the acceleration.
 a = w2R = 0.00272 m/s2 = .000278 g
Physics 151: Lecture 27, Pg 5
See text: 14.1
Gravitation
(Courtesy of Newton)
Newton found that amoon / g = .000278
 and noticed that RE2 / R2 = .000273

amoon
g
R

RE
This inspired him to propose the
Universal Law of Gravitation:
|FMm |= GMm / R2
G = 6.67 x 10 -11 m3 kg-1 s-2
Physics 151: Lecture 27, Pg 6
See text: 14.1
Gravity...


The magnitude of the gravitational force F12
exerted on an object having mass m1 by another
object having mass m2 a distance R12 away is:
m1m2
F12  G 2
R12
The direction of F12 is attractive, and lies along
the line connecting the centers of the masses.
m1
F12
F21
m2
R12
Physics 151: Lecture 27, Pg 7
Gravity...

Compact objects:
R12 measures distance between objects
R12

Extended objects:
R12 measures distance between centers
R12
Physics 151: Lecture 27, Pg 8
See text: 14.1

Gravity...
Near the earth’s surface:
R12 = RE
» Won’t change much if we stay near the earth's
surface.
» i.e. since RE >> h, RE + h ~ RE.
h
m
Fg
M Em
Fg  G 2
RE
M
RE
Physics 151: Lecture 27, Pg 9
See text: 14.3
Gravity...

Near the earth’s surface... Fg  G

So |Fg| = mg = ma

a=g
 ME 
M Em
 m G 2 
2
RE
 RE 

=g
All objects accelerate with
acceleration g, regardless
of their mass!
Where: g  G M E  9.81 m / s 2
RE2
Physics 151: Lecture 27, Pg 10
Example gravity problem:

What is the force of gravity exerted by the earth
on a typical physics student?
Typical student mass m = 55kg
g = 9.8 m/s2.
Fg = mg = (55 kg)x(9.8 m/s2 )
Fg = 539 N

Fg
The force that gravity exerts on any object is
called its Weight
W = 539 N
Physics 151: Lecture 27, Pg 11
Lecture 27, Act 1
Force and acceleration


Suppose you are standing on a bathroom scale in
Physics 203 and it says that your weight is W. What
will the same scale say your weight is on the surface
of the mysterious Planet X ?
You are told that RX ~ 20 REarth and MX ~ 300 MEarth.
(a)
0.75 W
(b)
1.5 W
X
(c)
2.25 W
E
Physics 151: Lecture 27, Pg 12
Lecture 27, Act 1
Solution


The gravitational force on a person
of mass m by another object (for instance
a planet) having mass M is given by:
Mm
F G 2
R
WX FX

Ratio of weights = ratio of forces:
WE FE
MXm
G 2
2
RX
M X  RE 




M Em
M E  RX 
G 2
RE
2
WX
1
 300     .75
WE
 20 
(A)
Physics 151: Lecture 27, Pg 13
See text: 14.3
Kepler’s Laws





Much of Sir Isaac’s motivation to deduce the laws of
gravity was to explain Kepler’s laws of the motions of the
planets about our sun.
Ptolemy, a Greek in Roman times, famously described a
model that said all planets and stars orbit about the
earth. This was believed for a long time.
Copernicus (1543) said no, the planets orbit in circles
about the sun.
Brahe (~1600) measured the motions of all of the
planets and 777 stars (ouch !)
Kepler, his student, tried to organize all of this. He came
up with his famous three laws of planetary motion.
Physics 151: Lecture 27, Pg 14
See text: 14.4
Kepler’s Laws
1st
All planets move in elliptical orbits with the sun
at one focal point.
2nd
The radius vector drawn from the sun to a
planet sweeps out equal areas in equal times.
3rd
The square of the orbital period of any planet is
proportional to the cube of the semimajor axis
of the elliptical orbit.

It was later shown that all three of these laws are a
result of Newton’s laws of gravity and motion.
Physics 151: Lecture 27, Pg 15
See text: 14.4
Kepler’s Third Law
Let’s start with Newton’s law of gravity and take the
special case of a circular orbit. This is pretty good
for most planets.
F G
G
M smp
M smp
R
2
R

2

m pv2
R
m p (2R / T ) 2
R
2

 3
4

2
 R
T  
 GM s 
Physics 151: Lecture 27, Pg 16
See text: 14.4
Kepler’s Second Law
This one is really a statement of conservation of
angular momentum.
 
M smp
  R  F  Rrˆ  G
rˆ
2
R
 G
M smp
R
rˆ  rˆ  0
  
 
L  R  p  mP R  v  Constant
Physics 151: Lecture 27, Pg 17
See text: 14.4
Kepler’s Second Law
  
 
L  R  p  mP R  v  Constant
dA
R
dR
2. The radius vector drawn from the sun to a planet
sweeps out equal areas in equal times.
dA

 Constant
dt
Physics 151: Lecture 27, Pg 18
See text: 14.4
Kepler’s Second Law
  
 
L  R  p  mP R  v  Constant
dA
R
dR
 1  
1 
1 L 
dA  R  dR  R  v dt   dt 
2
2
2 M 
dA
L

 Constant
dt 2M
Physics 151: Lecture 27, Pg 19
See text: 14.7
Energy of Planetary Motion
A planet, or a satellite, in orbit has some energy
associated with that motion.
Let’s consider the potential energy due to gravity in
general.
Mm
F  G
r2
s
p
R2
r2
M sm p
W   F(r)dr    G 2 dr
r
r1
r1

U
1 1
U  U f U i  W  GMsm p (  )
rf ri

Define ri as infinity

U 
GM s m p
RE
r
0
U
1
r
r
 151: Lecture 27, Pg 20
Physics
See text: 14.7
Energy of a Satellite
A planet, or a satellite, also has kinetic energy.
1 2 GM s m p
E  K  U  mv 
2
r
We can solve for v using Newton’s Laws,
GM s m p
r2
mv 2
 ma 
r
Plugging in and solving,
GM s m p GM s m p
GM s m p
E


2r
r
2r
Physics 151: Lecture 27, Pg 21
Recap of today’s lecture

Chapter 13
Gravity
 Planetary motion

Physics 151: Lecture 27, Pg 22