Transcript Rotational Motion

Rotational Motion
Reading: pp. 194 – 203 (sections 8.1 – 8.3)
HW #1
p. 217, question #1
p. 219, problem #1, 4, 5, 6, 15, 16, 17, 19
I. Introduction:
A rotating object is one that spins on a fixed axis. The position and direction
of the rotation axis will remain constant. The position of some part of the
object can be specified with standard cartesian coordinates, (x,y). All
objects will be assumed to rotate in a circular path of constant radius.
y
Both coordinates, (x, y),
change over time as the
object rotates.
(x, y)
R
The object’s position can
also be specified with
polar coordinates, (r, q).
q
x
For this last coordinate system,
only the angle changes. The
radius stays constant.
II. Definitions:
Since the polar coordinates only have one changing variable, the angle, we
will use this to simplify analysis of motion.
position is defined as where an object is in space. Here, we
A. The ___________
only need to specify the angle of the object with respect to some origin or
reference line.
The position of the object is measured by an angle, q, measured
counterclockwise from the positive x – axis.
The angle will be measured in
radians rather than
units of ___________
q
degrees .
___________
Radians are defined as the ratio of the length along the arc of a circle to the
position of an object divided by the radius of the circle.
s
q
R
R
s
q
s = length along the arc measured
counterclockwise from the +x – axis.
R = radius of the circular path.
Since q is the ratio of two lengths, the angle measurement really does not
have any units. The term “radians” is just used to specify how the angle is
measured.
1 revolution = 360 degrees = 2p radians
displacement of the object is just the difference in its position.
B. The _____________
angular displacement of the object as the difference in
We define the ____________________
its angular position.
q  q f  q i  q  q o
A positive q shows a counterclockwise {ccw} rotation, while a negative q
shows a clockwise {cw} rotation.
velocity
C. Motion can also be measured through a rate of rotation, a __________.
angular velocity
The ____________________
of an object is defined as the amount of
rotation of an object per time. The angular velocity is represented by the
greek letter w (lower case omega).
average angular velocity:
t = elapsed time.
q q  q o
w

t
t
rad
units 
s
D. Another measure of motion is the rate of change of velocity, called an
acceleration
________________.
angular acceleration of an object is defined as the amount of
The ____________________
change of the angular velocity of an object per time. The angular
acceleration is represented by the greek letter a (lower case alpha).
average angular acceleration:
w w  wo
a 

t
t
rad
units  2
s
In general, the motion may be complex, but we will again look at constant
angular acceleration cases, exactly the same way as we did back in Ch. 2.
The same equations of motion can be derived for circular motion.
Ch. 2
Ch. 8
Linear Motion
Rotational Motion
v  vo  at
w  wo  a t
x  vo t  12 at 2
q  wot  12 a t 2
v  vo  2ax
w  wo  2aq
2
2
v  vo  x

2
t
This is only true for constant accelerations.
Problem solving is the same as before.
2
2
w  wo
2
q

t
E. The motion of an object around a circle can also be represented as
actual distances along the circle and speeds tangent to the circle.
R
vt
tangential speed , v , of an object is defined as the angular
The _________________
t
velocity times the radius of the circle.
vt  Rw
The tangential speed measures the actual speed of the object as it
travels around the circle.
length along arc
average vt =
elapsed time
tangential acceleration
The _________________________
, at, of an object is defined as the
angular acceleration times the radius of the circle.
at  Ra
The tangential acceleration measures how the tangential speed
increases or decreases over time.
Example #1: A disk rotates from rest to an angular speed of 78.00 rpm in a
time of 1.300 seconds. a. What is the angular acceleration of the disk?
wo = 0, w = 78.00 revolutions per minute, t = 1.300 seconds. a = ?
w  wo  at
w  78.00
a
rev
min
 2p rad  1 min 

  8.168 rad s

 1 rev  60 sec 
w  wo
t
8.168 rad s  0

 6.283 rad s 2
1.300 s
b. Through what angle does the disk turn?
q  wot  12 at 2
q  0
rad
s
1.300 s   6.283
1
2
q  5.309 rad
rad
1.300 s 
2
s2
c. Through what angle will the disk turn if it were to maintain the same
angular acceleration up to 254.0 rad/s?
w  wo  2aq
2
2
w 2  wo 2 254.0 rad s 2  02
q 

2a
26.283 rad s 
2
q  5134 rad
d. The disk has a diameter of 12.00 inches. What is the tangential speed
and acceleration at the edge of the disk the moment the disk reaches 78.00
rpm?
 12.00 in  2.54 cm  1 m 

  0.1524 m
R

 2  1in  100 cm 
vt  Rw  0.1524 m8.168 rad s   1.245 m s
at  Ra  0.1524 m 6.283 rad s 2   0.9576 m s 2
Centripetal Force
HW: See Schedule
III. Centripetal Acceleration and Force.
When an object moves in a circle, the direction of its velocity is always
changing. This means the object is always accelerating! For an object
rotating at a constant angular speed, the acceleration of the mass is
always towards the center of the motion. This kind of acceleration is called
centripetal {“center seeking”} acceleration, and is represented as ac.
a ___________
The amount of acceleration depends on the radius of the circular path and
the speed around the circle.
v2
ac 
 w 2r
r
w = angular speed, v = tangential speed, r = radius of circular path.
Example #2: A wheel of a car has a diameter of 32.0 inches. A rock is
wedged into the grooves of the tire. a. What is the centripetal acceleration
on the rock if the wheel turns a rate equal to 70.0 mph?
v  70.0
mile
hour
 1609.344 m  1 hour 

  31.29 m s

 1 mile  3600 s 
 32.0 in  2.54 cm  1 m 

  0.4064 m
r 

 2  1in  100 cm 

v
31.29 s 
ac 

 2410 m s 2
r 0.4064 m 
2
m
2
b. What is the rotation rate of the tire, in rpm?
v 31.29 m s
w 
 77.0 rad s
r 0.4064 m
w  77.0
rad
 1 rev  60 s
s 
 2p rad  1 min



  735 rpm

If there is a centripetal acceleration making an object move in a circle, then
there must be an unbalanced force creating this acceleration.
centripetal force, and it also points towards
This force is called the _____________
the center of the circular motion. This force must be made by real forces
acting on an object. The centripetal force will be the sum of the radial
components of the forces acting on the object. A radial component points
center
towards the ___________
of the circle.
mv2
Fc  mac 
 mw 2 r
r
Example #3: A 0.475 kg mass is tied to the end of a 0.750 meter string and
the mass is spun in a horizontal circle. If the mass makes 22.0 revolutions
in a time of 2.50 seconds, what is the tension in the string holding the mass
to the circular motion?
rotation
R
T
q  22.0 rev  2p rad 

  55.3 rad s
w
 
t
 2.50 s  1 rev 
Fc  mw 2r  0.475 kg55.3 rad s  0.750 m
2
Fc  1090 N
m
Centripetal Force
HW #2: See Schedule
Last Chance for Make-up Tests!
Wednesday after school!
Example #4: A penny sits at the edge of a 12.00 inch diameter record. If
the coefficient of static friction is 0.222 between the penny and the record,
what is the maximum rotation rate of the record that will allow the penny to
remain on the record?
n
rotation
vertical forces balance.
n  mg
Fs
by definition friction is:
mg
Fs ,max   s n
Fs ,max   s mg
set the friction force equal
to the centripetal force
w
s g
r

Fc  mw 2 r  Fs ,max  s mg
0.2229.80 m s 
2
6.00 in  0.0254 m 

1in

w  3.78 rad s
Example #5: A mass m on a frictionless table is attached to a hanging
mass M by a cord through a hole in the table. Find the speed which m must
move in order for M to stay at rest. Evaluate the speed for m = 2.00 kg,
M = 15.0 kg, and r = 0.863 m.
Since M is at rest, the tension force
lifting it is equal to the weight of M:
T  Mg
This tension is also the centripetal
force on the mass m, causing it to
spin in a circular path:
mv2
T
r
Set the two equations equal to one another:
mv 2
T
 Mg
r
v
v
rMg
m
0.863 m15.0 kg9.80 m s 
2.00 kg
2
v  7.96 ms
Example #6: A common amusement park ride involves a spinning cylinder
with a floor that drops away. When a high enough rotation speed is
achieved, the people in the ride will stay on the side of the wall. A static
friction force holds each person up. Solve for the rotation rate of the room,
given the coefficient of friction for the wall and the radius of the room.
Example #6: A common amusement park ride involves a spinning cylinder
with a floor that drops away. When a high enough rotation speed is
achieved, the people in the ride will stay on the side of the wall. A static
friction force holds each person up. Solve for the rotation rate of the room,
given the coefficient of friction for the wall and the radius of the room.
set the friction force equal to the weight
as the vertical forces balance.
Fs ,max  mg
Fs
n
m
mg
Fs ,max   s n
forces balance
definition of force of
static friction
Horizontal direction: set the normal
force equal to the centripetal force.
n  Fc  mw 2 r
Combine all the information to solve for the rotation rate, w.
mw r  n 
2
w 
2
Fs , max
g
s

mg
s
g
w
s r
s r
What would be the rotation rate for a room 3.00 m wide and a carpeted wall
with a coefficient of friction of 0.750?
w
9.80 
m
s2
0.7501.50 m
 2.95 rad s  28.2 rpm
Example #7: (Banking Angle) Determine the angle of the roadway
necessary for a car to travel around the curve without relying on friction.
Assume the speed of the car and the radius of the curve are given.
n
q
q
mg
n cosq 
component of the normal
force that is vertical
n sin q 
component of the normal
force that is horizontal, also
becomes the centripetal force
balance the vertical forces:
n cos q  mg
set the net horizontal force equal to the centripetal force, with towards the
center of the circular path as the positive direction:
mv2
n sin q 
R
substitute in:
mg
n
cos q
mg
mv2
sin q 
cos q
R
v 2  Rg tan q
2
v
q  tan 1
Rg
Example #8: Conical Pendulum.
A mass of m = 1.5 kg is tied to the end of a cord whose length is L = 1.7 m.
The mass whirls around a horizontal circle at a constant speed v. The cord
makes an angle q = 36.9o. As the bob swings around in a circle, the cord
sweeps out the surface of a cone. Find the speed v and the period of
rotation T of the pendulum bob.
T cos q  mg
mv2
T sin q 
R
divide…
T sin q mv2

T cos q mgR
note:
R
sin q 
L
simplify…
2
v
tan q 
gR
v
v
v  Rg tan q
L sin q g  tan q
1.7 msin 36.9o 9.80 m s  tan 36.9o 
2
v  2.74 ms
the period, T, is the time for one revolution:
2p R
v
T
2p R 2p L sin q
T

v
v

2p 1.7 m sin 36.9o
T
2.74 ms
T  2.34 s

Example #9: Another common amusement park ride is a rollercoaster with
a loop. Determine the minimum speed at the top of the loop needed to
pass through the top of the loop.
There are two forces acting on the car: the
weight pulling straight downwards and the
normal force pushing perpendicular to the track.
As the car goes through the loop, the normal
force always points towards the center of the
loop. The vector sum of the normal force and
the radial component of the weight equals the
centripetal force.
The net force for the mass at the top of the loop is:
m
2
mv
Fc 
 Fnet   n  mg
r
The faster the car goes, the greater
the normal force to push the car into a
circular path. The minimum speed for
the car at the top of the loop is where
the normal force goes to zero.
2
mvmin
 mg
r
vmin  rg
n
mg
r
Example #9: (b) Use energy conservation to find the speed of the mass at
the bottom of the loop.
Etop  12 mv
2
min
Etop  KE  PE
 mg2r 
2
Ebottom  12 mvbot
0
Ebottom  KE  PE
Set the two energies equal:
1
2
2
2
mvbot
 12 mvmin
 mg 2r 
Divide by m and multiply by 2:
2
2
vbot
 vmin
 4 gr  rg  4rg  5rg
vbot  5rg
Example #9: (c) What is the necessary starting height of the mass if sliding
from rest down the ramp?
Etop ramp  KE  PE
Etoploop  KE  PE
Etop ramp  0  mgh
2
Etoploop  12 mvmin
 mg2r 
Set the two energies equal:
2
mgh  12 mvmin
 2mgr  12 mrg   2mgr  52 mgr
Divide both sides by m and g:
h  52 r  54 d
AP Summary Problems!
Example #10: {p. 135, problem #93} A circular curve of radius R in a new
highway is designed so that a car traveling at speed vo can negotiate the
turn safely on glare ice (zero friction). If the car travels too slowly, then it
will slip toward the center of the circle. If it travels too fast, then it will
slip away from the center of the circle. If the coefficient of static friction is
increased (from zero), a car can stay on the road while traveling at any
speed within a range from vmin to vmax. Derive formulas for vmin and vmax as
functions of s, vo, g and R.
Start with the too fast scenario. The car will want to slide up the ramp, so
friction will point to the bottom of the ramp.
n sin q
Ff cos q
F f sin q
Find horizontal and vertical components of
the forces:
n cosq
Vertical components of the forces balance:
n cos q  mg  Ff sin q
Definition of friction:
Ff   s n
n cos q  mg  s n sin q
mg
n
cos q   s sin q
mg
n
cos q   s sin q
Horizontal components of the forces make the centripetal force. Take towards
the left as the positive direction:
mv
n sin q  Ff cos q 
r
2
mv 2
n sin q  s n cos q 
r
mv 2
n  sin q   s cos q  
r
Ff   s n


mg
mv 2

  sin q  s cos q  
r
 cos q  s sin q 
 sin q  s cos q 
v  rg 

 cos q  s sin q 
2
Next is the too slow scenario. The car will want to slide down the ramp, so
friction will point to the top of the ramp.
n sin q
n cosq
F f sin q
Find horizontal and vertical components of
the forces:
Vertical components of the forces balance:
n cos q  Ff sin q  mg
Definition of friction:
Ff cos q
Ff   s n
n cos q  s n sin q  mg
mg
n
cos q   s sin q
mg
n
cos q   s sin q
Horizontal components of the forces make the centripetal force. Take towards
the left as the positive direction:
mv 2
n sin q  Ff cos q 
r
mv 2
n sin q  s n cos q 
r
mv 2
n  sin q  s cos q  
r
Ff   s n


mg
mv 2

  sin q  s cos q  
r
 cos q  s sin q 
 sin q  s cos q 
v  rg 

 cos q  s sin q 
2
Example #11: A mass is tied to the end of a string and spun in a vertical
circle. What is the difference in tension in the string between top and
bottom of the circle?
mvtop 2
mg
r
Ttop
  mg  Ttop
mvbottom 2
 Tbottom  mg
r
energy conservation:
Tbottom
1
2
mg
mvbottom2  12 mvtop 2  mg  2r 
mvbottom 2
 Tbottom  mg
r
mvtop 2
r
Tbottom
mvbottom 2

 mg
r
Ttop 
  mg  Ttop
mvtop 2
r
 mg
subtract:
Tbottom  Ttop
Tbottom  Ttop 
2
mvbottom

 mg 
r
mvbottom 2  mvtop 2
r
mvtop 2
 2mg
r
 mg
now use energy
conservation to
eliminate speeds
Tbottom  Ttop 
1
2
mvbottom 2  mvtop 2
r
 2mg
mvbottom2  12 mvtop 2  mg  2r 
mvbottom2  mvtop 2  2mg  2r 
Tbottom  Ttop
4mgr

 2mg
r
Tbottom  Ttop  6mg
Centripetal Force
HW check tomorrow!
Example #12: Sonam is seated on the top of a
frictionless hemispherical mound of ice of radius R.
Why she is there, I do not know. A small breeze upsets
the equilibrium and she starts sliding down the ice. At
what vertical height from the ground does she leave the
ice surface? Give the answer in terms of R.
When Sonam is at the side, as shown, the
sum of the radial components gives the
centripetal force:
mv
 mg cos q  n
R
0
Sonam leaves the surface when the
normal force becomes zero.
mg cos q
h = R cos q
2
n
mv 2  mgR cos q
mv 2  mgh
q
mg
Now use energy conservation to
find the speed of Sonam if she
starts at rest at the top.
Etop   mgR
Eside  12 mvside 2  mgh
From the previous slide:
mv 2  mgh
mgR  12 mgh  mgh  23 mgh
h  23 R
Newton’s Universal Law
of Gravity
HW #3 on schedule
Turn in Rotary Motion Lab
Newton’s Law of Universal Gravity:
Any two objects are attracted
to each other through the
force of gravity.
The force is proportional
to each mass.
F  m1
F  m2
The force is inversely proportional to the square of
the center to center distance between the masses.
1
F 2
r
“inverse square law”
These two equations can be combined into
a single equation, and a proportionality
constant can be introduced to make
an equality:
Gm1m2
F
r2
2
By Newton’s 3rd Law,
N
m
11
G

6
.
67384

10
forces are equal and
kg 2
opposite!
Example #1: Two students, each with mass 70 kg, sit 80 cm apart.
(a) What is the force of attraction between the two students?
m1  m2  70 kg
r  80 cm  0.80 m
Gm1m2
F
r2
2


N
m
11
 6.67384 10
70 kg 70 kg 
2 
kg 


2
0.80 m
F  5.110 7 N
(b) How does this compare to the weight of the students?
weight  m1 g  70 kg 9.80 m s 2 
weight  686 N
7
F
5.110 N
10

 7 10
weight
686 N
The force between you and your neighbor is less
than 1 part in a billion of your normal body weight.
Example #2: What is the direction of the net force on the mass at the
center of the image? Why?
All forces balance out
pairwise except for one:
Force on central mass
points upwards.
Example #3: What is the direction of the net force on the mass at the
center of the image? Why?
All forces balance
out pairwise
except for one:
Force on central
mass points
towards the left.
II. Compare the force of gravity to weight:
The weight of an object near the
surface of the Earth is just the
force of gravity exerted on the
object by the Earth.
GM E m
mg 
2
RE
Remember, r is the center to center (of the earth)
distance.
M E  5.98 1024 kg
RE  6.378 106 m
The mass of the object cancels out, and what is left is
a theoretical value for the acceleration due to gravity
near the surface of the Earth.
GM E
g
2
RE
Example #4: Estimate the value of ‘g’ for the Earth.
GM E
g
2
RE
2


N
m
11
24
 6.67384 10
 5.98 10 kg
2 
kg 


2
6
6.378 10 m


g  9.81 m s 2


Example #5: Estimate the value of ‘g’ near the surface of Peter Griffin.
Example #5: Estimate the value of ‘g’ near the surface of Peter Griffin.
width  1.4 m
Estimate volume
as sphere 1.4 m
in diameter.
4p
3


volume 
0.70 m
3
 1.44 m3
Multiply by density to get mass:
m  900 1.44 m   1293 kg
kg
m
3
3
 2800lbs!
Estimate ‘g’ at surface:
GM
g 2
R
2


N
m
11
 6.67384 10
1293 kg 
2 
kg 


2
0.70 m
g  1.8 10
7 m
s2
III. Satellite Motion.
A satellite is an object that
orbits (travels in a circular path
around) some gravity source
The force of gravity
provides the centripetal
force to keep the
satellite moving in a
circular path.
The equation of motion is:
2
mv
GMm
 2
r
r
m = satellite mass (divides out…)
M = mass of the gravity source
r = radius of the orbit (center to center distance!)
v = the tangential speed of the orbiting object
Example #6: The space shuttle orbits the Earth at an altitude of 400 km
above the surface of the Earth. (a) Determine the speed of the space
station around the Earth.
mv2 GMm
 2
r
r
r  6.378 106 m  4.00 105 m
2


N
m
11
24
 6.67384 10

5
.
98

10
kg
2 
GM
kg 

v

6
5
r
6.378 10 m  4.00 10 m


v  7,673 ms


(b) How much time will it take for the space station to orbit the Earth?
dist 2p r
T  time 

rate
v

2p 6.778 106 m
T
7673 ms

T  5550 s  92.5 min
Example #7: Write an equation that gives the period (time for one
revolution) of an orbit in terms of the radius of the orbit.
2p r
v
T
2
mv
GMm
 2
r
r
GM
 2p r 
2

 v 
r
 T 
2
r 3 GM

2
2
T
4p
Example #8: Determine the altitude (height above the surface) for a
geosynchronous satellite. The period of the satellite matches the length of
the Earth’s day, 24 hours.
 3600 s 
  8.64 104 s
T  24 h 
 1h 
GM 2
 T 
r 
T  GM 

2
4p
 2p 
2
3

11 N m
3
r   6.67384 10
2
kg

2

 8.64 10 s 
24
 5.98 10 kg 

2p





4
2

11 N m
r  3  6.67384 10
2
kg

2

 8.64 10 s 
24
 5.98 10 kg 

2p





4
r  4.23 107 m
This is the height from the center of the Earth.
Subtract the radius of the Earth to get the height from
the surface.
7
6
h  4.23 10 m  6.378 10 m
h  3.59 10 m
7
2
Example #9: {on your own} Determine the orbital velocity of Grog’s golf
ball.
r  REarth  6.378  106 m
Example #9: Determine the orbital velocity and period of Grog’s golf ball.
mv2 GMm
 2
r
r
2

11 N m 
24
6.67384

10
5.98

10
kg 

2 
kg 
GM  
v
6
6.378

10
m
r

v  7,910 ms
Example #10: What is your perceived weight
when you stand at the equator of the Earth?
n
Example #10: What is your perceived weight
when you stand at the equator of the Earth?
mw 2 r  mg  n
m
mg
Your perceived weight is the normal
force pushing on your feet to support
you.
r
n  mg  mw r
2
Some of the true gravity force is used to make
you move in a circle. The remainder of that
force is what you feel as your weight.
The percentage your
weight is reduced is: 
mw 2 r

mg
2

2p rad
6
6.378

10
m




s 
  24 h   3600 h  
9.80 m s2
 0.003
 0.3%
Example #11: {on your own!} What is maximum rotation rate for the Earth
so that objects at the equator will stay on the surface and not “fly off”?
Hint: The normal force goes to zero at this maximum rotation rate. The
objects are essentially in a low orbit.
mw r  mg  n
2
w
0
9.80 m s2
g
3 rad

1.24

10

s
6
r
6.378 10 m
T
2p
w
 5070 s  84.5 min
Potential Energy and
Newton’s Law of Gravity
?
Newton’s Law of Universal Gravity:
Any two objects are attracted
to each other through the
force of gravity.
Gm1m2
F
r2
2
N
m
11
G  6.67384 10
kg 2
This force also stores energy between the two
masses. The potential energy stored in the two
masses is:
Gm1m2
PE  
r
Example #12: What is the minimum speed that a spaceship at the surface
of the Earth must have to completely escape Earth’s gravity?
m
m
Total mechanical energy is conserved.
Esurface  KE  PE
1
2
E far away
0
mvsurface  PE  mv far away
2
1
2
1
2
mvsurface
2
0
 KE  PE
2 The least speed at the
surface corresponds to the
least speed far away.
GM Earth m

0
r
1
2
mvsurface
vsurface
vsurface
2
GM Earth m

0
r
2GM Earth

r
2


N
m
11
24
2  6.67384 10
5.98 10 kg 
2 
kg 


6
 6.378 10 m 
vsurface  1.12 10
4 m
s
 11.2 kms
Kepler’s Laws of Satellite
Motion
Handout
HW #4
IV. Kepler’s Laws.
Johannes Kepler was a German mathematician,
astronomer and astrologer. He is best known for his
eponymous laws of planetary motion, which he was
able to put together by painstakingly analyzing the
volumes of data collected by Tycho Brahe of the
planets motions in the sky.
Kepler
Brahe
Kepler’s 1st Law:
All planets (satellites) orbit the sun (or gravity
source) in elliptical orbits.
The sun sits at one focus
of the ellipse. The planet
will be closer to the sun
for part of its “year”, and
farther away for the other
part of its “year”.
Rp = perihelion = closest
distance of planet to sun
Ra = aphelion = farthest
distance of planet to sun
Kepler's first law. An ellipse is a closed curve such that the sum of
the distances from any point P on the curve to two fixed points
(called the foci, F1 and F2) remains constant. That is, the sum of
the distances, F1P + F2P, is the same for all points on the curve. A
circle is a special case of an ellipse in which the two foci coincide,
at the center of the circle.
Kepler’s 2nd Law:
As planets orbit around the sun, an imaginary line
from the planet to the sun will sweep out equal
areas in equal times.
Kepler's second law. The two shaded regions have equal areas. The
planet moves from point 1 to point 2 in the same time as it takes to move
from point 3 to point 4. Planets move fastest in that part of their orbit
where they are closest to the Sun. Exaggerated scale.
Potential energy from
gravity still has the same
behavior on this scale:
The farther the planet moves from the sun, the higher
the potential energy of the planet. But total energy is
still constant:
E  KE  PE  const
As the planet moves farther from the sun, it slows
down. The closer the planet to the sun, the faster it
moves in its orbit.
Measurable Application: Earth’s weather is due to
the tilt of the Earth’s axis relative to its orbital
plane.
Earth at perihelion
Earth at aphelion
Since summer (for the northern hemisphere) occurs
when Earth is farthest from the sun, the Earth
spends slightly more time in this part of the orbit as
compared to winter (again, for northern
hemisphere). The effect is that summer is about 3
days longer than winter. This is sometimes called
the “summer analemma”.
The analemma is the location
of the sun in the sky at the
same time each day of the
year plotted on a graph or
photograph. The uneven
shape is due to the elliptical
orbit of the Earth around the
sun.
Kepler’s 3rd Law:
The ratio of the cube of the length of the semi
major axis, to the square of the period of
revolution, is constant for all planetary satellites.
r3
K
2
T
r 3 GM

2
T
4p 2
r = length of semi major axis.
T = period of orbit.
Newton later figured out how this
constant fit into his gravitational
scheme.
Example #13: Earth’s orbital information is sometimes used as a standard:
The orbital period is one year and its semimajor axis is one AU
(astronomical unit). If the orbital period for Jupiter is 11.9 years, determine
the semimajor axis distance for Jupiter.
r 3 GM

 const
2
2
T
4p
3
3
aJ
aE
 2
2
TJ
TE
 TJ
TJ
 aE 
2
TE
 TE
2
a J  aE 3



2
3
 11.9 y 

a J  1.00 AU 
 1.00 y 
aJ  5.21 AU
2
3
Example #14: Earth’s orbit has a semimajor axis length of 1.50×108 km
and a period of 365.24 days. Determine the mass of the sun.
r 3 GM

2
2
T
4p
4p 2 r 3
M
2
GT


4p 1.50 10 m
M
2

2
11 N m 
h
s
 6.67384 10
365.24d 24 d 3600 h 
2 
kg 

2
11
M  2.00 10 kg
30
3
Example #15: A satellite moves in a circular orbit around Earth at a speed
of 5,000 m/s. Determine (a) the satellite’s altitude above the surface of
Earth and (b) the period of the satellite’s orbit.
mv2 GMm
 2
r
r
GM
r 2
v
2

11 N m 
24
6.67384

10
5.98

10
kg 

2 
kg 

r
m 2
 5000 s 
r  1.60 10 m
7
subtract the radius
of Earth to get the
altitude:
h  9.59 10 m
6
How much time will it take for the space station to orbit the Earth?
dist 2p r
T  time 

rate
v
T
2p 1.596 107 m 
5000 ms
T  2.01104 s  334 min  5.57 hours
Example #16: Io, a satellite of Jupiter, has an orbital period of 1.77 days
and an orbital radius of 4.22 × 105 km. From this data, determine the mass
of Jupiter.
r 3 GM

2
2
T
4p
M
4p 2 r 3
M
2
GT
4p
2
 4.22 10 m 
8
3
2
2


N
m
11
h
s 
1.77
d
24
3600




d
h 
 6.67384 10

2

kg 

M  1.90 10 kg
27
Example #17: Assume the Earth has a uniform density of r. Determine the
strength of Earth’s gravitational field as a function of its radius for the
interior of the Earth.
density is mass per
unit volume!
M
M
3M
r
 4

3
3
V
p
R
4
p
R
3
When inside a spherically symmetric object,
the pull of gravity at that point depends only on
the amount of mass contained in a ball whose
radius matches that location. The shell of mass
on the outside of that position does not
contribute to the pull of gravity.
GM inside
g r  
r2
Determine the mass located inside:
M inside
 3M  4 3 
 rV  
pr 
3 
 4p R  3

3
M inside
r
M 3
R
Plug into the formula for the acceleration due to gravity:
GM inside
g r  
r2
G
r3
 2M 3
r
R
GM
 3 r
R