CIRCULAR MOTION
and
GRAVITATION
Chapter 5 – Uniform Circular Motion and
Newton's Universal Law of Gravitation
Net Force causes Acceleration
That means that a net force causes an object’s
velocity to change…
Net force in direction of motion
- To speed up
(tangential)
- To slow down
v
t


 F  ma  m
- To change direction or move in a curve
Sideways Net force, perpendicular to direction of
motion (centripetal)
v2
r
c
r


 F  ma  m
Uniform Circular Motion
Magnitude of the velocity (speed) is constant, but the
direction of the velocity continuously changes as object
moves around in a circle
v1
w
r
q
v2
The velocity
vector at
each instant,
vt, is tangent
to the path.
Some terms for UCM:
Period, T, time it takes to make 1 revolution (s)
1
T
f
Frequency, f, number of revolutions in one
second (s-1 or Hz)
Tangential Speed, v, how
many meters per sec
v
w
1
r
q
d 2r

 2rf  wr
v2 vav  v 
t
T
Angular speed, w, how many
radians per sec
q 2
v
wav  w 

 2f 
t
T
r
Using Equations as Guide to Thinking
TANGENTIAL
Speed
2r
v
T
ANGULAR
Speed
2
w
T
If A is 1m from the center and D is 4
D
C
m from the center and all take 10 sec
B
A
to complete a revolution, what are
the velocities at A and D
0.63
vA = ______m/s
0.63
wA = ______rad/s
2.5
vD = ______m/s
0.63
wD = ______rad/s
How do the variables
relate to each other?
v  wr
Velocity is changing because the direction is changing,
so there must be an acceleration (and a force).
w
  
v v2  v1

aav 

t
t
v1
ac
-v1
v2
The acceleration points
towards the center of the
circle. It is perpendicular to
the velocity, in the radial
direction. It is called
centripetal acceration, ac
v3
(center-seeking)
Uniform Circular Motion
Centripetal acceleration
is the rate of change in
velocity of an object that
is due to the change in
DIRECTION of the
velocity.
It is always
perpendicular to the
velocity vector and
points toward the center
of the curve.
Uniform Circular Motion
What does the centripetal acceleration depend on?
displacement
v
s
w
r
q
r
q
similar triangles
v s
v
r
s
rt
v
r
Centripetal
Acceleration
ac
(magnitude)
(check units)
v

v
v

vt
ac

v
v2

 w 2r
r
Uniform Circular Motion
w
2r
 wr
v
T
v
ac
r
q
ac
v
2 v
w

T
r
v
4 r
2
ac 
w r  2
r
T
2
2
Special Cases of Acceleration
An object traveling in a straight line at constant speed:
a0
An unsupported object near the surface of the earth:
a  g  9.8m / s towards the center of the earth
2
An object moving along a circular path:
2
v
ac 
r
Example: Moon’s Centripetal Acceleration
The moon’s nearly circular orbit about the Earth has
a radius of about 384,000 km and a period T of 27.3
days. Determine the acceleration of the moon
toward the Earth in m/s2.
v
4 r 4 (3.84 x10 m)
ac 
 2 
6
2
r
T
(2.36 x10 s)


g
3
2

 2.72 x 10 m / s 
2 
 9.8 m / s 
4
 2.78 x 10 g  0.000278 g
2
2
2
8
Uniform Circular Motion
What causes centripetal acceleration?
w
v1
ac
r
Fc
q
v2
NET FORCE
The net force is centripetal, in the
direction of ac.
2
v2
4

r
2
ac 
w r  2
r
T
The centripetal force, FC, is
perpendicular to the velocity (radial
direction) and causes a change in the
direction only. FC is found by adding
all the forces in the radial direction
2
Newtons 2nd law:
mv
2
 Fr  FC  mac  r  mw r
NONuniform Circular Motion
Magnitude AND direction Acceleration and force vectors do not
point to the center in nonuniform
of the velocity change
v1
r
q
a
ac
circular motion. They have both radial
(centripetal) and tangential
components. The tangential component
of the net force causes object to speed
up/slow down and the radial component
causes a change in direction of velocity
aT
a
w
2
v
ac 
r
v2

 Fr  FC  mac
v
aT 
t

 FT  maT
a  a a
2
C
2
T
Uniform Circular Motion
What happens to an object moving in a circular path if the
centripetal force is removed?
Is there an
outward,
v
centrifugal
force?
v
ac
ac
F
Fc
r
Path of inertia
w
Doesn’t happen.
There is NO centrifugal force.
It is fake, fictitious.
r
w
This happens. Without the
centripetal force, ball continues in a
straight line with same speed it had
before string was cut.
Centripetal Force
Inwardly directed resultant force which causes a
body to turn; perpendicular to velocity. It is just
a fancy name for the net force in the radial
direction. Since it is a NET force, it is not drawn on
a force diagram.
Centripetal force always arises from the regular
forces we are familiar with such as gravity, friction,
tension, normal, electromagnetic.
It is not a new kind of force.
In our daily lives we come across many types of
curved or circular motions. Centripetal force is
necessary for any of these motions.
Car rounding a flat-curve
Car rounding a
banked-curve
Toy airplane on
a rope
FT
FTy
FTr
Fg
DEMO
What is the radius at which the coin will slide off the rotating
platform? Max coefficient of static friction between the
platform and coin is 0.36.
FN
fs
Fg
r
mv2
FC   Fr  mac 
r
Example: Force on a revolving ball (horizontal)
Estimate the force a person must exert on a string
attached to a 0.15 kg ball to make it revolve in a
horizontal circle of radius 0.600m. The ball makes 2
revolutions per second.
First: draw FBD and
FT
FG
Very small, so
neglect for now
resolve forces into
radial and non radial
components
Second: apply Newton's second
law. Set net force in radial
direction equal to mac.
Fc   Fr  mac
2
2
2
mv
m4 r (0.15)4 (0.6)
F T


 14 N
2
2
r
T
(1 / 2)
Approach for Analyzing the Circular Dynamics
1. Draw a picture. Position the object at a place in the circle
where all the forces acting on it can be visualized and drawn .
2. Select a coordinate system to analyze your system with one
axis in the radial direction (along the direction of acceleration)
and the other perpendicular to the radial direction (nonradial).
3. Identify all forces acting on the system by drawing a free body
diagram. Your FBD should be large and clear
4. Resolve force vectors into radial and non-radial
(perpendicular) components.
5. Apply Newton’s 2nd law to each axis or direction:
Fc = SFr = mac = mv2/r
and FNR = SFx or y = max or y
Force and acceleration directions for radial axis:
is pointing INTO the circle, is OUT of circle
+
6. Solve for relevant unknowns.
-
Homework #2: A
ferris wheel with a 20m
radius and tangential
speed of 4 m/s has all
70 kg of you riding it.
How big is the normal
force exerted on you at
a) the top b) the
bottom?
FN
FG
FN
FG
mv2
FC  mac 
 56 N
r
Top (FC=Fg-FN)
FC   Fr  ( mv 2 ) / r
mv 2
Fc  FG  FN 
r
FN  mg  ( mv 2 ) / r
FN  686  56  630 N
FN
FG
FN
Bottom (FC=FN-Fg)
Fc   Fr  mv / r
Fc  FN  FG  (mv2 ) / r
FN  mg  (mv2 ) / r
FN  686  56  742 N
2
FG
Top (FC=Fg-FN)
At max speed, FN=0
Fc  Fg 
mv 2max
FG
r
mg  (mv 2max ) / r
g  v 2max / r ( g  ac )
vmax  gr
FN
Bottom (FC=FN-Fg)
Fc  Fg
At top and bottom
Fc  FN  Fg  Fg
FN  2 Fg  2G ' s
FG
Top
FG FN
FN
Bottom
What speed must the roller coaster go for a passenger to
FG
experience 3G’s of force at the top of a 30m radius loop
de loop? At that speed, how many G’s do they
vtop  4 gr  34.3m / s
experience at the bottom?
vbottom  5G
Conical Pendulum: Tetherball. The game of tetherball is
played with a ball tied to a pole with a string. When the ball is
struck, it whirls around the pole. In what direction is the
acceleration of the ball and what force causes the acceleration?
Draw a FBD of the ball and resolve forces into radial and non
radial components
The net force points
horizontally toward the
center of rotation. The
horizontal-component of
the tension force is the
centripetal force and it
causes centripetal
acceleration.
y
FTy
FT
FTx
FG
x
Conical Pendulum: Tetherball. If the rope has a length
of 2.25 m and makes an angle of 21o with the pole, what
is the speed of the ball?
F
r
 Fc  mac
v2
 FTx  m
r
FTy
v2
FTy tan q  m
r
v2
mg tan q  m
r
v  gr sin q  gL(tan q )(sin q )
V = 1.74m/s
w =v/r =2.16 rad/s
FT
FTx
FG
How do you get a car to travel in a circle?
When car rounds a curve, there is centripetal acceleration.
The car must have an inward, radial force on it to move in
a curve. On a flat road, the centripetal force is supplied
by static friction between the tires and the road.
FN
FG
fs
r
If the friction force is not great enough, as in
icy conditions, the car will skid out of a circular
path into a more straight line path.
Skidding on a curve:
A 1000kg car rounds a curve on a
flat road of radius 50 m at a speed of 50 km/hr (14 m/s). The
coefficient of static friction between the tires and the dry road is
ms=0.6. What is the force of static friction between tires and road?
b) When the road is icy, the coefficient is ms=0.25. Will the car
make the same turn, or will it skid in icy conditions?
NO, the car will NOT make the turn
FN
fs
FG
FC   Fr  mac
2
r
fs is just the amount
to provide the
centripetal force. fs
is NOT fs max here.
2
mv
1000(14)
fs 

 3900 N
r
50
f s  m s FN  m s mg
f s  5880 N
Dry road
f s  2450 N
Icy road
Skidding on a curve:
On a dry road (ms=0.6), what is the
maximum speed a truck can travel around a 50m radius curve?
FN
fs
r
FG
FC   Fr  mac
f s max 
vmax
mv 2max
f s  m s FN  m s mg
r
rf s max
rm s mg


 m s rg  17.1m / s
m
m
For banked turns
cars and motorcycles on wall of death
globe of death
wall of death
Banking curves
The banking of curves can reduce the chance of skidding
because a component of the normal force of the road is in
the radial direction and supplies centripetal force. It
reduces the reliance on friction to make the curve.
FNy
q
FN
FNx
F
r
 mac
2
v
FNx  FN sin q  m
r
r
q
FG
For a given banking angle, q,
there will be a design speed for
which no friction is needed
Example: Banking angle
a) For a car traveling with a speed v around a curve of radius r,
find the formula for the angle at which a road should be
banked, q, so that no friction is required. The formula should
be in terms of v, r and q.
FNy
q
FN
FNx
q
FG
F
r
F
 mac
y
2
v
FN sin q  m
r
r
0
FN cos q  mg
mg
FN 
cos q
mg
v2
sin q  m
cos q
r
v2
tan q 
rg
Example: Banking angle
b) What is the angle for an expressway off-ramp curve of
radius 50 m at a design speed of 50 km/hr (14 m/s). In
other words, what is the angle of the ramp for which NO
friction is necessary to make the turn when going 14 m/s?
FNy
q
FN
FNx
q
FG
r
v2
142
tan q 

 0.4
rg (50)(9.8)
q  22
Example: Banking angle
b) What is the angle for a luge curve of radius 50 m at a
design speed of 130 km/hr (36 m/s). In other words,
what is the angle of the curve for which NO friction is
necessary to make the turn when going 36 m/s?
FNy
FN
q
v2
tan q 
rg
36 2

 2.645
(50)(9.8)
FNx
FG
q  69
q
Luge run
Example: Force on a revolving ball (vertical)
A ball on the end of a string is cleverly revolved at a uniform rate in
a vertical circle of radius 85.0 cm. If its speed is 4.15 m/s and its
mass is 0.300 kg, a) what is the centripetal force on the ball? b)
when is the tension minimum? Calculate the tension in the string
when the ball is c) at the top of its path, d) at the bottom of its
path, e) on the side
Fc=Fg+Tt =mv2/r
FG
Fc=Ts
=mv2/r
Tt
FC = 6.08 N
Ts
Tb
FG
FG
Fc=Tb-Fg
=mv2/r
Example: Force on a revolving ball (vertical)
A ball on the end of a string is cleverly revolved at a uniform rate in
a vertical circle of radius 85.0 cm. If its speed is 4.15 m/s and its
mass is 0.300 kg, Calculate the tension in the string when the ball
is b) at the top of its path, c) at the bottom, d) on the side
FC = 6.08 N
Top
Fc=Fg+Tt =mv2/r
FG
Side
Fc=Ts =mv2/r
Ts =6.08N
Ts
Tt
Tb
FG
FG
mv2
TT 
 mg Minimum
FT
r
TT =3.14N
Bottom
Fc=Tb-Fg
=mv2/r
mv2
TB 
 mg Maximum
FT
r
TB =9.02N
Example: Force on a revolving ball (vertical)
A ball on the end of a string is cleverly revolved at a uniform rate in
a vertical circle of radius 85.0 cm. A) if the max tension the string
can hold before breaking is 30N, what is the max speed of the ball?
B) what is the minimum speed the ball can go around at this radius
Top
F
Fc=Fg+Tt =mv2/r
FG
Side
Fc=Ts
FC  Fg 
mv
r
minimum
when
TT=0
vmin  gr
=mv2/r
Ts =6.08N
Tt
2
C
Ts
Tb
FG
FG
Bottom
Fc=Tb-Fg
Maximum FT
=mv2/r
mv 2
TB 
 mg
r
mv 2max
30 
 mg
r
vmax  8.8m / s
DO NOW
What is the minimum coefficient of static friction required to
keep a penny from sliding off a record rotating at 33.3 rpm
when it is placed 15 cm from the center of the record?
F
r
 mac
FN
fs
2
v
fs  m
r
r
FG
m4 r
m s FN  m s mg 
T2
2
(4 )(0.15)
m s (9.8) 
,
2
1.8
2
m s  0.186
Example: Ultracentrifuge
An ultracentrifuge rotor rotates at 50,000 rpm. The bottom of
a 4.00 cm long test tube is 10.0 cm from the rotation axis.
Calculate the centripetal acceleration in “g’s” at the bottom of
the tube. If the contents of the tube have a total mass of
12.0 g, what force must the bottom of the tube withstand?
2r
v
 2rf
T
 2 (0.1)(50000 / 60)  523.3 m / s
v 2 523.32
ac 

 2.74 x 10 6 m / s 2
r
0.1
 2.79 x 105 g
Fc  mac  (0.012)( 2.74 x106 )  32,880 N
(Equivalent to about 3.3 tons)
The Earth spins on its axis and we move with it in circular motion around
the center. (The radius of the Earth at the equator is RE = 6.38 x 106 m )
a) What is the centripetal acceleration at the equator?
b) What is the centripetal force acting on a 55 kg person standing at the
equator preventing that person from drifting off tangentially into space?
c) What provides that required centripetal force to keep us on the earth as
is rotates?
m = 55 kg
FN
FG
ac 
v2
RE

4 2 R E
T2
 0.034m / s 2  0.00347 g
F c mac  55(0.034)  1.86 N
Fc  1.86  FG  FN
FN  FG  1.86  537.1N
(your scale reads 0.34%
less than your true weight)
DO NOW
d) How fast does the Earth have to spin to make us
“weightless” at the equator (find the period)?
Fc  FG ( FN  0)
m = 55 kg
FG
ac  9.8 
v2
RE

4 2 R E
T2
T  5069.6s
 84.5 min  1.41hr
(you would drift off the
equator if the earth day
was <84.5min)
Connection between centripetal acceleration
and gravity
Yes
down
Feel a force? A
push or a pull?
Which way
is gravity?
Feel a force? A
push or a pull?
Yes
Which way
is gravity?
down
(Opposite the
push/pull)
Gravity always is opposite the push or pull
Connection between centripetal acceleration
and gravity
Feel a force? A
push or a pull?
Which way
is gravity?
He experiences perceived gravity due to
the centripetal acceleration and force. Its
direction is constantly changing.
Connection between centripetal acceleration and gravity
Space station in outer space where there is no gravity.
Can create gravity with centripetal force.
w
Feel a force? A
push or a pull?
Yes
Which way
is gravity?
down
http://www.youtube.com/watch?v=sk0ZahSHKsQ
Connection between centripetal acceleration and gravity
How fast do we need to rotate the space station to mimic
gravity? Lets say the space station has 500 m radius.
w
Yes
down
2
v
ac  9.8 m / s 2 
r
v  70 m / s
2r
T
 45 s
v
What about the center?
Connection between centripetal acceleration and gravity
Now swing vertically. Need a
certain ac to go around in a circle.
If that ac is exactly 9.8 m/s2, where
does the centripetal force come
from? does the string need to pull?
w
FG
FT
No, gravity can provide all
the centripetal force at the
top in this case.
Feel a force? A
push or a pull?
Which way is
perceived
gravity?
Radially
out
Newton’s Universal Law of Gravitation
Centripetal Acceleration of the Planets
Distance
from Sun
( x 109 m)
Period
(yr)
ac
(10-3 m/s2)
Mercury
57.9
0.241
39.48
Venus
108.2
0.615
11.33
Earth
149.6
1
5.92
Mars
227.9
1.881
2.55
Jupiter
778.3
11.86
0.219
Saturn
1427
29.46
0.065
Uranus
2871
84.07
0.016
Neptune
4497
164.81
0.0066
Pluto
5913
247.7
0.0038
4 2 r
ac  2
T
Centripetal Acceleration of the Planets
ac (x 10-3 m/s2)
100
ac = 131963(1/r1.999)
R² = 1
10
1
ac  2
r
1
0.1
0.01
0.001
10
100
1,000
10,000
r, Distance from the sun (109 m)
Newton’s Universal Law of Gravitation
m2
r
m1
Inverse
Square
Law

Fg ,m1,m 2

Fg ,m 2,m1


Gm1m2
FG ,m1 ,m2 


F
G , m2 , m1
2
r
(G = 6.6742 x 10-11 N m2/kg2)
Example: What is the magnitude of the
gravitational force between the Earth and the Sun?
r = 1.5 x 1011 m,
ME = 5.98 x 1024 kg,
MS = 2.0 x 1030 kg
FG
FG2=1/3FG
GM S M E
22
FG 
 3.6 x 10 N
2
r
Example: Spacecraft at 2 rE. On the surface of
the earth, a spacecraft weighs 20,000N. What is the
spacecraft’s weight (force of gravity acting on it)
when it orbits 2 earth radii from the earth’s center

FG  14 mg surface  5000 N
Gravitational Field Strength above the surface
Consider an object of mass m, at or near the surface of the
earth (a distance r=rE from the Earth’s center), for example an
apple. It is pulled by the force of gravity. According to
Newton’s 2nd law, the apple experiences an acceleration mg
due to the net force FG (in the absence of other forces).
r
GM Em
FG 
m
mg
2
r
FG GM E At Earths
g
 2
Surface, r=R
m
r
g = 9.8 N/kg
E
rE = 6.36 x 106 m
Gravitational field strength, g,
decreases with r2. In this example,
the field is created by the Earth.
Gravitational field strength at or below surface
Below the surface, gravitational field strength increases
with r and is proportional to the planet’s density, 
( = M/V).
m
M
FG
r
M(r)
GM (r )m
FG 
m
mg
2
r
Below surface, m is attracted to
center by M(r) which changes with
r, but the density is constant.
FG GM (r ) GV
g

 2
2
m
r
r
g  43 Gr
AT or Below surface, gravitational field strength, g,
increases with r and is proportional to density
Gravitational Field Strength of a sphere of
mass M and radius R
At or Below the
surface,
gravitational field
strength is
proportional to
Surface gravity GM
R2
Above the surface, gravitational
field strength is proportional to M
and decreases with r2

and increases
with r
FG
g
m
 43 Gr
g = ¼gsurface
FG GM
g
 2
m
r
g = 1/9 gsurface
R
2R
3R
Distance from center of sphere (m)
Gravitational Field
Find the surface gravitational field strength, gJ, of
Jupiter. Jupiter’s mass is 317.8 times that of the
Earth and its radius is about 11 times bigger than the
Earths’.
Gravitational
field strength
FG GM J
gJ 

2
m
rJ
317.8
2
gJ 
g E  2.626 g E  25.74m / s
2
11
value of g on other planets
Orbital Motion
In all orbital motion problems, the force of gravity is
the centripetal force
2r
v
T
v
r
Central body
mass = M
orbiting body
mass = m
FG
FG  FC
GMm mv

2
r
r
GM
2
v
r
2
Notice that the orbital motion
does not depend on the mass of
the orbiting body

Satellite in Circular Orbit
FC  Fr  mac
Calculate the velocity necessary to
2
F

(
mv
)/r
G
maintain a stable, circular orbit.
2
FG
GMm mv

2
r
r
GM
2
v
r
GM
v
 gr
r
Speed of a satellite
orbiting M
There is only one speed a
satellite can have if it is to
remain in a stable circular
orbit with a fixed radius
Geosynchronous satellite. A geosynchronous
satellite is one that stays above the same point on the
equator of the Earth. Such satellites are used for cable TV
transmission, weather forecasting and communication
relays. For a geosynchronous sat, determine
a) the height above the Earth’s surface b) orbital speed
2
mv
T = 1 day = 86,400 s
FC 
r
Orbital Radius = Distance
2
GM E m mv
from Earth’s center:

r2
r
r = 4.23 x 107 m
2
GM E GM ET
Height of satellite above Earth:
r 2 
v
4 2 r 2
h = r – rE = 36,000 km
2
GM
T
(6 rE)
3
E
r 
v = 2r/T
2
4
Example: Geosynchronous satellite. b) satellite’s
speed.
Distance of satellite from Earth’s center:
r = 4.23 x 107 m
2r
v
T
GM E
or v 
r
v = 3070 m/s
(6900 mph)
Example: The Sun’s Mass. Determine the mass of
the Sun given the Earth’s distance from the Sun is
rES=1.5 x 1011 m.
Fc   F r  mac
FG
FG  mac
r
GM S mE mE v

2
r
r
2
v = 2r/T
v r 4 r
MS 

2
G
GTE
2
(T E= 1 yr = 3.16 x 107 s)
2 3
M S  2.0 x 10 kg
30
Gravity holds many
objects together in orbits.
Moons…
… more
moons,
planets
….
satellites…
… and astronauts
and their space
shuttles.
But wait…
Why do astronauts in orbit seem to float?
Have they escaped the Earth’s gravity?
Example: Gravity in orbit. Calculate the force of
gravity on a 50 kg astronaut standing on Earth. Then
calculate the force of gravity on the astronaut in orbit.
The shuttle orbits the Earth at an orbital radius of
about 6.68 x 106 m.
r=6.68 x 106 m
ME=5.98 x 1024 kg
GM E m
FG 
2
r
G=6.67 x 10-11 N m2/kg2
Astronaut in Room S17
Astronaut in orbit:
FG =mg= 490 N
FG = 447N
According to your results, does the
floating make sense? Do the astronauts
feel a significantly different force of
gravity when in orbit? What’s going on?
Satellites and “Weightlessness”
(8333 m/s)
(7500 m/s)
(11,111 m/s)
KEPLER’S LAWS
Johannes Kepler (1571-1630)
-German astronomer
-Wrote DETAILED description of the motions of the
planets based on work of Tycho Brahe (1546-1601)
- Kepler’s work and his Laws of Planetary Motion were
indispencable to Newton.
KEPLER’S 3 LAWS OF PLANETARY MOTION
Keplers 1st Law: The path
of each planet about the Sun
is an ellipse with the Sun at
one focus.
Keplers 2nd Law: Each planet moves so that an
imaginary line drawn from the Sun to the planet
sweeps out equal areas in equal periods of time.
Keplers Laws animation
Elliptical Orbital Motion
SPEEDS UP
FGt
v
FG
FG
FG
FG
v
FGt
SLOWS DOWN
orbital motion sim
FGr
FGr
v
FGt The tangential
v
component of gravity
changes the planet’s
speed
FGr The radial component
of gravity
(perpendicular to the
velocity) changes the
planet’s direction
KEPLER’S 3 LAWS OF PLANETARY MOTION
Keplers 3rd Law: The ratio of the squares of the
periods of any 2 planets revolving about the Sun is
equal to the ratio of the cubes of their mean
distances from the Sun. That is,
2
 T1   r1 
    
 T2   r2 
3
r13 r23
 2
OR
2
T1 T2
should be the
same for each
planet.
Orbital Motion
F
r
 m E ac
Newton was able to show that
2
Kepler’s laws could be derived from FG  ( mE v ) / r
the Law of Gravitation and the
GM S mE mE v 2

Laws of Motion. He used Kepler’s
2
r
r
laws as evidence for his Law of
GM S
Universal Gravitation.
 v2
r
FG
Kepler’s
3rd Law
GM S  2r 


r
 T 
T2
4 2

3
r
GM S
2
Orbital Motion
Jupiter’s
Moons
Orbital
Radius
(km)
Io
421,700
Europa
671,034
Ganymede 1,070,412
Callisto
1,882,709
Orbital
Period
(days)
1.77
3.55
7.15
16.69
What is the
mass of Jupiter?
For all the moons orbiting Jupiter
r3
15
3
2

3
.
21
x
10
m
/
s
T2
GM J

4 2
M J  1.89 x10 27 kg
Jupiter’s Orbit
The Earth completes one orbit around the Sun in 1
year and has an orbital radius of 1.5 x 1011 m. If the
orbital radius of Jupiter is 7.78 x 1011 m, what is the
period of Jupiter’s orbit?
3
3
rJ
rE
 2
2
TE
TJ
Keplers
3rd Law
(1.5x10   (7.78x10 
11 3
(1yr 
2
TJ  12 years
11 3
(TJ 
2
The Supermassive Black Hole at the Center of Our
Galaxy: how we know it's there and what its mass is
1995.5
Started collecting data
Find the mass of the Black hole
and express it as a multiple of the
mass of our Sun (2 x 1030 kg)
Orbital
radius
(x 1014)
(m)
Orbital
Period
(yrs)
R3/T2
(x 1024)
(m3/s2)
SO-2
1.13
16.17
5.54
SO-38
1.61
18.3
12.5
SO-10
1.15
11.4
11.7
Star
MBlackHole = 1.6-3.7 x 106Msun
http://www.astro.ucla.edu/~ghezgroup/gc/pictures/orbitsMovie.shtml