Physics 350
Chapter 5
Circular Motion and the Law of Gravity
Centripetal Acceleration



Consider a car moving in a
circle with constant
velocity
Even though the car is
moving with constant
speed, it has an
acceleration
The centripetal
acceleration is due to
the change in the direction
of the velocity
Centripetal Acceleration
Using vectors,
rearrange to
determine the change
in velocity (direction)
 The vector change is
directed towards the
center of motion

Pictorial “Derivation” of Centripetal
Acceleration
a = Dv/Dt
v2
Top view:
a
a
a
a
a
a = v2/r (r is radius of curve)
v1
In uniform circular motion the
acceleration is constant, directed
towards the center. The velocity
has constant magnitude, and is
tangent to the path.
Centripetal Acceleration

The magnitude of the centripetal
acceleration is given by
2
v
ac 
r
◦ This direction is toward the center of the circle
Centripetal Acceleration

An object can have a
centripetal
acceleration only if
some external force
acts on it.
◦ In the case in the
figure, the force is
the tension in the
string
Centripetal Acceleration

For the car moving on
a flat circular track,
the force is the
friction between the
car and the track.
Centripetal Acceleration

Forces that act inward are considered to
be centripetal forces
◦ Examples:
 Tension in example above
 Gravity on a satellite orbiting the Earth
 Force of friction
Centripetal Acceleration

Applying Newton’s Second Law along the
radial direction we can determine the net
centripetal force Fc
Fc = mac = m (v2 / r)
Centripetal Acceleration

If the centripetal force were removed, the object
would leave its circular path and move in a
straight line tangent to the circle
◦ Merry go round
Curves, Centrifugal, Centripetal
Forces

Going around a curve smushes you against
window
◦ Understand this as inertia: you want to go
your body wants to
straight
keep going straight
but the car is accelerating
towards the center of the curve
Car acceleration is v2/r
 you think you’re being accelerated by v2/r relative to the car
Centripetal, Centrifugal Forces,
continued
The car is accelerated toward the center of the
curve by a centripetal (center seeking) force
 In your reference frame of the car, you experience
a “fake”, or fictitious centrifugal “force”

◦ Not a real force, just inertia relative to car’s acceleration
Centripetal Force
on car
velocity of car
(and the way you’d rather go)
Centripetal Forces

Fictitious Force - Centrifugal
 Driving in a car around a
curve feels like you are
applying a force to the car
outward
 Not really a force, the
force one feels is the car
applying a force on you
from the frictional force it
applies to the road
 Inertia keeps our bodies
wanting to move forward,
the car applies a force to
push it inwards
Rotating Drum Ride

Vertical drum rotates, you’re pressed against
wall
◦ Friction force against wall matches gravity
◦ Seem to stick to wall, feel very heavy
The forces real and perceived
Real Forces:
Friction; up
Centripetal; inwards
Gravity (weight); down
Perceived Forces:
Centrifugal; outwards
Gravity (weight); down
Perceived weight; down and out
Centripetal Acceleration

Gravitron
◦ Accelerating upwards?

Climbing car
Works in vertical direction too…

Roller coaster loops:
◦ Loop accelerates you downward (at top) with
acceleration greater than gravity
◦ You are “pulled” into the floor, train stays on track
 it’s actually the train being pulled into you!
Vertical Circular Motion
Consider the forces acting on a motorcycle
performing a loop-to-loop:
Can you think of other objects that undergo similar motions?
Old-Fashioned Swings




swing ropes:
what you feel
from your seat
gravity (mg)

resultant: centripetal
The angle of the ropes tells
us where the forces are:
Ropes and gravity pull on
swingers
If no vertical motions (level
swing), vertical forces cancel
Only thing left is horizontal
component pointing toward
center: centripetal force
Centripetal force is just mv2/r
(F = ma; a = v2/r)
What about our circular motions
on Earth?
Earth revolves on its axis once per day
 Earth moves in (roughly) a circle about the sun
 What are the accelerations produced by these
motions, and why don’t we feel them?

Earth Rotation
Velocity at equator: 2r / (86,400 sec) = 463
m/s
 v2/r = 0.034 m/s2

◦ ~300 times weaker than gravity, which is 9.8 m/s2
Makes you feel lighter by 0.3% than if not
rotating
 No rotation at north pole  no reduction in g
 If you weigh 150 pounds at north pole, you’ll
weigh 149.5 pounds at the equator

◦ actually, effect is even more pronounced than this (by
another half-pound) owing to stronger gravity at pole:
earth’s oblate shape is the reason for this
ConcepTest 5.1 Tetherball
1) toward the top of the pole
2) toward the ground
3) along the horizontal component of the
tension force
4) along the vertical component of the tension
force
5) tangential to the circle
In the game of tetherball,
the struck ball whirls
around a pole. In what
direction does the net force
on the ball point?
T
W
ConcepTest 5.1 Tetherball
In the game of tetherball,
the struck ball whirls
around a pole. In what
direction does the net force
on the ball point?
1) toward the top of the pole
2) toward the ground
3) along the horizontal component of the
tension force
4) along the vertical component of the tension
force
5) tangential to the circle
The vertical component of the tension
balances the weight. The horizontal
T
component of tension provides the
centripetal force that points toward the
W
T
center of the circle.
W
Centripetal Acceleration
The tangential component of the acceleration is
due to changing speed
 The centripetal component of the acceleration is
due to changing direction
 Total acceleration can be found from these
components

a 
a a
2
t
2
C
Centripetal Acceleration
Example
 A 1,000kg car rounds a curve on a flat
road of radius 50.0m at a speed of
50.0km/hr (14.0m/s). Will the car make
the turn if:
 a) the pavement is dry and the
coefficient of static friction is 0.800?
 b) the pavement is icy and the
coefficient of static friction is 0.200?


(Note: use max static friction here for the
extreme case of the tires almost slipping.)
Centripetal Acceleration

Example:

An engineer wishes to design a curved
exit ramp for a toll road in such a way
that a car will not have to rely on friction
to round the curve without skidding. He
does so by banking the road in such a way
that the force causing the centripetal
acceleration will be supplied by the
component of the normal force toward
the center of the circular path.
a) Show that curve must be banked at
tan θ = v2/rg.
b) Find the angle at which the curve
needs to be banked for a 50m radius and
a speed of 13.4 m/s.


Planetary Motion and Newtonian
Gravitation
Kepler’s Laws
1.
2.
3.
All planets move in elliptical orbits
with the Sun at one of the focal points.
A line drawn from the Sun to any
planet sweeps out equal areas in equal
time intervals.
The square of the orbital period of any
planet is proportional to the cube of
the average distance from the planet
to the Sun.
Kepler’s Laws


Kepler’s First Law
“All planets move in elliptical
orbits with the Sun at one of
the focal points.”
◦ Any object bound to
another by an inverse
square law will move in
an elliptical path
◦ Second focus is empty
Kepler’s Laws
Kepler’s Second Law

“A line drawn from the
Sun to any planet sweeps
out equal areas in equal
time intervals.”

Objects near the Sun will
need to cover more
distance per time
◦ Faster velocity

Kepler’s Laws
Kepler’s Third Law
 “The square of the orbital period of any
planet is proportional to cube of the
average distance from the Sun to the
planet”

◦ Orbital Period = time it takes a planet to
make one full orbit around the sun
◦ For orbit around the Sun, T2/r3 = K = KS =
2.97x10-19 s2/m3
◦ K is independent of the mass of the planet
 Therefore, all planets should have the same K
Kepler’s Laws

Kepler’s Third Law
◦ They do have the same K!
Kepler’s Laws

Planetary Data relative to Earth
Kepler's 3rd Law
T in years, a in astronomical units; then T2 = a3
Planet
Period T
Dist. a fr.
Sun
T2
a3
Mercury
0.241
0.387
0.05808
0.05796
Venus
0.616
0.723
0.37946
0.37793
Earth
1
1
1
1
Mars
1.88
1.524
3.5344
3.5396
Jupiter
11.9
5.203
141.61
140.85
Saturn
29.5
9.539
870.25
867.98
Uranus
84
19.191
7056
7068
Neptune
165
30.071
27225
27192
Pluto
248
39.457
61504
61429
Newtonian Mechanics

Kepler’s Laws described the kinematics of
the motion of the planets but didn’t
answer why the planets move the way
they do
The Universal Law of Gravity
• Any two bodies are attracting each
other through gravitation, with a force
proportional to the product of their
masses and inversely proportional to
the square of their distance:
F= G
m 1m 2
r2
(G is the Universal constant of gravity.)
Newtonian Gravitation

Newton’s Universal Law of Gravitation
m1m2
Fg = G ————
r2
where G is the gravitational constant,
G = 6.673 x 10-11 m3 kg-1 s-2

Example of the inverse-square law
Newtonian Gravitation

Applying Newton’s
third law to two
masses:
 Action-Reaction
Pair
F21 = -F12
“Every pair of particles exerts on one
another a mutual gravitational
force of attraction.”
Newtonian Gravitation

The gravitational force exerted by a
uniform sphere on a particle outside the
sphere is the same as the force exerted if
the entire mass of the sphere were
concentrated at its center

This is called Gauss’s Law.
◦ Applies to electric fields also
Why was the Law of Gravitation not obvious (except to Newton).
How big are gravitational forces between ordinary objects?
| F12 |
G m1m2
2
r12
1 Newton is about the force needed to
support 100 grams of mass on the Earth
About the weight of a small apple
m1
m2
r12
F12
1 kg
a liter of soda
1 kg sandwich
1 meter
6.67x10-11 N.
100 kg
a person
100 kg
another person
1 meter
6.67x10-7 N.
106 kg
a ship
106 kg
another ship
100 meters
0.67 N.
still hard to detect
Conclusion:
• G is very small, so…need huge masses to get perceptible forces
Does gravitation play a role in atomic physics & chemistry?
9.1x10-31 kg
electron
1.7x10-27 kg
proton
5x10-11 meter
orbit radius
4x10-47 N.
Circular Orbits
(a pretty good approximation for all the
planets because the eccentricities are
much less than 1.)
(velocity)
Centripetal acceleration
v2
ar  
r
(acceleration)
Newton’s second law
mv
GMm

r
r2
2
There is a subtle approximation here: we are approximating the center of
mass position by the position of the sun. This is a good approximation.
mv
GMm

2
r
r
2
Circular Orbits
The planetary mass m cancels out.
The speed is then
GM
v
r
Period of revolution
Time = distance / speed
i.e., Period = circumference / speed
2 r
r
4 r
2
T 
 2 r
, or T 
v
GM
GM
2
 Kepler’s third law: T 2  r 3
3
Generalization to elliptical orbits
(and the true center of mass!)
2 3
4 a
T 
G( M  m)
2
2 3
4 a

GM
where a is the semi-major axis of the ellipse
The calculation of elliptical orbits is difficult
mathematics.
Finding the Value of G





Henry Cavendish first
measured G directly (1798)
Two masses m are fixed at
the ends of a light horizontal
rod (torsion pendulum)
Two large masses M were
placed near the small ones
The angle of rotation was
measured
Results were fitted into
Newton’s Law G=6.67x10-11 N.m2/kg2
G versus g:
• G is the universal gravitational constant, the same everywhere
• g = ag is the acceleration due to gravity. It varies by location.
• g = 9.80 m/s2 at the surface of the Earth
Superposition:
The net force on a point mass when there are many others
nearby is the vector sum of the forces taken one pair at a time

Fon 1 




 Fi,1  F2,1  F3,1  F4,1
i1

F21
All gravitational effects are between
pairs of masses. No known effects
depend directly on 3 or more masses.
Example:
m2
m1

r'
m5

F31
m2 = m3

r'

r
m1

F41
m4

r
m2

r12

r13
m3

r14
m4
m4 = m5
m3

Fon
m1

 m1 gat
gi,1
 0 by symmetry
Gm
 agi,1  - 2 i
r1,i
m1
Newtonian Gravitation

Acceleration due to gravity
◦ Determined experimentally
◦ g value varies with altitude
ag = GME / r2
Altitude
(km)
ag (m/s2)
Altitude Example
0
9.83
Mean Earth Surface
8.8
9.80
Mt. Everest
36.6
9.71
Highest manned
balloon
400
8.70
Space shuttle orbit
35,700
0.225
Comm. Satellite
Gravitational “field” transmits the force
 A piece of mass m1 placed somewhere creates a “gravitational field” that
has values described by some function g1(r) everywhere in space.
 Another piece of mass m2 feels a force proportional to g1(r) and in the same
direction, also proportional to m2.

G m1m2

F12  m2g1 (r )  
r̂12
2
r12
Concepts for g-fields:
◦ No contact needed: “action at a distance”.
◦ Acceleration Field created by gravitational mass transmits the force as a distortion of
space that another (inertial) mass responds to.
 The field g1(r) is the gravitational force per unit mass
created by mass m1, present at all points whether or not
there is a test mass m2 located there
 The gravitational field vectors point in the direction of
the acceleration a particle would experience if placed in
the field at each point.
Field lines help to visualize strength and direction.

close together  strong field,

direction  force on test mass
g
Fg
m

GM
rˆ
2
r
Mass of the Earth
We can use the equation:
g = G*Mearth/Rearth2
to solve for Mearth since we know
◦ g = 9.8 m/s2 (from our lab experiment),
◦ G = 6.67 x 10-11 Nt-m2/kg2 (from precise
gravity force experiments), and
◦ Rearth = 6,400 km (since we know the
circumference of the earth = 25,000 miles).
Mass of the Earth
g = G*Mearth/Rearth2 or
Mearth = g*Rearth2/G
= 9.8 m/s2 * (6.4 x 106 m)2 / 6.67 x 10-11 Nt-m2/kg2
= 6.0 x 1024 kg .
Actual value = 5.9742 x 1024kg
This value is certainly large as we expect the
mass of the earth to be large.
ConcepTest Fly Me Away
You weigh yourself on a scale inside an
airplane that is flying with constant speed
at an altitude of 20,000 feet. How does your
measured weight in the airplane compare
with your weight as measured on the surface
of the Earth?
1) greater than
2) less than
3) same
ConcepTest Fly Me Away
You weigh yourself on a scale inside an
airplane that is flying with constant speed
at an altitude of 20,000 feet. How does your
measured weight in the airplane compare
with your weight as measured on the surface
1) greater than
2) less than
3) same
of the Earth?
At a high altitude, you are farther away from the center of
Earth. Therefore, the gravitational force in the airplane
will be less than the force that you would experience on the
surface of the Earth.
Kepler’s Laws

Example: