Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09
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Gravitation and
Circular Motion
Physics Rapid Learning Series
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www.RapidLearningCenter.com/
© Rapid Learning Inc. All rights reserved.
Wayne Huang, Ph.D.
Keith Duda, M.Ed.
Peddi Prasad, Ph.D.
Gary Zhou, Ph.D.
Michelle Wedemeyer, Ph.D.
Sarah Hedges,Ph.D.
© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 1

Learning Objectives
By completing this tutorial, you will:
„ Understand the nature of
the gravitational force
force.
„ Calculate the gravitational
force between objects.
„ Describe uniform circular
motion.
„ C l l t t i t l f
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Calculate centripetal force
and centripetal
acceleration.
„ Describe simulated
gravity situations.
Concept Map
Physics
Studies
Previous content
New content
Motion
Caused by
Motion Forces
F
Circular
M ti
Gravitation
Described by
Universal
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Gravitational
Constant
Centripetal
Force

Gravitation
Previously, we learned that gravity
accelerates falling objects at -9 8m/s2
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9.8m/s2.
Now we will learn more about the origin
of that acceleration.
Direction of Gravitation
Isaac Newton described this attractive force graivty
that acts between all pieces of matter in the universe.
Gravity is always attractive.
There is no “repulsive” gravity. So far, antigravity
is just science fiction.
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Law of Universal Gravitation
What factors do you think the gravitational
attraction of two bodies would depend on?
mass of objects, and distance between objects
F m1m2
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1 2
2
→
g d
Universal Gravitational Constant
The previous relationship only describes the
factors that influence gravity. To get a
numerically correct answer, with units, you need
a constant included in the equation.
This constant is called the universal gravitational
constant:
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G = 6.67 x 10 -11 N m2 /kg2

Universal Law of Gravitation
Universal
gravitational
constant
Two
masses,
kg
F = Gm m
1 2
2
g d
Force
from
Distance
between
bj t
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gravity, N
objects,
m
Because there is a quantity squared in the
denominator of the fraction, this formula may be
referred to as an inverse square law.
Gravitation Calculation Example
Calculate the gravitation force between the planet
Mars, 6.4 x 10 23 kg, and the sun, 2 x 10 30 kg. Assume
a distance of 2.0 x 10 11 m.
Fg
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Gravitation Example Solution
F = Gm m
1 2
2
g d
Substitute values, do math carefully!
(
23 30
) 6.4x10 kg)(2x10 kg)
(6.67x10 Nm
kg
F
−
=
g (2x10 m)
11 2
2
2
11
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Fg= 2.13 x 1021 N
Notice how the units cancel leaving the correct
force unit of Newtons.
Gravitational Fields
Field lines show which way an object will move
when exposed to some force.
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Thus, gravitational field lines on earth always
point to the center of the earth.

Thought Question
Imagine you are deep inside the earth in a cave.
Would you weigh more, less, or the same as on
the surface of the earth?
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Thought Question Answer
When on the surface of the earth, all of the mass
of the earth is pulling you down.
When underground, the mass above you actually
pulls you up, countering some of pull from the bulk
of the planet. You weigh less!
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Extension of Thought Question
What would happen if you were at a cave at the
center of the earth?
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Because the gravitational pull from the matter
equally all around you would cancel out, you
would be weightless! Fnet would be 0.
Planet Discovery
Gravity can have much more subtle effects.
The existence of Pluto and Neptune was predicted
before they were ever observed.
Their slight gravitational effects wobble the orbits of
the other planets, betraying their presence before
they were ever seen visually.
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Uniform
Circular
Motion
Obviously, not all objects move in a
linear path Another common occurrence
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path. is constant motion in a circular path.
Thought Question
Imagine you are swinging a ball on the end of a
string. The string suddenly breaks when the ball
is at the top.
Snap!
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Which direction will the ball initially fly?

Inertia in Action
When the string breaks, the ball will continue to
move tangent to the circle. Linear inertia at
work!
Snap!
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Obviously, if we were considering gravity, the ball
would begin to fall also.
Circular Motion
In a line, we usually measure a speed or velocity
in m/s, linear speed.
However, when things travel in a circle we could
describe a speed differently, rotational speed.
Example: 3 revolutions per second
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5 revolutions per minute (rpm)
2π radians per second

Linear Speed Example
If a tire on a car has a radius of .29m, and is
being rotated at 830 rpm, what is the speed at
the outer edge of the tire?
830 l ti
.29m
revolutions
per minute
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Linear Speed Solution
Use a previous formula:
v = d
t
However, the distance is now in a circle. So we
must find the distance around the circle…
v = 2 π r
t
Circumference of a
circle
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We also need to consider that it covers 830
circumferences in 1 minute
25m/s
v = 2π (.29m) 830 =
60 sec
830 rpm

Acceleration?
If an object is moving at a constant speed in a
circular path, is it accelerating?
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YES. Although its speed isn’t changing, it still
accelerates. Its direction is constantly changing,
this means its velocity is constantly changing, thus
it is accelerating.
Centripetal Acceleration
We can describe this type of acceleration in a
circular path by the following relationship:
Linear
Li
velocity,
m/s
a v
2
r
c =
radius of
centripetal
i l
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circular
path, m
acceleration
m/s2

Accelerations Require Forces
If an object is accelerating, it must have a net force
applied to it!
This is the force that is responsible for making an
object travel in the circular path instead of a
regular straight line.
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Instantaneous
velocity
Centripetal
force
Centripetal Force
Centripetal force is a “center seeking” force. It
always points towards the center for an object
moving in a circular path.
Centripetal force
Tangential or
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g
instantaneous
velocity
This isn’t some new force. Centripetal force can be
provided by tension from a string, gravitational pull,
friction on the road, etc.

Centrifugal Force?
Often, it seems like there is a force pushing
outward (not to the center).
There is no force like this. However, this apparent
force has been given a name: centrifugal force.
This is an often misunderstood and misused
term
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term.
Direction of Forces
Imagine you are driving along in a car,
you make a hard right turn.
You feel as if you are pushed into the left
side of your car. Actually, it is the left
side of the car that is pushing into you!
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Nothing is pushing you out (centrifugal).
You are being pulled inward (centripetal).

Fc Formula
Linear
velocity,
m/s
Mass, kg
2
F = mv
c r
Centripetal
Radius of
i l
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force, N
circular
path, m
Here is the formula for centripetal force.
Similarities
Notice how the formula for centripetal force
contains the formula for centripetal acceleration
within itself.
Fnet = Fc
2
r
ma = mv
Newton’s
2nd law:
Fnet=ma
Mass
cancels
out
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a v
2
r
c =

Centripetal Force Example
Imagine you whirl a 0.1 kg rubber stopper attached to
a string in a 1 m radius circle at 2 rev/sec. What is
the centripetal force on that stopper?
1m
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Example Solution
First change 2 rev/sec into m/s:
=2 (2πr) /sec Circumference
= 2(2 π 1m) / sec of f circle i l =2πr
2
= 12.6 m/s
Linear
velocity,
m/s
Then, substitute into the Fc formula:
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Fc = .1kg (12.6 m/s)2 / 1m
Fc = 15.8 kg m/s2
=15.8 N

Centripetal Acceleration Example
Imagine that a fighter pilot is traveling at 1000 km/hr,
when he decides to make a sharp turn. He can
withstand a maximum acceleration of 8g’s (8 times
the normal acceleration from gravity). What is the
smallest radius turn he can withstand?
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Centripetal Acceleration Solution
First, convert km/hr into m/s
277m/s
1000km× × 1hr
=
3600sec
1000m
1km
hr
Next, convert the acceleration he can withstand
2
2
78m/s
8g's× 9.8m/s =
1g
Finally, solve for the radius of the turn
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a v
2
r
c =
983m
2
r v 2
(277m/s)
= = =
78m/s
a
2
c

Banked Turns for a Car
Similarly, when a car is driving in a circular path,
on a banked road, the centripetal force comes
from a component of the normal force.
normal force θ
This component of
the normal force is Fc
weight
i ht
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θ
These can be calculated using simple trigonometry.
Example
At the Daytona 500 speedway, the turns in the
oval track have a maximum radius of 316 m and
are banked at an angle of 31°. Assuming no
friction, how fast could the cars make the turn?
Hint: use the previous diagram/example.
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Banked Turn Diagram
Use the right triangle to find the Fc:
tanθ = opp
adj
tanθ = Fc
W
tanθ = Fc
Fc
normal force
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31o
weight
mg
Fc mg tanθ =
Banked Turn Calculation
Next, substitute our expression for Fc back into the
formula to find v:
mv2
r
F mv
2
c =
mg tanθ mv
r
2
=
r g t tanθ θ = v2
2
Since the
mass
cancels
out it is
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out, irrelevant
v r g tanθ
= 2 =
316m(9.8m/s )(tan31 ) 43m/s
=
D

Reality Check
This number represents how fast the car could go
without any friction. Obviously there is friction so
the actual maximum speed on the track is much
higher.
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Simulated
Gravity
When objects are moving in a circle, the
centripetal force applied may mimic the
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usual gravitational force.

Orbits
When in a circular orbit, an object is continually
falling ( under the influence of the earth’s gravity).
However, it is continuing to move tangent to the earth,
so it continues in a circular path at a constant speed.
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Gravity Changes Direction
Fc
gravity
it
Tangential
velocity
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Notice that gravity does not pull the satellite
forward or backward. Gravity simply acts as the
centripetal force to keep it going in a circular orbit.

Gravity Equals Centripetal Force
Since the centripetal force is provided by gravity, we
can equate the two forces:
FG = FFc
2
GmM =
GM 2
mv
2 r
r
Notice the mass of
the satellite cancels
out!
The speed of an
i l biti
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v2
G =
r
v = GM
r
circular orbiting
satellite depends only
on the radius,
gravitational constant
and mass of the earth!
Correct Distance Value
The radius used in the previous equation is
measured from the center of the orbit ( center of the
earth).
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Don’t just plug in the distance above the surface
of the earth!

Importance of Mass
This means that satellites of any mass will have the
same orbital speed for any particular radius.
A giant satellite will have the same speed as a tiny
satellite in the same orbit.
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However, it can be much more difficult to get that
large satellite into orbit in the first place.
Energy Costs
It turns out that it takes 62,000,000 J of energy to
put 1kg outside of the Earth’s orbit. (62 MJ)
This is a large amount of energy, which is why it
is so costly and difficult to put people and
objects into space!
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Weight and Weightlessness
You can feel weightless
even though gravity is
acting on you.
Astronauts in free fall are
still being pulled around
the earth by gravity.
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Artificial Gravity
When astronauts live for long time periods in
space, it impacts their bodies. Bones may
weaken, muscles may lose mass, etc.
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In the future, humans may design space ships that
create “artificial” gravity.

Rotating Space Habitats
We can’t create a gravitational force, but we can use
centripetal force to act like gravity.
If a round space ship is large enough, and spins at
the correct rate, the centripetal force would simulate
gravity.
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Rotating Spaceship Example
If we want to simulate regular earth gravity in a
circular spaceship of radius 15m, how fast must it
rotate? Give m/s and rpm.
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Calculation of Linear Speed
Normal
a v
2
r
c =
Centripetal
force
formula
acceleration
from gravity
on Earth.
v acr =
v = (9.8m/s2 )(15m)
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v = 12.1m/s
This gives the linear speed (in m/s) at the edge.
Calculation of Rotational Speed
Next, consider the number of revolutions it makes
in 1 minute ( 60 sec).
d
Circumference
C cu e e ce
of a circle
Linear
speed
previously
found
t
v = v = 2 π r Xrevolutions
60sec
12 1m/s(60s)
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Xrevolutions
12.1m/=
2 π r
revolutions X = 7.7 rpm

Learning Summary
F = Gm m
1 2
d2
2
g d
Rotational
speed refers
to motion in a
In many
instances, the
G = 6.67 x 10 -11
N m2 /kg2
mass variable
cancels out.
2
Centripetal force
circular path,
rev/sec.
,
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F mv
r
c
=
a v
2
r
c
=
points inward and
forces objects to
maintain a circular
motion.
Congratulations
You have successfully completed
the core tutorial
Gravitation and Circular
Motion

Chemistry :: Biology :: Physics :: Math
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Circular Motion & Gravitation

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