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CF-1
Central Forces: Circular Motion and Gravitation
Circular motion: object moving in a circle of radius r, with
constant speed v.
r
T = period = time for 1 complete revolution, 1 cycle
v
( Don't confuse tension T with period T.)
speed v  v 
2 r
distance

time
T
An object moving in a circle is accelerating, because its velocity is changing, -- changing
direction. Recall the definition of acceleration:
a 
v
v  v1
, velocity v can change is two ways:
 2
t
t
Magnitude can change or
direction can change:
v
v
v1
v2
v2
v1
For circular motion with constant speed, one can show that
v2
1) the magnitude of the acceleration is a  a 
r
2) the direction of the acceleration is always towards the center of
motion. This is centripetal acceleration. "centripetal" = "toward center"
Notice that the direction of acceleration vector is always changing,
therefore this is not a case of constant acceleration (can't use the
"constant acceleration formulas")
r
v
a
a
v
Is claim (1) sensible?
 v2 
Check units:   
 r 
m s
m
2

 m2 
 s2 


m

m
s2

a 
Yep.
Think: to get a big a, we must have a rapidly changing velocity. Here, we need to rapidly change
the direction of vector v  need to get around circle quickly  need either large speed v or a
small radius r.  a = v2 / r makes sense. (Proof is given in the Appendix.)
Is claim (2) sensible?
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Observe that vector v is toward center of circle.
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CF-2
v1
v2
Direction of a = direction toward which
velocity is changing
v1
v
v2
Example: acceleration on a merry-go-round. Radius r = 5 m , period T = 3 s
v
2 r
T

2 (5)
 10.5 m/s
3
r
v
a
v2
10.5 
a =

r
5
2
 22 m/s 2  2.3 g's !
A human can withstand ~ 5 g's for a few minutes or ~10 g's for a few seconds without losing
consciousness.
Forces and circular motion
NII: Fnet  m a  To make something accelerate, we need a force in the same direction as
the acceleration.  Centripetal acceleration is always caused by centripetal force (force toward
center).
Example: Rock twirled on a string. (Assume no gravity)
m = 0.1 kg , T (period) = 1 s , radius r = 1 m
r
FT
What is tension FT in the string?
FT
FT is the only force acting. (No such thing as
"centrifugal force"!)
2 r
v2
FT  m a  m
, v
, 
r
T
( 2 r / T) 2
r
1
FT  m
 42 m 2  42 ( 0.1) 2  3.9 N  1 pound
r
T
1
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CF-3
What about the outward "centrifugal force"?
A person on a merry-go-round (or twirled on a rope by a giant) "feels" an outward force. This is
an illusion! There is no outward force on the person. Our intuition is failing us. Our intuition
about forces was developed over a lifetime of experiences in inertial (non-accelerating) reference
frames. If we are suddenly placed in an accelerating reference frame, our brains (wrongly)
interpret our sense impressions as if we were still in a non-accelerating frame.
The result is that the direction of the perceived force is exactly opposite the direction of the true
force. Example: A person in car accelerating forward. The chair pushes the driver forward. The
force on the driver is in the forward direction. But the driver "feels" herself pressed back into the
seat. It seems there is some force pushing the driver backward. WRONG!
a
"Centrifugal force" (not to be confused with "centripetal force") is also called a "pseudo-force"
or "fictitious force" . Newton's Laws are only valid in a non-accelerating reference frame (an
inertial frame). If we try to analyze motion in a non-inertial frame (for instance, in a rotating
frame) then Newton's Laws don't hold. However, we can pretend that Newton's Laws hold in an
accelerating frame if we pretend that "pseudo-forces" exist. That is, we can get the right answer
if we makes two mistakes. In my opinion, this is a Devil's bargain. Computational convenience
has come at the price of endless confusion of millions of physics students (and many
professional engineers!). My advice: If you have choice, NEVER do calculations in non-inertial
frames. Avoid using fictitious forces.
Consider the rock on the string again (still no gravity). If the string breaks, then there is no
longer any force on the rock and it will move in a straight line with constant velocity. [according
to NI: if Fnet = 0, then v = constant]. The reason the rock does not move in a straight line is
because the string keeps pulling it inward, turning it away from its straight-line path. There is
NO outward force on the rock.
v
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CF-4
Example: Rotation with friction. A car rounds a curve on a flat road (not banked). The radius
of the circular curve is r = 100 m, and the speed of the car is v = 30 m/s (  68 mph). How large
a static friction coefficient (S, not K !) is needed for the car to not skid off the road?
r
N
v
Ffric
a
a
mg
Ffric
View from rear of car
Top view
Fnet = Ffric = m a
N = mg
S 


S N = m v2 / r
S m g = m v2 / r
(car about to skid  Ffric = S N )
(m's cancel) 
v2
(30m/s)2

 0.92 (no units)
gr
(9.8m/s 2 )(100m)
So, need S  0.92 or else car will skid. (For rubber on dry asphalt S  1.0 , for rubber on wet
asphalt S  0.7. So car will skid if road wet.)
Gravity!
Newton's Universal Law of Gravitation (first stated by Newton): any two masses m1 and m2
exert an attractive gravitational force on each other according to
F  G
m1 F
m1 m 2
r2
F
m2
r
G = universal constant of gravitation = 6.67  10–11 N m2 / kg2 (G is very small, so it is very
difficult to measure!)
Don't confuse G with g.
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CF-5
Newton showed that the force of gravity must act according to this rule in order to produce the
observed motions of the planets around the sun, of the moon around the earth, and of projectiles
near the earth. He then had the great insight to realize that this same force acts between all
masses. [That gravity acts between all masses, even small ones, was experimentally verified in
1798 by Cavendish.]
Newton couldn't say why gravity acted this way, only how. Einstein (1915) General Theory of
Relativity, explained why gravity acted like this.
Example: Force of attraction between two humans. 2 people with masses m1  m2  70 kg a
distance r = 1 m apart.
F  G
m1 m2
r2

(6.67  1011)( 70)2
 3.3  107 N
2
1
This is a very tiny force! It is the weight of a 3.4 10–5 gram mass. A hair weighs 210–3 grams
– the force of gravity between two people talking is about 1/60 the weight of a single hair.
Computation of g
Important fact about the gravitational force from spherical masses: a spherical body exerts a
gravitational force on surrounding bodies that is the same as if all the sphere's mass were
concentrated at its center. This is difficult to prove (Newton worried about this for 20 years.)
test mass m
Fgrav
sphere, mass M
test mass m
Fgrav (same as with sphere)
point mass M
We can now compute the acceleration of gravity g ! (Before, g was experimentally determined,
and it was a mystery why g was the same for all masses.)
Fgrav = m a = m g
mass m,
dropped near
surface
Earth
RE
mass ME
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ME m
 mg
R E2
(since r = RE is distance from m to center of Earth)
G
m's cancel ! 
g 
G ME
R E2
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CF-6
Newton's Theory explains why all objects near the Earth's surface fall with the same
acceleration. Fnet = ma says that a bigger force is required to accelerate a bigger mass m. Fgrav =
GMm/r2 says that a bigger mass m feels a bigger force. So near the earth, bigger masses
experience a bigger force in a way that produces the same acceleration for all masses. Newton's
theory also makes a quantitative prediction for the value of g, which is correct.
Example: g on Planet X. Planet X has the same mass as earth (MX = ME) but has ½ the radius
(RX = 0.5 RE). What is gx , the acceleration of gravity on planet X?
Planet X is denser than earth, so expect gx larger than g.
gx 
G MX
RX
2

G ME
R 2 
2

E
G ME
1
 4 g . Don't need values of G, ME, and RE!
2
1/ 2 R E 2
g of earth
4
Or set up a ratio:
gx

gE
 G M X 

2 
 RX 
 G M E 

2 
 RE 

MX
ME
 RE 


 RX 
2
 1  22  4 ,
gX  4 gE
_________________ * __________________
At height h above the surface of the earth, g is less, since we are further from the surface, further
from the earth's center.
r = RE + h 
h
G ME
G ME
g 

r2
( R E  h) 2
earth
The space shuttle orbits earth at an altitude of about 200 mi  1.6 km/mi  320 km. Earth's
radius RE = 6380 km. So the space shuttle is only about 5% further from the earth's center than
we are. If r is 5% larger, then r2 is about 10% larger, and
Fgrav ( on mass m in shuttle)  G
ME m
( R E  h) 2
 about 10% less than on earth's surface
Astronauts on the shuttle experience almost the same Fgrav as when on earth. So why do we say
the astronauts are weightless??
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CF-7
"Weightless" does not mean "no weight".
"Weightless" means "freefall" means the only force acting is gravity.
If you fall down an airless elevator shaft, you will feel exactly like the astronauts. You will be
weightless, you will be in free-fall.
v
astronaut
An astronaut falls toward
the earth, as she moves
forward, just as a bullet
fired horizontally from a
gun falls toward earth.
N
Fgrav
Fgrav
Earth
Orbits
Consider a planet like Earth, but with no air. Fire projectiles horizontally from a mountain top,
with faster and faster initial speeds.
would go straight,
if no gravity
Planet
The orbit of a satellite around the
earth, or of a planet around the sun
obeys Kepler's 3 Laws.
Kepler, German (1571-1630). Before
Newton. Using observational data
from Danish astronomer Tycho Brahe
("Bra-hay"), Kepler discovered that
the orbits of the planets obey 3 rules.
orbits!
Planet
KI : A planet's orbit is an ellipse with the Sun at one focus.
KII : A line drawn from planet P to sun S sweeps out equal
areas in equal times.
Sun
same time intervals
same areas
S
slower
faster
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CF-8
KIII: For planets around the sun, the period T and the mean distance r from the sun are related
T2
T2
T2
by 3  constant . That is for any two planets A and B, A3  B3 . This means that
rA
rB
r
planets further from the sun (larger r) have longer orbital periods (longer T).
_________________ * __________________
Kepler's Laws were empirical rules, based on observations of the motions of the planets in the
sky. Kepler had no theory to explain these rules.
Newton (1642-1727) started with Kepler's Laws and NII (Fnet = ma) and deduced that
MS m P
Fgrav
 G
. Newton applied similar reasoning to the motion of the Earth-Moon
rSP 2
( Sun  planet )
ME m
 G
system (and to an Earth-apple system) and deduced that Fgrav
.
rEm 2
( Earth-mass m )
Newton then made a mental leap, and realized that this law applied to any 2 masses, not just to
the Sun-planet, the Earth-moon, and Earth-projectile systems.
Starting with Fnet = ma and Fgrav = G Mm / r2, Newton was able to derive Kepler's Laws (and
much more!). Newton could explain the motion of everything!
Derivation of KIII (for special case of circular orbits). Consider a small mass m in circular
orbit about a large mass M, with orbital radius r and period T. We aim to show that
T2 / r3 = const.
Start with NII: Fnet = m a
period T
M
The only force acting is gravity, and for circular motion
a = v2 / r 
r
G
m
v
Mm

r 2
 2 r 
v2
M
 m
 G
 v2  


r
r
 T 
2
[recall the v = dist / time = 2r / T ]

G
M
4 2 r 2

r
T2

T2
4 2

 constant, independent of m
r3
GM
(Deriving this result for elliptical orbits is much harder, but Newton did it.)
An extra result of this calculation is a formula for the speed v of a satellite in circular orbit:
GM
v 
. For low-earth orbit (few hundred miles up), this orbital speed is about 7.8 km/s
r
 5 miles/second.
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CF-9
Measurement of Big G
The value of G ("big G") was not known until 1798. In that year, Henry Cavendish (English)
measured the very tiny Fgrav between 2 lead spheres, using a device called a torsion balance.
Fgrav  G
m1 m2
r2
 G=
Fgrav r 2
( If Fgrav, r, and m's known, can compute G.)
m1m2
Before Cavendish's experiment, g and RE were known, so using g 
G ME
R E2
compute the product GME, but G and ME could not be determined separately.
, one could
With Cavendish's measurement of G, one could then compute ME. Hence, Cavendish "weighed
the earth".
Appendix: Proof of a = v2 / r for circular motion
The proof involves geometry (similar triangles). It is
mathematically simple, but subtle.
Consider the motion of a particle on a circle of radius r with
constant speed v. And consider the position of the particle at two
times separately by a short time interval t. (In the end we will
take the limit as t  0.) We can draw a vector diagrams
representing r1  r  r2 and v1  v  v2 :
r2
r1
v
r2
r = v t
r1
v2
v1
Notice that these are similar triangles (same angles, same length ratios) Also, note that
r1  r2  r and v1  v2  v .
Because the triangles are similar, we can write
v v
v
v2

, which is the same as a 
.
t
r
r
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v t
v
r


. A little algebra gives
v
r
r
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v2
v1