Transcript centripetal force gravitational potential

UNIFORM CIRCULAR MOTION
An object traveling with a constant speed in a circle is
accelerating because the object’s velocity is changing in
direction. The object is said to have “uniform” circular
motion and it is undergoing centripetal acceleration
(centrum- centre, petere- to seek (Newton) p. 123-124).

v2



r2
v1
v
v1
r

r2

r1




v2
r1
These two triangles are similar and for
any time interval this would be so.


v

r

This proof is not on any test.

v

r



v r

v 
v
lim

t 0
r

v r

a c  lim
t  0

ac 

v
lim
 t  0
r
 ac
v
ac 
r
[towards the center]
Notice on the first diagram that half
way through the interval v is pointing
at the center of the circle.
This is an instantaneous acceleration
so the direction of the acceleration is
constantly changing.
r t



t

2
r
This is velocity!
t
For the final equation remove the
absolute value notation.
For uniform circular motion . . .
2r
v
T
2
4 r
ac 
2
T
T is the period of revolution.
ac  4 rf
2
remember
2
1
f 
T
f is the frequency
An increase in the velocity of an object that is centripetally
accelerating (with no change to r) is called tangential acceleration. A parallel push causes tangential acceleration and
velocity’s magnitude changes. A perpendicular push causes
centripetal acceleration and only the direction of velocity changes.
What is the centripetal acceleration of a stone being whirled in
a circle, at the end of a 1.5 m string, on a smooth sheet of ice,
with a frequency of 1.25 Hz?
ac  4 2 rf 2
ac  4 2 (1.5 m)(1.25 Hz ) 2
m
ac  92.53 2
s
The planet Mercury moves in an approximately circular path at
an average distance of 5.8 x 1010 m, accelerating centripetally
at 0.04 m/s2. What is its period of rotation about the Sun?
4 2 r
ac  2
T
4 2 r
T 
ac
2
4 2 (5.8 x1010 m)
T 
m
0.04 2
s
T  7.566 106 s
2
CENTRIPETAL FORCE
Centripetal acceleration (center seeking) can be extended to
centripetal force. Remember centripetal acceleration is the
observed motion while centripetal force is responsible for this
acceleration. Finally, centripetal force is always a resultant force.

mv2
Fc 
[toward center]
r
This direction
is constantly
changing.
2r
v
T
4 mr
Fc 
2
T
2
Fc  4 mrf
2
2
textbook:
p. 133 2-8, p. 138 2-7,
p. 159 8, 11-13, 26-28
A 1000 kg car enters a level curve at 20 m/s. If the curve has
a radius of 80 m, what centripetal force must be supplied by
friction to keep the car from skidding?



FN

F net  F N  F g  F f
Fc  F f
Ff
2
mv
 Ff
r
m 2
1000kg(20 )
s
Ff 
80m
F f  5000 N
Notice this would give a m
Fg
s of 0.5
A 1000 kg ball travels around a frictionless, banked curve having
a radius of 80 m. If the bank is 20o to the horizontal, at what speed
must the ball travel to maintain a constant radius?




FR  FN  Fg

FNx

Fc  FNx

FN
2
mv
 tan 20o ( FNy )
r
mv 2
o
 tan 20 (mg )
80 m
20o
v  tan 20 (9.81 N / kg)(80 m)
m
v  16.90
s
2
o

FNy

Fg
A 3.5 kg steel ball is swung at a constant speed in a vertical circle of
radius 1.2 m on the end of a light, rigid steel rod. If the ball has a
frequency of 1.0 Hz, calculate the tension in the rod at the top and
bottom of the circle.



top
bottom
R
T
g
+ is up
F F F
top
Fc  FT  Fg

 4 mrf  FT  mg
2
2
FT
FT   4 2 mrf 2  mg
FT  131.5 N
bottom
Fc  FT  Fg
 4 2 mrf 2  FT  mg
FT  4 2 mrf 2  mg
FT  200.1 N

FT


Fg
Fg
LAW OF UNIVERSAL GRAVITATION
FG
Gm1m2

2
r
FG -gravitational force (in 2 directions)
G -universal gravitation constant
6.67x10-11 Nm2kg-2
r -distance between the objects
m1 -mass of the larger object
near the earth’s surface . . .
Gm1m2
m2 g 
2
r
Gm1
g 
r2
This equations could be applied to the surface of any planet or
to the acceleration you would experience at any distance from
on object.
What happens to the gravitational attraction between two particles
if one mass is doubled, the other tripled and the distance between
them cut in half?
G 2m1 3m2
r 2
F2
( )
 24
F2
2

Gm1m2
F1
F1
r2
Planet X has a radius that is 3.5 times the radius of the earth and
a mass that is 2.0 times the earth’s. Compare the acceleration due
to gravity at the surface of each planet.
gx 
Gm1
g 2
r
G 2me
(3.5re ) 2
2 Gm
gx  ( 2 ) 2 e
3.5
re
2
gx  ( 2 )g e
3.5
read p. 139-142 p. 141 1-6
gx

 0.163
ge
extra p. 143 8-13 p. 144 1-6
SATELLITES
A satellite is an object or a body that revolves around another
object, which is usually larger in mass.
Planets, moons, space shuttles, space stations, comets, and
“satellites” are satellites.
Satellites remain in a constant orbit because they are acted upon
by a centripetal force and display centripetal acceleration.
FG  Fc
Gm1m2
m2 v

2
r
r
Gm1
2
v 
r
2
remember m1 is the larger mass
and the central object
Calculate the orbital speed for a satellite at Earth’s surface, and two Earth radii above Earth’s
surface. Show the simpler calculation for the satellite at Earth’s surface.
v2
What is the radius of the orbit
of a geosynchronous satellite?
Gm1

r
Gm1
 2r 

 
r
 T 
2
Gm1T 2  4 2 r 3
2
Gm
T
1
r3
4 2
r
3
6.67 10
11
Nm2
5.98 10 24 kg (86400 s ) 2
2
kg
4 2
r  4.225 10 m
8
What is the period of rotation of the moon about the earth?
4 r
T
Gm1
2 3
4 2 (3.84 108 m)3
T
2
Nm
6.67 10 11 2 5.98 10 24 kg
kg
read 145-146 p. 151 1, 3-6 extra
T  2.367 10 s
6
T  27.40d
p. 147 2-4, 6 p. 160 14-20
GRAVITATIONAL FIELDS
A force field exists in the space surrounding an object in which a
force is exerted on objects (e.g. gravitational, electric, magnetic).
The strength of gravitational force fields is deter-mined by the Law
of Universal Gravitation. If two or more gravitational fields are
acting on an object then the net field is the sum of all the
individual fields.
read 274-275
p.276 2-6
p.277 1-8
KEPLER’S LAWS
In 1543 Copernicus proposes the heliocentric model of the solar
system in which planets revolve around the sun in circular orbits.
Slight irregularities show up over long periods of study.
Tycho Brahe takes painstaking observations for 20 years with large
precision instruments but dies (1600) before he can analyze them
properly. A young mathematician continues Brahe’s work.
From his analysis the kinematics of the planets is fully understood.
Kepler’s First Law of Planetary Motion
Each planet moves around the Sun in an orbit that is an ellipse,
with the Sun at one focus of the ellipse.
Kepler’s Second Law of Planetary Motion
The straight line joining a planet and the Sun sweeps out equal
areas in space in equal intervals of time.
Planets move faster when they are closer to the Sun (centripetal
force is stronger).
orbits are elliptical but are not very
elongated
equal areas
equal times
Kepler’s Third Law of Planetary Motion
The cube of the average radius of a planet is directly proportional
to the square of the period of the planet’s orbit.
We have already proved this a few slides back. Recall.
Gm1T
2
 4 r
2
3
3
r
Gm1

2
T
4 2
constant
For our solar system m1 is the mass of the sun.
Mars’ average distance from the sun is 2.28 x1011 m while its
period of rotation is 5.94 x 107 s. What is Jupiter’s average
distance from the sun if its period of rotation is 3.75 x 108 s ?
3
J
2
J
3
m
2
m
r
Gm1 r


2
T
4
T
read 278-283
rJ  7.79 10 m
11
p. 283 10-12
p. 284 4-7, 9
GRAVITATIONAL POTENTIAL ENERGY, AGAIN
Recall the Law of Universal Gravitation
FG
Gm1m2

2
r
for constant masses, a graph of force vs. radius would be . . .
The graph above is a F vs. d graph which means the shaded area
is the work required to move an object from r1 to r2.
The shaded area is not easy to calculate but can be done with a
geometric mean. In this case the work done by the lifter is equal to
Ep.
Another method involves calculus and integration over a range
from r1 to r2.
geometric mean
W  F1F2 (r2  r1 )
of force
Gm1m2 Gm1m2
E p 
(r2  r1 )
2
2
r1
r2
Gm1m2
E p 
(r2  r1 )
r1r2
Gm1m2 Gm1m2
E p 

r1
r2
Gm1m2
Gm1m2
E p  

r2
r1
Gm1m2
Ep  
r
Know these two equations,
you are not required to know
the previous development.
Which preceding equation can be simplified to mgh, the potential
energy change near the earth’s surface?
Potential energy is a negative
function!
It increases until it is zero.
PE stops here
because the objects
come into contact and
cannot get closer.
Recall
mv12
mv22
 mgh1 
 mgh2
2
2
or
mv22 mv12

 (mgh2  mgh1 )
2
2
so . . .
mv12 Gm1m2 mv22 Gm1m2



2
r1
2
r2
or
2
2
2
1
mv mv
Gm1m2
Gm1m2

 ( 

)
2
2
r2
r1
read p. 285-287
p. 287 1-5
Escape from a Gravitational Field
To escape a gravitational field an object must have at least a total
mechanical energy of zero!!
EM  EK  EP
for escape
EM  0
Escape energy - the minimum EK needed to project a mass (m2)
from the surface of another mass (m1) to escape the gravitational
force of m1
Escape speed - the minimum speed needed to project a mass
(m2) from the surface of another mass (m1) to escape the
gravitational force of m1
Binding energy - the additional EK needed by a mass (m2) to
escape the gravitational force of m1 (similar to escape energy but
applies to objects that possess Ek i.e. satellites).
To calculate the escape energy or the escape velocity of a mass
(m2):
EK  EP  0
To calculate the binding energy of a mass (m2):
EK  EP  EB  0
binding energy
Calculate the escape velocity of any object on the Earth’s surface.
EK  EP  0
2
m2v
Gm1m2

0
2
re
2Gm1
v 
re
2
2(6.67 1011 Nm2 kg 2 )(5.98 1024 kg)
v
6.38 106 m
m
v  11182
s
The escape velocity is the same
for all objects on the Earth’s
surface while the escape energy
is different for different massed
object.
What is Ek and EM of an orbiting body (satellite)?
FG  Fc
this is always true of satellites
2
Gm1m2
m2 v

2
r
r
Gm1m2
 m2 v 2
r
Gm1m2
m2 v 2

2r
2
 Ep
EK 
2
EM  E K  E P
EM  
EM 
Ep
2
Ep
 Ep
for an orbiting satellite !!
2
Note that the total energy is negative since the satellite is “bound”
to the central body.
read p.288-293
p. 293 6-11
extra p. 294 1-8 p. 300 1-17
#12 is interesting!
25,26 look fun
a) What is the speed of Earth in orbit about the Sun?
b) What is the total energy of Earth?
c) What is the binding energy of Earth?
d) If Earth was launched from the surface of the Sun to its present
orbit then what velocity must it be launched with (Ignore the radius
of Earth.)?
e) If Earth came to rest and fell to the Sun then what velocity
would it have when it hit the Sun (Ignore radius of Earth.)?
me= 5.98x1024 kg
ms= 1.99x1030 kg
re= 1.49x1011 m (of orbit)
rs= 6.96x108 m
(of the body)
G= 6.67x10-11 Nm2kg-2
a)
2r
4 m
v
v

2
.
966

10
T
s
7
T  3.156  10 s
or
b)
Gm1m2
 m2v 2
r
EM 
m
v  2.985  10
s
Ep
4
EM  2.664 10 J
2
Gms me
EM  
2re
c) The binding energy is 2.664 x 1033 J
33
d)
2
1
2
2
mv Gm1m2 mv Gm1m2



2
r1
2
r2
mv12 Gm1m2
Gm1m2


2
r1
2r2
mv12
Gm1m2 Gm1m2


2
2r2
r1
2
1
mv
36
 1.138 10 J
2
5 m
v1  6.169 10
s
mv12 Gm1m2 mv22 Gm1m2
e)



2
r1
2
r2
Gm1m2 Gm1m2 mv22



r1
r2
2
mv22
 1.135 1036 J
2
5 m
v2  6.161 10
s