Download unit 2 universal gravitation and circular motion

UNIVERSAL GRAVITATION
AND CIRCULAR MOTION
Two distinct types of motion have
been central to our understanding of
the universe.
Terrestrial motion - motion of
objects on the Earth.
Celestial motion - motion of objects
in the heavens.
Early theories of the universe
As far back as the 6th and 7thcenturies
B.C., Greek philosophers proposed
that the heavens were composed of
eight concentric, transparent spheres.
Each sphere rotated on a different
axis and at a different rate, but all
were centered about the Earth. This
geocentric, or Earth-centered, view
of the universe persisted with minor
modifications, for over 2000 years.
Nicolaus Copernicus thought that the
planetary position could be more
easily explained if the sun was the
center of the universe. Thus the
heliocentric system was born. His
theory predicted better the position of
the planets than did the geocentric
system.
Tycho Brahe (1546-1601) plotted the
paths of planets for more than 20
years to an accuracy of 1/1000th of a
degree.
Johann Kepler (1571-1630) was a
mathematician. He came up with
three laws:
1. The planets move about the sun in
elliptical paths with the sun at one
focus of the ellipse.
2. The straight line joining the sun
and a given planet sweeps out equal
areas in equal intervals of time.
3. The square of the period of
revolution of a planet about the sun is
proportional to the cube of its mean
distance from the sun. R3/T2 = k =
Kepler’s constant.
Newton’s Law of Universal
Gravitation
Newton determined that all bodies
(masses) anywhere in the universe
attract each other. The force of
attraction is directly proportional to
the product of the masses and
inversely proportional to the square of
the distance between them.
Formula,
Fg = GmM
R2
Fg = Force of
gravity (N)
G = Universal
Gravitational Constant
= 6.67x10-11
N.m2/kg2
M and m = masses in
kilograms.
R = distance
between the masses
from
center to
center in m.
Fg
M
R
m
Example
Find the force between two masses of
50 kg and 60 kg separated by a
distance of 50 cm.
Fg= GmM
R2
= (6.67x10-11)(50)(60)
0.502
= 8.0x10-7 N
Example
The force of attraction between two
masses of 500 kg is
6.0x10-7 N. Find the distance between
them.
Fg=GmM
R2 = GmM
R2
F
R2 = 6.67x10-11(500)(500)
6.0 x 10-7
R = 5.3 m
Example
A person whose mass is 70.0 kg stands
on the surface of the
Earth.
(a) What does the person weigh using
the law of universal gravitation?
(b) Use another formula to calculate
her
weight.
(a) F = GmM
11
(70)(5.98 x 1024)
R2
x 106)2
F = 6.67x10(6.37
F = 688 N
(b) F = W = mg = (70)(9.81) = 687 N
F = 687 N
Example
The force between two equal masses is
6.5x10-3 N. The distance between the
masses is 1.5 m. Find each mass.
F = GmM
R2
F = Gx2
14,808 kg
R2
m=x
M=x
x2 = FR2
G
Answer: 1.5x104 kg
Assignment Text. p.172 #8 - 13
p.694 #4 - 9
x=
Example
What is the gravitational force on a
35.0 kg object standing on the Earth’s
surface?
Example
The gravitational force between two
objects is 3.25x10-6 N. The mass
of one object is twice the mass of the
second object. If the distance between
the
masses is 2.00 m, find the mass of each
object.
Ratio method of calculating the
gravitational force
m1
m2
R
Fg
Example
The gravitational force between two
small
masses A and B when placed at a short
distance apart is 3.24x10-7 N. What is the
gravitational force between these objects
if the masses of both A and B are
doubled and distance between them is
tripled?
Example
The force between two masses is 3.5x106
N.
Find the force between them if the
masses
are tripled and the distance between
them is
halved.
Example
The force of gravity between two
equal masses separated by a distance
of 2.0 m is 5.0x10-7N.
Find the masses.
m=173kg
Cavendish Experiment
In 1798 Cavendish measured the
value of
G using a torsion balance.
.
Circular Motion
v
ac
Fc
 Characteristics of circular motion:
1. Object is pulled toward center of circle
by a force called CENTRIPETAL (centerseeking) FORCE
2. Acceleration is also directed to center
and
is called CENTRIPETAL
ACCELERATION.
3. The speed of the object is constant
4. Velocity of the object is constantly
changing.
5. Velocity is calculated from the tangent of
the circle at any given instant
Fc = mv2
(F = ma)
R
Where:
(tension) in N
ac = v2
R
Fc = centripetal force
ac = centripetal
acceleration in m/s2
m = mass in kg
R = radius of circle in
m
v = speed in m/s
Example
A 2.0 kg rubber stopper travels at 20 m/s in a
circular path of radius 10 m. Find the
centripetal acceleration and force.
ac = v2
Fc = mac
R
= (2)(40)
= (20)2
10
ac = 40 m/s2
Fc = 80 N
 The time for one revolution is called the
PERIOD (T)
 You know that speed = distance/time
 Distance for a circle is the circumference =
2R
v = 2R
T
Where:
v = speed in m/s
R = radius in m
T = period in s
T = Time for one
rotation or revolution
Example
If a car travels in a circular path with a radius
of 30 m and it takes 60 s, what is the speed of
the car?
R = 30 m v = 2R = 2(3.14)(30)
T = 60 s
T
60
 = 3.14
= 3.1 m/s
v=?
Example
An object with a mass of 2.0 kg travels in a
circular path of radius 20 m in 15 s. Find the
centripetal force.
m = 2.0 kg
2π(20) = 8.4 m/s
R = 20 m
15
Fc = mv2
R
v = 2πR =
T
= (2)(8.4)2
20
T = 15 s
Fc = ?
Fc = 7.0 N
 Period (T) = time for one revolution
 FREQUENCY (f) = number of revolutions
per second
f=1
T
T=1
f
f = frequency r/s or Hz (Hertz)
Example 1
A 10 kg object travels in a circular path at 30
rps. If the radius is 10 m find the centripetal
force.
m = 10 kg
Fc = mv2
v = 2πR =
1,903 m/s
R = 10 m
f = 30 rps
T = 1/30 = 0.033 s
Fc = ?
R
T
= (10)(1903)2
10
= 3.6  106 N
Example
What is the centripetal acceleration of a yo-yo
being swung in a horizontal circle with a period
of 0.20 s? The string is 10 cm long.
T = 0.20 s
ac = v2
v=?
R = 10 cm = 0.1 m
R
ac = ?
= 99 m/s2
Example
An 8.0 kg object travels in a circular path at
1,800 rpm. If the radius of the path is 10 m,
find
a) the centripetal acceleration
m = 8.0 kg
f = 1,800 rpm = 1,800/60 = 30 rps
T = 1/30 = 0.033 s
R = 10 m
ac = ?
ac = v2
v=
2(3.14)(10)
R
0.033
= 3.6  105 m/s2
b) the centripetal force
Fc = mac
= (8.0)(3.6 x 105)
= 2.9  106 N
p. 145 #9-11
p.151 – key terms
p. 166 #5
p. 173 #20 - 24
Banking Curves
 Banking curves often occur so that vehicles
do not need to rely only on friction to keep
them on the road
tan  = v2
Rg
where:  = banking angle
v = speed in m/s
R = radius of circle in m
g = acceleration due to gravity = 9.81
m/s2
Example 1
A car travels around a curve of radius 60.0 m at
a speed of 22.0 m/s. At what angle must the
curve be banked so the car doesn’t have to rely
on friction to stay on the road?
R = 60.0 m
tan  = v2
v = 22.0 m/s
Rg
g = 9.81 m/s2
= (22.0 m/s)2
(60.0 m)(9.81 m/s2)
= 0.8222901801
 = 39.43012526
 = 39.4
 we have looked at circular motion as
horizontal motion
 now lets look at vertical circular motion…we
need to incorporate gravity
v = Rg
where: v = speed in m/s
Equation #20
R = radius in m
g = 9.81 m/s2
Example 1
An amusement park ride spins in a vertical
circle. If the diameter of the ride is 5.80 m,
what minimum speed must the ride travel at the
top?
R = 5.80 m 2
v = Rg
= 2.90 m
= (2.90 m)(9.81 m/s2)
g = 9.81 m/s2
= 5.3337604 m/s
= 5.33 m/s
Vertical Circular Motion
 now we can look at the forces exerted when
an object is being swung in a vertical circle
on a
string orF rod:
g
Fc
FT
Fc
FT
Fg
 Fc = centripetal force (net
force) and it always acts
towards the centre
 FT = tension and it also
always acts towards the
centre
 Fg = force of gravity and it
always acts towards the
centre of the Earth
 Fc remains constant (as long as speed
remains constant) and Fg remains constant
 FT changes depending on where the mass is
in it’s circular path
 tension on the string/rod is greatest at the
bottom
Fc(net) = FT – Fg (vectors in opposite
direction)
FT = Fc + Fg
 tension on the string/rod is smallest at
the top
Fc(net) = FT + Fg (vectors in same
direction)
FT = Fc – Fg
Example 1
A 3.0 kg mass moves on a string in a vertical
circle of radius 2.5 m with a constant speed of
7.2 m/s.
a) Where is the tension in the string the
greatest? Calculate it.
Tension is the greatest at the bottom.
m = 3.0 kg
FT = Fc + Fg
r = 2.5 m
FT = mv2 + mg
v = 7.2 m/s
g = 9.81 m/s2
(3.0 kg)(9.81 m/s2)
r
= (3.0 kg)(7.2 m/s)2 +
2.5 m
= 91.638 N
= 92 N
b) Where is the tension in the string the
smallest? Calculate it.
Tension is smallest at the top.
m = 3.0 kg
FT = Fc – Fg
r = 2.5 m
FT = mv2 - mg
v = 7.2 m/s
r
g = 9.81 m/s2 = (3.0 kg)(7.2 m/s)2 - (3.0
kg)(9.81 m/s2)
2.5
= 32.778 N
= 33 N
B. Gravitation
 in 1666, Newton used mathematical
arguments to look at the force of attraction
between two objects
 the falling apple made him realize that the
earth was attracting the apple, and that this
attraction probably extended far above the
earth’s surface into space
 he recognized that the force acting on the
apple was proportional to its mass and that,
according to his own 3rd law, the apple must
also be attracting the earth
 if the earth was involved in the attraction,
then the force of attraction was also
proportional to the earth’s mass
m1
m2
r
 it all comes down to this:
1. all objects in the universe attract each
other
2. the force of attraction (Fg) between 2
masses (m1, m2)is directly proportional (if
one goes up the other goes up) to the
product of the masses: Fg  m1m2
3. the force of attraction between 2 masses
is indirectly proportional (if one goes up the
other goes down) to the square of the
distance (r) between them: Fg  1
r2
 combining these 2 proportionalities gives us:
Fg  m1m2
r2
 now, whenever you change from  to =,
there is a constant involved
 the formula is:
Fg = Gm1m2
r2
Equation #14
where: G = universal gravitation constant = 6.67
x 10-11 Nm2/kg2
m1m2 = mass 1 and mass 2 in kg
Fg = gravitational force of attraction in N
r = distance between centres in m
Example 1
What is the force of attraction between a 20 kg
mass and a 50 kg mass separated by a distance
of 50 cm?
m1 = 20 kg
Fg = Gm1m2
m2 = 50 kg
r2
r = 50 cm
= (6.67x10-11 Nm2/kg2)(20
kg)(50 kg)
= 0.50 m
(0.50 m)2
G = 6.67x10-11 Nm2/kg2
= 2.668 x 10-7 N
= 2.7 x 10-7 N
Example 2
A person has a mass of 60 kg.
a) what is his weight on the surface of the
earth?
m = 60 kg
Fg = mg
g = 9.81 m/s2
= (60 kg)(9.81 m/s2)
= 588.6 N
= 5.9 x 102 N
b) what is the gravitational force of attraction
between the person and the earth?
m1 = 60 kg
Fg = Gm1m2
m2 = 5.98x1024 kg
r2
r = 6.37 x 106 m
= (6.67x10-11
Nm2/kg2)(60 kg)(5.98 x 1024 kg)
G = 6.67x10-11 Nm2/kg2
(6.37 x
106 m)2
= 589.7927146 N
= 5.9 x 102 N
*****NOTE: This is the same number as part
a)!!!!!!!!!!!!!!!!!!
Example 3
Two masses, m1 = 5.0 x 106 kg and m2 = 3.0 x
108 kg, exert a force of 5.0 x 10-7 N. Find the
distance between them.
m1 = 5.0 x 106 kg
Fg = Gm1m2
m2 = 3.0 x 108 kg
r2
Fg = 5.0 x 10-7 N
r2 = Gm1m2
G = 6.67x10-11 Nm2/kg2
Fg
= (6.67x10-11
Nm2/kg2)(5.0x106 kg)(3.0x108 kg)
5.0 x 10-7 N
= 2.001 x 1011 N
 r = 447325.3849 N
= 4.5 x 105 m
Example 4
Two equal masses are attracted by a
gravitational force of 3.8 x 10-8 N. If they are
separated by 2.3 m, what is each mass?
m1 = m2
Fg = Gm1m2 = Gm2
G = 6.67x10-11 Nm2/kg2
r2
r2
r = 2.3 m
m2 = Fgr2
Fg = 3.8 x 10-8 N
G
= (3.8 x 10-8 N)(2.3 m)2
(6.67x10-11
Nm2/kg2)
= 3013.793103 kg2
m = 54.89802459 kg
= 55 kg
Example 5
Two masses exert a force of 3.8 x 10-6 N on
each other and they are 5.2 m apart. If one
mass is 2 times the size of the other, find the
mass of each object.
Fg = 3.8 x 10-6 N
Fg = Gm1m2
r = 5.2 m
r2
m1 = m
= G(m)(2m)
m2 = 2m
r2
G = 6.67x10-11 Nm2/kg2
= 2m2G
r2
m2 = Fgr2
2G
= (3.8 x 10-6 N)(5.2 m)2
2(6.67x10-11 Nm2/kg2)
= 770254.8726 kg2
 m1 = 877.6416538 kg
= 8.8 x 102 kg
 m2 = 2(877.6416538 kg)
= 1755.283308 kg
= 1.8 x 103 kg
Your Assignment: 1. pg 1 #1-10 in
workbook
2. pg 172 #8-12
 now we must look at the relationship between
force, mass and distance
 if you change one, what happens to the
others?
F  m1m2
r2
m1
X2
X2
X2
X3
X6
X8
m2
X2
X2
X2
X4
X4
r
X2
X2
X3
X4
F
10 N
20 N
40 N
10 N
30 N
27 N
20 N
X2
2.5 N
 you can use a chart like this for any formula
 let’s look at circular motion:
Fc = mv2
r
m
X2
X2
X4
X2
v
X2
X2
X3
r
X2
-
Fc
2.0 N
4.0 N
16 N
16 N
36 N
Satellites
 Newton’s “thought experiment”
 Cannon on top of mountain – fire it
horizontally
 Cannon is a projectile that falls 4.9 m every
second
 If the mountain is high enough, the cannon
will fall at the same rate that the earth
curves away….orbit!!!!!!
 Mountain would have to be at least 150 km
high to be above most of the atmosphere for
no air resistance
 Satellites are objects that are projected into
space
 They are at a constant height above the planet
and are in uniform circular motion
 The force of gravity keeps the satellite in
orbit around the earth (or any other planet)
Fg = Fc
 lets look at how to calculate the speed and
the period of an orbiting satellite:
mP
R
Fg
Fc
satellite
ms
earth
Speed:
Fg = Fc
GmsMP = msv2
R2
R
GMP = v2
R
solve for v
v = GMP
R
this formula is used for
satellites only.
where: v = speed in m/s
G = 6.67 x 10-11 Nm2/kg2
MP = mass of planet in kg
R = distance between centre of planet
and satellite (radius of planet + altitude
of the satellite above the surface of the
planet) in m
If orbital radius is given, it is the distance
from the centre of the planet to the satellite.
Example
***A satellite is 100 km above the surface of
the earth. What is the speed of the satellite?
G = 6.67 x 10-11 Nm2/kg2
v = GMP
MP = 5.98 x 1024 kg
R
R = 6.37 x 106 m + 1.0 x 105 m
= 6.47 x 106 m
v=?
= (6.67 x 10-11)(5.98 x 1024)
(6.47 x 106)
v = 7.85 x 103 m/s
Period:
 time taken to complete one revolution.
 some satellites are geosynchronous or
geostationary, which means they have a
period of 24 hours and they always maintain
a specific longitude and latitude
 used for global communications and data
collection and transmission eg. weather,
minerals, agriculture, astronomy, land and
ocean surveillance, solar radiation, the
environment etc.
 other satellites are not geosynchronous and
you can use a formula to calculate the period
(time for one revolution)
v = 2πR
T
T = 2πR
v
Example
*****Find the period of a satellite which is 200
km above the surface of the earth.
G = 6.67 x 10-11 Nm2/kg2
MP = 5.98 x 1024 kg
R = 6.37 x 106 m + 2.0 x 105 m
v2 = GM
R
v = 7791 m/s
T = 2πR
v
T = 5.3 x 103 s
D. Planetary Mechanics
 two different motions have been central to
our understanding of the universe:
1. Terrestrial motion – motion of objects on
earth
2. Celestial motion – motion of objects in
space
 Geocentric view – people used to think that
everything in the solar system revolved
around the earth; proposed by Greeks
 Heliocentric view – sun-centred motion of
planets; proposed by Copernicus (although
Galileo was probably the first to propose this,
he was thrown in jail)
 Tycho Brahe (1546 – 1601) – Danish
astronomer who plotted the paths of planets
for more than 20 years to an accuracy of
1/1000 of a degree
 Johannes Kepler (1571 – 1630) – German
astronomer who was one of Brahe’s
assistants
 he wanted to use a sun-centred system
(instead of geocentric) to explain Brahe’s
precise data
 he came up with 3 laws:
1. the planets move about the sun is
elliptical paths, with the sun at one focus
of the ellipse
planet
sun
focus
2. the straight line joining the sun and a
given planet sweeps out equal areas in
equal time intervals
planet
faster speed
closer to sun
slower speed
farther from sun
3. the square of the period of revolution of
a planet about the sun is proportional to
the cube of its mean distance from the
sun
K = R3
not on formula sheet
but
T2
comes from Equation
#12
where: R = mean orbital radius in m
T = orbital period in s
K = Kepler’s constant = 3.315 x 1018
m3/s2
Example 1
An asteroid has a period of 8.1 x 107 s.
What is its mean radius of orbit around the
sun?
T = 8.1 x 107 s
K = R3
K = 3.315 x 1018 m3/s2
T2
R3 = KT2
= (3.315 x 1018
m3/s2)(8.1 x 107 s)2
= 2.1749715 x 1034
R = 2.791372894 x
1011 m
= 2.8 x 1011 m
Example 2
A planets mean distance from the sun is 2.0
x 1011 m. What is the orbital period?
R = 2.0 x 1011 m
K = R3
K = 3.315 x 1018 m3/s2
T2
T 2 = R3
K
= (2.0 x 1011 m)3
(3.315 x 1018 m3/s2)
= 2.413273002 x 1015 s2
T = 49125075.08 s
= 4.9 x 107 s
Gravitational Field
 Gravitational field surrounds a mass of any
size as it is this field that provides the force
of attraction on another body
 Knowing the weight (force gravity) and
mass of an object, the gravitational field
strength can be calculated:
g = Fg
m
where: g = gravitational field in
N/kg or m/s2
Fg = gravitational force or
weight in N
m = mass in kg
Example 1
On a planet an object weighs 20 N. If the
mass of the object is 4.5 kg, what is the
gravitational field strength?
Fg = 20 N
g = Fg
m = 4.5 kg
m
= 20 N
1
N = 1 kg m/s2
4.5 kg
= 4.4 N/kg or m/s2
Example 2
What is the mass of a 40 N object
experiencing a gravitational field of 13
N/kg?
Fg = 40 N
g = Fg
g = 13 N/kg
m
13 = 40
m
13m=40
m = 3.1 kg
 Gravitational field strength is simply the
acceleration due to gravity at the location of
the mass e.g. on earth, in orbit, on another
planet etc.
 Strength varies inversely with the square of
the distance from the center of a planet
g = GM
r2
where: g = gravitational field strength in N/kg or
m/s2
G = 6.67 x 10-11 Nm2/kg2
M = mass of planet in kg
r = distance from center of planet in m
Example 1
Find the gravitational field strength on the
surface of Mercury (rM = 2.43 x 106 m and M =
3.2 x 1023).
G = 6.67 x 10-11 Nm2/kg2
r = 2.43 x 106 m
M = 3.2 x 1023 kg
10-11)(3.2 x 1023)
g = GM
r2
g = (6.67 x
(2.43
x 106)2
= 3.6 N/kg
or m/s2
Example 2
Find the gravitational field strength
experienced by astronauts if they are on the
space shuttle 500 km above the earth’s
surface.
G = 6.67 x 10-11 Nm2/kg2
GM
M = 5.98 x 1024 kg
r2
r = 6.37 x 106 + 500,000
10-11)(5.98 x 1024)
g=
= (6.67 x
(6.87 x 106)2
r = 6.87 x 106 m
g = 8.45
N/kg or m/s2
m (electron) = 9.11 x 10-31 kg
m (proton) = 1.67 x 10-27 kg
F. Einstein’s Theory of Relativity
 use pages 169 – 170 of your text to make
short point-form notes on Einstein’s Theory
of Relativity
 Einstein proposed that gravity is not a force
but an effect of space itself
 He believed that a mass changes the space
around it and causes it to be curved
 acceleration takes place because other bodies
move in a curved space (centripetal
acceleration)
 masses set up a gravitational field around
themselves which distorts the space around it
Some Cool Stuff to Think About
 the gravitational field strength of a black hole
is so strong that it bends light back into the
hole…nothing can leave it
 black holes are formed when a star collapses
in on itself
 when the radius is small enough and the
collapsing star is dense enough, light can no
longer escape
 the earth would have to collapse to a radius of
9 mm (0.009 m), without losing any mass, to
become a black hole
so…
*****that’s 50 trillion times higher than the g
we experience right now
g = 5.0 x 1013 x 9.81 m/s2
= 4.905 x 1014 m/s2