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
R
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 F = 6.67x10-11(70)(5.98 x 1024)
R2 (6.37 x 106)2
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 m = x
R2 M = x
F = Gx2 x2 = FR2 x = 14,808 kg
R2 G
Answer: 1.5x104 kg
Assignment Text. p.172 #8 - 13
p.694 #4 - 9
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 3.5x10-6 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
· Characteristics of circular motion:
1. Object is pulled toward center of circle by a force called CENTRIPETAL (center-seeking) 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 ac = v2 (F = ma)
R R
Where: Fc = centripetal force (tension) in N
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 = 2pR
v = 2pR
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 = 2pR = 2(3.14)(30)
T = 60 s T 60
p = 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 Fc = mv2 v = 2πR = 2π(20) = 8.4 m/s
R = 20 m R T 15
T = 15 s = (2)(8.4)2
Fc = ? 20
Fc = 7.0 N
· Period (T) = time for one revolution
· FREQUENCY (f) = number of revolutions per second
f = 1 T = 1
T 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 R T
f = 30 rps
T = 1/30 = 0.033 s = (10)(1903)2
Fc = ? 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 q = v2
Rg
where: q = 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 q = 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
q = 39.43012526°
q = 39.4°
· we have looked at circular motion as horizontal motion
· now lets look at vertical circular motion…we need to incorporate gravity
v = ÖRg Equation #20
where: v = speed in m/s
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 = ÖRg
= 2.90 m = Ö(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 or rod:
· 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 r
g = 9.81 m/s2 = (3.0 kg)(7.2 m/s)2 + (3.0 kg)(9.81 m/s2)
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
· 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 µ 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 µ 1
r2
· combining these 2 proportionalities gives us:
Fg µ m1m2
r2
· now, whenever you change from µ to =, there is a constant involved
· the formula is:
Fg = Gm1m2 Equation #14
r2
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.