Title: Chapter 2: Motion
1Chapter 2 Motion
2Speed
- The speed of something is the rate at which it
covers distance - If the speed is high, it travels faster and it
covers more distance in a given period of time - Speed (v) distance (d) / time (t)
- Common units to express speed are meters/second
(m/s) - When we dont know how something moved, its speed
is referred to as the average speed i.e. it
might have moved faster at some times and slower
at other times - For example, if a car has an average speed of 50
mi/h, it doesnt mean that it was moving at a
constant speed of 50 mi/h during the entire time
when it slowed down for traffic or stopped at a
stop sign, it was obviously moving slower than 50
mi/h - The instantaneous speed is how fast something is
going at any given moment - In a car, what the speedometer reads is the
instantaneous speed of the car
3Steps to Solving a Problem
- Step 1 Identify the information that is given in
the problem. - Step 2 Identify what information the problem is
looking for. - Step 3 Identify the equation that is needed to
obtain the information you are looking for, while
using the information you were given. - Step 4 Put the given information into the
equation found in Step 3 and solve for the
information you are looking for.
4Example Calculation of Speed
- Example What is the speed of an object that
traveled 34 meters in 65 seconds? - Answer
- 1. Given 34 meters, 65 seconds
- 2. Looking for speed
- 3. Equation v d/t
- 4. Solution v d/t 34 meters/65 seconds
0.52 m/s
5Distance and Time
- The equation for speed can be rewritten in order
to calculate - the distance (d) something travels at a given
speed (v) in a given period of time (t) - distance (d) speed (v) x time (t)
- the time (t) something takes to travel a given
distance (d) going a given speed (v) - time (t) distance (d) / speed (v)
- Distance is commonly expressed in meters (m) and
time is commonly expressed in seconds (s)
6Distance Solving the Equation and Example
Calculation
- Solving the Equation
- 1. Start with the equation for speed v d/t
- 2. Get d on a side by itself by multiplying both
sides by t, thus obtaining the equation for
distance d vt
- Example How far did an object move in 76 seconds
if it was traveling at a speed of 2.5 m/s? - Answer
- 1. Given 76 seconds, 2.5m/s
- 2. Looking for how far distance
- 3. Equation d vt
- 4. Solution d vt 2.5 m/s x 76 s 190 m
7Time Solving the Equation and Example Calculation
- Solving the Equation
- 1. Start with the equation for speed v d/t
- 2. Get t on top (it is now on the bottom) by
multiplying both sides by t, thus the equation
is vt d - 3. Get t on a side by itself by dividing both
sides by v, thus obtaining the equation for time
t d/v
- Example How long did it take for an object move
230 meters if it was traveling at a speed of 14.7
m/s? - Answer
- 1. Given 230 meters, 14.7m/s
- 2. Looking for how long time
- 3. Equation t d/v
- 4. Solution t d/v 230 m / 14.7 m/s 15.6 s
8Scalar Quantities and Vector Quantities
- Scalar quantities are quantities that contain a
number and a unit. Scalar quantities give only
the magnitude (how large it is), not the
direction - Examples of scalar quantities
- Speed
- Vector quantities, on the other hand, are
quantities that contain both a magnitude and a
direction - Examples of vector quantities
- Displacement, which is change in position
- Force
- Velocity, which is the speed of something and the
direction of this speed
9Representing Vector Quantities
- Vector quantities are represented by a vector, a
straight line that has an arrowhead at one end to
show the direction of the quantity - The length of the line is scaled for the
magnitude of the quantity - Vector quantities are usually printed in bold
(i.e. Force F), while scalar quantities are
usually printed in italics (i.e. Speed v). To
represent the magnitude of a vector quantity,
italics is also used (i.e. the magnitude of a
Force, F, is F). When a vector quantity is
handwritten, an arrow over the symbol is used to
indicate that it is a vector quantity.
10Adding Scalar Quantities and Vector Quantities
- To add scalar quantities, ordinary arithmetic is
used - To add vector quantities,
- if they are going in the same direction, ordinary
arithmetic is used - if they are not going in the same direction, the
following steps need to be followed - 1. Draw the vectors that are to be added together
tail to head. - In this example, to add vector B to vector A,
draw vector B with its tail at the head of vector
A - 2. Draw a vector from the tail of the first
vector to the head of the last vector to be added
this vector is the addition of all the vectors - 2. Draw a vector C from the tail of vector A to
the head of vector B vector C is the addition
of vector B to vector A
11Pythagorean Theorem
- A right triangle is a triangle in which two of
its sides are perpendicular to each other in
other words, its when the two sides meet at a
90 angle - The Pythagorean Theorem states that, in a right
triangle, the square of each of the short sides
added together is equal to the square of the
longest side (i.e. the hypotenuse) - If the short sides are A and B and the long side
is C, the Pythagorean Theorem can be shown as the
following equation - A2 B2 C2
- When the Pythagorean Theorem is applied to adding
vectors, A, B, and C are the magnitudes of the
vectors A, B, and C - The following equations can be used to find one
of the sides if the length of the other two sides
are known - A v(C2 B2)
- B v(C2 A2)
- C v(A2 B2)
12Example Calculations Using the Pythagorean Theorem
- Example 1 What is the length of side A if side B
is 2.3 m and side C is 7.6 m? - Answer
- 1. Given side B 2.3m, side C 7.6m
- 2. Looking for length of side A
- 3. Equation A v(C2 B2)
- 4. Solution A v(C2 B2) v(7.6m2 2.3m2)
- Example 2 What is the length of side B if side A
is 4.3 m and side C is 10.5 m? - Answer
- 1. Given side A 4.3m, side C 10.5m
- 2. Looking for length of side B
- 3. Equation B v(C2 A2)
- 4. Solution B v(C2 A2) v(10.5m2 4.3m2)
- Example 3 What is the length of side C if side A
is 8.7 m and side B is 3.1 m? - Answer
- 1. Given side A 8.7m, side B 3.1m
- 2. Looking for length of side C
- 3. Equation C v(A2 B2)
- 4. Solution C v(A2 B2) v(8.7m2 3.1m2)
13Acceleration
- An object that has an acceleration is one whose
velocity is changing - The change in velocity can be one of three
things - An increase in speed (i.e. going faster)
- A decrease in speed (i.e. going slower)
- A change in direction of the speed (i.e. turning)
- Acceleration is a vector quantity
- When acceleration is in a straight line
- Acceleration change in speed / time interval
- OR
- a (v2 - v1) / t, where a is the acceleration,
v2 is the final speed, v1 is the initial speed,
and t is the time interval - Common units to express acceleration are m/s2
- When acceleration is from a decrease in speed, it
is called negative acceleration or deceleration - We will assume that acceleration is constant,
although that is not always the case
14Example Calculations With Acceleration
- Example If an objects initial speed is 30m/s
and its final speed is 55m/s, what is its
acceleration over 571 seconds? - Answer
- 1. Given initial speed 30m/s, final speed
55m/s, 571 seconds - 2. Looking for acceleration
- 3. Equation a (v2 - v1) / t
- 4. Solution a (v2 - v1) / t (55m/s 30m/s)
/ 571 s 0.0438 m/s2
15Using Acceleration to Calculate Final Speed
- To calculate the final speed of something, the
equation for acceleration (a (v2 - v1) / t )
needs to be rewritten - 1. First bring t over to the other side of the
equation by multiplying it on both sides - 2. Then, separate v2 from v1 by adding v1 to both
sides - The equation to calculate final speed is v2 v1
at
- Example If an object has an initial speed of 60
m/s with an acceleration of 3.2m/s2, what is its
final speed after 43 seconds? - Answer
- 1. Given initial speed 60m/s, acceleration
3.2m/s2, 43 seconds - 2. Looking for final speed
- 3. Equation v2 v1 at
- 4. Solution v2 v1 at 60m/s
(3.2m/s2)(43s) 197.6m/s
16How Far?
- To calculate the average speed of an object
during uniform acceleration - average speed (v1 v2) / 2
- The equation to calculate the distance the object
moves is d vt - If the speed, v, is an average speed, then the
equation for distance is d (average speed)t - If we insert the equation for the average speed
into this equation, then the equation for
distance is d (v1 v2)/2t (v1t/2)
(v2t/2) - If we insert the equation for the final speed
into this equation, then the equation for
distance is d (v1t/2) ((v1 at)t/2)
(v1t/2) (v1t/2) (at2/2) v1t (at2/2) - So, d v1t (at2/2)
- If the object starts at rest, then v1 0, so d
(at2/2)
17Example Calculations for Average Speed and
Distance
- Average Speed Example What is the average speed
of an object that was moving at some times 45m/s
and at other times 35m/s? - Answer
- 1. Given speed 1 45m/s, speed 2 35m/s
- 2. Looking for average speed
- 3. Equation average speed (v1 v2) / 2
- 4. Solution average speed (v1 v2) / 2
(45m/s 35m/s)/2 40m/s
- Distance Example 1 If an object has an initial
speed of 10 m/s with an acceleration of 1.5m/s2,
how far did it travel in 76 seconds? - Answer
- 1. Given initial speed 10m/s, acceleration
1.5m/s2, 76 seconds - 2. Looking for how far distance
- 3. Equation d v1t (at2/2)
- 4. Solution d v1t (at2/2) (10m/s)(76s)
(1.5m/s2)(76s)2/2 5092m - Distance Example 2 If an object starts from rest
with an acceleration of 0.45m/s2, how far did it
travel in 5.6 seconds? - Answer
- 1. Given acceleration 0.45m/s2, 5.6 seconds,
initial speed 0 m/s (at rest) - 2. Looking for how far distance
- 3. Equation d (at2/2)
- 4. Solution d (at2/2) (0.45m/s2)(5.6s)2/2
7.06m
18Acceleration of Gravity
- Before Galileo, philosophers tried to answer
questions about the movement of objects due to
gravity by creating concepts that were so obvious
that there was no need to test them - Aristotle, an ancient Greek thinker, was one of
these famous philosophers - He thought that all falling bodies followed this
basic concept every material has a natural
place where it belonged and toward where it tried
to move - For example, stones are from the earth, so they
naturally fell downward, toward the earth
19Free Fall
- Galileo, an Italian physicist, experimented with
falling bodies 2000 years after Aristotle - He performed his experiments using objects
rolling down an inclined plane instead of falling
in free fall, but his results apply to objects in
free fall as well - He found that the higher an object is dropped,
the greater its speed when it reaches the ground,
meaning that the object has an acceleration - He also found that no matter the size of the
object, whether it is big or small, the
acceleration is the same - Galileo found that if an object that is falling
near the earths surface doesnt have to push its
way through any air, it has an acceleration of
9.8 m/s2 this acceleration is known as the
acceleration of gravity, and is represented by g - When an object is falling from rest, its downward
speed can be calculated by the following
equation vdownward gt, where g is 9.8 m/s2 and
t is the time
20How Far Does a Falling Object Fall?
- To determine how far (h) a falling object falls
from rest in a given time - 1. Start with d at2/2
- 2. If you replace d, distance, with h, height,
the equation is h at2/2 - 3. If you replace a, acceleration, with g, the
acceleration of gravity, the equation is h
gt2/2
21A Falling Object Speed vs. Height
- The downward speed is calculated by the following
equation vdownward gt - This equation shows that the speed is directly
proportional to the time i.e., the speed at 5
seconds is 5 times the speed at 1 second - The height is calculated by the following
equation h gt2/2 - This equation shows that the height is directly
proportional to the time squared (t2) - i.e., the
speed at 5 seconds is 25 times the speed at 1
second - The height thus increases faster with increasing
time than the downward speed
22Example Calculations of Downward Speed and Height
- Downward Speed Example What is the speed of an
object that is falling for 34 seconds? - Answer
- 1. Given 34 seconds, a g
- 2. Looking for speed
- 3. Equation vdownward gt
- 4. Solution vdownward gt (9.8m/s2)(34s)
333.2m/s
- Height Example How far does an object fall after
4.9 seconds? - Answer
- 1. Given 4.9 seconds, a g
- 2. Looking for how far height
- 3. Equation h gt2/2
- 4. Solution h gt2/2 (9.8m/s2)(4.9s)2/2
117.6m
23Thrown Objects
- The acceleration of gravity, g, is the same
whether an object is - just dropped
- thrown downward
- thrown upward
- thrown sideways
- If a ball is just dropped, it goes faster and
faster as it drops until it hits the ground - If a ball is thrown sideways, it has a curved
path that becomes steeper and steeper as it drops - If a ball is thrown upward, gravity first reduces
the upward speed of the ball until its upward
speed is zero. When its upward speed is zero, the
ball is at the top of its path and is momentarily
at rest. Then the ball starts falling toward the
ground, faster and faster, until it hits the
ground. - If a ball is thrown downward, the original speed
of the ball is steadily increased by the downward
acceleration (g)
24Parabolas
- A parabola is a curved path
- When a ball is thrown upward at an angle to the
ground, a parabola occurs - If there is no air resistance, the maximum
distance a ball can travel horizontally occurs
when the ball is thrown upward at an angle of 45
to the ground - If the angle is greater than 45 or less than
45, then the ball travels less than it would
have if it had be thrown at an angle of 45 - There is one angle above 45 and one angle below
45 that if the ball is thrown, it will land in
the same place
25Air Resistance
- Air resistance keeps things from developing the
full acceleration of gravity without air
resistance, anything, including a light shower,
would be dangerous - The faster something moves, the more air
resistance it encounters - i.e. the more the air
slows down its motion - For a falling object, the air resistance
increases with speed until the air resistance
equals the force of gravity on the object. Once
this occurs, the object falls at a constant
terminal speed. The terminal speed of an object
depends on its size, its shape, and how heavy it
is
26Air Resistance for Objects Thrown Upward at an
Angle
- If there is no air resistance (i.e., in a
vacuum), the maximum distance a ball can travel
horizontally occurs when the ball is thrown
upward at an angle of 45 to the ground - However, if there is air resistance, the maximum
distance a ball can travel horizontally doesnt
occur when the ball is thrown upward at an angle
of 45 to the ground it occurs when the ball is
thrown upward at an angle less than 45 to the
ground
27Falling Air vs. Vacuum
- In air, a stone falls faster than a feather
because the air resistance affects the feather
more than the stone - In a vacuum, since there is no air resistance,
both a stone and a feather fall with the same
acceleration, g (9.8 m/s2)
28The First Law of Motion
- If a ball is left alone on the floor, it will not
move unless someone pushes it - If there is a perfectly round ball and a
perfectly smooth floor, a ball that is pushed
will continue to roll forever in the same
direction with the same speed until someone
pushes to make it roll slower, to make it roll
faster, or to change its direction of motion - The First Law of Motion If no net force acts on
it, an object at rest remains at rest and an
object in motion remains in motion at constant
velocity (that is, at constant speed in a
straight line).
29Force
- A force is any influence that can change the
speed or direction of motion of an object. - Examples of forces the force of gravity (pulling
us downward), a magnet pulling a piece of iron, a
person lifting a box, a car pulling a trailer,
etc. - Just because a force is applied to an object,
doesnt mean that the object will move - a force
has to be of a certain amount to be able to move
an object - For example, if you push on a building, you will
not be able to move it because you dont have
enough force - Thus, every force does not result in accelerating
an object but every acceleration does result from
a force - An object continues to accelerate from the
application of a force only as long as the force
is not balanced out by any other forces i.e. it
continues to accelerate as long as there is a net
force
30Inertia and Mass
- Inertia is the reluctance of an object to change
its state of rest or of uniform motion in a
straight line i.e. inertia is what keeps an
object at rest or an object at constant speed in
a straight line - For example, when a car starts, you feel like you
are being pushed back in your seat because your
body wants to remain at rest, and, when a car
stops, you feel like you are being pushed forward
because your body wants to remain in motion - The mass of an object is the amount of matter it
contains. Mass is the property of matter that
shows itself as inertia if something has more
mass, it has more inertia (i.e., a bowling ball
has more inertia than a baseball because it has
more mass than a baseball). In the SI System, the
unit for mass is the kilogram (kg).
31The Second Law of Motion
- The greater the force applied to an object, the
greater the acceleration of that object from that
force AND the smaller the force applied to an
object, the smaller the acceleration of that
object from that force - Thus, the acceleration is directly proportional
to the force that is, if the force is doubled,
the acceleration is also doubled - If the same force is applied to two objects with
different masses, the object with the smaller
mass will have a greater acceleration than the
object with the larger mass - Thus, the acceleration is inversely proportional
to the mass that is, if the mass is doubled,
the acceleration is halved - These two statements give the following equation
- acceleration (a) Force (F) / mass (m) --- a
F/m - This equation can also be expressed in terms of
the force - Force (F) mass (m) x acceleration (a) --- F
ma - The direction of the acceleration is always in
the direction of the force applied to it i.e.,
so if the force is applied to the east, the
object the force is applied to will accelerate to
the east - The Second Law of Motion The net force on an
object equals the product of the mass and the
acceleration of the object. The direction of the
force is the same as that of the acceleration.
32The Newton
- The Force is equal to the mass times the
acceleration F ma - The units for force are therefore the units of
mass times the units of acceleration (kg) x
(m/s2) - The Newton (N) is the unit for force and it is
therefore equal to (kg) x (m/s2) - In the British System, the unit of force is the
pound - 1 N 0.225 lb OR 1 lb 4.45 N
33Example Calculations using the Second Law of
Motion
- Acceleration Example If a 430N force is applied
to an object, what is its acceleration if it has
a mass of 71kg? - Answer
- 1. Given 340N, 71kg
- 2. Looking for acceleration
- 3. Equation a F/m
- 4. Solution a F/m 340N/71kg 4.8m/s2
- Force Example What is the force on an object
that has an acceleration of 1.4m/s2 and a mass of
29kg? - Answer
- 1. Given 1.4 m/s2, 29kg
- 2. Looking for force
- 3. Equation F ma
- 4. Solution F ma 29kg x 1.4 m/s2 41N
34Weight and Mass
- The weight of an object is the force with which
it is attracted by the earths gravitational pull - For example, if you weigh 140 pounds, the earth
(i.e. gravity) is pulling you down with a force
of 140 pounds - Thus, weight is different than mass, which is the
amount of matter it contains i.e., the weight
of an object depends on the gravity, while the
mass doesnt thus, your mass is the same on
every planet, but your weight is different on
every planet because the gravity on every planet
is different - When an object is at the earths surface, an
objects weight (w) is the force that is exerted
on it. This force of gravity causes an object to
fall with constant acceleration due to gravity or
g 9.8 m/s2, as long as no other force acts on
the object. Thus, if we start with F ma and
substitute w for F and g for a, we obtain the
equation for weight - weight (w) mass (m) x acceleration of gravity
(g) - This equation shows weight is directly
proportional to mass i.e., if the mass is
greater, the weight is greater AND if the mass is
less, the weight is less - The weight of an object depends on where the
object is located i.e. the pull of gravity is
not the same everywhere on the earth - The pull of gravity is greater at sea level than
on a mountaintop AND the pull of gravity is
greater near the north and south pole than at the
equator
35Example Calculations using Weight
- Example What is the weight of an object that has
a mass of 54kg? - Answer
- 1. Given 54kg
- 2. Looking for weight
- 3. Equation w mg
- 4. Solution w mg 54kg x 9.8 m/s2 529N
36The Third Law of Motion
- If you push on a heavy object and it doesnt
move, it is because it is pushing back on you
with an equal force. The object stays in place
because there is an opposing force of friction
between the object and the floor it is on. You
stay in place because, just like on the object,
there is an opposing force of friction between
you and the floor you are on. - If you push on a heavy object that is on a
surface without friction, the object moves in the
direction that you push it. Since the object also
pushes back on you, you will move in the opposite
direction that you pushed the object in i.e. in
the direction that the object pushed on you. - The Third Law of Motion When one object exerts
a force on a second object, the second object
exerts an equal force in the opposite direction
on the second object. - In other words, the Third Law of Motion states
that NO force occurs singly i.e. all forces
occur in pairs - For example, you push downward on the earth with
the same force that the earth is pushing up on
you with. Also, a fruit that falls from a tree
due to the pull that the earth has on it has an
equal pull on the earth, even though it doesnt
appear so since the mass of the fruit is so small
when compared to the mass of the earth. - The Third Law of Motion applies to TWO forces
that are on TWO different objects - On one of the objects, there is the action force.
This is the force that the first object exerts on
the second object. - The force that the second object exerts on the
first object is called the reaction force, which
is opposite in direction to the action force. - This law allows us to walk walking is not due
to you pushing on the earth, but instead on the
earth pushing back on you i.e. as you move
forward, the earth is actually moving backward,
even though the amount is so small when you
compare it to its mass
37Circular Motion
- The moon circles around the earth and the planets
circle round the sun because of a force being
applied to them - An inward force must be applied to keep an object
moving in a curved path this force is called
centripetal force - The centripetal force points towards the center
of the objects curved path thus, the force is
perpendicular (at a right angle to) the direction
in which the object is moving
38Centripetal Force
- Centripetal force is calculated from the
following equation - Fc mv2/r
- where Fc is the centripetal force, m is the mass,
v is the speed, and r is the radius of the circle - The equation for centripetal force shows that it
is directly proportional to mass and speed and
inversely proportional to the radius of the
circle - The greater the mass of an object, the greater
the force needed to keep it in circular motion - The greater the speed of an object, the greater
the force needed to keep it in circular motion - The larger the radius of the circle (i.e., the
larger the circle), less force is needed to keep
it in circular motion
39Example Calculations using Centripetal Force
- Example What is the centripetal force on an
object that is moving in a circle with a radius
of 4.7m if its mass is 80kg and its velocity is
42m/s? - Answer
- Given 4.7m, 80kg, 42m/s
- Looking for centripetal force
- Equation Fc mv2/r
- Solution Fc mv2/r ((80kg)(42m/s)2)/4.7m
30026N
40Newtons Path to the Law of Gravity
- If the planets orbit is a circle, with the sun
at the center of the circle, the centripetal
force on the planet is from the sun. Thus, the
force of gravity between the planet and the sun
acts along a line between them. Keplers Second
Law, which concludes that a planet travels faster
when it is near the sun and slower when it is
away from the sun, was used by Newton to verify
that this conclusion works even though the
planets are in elliptical orbits. - To find the force of gravity between the planet
and the sun, Newton combined Keplers Third Law,
which states the following (period of a
planet)2/ (average orbit radius)3 is the same
value for every planet, with the equation for
centripetal force. He found that the force of
gravity (F) varies inversely with the square of
the distance (R) between the planet and the sun
(i.e. F varies with 1/R2). Thus, if a planet is
twice the distance from the sun, the force of
gravity or the gravitational attraction of the
planet is 1/22 or 1/4, AND if a planet is half
the distance from the sun, the force of gravity
or the gravitational attraction of the planet is
1/(1/2)2 or 1/(1/4) or 4. These conclusions are
supported by Keplers First Law, which states
that the planets move in ellipses around the sun,
with the sun at the center of the ellipse,
because only a force like the one described here
can keep the planets moving in a circular orbit - The final step was from Galileos work on falling
bodies, which states that an object in free fall
at the earths surface has the acceleration g.
Because of this, the objects weight, which is
the force of gravity on it, is proportional to
its mass since w mg. In his Third Law of
Motion, he states that every action force has a
reaction force i.e. if the earth attracts an
object, that means that the object is attracting
the earth. Since the earths attraction for the
object i.e. the objects weight depends on
the objects mass (w mg), the objects
attraction for the earth also depends on the
earths mass. Thus, the gravitational force
between two bodies is proportional to the masses
of both of the bodies. - To summarize Every object in the universe
attracts every other object with a force
proportional to the square of the distance
between them
41The Equation of Newtons Law of Gravity
- The Law of Gravity can be shown in the following
equation - F (Gm1m2)/R2
- where F is the gravitational force, G is a
constant of nature (i.e. the same value
everywhere), which equals 6.670 x 10-11 Nm2/kg2,
m1 is the mass of one of the objects, m2 is the
mass of the other object, and R is the distance
between the two objects - R is measured from the objects center of mass.
The location of the center of mass of an object
depends on its shape and how its mass is
distributed. - The equation from the Law of Gravity shows that
the force is directly proportional to the masses
of the objects and inversely proportional to the
distance between the two objects (R) - Thus, if the mass of the one of the objects
increases, then the force of gravity is larger - Also, if the distance between the two objects
increases, the force of gravity is smaller - On the earth, an astronaut weighs 600N. When the
astronaut is 100 times farther from the earth
(i.e. 640,000km), the astronaut weights 1/10O2 or
1/10,000 times as much as she did on earth, i.e.
0.06N, which is the weight of a cigar on the
earths surface
42Example Calculations using Newtons Law of Gravity
- Example What is the gravitational force between
an object that weighs 45kg and an object that
weighs 329kg if the distance between the two
objects is 43m? - Answer
- Given 45kg, 329kg, 43m
- Looking for gravitational force
- Equation F (Gm1m2)/R2
- Solution F (Gm1m2)/R2 ((6.670x10-11
Nm2/kg2)(45kg)(329kg))/(43m)2 5.3 x 10-10N
43Artificial Satellites
- The first artificial satellite, Sputnik I, was
launched in 1957 by the Soviet Union since
then, many other artificial satellites were
launched since then, mostly by the United States
and the former Soviet Union - In 1961, the first person, a Soviet cosmonaut,
circled the earth at a height of 240 km since
then, many other people have been in orbit - The satellites that are circling the earth range
from 130 km above the earth to 36,000 km above
the earth - The satellites that are closer to the earth are
called eyes in the sky. These satellites
monitor the surface of the earth for military
purposes and to provide information on weather
and earth resources (i.e. mineral deposits,
crops, water) - There are 24 satellites at 17,600 km above the
earth that are used for the Global Positioning
System (GPS) that was developed by the United
States. - GPS allows a person to find their position,
including their altitude, to within a few meters,
anywhere in the world at any time - The satellites that are farthest from the earth
circle the earth only once a day they remain in
place indefinitely over a particular location on
the earth. They are in what are called
geostationary orbits. These satellites can see
large areas of the earths surface. About 200 of
these satellites are used to relay telephone,
data, and television communications from one
place to another
44Why Satellites Dont Fall Down
- Satellites are actually falling down, but they
are falling down at such a rate that allows them
to maintain a stable orbit around the earth, just
like the moon, which is a natural satellite - The word stable for the orbit of a satellite is
a relative word because the satellite will
eventually fall down because there is friction
due to the extremely thin atmosphere present
where the satellite is located. The length of
time a satellite spends circling the earth ranges
from a matter of days to hundreds of years. - The gravitational force on a satellite is the
same as the gravitational force on us i.e. its
gravitational force is its weight, mg, where g is
the acceleration of gravity at the satellites
location above the earth. The value of g
decreases with increasing distance from the earth
i.e. a satellite that has an orbit father from
the earth has a lower g than a satellite that has
an orbit closer to the earth.
45The Speed Needed to Orbit the Earth
- For a satellite to circle the earth, it needs a
centripetal force (Fc mv2/r) and since the
force of gravity of the earth (w mg) provides
the centripetal force, the following can be
written - Fc w, so (mv2)/r mg
- If we solve this equation for v2, by multiplying
by r and dividing by m, we obtain the equation
v2 rg, so v v(rg), where, in this case, v is
the speed of the satellite - The second equation shows that the mass of the
satellite does not matter for the speed of the
satellite - For a satellite to circle the earth only a few
kilometers above the earth, the satellite needs a
speed of about 28,400 km/hour. If the speed of
the satellite is less than this, the satellite
would just fall back to earth. If the speed of
the satellite is more than this, the satellite
would orbit the earth in an elliptical orbit
instead of a circular one. For a satellite that
initially has an elliptical orbit to have a
circular orbit, a small rocket motor gives it a
push at the required distance from the earth. - If the speed of an object is greater than 40,000
km/hour, it can escape entirely from the earth.
This speed is the escape speed that an object
needs to leave the earth. The escape speed is the
speed that is required for something to leave the
gravitational influence of an astronomical body
permanently. The ratio between the escape speed
and the minimum orbit speed is v2 or 1.41.