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Energy
Physics 2013
Energy Intro
• Isaac Newton almost singe-handedly
invented the science of mechanics,
but there is one concept he missed!
• Energy comes in many
different forms and has
many definitions. But the
most basic definition of
energy comes in terms of work.
• Energy is the capacity to do work!
• The release of energy does work –
and doing work on something adds
energy to it.
• Energy and work are EQUIVALENT
CONCEPTS
and we can say that
 same units! (J)
http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Energy/Images/WhatIsWorkAnim.gif
Energy
Forms of Energy
Chemical
Electrical
Nuclear
Thermal
Copywrited by Holt, Rinehart, & Winston
Show Me the Money
Income
Liquid
Asset:
Cash
Transformatio
ns within
system
System
Saved
Asset:
Stocks
Expense
s
Transfers
into and
out of
system
The Basic Energy Model
Kinetic Energy
• If you throw a ball. You do work getting
the ball moving: You exert a force over
a distance.
• The ball has acquired some energy, the
energy of motion, or Kinetic Energy (KE)
• A “simple” mathematical derivation
shows that K.E. = ½ mv2
• Where m = mass (kg)
and v = velocity in (m/s)
Kinetic Energy Derivation
• Assume: Vo = 0
• Therefore,
KE = W = Fd = mad
• F = ma
vf
a
• vavg = vavg
vf
d

2
t
• Rearranging,
d
vft
2
t
• Substituting,
KE  m
vf
t

vft
2
KE = ½ mv2
• Kinetic Energy = ½
mass x speed2
• Joules = kg (m/s)2
Kinetic Energy
• Kinetic Energy Energy of Motion
Example Problem:
How much kinetic energy does a .25
kg baseball that is thrown with a
velocity of 9.3 m/s have?
m = .25 kg
v = 9.3 m/s
KE = 1/2mv2 =10.8 J
Work – Energy Theorem
• the net work done on an object
equals its change in kinetic energy.
1
1
2
2
W  K E f  K E o  m v f  m vo
2
2
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Example
A space probe of mass
m = 5.00 x 104 kg is
traveling at a speed of
vo = 1.10 x 104 m/s
through deep space. No
forces act on the probe
except that generated by
its own engine. The
engine exerts a constant
external force of
magnitude 4.00 x 105 N,
directed parallel to the
displacement of
magnitude 2.50 x 106 m.
Determine the final
speed of the probe.
m = 5.00 x 104 kg
vo = 1.10 x 104 m/s
F = 4.00 x 105 N
d = 2.50 x 106 m
vf = ?
1
1
2
2
W  K E f  K E o  m v f  m vo
2
2
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Potential Energy
• Suppose you lift an object to a certain height.
• As you exert a force (the object’s weight)
over the distance, you do work.
• The object isn’t moving in the end, but
energy has still has added because of where
it is in the Earth’s gravitational field.
• This type of energy is called Potential
Energy (energy of position) and is written
PE or U where
 U  W
Potential Energy
PE is divided into categories according
to its system:
• Mechanical
– Gravitational
– Elastic
•
•
•
•
Chemical
Electrical
Nuclear
Thermal
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Gravitational PE
• Work is required to elevate objects
against Earth’s gravitational field.
The potential energy due to elevated
positions is called gravitational PE.
GPE = PE = mgh
At A:
the body
possesses
G.P.E. because
the force of
gravity acts on
it.
At B:
the body
possesses more
G.P.E. than it did
at A because
work has been
done lifting it.
Gravitational PE
Calculated relative to a reference point
Path Independent
http://www.physicsclassroom.com/Class/energy
Calculate the PE at points A-E
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Copywrited by Holt, Rinehart, & Winston
Elastic Potential Energy
• Consider a spring being extended by a steadily
increasing force.
• When the force has a magnitude F,
the extension is x.
• This relationship is known as
Hooke’s Law
Fs=-kx
• Fs=restoring force of a spring
• k = spring constant
• x= displacement of the spring
Note: the negative sign indicates that the restoring
force of a spring always points in a direction
opposite to the displacement of the spring.
Copywrited by Holt, Rinehart, & Winston
Elastic Potential Energy
• A graph of force against extension would
be a straight line passing through the
origin, as long as the elastic limit of the
spring has not been exceeded (Hooke’s
law).
• The slope of the graph is k, the elastic
constant of the spring F = kx
Elastic Potential Energy
• The work done increasing the length of
the spring by x is given by:
• W = average force × extension
• The area under the curve or
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
W = ½ Fx = ½ kx²
• So, the elastic potential energy stored in
the spring is given by:
E.P.E. = ½ kx²
PEKE
• Potential energy is “potential” because it can
be gotten back as “real” kinetic energy.
• All you have to do is let the object fall!
• As the object falls faster and faster its potential
energy is slowly converted to kinetic energy.
• At the bottom, just before impact, its potential
energy is gone and its original potential energy
has become entirely kinetic.
• Mathematically, that is:
Example Problem
A 15 kg ball falls from a roof 8.0
meters about the ground. What is its
kinetic energy as it hits the ground?
K/U m=15 kg h = 8.0 m g = 9.81 m/s2
K E  P E  m g h  1 5 ( 9 . 8 1 )( 8 . 0 ) 
KE = 1177 J ~1200 J
Solving for speed  K E  P E 
v
2 gh 
1
2
m v2  m gh
2 ( 9 . 81)( 8 )
v = 12.53 m/s ~13 m/s
Clicker Understanding
• A child is on a playground swing,
motionless at the highest point of his
arc. As he swings back down to the
lowest point of his motion, what energy
transformation is taking place?
A. K  Ug
B. Ug  Eth
C. Us  Ug
D. Ug  K
E. K  E th
Clicker Understanding
• A skier is moving down a slope at a
constant speed. What energy
transformation is taking place?
A. K  Ug
B. Ug  Eth
C. Us  Ug
D. Ug  K
E. K  E th
Conservation of Energy
• The equality ½ mv2 = mgh is an example of
Conservation of Energy
• Energy can neither be created nor
destroyed, but can only be converted from
one form to another.
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Conservation of Energy
Look at the impact itself. Some of the energy is
• What into
happens
to angoes
object’s
energy
converted
sound. Some
into distorting
the floor
– and
distorting
the object
thatfloor
matter.
Some
– really
when
it finally
hitsfor
the
and
both
its
most
– goes and
into heat.
The object
and the
floor
are both
kinetic
potential
energy
are
zero?
a little warmer after the collision. The impact jiggles their
molecules – and heat is nothing both the kinetic energy of
billions of molecules!
Conservation of Energy
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Principle of Conservation of
Mechanical Energy
• The total mechanical energy (E = KE + PE)
of an object remains constant as the object
moves, provided that the net work done by
external nonconservative forces is zero.
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Conservative Forces
1. The work it does on a moving object is
independent of the path of the motion between the
object's initial and final position.
2. The work it does moving an object around a
closed path is zero
3. The work it does is stored in the form of energy
that can be released at a later time.
4. Work done by a conservative force is the negative
of the change in the potential energy W    U
c
Conservative Force
• Example: Gravity, Spring Force
Non-conservative Force
Non-conservative Forces
1. Path dependent
2. Work done on a closed path is not zero.
3. Work done by a nonconservative force
changes a system’s mechanical energy.
4. Work done by a nonconservative force equals
the change in a system’s mechanical energy.
Examples: Friction, Air resistance, tension, force
exerted by muscles, force exerted by motor
Principle of Conservation of
Mechanical Energy
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Principle of Conservation of
Mechanical Energy
http://www.physicsclassroom.com/mmedia/energy/ie.html
Principle of Conservation of
Mechanical Energy
http://www.physicsclassroom.com/mmedia/energy/hh.html
Principle of Conservation of
Mechanical Energy
http://www.physicsclassroom.com/mmedia/energy/pe.html
Conservation of Energy
Elastic Potential Energy
(Bungee)
Real Bungee
Principle of Conservation of
Mechanical Energy
1. Identify the forces
(conservative and
nonconservative)
– Any nonconsevative
force must do no work.
2. Choose the location
where the GPE is
zero.
3. Set the final total E of
the object equal to the
initial total E.
–
What is the speed of
the bicyclist at the
bottom of the hill
assuming that she is
coasting down the
hill?
The total E is the sum
of the KE and PE.
http://k12.albemarle.org/Instruction/Physics/energy2/test1c.htm
Principle of Conservation of
Mechanical Energy
1. Fweightconservative
2. GPE is zero at
bottom of the hill.
3.
E f  Eo
1
1
2
2
mv f  mgh f  mv o  mgho
2    
2   

Ef
vf 
What is the speed of
the bicyclist at the
bottom of the hill
assuming that she is
coasting down the
hill?
Eo
v o  2 gh o
2
v f  11 2  2 ( 9 . 8 )(15 )
Vf = 20.4 m/s
http://k12.albemarle.org/Instruction/Physics/energy2/test1c.htm
Work-Energy Theorem
(Expanded)
• The net work done by all the external
nonconservative forces equals the
change in the object's kinetic energy plus
the change in its gravitational potential
energy.
Wnc =  KE +  PE
Wnc = (KEf - KEo) + (PEf - PEo)
Wnc = (1/2mvf2 - 1/2mvo2) + (mghf - mgho)
Work-Energy Theorem
http://www.physicsclassroom.com/Class/energy
Work-Energy Theorem
• A hotwheels car sliding down a slope
into a cup.
http://www.physicsclassroom.com/mmedia/energy/hw.html
Work-Energy Theorem
http://www.physicsclassroom.com/mmedia/energy/se.html
Work-Energy Theorem
• Braking Problem
http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Energy/Images/BrakingDistance.gif
http://www.physicsclassroom.com/mmedia/energy/cs.html
Braking Distance
Law of Conservation of
Matter and Energy
• Energy cannot be created or destroyed;
it may be transformed from one form to
another, or transferred from one place to
another, but the total amount of energy
never changes.
• Matter and energy are interchangeable
only when the total amount of matter is
converted.
E = mc2
• c = 3 x 108 m/s (speed of light)
Planetary Motion &
Gravitational Fields
Historical
Background
Our ancestors envisioned the earth
at the center of the universe and
considered the stars to be fixed on
a great revolving crystal sphere.
The observation of planetary
motions depends on the frame of
reference of the viewer.
Early Theories
Geocentric Theory: Ptolemy~140 AD
• Motion of planets from the point of
view of the observer on Earth
Heliocentric Theory: Copernicus, 1543.
• Same motions but from the point of
view of the observer on the Sun
We use the heliocentric theory due to its
simpler mathematics.
Tycho Brahe (1546-1601)
The Royal Astronomer for King
Fredrick II of Denmark,
•
•
•
•
•
measured accurately the
positions of the planets
Charted planets motion for 20
years
Accurate to 1/60 of a degree
“Greatest Observational
Genius”.
35 years before the telescope.
Johannes Kepler
• Imperial Mathematician for Holy
Roman Emperor Rudolf II
• In 1600 became Brahe’s assistant.
• Stole Brahe’s data after his death in
1601, spent 15 years analyzing it.
• Deduced 3 laws of planetary motion
• Apply to any system composed of a
body revolving about a more massive
body
• Examples: moons, satellites, comets.
Kepler’s 1st Law
The Law of Orbits
Planets move in elliptical orbits
with the Sun at one of the focal
points.
Copyright ©2007 Pearson Prentice Hall, Inc.
Kepler’s
dA
k
dt
where k is a
constant
nd
2
Law
The Law of Areas
A line from the Sun to a
planet sweeps out equal
areas in equal lengths of
time.
Copyright ©2007 Pearson Prentice Hall, Inc.
Kepler’s
rd
3
Law
The Law of Periods
(the Harmonic Law)
The square of the orbital period of a
planet is directly proportional to the
cube of the average distance of the
planet from the sun.
2
3
T  r
• The ratio of the squares of the
periods of any two planets is equal to
the ratio of the cubes of their mean
distances from the sun.  T1  2  r1 3
    
 T2 
 r2 
Finding Kepler’s 3rd Law
with a Graphing Calculator
James Metz, The Physics Teacher, April 2000
Planet
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Kepler struggled
for 15 years to learn the
Kepler’s Values
relationship between the period (T) and the
The
calculator
return
y
= can
radius
(r)(r)
of
orbit
the
planets,
you
Radius
ofthe
orbit
of ofshould
Period
(T)
inbut
days
364.2740081x
easily
determine
relationship with a graphing
planet
in A.U. this 1.5
calculator.
0.389
87.77
Since the period is y and the
0.724
224.70
Using
a TI-83
3/2
radius
is
x,
T
=
364.2740081x
Press the STAT key
1.000
365.25
Select EDIT
Thus,
the of
period
the planet
a planet
Enter
the radius
orbit ofofeach
in the
1.524
686.98
table is
below
in L1 proportional to the 3/2
directly
4332.62
The 5.200
corresponding
period
in
L2
power of its orbital radius,
9.510
10
759.20
Kepler’s
Third
Law
of
Planetary
After entering the data
PressMotion.
STAT
Select CALC
Isaac Newton
• Recognized that a force must be
acting on the planets; otherwise, their
paths would be straight lines.
• Force  Sun (Kepler’s 2nd Law)
• Force decreases with the square of
the distance (Kepler’s 3rd Law)
Newton Continued…
Gravity
• Compared the fall of an apple with the fall of
the moon.
• The moon falls in the sense that it falls
away from the straight line it would
follow if there were no forces acting on it.
• Therefore, the motion of the moon and
the apple were the same motion.
• Showed, everything in the universe follows a
single set of physical laws.
• Law of Universal Gravitation.
Law of Universal Gravitation
(1666)
Every particle in the universe attracts every other
particle with a force.
This force is:
• proportional to the product of their masses
• inversely proportional to the square of the distance
between them.
• Acts on a line joining the two particles.
Gm1m 2
F
2
-11
2
2
r
G = 6.67 x 10 Nm /kg
Discovered by Henry Cavendish in 1798
Law of Universal Gravitation
What is the force of attraction between a 4.8
kg Physics book and a 2.4 kg Math book if
they are 5.0 cm apart?
Gm1m2
Fg 
2
r
G = 6.67 x 10-11 Nm2/kg2
Rule of 1’s Practice
Gm1m2
Fg 
r2
a. If m1 is doubled, how is F changed?
b. If neither of the masses were changed but r was
doubled, how would F change?
c. If r is not changed but both masses are doubled,
how would F change?
d. If r is halved and both masses doubled, how would F
change?
Acceleration due to Gravity
When only gravity is present,
Fg = Fnet where Fnet = ma (Newton’s 2nd law)
Fnet = ma =
So,
ma =
Note: a = g
GMm
2
r
Gm1m2
Fg 
2
r
GM
g 2
r
Changes in
Gravity
Acceleration due to Gravity on
different Planets
Consider an object thrown upward
v f  v o  2 ad
2
2
at the maximum height vf=0
v o   2 gd
2
If the same object was thrown upward with the same
initial speed gd  constant
So,
g 1d 1  g 2 d 2
g1 d 2

g 2 d1
Acceleration due to Gravity on
different Planets
• Example
A person can jump 1.5m on the earth.
How high could the person jump on a
planet having the twice the mass of the
earth and twice the radius of the earth?
Gravitational Field
Strength (g.f.s)
The force acting on a 1 kg mass at a
specific point in a gravitational field.
F
g
m
Units: N/kg
http://www.chrisbence.com/projects/powerplant/img/gravity_zoomed_mass.gif
From Newton’s 2nd law F/m = a
g.f.s is the acceleration due to gravity (g)
GM
g 2
r
Variation of g
With distance from a point mass
• g.f.s at a point p is
GM
g 2
r
• Variation of g with distance from the
center of a uniform spherical mass of
radius, R
Variation of g
GM
g 2
r
From
• Close to earth the acceleration due to gravity
changes slowly.
• As the distance from earth increases, the
change in the acceleration due to gravity
increases.
www.physicsclassroom.com/Class/energy/U5L1b.cfm
Variation of g
On a line joining the centers of two
point masses
• If m1 > m2 then
Note: when g = 0
the gravitational
fields cancel each
other out
Gravitational Potential Energy
(GPE) of two point sources
G P E   W   Fg d
• Gravitational force is not constant but
decreases as the separation of masses
increase.
• Requires calculus to compute the total
R
GMm
work done.
W 
G Mm
W 
r
R


G Mm

R
r
2
dr
G Mm
G PE  
R
Gravitational Potential (V)
Background:
The potential at a point in a gravitational field is
equal to the work done bringing a 1kg mass
from infinity to that point
 W G PE
V 

m
m
G Mm
V 
mr
Recall,
G Mm
G PE  
r
GM
V 
r
Units: J/kg
Gravitational Potential (V)
Potential (V) is the GPE per unit mass in a
gravitational field.
GM
V 
Why the negative sign?
r
• A body at infinity, has zero gravitational
potential.
• A body normally falls to its lowest state
of potential (energy)  as r decreases,
the potential decreases.
• We can do this by including a negative
sign in the above equation.
• As r decreases, V decreases
(becomes a greater negative quantity).
Source: http://www.saburchill.com/physics/chapters/0007.html
Equipotential Surfaces
Equipotential Surface – points that all
have the same potential (V) where
GM
V 
r
•
Formula shows that the
points will all be the same
distance from the mass or
lie on a sphere whose
center is that of a
spherical mass.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Equipotential Surfaces
Equipotential Surface – points that all
have the same potential (V)
• This can be shown two
dimensionally where the
blue lines represent the
equipotential lines
Similar to topographical maps
http://www.iop.org/activity/education/Projects/Teaching%20Advanced%20Physics/Fields/Images%20400/img_tb_4719.gif
Equipotential surfaces
due to two equal masses.
• Note that the field becomes
more spherical further away
from the masses
• At a far enough distance the field appears
to come form one mass with the combined
mass of the individual masses.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Equipotential Surfaces
Equipotential Surfaces
• Equipotential surfaces due to two
unequal masses
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Field Lines
Field Lines – Indicate the direction of
gravitational force that would act on a
point mass at that location.
1. For a gravitational field,
the lines will always
point inward as gravity
is always an attractive
force.
2. Note that the field lines
are perpendicular to
the equipotential lines.
http://www.intuitor.com/student/Planet%27s_g-field.jpg
http://www.chrisbence.com/projects/powerplant/img/gravity_zoomed_mass.gif
www.chrisbence.com/projects/powerplant/index.htm
Field Lines
Consider moving a test mass from point a to b as
shown where the distance between a and b is
r and the difference in potential is ΔV.
The work W required is
W  mV
W  Fg r  m gr
Setting these expressions
equal to each other
V
g
r
A more precise result based on using Calculus results in
dV
g
dr
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
The work done could also
be calculate as
Field Lines
V
g
r
This expression implies that on a graph of
the variation of gravitational potential with
distance (V vs r), the slope would be equal
to the gravitational field strength g
This also implies that the
equipotential surfaces are
always perpendicular to the
gravitational field lines.
• As shown for a single mass
• True for any shape field lines
and equipotential surfaces
http://www.intuitor.com/student/Planet%27s_g-field.jpg
Escape Velocity
• The velocity needed to overcome
(escape) the gravitational pull of a
planet.
• As the body is moving away from the
planet, it is losing kinetic energy and
gaining potential energy.
• To completely escape from the
gravitational attraction of the planet, the
body must be given enough kinetic
energy to take it to a position where its
potential energy is zero.
Escape Velocity
• The potential energy possessed by a body
of mass m, in a gravitational field is given
by
GPE = Vm
• If the field is due to a planet of mass M and
radius R, then the escape velocity can be
calculated as follows:
ΔKE = ΔGPE
• So,
2GM
vesc 
R
• as g = GM/R²
v esc 
2 gR
1 mv 2 esc  GMm
2
rE
Satellites: Orbits & Energy
G Mm using Newton’s second law (F=ma)
U 
r
G Mm
v2
F 
m
2
r
r
Solving for v2
where v2/r is the
centripetal acceleration.
GM
v 
r
2
G Mm
KE  mv 
So,
2r
1
2
2
U
KE  
2
Total Mechanical Energy of a satellite is:
G Mm G Mm
ME  KE  U 

2r
r
ME  
G Mm
2r
Satellites: Orbits & Energy
ME = -KE
This is for a circular orbit.
• For a satellite in an elliptical orbit with a
semi-major axis a
G Mm
E
2a
where a was substituted for r
Energy Graphs
• As shown the energies for a satellite are,
G Mm
KE 
2r
• Graphing these,
http://www.opencourse.info/astronomy/introduction/06.motion_gravity_laws/energy_elliptical.gif
G Mm
U 
r
G Mm
ME  
2r
Energy Wells
• If we rotate the potential energy
graph, we can generate a 3D picture
known as an energy well.
http://www.opencourse.info/astronomy/introduction/06.motion_gravity_laws/
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Energy Changes in a Gravitational
Field
• A mass placed in a gravitational field
experiences a force.
– If no other force acts, the total energy will
remain constant but energy might be
converted from g.p.e. to kinetic energy.
• If the mass of the planet is M and the
radius of the orbit of the satellite is r, then
it can easily be shown that the speed of
the satellite, v, is given by
– if r decreases, v must increase.
Energy Changes in a Gravitational
Field
• If the satellite’s mass is m, then the kinetic
energy, K, of the satellite is K  1 m GM 
2  r 
• the potential energy of the satellite, U, is
These equations show:
 GM 
U  m

 r 
– that if r decreases, K increases but U decreases
(becomes a bigger negative number)
– the decrease in U is greater than the increase in K.
• Therefore, to fall from one orbit to a lower orbit,
the total energy must decrease.
– work must be done to decrease the energy of the
satellite if it is to fall to a lower orbit.
Energy Changes in a Gravitational
Field
• The work done, W, is equal to the change in the
total energy of the satellite, W = ΔK + ΔP.
• This work results in a conversion of energy
from gravitational potential energy to
internal energy of the satellite (it makes it
hot!).
• Air resistance reduces the speed of the satellite
along its orbit. This allows the satellite to fall
towards the planet. As it falls, it gains speed.
• So, if a viscous drag (air resistance) acts on a
satellite, it will
– decrease the radius of the orbit
– increase the speed of the satellite in it’s new orbit.
Energy Changes in a Gravitational
Field
• In principle, the satellite could settle in a
lower, faster orbit.
• In practice it will usually be falling to a
region where the drag is greater. It will
therefore continue to move towards the
planet in a spiral path.
Energy Changes in a
Gravitational Field
• Looking at this with a fbd of the satellite
spiraling downward due to air resistance.
– Note that the velocity is no longer perpendicular to
the weight.
– Due to this there is a component of gravity which is
now acting in the direction of the velocity providing
a net force which is accelerating the satellite.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Weightlessness
Astronauts on the orbiting space station are
weightless because...
a. there is no gravity in space and they do
not weigh anything.
b. space is a vacuum and there is no gravity
in a vacuum.
c. space is a vacuum and there is no air
resistance in a vacuum.
d. the astronauts are far from Earth's surface
at a location where gravitation has a
minimal affect.
e. None of the above
Source: http://www.physicsclassroom.com/Class/circles/u6l4d.cfm
Weightlessness
Source: Kirk, Tim, Physics for the IB Diploma, Oxford University Press 2007
Astronauts on the orbiting space station are weightless
because...
The astronaut and the space station are both free-falling
together - Apparent Weightlessness
• The gravitational force on the astronaut provided the
needed centripetal acceleration for the astronaut to
stay in orbit.
• The space station remains in orbit because of the
gravitational force on it.
• However, since there is no contact force between the
satellite and the astronaut here is an apparent
weightlessness
Weightlessness
Source: Kirk, Tim, Physics for the IB Diploma, Oxford University Press 2007
Weightlessness
Astronauts on the orbiting space station are
weightless because...with calculations
• Summing the forces on the astronaut
F  F
g
• For circular motion
(Will learn in the next unit)
•
•
•
•
 FN  Fnet
Fnet
v2
m
r
Mm
v2 m  M

N G
 m  G
 v2 
r2
r
r r

Mm
v2
 F G r 2  FN  m r
So,
or
Recall for a satellite, v  GM
r
Substituting, N  m  G M  G M   0
r r
r 
With no reaction force (normal) the astronaut
feels weightless
2
Source: Physics for the IB Diploma 5th Edition (Tsokos) 2008