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HSC Physics Topic 1
SPACE
What is this topic about?
To keep it as simple as possible, (K.I.S.S.) this topic involves the study of:
1. GRAVITY & GRAVITATIONAL POTENTIAL ENERGY
2. PROJECTILES & SATELLITES
3. NEWTON’S LAW OF UNIVERSAL GRAVITATION
4. EINSTEIN’S THEORY OF RELATIVITY
...all in the context of the universe and space travel
but first, an introduction...
Mass, Weight & Gravity
You will study how Gravity is responsible for
holding the Solar System together...
were covered briefly in the Preliminary Course.
In this topic you will revise these concepts, and
be introduced to the concept of “Gravitational
Potential Energy”.
Then, you move on to study two important forms
of motion that are controlled by gravity...
Projectiles...
Earth is in a
gravitational orbit
around the Sun
Once launched,
the path of a
projectile is
entirely
determined by
gravity.
and study a variety of aspects of Physics that
relate to Space Travel
...and Satellites in Orbit.
1970’s Apollo mission
to the Moon
There are over 1,000
artificial satellites in
Earth orbit.
Some provide
communication links
for telephone,
internet and TV.
Launch &
Re-e
entry
are the
tricky bits...
Orbiting is
simple
Physics!
Others watch the
weather, or study
patterns of land use,
or search for natural
resources.
Some are for
military
surveillance.
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In the final section you will study one of the most
famous (and least understood) theories of Science:
Einstein’s Theory of Relativity
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CONCEPT DIAGRAM (“Mind Map”) OF TOPIC
Some students find that memorising the OUTLINE of a topic helps them learn and
remember the concepts and important facts. As you proceed through the topic,
come back to this page regularly to see how each bit fits the whole.
At the end of the notes you will find a blank version of this “Mind Map” to practise on.
Gravitational
Acceleration
“g”
Mass
&
Weight
Projectile
Motion
Height, Range,
Time of Flight, etc
Gravitational
Potential
Energy
Satellites
&
Orbits
Gravity &
Gravitational
Fields
Projectiles
&
Satellites
Circular
Motion
Kepler’s
Law of Periods
SPACE
Gravitational
Fields
Newton’s
Law of
Universal
Gravitation
Einstein’s
Theory of
Relativity
Michelson-Morley
Experiment
& its Significance
Evidence
Supporting
Relativity
Einstein’s
Idea
and the
Consequences
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Law of
Universal
Gravitation
Gravity
&
Space
Probes
Frames of
Reference &
Relativity
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1. GRAVITY & GRAVITATIONAL POTENTIAL ENERGY
Gravitational Field
Weight & Gravity
In one way, Gravity resembles electrical charge
and magnetism... it is able to exert a force on
things without touching them. Such forces are
explained by imagining that there is an invisible
“Force Field” reaching through space.
You should already be aware that the “Weight” of
an object is the Force due to gravity, attracting the
object’s mass toward the Earth. You also know that
(ignoring air resistance) all objects near the Earth
will accelerate downwards at the same rate. This
acceleration rate is known as “g”, and is
approximately 10ms-2.
Gravitational fields are imagined to surround
anything with mass... that means all matter, and
all objects. The field exerts a force on any other
mass that is within the field.
Weight = Mass x Acceleration due toGravity
W = mg
Unlike electro-magnetism, gravity can only
attract; it can never repel.
Weight is in newtons (N)
Mass in kilograms (kg)
“g” is acceleration in ms-2.
Of the various “field forces”, Gravity is by far
the weakest, although when enough mass is
concentrated in one spot (e.g. the Earth) it
doesn’t seem weak!
Measuring “g”
How This Relates
to “g”
One of the first activities you may have done in
class would have been to determine the value
of “g”, the acceleration due to gravity.
By accurately timing (say) 10 swings of the
pendulum, and then dividing by 10, the Period
(T) can be measured. This value needs to be
squared for graphing.
Len
gth
in m
etre
s
A common experimental method to do this
involves using a pendulum.
It turns out that the rate at
which a pendulum swings
(its Period) is controlled
by only 2 things:
• its length, and
• the acceleration due to
gravity
Mathematically,
The length of the pendulum (L) is also
measured as accurately as possible.
Typically, the measurements are repeated for
several different lengths of pendulum, then the
results are graphed as shown.
T2 = 4π2
L
g
You are NOT required to
know this equation.
3.0
Lin
eo
fb
es
tf
it
• The straight line graph shows there is a direct
relationship between the Length (L) and the (Period)2.
(s2)
2.0
so,
Time taken for 1 complete
(back-a
and-fforth) swing is
called the “Period” of the
pendulum (“T”)
Analysis
• Gradient,
T2 = 4π2 ≅ 4.0
L
g
Therefore,
g ≅ 4π2/4.0 = 9.9 ms-2.
1.0
Accepted value, g = 9.81ms-2
Gradient =
T2 ≅ 4.0
L
0
(Period)2
T2 = 4π2L
g
0
0.5
1.0
Length of Pendulum (m)
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Explanations for Not Getting Exact Value:
The main causes of experimental error are any jerking,
stretching or twisting in the string, which causes the
pendulum swing to be irregular. This is why the most
accurate results will be obtained with very small, gentle
swings.
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Gravity and Weight
on Other Planets
Gravitational Potential Energy(GPE)
Potential Energy is commonly defined as the
energy “stored” in an object. In the case of any
object on or near the Earth, the amount of GPE
it contains depends on
We are so used to the gravity effects on Earth
that we need to be reminded that “g” is different
elsewhere, such as on another planet in our
Solar System.
• its mass
• its height above the Earth
Since “g” is different, and
W = mg
it follows that things have a different weight if
taken to another planet.
If that object is allowed to fall down, it loses
some GPE and gains some other form of energy,
such as Kinetic or Heat. To raise the object
higher, you must “do work” on it, in order to
increase the amount of GPE it contains.
Values of “g” in Other Places in the Solar
System
Planet
However, for mathematical reasons, the point
where an object is defined to have zero GPE is
not on Earth, but at a point an infinite distance
away. So GPE is defined as follows:
Earth
Mars
Jupiter
Neptune
Moon
Gravitational Potential Energy
is a measure of the work done
to move an object from infinity,
to a point within the gravitational field.
g
(ms-2)
9.81
3.8
25.8
10.4
1.6
g
(as multiple of Earth’s)
1.00
0.39
2.63
1.06
0.17
This definition has an important consequence:
it defines GPE as the work done to bring an
object towards the Earth, but we know that you
need to do work to push an object (upwards)
away from Earth.
Therefore, GPE is, by definition, a negative
quantity!
GPE = -GmM
R
G = Gravitational Constant (= 6.67x10-11)
m = mass of object (kg)
M = mass of Earth, or other planet (kg)
R = distance (metres) of mass “m” from the
centre of the Earth
Calculating a Weight
on another Planet
Example
If an astronaut in his space
suit weighs 1,350N on Earth,
what will he weigh on Mars
where g=3.84ms-2?
Note: the HSC Syllabus does NOT require you to
carry out calculations using this equation. You ARE
required to know the definition for GPE.
In the interests of better understanding, here is an
example of how the equation could be used:
Solution
W = mg
On Earth, 1,350 = m x 9.81
∴ mass = 1,350/9.81
= 137.6 kg
How much GPE does a 500kg satellite have when in
orbit 250km (= 250,000m) above the Earth’s surface?
(Earth’s mass = 5.98x1024kg,
Earth radius = 6.38x106m)
Solution
So on Mars, W = mg
= 137.6x3.84
= 528kg.
GPE = -GmM
R
= -6.67x10-11x500x5.98x1024
(6.38x106 + 250,000)
= -3.00x1010 J.
WORKSHEET at end of this section
The negative value is due to the definition of GPE.
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Worksheet 1
Gravity & GPE
Fill in the blank spaces.
Student Name...........................................
Any mass within a gravitational field
possesses “Gravitaional Potential
Energy” (GPE). This is defined as “the
amount of l)................................ to move
an object from m)................................ to
a point within the field.” In reality, work
must be done to move any mass in the
opposite direction, so the definition
means that the value for GPE is always
a n)................................... quantity.
The weight of an object is the
a)...................... due to b)..........................
Near the Earth, all objects will
c)................................... at the same rate,
approximately d)..................ms-2
Experimentally, “g” can be easily
determined by measuring the length
and e).............................. of a pendulum.
When the results are graphed
appropriately, the f)................................
of the graph allows calculation of “g”.
The value of “g” at the surface of the
Earth is o)...................ms-2, but has a
different value in other places, so the
p)................................. of any object will
be different on a different planet.
However, the q)........................ will
remain the same.
Gravity acts at a distance by way of a
g)...................... ........................ the same
as electro-magnetism, but the force only
h).......................
and
can
never
i)....................... Gravity is a property of
“mass”; every object is surrounded by a
j).................................... which will attract
any other k)........................... within the
field.
Worksheet 2
Mass & Weight
COMPLETED WORKSHEETS
BECOME SECTION SUMMARIES
Practice Problems
Student Name ...........................................
1.
A small space probe has a mass of 575kg.
2.
If a martian weighs 250N when at home, what
will he/she/it weigh:
a) on Earth? (hint: firstly find the mass)
a) What is its mass
i) in orbit?
ii) on the Moon?
b) on Neptune?
iii) on Jupiter?
c) on the Moon?
b) What is its weight
i) on Earth?
3.
A rock sample, weight 83.0N, was collected by a
space probe from the planet Neptune.
a) What is its mass?
ii) on the Moon?
b) What will it weigh on Earth?
iii) on Jupiter?
c) On which planet would it weigh 206N?
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2. PROJECTILES & SATELLITES
What is a Projectile?
Projectile Motion
A projectile is any object that is launched, and
then moves only under the influence of gravity.
By simple observation of a golf ball trajectory, or
a thrown cricket ball, the motion of any
projectile can be seen to be a curve. It is in fact
a parabola, and you might think the Physics of
this is going to be difficult. NOT SO... it is really
very simple. Just remember the following:
Examples:
Once struck, kicked or
thrown, a ball in any sport
becomes a projectile.
Horizontal Motion
is CONSTANT VELOCITY
Vertical Motion
is CONSTANT ACCELERATION
at “g”, DOWNWARDS
Projectiles
You must analyse projectile motion as 2
separate motions; horizontal (x-axis) and
vertical (y-axis) must be dealt with separately,
and combined as vectors if necessary.
Any bullet,
shell or bomb
is a projectile
once it is fired,
launched or
dropped.
The Trajectory (Path)
of a Projectile
At any instant, the projectile’s
position or velocity is the vector sum
of horizontal + vertical components
An example which is
NOT a Projectile:
Not a
Projectile
Uy
U
θ angle of launch
A rocket or guided missile,
while still under power, is
NOT a projectile.
Maximum Height
The Intitial Launch
Velocity has horizontal &
vertical components
Horizontal
Velocity
Vx
Vertical
Velocity
Vy
Ux
“Range” = Total Horizontal Displacement
Equations for Projectile Motion
Once the engine stops
firing it becomes a
projectile.
1. Resolve the Initial Launch Velocity into
Vertical & Horizontal Components
U
Sin θ = Uy & Cos θ = Ux
U
U
θ
Projectiles are subject to
only one force...
Gravity!
∴ Uy = U.Sin θ,
Ux = U.Cos θ
Ux
When a projectile is travelling through air, there
is, of course, an air-resistance force acting as
well. For simplicity, (K.I.S.S. Principle) airresistance will be ignored throughout this topic.
2. Horizontal Motion is constant velocity, so
In reality, a projectile in air, does not
behave the way described here
because of the effects of air-resistance.
3. Vertical Motion is constant acceleration at “g”
Vx = Sx
t
is all you need
To find vertical velocity:
Vy = Uy + g.t
(from v=u+at)
To find vertical displacement:
Sy = Uy.t + 1.g.t2 (from S=ut+ 1at2)
2
2
The exact motion depends on many
factors and the Physics becomes very
complex, and beyond the scope
of this course.
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Uy
The syllabus specifies a 3rd equation as well, but
its use can be avoided. (K.I.S.S. Principle)
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Analysing Projectile Motion
Example 1
The cannon shown fires a shell at an initial velocity of 400ms-1.
If it fires at an angle of 20o, calculate:
a) the vertical and horizontal components of the initial velocity.
b) the time of flight. (assuming the shell lands at
the same horizontal level)
U=400ms-11
θ = 20o
c) the range. (same assumption)
d) the maximum height it reaches.
a)
Uy = U.Sin θ
= 400..Sin20
=136.8ms-11
(upwards)
Ux = U.Cos θ
=400Cos20
=375.9ms-11
(horizontal)
Point to Note:
The mass of the projectile does NOT enter
into any calculation. The trajectory is
determined by launch velocity & angle, plus
gravity. Mass is irrelevant!
c) Range is horizontal displacement
b) The shell is fired upwards, but
acceleration due to gravity is downwards.
You must assign up = (+ve), down = ( -v
ve).
Remember
Vx= Ux= constant velocity
At the top of its arc, the shell will have an
instantaneous vertical velocity= zero.
Vx = Sx
t
∴ Sx = Vx.t
Vy = Uy + g.t
0 = 136.8 + (-9
9.81)xt
∴ t = -1
136.8/-9
9.81
= 13.95 s
(use time of flight)
= 375.9 x 27.9
= 10,488m
Range = 1.05x104m
(i.e. 10.5 km)
This means it takes 13.95s to reach the top
of its arc. Since the motion is symmetrical,
it must take twice as long for the total
flight.
∴ time of flight = 27.9s
d) Vertical Height
(up =(+ve), down =( -v
ve))
Sy = Uy.t + 1.g.t2
2
= 136.8x13.95 + 0.5x(-9
9.81)x(13.95)2
= 1901.5 + (-9
947.7)
= 953.8m = 9.54x102m.
Note: the time used is the time to reach the top of
the arc... the time at the highest point.
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Analysing Projectile Motion
Example 2
The batsman has just hit the ball upwards at an angle of 55o, with an
intial velocity of 28.0ms-1. The boundary of the field is 62.0m away from
the batsman.
Resolve the velocity into vertical and horizontal components, then use
these to find:
Remember to let UP = (+ve)
a) the time of flight of the ball.
b) the maximum height reached.
DOWN = ( -v
ve)
acceleration = “g” = -9
9.81ms-22
c) whether or not he has “hit a 6” by clearing the boundary.
d) the velocity of the ball (including direction) at the instant t = 3.50s.
Vertical & Horizontal
Components of Velocity
a) Time of Flight
At highest point Vy=0, so
Uy = U.Sin θ, Ux = U.Cos θ
=28Sin55
=28Cos55
=22.9ms-11 =16.1ms-11
Vy = Uy + g.t
0 = 22.9 + (-9
9.81)xt
∴ t = -2
22.9/-9
9.81
= 2.33s
b) Maximum Height
is achieved at t = 2.33s, so
This is the mid-p
point of the
arc, so
time of flight = 4.66s
Sy = Uy.t + 1.g.t2
2
= 22.9x2.33+0.5x(-9
9.81)x(2.33)2
= 53.5 + (-2
26.6)
= 26.9m
c) Range will determine if he’s “hit a 6”.
Vx= Ux= constant velocity
Sx = Vx.t (use total time of flight)
= 16.1 x 4.66
= 75.0m
That’ll be 6 !
d) Velocity at t = 3.50s ?
Vertical
Horizontal
16.1
=22.9+(-9
9.81)x3.50
= -1
11.4ms-11
(this means it is downwards)
Vx= Ux= constant
=
θ
16.1ms-11
Re
su
lta
nt
Ve
loc
ity
11.4
Vy = Uy + g.t
By Pythagorus,
Tan θ = 11.4/16.1
V2 = Vy2 + Vx2
∴ θ ≅ 35o
= (-1
11.4)2 + 16.12
∴ V = Sq.root(389.17) = 19.7ms-11 at an angle 35o below horizontal
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Analysing Projectile Motion (cont)
Projectiles Launched Horizontally
If you find solving Projectile Motion problems is
difficult, try to learn these basic rules:
A common situation with projectile motion is
when a projectile is launched horizontally, as in
the following example. This involves half the
normal trajectory.
• The “launch velocity” must be resolvedinto a
horizontal velocity (Ux) and a vertical velocity
(Uy). Once you have these, you can deal with
vertical and horizontal motion as 2 separate
things.
Plane flying horizontally,
at constant 50.0ms-11
Releases a bomb from
Altitude = 700m
• The motion is symmetrical, so at the highest
point, the elapsed time is exactly half the total
time of flight.
Questions
a) How long does it take for the
bomb to hit the ground?
b) At what velocity does it hit?
c) If the plane continues flying
straight and level, where is it when
the bomb hits?
Uy
U
θ angle of launch
Maximum Height
The top of the arc is the mid-p
point.
At this point Vy = zero
Horizontal
Velocity
Vx
Vertical
Velocity
Vy
Solution
Because the plane is flying horizontally, the
intitial velocity vectors of the bomb are:
Horizontal, Ux= 50.0ms-1,
Vertical,
Uy= zero
a) Time to hit the ground
We know the vertical distance to fall (-700m
(down)), the acceleration rate (g= -9.81ms-2)
and that Uy=0.
Ux
“Range” = Total Horizontal Displacement
• Also, at the highest point, Vy = zero.
The projectile has been rising to this point.
After this point it begins falling.
For an instant Vy = 0. Very useful knowledge!
Sy = Uy.t + 1.g.t2
2
-700 = 0xt + 0.5 x(-9.81)x t2
-700 = -4.905xt2
∴ t2 = -700/-4.905
t = 11.9s
b) Final Velocity at impact
Vertical
Horizontal
Vy = Uy + g.t
Vx= Ux
• Maximum Rangeis achieved at a launch angle
of 45o.
Angle greater than 45o
LAUNCH ANGLE 45o
GIVES MAXIMUM
RANGE
Angles less than 45o
= 0 + (-9.81)x11.9
Vy= -117ms-1. (down)
Vx= 50.0ms-1.
50.0
Tan θ = 117/50
∴ θ ≅ 67o.
• Vertical Motion is constant acceleration
at g= -9.81ms-2, so use Vy = Uy + g.t
to find “t” at the max.height (when Vy=0)
or, find Vy at a known time.
Bomb hits the ground at 127ms-1,
at angle 67o below horizontal.
c) Where is the Plane?
Since both plane and bomb travel at the same
horizontal velocity, it follows that they have
both travelled exactly the same horizontal
distance when the bomb hits. i.e. the plane is
directly above the bomb at impact.
Sy = Uy.t + 1.g.t2
2
to find vertical displacement (Sy) at a known
time, or find the time to fall through a known
height (if Uy=0)
Use
(In warfare, this is a problem for low-level
bombers... the bombs must have
delayed-action fuses)
WORKSHEET at the end of this section
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117
• Horizontal Motion is constant velocity... easy.
Use Vx = Ux
and Sx = Ux.t
θ
ity
loc
Ve
al
Fin
V2=Vy2 + Vx2
= 1172 + 50.02
∴ V = Sq.Root(16,189)
= 127ms-1.
PROJECTILES LAUNCHED AT
SAME VELOCITY
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Isaac Newton and Orbiting
Galileo and Projectile Motion
Once Isaac Newton had developed the Maths
and discovered the laws of motion and gravity,
he too looked at Projectile Motion.
Notice that NONE of the equations used to
analyse Projectile Motion ever use the mass of
the projectile. This is because all objects,
regardless of mass, accelerate with gravity at
the same rate (so long as air-resistance is
insignificant).
Newton imagined a cannon on a very high
mountain, firing projectiles horizontally with
ever-increasing launch velocities:
It was Galileo, (1564-1642) who you learned
about in “The Cosmic Engine”, who first
discovered this.
If launch velocity is high
enough, the projectile
escapes from the Earth’s
gravity
EARTH
At the right velocity, the projectile
curves downwards at the same
rate as the Earth curves... it will
circle the Earth in orbit!
Newton had discovered the concept of a
gravitational orbit, and the concept of “escape
velocity”.
His famous experiment was to drop objects of
the same size and shape, but of different weight,
from the leaning tower in Pisa. He found that all
objects hit the ground at the same time, thereby
proving the point.
Escape Velocity is defined as the launch
velocity needed for a projectile to escape from
the Earth’s gravitational field.
Mathematically, it can be shown that
He also studied projectile motion. In his day,
cannon balls were the ultimate weapon, but
trajectories were not understood at all. To slow
the motion down for easier study, Galileo rolled
balls down an incline:
Escape Velocity, Ve =
2GME / RE
G= Gravitational Constant (later in topic)
ME= Mass of the Earth
RE= Radius of Earth
You are NOT required to learn, nor use,
this equation.
Although not
falling freely, the balls
accelerated uniformly, and Galileo was
able to see that the motion was a combination
of 2 motions:
• horizontal, constant velocity
and
• vertical, constant acceleration
What you should learn is that:
• The mass of the projectile is not a factor.
Therefore, all projectiles, regardless of mass,
need the same velocity to escape from Earth,
about 11km per second!
Galileo had discovered the basic principles of
Projectile Motion.
• The Escape Velocity depends only on the mass
and radius of the Earth.
Unfortunately, he lacked the mathematical
formulas to go any further with his analysis.
It follows that different planets have different
escape velocities. Here are a few examples...
That only became possible after the work of
Isaac Newton, and his 3 Laws of Motion, and
Theory of Gravitation.
PLANET
Earth
Moon
Mars
Jupiter
Coincidentally, Newton was born in the same
year that Galileo died.
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10
ESCAPE VELOCITY
in km/sec
(ms-1)
11.2
1.12 x104
2.3
2.3 x103
5.0
5.0 x103
60.0
6.0 x104
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Rockets Achieve Orbit
Placing a Satellite in Earth Orbit
A projectile needs an enormous velocity to
escape from the Earth’s gravitational field...
about 11 km per second. Think of a place 11 km
away from you, and imagine getting there in 1
second flat!
To keep the g-forces low while accelerating to
the velocity required for orbit, AND then to
operate in the airless conditions of space, the
rocket is the only practical technology
developed so far.
What about Newton’s idea of an orbiting
projectile? If it is travelling at the right velocity,
a projectile’s down-curving trajectory will match
the curvature of the Earth, so it keeps falling
down, but can never reach the surface. A
projectile “in orbit” like this is called a
“satellite”.
A Brief History of Rocketry
Simple solid-fuel (e.g. gunpowder) rockets have
been used as fireworks and weapons for over
500 years.
About 100 years ago, the Russian Tsiolkovsky
(1857-1935) was the first to seriously propose
rockets as vehicles to reach outer space. He
developed the theory of multi-stage, liquid-fuel
rockets as being the only practical means of
achieving space flight.
It can be shown that to achieve orbit, the launch
velocity required is less than escape velocity,
but still very high... about 8 km per second. How
is this velocity possible?
The American Robert Goddard (18821945) developed rocketry theory futher,
but also carried out practical
experiments including the first liquidfuel rocket engine.
In a 19th century novel, author Jules Verne
proposed using a huge cannon to fire a space
capsule (including human passengers) into
space. Let’s consider the Physics:
V2
The “g-Forces” in a Space Launch
Goddard’s experiments were the basis
of new weapons research during
World War II, especially by Nazi
Germany. Wernher von Braun (19121977) and others developed the
liquid-fuel “V2” rocket to deliver
explosive warheads at supersonic
speeds from hundreds of kilometers away.
To accelerate a capsule (and astronauts)
upwards to orbital velocity requires a force. The
upward “thrust” force must overcome the
downward weight force AND provide upward
acceleration.
Astronaut During Acceleration to Orbital Velocity
Total Net Force
causes acceleration
Net
Force= ma
ΣF = ma
Weight = mg
Force
“THRUST” Force = T
At the end of the war many V2’s, and the German
scientists who developed them, were captured
by either the Russians or the Americans. They
continued their research in their “new”
countries, firstly to develop rockets to carry
nuclear weapons (during the “Cold War”) and
later for space research.
Greek letter
Sigma ( Σ )
means total
If up = (+ve), down ( -v
ve)
then
ΣF = T - mg = ma
The Russians
achieved the
first satellite
(“Sputnik”
1957) and the
first human
in orbit, and
the
Americans
the first
manned
missions to the Moon (1969).
∴ T = ma + mg
So, if the Thrust
This means the astronaut
force causes
will “feel” the thrust as an
acceleration of (say)
increase in weight.
about 10ms-2, as well
as overcoming his weight force, the 80kg
astronaut will feel a pushing force of;
T = ma + mg
= 80x10 + 80x10
= 1,600N
≅10ms-2 )
( g≅
Space
Shuttle
launch
This is twice his normal weight of 800N...
we say the force is “2g”.
A fit, trained astronaut can tolerate forces of
“5g”, but anything above about “10g” is lifethreatening. Jules Verne’s cannon astronauts
would have suffered forces of about 200g...
instantly fatal.
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The business
end of a
1970’s liquidfuel rocket
engine
Since then, the use of
satellites has become
routine and essential to our
communications, while
(unmanned) probes have
visited nearly every other
planet in the Solar System.
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Direction of Launch Physics of a Rocket Launch
Straight upwards, right?
Wrong!
Conservation of
Momentum
Why a rocket moves was dealt with in the
Preliminary topic “Moving About”.
To reach Earth orbit, rockets are aimed toward
the EAST to take advantage of the Earth’s
rotation. The rocket will climb vertically to clear
the launch pad, then be turned eastward.
Newton’s 3rd Law
Launch
Trajectory
Reaction force pushes
rocket forward
Force on = Force on
Exhaust
Rocket
Gases
Orbit path
It can also be shown that
Earth, viewed
from above
North Pole
Action Force
pushes on
exhaust gasses,
accelerating
them
backwards
Change of Momentum = Change of Momentum
of Exhaust Gases
of Rocket
Rotation
backwards ( -v
ve)
At the equator, the Earth is rotating eastwards at
about 1,700km/hr (almost 0.5km/sec) so the
rocket already has that much velocity towards
its orbital speed.
forwards (+ve)
( -)Mass x velocity = Mass x velocity
The mass x velocity (per second) of the
exhaust gases stays fairly constant during the
lift-off. However, the mass of the rocket
decreases as its fuel is burnt. Therefore, the
rocket’s velocity must keep increasing in order
to maintain the Conservation of Momemtum.
Rocket launch facilities are always sited as
close to the equator as possible, and usually
near the east coast of a continent so the launch
is outwards over the ocean.
Forces Experienced by Astronauts
If the “Thrust” force from the rocket engine
remains constant throughout the “burn”, but the
total rocket mass decreases due to consumption of
the fuel, then the acceleration increases.
The concept of “g-forces” was explained on the
previous page.
Thrust Force,
T = ma + mg
If “T” remains constant, but “m” keeps decreasing,
then “a” must keep increasing.
(This assumes “g” is constant...
Actually it decreases with altitude,
so “a” must increase even more)
Not only does the rocket accelerate upwards, but
even the acceleration keeps accelerating!
Photo by Shelley Kiser
The astronauts will feel increasing “g-forces”. At
lift-off, they will experience perhaps only “2g”, but
over several minutes this will increase to perhaps
“7g” as the rocket burns thousands of tonnes of
fuel and its mass decreases.
The Space Shuttle’s engines are throttled-back
during the launch to counteract this, so the
astronauts are not injured by increasing “g-force”.
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Photo: Russian Soyez lift-off,
courtesy Ali Cimen, senior
reporter, Zaman Daily, Istanbul.
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Types of Orbits
There are 2 main types of
satellite orbits:
Satellites and Orbits Orbits & Centripetal Force
The orbit of a satellite is often an ovalshape, or “ellipse”. However, in this
topic we will always assume the orbits are
circular... K.I.S.S. Principle.
Low-Earth Orbit
As the name suggests, this type of orbit is
relatively close to the Earth, generally from
about 200km, out to about 1,000km above the
surface.
Circular Motion was introduced in a Preliminary
topic. To maintain motion in a circle an object
must be constantly acted upon by “Centripetal
Force”, which acts towards the centre of the
circle.
For any satellite, the closer it is, the faster it
must travel to stay in orbit. Therefore, in a LowEarth Orbit a satellite is travelling quickly and
will complete an orbit in only a few hours.
Fc
A common low orbit is a “Polar Orbit” in which
the satellite tracks over the north and south
poles while the Earth rotates underneath it.
N
V
Object in
Circular
Motion
Instaneous Velocity
vector is a tangent to
the circle
Fc
Centripetal Force
Vector
always towards centre
This type of orbit is
ideal for taking photos
or Radar surveys of
Earth.
Polar Orbit
Earth’s
Rotation
Equator
S
The satellite only “sees”
a narrow north-s
south
strip of the Earth, but
as the Earth rotates,
each orbit looks at a
new strip.
The object is constantly
accelerating. The
“centripetal acceleration”
vector is towards the
centre.
What Causes Centripetal Force?
Example
Swinging an object
around on a string.
Eventually, the entire
Earth can be surveyed.
Being a close orbit, fine
details can be seen.
Centripetal Force caused by...
Tension Force in the string.
Vehicle turning a
circular corner.
Friction Force between tyres
and road.
Satellite in orbit
around Earth.
Gravitational Force between
satellite mass and Earth’s
mass.
Fc = mv2
R
Geo-stationary Orbits are
those where
the period of the satellite (time taken for one orbit)
is exactly the same as the Earth itself... 1 day.
Fc = Centripetal Force, in newtons (N)
m = mass of object in orbit, in kg
v = orbital velocity, in ms-1
R = radius of orbit, in metres (m)
This means that the satellite is always directly
above the same spot on the Earth, and seems to
remain motionless in the same position in the
sky. It’s not really motionless, of course, but
orbiting around at the same angular rate as the
Earth itself.
When considering the radius of a satellite orbit, you
need to be aware that the orbital distance is often
described as the height above the surface. To get
the radius, you may need to add the radius of the
Earth itself... 6,370km (6.37 x 106 m)
Geo-stationary orbits are usually above the
equator, and have to be about 36,000km above the
surface in order to have the correct orbital speed.
Calculating Velocity from Radius & Period
Satellite motion is often described by the
radius of the orbit, and the time taken
for 1 orbit = the Period (T)
Being so far out, these satellites are not much
good for photographs or surveys, but are ideal
for communications. They stay in the same
relative position in the sky and so radio and
microwave dishes can be permanently aimed at
the satellite, for continuous TV, telephone and
internet relays to almost anywhere on Earth.
Now, circumference of a circle = 2πR
Therefore, the orbital velocity V = 2πR
T
Three geo-stationary satellites, spaced evenly
around the equator, can cover virtually the
whole Earth with their transmissions.
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V
distance
traveled
time taken
Example Problem next page...
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Kepler’s “Law of Periods”
Centripetal Force and Satellites
Example Problem
A 250kg satellite in a circular
orbit 200km above the
Earth, has an orbital
period of 1.47hours.
When Johannes Kepler (1571-1630) studied the
movement of the planets around the Sun (see
Preliminary topic “Cosmic Engine”) he
discovered that there was always a
mathematical relationship between the Period of
the orbit and its Radius:
R
200km
a) What is its orbital velocity?
b) What centripetal force acts
on the satellite?
(Radius of Earth = 6.37x106m)
R3
“proportional to”)
This means that
Solution
a) First, find the true radius of the orbit, and get
everything into S.I. units:
Radius of orbit = 200,000 + 6.37x106 = 6.57x106m
Period = 1.47hr = 1.47 x 60 x 60 = 5.29x103 seconds
V = 2πR = 2 x
T
b)
Fc =
α T2 (Greek letter alpha (α ) means
π x 6.57x106/5.29x103 = 7.80x103ms-1.
mv2
=
R3 = constant
T2
This means that for every satellite of the Earth,
the (Radius)3 divided by (Period)2 has the same
value.
This is a very useful relationship...
see Example Problem at bottom left
250x(7.80x103)2/6.57x106
R
= 2,315 = 2.32 x 103 N.
At this point, the HSC Syllabus is rather
vague about whether you need to learn
and know the following mathematical
development.
The satellite is travelling at about 8 km/sec,
held in orbit by a gravitational force of about 2,300N.
WORKSHEET at the end of this section
You may be safe to ignore it... (K.I.S.S.) but follow it if
you can. Either way, you DO need to be able to use the
final equation shown below.
Kepler’s Law of Periods was discovered empirically...
that is, it was discovered by observing the motion of
the planets, but Kepler had no idea WHY it was so.
Kepler’s Law of Periods
When Isaac Newon developed his “Law of Universal
Gravitation” (next section) he was able to prove the
theoretical basis for Kepler’s Law, as follows:
Example Problem
A geo-stationary satellite has a period of 24.0 hours.
Use Kepler’s Law of Periods to find its orbital radius.
Use data from the example above.
Solution
For the satellite above,
(units are km & hours)
R3 = 6,5703 = 1.31 x 1011
T2
1.472
According to the law of periods, ALL satellites of Earth
must have the same value for R3/T2
So, for the geo-stationary satellite:
R3 = 1.31 x 1011
T2
So R3 = 1.31x1011x(24.0)2
∴R = CubeRoot(7.55x1013)
= 4.23 x 104 km
This is approx. 42,000km from Earth’s centre, or about
36,000km above the surface.
Note: When using Kepler’s Law in this way
it doesn’t matter which units are used,
as long as you are consistent.
Centripetal Force = Gravitational Force
Fc = mv2 = FG = GMm
R
R2
∴ v2 = GM
R
So, 4π2R2 = GM
T2
R
re-arranging, R3 = GM
T2
4π2
but v = 2πR
T
Since the right hand side are all constant values,
this proves Kepler’s Law and establishes the Force
of Gravity as the controlling force for all orbiting
satellites, including planets around the Sun.
In the above,
G = Universal Gravitational Constant
M = mass of the Earth (or body being orbited)
m = mass of satellite... notice that it disappears!
In this example, km & hrs were used. The same result
will occur if metres & seconds are used.
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The Centripetal Force of orbiting is provided by
the Gravitational Force between the satellite and
the Earth, so
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Re-Entry From Orbit
Kepler’s Law of Periods (Again!)
Getting a spacecraft into orbit is difficult enough, but
the most dangerous process is getting it down again
in one piece with any astronauts on board alive and
well.
On the previous page, the sample problem was able to
calculate the orbital radius for a geo-stationary
satellite by comparing the ratio of R3/T2 for 2 satellites.
With Newton’s development of Kepler’s Law, we can
do it again a different way...
In orbit, the satellite and astronauts have a high
velocity (kinetic energy) and a large amount of GPE
due to height above the Earth. To get safely back to
Earth, the spacecraft must decelerate and shed all
that energy.
Example Problem
Find the orbital radius of a geo-stationary satellite,
given that its period of orbit is 24.0 hours.
(24.0hr = 24.0x60x60 = 8.64 x 104 sec)
Doing this way, you MUST use S.I. units!!
(G= Gravitational Constant = 6.67 x
M = Mass of Earth = 5.97 x 1024kg)
It is impossible to carry enough fuel to use rocket
engines to decelerate downwards in a reverse of the
lift-off, riding the rocket back down at the same rate it
went up.
10-11
R3 = GM
T2
4π2
Instead, the capsule is slowed by “retro-rockets” just
enough to cause it to enter the top of the atmosphere
so that friction with the air does 2 things:
R3 = 6.67x10-11 x 5.97x1024 x (8.64x104)2
4π2
∴ R = CubeRoot (7.5295x1022)
= 4.22 x 107m.
• cause deceleration of the capsule at a survivable
rate of deceleration not more than (say) “5-g”, and
• convert all the Ek and GPE into heat energy.
This is about 42,000km, or about 36,000km above the
surface... the same answer as before. (It better be!)
The trick is to enter the atmosphere at the
correct angle:
WORKSHEET at the end of the section
Angle too shallow...
Spacecraft bounces off upper air
layers, back into space
Decay of Low-Earth Orbits
Where does “Space” begin?
It’s generally agreed that by 100km above the
surface of the Earth the atmosphere has ended,
and you’re in outer space. However, although
this seems to be a vacuum, there are still a few
atoms and molecules of gases extending out
many hundreds of kilometres.
Upper Atmosphere
Earth’s Surface
Angle correct...
Spacecraft decelerates safely along
a descent path of about 1,000km
of “Atmospheric
Braking”
Therefore, any satellite in a low-Earth orbit will
be constantly colliding with this extremely thin
“outer atmosphere”. The friction or airresistance this causes is extremely small, but
over a period of months or years, it gradually
slows the satellite down.
Correct angle is
between 5-7
7o
As it slows, its orbit “decays”. This means it
loses a little altitude and gradually spirals
downward. As it gets slightly lower it will
encounter even more gas molecules, so the
decay process speeds up.
Earth’s Surface
Angle too steep...
“g-fforces” may kill astronauts.
Heat may cause craft to burn-u
up.
Once the satellite reaches about the 100km level
the friction becomes powerful enough to cause
heating and rapid loss of speed. At this point the
satellite will probably “burn up” and be
destroyed as it crashes downward.
Earth’s Surface
Modern satellites are designed to reach their
low-Earth orbit with enough fuel still available to
carry out short rocket engine “burns” as
needed to counteract decay and “boost”
themselves back up to the correct orbit. This
way they can remain in low-Earth orbits for
many years.
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Early spacecraft used “ablation shields”, designed to
melt and carry heat away, with the final descent by
parachute. The Space Shuttle uses high temperature
tiles and high-tech insulation for heat protection, and
glides in on its wings for final landing like an aircraft.
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Worksheet 3
Projectiles & Satellites
Fill in the blank spaces.
Student Name...........................................
A projectile is any object which is launched, and then
moves a).................................. The path of a projectile
is called its b)................................, and is a curve.
Mathematically, the curve is a c)..................................
The only feasible technology (so far) for achieving the
necessary af)................... ........................., while
keeping the ag).................................. reasonably low,
is the use of ah)............................ One of the important
steps in the history of rocketry was achieved by
Robert Goddard, who built and tested the first
ai).............................-fuelled rocket.
To analyse projectile motion it is essential to treat the
motion as 2 separate motions; d)..................................
and .................................... If the launch velocity and
the e)............................ of launch are known, you
should always start by f)...................................... the
initial velocity into horizontal and vertical
g)......................................
Rockets are always launched towards the
aj)...................... to take advantage of the Earth’s
ak)................................. Rocket propulsion is a
consequence of Newton’s al)........... Law. During the
launch, momentum is am).............................. The
backward momentum gained by the exhaust gases is
matched by the an).......................... momentum gained
by the ao)............................. However, the mass of the
rocket ap)....................... rapidly as is burns huge
amounts of fuel. This means that even with constant
thrust, the acceleration rate aq).........................., and
the astronauts feel increasing ar)..............................
The horizontal motion is always h)........................
....................... and the vertical is constant
i)................................... due to j)............................. The
usual strategy is to find the k)...................... of flight,
by using the fact that at the top of the projectile’s arc
its vertical velocity is l)......................... Once this is
known, it becomes possible to calculate the
maximum m)....................... attained, and the
n)........................ (total horizontal displacement.). The
projectile’s position and velocity at any instant can be
found by combining the o)............................. and
....................................... vectors. Maximum range of
any projectile occurs when the angle of launch is
p).................... degrees upwards.
There are basically 2 different types of orbit for a
satellite: as)........................................ orbits are when
the satellite is at).......................... km from Earth and
travelling very au).............................. This is ideal for
satellites used for av)...................................... and
........................................... The other type of orbit is
called aw)..................................... For this the satellite
is positioned so its ax)......................... is exactly 24
hours. This means it orbits at the same relative rate
as the Earth’s ay)........................., and seems to stay
in the az).................................................... This is ideal
for ba)................................ satellites.
Historically, it was q)...................................... who first
proved that (ignoring air-resistance) all objects
accelerate
under
gravity
r).....................
................................... He also investigated projectile
motion and was the first to see that the horizontal
motion is constant s).......................... while the
t)................................. is constant acceleration.
Any object undergoing Circular Motion is being acted
upon by bb).............................. force, which is always
directed towards the bc)............................................
For an object twirled on a string, the centripetal force
is provided by the bd)...................... ..............................
For a car turning a corner, it’s the force of
be)..................... between tyres and road. For a
satellite, it’s the force of bf).....................................
Later, u)...................................... developed the
mathematics of both gravity and motion, which
allowed projectile motion to be understood and
ananalysed. He also discovered the concept of
v)...................... velocity, and of objects being in
w)..........................., by imaging what would happen to
cannon balls being fired horizontally at increasing
velocities from a high mountain.
Johannes bg)............................... discovered the “Law
of Periods” for satellites. Later, Newton was able to
shown that this was a consequence of
bh)............................ attraction between the satellite
and whatever it is orbiting.
“Escape Velocity” is defined as the velocity a
projectile needs in order to x)......................
...............................................
Satellites & Orbits
Low-Earth orbits will eventually “bi)............................”
due to the satellite gradually losing speed by collision
with bj)...................................................
If a projectile is travelling horizontally at the correct
y)..................................., then its down-curving
trajectory will match the z)................................. of the
Earth. The projectile will continue to “fall down” but
never reach the surface... it is a aa)...............................
which is ab)................................. around the Earth. To
place
a
satellite
in
orbit,
it
must
be
ac).................................... up to orbital speeds.
Re-entry of a spacecraft from orbit is extremely
dangerous: bk)....................... from high velocity can
cause high g-forces, and friction causes production
of bl)........................... energy which can cause the
craft to burn-up. The trick is to enter the atmosphere
at exactly the correct bm)...........................
During upward acceleration, an astronaut will
experience “ad)..........................” which feel like an
increase in ae)......................... and can be lifethreatening if too high.
COMPLETED WORKSHEETS
BECOME SECTION SUMMARIES
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Worksheet 4
Projectiles
Practice Problems
Student Name ...........................................
3. The bullet in Q1(b), was fired from a height of
2.00m, across a level field. Calculate:
a) how long it takes to hit the ground.
1. For each of the following projectiles, resolve
the initial launch velocity into horizontal and
vertical components.
a) A rugby ball kicked upwards at an angle of
60o, with velocity 20.5ms-1.
b) how far from the gun it lands.
b) A bullet fired horizontally at 250ms-1.
c) At the same instant that the bullet left the
barrel, the empty bullet cartridge dropped (from
rest) from the breech of the gun, 2.00m above
the ground. How long does it take to hit the
ground? Comment on this result, in light of the
answer to (a).
c) A baseball thrown at 15.0ms-1, and an up
angle of 25o.
d) An artillery shell fired at 350ms-1, upwards at 70o.
4. For the artillery shell in Q1(d), calculate:
a) the time to reach the highest point of its arc.
e) An arrow released from the bow at 40.0ms-1,
at 45o up.
b) the maximum height reached.
2. For the arrow in Q1(e), find
a) the time to reach the highest point of its arc.
c) its range (on level ground).
b) the maximum height reached.
5. The rugby ball in Q1(a) was at ground level
when kicked.
a) Find its exact position 2.50s after being
kicked.
c) its range (on level ground).
b) What is its instantaneous velocity at this
same time?
Remember that for full marks
in calculations, you need to show
FORMULA, NUMERICAL SUBSTITUTION,
APPROPRIATE PRECISION and UNITS
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Worksheet 5
Practice Problems
Orbits & Centripetal Force
Student Name ...........................................
2. (cont)
b) What is its altitude above the earth’s surface?
1.
A satellite orbiting 1,000km above the Earth’s
surface has a period of 1.74 hours. (Radius of
Earth=6.37x106m)
a) Find its orbital velocity, using V=2πR/T
c) What is the period of its orbit?
b) If the satellite has a mass of 600kg, find the
centripetal force holding it in orbit.
3. A satellite is being held in Earth orbit by a
centripetal force of 2,195N. The orbit is 350km
above the Earth, and the satellite’s period is 1.52
hours.
a) Find the orbital velocity.
2.
A 1,500kg satellite is in a low-Earth orbit travelling
at a velocity of 6.13 km/s (6.13x103ms-1). The
Centripetal force acting on it is 5.32x103N.
a) What is the radius of its orbit?
b) What is the satellite’s mass?
Worksheet 6
Practice Problems
Kepler’s Law of Periods
Student Name ...........................................
3.
a) Planet Mars has mass= 6.57x1023kg.
Calculate the “orbital constant” GM/4π2 for
Mars. (G=Gravitational Constant = 6.67x10-11)
1. Fill in the table using data for each of the
satellites in Q’s 1, 2 & 3 in Worksheet 4.
Radius (m)
Period (s)
R3/T2
Q1.
Q2.
b) Find the orbital Radius of a satellite orbitting
Mars, if its Period is 1.60x103s.
Q3.
Explain how this data supports Kepler’s Law of
Periods.
2. Use the average value of R3/T2 from the table
above to calculate the following:
a) Find the Radius of an Earth orbit if the
Period is 1.60x103s.
c) Find the period of a Mars satellite when
R=2.56x107m.
b) What is the radius of orbit if T=1.15x104s?
d) In Q2(c) you calculated the period of an Earth
satellite with the same orbital radius.
c) Find the period of a satellite if R= 2.56x107m.
Compare the answers to Q2(c) and Q3(c). Which
satellite travels at the highest orbital velocity?
d) Find T when the satellite orbit is 2,000km
above the Earth’s surface.
e) Complete the blanks in this general
statement:
At a given orbital radius, a satellite orbiting a
smaller planet needs to travel at a
............................................... velocity. The bigger
the planet, the ............................................. the
velocity would need to be.
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Worksheet 7 Test Questions
Multiple Choice
sections 1&2
1. According to the formal definition of “Gravitational
Potential Energy” (GPE)
A. There is zero GPE on the surface of the Earth.
B. GPE depends on mass, height and velocity.
C. There is zero GPE at an infinite distance
from Earth.
D. GPE depends only on the weigh of the object.
8.
To get maximum advantage from the rotation of the
Earth, a space launch is always directed toward the
A. north
B. south
C. east
D. west
≅10ms-2 while on Mars g≅
≅4ms-2. A 100kg
2. On Earth g≅
object transported to Mars would have mass and
weight (respectively) of:
A. 100kg and 400N.
B. 400kg and 1,000N.
C. 100kg and 1,000N.
D. 400kg and 400N.
9.
The chief of the C.I.A. has asked you to plan the
deployment of a spy-satellite in an orbit suitable for
taking photos of suspected terrorist bases, in many
locations, world-wide. You would be best to plan for
the satellite to be in a:
3. The diagram shows the trajectory of a projectile,
and 2 points X & Y.
X
Y
A.
B.
C.
D.
Which pair of vectors below correctly identifies the
total acceleration vector of the projectile at points X
and Y?
Point X
Point Y
A.
Low-Earth Orbit, around the Equator.
Geo-stationary Orbit, above the Equator.
Geo-stationary orbit, allowing full earth coverage.
Low-Earth, Polar Orbit.
10.
Once the Space Shuttle reaches its orbital velocity the
engines are turned off, and the craft orbits in freefalling circular motion. This motion is characterized
by:
A. constant acceleration directed at a tangent to the
circle.
B. constant velocity, with no forces acting.
C. constant centrifugal force, pushing things
outwards.
D. constant acceleration directed at the centre of the
circle.
B.
C.
D.
4.
To analyse projectile motion mathematically, usually
the first thing to do is to:
A. find the time of flight.
B. calculate the range.
C. calculate the maximum height reached.
D. resolve the initial velocity into vertical & horizontal
components.
11.
The data below relates to 3 of the moons of Jupiter
and one of the moons of Mars. The units of
measurement are arbitrary. Which moon (A,B,C or D)
belongs to Mars?
5.
Ignoring air-resistance, the maximum range for any
projectile (for the same launch velocity) will occur
when:
A. it is launched horizontally.
B. it is launched at 45o upwards.
C. it is launched to achieve a greater height.
D. its vertical acceleration is increased.
Moon
A.
B.
C.
D.
Orbital Radius
9.2
8.5
13.3
10.0
Orbital Period
12.5
10.0
19.6
12.8
12.
All Low-Earth orbits are prone to “decay” because of:
A. the satellite running out of fuel to maintain forward
speed.
B. friction with the few gas molecules in its orbital
path.
C. Earth’s gravity gradually pulling it downwards.
D. magnetic effects of a Polar orbit.
6.
It is known that the value of “escape velocity” for any
planet is
• proportional to the mass of the planet, and
• inversely proportional to the planet’s radius.
Therefore, the planet which would definitely have a
lower escape velocity than Earth would have
(compared to Earth)
A.
B.
C.
D.
Student Name...........................................
7.
During a launch, the acceleration of a rocket:
A. increases, because mass decreases.
B. increases, because kinetic energy increases.
C. decreases, because momentum increases.
D. remains constant, because of constant thrust
force.
more mass, smaller radius.
less mass, larger radius.
more mass, larger radius.
less mass, smaller radius.
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Worksheet 7 Test Questions
Longer Response Questions
sections 1&2
Student Name...........................................
17. (3 marks)
It was Isaac Newton who discovered the concept of
“orbiting” the Earth by thinking about projectiles.
Outline Newton’s scenario for how a projectile could
end up in orbit.
Mark values shown are suggestions only, and are to
give you an idea of how detailed an answer is
appropriate. Answer on reverse if insufficient space.
13. (3 marks)
An alien creature weighs 7.25x103N on its home
planet. On Earth, the creature weighs 5.84x103N.
a) What is the creature’s mass?
18. (4 marks)
Author Jules Verne wrote a novel in which a space
ship was launched by firing it from a cannon, on a
journey to the Moon. The required velocity would be
1.05x104ms-1.
With the cannon barrel 200m long, it would be
reasonable for the time of launch (i.e. duration of
acceleration) to be 7.50s.
b) What is the value of “g” on the creature’s home
planet?
a) Calculate the acceleration rate of the capsule.
14. (4 marks)
A ball was rolled along a
horizontal table at 5.45ms-1.
If the table is 1.20m high, where will
the ball hit the ground?
b) What is the value (in terms of “g”) of the “g-force”
that the passengers would experience?
c) Comment on the feasibility of such a launch.
19. (3 marks)
Give a brief outline of the contributions of one of
these men to the science of rocketry.
Tsiolkovsky, Goddard or von Braun
15. (8 marks)
An arrow was released from the bow at an upward
angle of 60o and an initial velocity of 42.0ms-1. It hits
its target at the same horizontal level from which it
was released.
a) Find the time of flight.
20. (4 marks)
The early space rocket engines produced a constant
“thrust force” throughout their “burn”. One of the
advantages of the Space Shuttle engine is that it has
a throttle control. During launch, the engine is
gradually throttled back and thrust reduced during
the ascent into orbit. Explain how this contributes to
the safety and comfort of the astronauts on board.
b) Find the maximum height reached.
c) Calculate the distance from bow to target.
16. (6 marks)
These military bombs
are designed to be
dropped from the
aircraft at an altitude of
15,000m when the
plane is in level flight at
a velocity of 300ms-1.
Photo: Arian Kulp
b) Calculate the orbital velocity of this satellite.
c) Find the strength of the force holding it in orbit,
given that the satellite has a mass of 2,650kg.
a) Ignoring airresistance, how far in
front of the target must
the bombs be released?
22. (5 marks)
Discuss the problems of recovering a spacecraft from
orbit and outline the process of “atmospheric
braking”, with reference to the importance of the
angle of descent.
b) How fast will they be going (magnitude only) when
they hit the ground?
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21. (6 marks)
a) Use the relationship R3/T2 =GM/4π2 to find the
radius of orbit of an Earth satellite with a period of
2.00hours.
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3. NEWTON’S LAW OF UNIVERSAL GRAVITATION
Effects of Mass & Distance on FG
Gravitational Fields
How does the Gravitational Force change for
different masses, and different distances?
The concept of the Gravitational Field was introduced
in section 1. Every mass acts as if surrounded by an
invisible “force field” which attracts any other mass
within the field. Theoretically, the field extends to
infinity, and therefore every mass in the universe is
exerting some force on every other mass in the
universe... that’s why it’s called Universal Gravitation.
Imagine 2 masses, each 1kg, separated by a
distance of 1 metre.
FG = GMm = G x 1 x 1 = G
d2
12
Effect of masses
Newton’s Gravitation Equation
It was Isaac Newton who showed that the
strength of the gravitational force between 2
masses:
Now imagine doubling the mass of one object:
FG = GMm = G x 2 x 1 = 2G (Twice the force)
d2
12
• is proportional to the product of the masses,
and
• inversely proportional to the square of the
distance between them.
What if both masses are doubled?
FG = GMm = G x 2 x 2 = 4G (4X the force)
d2
12
Effect of Distance
Go back to the original masses, and double the
distance:
FG = GMm = G x 1 x 1 = G
( 1/4 the force)
d2
22
4
FG = GMm
d2
FG = Gravitational Force, in N.
G = “Universal Gravitational Constant” = 6.67 x 10-11
M and m = the 2 masses involved, in kg.
d = distance between M & m (centre to centre) in metres.
Gravitational Force shows the “Inverse Square”
relationship...
triple the distance = one ninth the force
10 x the distance = 1/100 the force, etc.
In the previous section on satellite orbits, you were
already using equations derived from this.
Universal Gravitation and
Orbiting Satellites
It should be obvious by now that it is FG which
provides the centripetal force to hold any
satellite in its orbit, and is the basis for Kepler’s
Law of Periods.
Example Calculation 1
Find the gravitational force acting between the Earth
and the Moon.
Earth mass = 5.97 x 1024kg
Moon mass = 6.02 x 1022kg.
Distance Earth-Moon = 248,000km = 2.48x108m.
Solution
Not only does this apply to artificial satellites
launched into Earth orbit, but for the orbiting of
the Moon around the Earth, and of all the planets
around the Sun.
FG = GMm
d2
= 6.67x10-11x5.97x1024x6.02x1022
(2.48x108)2
= 3.90 x 1020N.
Example 2
Find the gravitational force acting between the Earth,
and an 80kg person standing on the surface, 6,370km
from Earth’s centre (d=6.37 x 106m).
Solution
FG = GMm
d2
= 6.67x10-11x5.97x1024x 80
(6.37x106)2
= 785 N.
Our entire Solar System is orbiting the Galaxy
because of gravity, and whole galaxies orbit
each other. Ultimately, gravity holds the entire
universe together, and its strength, compared to
the expansion of the Big Bang, will determine
the final fate of the Universe.
This is, of course, the person’s weight!... and sure
enough
W = mg = 80 x 9.81 = 785N.
Gravitational Force = Weight Force
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“Slingshot Effect” for
Space Probes
Planet orbit
One of the more interesting aspects of gravity
and its effects on space exploration is called the
“Slingshot Effect”.
2nd
planet
visited
Slingshot
Trajectory
• Scientists wish to explore and learn about all the
planets, comets, etc, in the Solar System, but...
• It costs billions of dollars to send a space
probe to another planet, so...
Planet orbit
1st planet
visited
• It makes sense to send one probe to several
planets, rather than a separate spacecraft to
each planet, but...
• the distances are enormous. Even at the high
speed of an inter-planetary probe (50,000 km/hr)
it still takes years to reach other planets.
• Furthermore, having reached and done a “flyby” to study one planet, the probe may need to
change direction and speed to alter course for
the next destination, and...
• It may be impossible to carry enough fuel to
make the necessary direction changes by using
rocket engines alone.
Got all that?
The solution to all these factors is to fly the
spacecraft close enough to a planet so that the
planet’s gravity causes it to swing around into a
new direction AND gain velocity (without
burning any fuel).
Worksheet 8
So how can the the spacecraft gain extra
velocity (and kinetic energy) from nothing?
The answer is that whatever energy the
spacecraft gains, the planet loses. Energy is
conserved. The planet’s spin will be slowed
down slightly by the transfer of energy to the
spacecraft.
Of course, the huge mass of a planet means that
the energy it loses is so small to be totally
insignificant.
Practice Problems
Universal Gravitation
1.
Fred (75kg) and girlfriend Sue (60kg) are very
much attracted to each other. How much?
Find the gravitational force attracting them when
they are 0.5m apart.
Fill in the blanks.
The strength of the gravitational force of
attraction between 2 masses is proportional to
a)................................. and inversely proportional
to b).................................................................... So,
if one mass is doubled the force will
c)............................., but if the distance is
doubled, then the force will d)....................
................................................
2.
What is the gravitational force of attraction
between 2 small asteroids with masses of
6.75x108kg and 2.48x109kg separated by 425m?
The force due to gravity provides the
e).............................................. force for all
satellites,
including
the
Moon
and
f).................................... orbiting the Sun. In space
exploration, gravity can be used to alter a
spacecraft’s g)................................. and to gain
h)............................................ This is known as the
i)...................................... Effect. The spacecraft
gains energy, while j).............................................
loses an k)................................. amount.
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Spacecraft
3.
The mass of the Moon is 6.02x1022kg. A comet
with mass 5.67x1010kg is attracted to the Moon
by a force of 6.88x1010N. How far apart are the 2
bodies?
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4. EINSTEIN’S THEORY OF RELATIVITY
In Michelson & Morley’s experiment the “boats”
were beams of light from the same source, split
and reflected into 2 right-angled beams sent out
to mirrors and reflected back. The “current” was
the “aether wind” blowing through the
laboratory due to the movement of the Earth
orbiting the Sun at 100,000km/hr.
The Aether Theory
The idea of the universal “aether” was a theory
developed to explain the transmission of light
through empty space (vacuum) and through
transparent substances like glass or water.
The basic idea was this:
Sound waves are vibrations in air.
Water waves travel as disturbances in water.
Sounds and shock waves travel
through the solid Earth.
It seems that all waves have a “medium” to travel
through, so what is the medium for light waves?
Earth is hurtling through
the Aether while orbiting
the Sun.
From the 17th to 19th centuries, as modern Science
developed, it became the general belief that there was
a substance called the “aether” which was present
throughout the universe as the medium for light
waves to be carried in. The aether was invisible,
weightless and present everywhere, even inside
things like a block of glass, so light could travel
through it. The vacuum of space was actually filled by
the universal aether.
This creates a “current” or
“aether wind”
In the laboratory, this
light beam travels
across the current
The Michelson-Morley Experiment
In 1887, American scientists A.A. Michelson and
E.W. Morley attempted to detect the aether by
observing the way that the movement of the
Earth through the aether would affect the
transmission of light.
Equipment is able
to be rotated
This one travels with
the aether wind
On arrival back at the start, the beams were recombined in an “interferometer”, producing an
interference pattern as the light waves recombined.
An Analogy to their experiment...
Imagine 2 identical boats, capable
of exactly the same speed. They
both travel a course out and back
over exactly the same distance, but
at right angles to each other.
Travels out and back
across the current
Stationary Aether
throughout the
Universe
The entire apparatus was mounted on a rotating
table. Once the apparatus was working, and the
interference pattern appeared, the whole thing
was rotated 90o, so that the paths of the light
rays in the aether wind were swapped.
Theoretically, this should have created a change
in the interference pattern, as the difference
between the beams was swapped.
In still water, they will get back at
the same time.
But what if there is a current?
Now, they will NOT arrive back at
the same time, because the current
will alter their relative speeds.
(The one moving across the current
will arrive later.)
The Result...
There was NO CHANGE in the interference
pattern.
The experiment was repeated in many other
laboratories, with more sensitve interferometers
and all sorts of refinements and adjustments.
The result remained negative... no effect of the
aether wind could be detected.
water current
Either the experiment has
something wrong with it
or
the theory of the “Aether” is wrong!
water current
water current
Enter Albert Einstein...
Travels with, and then against, the current
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Relative Motion and
Frames of Reference
How Science Works
The Michelson-Morley Experiment is probably
the most famous “failed experiment” in the
history of Science. It’s importance is not just
historical interest, but a lesson in how Science
works.
Ever been sitting in a train at a station looking at
another train beside you? Suddenly, the other train
begins moving. Or is it your train beginning to move
the other way?
The only way to be sure is to look out the other side
at the station itself, in order to judge which train is
really moving. You are using the railway station as
your “Frame of Reference” in order to judge the
relative motion of the 2 trains.
There is no such thing as a “failed experiment”!
Scientists produce hypotheses in an attempt to
explain the universe and its phenomena. There
can be 2 or more totally different hypotheses
attempting to explain the same thing.
We often use the Earth itself (or a railway station
attached to it) as our frame of reference. The Earth
seems fixed and immovable, so everything else can
be judged as moving relative to the fixed Earth... but
we also know it’s NOT really fixed and unmoving, but
orbiting around the Sun.
Natural Phenomenon
to be explained
Hypothesis 1
Hypothesis 2
e.g the Aether
e.g. Relativity
Predictions arise
from this idea.
Predictions arise
from this idea.
These can be
tested by
experiment
These can be
tested by
experiment
Experimental
Results
DO NOT
agree with
predictions
Experimental
Results
DO
agree with
predictions
Hypothesis
Rejected as
Wrong
Hypothesis
Accepted as
Correct Theory
Astronomers use the background of “fixed stars” as
their frame of reference to judge relative planetary
movements, but we know that these aren’t really fixed
either.
In fact, there is no point in the entire Universe
that is truly “fixed” that could be used as an
“absolute reference” to judge and measure all
motion against.
Sir Isaac Newton was aware of this idea, and
figured out that it really doesn’t matter whether
your frame of reference is stationary or moving
at a constant velocity. So long as it is not
accelerating,
the
observations,
and
measurements of motion will come out the same
anyway. This raises the idea of an “Inertial
Frame of Reference”.
An Inertial Frame of Reference
is not accelerating
Within any Inertial Frame of Reference
all motion experiments
(and all “Laws of Physics”)
will produce the same results
Distinguishing Inertial & Non-Inertial
Frames of Reference
Imagine you are inside a closed vehicle and cannot
see out. How can you tell if your “Frame of
Reference” is “Inertial” or not?
This is exactly what happened. In the 30 years
after the Michelson-Morley Experiment, a new
Hypothesis was proposed which did not require
any “aether”. From it arose many predictions
which have all been spectacularly confirmed by
experiment, so we believe the “Aether Theory”
is wrong, and “Relativity Theory” correct.
A simple indication would be to hang a mass on a
string from the ceiling. If it hangs straight down
there is no acceleration. If it hangs at an angle, (due
to its inertia) then your vehicle is accelerating.
Does it matter whether your vehicle is stationary or
moving at constant velocity? Not at all! The mass
still hangs straight down, and any Physics
experiments will give the same result as any other
observer in any other Inertial Frame of Reference.
The Michelson-Morley Experiment was not a
failure... it was a vital link in the scientific search
for truth.
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Albert Einstein’s Strange Idea
The Principle of Relativity
Albert Einstein (1879-1955) has gone down in
the History of Science as one of the “Greats”,
and just about the only scientist to ever match
the achievements of the great Sir Isaac Newton.
was already well known before Einstein, and stated in
various forms by Galileo, Newton and many others.
1. In an Inertial Frame of Reference
all measurements and experiments
give the same results
Einstein’s “Theory of Relativity” is famous as a
great achievement, (true!) and as something
incredibly complicated that hardly anyone can
understand (false! It’s a dead-simple idea, but it
defies “common sense”.)
2. It is impossible to detect the motion
of an Inertial Frame of Reference by
experiment within that frame of reference
3. The only way to measure the motion
of your frame of reference
is by measuring it against
someone else’s frame of reference
Einstein once declared “common sense” as “a
deposit of prejudice laid down in the mind prior
to the age of 18”. To understand “Einstein’s
Relativity” you need to ignore “common sense”
and have a child-like open-mind to fantasy and
the K.I.S.S. Principle...
These are all statements of the “Principle of
Relativity”.
Einstein’s Gedanken (a “Thought Experiment”)
Einstein had, in some ways, a child-like imagination. He wondered what it would be like to travel on
a train moving at the speed of light. (100 years ago a train was the ultimate in high-speed travel).
Train Velocity at, or near,
the speed of light
Relativity Train Tours
What if you tried to look in a mirror?
Classical Physics would suggest that light
(trying to travel in the aether wind) from
your face could not catch up to the
mirror to reflect off it.
So, vampire-llike, you have no reflection!
But Einstein remembered Michelson & Morley’s failure to
measure the “aether wind” and applied the Principle of
Relativity...
In a non-accelerating, Inertial Frame of Reference, you
would measure the speed of light (and anything else, like
reflection) exactly the same as anyone else... you would
see your reflection, and everything appears normal.
What Would Another Observer See?
What about a person standing in the train station
as you flash (literally!) through at the speed of
light? What would they see through the train
window as you zap by?
Again, according to the results of the MichelsonMorley experiment, these observers will measure
light waves from you as travelling at the same
speed of light as you measure inside the train,
because everyone is in an Inertial F. of R.
(Naturally, both train and platform are fully
equipped with interferometers and high-tech ways
to do this)
But, if you are travelling at the speed
of light, how is it possible for you,
and the stationary observers on the
platform, to both measure the same
light wave as having the
same velocity?
Well, says Einstein,
if THE SPEED OF LIGHT is FIXED,
then SPACE and TIME must be
RELATIVE.
What does this mean?
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Einstein’s “Thought Experiment” continued...
You are on a train travelling at, or near, the speed of light. You carry out some Physics experiments
and measure the speed of light, and the law of reflection as being perfectly normal.
Train Velocity at, or near,
the speed of light
Relativity Train Tours
Meanwhile, stationary observers are standing
on the platform as your train flashes by. They
also measure the speed of light and get the
exact same answer.
Seen and measured by them,
YOUR LENGTH & TIME HAS CHANGED!
And you see them the same way!
Einstein’s conclusion from the
Principle of Relativity and the
Michelson-Morley experiment is that:
However, the people
on the platform see
you as
compressed in
space like this:
The Speed of Light is Always the Same
(for observers in Inertial Frames of Reference)
and therefore,
LENGTH & TIME must change
as measured by another observer
who is in relative motion
Furthermore, when
they study your
clock they see it is
running much slower
than their own is.
Length Contraction & Time Dilation
If you can ignore “common sense” and accept the fantasy of a train moving
at 300,000 km/sec then Einstein’s proposal makes sense:
If everyone (in any Inertial F. of R.) measures the speed of light as being the same, then the measurements of
SPACE and TIME must be relative, and different as seen by an observer in another F. of R.
L = Lo 1 - v 2
c2
L = Length observed by outside observer
Lo= “rest length” measured within F.of R.
v = relative velocity of observer
c = speed of light = 3.00 x 108ms-1
It turns out that the measurement of length must
get shorter as your velocity increases...
(as seen by an observer in another Inertial F. of R.)
...and time gets longer. Time goes slower!
Example Calculation
On board a spacecraft travelling at “0.5c” (i.e. half
the speed of light = 1.50x108ms-1) relative to the
Earth, you measure your craft as being 100 metres
long. Carrying out this measurement takes you 100
seconds.
THIS IS LENGTH CONTRACTION.
IT OCCURS ONLY IN THE DIRECTION OF
THE RELATIVE MOTION
t=
Observers on Earth (with an amazing telescope) are
watching you. How much time elapses for them, and
what is their measurement of your spacecraft?
to
1 - v2
c2
Solution
The factor
t = time observed by outside observer
to= time measured within F.of R.
v = relative velocity of observer
c = speed of light = 3.00 x 108ms-1
= Sq.Root(1- (1/2)2/12)
= 0.866
So Length, L=Lox 0.866 = 100x0.866 = 86.6m.
Time, t = to/ 0.866 = 100/0.866 = 115s.
They see your craft as being shorter,
and your time as going slower!
THIS IS TIME DILATION
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c2
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Mass Changes Too
Relativity and Reality
Not only does length contract, and time stretch,
but mass changes too.
Do these alterations to time and space really
happen? Yes they do, and they have been
measured!
m=
• Extremely accurate “atomic clocks” have been
synchronized, then one flown around the world in a
high speed aircraft. When brought back together,
the clock that travelled was slightly behind the
other... while travelling at high speed it’s time had
slowed down a little, relative to the other.
• Certain unstable sub-atomic particles always
“decay” within a precise time. When these
particles are travelling at high speeds in a
particle accelerator, their decay time is much
longer (as measured by the stationary
scientists). At high speed the particle’s time has
slowed down relative to the scientists’ time.
It’s important for you to realise that, if this particle
could think, it would not notice any slow-down in
time... its own “feeling” of time and its little digital
watch would seem perfectly normal to it. But, from
the relative viewpoint of the scientists measuring
the particle’s decay, its time has slowed down
relative to laboratory time.
mo
1 - v2
c2
m = mass observed by outside observer
mo= “rest mass” measured within F.of R.
v = relative velocity of observer
c = speed of light = 3.00 x 108ms-1
THIS IS MASS DILATION
Two of the most fundamental laws ever
discovered by Science are the “Law of
Conservation of Energy” and the “Law of
Conservation of Matter”. These state that
energy and matter (mass) cannot be created nor
destroyed.
Einstein found that the only way to avoid
breaking these laws under “Relativity” was to
combine them. Hence, the most famous
equation of all:
Confirmation of Relativity
E = mc2
Einstein published his theory in 2 parts, in 1905 and
1915. At that time there was no way to test the
predictions of Relativity to find supporting evidence.
E = Energy, in joules
m = Mass, in kg
c = speed of light = 3.00 x 108ms-1
The Michelson-Morley experiment had failed to find
supporting evidence for the existence of the “aether”,
so maybe “Relativity” would fail too, but first
scientists had to find testable predictions.
THIS IS THE EQUIVALENCE OF
MASS & ENERGY
The first test was that, according to Relativity, light
from a distant star passing close to the Sun should be
bent by a measurable amount, making the star appear
to change position in the sky. The only way to test this
prediction was during a solar eclipse.
Within 30 years of Einstein’s ideas being
published, the “Equivalence of Mass & Energy”
was dramatically confirmed by the release of
nuclear energy from atomic fission.
At the next occurrence of an eclipse, the
observations were made, and showed results
exactly as predicted by Relativity.
In the following years, experiments with nuclear
reactions (which led to the development of the
“atom bomb”, and nuclear power) were able to
confirm the conversion of matter into energy
according to E=mc2.
Photo of the atom bomb
explosion on the
Japanese city of Nagasaki,
1945.
Later still came the measurements of time
dilation (described above) and mass dilation has
also been measured for high-speed particles in
a particle accelerator.
This is the process occurring in a nuclear
reactor used to generate electricity in many
countries. It is also the energy source in an
“atomic bomb”.
EVERY RELATIVITY PREDICTION THAT CAN BE
TESTED HAS SHOWN RESULTS SUPPORTING
& CONFIRMING THE THEORY...
that’s why we believe it to be correct.
HSC Physics Topic 1 “Space”
Copyright © 2005-2
2009 keep it simple science
www.keepitsimplescience.com.au
Einstein, a life-long pacifist, was appalled by the
destructive uses of the technology which grew
from his discoveries.
27
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Some Implications of Relativity
This all means that as your spacecraft
accelerates and approaches the speed of light,
your faithful observer sees your length
approach zero, your time slowing down and
approaching being totally stopped, and your
mass increasing to approach infinity.
What Happens as
Speed Approaches “c”
Several of the Relativity equations
contain the factor:
1 - v2
c2
This is known as the “Lorentz-FitzGerald
Contraction”. In the following explanations it
will be referred to as the “LFC”.
At the speed of light, the calculations for time
and mass dilation become mathematically
“undefined”... this is generally taken to mean
that no object can ever be accelerated up to the
speed of light.
Consider firstly, what happens to the value of
the LFC at different relative velocities:
If V=zero:
Another way to reach this conclusion is that as
you speed up, your mass increases. To
accelerate more, greater force is needed
because your increased mass resists
acceleration. As your mass approaches infinity,
an infinite amount of force is needed to
accelerate you more...it’s impossible to reach c.
LFC = Sq.Root(1-0) = 1
This means that if you (in your spacecraft) and
the observer watching you have zero relative
velocity (i.e. you are travelling at the same
relative speed) then both of you will measure
the same length, time and mass... no relativistic
effects occur.
All the energy put into trying to accelerate goes
into increasing your mass, according to E=mc2.
As V increases, the value of the LFC decreases:
Value of
Relative Velocity
(as fraction of c)
LFC
0.1c
0.995
0.5c
0.886
0.9c
0.436
0.99c
0.141
0.999c
0.045
Another consequence of Relativity is that you,
and your observer, will not agree on
simultaneous events. You may see 2 things
occur at the same instant, but the relativistic
observer will see the 2 events occurring at
different times.
It is even possible that the observer could see
an effect (e.g. spilt milk) before seeing the cause
(e.g. glass tipped over). Curiouser & curioser!
Approaching zero
Approaching c
If V = c:
Simultaneous Events
LFC = Sq.Root( 1 - 1) = zero
How We Define Length & Time
Our S.I. unit of length, the metre, was originally
defined by the French as “One ten-millionth of
the distance from the Equator to the Earth’s
North Pole”.
Our defintion of length is actually based on the
measurement of time! (What’s even more
amazing is that we actually have ways to
measure such a fraction of a second!)
Based on this,
special metal bars
were carefully
made to be used as
the “standard”
metre from which
all other measuring
devices were
One of the precisely-m
made
made.
platinum bars used to define the
So how do we define “a second” of time?
The modern definition involves a multiple of the
time it takes for a certain type of atom to
undergo an atomic “vibration”, which is
believed to be particularly regular and is, of
course, measurable.
metre up until 1960.
As our technology
improved, so did our ability to measure time and
distance. Today we define the metre as “the
distance travelled by light during a time interval
of 1/299,792,458th of a second.”
HSC Physics Topic 1 “Space”
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2009 keep it simple science
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28
Part of the
mechanism of
an “atomic
clock” which
measures
atomic
vibrations to
give our
standards of
time and
distance.
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Worksheet 9
Relativity
Fill in the blank spaces.
Student Name...........................................
The theory of the “aether” was invented to
explain a).........................................................
because it was thought that all waves needed a
b)............................. to travel through. The aether
was invisible and c)................................, and was
present throughout the d).....................................
Albert Einstein applied this principle to the
Michelson-Morley result. He concluded that all
observers will always measure the speed of light
as being p)..................................... For this to
happen,
then
q).........................
and
.............................. must be relative. This means
that the measurements of length and time as
seen
by
r)....................
.......................................................... ................ will
be different.
The American scientists e).............................. &
............................... attempted to detect the
aether by experiment. Their apparatus used 2
f)..................................., travelling at right angles.
When brought together by mirrors, the beams
produced an g).......................................... pattern.
The idea was that the pattern should change
when the apparatus was h)...................................,
because one beam should be travelling with the
“aether wind” and the other i)...............................
it. This “aether wind” would be caused by
j)...................... ........................... through space.
The result was that k)...........................
..............................................................
Relativity Theory predicts that Length will
s)....................... while time will t)..........................
Also, mass will u)............................., thereby
making
it
impossible
to
actually
v).............................................. Relativity also
predicts that mass can be converted into
w).............................. and vice-versa.
An “l)......................... Frame of Reference” is one
which is not m)....................................... Within
such a place, all measurements and
experiments will give the n)................
.......................... This idea is known as the
“Principle of o)......................................”.
Although it defies common sense, many aspects
of Relativity have been confirmed by
x).........................................
For
example,
synchronised clocks have been found to
disagree
if
one
of
them
is
y)..............................................................
The
conversion of mass into energy has been
observed
(many
times)
during
z)............................ reactions.
Worksheet 10
Relativity
Student Name ...........................................
Practice Problems
2.(cont.)
b) What relativistic mass will it have if accelerated up
to 0.9999c? (99.99% of “c”)
1. A spacecraft is travelling at 95% of the speed of
light relative to an observer on Earth. On board is a
fluorescent light tube which is 0.95m long and is
switched on for 1 hour ship-time.
a) How long is the fluoro tube as seen be the Earth
observer?
3. In a nuclear reactor, over a perio of time, a total of
2.35kg of “mass deficit” occurs. This mass has
“disappeared” during the nuclear reactions.
Calculate the amount of energy this has released.
b) The Earth observer measures the time for which
the light was on. What time does he measure?
4. According to the “Big Bang” Theory, in the first
moments of the Universe there was nothing but
energy. Later, matter formed by conversion from the
energy.
2. A sub-atomic particle has a “rest mass” of
5.95x10-29kg. The particle was accelerated by a
particle accelerator up to a velocity of 0.99c. (99% “c”)
Calculate how much energy was needed to produce
enough matter to form the Earth
(mass= 5.97 x 1024kg).
a) What relativistic mass will the particle now have, if
measured by the scientists in the laboratory?
HSC Physics Topic 1 “Space”
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Worksheet 11 Test Questions
Multiple Choice
sections 3 & 4
Longer Response Questions
Mark values shown are suggestions only, and are to
give you an idea of how detailed an answer is
appropriate. Answer on reverse if insufficient space.
1. The gravitational force between 2 masses is
“F” units when they are distance “d” apart.
If these masses were brought closer, to a
distance “0.25d”, then the force between them
would be:
A. 0.25 F
B. 4F
C. F/16
D. 16F
7. (5 marks)
An astronaut with mass (including spacesuit) 120kg
is standing on the surface of the planet Mercury. The
planet has a mass of 2.99x1023kg and radius
2.42x106m.
a) Calculate the gravitational force acting on the
astronaut.
2. The “Aether” was an idea
A. for a new anaesthetic.
B. used to explain the Principle of Relativity.
C. to explain how light could travel in a vacuum.
D. to explain the interference of light waves.
b) From your answer to (a) calculate the value of “g”
on the surface of Mercury.
3. In an “Inertial Frame of Reference” a mass
hanging from the ceiling by a string would
probably.
A. swing back and forth.
B. hang straight down.
C. hang at an angle from the vertical.
D. undergo mass dilation due to relativistic
effects.
8. (5 marks)
a) Explain what is meant by the “Slingshot Effect”.
b) Why is it useful in space exploration.
c) State how energy is conserved in the process.
9. (7 marks)
Give a brief description of the famous MichelsonMorley experiment, including:
a) their aim or purpose in doing the experiment.
4. Two cosmonauts in separate spacecraft
travelling at different relativistic velocities are
able to make a series of observations and
measurements of their own spacecraft, and of
each other’s spacecraft. The one thing they
would agree with each other about is:
A.
B.
C.
D.
the
the
the
the
b) an outline of the method used.
c) the results.
simultaneity of 2 events occurring together.
value of the velocity of light.
passage of time in each other’s spacecraft.
length of each other’s spacecraft.
10. (3 marks)
What did Albert Einstein conclude about the
measurement of space and time, taking into account
the results of the Michelson-Morley experiment?
5. The first
experimental confirmation of
Einstein’s Theory of Relativity was:
A. the change in the apparent position of a star
due to the gravity of the Sun.
B. the result of the Michelson-Morley
Experiment.
C. the release of energy from the first
atom bomb.
D. measurements made with a mirror on a
high speed train.
11. (8 marks)
A sub-atomic particle, at rest in the laboratory, has the
following characteristics:
Mass = 3.22x10-27kg.
Diameter= 7.38x10-16m.
Half-Life (Time to “decay”)= 2.58x10-2s.
The particle is now accelerated up to a velocity of
0.999c (i.e. 99.9% of the speed of light) in a particle
accelerator. Calculate:
a) its relativistic
i) diameter
ii) half-life
iii) mass
as measured by the scientists in the laboratory.
6.
From the Earth, you are able to observe and
measure several features of an alien spacecraft
as it flies by at 90% of the speed of light.
Compared to the measurements made by the
alien on board, your measurements would
show:
Craft mass
Craft length
Craft time
A.
less
shorter
faster
B.
more
longer
slower
C.
more
shorter
slower
D.
less
longer
faster
HSC Physics Topic 1 “Space”
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2009 keep it simple science
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Student Name...........................................
b) While travelling at 0.999c, the particle undergoes a
nuclear reaction which results in its total annihilation
by conversion to energy. What energy release is
measured by the scientists? (The scientists observe
the conversion of its relativistic mass, not rest-mass)
30
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CONCEPT DIAGRAM (“Mind Map”) OF TOPIC
Some students find that memorising the OUTLINE of a topic
helps them learn and remember the concepts and important facts.
Practise on this blank version.
SPACE
HSC Physics Topic 1 “Space”
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2009 keep it simple science
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Worksheet 4
Answer Section
Worksheet 1
a) force
c) accelerate
e) period
g) force field
i) repel
k) mass
m) infinity
o) 9.81
q) mass
1.
Uy = U.Sinθ
Ux = U.Cosθ
a) = 20.5xSin60
=20.5xCos60
= 17.8ms-1.
= 10.3ms-1.
b) vertical = zero
horizontal = 250ms-1.
c) Uy = 15.0xSin25
Ux = 15.0xCos25
= 6.34ms-1.
= 13.6ms-1.
d)
350xSin70
350xCos70
= 329ms-1.
= 120ms-1.
e) 40.0xSin45
40.0xCos45
= 28.3ms-1.
= 28.3ms-1.
2.
a) At highest point, Vy=0, and Vy = Uy + g.t
0 = 28.3 + (-9.81x t)
t = -28.3/-9.81
= 2.88s.
b) Sy = Uy.t + 1.g.t2
2
= 28.3x2.88 + (0.5x (-9.81) x 2.882)
= 81.5 + ( -40.7) = 40.8m.
c) Sx = Vx.t = 28.3 x (2.88x2)
(twice the time to reach max.ht.)
= 163m.
3.
a) It is fired from max height,
so Sy = -2.00 (down, so -ve)
Sy = Uy.t + 1.g.t2
2
-2.00 = 0xt +(0.5x( -9.81)x t2)
-2.00 = 0 - 4.905 x t2
t2 = -2.00/-4.905
t = 0.639s.
b) Sx = Vx.t = 250x0.639 = 160m.
c) see working for (a).
Empty cartridge takes 0.639s to hit the ground. It
falls down at exactly the same rate as the bullet. The
difference is where each lands horizontally.
4.
a) At highest point, Vy=0, and Vy = Uy + g.t
0 = 329 + (-9.81)x t
t = -329/-9.81
= 33.5s.
b) Sy = Uy.t + 1.g.t2
2
= 329x33.5 + (0.5x( -9.81)x33.52)
= 11,022 - 5,505
= 5,517 = 5.52x103m.
c) Sx = Vx.t = 120x(33.5x2)
(twice the time to reach max.ht.)
= 8,040 = 8.04x103m.
5.
Horizontal Displ.
a) Vertical displacement
Sy = Uy.t + 1.g.t2
Sx = Vx.t
2
= 10.3 x 2.50
= 17.8x2.50 + (0.5x(-9.81)x2.502) = 25.8m
= 44.5 + (-30.65)
=13.4m (+ve, therefore up)
Ball is 25.8 metres down-field and 13.4 m high.
Horizontal velocity
b) Vertical velocity
Vy = Uy + g.t
Vx = Ux = 10.3 ms-1
= 17.8 + (-9.81)x2.50
10.3
= -6.725ms-1 (downwards)
b) gravity
d) 10 (9.81)
f) gradient
h) attracts
j) gravitational field
l) work done
n) negative
p) weight
Worksheet 2
1.
a) i) 575kg
ii) 575kg
iii) 575kg.
b) i) W=mg = 575x9.81 = 5,641 = 5.64x103N.
ii) W=mg = 575x1.6 = 920 = 9.2x102N.
iii) W=mg = 575x25.8 = 14,835 = 1.48x104N.
2.
a) On Mars; W=mg, so m=W/g = 250/2.8 = 65.8kg
On Earth; W=mg = 65.8x9.81 = 645 = 6.5x102N.
b) On Neptune; W=mg = 65.8x10.4= 684 = 6.8x102N.
c) On Moon; W=mg = 65.8x1.6 = 105 = 1.1x102N.
3.
a) On Neptune; W=83.0 =mg, so m= 83.0/10.4 = 7.98kg.
b) On Earth; W=mg = 7.98x9.81 = 78.3N.
c) W=206=mg, so g=206/7.98 = 25.8ms-2.
matches Jupiter
Worksheet 3
a) only under gravity
b) trajectory
c) parabola
d) horizontal & vertical
e) angle
f) resolving
g) components
h) constant velocity
i) acceleration
j) gravity
k) time
l) zero
m) height
n) range
o) horizontal & vertical p) 45
q) Galileo
r) at the same rate
s) velocity
t) vertical
u) Newton
v) escape
w) orbit
x) escape from the Earth’s gravitational field
y) velocity
z) curvature
aa) satellite
ab) in orbit
ac) accelerated
ad) g-forces
ae) weight
af) velocity
ag) g-forces
ah) rockets
ai) liquid
aj) east
ak) rotation
al) 3rd
am) conserved
an) forward
ao) rocket
ap) decreases
aq) increases
ar) g-forces
as) low-Earth
at) 200-1,000
au) quickly/fast
av) photos & surveys
aw) geo-stationary
ax) period
ay) rotation
az) same position in the sky
ba) communication
bb) centripetal
bc) centre of the circle bd) tension in the string
be) friction
bf) gravity
bg) Kepler
bh) gravitational
bi) decay
bj) gas molecules/upper atmosph.
bk) deceleration
bl) heat
bm) angle
HSC Physics Topic 1 “Space”
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V2 = Vy2 + Vx2 = 10.32 + 6.7252
∴ V = sq.Root(151.32) = 12.3ms-1.
Tan θ = 6.725/10.3,
∴ θ ≅ 33o below horizontal
32
θ
Re
sul
tan
t
6.725
vel
oci
ty
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Worksheet 5
Worksheet 7
1.
a) T=1.74 hours = 1.74x60x60= 6,264s
R= 1,000 km (=106m) + 6.37x106 = 7.37x106m
V = 2πR/T
= 2xπx7.37x106/6,264
= 7,393 = 7.39x103ms-1.
b) Fc=mv2/R = 600x(7.39x103)2/7.37x106
= 4.45x103N.
2.
a)Fc=mv2/R, so R = mv2/F = 1,500x(6.13x103)2/5.32x103
= 1.06x107m.
b)Altitude =1.06x107 - 6.37x106 = 4.23x106m (4,230km)
c) V=2πR/T, so T = 2πR/V = 2xpx1.06x107/6.13x103
= 1.09x104s. (3.02 hours)
3.
R = 350km + 6.37x106m = 6.72x106m
T= 1.52 hrs = 1.52x60x60 = 5.47x103s.
a) V=2πR/T = 2xπx6.72x106/5.47x103
= 7.72x103ms-1.
b)Fc=mv2/R, so m=F.R/v2 = 2,195x6.72x106/(7.72x103)2
= 247kg.
1. C
7. A
2
3
Radius(m)
7.37x106
1.06x107
6.72x106
Period(s)
6.26x103
R3/T2
1.02x1013
1.09x104
1.00x1013
5.47x103
1.01x1013
4. D
10. D
5. B
11. A
6. B
12. D
W=mg, so m=W/g = 5.84x103/9.81
= 595kg.
b) On home planet: W=mg, so g=W/m = 7.25x103/595
= 12.2ms-2.
14.
Uy=0, Ux=5.45ms-1, Sy = -1.20m (down (-ve))
Time of flight: Sy = Uy.t + 0.5.g.t2
-1.20 = 0xt + (0.5x(-9.81)xt2)
t = sq.root( -1.20/-4.905)
= 0.495s.
Horizontal distance: Sx = Ux.t = 5.45x0.495 = 2.95m.
The ball lands 2.95m from the base of the table.
15.
Uy = U.Sinθ
= 42.0xSin60
= 36.4ms-1
Ux = U.Cosθ
= 42.0xCos60
= 21.0ms-1.
a) At max.height, Vy = 0,
and Vy = Uy + g.t
0 = 36.4 x (-9.81)x t
t = -36.4/-9.81
= 3.71s (to highest point)
Time of flight = 3.71x2 = 7.42s.
b) Sy = Uy.t + 0.5.g.t2 (use time to highest point)
= 36.4x3.71 + (0.5x(-9.81)x3.712)
= 135 + ( -67.5) = 67.5m.
c) Range: Sx = Ux.t = 21.0x7.42 (Time for entire flight)
= 156m.
16.
a) Uy=0, Ux=300ms-1, Sy = -15,000m (down (-ve))
Time of flight: Sy = Uy.t + 0.5.g.t2
-15,000 = 0xt + (0.5x(-9.81)xt2)
t = sq.root( -15,000/-4.905)
= 55.3s.
Horizontal distance: Sx = Ux.t = 300x55.3 = 16,590m
= 1.66x104m.
Bombs must be released over 16km before the target.
b) Vy = Uy + g.t
Ux=300ms-1.
= 0 + ( -9.81)x55.3
300
θ
= 542ms-1.
Re
V2 = Vy2 + Vx2 = 5422 + 3002
sul
tan
∴ V = sq.Root(383,764) = 619ms-1.
tv
elo
city
(almost twice the speed of sound!)
The ratio R3/T2 is the same for all 3 satellites.
(slight differences are due to rounding-off errors in
calculations)
Keplers Law states that this ratio should be the same
for the satellites of any planet.
2.
Average value from table = 1.01x1013= constant
a) R3 = constant x T2
R=Cube.Root(1.01x1013x(1.60x103)2) = 2.96x106m.
b) R=Cube.Root(1.01x1013x(1.15x104)2) = 1.10x107m.
c) T2 = R3/constant
T =Sq.Root(2.56x107)3/1.01x1013) = 4.08x104s.
d) R = 2,000km + 6.37x106m = 8.37x106m.
T =Sq.Root(8.37x106)3/1.01x1013) = 7.62x103s.
3.
a) constant = GM/4π2 = 6.67x10-11x6.57x1023/4xπ2
= 1.11x1012
b) R3/T2 = constant, so R=Cube.Root(constant x T2)
=
Cube.root(1.11x1012x(1.60x103)2)
= 1.42x106m.
c) T =sq.Root(R3/constant) = (2.56x107)3/1.11x1012
= 1.23x105s.
d) Earth satellite, T = 4.08x104s.
Mars satellite, T = 1.23x105s.
Earth satellite’s period is shorter, therefore it travels
faster.
17.
He imagined a cannon firing projectiles horizontally
from a very high mountain. As the launch velocity is
increased, the shot travels further before reaching the
ground. At a high-enough velocity, the downward
curve of its trajectory could match the curvature of
the Earth. This projectile would travel right around the
Earth, constantly falling towards it, but never getting
any closer.
18.
a) a = (v-u)/t = (1.05x104 - 0)/7.50
= 1,400 (1.40x103) ms-2.
b) multiple of “g”: 1,400/9.81 =143 times “g”
Passengers experience g-force equivalent of 143
times their normal weight.
c) This amount of g-force would be instantly fatal.
This is not a feasible method for human space launch.
e) At a given orbital radius, a satellite orbiting a
smaller planet needs to travel at a ....lower.....velocity.
The bigger the planet, the ....faster (higher).... the
velocity would need to be.
HSC Physics Topic 1 “Space”
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3. C
9. D
13.
a) On Earth:
Worksheet 6
1.
Question
1
2. A
8. C
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Worksheet 9
Worksheet 7 (cont.)
a) transmission of light in vacuum
b) medium
c) massless / weightless
d) Universe
e) Michelson & Morley
f) beams of light
g) interference
h) rotated 90 degrees
i) across
j) the Earth’s motion
k) no change to the interference pattern-no aether
wind detected.
l) Inertial
m) accelerating
n) same results
o) Relativity
p) the same
q) space & time
r) an observer travelling at a different relative velocity
s) shorten
t) lengthen / slow down
u) increase
v) accelerate to the speed of
light
w) energy
x) observation/experiment
y) transported at high speed
z) nuclear
19.
(example) For 3 marks, try to make 3 points.
Robert Goddard built and tested the first liquid-fuel
rocket engine, an essential step for practical rocketry.
His experiments in the 1920’s-1930’s were the basis
for later research during World War II which led to the
first long-range rockets. Goddard also advanced the
theory of multi-stage rockets as the way to reach
outer space.
20.
Although a rocket may produce constant thrust, the
mass decreases rapidly as the fuel is burned. This
means that the rate of acceleration increases during
the ascent. This means that the g-forces keep
increasing.
By being able to throttle back the engines, the Space
Shuttle can reduce this effect and maintain more
constant g-forces during the acent. This is much
safer and more comfortable for the astronauts.
Worksheet 10
1 - v2
c2
Using the abbreviation “LFC”=
21.
a) period, T= 2.00hr = 2.00x60x60 = 7,200s.
R3/T2 = GM/4π2
R = Cube.Root(GMT2/4π2)
= Cube.Root(6.67x10-11x5.97x1024x(7,200)2/4π2
= 8.06x106m.
b) V = 2πR/T = 2xpx8.06x106/7,200
= 7.03x103ms-1.
2
c) Fc=mv /R = 2,650x(7.03x103)2/8.06x106
= 1.62x104N.
22.
It is impossible for a spacecraft to carry enough fuel
to use its rocket engines for the complete
deceleration and descent from the high speed of
orbit. So atmospheric braking is used instead. If the
capsule enters the atmosphere at just the right angle,
friction can slow it down gradually, with the energy
being converted to heat.
1.
a) At 0.95c, LFC = Sq.root( 1 - (0.952/12))
= 0.31
L = L0 x LFC = 0.95 x 0.31 = 0.29m.
b) t = t0 /LFC = 1/0.31 = 3.2 hours.
2.
a) At 0.99c, LFC = Sq.root(1 - (0.992/12))
= 0.14
m = m0 / LFC = 5.95x10-29/0.14 = 4.25x10-28kg.
b) At 0.9999c, LFC = Sq.root(1 - (0.99992/12))= 0.01414
m = m0 / LFC = 5.95x10-29/0.01414 = 4.21x10-27kg.
3.
E=mc2 = 2.35x(3.00x108)2 = 2.12x1017J.
4.
E=mc2 = 5.97x1024x(3.00x108)2 = 5.37x1041J.
If the re-entry angle is too steep, the g-forces will be
too high, and the craft may burn up. If the angle is too
shallow, the craft may bounce off the atmosphere and
go into an un-recoverable orbit.
Worksheet 11
1. D
Worksheet 8
3. B
4. B
5. A
6. C
7.
a)FG=GMm/d2 = 6.67x10-11x2.99x1023x120/(2.42x106)2
= 409N.
b) This gravitational force is the astronaut’s weight,
and W= mg, so g=W/m = 409/120 = 3.41ms-2.
a) the product of the masses
b) the square of the distance between them
c) double
d) decrease by a factor of 4.
e) centripetal
f) the planets
g) direction/trajectory
h) speed
i) Slingshot
j) planet
k) equal
8.
a) The Slingshot Effect is the technique of flying a
spacecraft near a planet so that the planet’s gravity
accelerates the craft and alters its trajectory onto a
new, desired heading.
Practice Problems
1.
FG = GMm/d2 = 6.67x10-11x75x60/0.52
= 1.20x10-6N. (about 1 millionth of a newton)
2.
FG = GMm/d2 = 6.67x10-11x6.75x108x2.48x109/425
= 2.63x105N.
3.
d =Sq.Root(GMm/F)
= Sq.root(6.67x10-11x6.02x1022x5.67x1010/6.88x1010)
= 1.82x106m.
(Since this equals 1,820km, and the radius of the
Moon is 1,738km, then the comet is just 82km from
the surface... DEEP IMPACT about to happen!)
HSC Physics Topic 1 “Space”
Copyright © 2005-2
2009 keep it simple science
www.keepitsimplescience.com.au
2. C
b) It is useful for enabling a craft to visit several
planets on one flight, gaining speed and a new
direction without the need for large amounts of fuel.
c) The gain of energy by the spacecraft is exactly
equal to the loss of energy by the planet (which loses
an insignificant amount of rotational energy).
34
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11.
a) At 0.999c, the “LFC” factor
Worksheet 11 (cont)
9.
a) The M-M experiment was designed to detect the
“universal aether” which had been hypothesised as a
medium for transmission of light waves.
1 - v2
c2
= Sq.Root(1 - 0.9992) = 0.0447
12
i) L = L0 x LFC
= 7.38x10-16x0.0447
= 3.30x10-17m. (observed diameter)
ii) t = t0 /LFC
= 2.58x10-2 / 0.0447
= 5.77x10-1s (observed half-life)
iii) m = m0 / LFC
= 3.22x10-27/ 0.0447
= 7.20x10-26kg (observed mass)
b) E = mc2 = 7.20x10-26x(3.00x108)2 = 6.48x10-9J.
b) The method involved a beam of light, split into 2
beams which travelled at right angles. Mirrors then
recombined the beams in an interferometer to
produce an interference pattern. The apparatus could
then be rotated 90 degrees. Since one beam traveled
along the “aether wind” (caused by the Earth’s
movement) and the other across it, there should have
been a change in the interference pattern when the
apparatus was rotated.
c) The result was negative. No change in the
interference pattern occurred.
10.
Einstein concluded that the speed of light is constant
for all observers in any Inertial Frame of Reference
(IFR), regardless of relative motion. This why the M-M
experiment failed to detect a difference between the
speed of light in 2 different directions.
However, this requires that the measurements of
space and time must be different relative to an
observer’s IFR. This means that measurements of
length and time taken in one IFR, will be different to
what a relativistic observer sees.
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