hsc physics notes kiss

KCiC Physics 5 Spacecopyright © 2009 keep it simple sciencewww.keepitsimplescience.com.au

Slide 1

keep it simple scienceKey Concepts in Colour

HSC Physics Topic 1SPACE

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keep it simple science


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keep it simple science SPACEFirst, Some Revision:

Mass, Weight & Gravitywere covered briefly in the Preliminary Course.

In this topic you will revise these concepts, and beintroduced to the concept of “Gravitational PotentialEnergy”.

Then, you move on to studytwo important forms ofmotion that arecontrolled bygravity...

Projectiles......and Satellites in Orbit.

You will studyhow Gravity isresponsible for

holding theSolar System

together...

and study avariety ofaspects of

Physicsthat relate

to SpaceTravel

Launch & Re-entryare the

tricky bits...

Orbiting issimple

Physics!

Once launched,the path of aprojectile is

entirelydetermined by

gravity.

There are over 1,000 artificialsatellites in Earth orbit.

Some provide communication linksfor telephone, internet and TV.

Others watch the weather, or studypatterns of land use, or search for

natural resources.

Some are for military surveillance.

In the final section you will study one of the mostfamous (and least understood) theories of Science:

Einstein’s Theory of Relativity

Earth is in agravitational orbitaround the Sun

1970’s Apollo missionto the Moon

HSC Physics Topic 1



Gravity &Space Probes

Projectiles

Einstein’sTheory

Launch &Re-Entry ofSpace Craft

Frames ofReference &

Relativity

Satellites &Orbits

Circular Motion

Evidence forRelativityTheory

The AetherTheory

UniversalGravitation

Mass, Weightand “g”

Kepler’s Lawof Periods

GravitationalPotentialEnergy

SPACE

1. Gravity &Gravitational

Fields2. Projectiles& Satellites

3. Newton’sLaw of

Gravitation4. Einstein’s

Relativity

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Weight & GravityYou should already be aware that the “Weight” of anobject is the Force due to gravity, attracting theobject’s mass toward the Earth. You also know that(ignoring air resistance) all objects near the Earth willaccelerate downwards at the same rate. Thisacceleration rate is known as “g”, and isapproximately 10ms-2.

Gravitational FieldIn one way, Gravity resembles electrical charge andmagnetism... it is able to exert a force on thingswithout touching them. Such forces are explained byimagining that there is an invisible “Force Field”reaching through space.

Gravitational fields are imagined to surroundanything with mass... that means all matter, and allobjects. The field exerts a force on any other massthat is within the field.

Unlike electro-magnetism, gravity can only attract; itcan never repel.

Of the various “field forces”, Gravity is by far theweakest, although when enough mass isconcentrated in one spot (e.g. the Earth) it doesn’tseem weak!

1. GRAVITY & GRAVITATIONAL POTENTIAL ENERGY

Weight = Mass x Acceleration due to Gravity

W = mgWeight is in newtons (N)Mass in kilograms (kg)

“g” is acceleration due to gravity = 9.81 ms-2.



Measuring “g”One of the first activities you may have

done in class would have been to determine the valueof “g”, the acceleration due to gravity.

A common experimental method to do this involvesusing a pendulum.

By accurately timing (say) 10 swings of the pendulum,and then dividing by 10, the Period (T) can bemeasured. This value needs to be squared forgraphing.

The length of thependulum (L) is alsomeasured asaccurately aspossible.

Typically, themeasurements arerepeated for severaldifferent lengths ofpendulum, then theresults are graphedas shown.

Time taken for 1complete (back-aand-forth) swing is called

the “Period” of thependulum (“T”)

LLeenngg

tthh iinn

mmeett

rreess

How a Pendulum Relates to “g”

It turns out that the rate atwhich a pendulum swings (itsPeriod) is controlled by only 2things:

• its length, and• the acceleration due to gravity

Mathematically,

T2 = 4ππ2Lg

so, T2 = 4ππ2

L g

Analysis• The straight line graph shows there is a directrelationship between the Length (L) and the (Period)2.

• Gradient, T2 = 4ππ2 ≅≅ 4.0L g

Therefore, g ≅≅ 4ππ2/4.0 = 9.9 ms-2.0 0.5 1.0Length of Pendulum (m)

(Per

iod)

2(s

2 )00

11..00

22..00

33..00

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ooff bb

eesstt ff

iitt

Accepted value,g = 9.81ms-2

Gradient = T22 ≅≅ 4.0 L

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You are NOTrequired toknow thisequation.

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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Potential Energy is commonly defined as the energy“stored” in an object. In the case of any object on ornear the Earth, the amount of GPE it containsdepends on

• its mass• its height above the Earth

If that object is allowed to fall down, it loses someGPE and gains some other form of energy, such asKinetic or Heat. To raise the object higher, you must“do work” on it, in order to increase the amount ofGPE it contains.

However, for mathematical reasons, the point wherean object is defined to have zero GPE is not on Earth,but at a point an infinite distance away. So GPE isdefined as follows:

This definition has an important consequence: it defines GPE as the work done to bring an objecttowards the Earth, but we know that you need to dowork to push an object (upwards) away from Earth.

Therefore, GPE is, by definition, a negative quantity!

Gravitational Potential Energy is a measure of thework done to move an object from infinity,

to a point within the gravitational field.

GPE = -GmMR

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

Gravitational Potential Energy (GPE)

Note: the HSC Syllabus does NOT require you to carryout calculations using this equation. You ARE requiredto know the definition for GPE.

In the interests of better understanding, here is anexample of how the equation could be used:

How much GPE does a 500kg satellite have when inorbit 250km (= 250,000m) above the Earth’s surface?(Earth’s mass = 5.98x1024kg, Earth radius = 6.38x106m)

Solution GPE = -GmMR

= -6.67x10-11x500x5.98x1024

(6.38x106 + 250,000)= -3.00x1010 J.

The negative value is due to the definition of GPE.



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We are so used to the gravity effects on Earth that weneed to be reminded that “g” is different elsewhere,such as on another planet in our Solar System.

Since “g” is different, and W = mgit follows that things have a different weight if taken toanother planet.

Values of “g” in Other Places in the Solar System:

Planet g g (ms-2) (as multiple of Earth’s)

Earth 9.81 1.00Mars 3.8 0.39Jupiter 25.8 2.63Neptune 10.4 1.06Moon 1.6 0.17

Gravity and Weight on Other Planets

Calculating a Weight on another PlanetExampleIf an astronaut in his space suit weighs 1,350N onEarth, what will he weigh on Mars where g=3.84ms-2?

Solution W = mg On Earth, 1,350 = m x 9.81

∴∴ mass = 1,350/9.81 = 137.6 kg

So on Mars, W = mg = 137.6x3.84 = 528kg.



Activity 1The following activity might be completed by class discussion,

or your teacher may have paper copies for you to do.

Gravity Student Name .................................

1. A student tried to measure “g”, the acceleration due to gravity, using a“ticker-timer” measurement on a falling object. No matter how carefully it wasdone, the final result was always very low... 8 ms-2 or less.Suggest why this method always results in a large error.

2.a) What is the definition of Gravitational Potential Energy”?

b) What unusual feature arises from this definition?

3. Explain why you would have a different weight on another planet.

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What is a Projectile?A projectile is any object that is launched, and then

moves only under the influence of gravity.

Examples:Once struck, kicked or thrown, a ball in any sport becomes aprojectile.

Any bullet, shell or bombis a projectile once it is

fired, launched ordropped.

2. PROJECTILES & SATELLITES

Projectiles

An example which is NOT a Projectile:

A rocket or guided missile,while still under power, isNOT a projectile.

Once the engine stops firing itbecomes a projectile.

Projectiles are subject to only one force... Gravity!

When a projectile is travelling through air, there is, ofcourse, an air-resistance force acting as well. Forsimplicity, (K.I.S.S. Principle) air-resistance will beignored throughout this topic.

In reality, a projectile in air, does not behave the way described here because of the effects

of air-resistance.The exact motion depends on many factors and the

Physics becomes very complex, and beyond thescope of this course.

Not aProjectile



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By simple observation of a golf ball trajectory, or athrown cricket ball, the motion of any projectile can beseen to be a curve. It is in fact a parabola, and youmight think the Physics of this is going to be difficult.NOT SO... it is really very simple.

You must analyse projectile motion as 2 separatemotions; horizontal (x-axis) and vertical (y-axis) must be dealt with separately, and combined as

vectors if necessary.

Horizontal Motion is CONSTANT VELOCITY

Vertical Motion is CONSTANT ACCELERATION at “g”, DOWNWARDS

θθ angle of launch

The Intitial LaunchVelocity has

horizontal & verticalcomponents

MMaaxx

iimmuumm

HHeeii

gghhtt

“Range” = Total Horizontal Displacement

HorizontalVelocity Vx

VerticalVelocity Vy

At any instant, the projectile’sposition or velocity is the

vector sum of horizontal +vertical components

UUy

Ux

1. Resolve the Initial Launch Velocity intoVertical & Horizontal Components

Sin θθ = Uy & Cos θθ = UxU U

∴∴ Uy = U.Sinθθ,, Ux = U.Cosθθ

2. Horizontal Motion is constant velocity, so

Vx = Sx is all you needt

3. Vertical Motion is constant acceleration at “g”

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 syllabus specifies a 3rd equation as well, butits use can be avoided. (K.I.S.S. Principle)

θθ

U Uy

Ux

Projectile MotionEquations for Projectile Motion

The Trajectory (Path) of a Projectile


Slide 11

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b) The shell is fired upwards, butacceleration due to gravity is downwards.You must assign up = (+ve), down = ( -vve).

At the top of its arc, the shell will have aninstantaneous vertical velocity = zero.

Vy = Uy + g.t So, 0 = 136.8 + (-99.81)xt

∴∴ t = -1136.8/-99.81= 13.95s

This means it takes 13.95s to reach the topof its arc. Since the motion is symmetrical,it must take twice as long for the totalflight. ∴∴ time of flight = 27.9s

c) Range is horizontal displacement

Remember Vx= Ux= constant velocity

Vx = Sx t∴∴ Sx = Vx.t (use time of

flight)

= 375.9 x 27.9= 10,488m

Range = 1.05x104m (i.e. 10.5 km)

d) Vertical Height up =(+ve), down =( -vve)

Sy = Uy.t + 1.g.t22

= 136.8x13.95+ 0.5x(-99.81)x(13.95)2= 1908.4 + (-9954.5)= 953.9m = 9.54x102m

Note: the time used is the time toreach the top of the arc... the time at the highest point

Analysing Projectile MotionExample 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)c) the range. (same assumption)d) the maximum height it reaches.

U=400ms-11

θθ = 20o

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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!

a) Uy = U.Sinθθ Ux = U.Cosθθ

= 400..Sin20 =400Cos20=136.8ms-11 =375.9ms-11

((upwards) (horizontal)


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Site Licence Conditions only

d) Velocity at t = 3.50s ?

VerticalVy = Uy + g.t

=22.9+(-99.81)x3.50 = -111.4ms-11 (downwards)

Horizontal Vx= Ux= constant = 16.1ms-11

By Pythagorus, V2 = Vy

2 + Vx2

= (-111.4)2 + 16.12

∴∴ V = Sq.root(389.17) = 19.7ms-11

16.1RRessullttaantt Velloociittyy

11.4

θb) Maximum Height

is achieved at t = 2.33s, so

Sy = Uy.t + 1.g.t22

= 22.9x2.33+0.5x(-99.81)x(2.33)2= 53.5 + (-226.6)= 26.9m

Resolve the velocity into vertical and horizontalcomponents, then use these to find:

a) the time of flight of the ball.

b) the maximum height reached.

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.

Remember to let UP = (+ve)DOWN = ( -vve)

acceleration = “g” = -99.81ms-22

Vertical & HorizontalComponents of Velocity

Uy = U.Sinθθ,, Ux = U.Cosθθ=28Sin55 = 28Cos55=22.9ms-11 = 16.1ms-11

a) Time of Flight

At highest point Vy=0, so

Vy = Uy + g.t 0 = 22.9 + (-99.81)xt

∴∴ t = -222.9/-99.81= 2.33s

This is the mid-ppoint of the arc, so

time of flight = 4.66s

c) Range will determine if he’s “hit a 6”.

Vx= Ux= constant velocitySx = Vx.t (use total time of

flight)= 16.1 x 4.66= 75.0m

Analysing Projectile Motion Example 2The 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.

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Tan θθ = 11.4/16.1∴∴ θθ ≅≅ 35o

at an angle 35o below horizontal

SIX !



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If you find solving Projectile Motion problems isdifficult, try to learn these basic rules:

• The “launch velocity” must be resolved into ahorizontal velocity (Ux) and a vertical velocity (Uy).Once you have these, you can deal with vertical andhorizontal motion as 2 separate things.

• The motion is symmetrical, so at the highest point,the elapsed time is exactly half the total time of flight.

• 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!

Analysing Projectile Motion (cont)

• Maximum Range is achieved at a launch angle of 45o.

• Horizontal Motion is constant velocity... easy.Use Vx = Ux and Sx = Ux.t

• 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.

Use Sy = Uy.t + 1.g.t2

2 to find vertical displacement (Sy) at a known time, orfind the time to fall through a known height (if Uy=0)

The top of the arc is the mid-ppoint.At this point Vy = zero

θθ angle of launch MMaaxx

iimmuumm

HHeeii

gghhtt

“Range” = Total Horizontal Displacement

HorizontalVelocity Vx

VerticalVelocity Vy

UUy

Ux

PROJECTILES LAUNCHEDAT SAME VELOCITY

LAUNCH ANGLE45o GIVES

MAXIMUM RANGE

Angle greater than 45o

Angles less than 45o

KCiC Physics 5 Spacecopyright © 2009 keep it simple sciencewww.keepitsimplescience.com.au Slide 14 Usage & copying is permitted according to the


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keep it simple scienceProjectiles Launched HorizontallyA common situation with projectile motion is when a projectile is launched horizontally, as inthe following example. This involves half the normal trajectory.

Plane flying horizontally, atconstant 50.0ms-11

Releases a bomb fromAltitude = 700m

Questionsa) 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?SolutionBecause the plane is flying horizontally, the intitialvelocity 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.

Sy = Uy.t + 1.g.t2

2 -700 = 0xt + 0.5 x(-9.81)x t2

-700 = -4.905xt2

∴∴ t2 = -700/-4.905t = 11.9s

c) Where is the Plane?

Since both plane and bombtravel at the same horizontalvelocity, it follows that theyhave both travelled exactlythe same horizontal distancewhen the bomb hits. i.e. theplane is directly above thebomb at impact.

(In warfare, this is a problemfor low-level bombers... the

bombs must havedelayed-action fuses)

b) Final Velocity at impactVertical HorizontalVy = Uy + g.t Vx= Ux

= 0 + (-9.81)x11.9 Vx= 50.0ms-1.Vy= -117ms-1. (down)

V2=Vy2 + Vx

2

= 1172 + 50.02

∴∴ V = Sq.Root(16,189)= 127ms-1.

Tan θθ = 117/50∴∴ θθ ≅≅ 67o.

Bomb hits the ground at 127ms-1, at angle 67o below horizontal.

50.0

FFinal

VVeelooccity

θθ

117



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Notice that NONE of the equations used toanalyse Projectile Motion ever use the mass ofthe projectile. This is because all objects,regardless of mass, accelerate with gravity atthe same rate (so long as air-resistance isinsignificant).

It was Galileo,(1564-1642) who youlearned about in “TheCosmic Engine”, whofirst discovered this.

His famousexperiment was todrop objects of thesame size and shape,but of differentweight, from theleaning tower in Pisa.He found that allobjects hit the groundat the same time,thereby proving thepoint.

Galileo and Projectile MotionHe also studied projectile motion. In his day, cannon ballswere the ultimate weapon, but trajectories were notunderstood at all. To slow the motion down for easierstudy, Galileo rolled balls down an incline:

Although notfalling freely, the ballsaccelerated uniformly, and Galileo was able to see that the motion was a combinationof 2 motions:

• horizontal, constant velocityand • vertical, constant acceleration

Galileo had discovered the basic principles of ProjectileMotion. Unfortunately, he lacked the mathematicalformulas to go any further with his analysis.

That only became possible after the work of Isaac Newton,and his 3 Laws of Motion, and Theory of Gravitation.

Coincidentally, Newton was born in the same year thatGalileo died.



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Once Isaac Newton had developed theMaths and discovered the laws of motionand gravity, he also looked at ProjectileMotion.

Newton imagined a cannon on a very highmountain, firing projectiles horizontally withever-increasing launch velocities:

Newton had discovered the concept of agravitational orbit, and the concept of“escape velocity”.

EARTH At the rightvelocity, the

projectile curvesdownwards at thesame rate as theEarth curves... it

will circle the Earthin orbit!

If launch velocity ishigh enough, theprojectile escapesfrom the Earth’s

gravity

Isaac Newton and OrbitingEscape Velocity is the launch velocity needed for aprojectile to escape from the Earth’s gravitational field.

Mathematically, it can be shown that

Escape Velocity, Ve = 2GME / RE

G= Gravitational Constant (later in topic)ME= Mass of the EarthRE= Radius of Earth

You are NOT required to learn, nor use, this equation.

What you should learn is that:

The mass of the projectile is not a factor. Therefore, allprojectiles, regardless of mass, need the same velocity to

escape from Earth, about 11km per second!

The Escape Velocity depends only on the mass and radius of the Earth.

It follows thatdifferent planetshave differentescape velocities.

PLANET ESCAPE VELOCITYin km/sec (ms-1)

Earth 11.2 1.12 x104

Moon 2.3 2.3 x103

Mars 5.0 5.0 x103

Jupiter 60.0 6.0 x104





Projectiles Student Name .................................

1.a) In the absence of friction, what is the only force acting on a projectile?

b) Describe the horizontal component of a projectile’s motion.

c) Describe the vertical component of a projectile’s motion.

d) At what launch angle is maximum range achieved?

2.a) Who was the first person to discover the basics of projectile motion

(as described in Q1 (b) & (c))?b) Who was the first to analyse projectile motion mathematically?

c) This person also discovered 2 new concepts with huge significance for spaceflight. What are these 2 concepts?

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A projectile needs an enormous velocity toescape from the Earth’s gravitational field...about 11 km per second. Think of a place 11km away from you, and imagine getting therein 1 second flat!

What about Newton’s idea of an orbitingprojectile? If it is travelling at the right velocity,a projectile’s down-curving trajectory willmatch the curvature of the Earth, so it keepsfalling down, but can never reach the surface.A projectile “in orbit” like this is called a“satellite”.

It can be shown that to achieve orbit, thelaunch velocity required is less than escapevelocity, but still very high... about 8 km persecond. How is this velocity possible?

In a 19th century novel, author Jules Verneproposed using a huge cannon to fire a spacecapsule (including human passengers) intospace. Let’s consider the Physics:

Placing a Satellite in Earth OrbitThe “g-Forces” in a Space LaunchTo accelerate a capsule (and astronauts) upwards to orbitalvelocity requires a force. The upward “thrust” force mustovercome the downward weight force AND provide upwardacceleration.

So, if the Thrust force causes acceleration of(say) about 10ms-2, as well as overcoming his weight force, the 80kg astronaut will feela pushing force of;

T = ma + mg= 80x10 + 80x10 ( g≅≅10ms-2 )= 1,600N

This is twice his normal weight of 800N... the force is “2g”.

Astronaut During Acceleration to Orbital Velocity

Weight = mgForce

Net Force= ma

Total Net Force causes acceleration

Greek letterΣΣF = ma Sigma ( ΣΣ )

means totalIf up = (+ve), down ( -vve) then

ΣΣF = T - mg = ma

∴∴ T = ma + mg

This means the astronaut will“feel” the thrust as an

increase in weight.

“THRUST” Force = T

A fit, trained astronaut can tolerate forces of “5g”, but anything above about “10g” is life-threatening.Jules Verne’s cannon astronauts would have suffered forces of about 200g... instantly fatal.



A Brief History of RocketrySimple solid-fuel (e.g. gunpowder) rockets have beenused as fireworks and weapons for over 500 years.

About 100 years ago, the Russian Tsiolkovsky (1857-1935) was the first to seriously propose rockets asvehicles to reach outer space. He developed thetheory of multi-stage, liquid-fuel rockets as being theonly practical means of achieving space flight.

The American Robert Goddard (1882-1945)developed rocketry theory futher, but alsocarried out practical experiments includingthe first liquid-fuel rocket engine.

Goddard’s experiments were the basis ofnew weapons research during World War II,especially by Nazi Germany. Wernher vonBraun (1912-1977) and others developedthe liquid-fuel “V2” rocket to deliverexplosive warheads at supersonic speedsfrom hundreds of kilometers away.

V2

Rockets Achieve OrbitTo 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.

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At the end of the war many V2’s, and the Germanscientists who developed them, were captured byeither the Russians or the Americans. They continuedtheir research in their “new” countries, firstly todevelop rockets to carry nuclear weapons (during the“Cold War”) and later for space research.

The Russians achievedthe first satellite(“Sputnik” 1957) andthe first human inorbit, and theAmericans the firstmanned missions tothe Moon (1969).

Since then, the use ofsatellites has become routineand essential to ourcommunications, while(unmanned) probes havevisited nearly every otherplanet in the Solar System.

Space Shuttlelaunch



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Direction of LaunchStraight upwards, right? Wrong!

To reach Earth orbit, rockets are aimed toward theEAST to take advantage of the Earth’s rotation. Therocket will climb vertically to clear the launch pad,then be turned eastward.

Earth, viewedfrom aboveNorth Pole

Rotation

Orbitpath

LaunchTrajectory

At the equator, the Earth is rotating eastwards at about1,700km/hr (almost 0.5km/sec) so the rocket alreadyhas that much velocity towards its orbital speed.

Rocket launch facilities are always sited as close tothe equator as possible, and usually near the eastcoast of a continent so the launch is outwards over theocean.

Conservation of MomentumWhy a rocket moves was dealt with in the Preliminarytopic “Moving About”.

Newton’s 3rd Law

Force on = Force onExhaust RocketGases

It can also be shown that

Change of Momentum = Change of Momentumof Exhaust Gases of Rocket

( -)Mass x velocity = Mass x velocity

The mass x velocity (per second) of the exhaustgases stays fairly constant during the lift-off.

However, the mass of the rocket decreases as its fuel isburnt. Therefore, the rocket’s velocity keeps increasingto maintain the Conservation of Momentum.

Action Forceaccelerates

exhaustgasses

backwards

bbaacckkwwaarrddss (( -vvee)) ffoorrwwaarrddss ((++vvee))

Physics of a Rocket Launch

Reaction forcepushes rocketforward



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If the “Thrust” force from the rocketengine remains constant throughout the“burn”, but the total rocket massdecreases due to consumption of the fuel,then the acceleration increases.

The concept of “g-forces” was explainedin an earlier slide.

Thrust Force, T = ma + mg

If “T” remains constant, but “m” keepsdecreasing, then “a” must keepincreasing.

(This assumes “g” is constant... Actually it decreases with altitude, so “a” must increase even more)

Photo: RussianSoyez lift-ooff,courtesy Ali

Cimen, seniorreporter, ZamanDaily, Istanbul. Ph

oto

by

Shel

ley

Kise

r

Forces Experienced by Astronauts

Not only does the rocket accelerateupwards, but even the acceleration keepsaccelerating!

The astronauts will feel increasing “g-forces”. At lift-off, they will experienceperhaps only “2g”, but over several minutesthis will increase to perhaps “7g” as therocket burns thousands of tonnes of fueland its mass decreases.

The Space Shuttle’s engines are throttled-back during the launch to counteract this,so the astronauts are not injured byincreasing “g-force”.



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Low-Earth OrbitAs the name suggests, this type of orbit is relativelyclose to the Earth, generally from about 200km, out toabout 1,000km above the surface.

For any satellite, the closer it is, the faster it musttravel to stay in orbit. Therefore, in a Low-Earth Orbita satellite is travelling quickly and will complete anorbit in only a few hours.

A common low orbit is a “Polar Orbit” in which thesatellite tracks over the north and south poles whilethe Earth rotates underneath it.

Satellites and OrbitsThere are 2 main types of satellite orbits:

PolarOrbit

N

S

This type of orbit is ideal fortaking photos or Radar surveys

of Earth.

The satellite only “sees” anarrow north-ssouth strip of theEarth, but as the Earth rotates,each orbit looks at a new strip.

Eventually, the entire Earth canbe surveyed.

Being a close orbit, fine detailscan be seen.

Geo-stationary Orbits are those where theperiod of the satellite (time taken for one orbit) is exactlythe same as the Earth itself... 1 day.

This means that the satellite is always directly abovethe same spot on the Earth, and seems to remainmotionless in the same position in the sky. It’s notreally motionless, of course, but orbiting around atthe same angular rate as the Earth itself.

Geo-stationary orbits are usually above the equator, andhave to be about 36,000km above the surface in order tohave the correct orbital speed.

Being so far out, these satellites are not much goodfor photographs or surveys, but are ideal forcommunications. They stay in the same relativeposition in the sky and so radio and microwavedishes can be permanently aimed at the satellite, forcontinuous TV, telephone and internet relays toalmost anywhere on Earth.

Three geo-stationary satellites, spaced evenly aroundthe equator, can cover virtually the whole Earth withtheir transmissions.

Earth’sRotation

Equator


Slide 23Usage & copying is permitted according to the


Orbits & Centripetal ForceThe orbit of a satellite is often an oval-shape, or “ellipse”. However, in this topic we will always assume the orbits are circular... K.I.S.S. Principle.

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Circular MotionTo maintain motion in a circle an object must be constantlyacted upon by “Centripetal Force”, which acts towards thecentre of the circle.

Centripetal ForceVector is alwaystowards centre

Instaneous Velocityvector is a tangent

to the circle

Object inCircularMotion

FcFc

V

V

The object isconstantly

accelerating. The“centripetal

acceleration” vector istowards the centre.

What Causes Centripetal Force?Example Centripetal Force caused by...Swinging an object Tension Force in the string.around on a string.

Vehicle turning a Friction Force between tyrescircular corner. and road.

Satellite in orbit Gravitational Force betweenaround Earth. satellite mass and Earth’s mass.

Centripetal Force & SatellitesExample ProblemA 250kg satellite in a circular orbit 200km abovethe Earth, has an orbital period of 1.47hours.

a) What is its orbital velocity?b) What centripetal force acts on the satellite?(Radius of Earth = 6.37x106m)

Solutiona) First, find the true radius of the orbit, and get everything into S.I.units: Radius of orbit = 200,000 + 6.37x106 = 6.57x106mPeriod = 1.47hr = 1.47 x 60 x 60 = 5.29x103 seconds

V = 2ππR = 2 x ππ x 6.57x106/5.29x103 = 7.80x103ms-1.T

b) Fc = mv2 = 250x(7.80x103)2/6.57x106

R= 2,315 = 2.32 x 103 N.

The satellite is travelling at about 8 km/sec, held in orbit by a gravitational force of about 2,300N.

Fc = mv2

RFc = Centripetal Force, in newtons (N)m = mass of object in orbit, in kgv = orbital velocity, in ms-1

R = radius of orbit, in metres (m)

When considering the radius of a satellite orbit, be awarethat the orbital distance is often described as the heightabove the surface. To get the radius, you may need to

add the radius of the Earth itself... 6,370km (6.37 x 106 m)

Calculating Velocity from Radius & PeriodSatellite motion is often described by the radius of the

orbit, and the time taken for 1 orbit = the Period (T)

Now, circumference of a circle C = 2ππRTherefore, the orbital velocity V = 2ππR

T

distancetraveled

time taken



When Johannes Kepler (1571-1630) studied themovement of the planets around the Sun (seePreliminary topic “Cosmic Engine”) he discoveredthat there was always a mathematical relationshipbetween the Period of the orbit and its Radius:

R3 αα T2 (Greek letter alpha (αα ) means “proportional to”)

This means that

R3 = constantT2

This means that for every satellite of the Earth, the(Radius)3 divided by (Period)2 has the same value.

At this point, the HSC Syllabus is rather vague about whether youneed to learn and know the following mathematical development.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 shownat right.

Kepler’s Law of Periods was discovered empirically...that is, it was discovered by observing the motion ofthe planets, but Kepler had no idea WHY it was so.

When Isaac Newon developed his “Law of UniversalGravitation” (next section) he was able to prove thetheoretical basis for Kepler’s Law, as follows:

Kepler’s “Law of Periods”®


The Centripetal Force of orbiting is provided by theGravitational Force between the satellite and theEarth, so

Centripetal Force = Gravitational ForceFc = mv2 = FG = GMm

R R2

∴∴ v2 = GM but v = 2ππRR T

So, 4ππ2R2 = GMT2 R (Re-arrange this)R3 = GMT2 4ππ2

Since the right hand side are all constant values, thisproves Kepler’s Law and establishes the Force of Gravityas the controlling force for all orbiting satellites,including planets around the Sun.

In the above, G = Universal Gravitational ConstantM = mass of the Earth (or body being orbited)m = mass of satellite... notice that it disappears!

RPeriod=T

Time takenfor 1 orbit



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In problem 1, we could calculate the orbital radiusfor a geo-stationary satellite by comparing theratio of R3/T2 for two different satellites.

With Newton’s development of Kepler’s Law, wecan do it again a different way...

Problem 2Find the orbital radius of a geo-stationarysatellite, given that its period of orbit is 24.0hours.

(24.0hr = 24.0x60x60 = 8.64 x 104 sec)Doing this way, you MUST use S.I. units!!(G= Gravitational Constant = 6.67 x 10-11

M = Mass of Earth = 5.97 x 1024kg)

SolutionR3 = GMT2 4ππ2

R3 = 6.67x10-11 x 5.97x1024 x (8.64x104)2

4ππ2

∴∴ R = CubeRoot (7.5295x1022)= 4.22 x 107m.

This is about 42,000km, or about 36,000kmabove the surface... the same answer as before.

(It better be!)

Problem 1In a previous slide, a low-Earth satellite was found tohave be orbiting with R = 6,570 km (from Earth centre)and an a period, T = 1.47 hours.

A geo-stationary satellite has a period of 24.0 hours.Use Kepler’s Law of Periods to find its orbital radius,using this data.

SolutionFor the low-Earth satellite , R3 = 6,5703 = 1.31 x 1011

(units are km & hours) T2 1.472

According to the law of periods, ALL satellites ofEarth 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 the Earth’s centre, or about 36,000km above the surface.

Note: When using Kepler’s Law in this way it doesn’tmatter which units are used, as long as you are

consistent. In this example, km & hrs were used. Thesame result will occur if metres & seconds are used.

Kepler’s Law of Periods Example Problems



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Where does “Space” begin?

It’s generally agreed that by 100km above thesurface of the Earth the atmosphere hasended, and you’re in outer space. However,although this seems to be a vacuum, there arestill a few atoms and molecules of gasesextending out many hundreds of kilometres.

Therefore, any satellitein a low-Earth orbit willbe constantly collidingwith this extremely thin“outer atmosphere”.The friction or air-resistance this causesis extremely small, butover a period ofmonths or years, itgradually slows thesatellite down.

Decay of Low-Earth OrbitsAs it slows, its orbit “decays”. This means itloses a little altitude and gradually spiralsdownward. As it gets slightly lower it willencounter even more gas molecules, so thedecay process speeds up.

Once the satellite reaches about the 100kmlevel the friction becomes powerful enough tocause heating and rapid loss of speed. At this

point the satellite will probably“burn up” and be destroyed as itcrashes downward.

Modern satellites are designed toreach their low-Earth orbit withenough fuel still available tocarry out short rocket engine“burns” as needed to counteractdecay and “boost” themselvesback up to the correct orbit. Thisway they can remain in low-Earth

orbits for many years.


Slide 27

In orbit, the satellite and astronauts have a highvelocity (kinetic energy) and a large amount of GPEdue to height above the Earth. To get safely back toEarth, the spacecraft must decelerate and shed all thatenergy.

It is impossible to carry enough fuel to use rocketengines to decelerate downwards in a reverse of thelift-off, riding the rocket back down at the same rate itwent up.

Instead, the capsule is slowed by “retro-rockets” justenough to cause it to enter the top of the atmosphereso that friction with the air does 2 things:

• 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.

The trick is to enter the atmosphere at the correctangle, as shown in the diagrams.

Upper Atmosphere

Angle too shallow... Spacecraft bouncesoff upper air layers, back into space

Earth’s Surface

Angle correct...Spacecraft decelerates safelyalong a descent path of about1,000km of “Atmospheric

Braking”

Earth’s Surface

Angle too steep...“g-fforces” may kill astronauts.Heat may cause craft to burn-uup.

Earth’s Surface

Early spacecraft used “ablation shields”, designed to meltand carry heat away, with the final descent by parachute.

The Space Shuttle uses high temperature tiles for heat protection, then glides in on its wings

for a final landing like an aircraft.

Correct angle isbetween 5-77oo

Re-Entry From OrbitGetting a spacecraft into orbit is difficult enough, but the most dangerous process is gettingit down again in one piece with any astronauts on board alive and well.

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Satellites & Orbits Student Name .................................

1. In the history of rocketry, what was the contribution of:a) Tsiolkovsky?b) Goddard?c) von Braun?

2. Why is Jules Verne’s (fictional) idea of launching a spacecraft using a “spacecannon” not a practical way to do it?

3. What is the basic principle of rocket motion?

4.a) What is a “Geo-Stationary” satellite orbit?

b) Why is this type of orbit ideal for communication satellites?

c) What is a Polar, Low-Earth satellite orbit?

d) Why is this orbit ideal for (for example) photographic surveys of the Earth?

5. What is “atmospheric braking” during re-entry of a spacecraft?

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Gravitational FieldsThe concept of the Gravitational Field was introducedin section 1. Every mass acts as if surrounded by aninvisible “force field” which attracts any other masswithin the field. Theoretically, the field extends toinfinity, and therefore every mass in the universe isexerting some force on every other mass in theuniverse... that’s why it’s called Universal Gravitation.

Newton’s Gravitation EquationIt was Isaac Newton who showed that the strength ofthe gravitational force between 2 masses:

• is proportional to the product of the masses, and• inversely proportional to the square of the

distance between them.

3. NEWTON’S LAW OF UNIVERSAL GRAVITATION

FG = GMmd2

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.

In the previous section on satellite orbits, you werealready using equations derived from this.

Example Calculation 1Find the gravitational force acting between the Earthand the Moon.

Earth mass = 5.97 x 1024kg.Moon mass = 6.02 x 1022kg.

Distance Earth-Moon = 248,000km = 2.48x108m.

Solution FG = GMmd2

= 6.67x10-11x5.97x1024x6.02x1022

(2.48x108)2

= 3.90 x 1020N.

Example 2Find 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 = GMmd2

= 6.67x10-11 x 5.97 x 1024 x 80(6.37x106)2

= 785 N.This is, of course, the person’s weight!... and sureenough:

W = mg = 80 x 9.81 = 785N.∴∴ Gravitational Force = Weight Force


Slide 30


Effects of Mass & Distance on FGHow does the Gravitational Force change for differentmasses, and different distances?

Imagine 2 masses, each 1kg, separated by a distance of 1metre.

FG = GMm = G x 1 x 1 = Gd2 12

Effect of massesNow imagine doubling the mass of one object:FG = GMm = G x 2 x 1 = 2G (Twice the force)

d2 12

What if both masses are doubled?FG = GMm = G x 2 x 2 = 4G (4X the force)

d2 12

Effect of DistanceGo back to the original masses, and double thedistance:FG = GMm = G x 1 x 1 = G ( 1/4 the force)

d2 22 4

Gravitational Force shows the “Inverse Square”relationship...

triple the distance = one ninth the force10 x the distance = 1/100 the force, etc.

Universal Gravitation and Orbiting Satellites

It should be obvious by now that it is FG whichprovides the centripetal force to hold any satellitein its orbit, and is the basis for Kepler’s Law ofPeriods.

Not only does this apply to artificial satelliteslaunched into Earth orbit, but for the orbiting ofthe Moon around the Earth, and of all the planetsaround the Sun.

Our entire SolarSystem is

orbiting theGalaxy becauseof gravity, andwhole galaxies

orbit each other.Ultimately,

gravity holds theentire universe

together, and itsstrength,

compared to theexpansion of the Big Bang, will determine the

final fate of the Universe.

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®


One of the more interesting aspects of gravity and itseffects on space exploration is called the “SlingshotEffect”.

1. Scientists wish to explore and learn about all the planets,comets, etc, in the Solar System, but...

2. It costs billions of dollars to send a space probe toanother planet, so...

3. It makes sense to send one probe to several planets,rather than a separate spacecraft to each planet, but...

4. the distances are enormous. Even at the high speedof an inter-planetary probe (50,000 km/hr) it still takesyears to reach other planets.

5. Furthermore, having reached and done a “fly-by” tostudy one planet, the probe may need to changedirection and speed to alter course for the nextdestination, and...

6. It may be impossible to carry enough fuel to make thenecessary direction changes by using rocket enginesalone.

Got all that?The solution to all these factors is to fly the spacecraftclose enough to a planet so that the planet’s gravitycauses it to swing around into a new direction ANDgain velocity (without burning any fuel).

So how can the the spacecraft gain extra

velocity (and kinetic energy) from nothing?

The answer is that whatever energy thespacecraft gains, the planet loses. Energy

is conserved. The planet’s spin will beslowed down slightly by the transfer of

energy to the spacecraft.

Of course, the huge mass of a planetmeans that the energy it loses is so small

to be totally insignificant.

Spacecraft

Planet orbit

Planet orbit

1stplanetvisited

2nd planetvisited

SlingshotTrajectory

To 3rd planetaryfly-bby

“Slingshot Effect” for Space Probes



4. EINSTEIN’S THEORY OF RELATIVITYThe Aether Theory

The idea of the universal “aether” was a theory developed to explain the transmission of light through emptyspace (vacuum) and through transparent substances like glass or water. The basic idea was:Sound waves are vibrations in air. Water waves travel as disturbances in water. Sounds and shock wavestravel through the solid Earth. It seems that all waves have a “medium” to travel through, so there must be amedium for light waves.

From the 17th to 19th centuries, as modern Science developed, it became the general belief that there was asubstance called the “aether” which was present throughout the universe as the medium for light waves to becarried 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.

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The Michelson-Morley ExperimentIn 1887, American scientists A.A. Michelson and E.W. Morley attempted to detect the aether by observing theway that the movement of the Earth through the aetherwould affect the transmission of light.

Here is an analogy to their experiment...Imagine 2 identical boats, capable of exactly the samespeed. They both travel a course out and back over exactlythe same distance, but at right angles to each other.

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, becausethe current will alter their relative speeds. (The one moving across the current will arrive later.)

This boat travels with, and then against, the current

This boattravels outand backacross the

current

water current

water current

Descriptionof this

experimentis continuednext slide.



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In Michelson & Morley’s experiment the“boats” were beams of light from the samesource, split and reflected into 2 right-angledbeams sent out to mirrors and reflected back.The “current” was the “aether wind” blowingthrough the laboratory due to the movement ofthe Earth orbiting the Sun at 100,000km/hr.

The Michelson-Morley Experiment (continued)The entire apparatus was mounted on arotating table. Once the apparatus wasworking, and the interference patternappeared, the whole thing was rotated 90o, sothat the paths of the light rays in the aetherwind were swapped. Theoretically, this shouldhave created a change in the interferencepattern, as the difference between the beamswas swapped.

The Result...There was NO CHANGE in the

interference pattern.

The experiment was repeated in many otherlaboratories, with more sensitve

interferometers and all sorts of refinementsand adjustments.

The result remained negative... no effect of theaether wind could be detected.

Enter Albert Einstein...

On arrival back at the start, the beams werere-combined in an “interferometer”,producing an interference pattern as the lightwaves re-combined.

Either the experiment has somethingwrong with it or the theory of the

“Aether” is wrong!

Stationary Aetherthroughout the Universe

Earth is hurtling throughthe Aether while orbiting

the Sun.

This creates a “current” or“aether wind”

The opticalequipmentis able tobe rotated

In the laboratory, thislight beam travelsacross the current This one travels with

the aether wind



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The Michelson-Morley Experiment is probably themost famous “failed experiment” in the history ofScience. It’s importance is not just historicalinterest, but a lesson in how Science works.

There is no such thing as a “failed experiment”!

Scientists produce hypotheses in an attempt toexplain the universe and its phenomena. Therecan be 2 or more totally different hypothesesattempting to explain the same thing.

This is exactly what happened. In the 30 yearsafter the Michelson-Morley Experiment, a newHypothesis was proposed which did not requireany “aether”. From it arose many predictionswhich have all been spectacularly confirmed byexperiment, so we believe the “Aether Theory” iswrong, and “Relativity Theory” correct.

The Michelson-Morley Experiment was not afailure... it was a vital link in the scientific searchfor truth. Hypothesis

Accepted asCorrect Theory

HypothesisRejected as

Wrong

Natural Phenomenonto be explained

Hypothesis 1

e.g the Aether

Hypothesis 2

e.g. Relativity

ExperimentalResultsDO NOT

agree withpredictions

ExperimentalResults

DOagree withpredictions

Predictions arisefrom this idea.

These can betested by

experiment

Predictions arisefrom this idea.

These can betested by

experiment

How Science Works



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Ever been sitting in a train at a station lookingat another train beside you? Suddenly, the

other train begins moving. Or is it your trainbeginning to move the other way?

The only way to be sure is to look out theother side at the station itself, in order to

judge which train is really moving. You areusing the railway station as your

“Frame of Reference” in order to judge the relative motion of the 2 trains.

We often use the Earth itself (or a railwaystation attached to it) as our frame ofreference. The Earth seems fixed and

immovable, so everything else can be judgedas moving relative to the fixed Earth... but wealso know it’s NOT really fixed and unmoving,

but orbiting around the Sun.

Astronomers use the background of “fixedstars” as their frame of reference to judge

relative planetary movements, but we knowthat these aren’t really fixed either.

In fact, there is no point in the entire Universethat is truly “fixed” that could be used as an

“absolute reference” to judge and measure allmotion against.

An Inertial Frame of Referenceis not accelerating.

Within any Inertial Frame of Reference all motionexperiments (and all “Laws of Physics”)

will produce the same results.

Distinguishing Inertial & Non-InertialFrames of Reference

Imagine you are inside a closed vehicle and cannot see out. Howcan you tell if your “Frame of Reference” is “Inertial” or not?

A simple indication would be to hang a mass on a string from theceiling. If it hangs straight down there is no acceleration. If it hangsat an angle, (due to its inertia) then your vehicle is accelerating.

Does it matter whether your vehicle is stationary or moving atconstant velocity? Not at all! The mass still hangs straight down,and any Physics experiments will give the same result as any otherobserver in any other Inertial Frame of Reference.

Relative Motion and Frames of ReferenceSir Isaac Newton was aware of this idea, and figuredout that it really doesn’t matter whether your frame ofreference is stationary or moving at a constantvelocity. So long as it is not accelerating, theobservations, and measurements of motion will comeout the same anyway. This raises the idea of an“Inertial Frame of Reference”.



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Albert Einstein (1879-1955) has gone down in theHistory of Science as one of the “Greats”, and justabout the only scientist to ever match theachievements of the great Sir Isaac Newton.

Einstein’s “Theory of Relativity” is famous as a greatachievement, (true!) and as something incrediblycomplicated that hardly anyone can understand(false! It’s a dead-simple idea, but it defies “commonsense”.)

Einstein once declared “common sense” as “adeposit of prejudice laid down in the mind prior to theage of 18”. To understand “Einstein’s Relativity” youneed to ignore “common sense” and have a child-likeopen-mind to fantasy and the K.I.S.S. Principle...

The Principle of Relativitywas already well known before Einstein, and statedin various forms by Galileo, Newton and many others

These are all statements of the “Principle of Relativity”.

1. In an Inertial Frame of Reference allmeasurements and experiments

give the same results.

2. It is impossible to detect the motionof an Inertial Frame of Reference by experiment

within that frame of reference.

3. The only way to measure the motion of yourframe of reference is by measuring it against

someone else’s frame of reference.

Albert Einstein’s Strange Idea

RReellaattiivviittyy TTrraaiinn TToouurrss

Train Velocity at, or near,the speed of light

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).

Continuednext slide...





The Aether Student Name .................................

1.a) What were the properties of the “universal aether”?

b) Why was the aether theory proposed?

2.a) What was the purpose of the Michelson-Morley experiment?

b) In what sense was the experiment a failure?

c) In what sense was it a scientific triumph?

3.a) What is an “Inertial Frame of Reference”?

b) What is the “Principle of Relativity”?

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®

keep it simple scienceEinstein’s Gedanken (a “Thought Experiment”) continued

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.



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 themirror to reflect off it.

So, vampire-llike, you have no reflection!

But Einstein remembered Michelson & Morley’s failure tomeasure the “aether wind” and applied the Principle of Relativity...

In a non-accelerating, Inertial Frame of Reference, you wouldmeasure the speed of light (and anything else, like reflection)exactly the same as anyone else... you would see your reflection,and everything appears normal.

But, if you are travelling at the speed oflight, how is it possible for you, and thestationary observers on the platform, to

both measure the same light wave ashaving the same velocity?

Well, says Einstein, if THE SPEED OF LIGHT is FIXED, thenSPACE and TIME must be RELATIVE.

What does this mean?

Continued...

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 seethrough the train window as you zap by?

Again, according to the results of the Michelson-Morleyexperiment, these observers will measure light waves from youas travelling at the same speed of light as you measure insidethe train, because everyone is in an Inertial F. of R.

(Naturally, both train and platform are fully equipped withinterferometers and high-tech ways to do these measurements.)


Slide 39


Meanwhile, stationary observers arestanding on the platform as your train

flashes by. They also measure the speedof light and get the exact same answer.

However, the people on the platform seeyou as compressed in space like this:

Furthermore, when they study your clock they see it is

running much slower than their own is.

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:

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 observerwho is in relative motion.



Einstein’s “Thought Experiment” continued...You are on a train travelling at, or near, the speed of light. You carry out some experiments

and measure the speed of light, and the law of reflection as being perfectly normal.

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If everyone (in any Inertial F. of R.) measures the speedof light as being the same, then the measurements ofSPACE and TIME must be relative, and different asseen by an observer in another F. of R.

It turns out that the measurement of length must getshorter as your velocity increases... (as seen by anobserver inanotherInertial F. ofR.) ...andtime getslonger. Timegoes slower!

Length Contraction & Time DilationIf you can ignore “common sense” and accept the fantasy of a train moving at 300,000 km/sec then Einstein’s proposal makes sense:

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

THIS IS LENGTH CONTRACTION.

IT OCCURSONLY IN THE

DIRECTION OFTHE RELATIVE

MOTION

1 - v2c2

L = Lo

Example CalculationOn board a spacecraft travelling at “0.5c” (i.e.half the speed of light = 1.50x108ms-1) relative tothe Earth, you measure your craft as being 100metres long. Carrying out this measurementtakes you 100 seconds.

Observers on Earth (with an amazing telescope)are watching you. How much time elapses forthem, and what is their measurement of yourspacecraft?

Solution

= Sq.Root(1- 0.52/12)= 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!

The factor 1 - v2

c2

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

THIS IS TIME DILATION

t = to

1 - v2c2

Slide 40



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Extremely accurate “atomic clocks” havebeen synchronised, then one flown around

the world in a high speed aircraft. Whenbrought back together, the clock that

travelled was slightly behind the other...while travelling at high speed it’s time hadslowed down a little, relative to the other.

Certain unstable sub-atomic particlesalways “decay” within a precise time.When these particles are travelling at

high speeds in a particle accelerator, theirdecay time is much longer (as measured

by the stationary scientists). At highspeed the particle’s time has sloweddown relative to the scientists’ time.

It’s important for you to realise that, if aparticle could think, it would not notice any

slow-down in time... its own “feeling” oftime and its little digital watch would seemperfectly normal to it. But, from the relativeviewpoint of the scientists measuring theparticle’s decay, its time has slowed down

relative to laboratory time.

Relativity and RealityDo these alterations to time and space really happen? Yes, and they have been measured!

Confirmation of RelativityEinstein published his theory in 2 parts, in 1905 and 1915. At thattime there was no way to test the predictions of Relativity to findsupporting evidence.

The Michelson-Morley experiment had failed to find supportingevidence for the existence of the “aether”, so maybe “Relativity”would fail too, but first scientists had to find testablepredictions.

The first test was that, according to Relativity, light from adistant star passing close to the Sun should be bent by ameasurable amount, making the star appear to change positionin the sky. The only way to test this prediction was during a solareclipse.

At the next occurrence of an eclipse, the observations weremade, 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 nuclearpower) were able to confirm the conversion of matter into energyaccording to E=mc2.

Later still came the measurements of time dilation (describedabove left) and mass dilation has also been measured for high-speed particles in a particle accelerator.

EVERY RELATIVITY PREDICTION THAT CAN BE TESTED HAS SHOWNRESULTS SUPPORTING & CONFIRMING THE THEORY...

that’s why we believe it to be correct.



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Photo of the atombomb explosion onthe Japanese city of

Nagasaki, 1945.

Mass Changes As WellNot only does length contract, and time stretch, but mass changes too.

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

1 - v2c2

m = mo

This is the process occurring in a nuclearreactor used to generate electricity inmany countries. It is also the energysource in an “atomic bomb”.

Einstein, a life-long pacifist, was appalledby the destructive uses of the technologywhich grew from his discoveries.

Within 30 years of Einstein’s ideas being published, the“Equivalence of Mass & Energy” was dramaticallyconfirmed by the release of nuclear energy from atomicfission.

Two of the most fundamental laws ever discovered by Scienceare the “Law of Conservation of Energy” and the “Law ofConservation of Matter”. These state that energy and matter(mass) cannot be created nor destroyed.

Einstein found that the only way to avoid breaking these lawsunder “Relativity” was to combine them. Hence, the mostfamous equation of all:

E = mc2

E = Energy, in joules m = Mass, in kgc = speed of light = 3.00 x 108ms-1

THIS IS THE EQUIVALENCE OF MASS & ENERGY



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What Happens as Speed Approaches “c”

Several of the Relativity equations contain the factor:

This is known as the “Lorentz-FitzGeraldContraction”. In the following explanations it will bereferred to as the “LFC”.

Consider firstly, what happens to the value of the LFCat different relative velocities:

If V=zero LFC = Sq.Root(1-0) = 1

This means that if you (in your spacecraft) and theobserver watching you have zero relative velocity (i.e.you are travelling at the same relative speed) thenboth of you will measure the same length, time andmass... no relativistic effects occur.

As V increases, the value of the LFC decreases:Relative Velocity Value of(as fraction of c) LFC

0.1c 0.9950.5c 0.8860.9c 0.4360.99c 0.1410.999c 0.045

If V = c LFC = Sq.Root( 1 - 1) = zero

Appr

oach

ing

c

Appr

oach

ing

zero

1 - v2

c2

Some Implications of RelativityThis all means that as your spacecraft acceleratesand approaches the speed of light, your faithfulobserver sees your length approach zero, your timeslowing down and approaching being totally stopped,and your mass increasing to approach infinity.

At the speed of light, the calculations for time andmass dilation become mathematically “undefined”...this is generally taken to mean that no object can everreach the speed of light.

Another way to reach this conclusion is that as youspeed up, your mass increases. To accelerate more,greater force is needed because your increased massresists acceleration. As your mass approachesinfinity, an infinite amount of force is needed toaccelerate you more... it’s impossible to reach c.

All the energy put into trying to accelerate goes intoincreasing your mass, according to E=mc2.

Simultaneous EventsAnother consequence of Relativity is that you, andyour observer, will not agree on simultaneous events.You may see 2 things occur at the same instant, butthe relativistic observer will see the 2 eventsoccurring at different times.

It is even possible that the observer could see aneffect (e.g. spilt milk) before seeing the cause(e.g. glass tipped over). Curiouser & curioser!



Activity The following activity might be completed by class discussion,


Theory of Relativity Student Name .................................

1.a) In his famous “Gedanken” (thought experiment) what did Einstein imagine?

b) He imagined 2 different groups of people measuring something.Who, and what?

c) Einstein decided that these observers will get the same result for theirmeasurements. Outline his reasoning for this.

d) If this measurement comes out the same for each observer, what did Einsteinconclude must be true?

2. For observers in “relativistic” frames of reference, what do they observehappens to the measurement of:a) length? b) time? c) mass?

3. What is the famous equation which arises from the “mass dilation” effect?

4. Outline a piece of evidence which supports Einstein’s theory.

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How We Define Length & TimeOur S.I. unit of length, the metre, wasoriginally defined by the French as “One ten-millionth of the distance from the Equator tothe Earth’s North Pole”.

Based on this, specialmetal bars were carefullymade to be used as the“standard” metre fromwhich all othermeasuring deviceswere made.

As our technology improved, so did ourability to measure time and distance. Todaywe define the metre as “the distancetravelled by light during a time interval of1/299,792,458th of a second.”

Our defintion of length is actually based onthe measurement of time! (What’s even moreamazing is that we actually have ways tomeasure such a fraction of a second!)

So how do we define “a second” of time? The modern definition involves a multiple ofthe time it takes for a certain type of atom toundergo an atomic “vibration”, which isbelieved to be particularly regular and is, ofcourse, measurable.One of the precisely-mmade

platinum bars used to definethe metre up until 1960. Part of the

mechanism of an“atomic clock”

which measuresatomic vibrations

to give ourstandards of time

and distance.

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