Phys5.SpaceU

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keep it simple science®

1

but first, an introduction...

HSC Physics Topic 1

SPACEWhat 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 & SATELLITES3. NEWTON’S LAW OF UNIVERSAL GRAVITATION

4. EINSTEIN’S THEORY OF RELATIVITY...all in the context of the universe and space travel

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

In this topic you will revise these concepts, andbe introduced to the concept of “GravitationalPotential Energy”.

Then, you move on to study two important formsof motion that are controlled by gravity...

Projectiles...

...and Satellites in Orbit.

You will study how Gravity is responsible forholding the Solar System together...

and study a variety of aspects of Physics thatrelate to Space Travel

Launch & Re-eentryare thetricky bits...

Orbiting issimplePhysics!

Once launched,the path of aprojectile is

entirelydetermined by

gravity.

There are over 1,000artificial satellites in

Earth orbit.

Some providecommunication links

for telephone,internet and TV.

Others watch theweather, or study

patterns of land use,or search for natural

resources.

Some are formilitary

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




2

Mass&

Weight

Height, Range,Time of Flight, etc

GravitationalPotentialEnergy

GravitationalAcceleration

“g”

CircularMotion

GravitationalFields

Law ofUniversal

Gravitation

Gravity&

SpaceProbes

Frames ofReference &

Relativity

Evidence SupportingRelativity

Michelson-MorleyExperiment

& its Significance

SPACE

Gravity &Gravitational

Fields

Projectiles&

Satellites

Newton’sLaw of

UniversalGravitationEinstein’s

Theory ofRelativity

CONCEPT DIAGRAM (“Mind Map”) OF TOPICSome 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.

Kepler’s Law of Periods

ProjectileMotion

Satellites&

Orbits

Einstein’sIdea

and theConsequences




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

Gravitational FieldIn one way, Gravity resembles electrical chargeand magnetism... it is able to exert a force onthings without touching them. Such forces areexplained by imagining that there is an invisible“Force Field” reaching through space.

Gravitational fields are imagined to surroundanything with mass... that means all matter, andall objects. The field exerts a force on any othermass that is within the field.

Unlike electro-magnetism, gravity can onlyattract; it can never repel.

Of the various “field forces”, Gravity is by farthe weakest, although when enough mass isconcentrated in one spot (e.g. the Earth) itdoesn’t seem weak!

3

1. GRAVITY & GRAVITATIONAL POTENTIAL ENERGY

Weight = Mass x Acceleration due toGravity

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

“g” is acceleration in ms-2.

Measuring “g”One of the first activities you may have done inclass would have been to determine the valueof “g”, the acceleration due to gravity.

A common experimental method to do thisinvolves using a pendulum.

By accurately timing (say) 10 swings of thependulum, and then dividing by 10, the Period(T) can be measured. This value needs to besquared for graphing.

The length of the pendulum (L) is alsomeasured as accurately as possible.

Typically, the measurements are repeated forseveral different lengths of pendulum, then theresults are graphed as shown.

Time taken for 1 complete(back-aand-fforth) swing iscalled the “Period” of the

pendulum (“T”)

Leng

tth iin

mett

rres

How This Relates to “g”

It turns out that the rate atwhich a pendulum swings(its Period) is controlledby only 2 things:

• its length, and• the acceleration due to

gravity

Mathematically,

T2 = 4ππ2Lg

so, T2 = 4ππ2

L g

You are NOT required toknow this equation.

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.

Explanations for Not Getting Exact Value:The main causes of experimental error are any jerking,stretching or twisting in the string, which causes thependulum swing to be irregular. This is why the mostaccurate results will be obtained with very small, gentleswings.0 0.5 1.0

Length of Pendulum (m)

(Per

iod)

22(s

22 )0

1.0

2.0

3.0

Linne

off b

est ff

it

Accepted value, g = 9.81ms-2

Gradient = T22 ≅≅ 4.0 L

4

Gravitational Potential Energy(GPE)Potential Energy is commonly defined as theenergy “stored” in an object. In the case of anyobject on or near the Earth, the amount of GPEit contains depends on

• its mass• its height above the Earth

If that object is allowed to fall down, it losessome GPE and gains some other form of energy,such as Kinetic or Heat. To raise the objecthigher, you must “do work” on it, in order toincrease the amount of GPE it contains.

However, for mathematical reasons, the pointwhere an object is defined to have zero GPE isnot on Earth, but at a point an infinite distanceaway. So GPE is defined as follows:

This definition has an important consequence: it defines GPE as the work done to bring anobject towards the Earth, but we know that youneed to do work to push an object (upwards)away from Earth.

Therefore, GPE is, by definition, a negativequantity!

Note: the HSC Syllabus does NOT require you tocarry out calculations using this equation. You ARErequired to 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.

Gravity and Weight on Other Planets

We are so used to the gravity effects on Earththat we need to be reminded that “g” is differentelsewhere, such as on another planet in ourSolar System.

Since “g” is different, and W = mgit follows that things have a different weight iftaken to another planet.

Values of “g” in Other Places in the SolarSystem

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

Calculating a Weight on another Planet

ExampleIf an astronaut in his spacesuit weighs 1,350N on Earth,what will he weigh on Marswhere 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.

WORKSHEET at end of this section




Gravitational Potential Energy is a measure of the work 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

The weight of an object is thea)...................... due to b)..........................

Near the Earth, all objects willc)................................... at the same rate,approximately d)..................ms-2

Experimentally, “g” can be easilydetermined by measuring the lengthand e).............................. of a pendulum.When the results are graphedappropriately, the f)................................of the graph allows calculation of “g”.

Gravity acts at a distance by way of ag)...................... ........................ the sameas electro-magnetism, but the force onlyh)....................... and can neveri)....................... Gravity is a property of“mass”; every object is surrounded by aj).................................... which will attractany other k)........................... within thefield.

1. A small space probe has a mass of 575kg.

a) What is its massi) in orbit?

ii) on the Moon?

iii) on Jupiter?

b) What is its weighti) on Earth?

ii) on the Moon?

iii) on Jupiter?

5




COMPLETED WORKSHEETSBECOME SECTION SUMMARIES

Worksheet 1 Gravity & GPEFill in the blank spaces. Student Name...........................................

Any mass within a gravitational fieldpossesses “Gravitaional PotentialEnergy” (GPE). This is defined as “theamount of l)................................ to movean object from m)................................ toa point within the field.” In reality, workmust be done to move any mass in theopposite direction, so the definitionmeans that the value for GPE is alwaysa n)................................... quantity.

The value of “g” at the surface of theEarth is o)...................ms-2, but has adifferent value in other places, so thep)................................. of any object willbe different on a different planet.However, the q)........................ willremain the same.

Worksheet 2 Practice Problems Mass & Weight Student Name ...........................................

2. If a martian weighs 250N when at home, whatwill he/she/it weigh:a) on Earth? (hint: firstly find the mass)

b) on Neptune?

c) on the Moon?

3. A rock sample, weight 83.0N, was collected by aspace probe from the planet Neptune.a) What is its mass?

b) What will it weigh on Earth?

c) On which planet would it weigh 206N?

What is a Projectile?A projectile is any object that is launched, andthen moves only under the influence of gravity.

Examples:

Once struck, kicked orthrown, a ball in any sportbecomes a projectile.

Any bullet,shell or bombis a projectile

once it is fired,launched or

dropped.

An example which is NOT a Projectile:

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

Once the engine stopsfiring it becomes aprojectile.

Projectiles are subject toonly one force...Gravity!

When a projectile is travelling through air, thereis, of course, an air-resistance force acting aswell. For simplicity, (K.I.S.S. Principle) air-resistance will be ignored throughout this topic.

In reality, a projectile in air, does notbehave the way described here

because of the effects of air-resistance.

The exact motion depends on manyfactors and the Physics becomes very

complex, and beyond the scope of this course.

Projectile MotionBy simple observation of a golf ball trajectory, ora thrown cricket ball, the motion of anyprojectile can be seen to be a curve. It is in facta parabola, and you might think the Physics ofthis is going to be difficult. NOT SO... it is reallyvery simple. Just remember the following:

You must analyse projectile motion as 2separate motions; horizontal (x-axis) andvertical (y-axis) must be dealt with separately,and combined as vectors if necessary.

The Trajectory (Path) of a Projectile

Equations for Projectile Motion

6

2. PROJECTILES & SATELLITESkeep it simple science

®



Horizontal Motionis CONSTANT VELOCITY

Vertical Motionis CONSTANT ACCELERATION

at “g”, DOWNWARDS

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vveerrttiiccaall ccoommppoonneennttss

MMaaxx

iimmuumm

HHeeii

gghhtt

““RRaannggee”” == TToottaall HHoorriizzoonnttaall DDiissppllaacceemmeenntt

HHoorriizzoonnttaallVVeelloocciittyy

VVx

VVeerrttiiccaallVVeelloocciittyy

VVy

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UUyy

Uxx

1. Resolve the Initial Launch Velocity intoVertical & Horizontal Components

Sin θθ = Uyy & Cos θθ = UxxU U

∴∴ Uyy = U.Sinθθ,, Uxx = U.Cosθθ

2. Horizontal Motion is constant velocity, so

Vxx = Sxx is all you needt

3. Vertical Motion is constant acceleration at “g”

To find vertical velocity:Vyy = Uyy + g.t (from v=u+at)

To find vertical displacement:Syy = Uyy.t + 1.g.t22 (from S=ut+ 1at22)

2 2

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

θθ

U Uyy

Uxx

PPrroojjeeccttiilleess

NNoott aaPPrroojjeeccttiillee




7

Analysing Projectile MotionExample 1The 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 atthe same horizontal level)

c) the range. (same assumption)

d) the maximum height it reaches.

a) Uyy = U.Sinθθ Uxx = U.Cosθθ

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

((upwards) (horizontal)

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.

Vyy = Uyy + g.t 0 = 136.8 + (-99.81)xt

∴∴ t = -1136.8/-99.81= 13.95 s

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 Vxx= Uxx= constant velocity

Vxx = Sxx t

∴∴ Sxx = Vxx.t (use time of flight)

= 375.9 x 27.9= 10,488m

Range = 1.05x1044m (i.e. 10.5 km)

d) Vertical Height (up =(+ve), down =( -vve))Syy = Uyy.t + 1.g.t22

2 = 136.8x13.95 + 0.5x(-99.81)x(13.95)22= 1901.5 + (-9947.7)= 953.8m = 9.54x1022m.

Note: the time used is the time to reach the top ofthe arc... the time at the highest point.

Point to Note:The mass of the projectile does NOT enter

into any calculation. The trajectory isdetermined by launch velocity & angle, plus

gravity. Mass is irrelevant!

U=400ms-11

θθ = 20oo




8

d) Velocity at t = 3.50s ?Vertical Horizontal

Vyy = Uyy + g.t Vxx= Uxx= constant

=22.9+(-99.81)x3.50 = 16.1ms-11

= -111.4ms-11

(this means it is downwards)

Tan θθ = 11.4/16.1∴∴ θθ ≅≅ 35oo

at an angle 35oo below horizontal

By Pythagorus, V22 = Vyy

22 + Vxx22

= (-111.4)22 + 16.122

∴∴ V = Sq.root(389.17) = 19.7ms-11

1166..11

RReessuullttaanntt VVeelloocciittyy

1111..44

θ

b) Maximum Heightis achieved at t = 2.33s, so

Syy = Uyy.t + 1.g.t222

= 22.9x2.33+0.5x(-99.81)x(2.33)22= 53.5 + (-226.6)= 26.9m

Analysing Projectile MotionExample 2The batsman has just hit the ball upwards at an angle of 55o, with anintial velocity of 28.0ms-1. The boundary of the field is 62.0m away fromthe batsman.

Resolve the velocity into vertical and horizontal components, then usethese 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

Uyy = U.Sinθθ,, Uxx = U.Cosθθ=28Sin55 =28Cos55=22.9ms-11 =16.1ms-11

a) Time of Flight

At highest point Vyy=0, so

Vyy = Uyy + g.t 0 = 22.9 + (-99.81)xt

∴∴ t = -222.9/-99.81= 2.33s

This is the mid-ppoint of thearc, so time of flight = 4.66s

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

Vxx= Uxx= constant velocitySxx = Vxx.t (use total time of flight)

= 16.1 x 4.66= 75.0m That’ll be 6 !

9

Analysing Projectile Motion (cont)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 withvertical and horizontal motion as 2 separatethings.

• The motion is symmetrical, so at the highestpoint, the elapsed time is exactly half the totaltime 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!

• Maximum Range is achieved at a launch angleof 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 knowntime, or find the time to fall through a knownheight (if Uy=0)

WORKSHEET at the end of this section

Projectiles Launched HorizontallyA common situation with projectile motion iswhen a projectile is launched horizontally, as inthe following example. This involves half thenormal trajectory.




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iimmuumm

HHeeii

gghhtt

““RRaannggee”” == TToottaall HHoorriizzoonnttaall DDiissppllaacceemmeenntt

HHoorriizzoonnttaallVVeelloocciittyy

VVx

VVeerrttiiccaallVVeelloocciittyy

VVy

TThhee ttoopp ooff tthhee aarrcc iiss tthhee mmiidd-ppooiinntt..AAtt tthhiiss ppooiinntt VVyy == zzeerroo

UUyy

Uxx

PPRROOJJEECCTTIILLEESS LLAAUUNNCCHHEEDD AATTSSAAMMEE VVEELLOOCCIITTYY

LLAAUUNNCCHH AANNGGLLEE 4455o

GGIIVVEESS MMAAXXIIMMUUMMRRAANNGGEE

Plane flying horizontally,at constant 50.0ms-11

Releases a bomb fromAltitude = 700m

Questionsa) How long does it take for thebomb to hit the ground?b) At what velocity does it hit?c) If the plane continues flyingstraight and level, where is it whenthe bomb hits?

SolutionBecause the plane is flying horizontally, theintitial velocity vectors of the bomb are:

Horizontal, Ux= 50.0ms-1,Vertical, Uy= zero

a) Time to hit the groundWe 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

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.

c) Where is the Plane?Since both plane and bomb travel at the samehorizontal velocity, it follows that they haveboth travelled exactly the same horizontaldistance when the bomb hits. i.e. the plane isdirectly above the bomb at impact.

(In warfare, this is a problem for low-levelbombers... the bombs must have

delayed-action fuses)

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10

Galileo and Projectile MotionNotice 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 you learnedabout in “The Cosmic Engine”, who firstdiscovered this.

His famous experiment was to drop objects ofthe same size and shape, but of different weight,from the leaning tower in Pisa. He found that allobjects hit the ground at the same time, therebyproving the point.

He also studied projectile motion. In his day,cannon balls were the ultimate weapon, buttrajectories were not understood at all. To slowthe motion down for easier study, Galileo rolledballs 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 ofProjectile Motion.

Unfortunately, he lacked the mathematicalformulas to go any further with his analysis.

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

Coincidentally, Newton was born in the sameyear that Galileo died.

Isaac Newton and OrbitingOnce Isaac Newton had developed the Mathsand discovered the laws of motion and gravity,he too looked at Projectile Motion.

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 “escapevelocity”.

Escape Velocity is defined as the launchvelocity needed for a projectile to escape fromthe 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, all projectiles, regardless of mass,need the same velocity to escape from Earth,about 11km per second!

• The Escape Velocity depends only on the massand radius of the Earth.

It follows that different planets have differentescape velocities. Here are a few examples...

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




EARTH

At the right velocity, the projectilecurves downwards at the samerate as the Earth curves... it will

circle the Earth in orbit!

If launch velocity is highenough, the projectile

escapes from the Earth’sgravity

11

Placing a Satellite in Earth OrbitA projectile needs an enormous velocity toescape from the Earth’s gravitational field...about 11 km per second. Think of a place 11 kmaway from you, and imagine getting there in 1second flat!

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

It can be shown that to achieve orbit, the launchvelocity required is less than escape velocity,but still very high... about 8 km per second. Howis 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:

The “g-Forces” in a Space LaunchTo accelerate a capsule (and astronauts)upwards to orbital velocity requires a force. Theupward “thrust” force must overcome thedownward weight force AND provide upwardacceleration.

So, if the Thrust force causesacceleration of (say)about 10ms-2, as well as overcoming his weight force, the 80kgastronaut will feel a pushing force of;

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

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 life-threatening. Jules Verne’s cannon astronautswould have suffered forces of about 200g...instantly fatal.

Rockets Achieve OrbitTo keep the g-forces low while accelerating tothe velocity required for orbit, AND then tooperate in the airless conditions of space, therocket is the only practical technologydeveloped so far.

A Brief History of RocketrySimple solid-fuel (e.g. gunpowder) rockets havebeen used as fireworks and weapons for over500 years.

About 100 years ago, the Russian Tsiolkovsky(1857-1935) was the first to seriously proposerockets as vehicles to reach outer space. Hedeveloped the theory of multi-stage, liquid-fuelrockets as being the only practical means ofachieving space flight.

The American Robert Goddard (1882-1945) developed rocketry theory futher,but also carried out practicalexperiments including the first liquid-fuel rocket engine.

Goddard’s experiments were the basisof new weapons research duringWorld War II, especially by NaziGermany. Wernher von Braun (1912-1977) and others developed theliquid-fuel “V2” rocket to deliverexplosive warheads at supersonicspeeds from hundreds of kilometers away.

At the end of the war many V2’s, and the Germanscientists who developed them, were capturedby either the Russians or the Americans. Theycontinued their research in their “new”countries, firstly to develop rockets to carrynuclear weapons (during the “Cold War”) andlater for space research.

The Russiansachieved thefirst satellite(“Sputnik”1957) and thefirst humanin orbit, andtheAmericansthe firstmannedmissions to the Moon (1969).

Since then, the use ofsatellites has become

routine and essential to ourcommunications, while

(unmanned) probes havevisited nearly every other

planet in the Solar System.




Astronaut During Acceleration to Orbital Velocity

Weight = mgForce

Net Force= ma

Total Net Force causes acceleration

GGrreeeekk lleetttteerrΣΣF = ma SSiiggmmaa (( ΣΣ ))

mmeeaannss ttoottaall If up = (+ve), down ( -vve)then

ΣΣF = T - mg = ma

∴∴ T = ma + mg

This means the astronautwill “feel” the thrust as an

increase in weight.

SpaceShuttlelaunch

“THRUST” Force = T

The businessend of a

1970’s liquid-fuel rocket

engine

V2




12

Direction of LaunchStraight upwards, right?Wrong!

To reach Earth orbit, rockets are aimed towardthe EAST to take advantage of the Earth’srotation. The rocket will climb vertically to clearthe launch pad, then be turned eastward.

Earth, viewedfrom aboveNorth Pole

Rotation

OOrrbbiitt ppaatthh

LLaauunncchhTTrraajjeeccttoorryy

At the equator, the Earth is rotating eastwards atabout 1,700km/hr (almost 0.5km/sec) so therocket already has that much velocity towardsits orbital speed.

Rocket launch facilities are always sited asclose to the equator as possible, and usuallynear the east coast of a continent so the launchis outwards over the ocean.

Conservation of Momentum

Why a rocket moves was dealt with in thePreliminary topic “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 theexhaust gases stays fairly constant during thelift-off. However, the mass of the rocketdecreases as its fuel is burnt. Therefore, therocket’s velocity must keep increasing in orderto maintain the Conservation of Momemtum.

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Physics of a Rocket Launch

Forces Experienced by AstronautsIf the “Thrust” force from the rocket engineremains constant throughout the “burn”, but thetotal rocket mass decreases due to consumption ofthe fuel, then the acceleration increases.

The concept of “g-forces” was explained on theprevious 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, buteven the acceleration keeps accelerating!

The astronauts will feel increasing “g-forces”. Atlift-off, they will experience perhaps only “2g”, butover several minutes this will increase to perhaps“7g” as the rocket burns thousands of tonnes offuel and its mass decreases.

The Space Shuttle’s engines are throttled-backduring the launch to counteract this, so theastronauts are not injured by increasing “g-force”. Photo: Russian Soyez lift-off,

courtesy Ali Cimen, seniorreporter, Zaman Daily, Istanbul.

Photo by Shelley Kiser

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13

Types of OrbitsThere are 2 main types of

satellite orbits:

Low-Earth OrbitAs the name suggests, this type of orbit isrelatively close to the Earth, generally fromabout 200km, out to about 1,000km above thesurface.

For any satellite, the closer it is, the faster itmust travel to stay in orbit. Therefore, in a Low-Earth Orbit a satellite is travelling quickly andwill complete an orbit in only a few hours.

A common low orbit is a “Polar Orbit” in whichthe satellite tracks over the north and southpoles while the Earth rotates underneath it.

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

This means that the satellite is always directlyabove the same spot on the Earth, and seems toremain motionless in the same position in thesky. It’s not really motionless, of course, butorbiting around at the same angular rate as theEarth itself.

Geo-stationary orbits are usually above theequator, and have to be about 36,000km above thesurface in order to have the correct orbital speed.

Being so far out, these satellites are not muchgood for photographs or surveys, but are idealfor communications. They stay in the samerelative position in the sky and so radio andmicrowave dishes can be permanently aimed atthe satellite, for continuous TV, telephone andinternet relays to almost anywhere on Earth.

Three geo-stationary satellites, spaced evenlyaround the equator, can cover virtually thewhole Earth with their transmissions.

Orbits & Centripetal ForceThe orbit of a satellite is often an oval-shape, or “ellipse”. However, in this

topic we will always assume the orbits arecircular... K.I.S.S. Principle.

Circular Motion was introduced in a Preliminarytopic. To maintain motion in a circle an objectmust be constantly acted upon by “CentripetalForce”, which acts towards the centre of thecircle.

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.

Example Problem next page...




Satellites and Orbits

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N

S

This type of orbit isideal for taking photos

or Radar surveys ofEarth.

The satellite only “sees”a narrow north-ssouthstrip of the Earth, butas the Earth rotates,each orbit looks at a

new strip.

Eventually, the entireEarth can be surveyed.

Being a close orbit, finedetails can be seen.

Centripetal ForceVector

always towards centre

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FccFcc

V

V

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, youneed to be aware that the orbital distance is oftendescribed as the height above the surface. To getthe 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 = 2ππR

Therefore, the orbital velocity V = 2ππRT

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Earth’sRotation

Equator

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14

Kepler’s Law of PeriodsExample ProblemA 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.

SolutionFor the satellite above, R3 = 6,5703 = 1.31 x 1011

(units are km & hours) T2 1.472

According to the law of periods, ALL satellites of Earthmust 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 about36,000km 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.

In this example, km & hrs were used. The same resultwill occur if metres & seconds are used.

Kepler’s “Law of Periods”When Johannes Kepler (1571-1630) studied themovement of the planets around the Sun (seePreliminary topic “Cosmic Engine”) hediscovered that there was always amathematical relationship between the Period ofthe 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 samevalue.

At this point, the HSC Syllabus is rathervague about whether you need to learnand know the following mathematicaldevelopment.

You may be safe to ignore it... (K.I.S.S.) but follow it ifyou can. Either way, you DO need to be able to use thefinal equation shown below.

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:

The Centripetal Force of orbiting is provided bythe Gravitational Force between the satellite andthe Earth, so

Centripetal Force = Gravitational ForceFc = mv2 = FG = GMm

R R2

∴∴ v2 = GM but v = 2ππRR T

So, 4ππ2R2 = GMT2 R

re-arranging, R3 = GMT2 4ππ2

Since the right hand side are all constant values,this proves Kepler’s Law and establishes the Forceof Gravity as the controlling force for all orbitingsatellites, 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!




Centripetal Force and SatellitesExample ProblemA 250kg satellite in a circularorbit 200km above the 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 geteverything 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.

WORKSHEET at the end of this section

R

200km

This is a very useful relationship...see Example Problem at bottom left


15

Kepler’s Law of Periods (Again!)On the previous page, the sample problem was able tocalculate the orbital radius for a geo-stationarysatellite by comparing the ratio of R3/T2 for 2 satellites.

With Newton’s development of Kepler’s Law, we cando it again a different way...Example ProblemFind 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 10-11

M = Mass of Earth = 5.97 x 1024kg)

R3 = 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 36,000km above thesurface... the same answer as before. (It better be!)

WORKSHEET at the end of the section

Re-Entry From OrbitGetting a spacecraft into orbit is difficult enough, butthe most dangerous process is getting it down againin one piece with any astronauts on board alive andwell.

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 allthat energy.

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 survivablerate 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 thecorrect angle:



Decay of Low-Earth OrbitsWhere does “Space” begin?

It’s generally agreed that by 100km above thesurface of the Earth the atmosphere has ended,and you’re in outer space. However, althoughthis seems to be a vacuum, there are still a fewatoms and molecules of gases extending outmany hundreds of kilometres.

Therefore, any satellite in a low-Earth orbit willbe constantly colliding with this extremely thin“outer atmosphere”. The friction or air-resistance this causes is extremely small, butover a period of months or years, it graduallyslows the satellite down.

As 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 100km levelthe friction becomes powerful enough to causeheating and rapid loss of speed. At this point thesatellite will probably “burn up” and bedestroyed as it crashes downward.

Modern satellites are designed to reach theirlow-Earth orbit with enough fuel still available tocarry out short rocket engine “burns” asneeded to counteract decay and “boost”themselves back up to the correct orbit. Thisway they can remain in low-Earth orbits formany years.

Upper Atmosphere

Angle too shallow...Spacecraft bounces off upper air

layers, back into space

Earth’s Surface

Angle correct...Spacecraft decelerates safely alonga descent path of about 1,000km

of “AtmosphericBraking”

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 tomelt and carry heat away, with the final descent byparachute. The Space Shuttle uses high temperaturetiles and high-tech insulation for heat protection, andglides in on its wings for final landing like an aircraft.

Correct angle isbetween 5-77oo

A projectile is any object which is launched, and thenmoves a).................................. The path of a projectileis called its b)................................, and is a curve.Mathematically, the curve is a c)..................................

To analyse projectile motion it is essential to treat themotion as 2 separate motions; d)..................................and .................................... If the launch velocity andthe e)............................ of launch are known, youshould always start by f)...................................... theinitial velocity into horizontal and verticalg)......................................

The horizontal motion is always h)............................................... and the vertical is constanti)................................... due to j)............................. Theusual strategy is to find the k)...................... of flight,by using the fact that at the top of the projectile’s arcits vertical velocity is l)......................... Once this isknown, it becomes possible to calculate themaximum m)....................... attained, and then)........................ (total horizontal displacement.). Theprojectile’s position and velocity at any instant can befound by combining the o)............................. and....................................... vectors. Maximum range ofany projectile occurs when the angle of launch isp).................... degrees upwards.

Historically, it was q)...................................... who firstproved that (ignoring air-resistance) all objectsaccelerate under gravity r)........................................................ He also investigated projectilemotion and was the first to see that the horizontalmotion is constant s).......................... while thet)................................. is constant acceleration.

Later, u)...................................... developed themathematics of both gravity and motion, whichallowed projectile motion to be understood andananalysed. He also discovered the concept ofv)...................... velocity, and of objects being inw)..........................., by imaging what would happen tocannon balls being fired horizontally at increasingvelocities from a high mountain.

“Escape Velocity” is defined as the velocity aprojectile needs in order to x).....................................................................

Satellites & OrbitsIf a projectile is travelling horizontally at the correcty)..................................., then its down-curvingtrajectory will match the z)................................. of theEarth. The projectile will continue to “fall down” butnever reach the surface... it is a aa)...............................which is ab)................................. around the Earth. Toplace a satellite in orbit, it must beac).................................... up to orbital speeds.

During upward acceleration, an astronaut willexperience “ad)..........................” which feel like anincrease in ae)......................... and can be life-threatening if too high.

The only feasible technology (so far) for achieving thenecessary af)................... ........................., whilekeeping the ag).................................. reasonably low,is the use of ah)............................ One of the importantsteps in the history of rocketry was achieved byRobert Goddard, who built and tested the firstai).............................-fuelled rocket.

Rockets are always launched towards theaj)...................... to take advantage of the Earth’sak)................................. Rocket propulsion is aconsequence of Newton’s al)........... Law. During thelaunch, momentum is am).............................. Thebackward momentum gained by the exhaust gases ismatched by the an).......................... momentum gainedby the ao)............................. However, the mass of therocket ap)....................... rapidly as is burns hugeamounts of fuel. This means that even with constantthrust, the acceleration rate aq).........................., andthe astronauts feel increasing ar)..............................

There are basically 2 different types of orbit for asatellite: as)........................................ orbits are whenthe satellite is at).......................... km from Earth andtravelling very au).............................. This is ideal forsatellites used for av)...................................... and........................................... The other type of orbit iscalled aw)..................................... For this the satelliteis positioned so its ax)......................... is exactly 24hours. This means it orbits at the same relative rateas the Earth’s ay)........................., and seems to stayin the az).................................................... This is idealfor ba)................................ satellites.

Any object undergoing Circular Motion is being actedupon by bb).............................. force, which is alwaysdirected towards the bc)............................................For an object twirled on a string, the centripetal forceis provided by the bd)...................... ..............................For a car turning a corner, it’s the force ofbe)..................... between tyres and road. For asatellite, it’s the force of bf).....................................

Johannes bg)............................... discovered the “Lawof Periods” for satellites. Later, Newton was able toshown that this was a consequence ofbh)............................ attraction between the satelliteand whatever it is orbiting.

Low-Earth orbits will eventually “bi)............................”due to the satellite gradually losing speed by collisionwith bj)...................................................

Re-entry of a spacecraft from orbit is extremelydangerous: bk)....................... from high velocity cancause high g-forces, and friction causes productionof bl)........................... energy which can cause thecraft to burn-up. The trick is to enter the atmosphereat exactly the correct bm)...........................

16




COMPLETED WORKSHEETSBECOME SECTION SUMMARIES

Worksheet 3 Projectiles & SatellitesFill in the blank spaces. Student Name...........................................


1. For each of the following projectiles, resolvethe initial launch velocity into horizontal andvertical components.

a) A rugby ball kicked upwards at an angle of60o, with velocity 20.5ms-1.

b) A bullet fired horizontally at 250ms-1.

c) A baseball thrown at 15.0ms-1, and an upangle of 25o.

d) An artillery shell fired at 350ms-1, upwards at 70o.

e) An arrow released from the bow at 40.0ms-1,at 45o up.

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).

17



Remember that for full marksin calculations, you need to show

FORMULA, NUMERICAL SUBSTITUTION,APPROPRIATE PRECISION and UNITS

Worksheet 4 Practice ProblemsProjectiles Student Name ...........................................

3. The bullet in Q1(b), was fired from a height of2.00m, across a level field. Calculate:a) how long it takes to hit the ground.

b) how far from the gun it lands.

c) At the same instant that the bullet left thebarrel, the empty bullet cartridge dropped (fromrest) from the breech of the gun, 2.00m abovethe ground. How long does it take to hit theground? Comment on this result, in light of theanswer to (a).

4. For the artillery shell in Q1(d), calculate:a) the time to reach the highest point of its arc.


c) its range (on level ground).

5. The rugby ball in Q1(a) was at ground levelwhen kicked.a) Find its exact position 2.50s after beingkicked.

b) What is its instantaneous velocity at thissame time?

1. A satellite orbiting 1,000km above the Earth’ssurface has a period of 1.74 hours. (Radius ofEarth=6.37x106m)a) Find its orbital velocity, using V=2ππR/T

b) If the satellite has a mass of 600kg, find thecentripetal force holding it in orbit.

2. A 1,500kg satellite is in a low-Earth orbit travellingat a velocity of 6.13 km/s (6.13x103ms-1). TheCentripetal force acting on it is 5.32x103N.

a) What is the radius of its orbit?

2. (cont)b) What is its altitude above the earth’s surface?

c) What is the period of its orbit?

3. A satellite is being held in Earth orbit by acentripetal force of 2,195N. The orbit is 350kmabove the Earth, and the satellite’s period is 1.52hours.a) Find the orbital velocity.

b) What is the satellite’s mass?

18




1. Fill in the table using data for each of thesatellites in Q’s 1, 2 & 3 in Worksheet 4.

Radius (m) Period (s) R3/T2

Q1.

Q2.

Q3.

Explain how this data supports Kepler’s Law ofPeriods.

2. Use the average value of R3/T2 from the tableabove to calculate the following:a) Find the Radius of an Earth orbit if the

Period is 1.60x103s.

b) What is the radius of orbit if T=1.15x104s?

c) Find the period of a satellite if R= 2.56x107m.

d) Find T when the satellite orbit is 2,000kmabove the Earth’s surface.

Worksheet 5 Practice ProblemsOrbits & Centripetal Force Student Name ...........................................

Worksheet 6 Practice ProblemsKepler’s Law of Periods Student Name ...........................................

3.a) Planet Mars has mass= 6.57x1023kg.Calculate the “orbital constant” GM/4ππ2 forMars. (G=Gravitational Constant = 6.67x10-11)

b) Find the orbital Radius of a satellite orbittingMars, if its Period is 1.60x103s.

c) Find the period of a Mars satellite whenR=2.56x107m.

d) In Q2(c) you calculated the period of an Earthsatellite with the same orbital radius.

Compare the answers to Q2(c) and Q3(c). Whichsatellite travels at the highest orbital velocity?

e) Complete the blanks in this generalstatement:At a given orbital radius, a satellite orbiting asmaller planet needs to travel at a............................................... velocity. The biggerthe planet, the ............................................. thevelocity would need to be.


Multiple Choice1. According to the formal definition of “GravitationalPotential 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.

2. On Earth g≅≅10ms-2 while on Mars g≅≅4ms-2. A 100kgobject transported to Mars would have mass andweight (respectively) of:A. 100kg and 400N.B. 400kg and 1,000N.C. 100kg and 1,000N.D. 400kg and 400N.

3. The diagram shows the trajectory of a projectile,and 2 points X & Y.

Which pair of vectors below correctly identifies thetotal acceleration vector of the projectile at points Xand Y?

Point X Point YA.

B.

C.

D.

4. To analyse projectile motion mathematically, usuallythe 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.

5.Ignoring air-resistance, the maximum range for any

projectile (for the same launch velocity) will occurwhen: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.

6.It is known that the value of “escape velocity” for anyplanet is

• proportional to the mass of the planet, and• inversely proportional to the planet’s radius.

Therefore, the planet which would definitely have alower escape velocity than Earth would have(compared to Earth)

A. more mass, smaller radius.B. less mass, larger radius.C. more mass, larger radius.D. less mass, smaller radius.

19



Worksheet 7 Test Questions sections 1&2 Student Name...........................................

XX YY

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.

8. To get maximum advantage from the rotation of theEarth, a space launch is always directed toward the A. northB. southC. eastD. west

9. The chief of the C.I.A. has asked you to plan thedeployment of a spy-satellite in an orbit suitable fortaking photos of suspected terrorist bases, in manylocations, world-wide. You would be best to plan forthe satellite to be in a:

A. Low-Earth Orbit, around the Equator.B. Geo-stationary Orbit, above the Equator.C. Geo-stationary orbit, allowing full earth coverage.D. Low-Earth, Polar Orbit.

10. Once the Space Shuttle reaches its orbital velocity theengines are turned off, and the craft orbits in free-falling circular motion. This motion is characterizedby: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.

11. The data below relates to 3 of the moons of Jupiterand one of the moons of Mars. The units ofmeasurement are arbitrary. Which moon (A,B,C or D)belongs to Mars?

Moon Orbital Radius Orbital PeriodA. 9.2 12.5B. 8.5 10.0C. 13.3 19.6D. 10.0 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.

Longer Response QuestionsMark values shown are suggestions only, and are togive you an idea of how detailed an answer isappropriate. Answer on reverse if insufficient space.

13. (3 marks)An alien creature weighs 7.25x103N on its homeplanet. On Earth, the creature weighs 5.84x103N.a) What is the creature’s mass?

b) What is the value of “g” on the creature’s homeplanet?

14. (4 marks)A ball was rolled along ahorizontal table at 5.45ms-1.If the table is 1.20m high, where willthe ball hit the ground?

15. (8 marks)An arrow was released from the bow at an upwardangle of 60o and an initial velocity of 42.0ms-1. It hitsits target at the same horizontal level from which itwas released.a) Find the time of flight.

b) Find the maximum height reached.

c) Calculate the distance from bow to target.

16. (6 marks)These military bombsare designed to bedropped from theaircraft at an altitude of15,000m when theplane is in level flight ata velocity of 300ms-1.

a) Ignoring air-resistance, how far infront of the target mustthe bombs be released?

b) How fast will they be going (magnitude only) whenthey hit the ground?

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 couldend up in orbit.

18. (4 marks) Author Jules Verne wrote a novel in which a spaceship was launched by firing it from a cannon, on ajourney to the Moon. The required velocity would be1.05x104ms-1.With the cannon barrel 200m long, it would bereasonable for the time of launch (i.e. duration ofacceleration) to be 7.50s.

a) Calculate the acceleration rate of the capsule.

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 ofthese men to the science of rocketry.

Tsiolkovsky, Goddard or von Braun

20. (4 marks)The early space rocket engines produced a constant“thrust force” throughout their “burn”. One of theadvantages of the Space Shuttle engine is that it hasa throttle control. During launch, the engine isgradually throttled back and thrust reduced duringthe ascent into orbit. Explain how this contributes tothe safety and comfort of the astronauts on board.

21. (6 marks)a) Use the relationship R3/T2 =GM/4ππ2 to find theradius of orbit of an Earth satellite with a period of2.00hours.

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.

22. (5 marks)Discuss the problems of recovering a spacecraft fromorbit and outline the process of “atmosphericbraking”, with reference to the importance of theangle of descent.

20




Worksheet 7 Test Questions sections 1&2 Student Name...........................................

Photo: Arian Kulp




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 thestrength of the gravitational force between 2masses:

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

distance between them.

21

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 1024kgMoon 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, 6,370kmfrom Earth’s centre (d=6.37 x 106m).

Solution FG = GMmd2

= 6.67x10-11x5.97x1024x 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

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

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

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 anysatellite in its orbit, and is the basis for Kepler’sLaw of Periods.

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 Solar System is orbiting the Galaxybecause of gravity, and whole galaxies orbiteach other. Ultimately, gravity holds the entireuniverse together, and its strength, compared tothe expansion of the Big Bang, will determinethe final fate of the Universe.




22

“Slingshot Effect” for Space Probes

One of the more interesting aspects of gravityand its effects on space exploration is called the“Slingshot Effect”.

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

• It costs billions of dollars to send a spaceprobe to another planet, so...

• It makes sense to send one probe to severalplanets, rather than a separate spacecraft toeach planet, but...

• the distances are enormous. Even at the highspeed of an inter-planetary probe (50,000 km/hr)it still takes years to reach other planets.

• Furthermore, having reached and done a “fly-by” to study one planet, the probe may need tochange direction and speed to alter course forthe next destination, and...

• It may be impossible to carry enough fuel tomake the necessary direction changes by usingrocket engines alone.

Got all that?

The solution to all these factors is to fly thespacecraft close enough to a planet so that theplanet’s gravity causes it to swing around into anew direction AND gain velocity (withoutburning any fuel).

So how can the the spacecraft gain extravelocity (and kinetic energy) from nothing?

The answer is that whatever energy thespacecraft gains, the planet loses. Energy isconserved. The planet’s spin will be sloweddown slightly by the transfer of energy to thespacecraft.

Of course, the huge mass of a planet means thatthe energy it loses is so small to be totallyinsignificant.

Spacecraft

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SlingshotTrajectory

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Worksheet 8Universal Gravitation

Fill in the blanks.

The strength of the gravitational force ofattraction between 2 masses is proportional toa)................................. and inversely proportionalto b).................................................................... So,if one mass is doubled the force willc)............................., but if the distance isdoubled, then the force will d)....................................................................

The force due to gravity provides thee).............................................. force for allsatellites, including the Moon andf).................................... orbiting the Sun. In spaceexploration, gravity can be used to alter aspacecraft’s g)................................. and to gainh)............................................ This is known as thei)...................................... Effect. The spacecraftgains energy, while j).............................................loses an k)................................. amount.

Practice Problems1. Fred (75kg) and girlfriend Sue (60kg) are verymuch attracted to each other. How much?Find the gravitational force attracting them whenthey are 0.5m apart.

2. What is the gravitational force of attractionbetween 2 small asteroids with masses of6.75x108kg and 2.48x109kg separated by 425m?

3. The mass of the Moon is 6.02x1022kg. A cometwith mass 5.67x1010kg is attracted to the Moonby a force of 6.88x1010N. How far apart are the 2bodies?

The Aether TheoryThe idea of the universal “aether” was a theorydeveloped to explain the transmission of lightthrough empty space (vacuum) and throughtransparent 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 travelthrough, so what is the medium for light waves?

From the 17th to 19th centuries, as modern Sciencedeveloped, it became the general belief that there wasa substance called the “aether” which was presentthroughout the universe as the medium for lightwaves to be carried in. The aether was invisible,weightless and present everywhere, even insidethings like a block of glass, so light could travelthrough it. The vacuum of space was actually filled bythe universal aether.

The Michelson-Morley ExperimentIn 1887, American scientists A.A. Michelson and E.W. Morley attempted to detect the aether byobserving the way that the movement of theEarth through the aether would affect thetransmission of light.

An Analogy to their experiment...

In Michelson & Morley’s experiment the “boats”were beams of light from the same source, splitand reflected into 2 right-angled beams sent outto mirrors and reflected back. The “current” wasthe “aether wind” blowing through thelaboratory due to the movement of the Earthorbiting the Sun at 100,000km/hr.

On arrival back at the start, the beams were re-combined in an “interferometer”, producing aninterference pattern as the light waves re-combined.

The entire apparatus was mounted on a rotatingtable. Once the apparatus was working, and theinterference pattern appeared, the whole thingwas rotated 90o, so that the paths of the lightrays in the aether wind were swapped.Theoretically, this should have created a changein the interference pattern, as the differencebetween the beams was swapped.

The Result...

There was NO CHANGE in the interferencepattern.

The experiment was repeated in many otherlaboratories, with more sensitve interferometersand all sorts of refinements and adjustments.

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

Enter Albert Einstein...

23

4. EINSTEIN’S THEORY OF RELATIVITYkeep it simple science

®



Imagine 2 identical boats, capableof exactly the same speed. They

both travel a course out and backover exactly the same distance, but

at right angles to each other.

In still water, they will get back atthe same time.

But what if there is a current?Now, they will NOT arrive back at

the same time, because the currentwill alter their relative speeds.

(The one moving across the currentwill arrive later.)

Stationary Aetherthroughout the

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tthhee SSuunn..

TThhiiss ccrreeaatteess aa ““ccuurrrreenntt”” oorr““aaeetthheerr wwiinndd””

In the laboratory, thislight beam travelsacross the current

This one travels withthe aether wind

Either the experiment has something wrong with it

orthe theory of the “Aether” is wrong!

Equipment is ableto be rotated

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24

How Science WorksThe Michelson-Morley Experiment is probablythe most famous “failed experiment” in thehistory of Science. It’s importance is not justhistorical interest, but a lesson in how Scienceworks.

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”is wrong, and “Relativity Theory” correct.

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

Relative Motion and Frames of Reference

Ever been sitting in a train at a station looking atanother train beside you? Suddenly, the other trainbegins moving. Or is it your train beginning to movethe other way?

The only way to be sure is to look out the other sideat the station itself, in order to judge which train isreally moving. You are using the railway station asyour “Frame of Reference” in order to judge therelative motion of the 2 trains.

We often use the Earth itself (or a railway stationattached to it) as our frame of reference. The Earthseems fixed and immovable, so everything else canbe judged as moving relative to the fixed Earth... butwe also know it’s NOT really fixed and unmoving, butorbiting around the Sun.

Astronomers use the background of “fixed stars” astheir frame of reference to judge relative planetarymovements, but we know that these aren’t really fixedeither.

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.

Sir Isaac Newton was aware of this idea, andfigured out that it really doesn’t matter whetheryour frame of reference is stationary or movingat a constant velocity. So long as it is notaccelerating, the observations, andmeasurements of motion will come out the sameanyway. This raises the idea of an “InertialFrame of Reference”.




HypothesisAccepted as

Correct Theory

HypothesisRejected as

Wrong

An Inertial Frame of Referenceis not accelerating

Within any Inertial Frame of Referenceall motion experiments

(and all “Laws of Physics”) will produce the same results

Distinguishing Inertial & Non-InertialFrames of Reference

Imagine you are inside a closed vehicle and cannotsee out. How can you tell if your “Frame ofReference” is “Inertial” or not?

A simple indication would be to hang a mass on astring from the ceiling. If it hangs straight downthere is no acceleration. If it hangs at an angle, (dueto its inertia) then your vehicle is accelerating.

Does it matter whether your vehicle is stationary ormoving at constant velocity? Not at all! The massstill hangs straight down, and any Physicsexperiments will give the same result as any otherobserver in any other Inertial Frame of Reference.

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




25

Albert Einstein’s Strange IdeaAlbert Einstein (1879-1955) has gone down inthe History of Science as one of the “Greats”,and just about the only scientist to ever matchthe achievements of the great Sir Isaac Newton.

Einstein’s “Theory of Relativity” is famous as agreat achievement, (true!) and as somethingincredibly complicated that hardly anyone canunderstand (false! It’s a dead-simple idea, but itdefies “common sense”.)

Einstein once declared “common sense” as “adeposit of prejudice laid down in the mind priorto the age of 18”. To understand “Einstein’sRelativity” you need to ignore “common sense”and have a child-like open-mind to fantasy andthe K.I.S.S. Principle...

But, if you are travelling at the speedof light, how is it possible for you,

and the stationary observers on theplatform, 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 beRELATIVE.

What does this mean?

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).

The Principle of Relativitywas already well known before Einstein, and stated invarious forms by Galileo, Newton and many others.

These are all statements of the “Principle ofRelativity”.

1. In an Inertial Frame of Referenceall measurements 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 your frame of reference is by measuring it against

someone else’s frame of reference

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Train Velocity at, or near,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 the mirror 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 ofRelativity...

In a non-accelerating, Inertial Frame of Reference, youwould measure the speed of light (and anything else, likereflection) exactly the same as anyone else... you wouldsee your reflection, and everything appears normal.

What Would Another Observer See?What about a person standing in the train stationas you flash (literally!) through at the speed oflight? What would they see through the trainwindow as you zap by?

Again, according to the results of the Michelson-Morley experiment, these observers will measurelight waves from you as travelling at the samespeed of light as you measure inside the train,because everyone is in an Inertial F. of R.

(Naturally, both train and platform are fullyequipped with interferometers and high-tech waysto do this)

26

Meanwhile, stationary observers are standingon the platform as your train flashes by. Theyalso measure the speed of light and get theexact same answer.

However, the peopleon the platform seeyou ascompressed inspace like this:

Furthermore, when they study yourclock they see it isrunning much slowerthan 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

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Train Velocity at, or near,the speed of light

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.

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

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!

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 observerLo= “rest length” measured within F.of R.v = relative velocity of observerc = speed of light = 3.00 x 108ms-1

THIS IS LENGTH CONTRACTION.IT OCCURS ONLY IN THE DIRECTION OF

THE RELATIVE MOTION

1 - v2

c2L = Lo

Example CalculationOn board a spacecraft travelling at “0.5c” (i.e. halfthe speed of light = 1.50x108ms-1) relative to theEarth, you measure your craft as being 100 metreslong. Carrying out this measurement takes you 100seconds.

Observers on Earth (with an amazing telescope) arewatching you. How much time elapses for them, andwhat is their measurement of your spacecraft?

= 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!

SolutionThe factor 1 - v2

c2t = time observed by outside observerto= time measured within F.of R.v = relative velocity of observerc = speed of light = 3.00 x 108ms-1

THIS IS TIME DILATION

t = to

1 - v2

c2


27

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

Two of the most fundamental laws everdiscovered by Science are the “Law ofConservation of Energy” and the “Law ofConservation of Matter”. These state thatenergy and matter (mass) cannot be created nordestroyed.

Einstein found that the only way to avoidbreaking these laws under “Relativity” was tocombine them. Hence, the most famousequation of all:

Within 30 years of Einstein’s ideas beingpublished, the “Equivalence of Mass & Energy”was dramatically confirmed by the release ofnuclear energy from atomic fission.

This is the process occurring in a nuclearreactor used to generate electricity in manycountries. It is also the energy source in an“atomic bomb”.

Einstein, a life-long pacifist, was appalled by thedestructive uses of the technology which grewfrom his discoveries.



m = mass observed by outside observermo= “rest mass” measured within F.of R.v = relative velocity of observerc = speed of light = 3.00 x 108ms-1

THIS IS MASS DILATION

E = mc2

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

THIS IS THE EQUIVALENCE OFMASS & ENERGY

1 - v2

c2

m = mo

Relativity and RealityDo these alterations to time and space reallyhappen? Yes they do, and they have beenmeasured!

• Extremely accurate “atomic clocks” have beensynchronized, then one flown around the world in ahigh speed aircraft. When brought back together,the clock that travelled was slightly behind theother... while travelling at high speed it’s time hadslowed down a little, relative to the other.

• Certain unstable sub-atomic particles always“decay” within a precise time. When theseparticles are travelling at high speeds in aparticle accelerator, their decay time is muchlonger (as measured by the stationaryscientists). At high speed the particle’s time hasslowed down relative to the scientists’ time.

It’s important for you to realise that, if this particlecould think, it would not notice any slow-down intime... its own “feeling” of time and its little digitalwatch would seem perfectly normal to it. But, fromthe relative viewpoint of the scientists measuringthe particle’s decay, its time has slowed downrelative to laboratory time.

Confirmation of RelativityEinstein published his theory in 2 parts, in 1905 and1915. At that time there was no way to test thepredictions of Relativity to find supporting evidence.

The Michelson-Morley experiment had failed to findsupporting evidence for the existence of the “aether”,so maybe “Relativity” would fail too, but firstscientists had to find testable predictions.

The first test was that, according to Relativity, lightfrom a distant star passing close to the Sun should bebent by a measurable amount, making the star appearto change position in the sky. The only way to test thisprediction was during a solar eclipse.

At the next occurrence of an eclipse, theobservations were made, and showed resultsexactly as predicted by Relativity.

In the following years, experiments with nuclearreactions (which led to the development of the“atom bomb”, and nuclear power) were able toconfirm the conversion of matter into energyaccording to E=mc2.

Later still came the measurements of timedilation (described above) and mass dilation hasalso been measured for high-speed particles ina particle accelerator.

EVERY RELATIVITY PREDICTION THAT CAN BETESTED HAS SHOWN RESULTS SUPPORTING

& CONFIRMING THE THEORY... that’s why we believe it to be correct.

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28

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 itwill be referred to as the “LFC”.

Consider firstly, what happens to the value ofthe LFC at different relative velocities:

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

This means that if you (in your spacecraft) andthe observer watching you have zero relativevelocity (i.e. you are travelling at the samerelative speed) then both of you will measurethe same length, time and mass... no relativisticeffects 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



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cc

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1 - v2

c2

Some Implications of RelativityThis all means that as your spacecraftaccelerates and approaches the speed of light,your faithful observer sees your lengthapproach zero, your time slowing down andapproaching being totally stopped, and yourmass increasing to approach infinity.

At the speed of light, the calculations for timeand mass dilation become mathematically“undefined”... this is generally taken to meanthat no object can ever be accelerated up to thespeed of light.

Another way to reach this conclusion is that asyou speed up, your mass increases. Toaccelerate more, greater force is neededbecause your increased mass resistsacceleration. As your mass approaches infinity,an infinite amount of force is needed toaccelerate you more...it’s impossible to reach c.

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

Simultaneous EventsAnother consequence of Relativity is that you,and your observer, will not agree onsimultaneous events. You may see 2 thingsoccur at the same instant, but the relativisticobserver will see the 2 events occurring atdifferent times. It is even possible that the observer could seean effect (e.g. spilt milk) before seeing the cause(e.g. glass tipped over). Curiouser & curioser!

How We Define Length & TimeOur S.I. unit of length, the metre, was originallydefined by the French as “One ten-millionth ofthe distance from the Equator to the Earth’sNorth Pole”.

Based on this,special metal barswere carefullymade to be used asthe “standard”metre from whichall other measuringdevices weremade.

As our technologyimproved, so did our ability to measure time anddistance. Today we define the metre as “thedistance travelled by light during a time intervalof 1/299,792,458th of a second.”

Our defintion of length is actually based on themeasurement 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 of thetime 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-mmadeplatinum bars used to define the

metre up until 1960.

Part of themechanism of

an “atomicclock” which

measuresatomic

vibrations togive our

standards oftime anddistance.


The theory of the “aether” was invented toexplain a).........................................................because it was thought that all waves needed ab)............................. to travel through. The aetherwas invisible and c)................................, and waspresent throughout the d).....................................

The American scientists e).............................. &............................... attempted to detect theaether by experiment. Their apparatus used 2f)..................................., travelling at right angles.When brought together by mirrors, the beamsproduced an g).......................................... pattern.The idea was that the pattern should changewhen 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 byj)...................... ........................... through space.The result was that k).........................................................................................

An “l)......................... Frame of Reference” is onewhich is not m)....................................... Withinsuch a place, all measurements andexperiments will give the n).......................................... This idea is known as the“Principle of o)......................................”.

2.(cont.)b) What relativistic mass will it have if accelerated upto 0.9999c? (99.99% of “c”)

3. In a nuclear reactor, over a perio of time, a total of2.35kg of “mass deficit” occurs. This mass has“disappeared” during the nuclear reactions.Calculate the amount of energy this has released.

4. According to the “Big Bang” Theory, in the firstmoments of the Universe there was nothing butenergy. Later, matter formed by conversion from theenergy.

Calculate how much energy was needed to produceenough matter to form the Earth (mass= 5.97 x 1024kg).

29



Worksheet 9 RelativityFill in the blank spaces. Student Name...........................................

Albert Einstein applied this principle to theMichelson-Morley result. He concluded that allobservers will always measure the speed of lightas being p)..................................... For this tohappen, then q)......................... and.............................. must be relative. This meansthat the measurements of length and time asseen by r).............................................................................. ................ willbe different.

Relativity Theory predicts that Length wills)....................... while time will t)..........................Also, mass will u)............................., therebymaking it impossible to actuallyv).............................................. Relativity alsopredicts that mass can be converted intow).............................. and vice-versa.

Although it defies common sense, many aspectsof Relativity have been confirmed byx)......................................... For example,synchronised clocks have been found todisagree if one of them isy).............................................................. Theconversion of mass into energy has beenobserved (many times) duringz)............................ reactions.

Worksheet 10 Practice ProblemsRelativity Student Name ...........................................1. A spacecraft is travelling at 95% of the speed oflight relative to an observer on Earth. On board is afluorescent light tube which is 0.95m long and isswitched on for 1 hour ship-time.

a) How long is the fluoro tube as seen be the Earthobserver?

b) The Earth observer measures the time for whichthe light was on. What time does he measure?

2. A sub-atomic particle has a “rest mass” of 5.95x10-29kg. The particle was accelerated by aparticle accelerator up to a velocity of 0.99c. (99% “c”)

a) What relativistic mass will the particle now have, ifmeasured by the scientists in the laboratory?

Multiple Choice1. The gravitational force between 2 masses is“F” units when they are distance “d” apart.If these masses were brought closer, to adistance “0.25d”, then the force between themwould be:A. 0.25 F B. 4FC. F/16 D. 16F

2. The “Aether” was an ideaA. 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.

3. In an “Inertial Frame of Reference” a masshanging from the ceiling by a string wouldprobably.A. swing back and forth.B. hang straight down.C. hang at an angle from the vertical.D. undergo mass dilation due to relativisticeffects.

4. Two cosmonauts in separate spacecrafttravelling at different relativistic velocities areable to make a series of observations andmeasurements of their own spacecraft, and ofeach other’s spacecraft. The one thing theywould agree with each other about is:A. the simultaneity of 2 events occurring together.B. the value of the velocity of light.C. the passage of time in each other’s spacecraft.D. the length of each other’s spacecraft.

5. The first experimental confirmation ofEinstein’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.

6. From the Earth, you are able to observe andmeasure several features of an alien spacecraftas it flies by at 90% of the speed of light.Compared to the measurements made by thealien on board, your measurements wouldshow:

Craft mass Craft length Craft timeA. less shorter fasterB. more longer slowerC. more shorter slowerD. less longer faster

Longer Response QuestionsMark values shown are suggestions only, and are togive you an idea of how detailed an answer isappropriate. Answer on reverse if insufficient space.

7. (5 marks)An astronaut with mass (including spacesuit) 120kgis standing on the surface of the planet Mercury. Theplanet has a mass of 2.99x1023kg and radius2.42x106m.a) Calculate the gravitational force acting on theastronaut.

b) From your answer to (a) calculate the value of “g”on the surface of Mercury.

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 Michelson-Morley experiment, including:a) their aim or purpose in doing the experiment.

b) an outline of the method used.

c) the results.

10. (3 marks)What did Albert Einstein conclude about themeasurement of space and time, taking into accountthe results of the Michelson-Morley experiment?

11. (8 marks)A sub-atomic particle, at rest in the laboratory, has thefollowing 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 of0.999c (i.e. 99.9% of the speed of light) in a particleaccelerator. Calculate:a) its relativistici) diameterii) half-lifeiii) massas measured by the scientists in the laboratory.

b) While travelling at 0.999c, the particle undergoes anuclear reaction which results in its total annihilationby conversion to energy. What energy release ismeasured by the scientists? (The scientists observethe conversion of its relativistic mass, not rest-mass)

30




Worksheet 11 Test Questions sections 3 & 4 Student Name...........................................

31




CONCEPT DIAGRAM (“Mind Map”) OF TOPICSome 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

32

Answer SectionWorksheet 1a) force b) gravityc) accelerate d) 10 (9.81)e) period f) gradientg) force field h) attractsi) repel j) gravitational fieldk) mass l) work donem) infinity n) negativeo) 9.81 p) weightq) mass

Worksheet 21. 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 3a) only under gravity b) trajectoryc) parabola d) horizontal & verticale) angle f) resolvingg) components h) constant velocityi) acceleration j) gravityk) time l) zerom) height n) rangeo) horizontal & vertical p) 45q) Galileo r) at the same rates) velocity t) verticalu) Newton v) escapew) orbitx) escape from the Earth’s gravitational fieldy) velocity z) curvatureaa) satellite ab) in orbitac) accelerated ad) g-forcesae) weight af) velocityag) g-forces ah) rocketsai) liquid aj) eastak) rotation al) 3rdam) conserved an) forwardao) rocket ap) decreasesaq) increases ar) g-forcesas) low-Earth at) 200-1,000au) quickly/fast av) photos & surveysaw) geo-stationary ax) perioday) rotation az) same position in the skyba) communication bb) centripetalbc) centre of the circle bd) tension in the stringbe) friction bf) gravitybg) Kepler bh) gravitationalbi) decaybj) gas molecules/upper atmosph.bk) deceleration bl) heatbm) angle

Worksheet 41. Uy = U.Sinθθ Ux = U.Cosθθa) = 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/-4.905t = 0.639s.

b) Sx = Vx.t = 250x0.639 = 160m.c) see working for (a). Empty cartridge takes 0.639s to hit the ground. Itfalls down at exactly the same rate as the bullet. Thedifference is where each lands horizontally.4.a) At highest point, Vy=0, and Vy = Uy + g.t

0 = 329 + (-9.81)x tt = -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.a) Vertical displacement Horizontal Displ.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.b) Vertical velocity Horizontal velocityVy = Uy + g.t Vx = Ux = 10.3 ms-1

= 17.8 + (-9.81)x2.50= -6.725ms-1 (downwards)

V2 = Vy2 + Vx

2 = 10.32 + 6.7252

∴∴ V = sq.Root(151.32) = 12.3ms-1.Tan θθ = 6.725/10.3,

∴∴ θθ ≅≅ 33o below horizontal




RReessuullttaanntt vveelloocciittyy

1100..33θ

66..772255


33

Worksheet 51.a) T=1.74 hours = 1.74x60x60= 6,264s

R= 1,000 km (=106m) + 6.37x106 = 7.37x106mV = 2ππR/T

= 2xππx7.37x106/6,264= 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=2ππR/T, so T = 2ππR/V = 2xpx1.06x107/6.13x103

= 1.09x104s. (3.02 hours)3.R = 350km + 6.37x106m = 6.72x106mT= 1.52 hrs = 1.52x60x60 = 5.47x103s.a) V=2ππR/T = 2xππx6.72x106/5.47x103

= 7.72x103ms-1.b)Fc=mv2/R, so m=F.R/v2 = 2,195x6.72x106/(7.72x103)2

= 247kg.

Worksheet 61.Question Radius(m) Period(s) R3/T2

1 7.37x106 6.26x103 1.02x1013

2 1.06x107 1.09x104 1.00x1013

3 6.72x106 5.47x103 1.01x1013

The ratio R3/T2 is the same for all 3 satellites.(slight differences are due to rounding-off errors incalculations)Keplers Law states that this ratio should be the samefor the satellites of any planet.

2. Average value from table = 1.01x1013= constanta) 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/4ππ2 = 6.67x10-11x6.57x1023/4xππ2

= 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 travelsfaster.

e) At a given orbital radius, a satellite orbiting asmaller planet needs to travel at a ....lower.....velocity.The bigger the planet, the ....faster (higher).... thevelocity would need to be.



Worksheet 71. C 2. A 3. C 4. D 5. B 6. B7. A 8. C 9. D 10. D 11. A 12. D

13.a) On Earth: 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.Sinθθ Ux = U.Cosθθ

= 42.0xSin60 = 42.0xCos60= 36.4ms-1 = 21.0ms-1.

a) At max.height, Vy = 0, and Vy = Uy + g.t

0 = 36.4 x (-9.81)x tt = -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= 542ms-1.

V2 = Vy2 + Vx

2 = 5422 + 3002

∴∴ V = sq.Root(383,764) = 619ms-1.(almost twice the speed of sound!)

17. He imagined a cannon firing projectiles horizontallyfrom a very high mountain. As the launch velocity isincreased, the shot travels further before reaching theground. At a high-enough velocity, the downwardcurve of its trajectory could match the curvature ofthe Earth. This projectile would travel right around theEarth, constantly falling towards it, but never gettingany 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 143times their normal weight.c) This amount of g-force would be instantly fatal.This is not a feasible method for human space launch.

RReessuullttaanntt vveelloocciittyy

330000θθ

554422

34

Worksheet 7 (cont.)19.(example) For 3 marks, try to make 3 points.Robert Goddard built and tested the first liquid-fuelrocket engine, an essential step for practical rocketry.His experiments in the 1920’s-1930’s were the basisfor later research during World War II which led to thefirst long-range rockets. Goddard also advanced thetheory of multi-stage rockets as the way to reachouter space.

20.Although a rocket may produce constant thrust, themass decreases rapidly as the fuel is burned. Thismeans that the rate of acceleration increases duringthe ascent. This means that the g-forces keepincreasing.

By being able to throttle back the engines, the SpaceShuttle can reduce this effect and maintain moreconstant g-forces during the acent. This is muchsafer and more comfortable for the astronauts.

21.a) period, T= 2.00hr = 2.00x60x60 = 7,200s.R3/T2 = GM/4ππ2

R = Cube.Root(GMT2/4ππ2)= Cube.Root(6.67x10-11x5.97x1024x(7,200)2/4ππ2

= 8.06x106m.b) V = 2ππR/T = 2xpx8.06x106/7,200

= 7.03x103ms-1.c) Fc=mv2/R = 2,650x(7.03x103)2/8.06x106

= 1.62x104N.22.It is impossible for a spacecraft to carry enough fuelto use its rocket engines for the completedeceleration and descent from the high speed oforbit. So atmospheric braking is used instead. If thecapsule enters the atmosphere at just the right angle,friction can slow it down gradually, with the energybeing converted to heat.

If the re-entry angle is too steep, the g-forces will betoo high, and the craft may burn up. If the angle is tooshallow, the craft may bounce off the atmosphere andgo into an un-recoverable orbit.

Worksheet 8a) the product of the massesb) the square of the distance between themc) double d) decrease by a factor of 4.e) centripetal f) the planetsg) direction/trajectory h) speedi) Slingshot j) planetk) equalPractice Problems1.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 theMoon is 1,738km, then the comet is just 82km fromthe surface... DEEP IMPACT about to happen!)

Worksheet 9a) transmission of light in vacuumb) medium c) massless / weightlessd) Universe e) Michelson & Morleyf) beams of light g) interferenceh) rotated 90 degrees i) acrossj) the Earth’s motionk) no change to the interference pattern-no aetherwind detected.l) Inertial m) acceleratingn) same results o) Relativityp) the same q) space & timer) an observer travelling at a different relative velocitys) shorten t) lengthen / slow downu) increase v) accelerate to the speed oflightw) energy x) observation/experimenty) transported at high speed z) nuclear

Worksheet 10Using the abbreviation “LFC”=

1.a) At 0.95c, LFC = Sq.root( 1 - (0.952/12))

= 0.31L = 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))

= 0.14m = m0 / LFC = 5.95x10-29/0.14 = 4.25x10-28kg.b) At 0.9999c, LFC = Sq.root(1 - (0.99992/12))= 0.01414m = 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.

Worksheet 111. D 2. C 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.

8.a) The Slingshot Effect is the technique of flying aspacecraft near a planet so that the planet’s gravityaccelerates the craft and alters its trajectory onto anew, desired heading.

b) It is useful for enabling a craft to visit severalplanets on one flight, gaining speed and a newdirection without the need for large amounts of fuel.

c) The gain of energy by the spacecraft is exactlyequal to the loss of energy by the planet (which losesan insignificant amount of rotational energy).




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35

Worksheet 11 (cont)9.a) The M-M experiment was designed to detect the“universal aether” which had been hypothesised as amedium for transmission of light waves.

b) The method involved a beam of light, split into 2beams which travelled at right angles. Mirrors thenrecombined the beams in an interferometer toproduce an interference pattern. The apparatus couldthen be rotated 90 degrees. Since one beam traveledalong the “aether wind” (caused by the Earth’smovement) and the other across it, there should havebeen a change in the interference pattern when theapparatus was rotated.

c) The result was negative. No change in theinterference pattern occurred.

10.Einstein concluded that the speed of light is constantfor all observers in any Inertial Frame of Reference(IFR), regardless of relative motion. This why the M-Mexperiment failed to detect a difference between thespeed of light in 2 different directions.However, this requires that the measurements ofspace and time must be different relative to anobserver’s IFR. This means that measurements oflength and time taken in one IFR, will be different towhat a relativistic observer sees.

11.a) At 0.999c, the “LFC” factor

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.



= Sq.Root(1 - 0.9992) = 0.044712

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