module 1 space

HSC Physics Module 1: Space

Summary

Robert Lee Chin 1

Space: 1. The Earth‟s gravitational field Define weight as the force on an object due to a gravitational field

Weight is defined as the force of an object due to a gravitational field. It is a vector quantity

with the unit Newton (N). Mathematically, weight can be expressed as W=mg, where m is the

mass of the object and g is the acceleration due to gravity.

Explain that a change in gravitational potential energy is related to work done

Consider the work done to move an object from the Earth‟s surface to a height, h:

(where forcegravity is the weight)

(as the displacement over which the work is

done is equal to the height of the lift)

Mathematically, W = mgh

Perform an investigation and gather information to determine a value for

acceleration due to gravity using pendulum motion or computer assisted technology

and identify reason for possible variations from the value 9.8ms-2

Investigation: Determining a value for gravity

Aim: To determine a value for g, by observing the motion of a swinging pendulum

Equipment:

-50g mass or large nut to act as pendulum

-about 1.2m length of string

-a support at least one metre above the ground (e.g. hook on ceiling)

-stopwatch

-another person to assist

Method:

1/ Adjust the length of the pendulum, l, to one metre.

2/ Set up equipment as shown:

Work done = force x displacement

Work done = forcegravity x displacement

Work done = [mass x gravity] x displacement

Work done = [mass x gravity] x height


Summary

Robert Lee Chin 2

3/ Set pendulum gently by pulling it back 10˚ from vertical. Starting at the extreme of

the motion, time 10 full periods, finishing back at the same side.

4/ Divide time by 10 and record in table. Calculate a value for g, using 2

24

Tg

5/ Repeat steps 3-4 four more times, shortening string by 5cm each time.

6/ Calculate an average value for g, using all 5 trials

Results:

Trial Pendulum length (m) Period (s) g (ms-2

) T2

1 1.01 2.00 9.9680 4.00

2 0.97 1.93 10.281 3.72

3 0.91 1.97 9.256 3.88

4 0.85 1.89 9.394 3.57

5 0.81 1.88 9.041 3.53

Average 9.59 n/A

Alternatively, plot T2 against l on an x-y graph and draw the line of best fit:

T squared (s2) length (m)

4 1.01

3.72 0.97

3.88 0.91

3.57 0.85

3.53 0.81

0.00 0.00

l

Time taken for one

complete swing is

called the period, T


Summary

Robert Lee Chin 3

Calculate „g‟ using the formula: gradient

4πg

2

Gradient ≈ 4.10 (3 s.f.)

s.f.) (3 63.9...62888.910.4

4''

2

g

Discussion:

Variations in g were due to reaction time in timing the swing- the accuracy of the stopwatch

is much more accurate than reaction time. Also, friction between the string and its attachment

as well as air resistance, which would have slowed the swing of the pendulum.

The reliability of the results could be improved by performing multiple trials for the same

length. Also, any results that do not conform with the majority of the results e.g. a value of

7ms-2

for „g‟, should be ignored and the trial repeated until a measurement that conforms to

the majority is obtained.

The validity of the experiment could be improved by using an attachment for the string that

would produce as little friction as possible, without causing the pendulum to swing

erratically. The experiment was conducted slightly above sea level (approximately 50

metres). While this altitude may not seem like much, it could have has some effect on the

results. Finally, two people were involved with performing the experiment- one to time the

pendulum and another to let go of the pendulum. It would be better to have only one person

to perform both of these steps because there are differences in reaction times when two or

more people are involved..

The first method gave a fairly reliable value for g, (9.59ms-2

), only 0.21ms-2

less than the

published value of 9.8ms-2

. Using the second method gave a slightly more accurate value of

9.63ms-2

. The second method is more valid because it relies on the line of best fit, which

helps to rule out any discrepant results

T squared vs L

y = 4.0965x

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 0.2 0.4 0.6 0.8 1 1.2

length (m)

T s

qu

are

d (

s s

qu

are

d)


Summary

Robert Lee Chin 4

Gather secondary information to predict the value of acceleration due to gravity on

other planets

The value of g on a planet is given by:

metresin planet of radius

kgin planet of mass

10676constant nalgravitatio UniversalG

where,

21311

2

Planet

Planet

Planet

Planet

r

m

skgm

r

mGg

.

)(

Planet Value

of g

(ms-2

)

Calculations

Mercury

3.70 s.f.) (3 70369709310442

1030310676

10442 10303

2

26

2311

2

623

msr

mGg

mrkgm

Planet

Planet

PlanetPlanet

.....).(

).(.

)(

..

Venus

8.87 s.f.) (3 87887450810056

1087410676

10056 10874

2

26

2411

2

624

msr

mGg

mrkgm

Planet

Planet

PlanetPlanet

....).(

).(.

)(

..

Earth 9.80

Mars

3.70 s.f.) (3 70370427310403

1042610676

10403 10426

2

26

2311

2

623

msr

mGg

mrkgm

Planet

Planet

PlanetPlanet

....).(

).(.

)(

..

Jupiter

24.8 s.f.) (3 82478942410157

1090110676

10157 10901

2

27

2711

2

727

msr

mGg

mrkgm

Planet

Planet

PlanetPlanet

....).(

).(.

)(

..

Saturn

10.4 s.f.) (3 410437661010036

1069510676

10036 10695

2

27

2611

2

726

msr

mGg

mrkgm

Planet

Planet

PlanetPlanet

....).(

).(.

)(

..

Uranus

8.83

s.f.) (3 83883416810562

1068810676

10562 10688

2

27

2511

2

725

msr

mGg

mrkgm

Planet

Planet

PlanetPlanet

....).(

).(.

)(

..

Pluto

0.623

s.f.) (3 632062273010181

1030110676

10181 10301

2

26

2211

2

622

msr

mGg

mrkgm

Planet

Planet

PlanetPlanet

....).(

).(.

)(

..


Summary

Robert Lee Chin 5

Analyse information using the expression F=mg to determine the weight force for a

body on Earth and for the same body on other planets

The magnitude of the weight force on an object can be calculated using a slightly altered

form of Newton‟s second law: F = mg. The direction of the weight force is the same as the

direction of the gravitational field at the location of the object i.e. towards the centre of the

Earth.

The acceleration due to gravity at the Earth‟s surface is 9.8ms-2

. The weight of a 465kg llama

can be calculated as follows:

s.f.) (2 4600455789465

9.8msg 465 -2

NmgW

kgm

.

Acceleration due to gravity for the same 465kg llama on different planets:

Planet Value of g

(ms-2

)

Weight Force Calculations

Mercury 3.70 1700

s.f.) (2 170051720703465 NmgW ..

Venus 8.87 4100

s.f.) (2 4100554124878465 NmgW ..

Earth 9.8 4600 -

Mars 3.70 1700

s.f.) (2 170051720703465 NmgW ..

Jupiter 24.8 12000

s.f.) (2 1200011532824465 NmgW .

Saturn 10.4 4800

s.f.) (2 48004836410465 NmgW .

Uranus 8.83 4100 s.f.) (2 4100954105838465 NmgW ..

Pluto 0.623 290 s.f.) (2 29088293410465 NmgW ..


Summary

Robert Lee Chin 6

Define gravitational potential energy as the work done to move an object from a

very large distance away to a point in a gravitational field, r

mmGE p

21

The gravitational potential energy of an object at some point within a gravitational field is

defined as: the work done to moving the object from an infinite distance to a point in a

gravitational field

At an infinite distance away from a gravitational field, the gravitational potential energy is

zero. As an object moves towards the source of the field i.e. the centre of a planet,

gravitational potential energy becomes negative. Although this value is negative, it represents

a change in Ep i.e. work done. Work is therefore required to push an object away from, as

well as towards the Earth.

Mathematically:

)2

2

1

21311-

21

(mplanet of radius

planet of mass

field e within thmovingobject an

106.67 constant grav. universalG

energy potential grav.

where,

r

m

m

skgm

E

r

mmGE

p

p

Ep

r

d

+

-

Earth


Summary

Robert Lee Chin 7

Space: 2. A successful rocket launch

Solve problems and analyse information to calculate the

actual velocity of a projectile from its horizontal and

vertical components using:

A projectile is any object that is launched, and then moves only under the influence of

gravity. Examples are a ball that is struck, a bullet shell once it is fired. A rocket or missile is

NOT a projectile.

The trajectory of a projectile can be analysed as two separate motions:

Horizontal (x-axis) motion, representing constant velocity

Vertical (y-axis) motion, representing constant acceleration at “g”, downwards

Equations for Projectile motion:

1) Resolve initial launch velocity into vertical and horizontal component

2

22

22

2

1

2

tatuy

tux

yauv

atuv

uv

yy

x

yyy

xx

Ux

Δy

Vertical

velocity,

Vy

Horizontal

velocity,

Vx

Uy

Ux

U

θ Angle of launch

Δx= range = total horizontal distance

Ux

Uy

θ

U

cos.sin. uuuu xy &


Summary

Robert Lee Chin 8

2) Horizontal motion is constant, so all you need is tuxuv xxx and

3) Vertical motion is constant acceleration at “g”

To find vertical velocity, use

To find vertical displacement, use

Points to remember:

-the motion is symmetrical, so the elapsed time is half the full flight

-vertical displacement, y has a negative sign

-at its peak height, 0yv

-Maximum range is achieved with a launch angle of 45˚, i.e. 45

-horizontal velocity is constant i.e. xx uv

-Use tauv yyy to find “t” at max height (when vy=0) or find vy at a given time

-Use 2

2

1tatuy yy to find Δy at a given time, or find the time to fall through a given

height (if uy=0)

Example Problems:

1) A projectile is fired with a velocity of 50m/s at an angle of 30° to the horizontal.

Determine the range of the projectile.

s.f.) (3 220...924847.2208.9

)30sin50(230cos50

8.9

)30sin50(2

8.9

30sin50

8.9

30sin500

0 flight,-half heConsider t

8.9 30cos50 30sin50

2

1

1

2

mtux

t

a

uvt

msv

msauu

x

y

yy

y

yxy

2) Military bombs must be dropped from an altitude of 15 000m when the plane is

flying level at 300ms-1

.

)9.8ms( -2

yyyy atauv

)9.8ms( 2

1 2-2

yyy atatuy


Summary

Robert Lee Chin 9

a) How far in front must the bombs be released?

mtux

fsst

tt

tatuy

msumsumy

x

yy

yx

16590355300

33552383335594

15000

892

1015000

2

1

030015000

2

2

11

).(

.).(......

).(

b) How fast will they be going (in magnitude) when they hit the ground?

s.f.) (3 101964343961994541300

94541355890

300

890300355

1212222

1

1

211

msmsvvv

mstauv

msuv

msamsumsust

yx

yyy

xx

yyx

.....).()(

.).(.

..

Describe the trajectory of an object undergoing projectile motion within the Earth‟s

gravitational field in terms of horizontal and vertical components.

The trajectory of a projectile (ignoring air resistance) is parabolic, with constant downward

acceleration at “g”. The trajectory can be analysed by considering its horizontal and vertical

components at particular instances during the flight. The horizontal motion of the projectile is

a constant velocity. Its vertical motion is changing all the time due to gravity, which causes

the projectile to accelerate at 9.8 m s-2

downwards.

Describe Galileo‟s analysis of projectile motion

Notice that none of the equations used for projectile motion ever use the mass of the

projectile. All objects, regardless of their mass, accelerate with gravity at the same rate.

Galileo (1564-1642) performed an experiment to verify this. He dropped objects of the same

size and shape, but with different mass, from the leaning tower in Pisa. He found that

regardless of mass, all objects hit the ground at the same time.

He also rolled cannon balls down an incline (there was friction) and was able to see 2

motions: constant horizontal velocity and constant vertical acceleration.


Summary

Robert Lee Chin 10

Perform a first-hand investigation, gather secondary information and analyse data

to calculate initial and final velocity, maximum height reached, range, time of flight

of a projectile for a range of situations by using simulations, data loggers and

computer analysis

Experiment: Projectile launcher

Aim: To construct a working model of a projectile launcher and to launch it at a target to gather

data for calculating various aspects of the projectile‟s flight including force, velocity and

range.

Equipment:

- projectile launcher

- steel ball bearings (small enough to fit in projectile launcher barrel)

- scientific scales

- digital video camera

- target (circle of diameter 5 % of range)

Note: The experiment requires the assistance of another person during the launching of the

projectile.

Procedure:

1) Weigh and record the mass of the steel ball with a scientific scale.

2) Set up equipment as shown below:

3) Place steel ball in barrel of projectile launcher. Pull back end of solid plunger rod

at least 20cm until a strong tension force can be felt. Get a friend to help hold

down the back end of the launcher base.

Target

Range


Summary

Robert Lee Chin 11

4) Ask an assistant to set up the video camera ready to record the projectile‟s flight

when launched.

5) Aim the launcher at the target and adjust the angle of launch according to the

range of the target.

6) Release the end of the rod and watch the ball bearing go!

7) Repeat steps 1-6 as many times as possible until accuracy is improved. Once the

desired amount of stretch and angle of launch is determined, keep these variables

controlled (constant) in each trial to provide reliable results.

8) Measure the displacement of the projectile in the launcher (this will depend on

how far back the plunger rod is pulled) and the angle of launch for each trial.

9) Use gathered data to calculate/determine values for:

-horizontal and vertical components of the initial velocity

-the resultant initial velocity

-average acceleration & net force of projectile while in launcher.

-range if launched from 50m cliff

-velocity one quarter through its flight


Summary

Robert Lee Chin 12

Results:

Gathered Data

Displacement of projectile in launcher, s = 0.284 m

Angle of launch, = 31˚

Mass of projectile, m = 0.015 kg

Discussion:

Initial horizontal velocity, ux = 5.6 ms-1

Initial vertical velocity, uy = 3.4 ms-1

Initial resultant velocity, v = 6.6 ms-1

, 031˚ above horizontal

Average acceleration of projectile while in launcher, aav = 76.7 ms-2

Net Force on projectile while in launcher, Fnet = 1.15 N

Range if launched from 50 m cliff = 20.1 m

Velocity one quarter through time of flight = 5.8 ms-1

, 016˚ above horizontal

Trial No. Time of

flight t, (s)

Range x ,

(m)

Hit on target/distance

off from target (m)

1 0.70

4.0 Off target- 0.09

2 0.67

4.0 Off target - 0.11

3 0.78

4.0 Off target - 0.15

Average 0.72

4.0 n/a


Summary

Robert Lee Chin 13

Quantity Data Calculations (all answers to 2 s.f.)

Initial

horizontal

velocity (ux) st

msx

av 720

04

.

.

165720

04ms

t

xux .

.

.

Initial

Vertical

Velocity (uy) 31

65 1msux .

1433165 msuu xy .tan.tan

Resultant

initial

velocity (v)

31

65

43

1

1

msu

msu

x

y

.

.

horizontal above 031,6.6

6.64.36.5

1

12222

msv

msuuv yx

Average

acceleration,

(aav)

ms

msv

2840

66 1

.

. 2

22

7728402

66

2ms

s

vaav

.

.

Net Force,

(Fnet)

ms

msv

kgm

2840

66

0150

1

.

.

.

Ns

mvFnet 21

28402

660150

2

22

..

..

Range if

launched

from 50m

cliff

1

2

1

6.5

8.9

4.3

50

msu

msa

msu

my

x

y

y

0504.39.4 8.92

14.350

2

1

22

2

tttt

tatuy yy

Equation is in the form ax2+bx+c=0. Using the quad. formula, t is given

by +ve value of x:

mtux

st

a

acbbx

x 1.206.36.5

6.3

9.26.39.42

509.444.34.3

2

422

Velocity one

quarter of

time

through

flight

1

1

4.3

18.04

72.0

4

6.5

msu

s

tt

msv

y

av

x

horizontal above 016,8.5 0166.5

6.1tantan

82.56.16.5

6.118.08.94.3

111

12222

1

msvv

v

msvvv

mstauv

x

y

yx

yyy


Summary

Robert Lee Chin 14

Explain the concept of escape velocity in terms of the:

-gravitational constant

-mass and radius of the planet

Escape velocity is defined as: the launch velocity required for a projectile to escape from the

Earth‟s gravitational field.

Mathematically, Escape

velocity, where,

The escape velocity is therefore proportional to the gravitational constant, G and mass of the

planet i.e. the greater the mass of the planet, the higher the escape velocity .It is inversely

proportional to the radius of the planet i.e. the larger the planet, the lower the escape velocity.

Note: The escape velocity of a planet is the same for all objects, independent of their mass.

The escape velocity on earth is approximately 11.2km/s.

Outline Newton‟s concept of escape velocity

Sir Isaac Newton, the scientist who developed the mathematics behind projectile motion,

created the following thought experiment: a person climbed a very high mountain and

launched a cannon ball from the peak. The cannon ball follows a parabolic trajectory before

hitting Earth. If the cannon ball were to be launched with increasing velocities, it would

travel around the Earth because, as it falls, the curvature of the Earth curves away from it.

Thus, the projectile would move in a circular orbit at a fixed height above the Earth‟s surface.

If the cannon ball were to be launched even faster than this, its orbit would changes from

being a circle to an ellipse. Faster still and it would follow a parabolic or hyperbolic path

away from the Earth, escaping it entirely.

21311-106.67 constant grav. universalG

2

skgm

r

GmV

E

Ee

Earth


Summary

Robert Lee Chin 15

Identify why the term „g forces‟ is used to explain the forces acting on an astronaut

during launch

The term “g-force” refers to a mass‟s apparent rate as a ratio of its true weight.

True weight refers to the weight as mass experiences when stationary (or in relative motion)

on the ground e.g. sitting in a vehicle moving horizontally across a flat surface at constant

velocity, standing still at ground level. True weight cannot be felt because it is the force due

to gravity; gravity being a force-at-a-distance.

Apparent weight is the weight a mass experiences due to contact forces or an acceleration

(other than gravity). These forces are also known as inertial forces because they arise from

the body‟s inertia or resistance to having its momentum changed.

Humans and g-force:

-the maximum g-force a properly supported human can withstand is 20g (using fiberglass

seating in spacecraft)

-most people experience tunnel vision and lose colour perception at 4g

G-force experienced by astronaut during a launch into low orbit:

weighttrue

eightapparent wforce-g

2nd

stage jettisoned

(decreasing mass)

Ignition of

2nd

stage

1st stage

(decreasing mass)

time

g-force

3

1

2

4 1st

stage

jettisoned


Summary

Robert Lee Chin 16

Comparison of forces of a launch to a roller coaster ride:

g-force Period during rocket launch Period during roller coaster ride

>1g Rockets burning Accelerating upwards

1g At point of lift off on horizontal surfaces and when it reaches peak

<1g Just after fuel is spent Accelerating downhill

Identify data sources, gather, analyse and present information on the contribution of

one of the following to the development of space exploration: Tsiolkovsky, Oberth,

Goddard, Esnault-Pelterie, O„Neill or von Braun

Robert Goddard

Goddard was a pioneer of modern, liquid-fuel rocketry. As an undergraduate, he wrote a

paper proposing the use of gyroscopes for rocket stabilization. Later, he began studying the

possible use of liquid fuels in rockets to increase fuel efficiency. He also used calculus to

make theoretical calculations of the velocity and position of rockets. Later, he patented the

idea of a rocket fueled by gasoline and nitrous oxide- a milestone in rocketry. Goddard

designed an experiment that allowed him to confirm that rockets can perform in a complete

vacuum i.e. space. During his lifetime, he bought over 200 patents relating to rocket design.

>1g

>1g

<1g <1g

1g 1g


Summary

Robert Lee Chin 17

Discuss the effect of the Earth„s orbital motion and its rotational motion on the

launch of a rocket

The rotation of the Earth on its axis and the orbital motion of the Earth around the sun can be

used to provide a launched rocket with a “velocity boost”. This saves fuel whilst giving the

rocket the required kinetic energy to achieve the target velocity.

To reach Earth orbit, rockets are aimed towards the EAST to take advantage of the Earth‟s

rotation (about 1700km/hr or 500m/s). They receive this 1700km/h velocity boost which

contributes to their target velocity of about 30 000km/h for a low earth orbit.

Rockets destined to go beyond the Earths orbit are NOT launched until the direction of the

Earth‟s orbit around the sun corresponds with the desired direction of the rocket. The rocket

is first launched into a low-Earth orbit, as shown above. Rockets are then fired to accelerate it

past the Earths‟ orbit. The rocket gains the velocity of the Earth relative to the Sun which is

about 107 000 km/h.

Earth rotation

from

North Pole

Orbital path

Launch trajectory

Earth Sun

Launch trajectory


Summary

Robert Lee Chin 18

Analyse the changing acceleration of a rocket during launch in terms of the:

-Law of Conservation of Momentum

-forces experienced by astronauts

Rocket propulsion is a force pair derived from Newton‟s 3rd

law i.e.,

conserved is Momentum

mv)(mv)(

pp

FF-

Rocketgases

Rocketgases

rocketon gasesgaseson Rocket

tt

This means the rockets momentum remains constant at all times. During liftoff, the rockets

mass decreases as fuel is burnt. Therefore, to maintain Conservation of momentum, the

rocket‟s velocity must continue to increase i.e. the rocket‟s acceleration will increase.

Acceleration of rocket vs. Time:

Reaction force

pushes rocket

forwards

Action force pushes on

gases, accelerates them

backwards

Acceleration

Time


Summary

Robert Lee Chin 19

Two forces act on an astronaut during launch: The downward weight force and the upwards

thrust force. Newton‟s 2nd

law states that maF . Therefore, m

mgT

m

Fa

)(. If the

thrust force, „T’, remains constant, but „m’ keeps decreasing as fuel is burnt, then „a’ must

continue to increase.

What does this mean for astronauts? It means they will experience increasing g-forces. Also

note that as the rocket moves higher, the atmosphere thins, meaning the rocket will accelerate

even faster! However, modern rockets have the ability to throttle back the thrust, ensuring a

safer flight for the astronauts.

Analyse the forces involved in uniform circular motion for a range of objects,

including satellites orbiting the Earth

Objects in uniform circular motion are always subject to a centripetal (centre-seeking) force.

The force is tangent to the circle at a particular instant and is always directed perpendicular to

the velocity at that instant.

What causes centripetal force?

Example Fc caused by…

Swinging an object on a string Tension force in string

Vehicle turning a corner Frictional force between tyres and ground

Satellite orbiting Earth Gravitational force between satellite and

Earth

A gear turning inside a motor Torque of gears turning on eachother

Electron moving perpendicular to a uniform

magnetic field

Magnetic force between magnetic field and

electron

V

FC


Summary

Robert Lee Chin 20

Solve problems and analyse information to calculate centripetal force acting on a

satellite undergoing uniform circular motion about the Earth using: r

mvF

2

For a satellite orbiting the Earth, the centripetal force is supplied by the gravitational force

between the satellite and Earth. Hence, use:

metresin surface, aboveheight orbit/ of radius

kgin velocity,orbitalv

kgin object, of mass

(N) Newtonsin force, lcentripeta

where,2

r

m

F

r

mvF

C

C

If the orbital velocity is not known but the masses of the two bodies (e.g. satellite and Earth)

are known, the centripetal force can be found because it is supplied by the gravitational force:

constant grav. universal

body centralor mass

body orbiting of mass

surface aboveht orbit/heig of radius

where,

2

1

2

21

G

m

m

r

r

mmGFG

Orbital velocity can be calculated using the same values as above:



planetother or Earth of mass

where,

G

r

m

r

Gmv

The velocity can also be calculated if the radius and period are known:

secondsin orbit, onefor taken timeT

where,2

T

rv

Example Problems:


Summary

Robert Lee Chin 21

1) The Hubble telescope has a mass of 1 200 kg and orbits at an altitude of 600 km.

Calculate the centripetal force acting on this satellite as it performs circular motion

around the Earth.

The Centripetal force is the gravitational force between the satellite and Earth i.e. FC = FG.

s.f.) (3 108499396983510976

109751200 )10676

10976103761006 10676 10975 1200

3

26

2411

2

21

665213112421

Nr

mmGF

mrskgmGkgmkgm

G

--

.....).(

).(.(

.....

Hence, the centripetal force acting on the satellite is 9.84 x 103

N.

2) A car with a mass of 800 kg travels at a constant speed of 7.5m/s on a roundabout so

that it follows a circular path with a radius of 16m.

A person observing makes the statement: „there is not net force acting on the car

because the speed is constant and the friction between the tyres and the road

balances the centripetal force acting on the car.‟

Assess this statement. Support your answer with an analysis of the horizontal force

acting on the car, using the numerical data provided.

The car is being acted upon by a net force which is the centripetal (centre-seeking) force. The

centripetal force is directed to the centre of the circle, perpendicular to the motion of the car

at any instant. The friction is not balancing the centripetal force; the friction IS the centripetal

force that is keeping the car in uniform circular motion. If there were no centripetal force, that

is, no friction (e.g. on a very smooth surface such as ice), the car would go off at a tangent to

the circle.

N.

r

mvF

ms.vmrkgm

c

-

5.281216

)57(800

57 16 800

22

1

Hence, the car is being acted on by a centripetal net force of 2812.5N


Summary

Robert Lee Chin 22

Compare qualitatively low Earth and geo-stationary orbits

Low-Earth Geo-stationary

Altitude (km) Above the Earth‟s atmosphere but

below the van Allen radiation

belts

Outer van Allen belts

(35 800 km)

250 1000

Orbital period (hours) 1.49 1.75 23.9 (aka “sidereal day”)

Orbital velocity (m/s) 7750 7350 3070

Uses Spy, surveying and weather

satellites, remote sensing, military

purposes and for human

spaceflight

Communications, weather

satellites, scientific studies

Additional information Other sub-types include polar

orbits(spying) and equatorial low-

earth orbits

From Earth, appears to be

fixed in the sky.

All occupy a single ring

positioned above the equator

35 800km Geostationary orbit is much

higher, with a period ≈24 hrs.

A communications receiving

dish needs only to be aimed

directly at it to receive optimal

signal strength- no further

adjustments are needed.

Low-Earth orbit has shorter

period ≈1.5hours, but faster

velocity. Used by the Hubble

space telescope.

Earth, viewed

from North

pole

250-1000 km


Summary

Robert Lee Chin 23

Define the term orbital velocity and the quantitative and qualitative relationship

between orbital velocity, the gravitational constant, mass of the central body, mass

of the satellite and the radius of the orbit using Kepler's Law of Periods

Orbital velocity is defined as: the instantaneous speed (magnitude) in the direction indicated

by an arrow (directional) drawn as a tangent to a particular point on the orbital path.

Johannes Kepler discovered the mathematical relationship between the period of an orbit and

its radius:

constant

i.e.

2

3

23

T

R

TR

We know that for an orbiting satellite, FC = FG, so:

body central of mass


orbit of period


where,4

4

2 velocity,orbitalbut

22

3

2

22

2

2

21

M

G

T

R

GM

T

RHence

r

GM

T

rSo

T

rv

r

GMv

r

GmM

r

vm

,

,

Kepler‟s law of periods also applies to planets. When comparing any number of planets in the

solar system, 2

3

T

R will always give a constant.


Summary

Robert Lee Chin 24

Solve problems and analyse information using: 22

3

4

GM

T

R

Example Problems:

1) Two planets, X and Y, travel around a star in the same direction, in circular orbits.

Planet X completes one revolution around the star in time T. The radii of the orbits are

in the ratio 1:4.

How many revolutions does planet Y make about the star in the same time T?

srevolution 8

1 makemust Y Hence,

s.revolution 8 makes X ,revolution one make toY it takes timesame In the i.e. 8

64

41

4

4 and

constant, Using

22

2

3

2

2

3

2

3

2

3

2

3

2

3

XY

XY

YX

Y

X

X

X

XY

Y

Y

X

X

TT

TT

TT

T

R

T

R

RRT

R

T

R

T

R

)(

)(

)(

)(

)(

)(

)(

)(

2) In June 1969 the Apollo 11 Command Module with Michael Collins on board orbited

the moon waiting for the ascent Module to return from the Moon‟s surface. The mass of

the Command module was 9.98 x 103 kg, its period was 119 minutes and the radius of its

orbit from the moon was 1.85 x 106 metres

a) Assuming the command module was in circular orbit, calculate

i) The mass of the moon

s.f.) (3 1035710351107714010676

1085144

4

10851 714060119

2222

211

362

2

32

22

3

6

kg.

GT

RM

GM

T

R

m.RsT

.....))(.(

)(


Summary

Robert Lee Chin 25

ii) the magnitude of the orbital velocity of the command module

needed.not is module command theof mass The

planet! theof mass the torefers here mass the:Remember

s.f.) (3 1630...87373.162710851

)1035.7)(1067.6(

1035.7

1

6

2211

22

ms.R

GmV

M

3) From nearest to furthest, the four satellite moons of Jupiter first observed by Galileo

in the year 1610 are called Io, Europa, Ganymede and Callisto. For the first three

moons, the orbital period T of each is exactly twice that of the one orbiting immediately

inside it. That is, TEuropa=2xTIo

TGanymede=2xTEuropa

The mass of Jupiter is 1.90x1027

kg, and the orbital radius of Io is 421 600 km.

a) Use Kepler‟s law of periods to calculate Ganymede‟s orbital radius.

s.f.) (3 m1076.1...3.175587433

4

)1090.1)(1067.6)(10216.44(

4

)Gm)(T4(R

4

)Gm)(T4()R(

4

Gm

)T4(

)R(

)T4(

)R(

)T(

)R(

T4T and )T(

)R(

)T(

)R(

m10216.4km421600R

8

32

27118

32

IoGany mede

2

Io3

Gany mede

22

Io

3

Gany mede

2

Io

3

Gany mede

2

Io

3

Io

IoGany mede2

Gany mede

3

Gany mede

2

Io

3

Io

8

Io

b) Calculate Ganymede‟s orbital speed

s.f.) (3 ms1068.2...87445.268331076.1

)1090.1)(1067.6(

R

GmV

kg1090.1m m1076.1R

14

8

2711

278


Summary

Robert Lee Chin 26

Account for the orbital decay of satellites in low Earth orbit

A satellite in a stable orbit around the Earth possesses a certain amount of mechanical energy,

which is the sum of its kinetic energy (due to its high speed) and its gravitational energy (due

to its altitude). The lower the altitude of the orbit, the lower the total mechanical energy is.

Orbital decay affects satellites within the Earth‟s atmosphere (up to 1000 km above Earth‟s

surface). The force due to friction between the satellite and the air particles of the atmosphere

creates heat, resulting in the satellite losing kinetic energy. Gravity causes the satellite to lose

altitude and as the satellite falls to lower altitudes, the atmosphere becomes denser and

frictional forces increase.

This decay is cyclic and speeds up as time passes, causes it to plunge closer to Earth before

vaporizing due to enormous heat build up. Note that few satellites actually make it back to

Earth intact.

Discuss issues associated with safe re-entry for a manned spacecraft into the Earth‟s

atmosphere and landing on the Earth‟s surface

Issues with re-entry include:

-heat build-up

-large g-forces produced

-ionisation blackout

-landing

In orbit, a spacecraft has a high kinetic energy (due to high velocity) and GPE due to its

height above the Earth‟s surface. During re-entry, the craft must decelerate to lose all this

energy.

As the atmosphere decelerates the spacecraft, much of this energy is converted to heat. This

extreme heat must be tolerated or minimised otherwise the craft risks vaporising. This can be

minimised by using a blunt shape for the nose cone. A blunt shape creates a shockwave ahead

of itself that absorbs heat generated by friction. Another method is to use protective layers

e.g. sacrificial layers, insulating panels or heat-resistant alloys. Finally, the craft can reduce it

speed using a series of braking ellipses, in which the craft uses the atmosphere to slow it

down before passing into an elliptical orbit to cool down.

The deceleration of the craft also results in g-forces, much greater than those experienced

during launching due to. Humans are better able to tolerate g-forces lying down, directed

“eyes in” rather than “eyes out”. Space shuttles are designed so that astronauts are reclined

back and experience g-forces “eyes in”. Use of breaking ellipses also reduces the g-forces

Ionisation blackout is a radio blackout that occurs as ions accumulate on the craft as heat

builds-up. The Tracking and Data relay Satellite and new space shuttle designs have

eliminated this problem. The shuttle design creates a hole in the layer of ions at the tail end of

the craft. Radio signals are sent from this opening to the Tracking and Data relay Satellite,

then back down to ground station.

The spacecraft still have considerable velocity when landing- enough to kill the occupants.

Safe landing can be achieved using a separate re-entry capsule with a parachute and landing


Summary

Robert Lee Chin 27

in the ocean. Another method is for the craft to reach a specific altitude before the occupants

complete the descent via parachute. Lastly, the craft can be fitted with landing gear and a

drag chute, allowing it to land on the airfield and be reused.

Identify that there is an optimum angle for re-entry into the Earth‟s atmosphere and

the consequences of failing to achieve this angle

The optimum angle of re-entry depends on the craft, but is usually between 5˚ to 7˚- a 2˚ entry

“window”. A spacecraft following the correct angle will decelerate safely along a descent

path of about 1000 km. An angle of 7˚ to 9˚ will subject the astronauts to dangerously high g-

forces. Any steeper than 9˚ will cause the craft to burn up. Re-entering at angles shallower

than 5˚ will cause the craft to rebound off the atmosphere back into space.

˚

˚

Space: 3. The Solar System is held together by gravity

Present information and use available evidence to discuss the factors affecting the

strength of the gravitational force

The strength of the gravitational force is proportional to the product of the masses of the two

objects i.e. mMFG and inversely proportional to their distances apart i.e. 2

1

rFG . So the

greater the masses of the two objects, the greater the gravitational force and the greater their

distance apart, the weaker the gravitational force.

Angle < 5˚

– too shallow; craft will

“bounce off”

Angle between 5˚

an 7˚

– optimal range

Angle > 7˚

– too sharp; craft

will “burn up”


Summary

Robert Lee Chin 28

Given that M and m are constant, g decreases by a factor of the square of r:

The distribution of mass in the two objects also affects the uniformity of the gravitational

field. For example, areas of higher or lower density such as mineral, oil and gas deposits will

affect g and variations in the thickness of the Earth‟s crust due to tectonic plate boundaries.

The shape of the two objects will also affect g. For example, the Earth is not a perfect sphere,

but flattened at the poles- g will have greater value here.

Variations in altitude affect g i.e. the higher you are, the lower g is. Finally the rotation of the

Earth creates a centrifuge effect that effectively lowers the measured value of g.

Describe a gravitational field in the region surrounding a massive object in terms of

its effect on other masses in it

From the mathematical formula:

2)( Planet

Planet

r

mGg

, the gravitational field surrounding a

massive object is proportional to its mass but inversely proportional to its radius squared.

Every mass acts as if surrounded by a “force field” which attracts any other masses within

this field. This field extends on to infinity, meaning that every mass in the universe is

exerting a force an every other mass, hence the term Universal gravitation.

Define Newton's Law of Universal Gravitation

Sir Isaac Newtons law of Universal gravitational showed that the gravitational force between

two masses:

- is proportional to the product of their masses

- is inversely proportional to the square of the distance from their centres apart

Mathematically, this is given by:

metresin apart, centres their from distance

kgin inviolved, objects of masses and

106.67constant grav. universal

where,

21311-

2

r

Mm

skgmG

r

mMGFG

gFG

M

3r

m

m 2r

m r

m 4r

9

gFG

4

gFG

16

gFG


Summary

Robert Lee Chin 29

Solve problems and analyse information using 2

21

r

mmGF

Example problems:

1) Describe the way in which gravitational force varies with distance for any two

objects.

For any two objects, (assuming their masses remain constant), the gravitational force between

them is inversely proportional to the square of their distance apart i.e. 2r

kF , where k is a

constant. Say the force is equal to g at a distance of r. Then the force diminishes by half at a

distance of 2r. At a distance of 3r, it is one-ninth and at a distance of 4r, it is only one-

sixteenth. The force diminishes rapidly at first, then more slowly as the distance increases.

However, the force never reaches zero unless the distance is infinity.

3) A geostationary satellite of mass 860kg orbits the Earth with a orbital radius of 35

800km. Calculate the force of gravity keeping it in orbit, assuming mass of Earth is

5.97x1024

kg.

s.f.) (3 100141000596410583

1097586010676

10975 10583 860

1010

28

2411

2

21

242

81

Nr

mmGF

kgmmrkgm

.....).(

).().(

..

Discuss the importance of Newton's Law of Universal Gravitation in understanding

and calculating the motion of satellites

Once a launched rocket has gained sufficient altitude, it can be accelerated into orbit. It must

attain a specific orbital speed to achieve orbit. If the speed is too low, it will spiral back into

the atmosphere, eventually burning up or crashing back down to Earth. If the speed is too

high, it will leave the Earth‟s orbit altogether. For satellites orbiting the Earth or other

planets, a centripetal force maintains uniform circular motion. Without this, it would fly off at

a tangent to the circle. The centripetal force keeping satellites in orbit is gravity.

By equating the expression for gravitational force with the expression for uniform circular

motion, we can find the velocity required to maintain uniform circular motion around the

Force

Distance


Summary

Robert Lee Chin 30

Earth (orbital velocity), r

Gmv E . Given that mass and radius of the Earth are constants,

and noting that r = rEarth + altitude, the only variable is the altitude of the satellite above the

Earth. The greater the radius of orbit, the lower the velocity required to maintain orbit. At a

given orbital radius, a satellite orbiting a smaller planet would need to travel at a lower

velocity. The bigger the planet, the greater the velocity would need to be.

Furthermore, the above expression for orbital velocity can be used to derive Kepler‟s law of

periods i.e. constant4 22

3EGm

T

R

Identify that a slingshot effect can be provided by planets for space probes

When scientists wish to send out probes into deep space for exploration, there are

considerable implications:

-it costs billions of $ to send a space probe to a single planet…

-so it makes more sense to send one probe to several planets…

-but the distances are enormous and even at high speeds, it takes years to reach planets…

-probes may also need to change direction when travelling between planets…

The solution is to fly the space probe close to a planet. This causes the planet to give some of

its kinetic energy to the probe which then swings around in a new direction with increased

velocity. The planet will actually slow down as it loses a small amount of energy, but because

of the huge mass of the planet compared to the probe, the energy loss is insignificant.

“slingshot” trajectory

Planet orbit

Planet orbit

Probe


Summary

Robert Lee Chin 31

Space: 4. Current and emerging understanding about time and

space

Outline the features of the aether model for the transmission of light

During the 19th

century, physicists knew that all waves required a medium to travel through.

Since light was a waveform, it was hypothesised that there was a universal “aether” aka

luminiferous aether) which transmitted light through the vacuum of space.

Features of the aether:

-completely transparent

-fill all of space

-be stationery in space i.e. act as an absolute frame of reference for all other bodies

-possess extremely low density

-have no mass

-have great elasticity to propagate light waves

Describe and evaluate the Michelson-Morley attempt to measure the relative velocity of

the Earth through the aether

It was reasoned that if the aether existed, the Earth should move through the aether as it

orbited the Sun. This is analogous to putting your hand out of the window of a moving car on

a windless day- you will feel an apparent wind.

The Michelson-Morley experiment in 1887 tried to detect the aether using the principle that a

beam of light travelling into the aether would be slowed down compared to a beam of light

travelling across the aether and an effect called interference.

The apparatus was set up on a large stone block floating on mercury. A single light source

sends a beam of light to a half-silvered mirror and is split in two. One beam (red) heads into

the aether while the other (blue) heads across it. Both beams travel the same distance, reflect

against a mirror and finish their journey at a telescope. The interference pattern of the two

light waves is compared using an interferometer. The entire apparatus is then rotated by 90˚

so that the rays reverse positions. If the interference pattern changed, it would provide

evidence for the aether‟s existence.


Summary

Robert Lee Chin 32

The experiment concluded with a null result i.e. there was never any change in interference

pattern detected. Even though it was repeated extensively with various adjustments and

refinements e.g. different times of year, different altitudes, it was always a null result.

Gather and process information to interpret the results of the Michelson-Morley

experiment

Michelson and Morley attempted to prove the existence of the aether using the assumption

that the speed light waves would be slowed down when travelling against the “aether wind”

compared to light waves travelling perpendicular to it. The shone monochromatic light as a

half-silvered mirror inclined at 45 o to the light beam, so that half the light passed through the

mirror while the other travelled perpendicular to it. Both light beams travelled the same

distance and were reflected back by mirrors placed perpendicular to the beams. The two

beams formed an interference pattern of alternating light and dark patterns. When the

apparatus was rotated, they expected to see a change in interference pattern as the two beams

had switched positions. No such change was ever found even though the equipment was

sufficiently sensitive enough. Hence, they concluded that the aether probably did not exist

Light

source

Mirror

Aether wind

Half-

silvered

Mirror

Telescope

Mirror

Apparatus

is rotated

by 90˚


Summary

Robert Lee Chin 33

Discuss the role of the Michelson-Morley experiments in making determinations

about competing theories

A hypothesis gives rise to predictions that can be tested by experiment. From the results of

the experiment, judgements can be made as to the validity of the hypothesis. If the results of

the experiment do agree with the predictions, then the hypothesis can be accepted as true. If

the results do NOT agree with the predictions, they can be rejected as wrong.

The Michelson-Morley experiments were performed to test the prediction that an aether wind

should exist. The experiment had null results in all instances, despite the equipment being

adequately sensitive. However, this did NOT disprove the theory- it merely failed to find any

evidence for it.

Various modifications were made to the aether model. Each modified theory resulted in new

predictions, yet every test failed.

20 years after the Michelson-Morley experiment, Einstein proposed his theory of relativity

which did not require an aether model. This theory came with its own predictions which

could not be tested at that time. As technology improved, the predictions were tested all

found to be true.

However, the Michelson-Morley experiment was NOT a failure. It provided supporting

evidence for the theory of relativity by allowing a choice to be made between the two

conflicting theories.

Perform an investigation to distinguish between inertial and non-inertial frames of

reference.

Investigation: inertial and non-inertial frames of reference

Aim: to distinguish between an inertial and a non-inertial frame of reference

You will need:

-the help of another person who can drive

-a plumb bob, or any mass e.g. a nut hanging on a short length of string.

Method:

1/ Once in car, ask driver to take short trips across some smooth straight sections of road

travelling at constant speed and accelerating. Do the same when travelling along

bends. Do NOT look out the window; look only at the plumb bob.

2/ Note when the bob hangs directly down and when it does not. The plumb bob

effectively acts as an accelerometer.


Summary

Robert Lee Chin 34

Results:

Events when bob hangs straight down Events when bob does not hang straight down

Car is at rest

Car is turning a corner

Travelling at constant velocity on smooth

straight

Car is accelerating/decelerating

The plumb bob is on an angle when in a non-inertial frame of reference i.e. accelerating,

decelerating or changing direction. The bob hangs straight when in an inertial frame of

reference i.e. at rest or when moving along a flat straight at constant speed.

Conclusion: In an inertial frame of reference, there is no way of determining velocity without reference to

an outside point. In anon-inertial frame of reference, objects will experience a net force,

indicated in this case when the bob hangs on an angle.

Constant velocity/at

rest: inertial frame of

reference

Turning a corner, accelerating or

decelerating: non-inertial frame

of reference


Summary

Robert Lee Chin 35

Outline the nature of an inertial frame of reference

An inertial frame of reference is a rigid framework relative to which measurement such as

position, velocity and displacement can be measured. It is defined as one that is moving with

constant velocity (including direction), or is stationery. One cannot distinguish between two

inertial frames of reference.

A non-inertial frame of reference involves a change in velocity i.e. an acceleration or

deceleration.

Discuss the principle of relativity

The principle of relativity states that all uniform, steady motion is relative to an observer i.e.

undetectable without reference to another frame of reference. For example, if travelling on a

plane that is flying horizontally at constant velocity at night, you cannot tell the plane is

moving at all!

In an inertial frame of reference, all measurements and experiments give the same results.

The principle of relativity:

-only applies for inertial frames of reference

-does not apply for non-inertial frames of reference

Analyse and interpret some of Einstein‟s thought experiments involving mirrors and

trains and discuss the relationship between thought and reality

Einstein’s “Thought experiment”:

Imagine you are in a train that is travelling at the speed of light. You are holding a mirror in

front of you. Will you be able to see your reflection?

There are two possible outcomes:

No, you will not see your reflection. This is because you are already travelling at the speed of

light. This would support the aether model by implying that light has a fixed velocity relative

to it. But… it would violate the principle of relativity because you would be able to tell how

fast you were going.

Yes, you would see your reflection. This would support the principle of relativity because

you would not be able to distinguish your speed. BUT… it would mean that a person

observing your reflection from outside the train would see it moving at twice the speed of

light!

Einstein concluded that…

Yes, he would be able to see his reflection because the principle of relativity is NEVER

violated. But… the person outside the train would observe your reflection to be travelling

normally. This meant that you would both perceive time to be passing differently.

The aether model could be regarded as superfluous.


Summary

Robert Lee Chin 36

The main limitation of these experiments is that they are based on your common sense

because most such situations cannot be test in reality. The common goal of a thought

experiment is to explore the potential consequences of the principle in question

Describe the significance of Einstein‟s assumption of the constancy of light

What Einstein assumed from his theory of relativity was all observers see light travelling at

the same velocity, (300 000 000 m/s), regardless of their motion. His assumption explained

the null result of the Michelson-Morley experiment and proved there was no need for an

aether model.

The significance of the constancy of light is that there is no such thing as an absolute frame of

reference i.e. the principle of relativity is NEVER violated and that space and time are

relative values.

Identity that if c is constant, then space and time are relative

In classical physics, space (distance, position, velocity etc.) were relative values but time was

an absolute, passing identically for everyone.

The theory of relativity challenged classical physics by assuming that c is a constant. Since

time

distancec , then space and time pass differently for different observers, depending on how

fast they are moving.

Discuss the concept that length standards are defined in terms of time in contrast to

the original metre standard

The metre was first defined in 1793 by the French government as “one-millionth of the length

of the Earth‟s quadrant passing through Paris”. After this arc was (incorrectly) measured,

platinum and iron standards were made.

When the Systeme Internationale (SI) was established in 1875, it was defined as “the distance

scribed between two lines on a single bar of platinum-iridium alloy”.

The current definition takes advantage of better technologies and the constancy of light to

give a more precise definition. One metre is defined as “the length of the path travelled by

light in a vacuum during the time interval of th458 792 299

1of a second”.

Other measurements defined in terms of the speed of light include the light-year

(approximately 9.47 x 1012

km)


Summary

Robert Lee Chin 37

Analyse information to discuss the relationship between theory and the evidence

supporting it, using Einstein‟s predictions based on relativity that were made many

years before evidence was available to support it

Einstein‟s theory of relativity predicted alterations to time and space including time dilation,

mass dilation and length contraction. When Einstein proposed his theory in 1905, these

predictions could not be tested because technology was not yet advanced enough. These

predictions can now be tested, because of technological developments.

Atomic clocks were developed in the latter half of the 21st century. Extremely accurate

atomic clocks have been synchronised, then flown around the world in high-speed aircraft.

When brought back down, the clock that was in the aircraft had slowed down slightly.

The strongest evidence comes from particle accelerator experiments. Subatomic particles

such as electrons and protons are accelerated to speeds very close to the speed of light. Their

masses are observed to increase exponentially and infact, newer, heavier particles can be

created by colliding subatomic particles together. The half-life decay of radioactive particles

travelling at relativistic speeds has been shown to take much longer from the perspective of

the scientists.

Special relativity has played a key role in the development and design of particle accelerators.

One of the earliest particle accelerators, the Cyclotron, reached an energy limit and hade to be

modified due to relativistic effects.

Explain qualitatively and quantitatively the consequence of special relativity in

relation to:

-the relativity of simultaneity

-the equivalence between mass and energy

-length contraction

-time dilation

-mass dilation

The relativity of simultaneity refers to the idea that if an observer sees any two events to be

simultaneous, then another observer in a different frame of reference will not judge them to

be simultaneous i.e. whether you judge two events to be simultaneous depends on the framer

of reference.

As an example, say there are two people, A and B. Both are equidistant from two identical

light sources and A remains stationary. At the moment when B begins to move right with

velocity, v, both lights flash. A, who remains stationery will judge both flashes of light to

reach him/her at the same time. B, however, has moved towards the right slightly so that the

light from flash 2 reaches them slight before flash 1 (remembering that the speed of light is

not infinite). Hence, B will judge the two flashes of light as not simultaneous.


Summary

Robert Lee Chin 38

The rest mass of an object is equivalent to a certain amount of energy. Mass can be converted

into energy and vice versa. For example, nuclear fission reactions that occur in the Sun

involve mass being converted into energy and when water is heated, its mass will increase

slightly.

Einstein‟s famous equation expresses this equivalence of mass and energy: 2mcE , where

E is the amount of energy, m is the rest mass and c is the speed of light. Since c is squared,

the amount of energy released from even a small mass is enormous.

For length contraction, time dilation and mass dilation relativistic effects, we use the factor:

2

2

1c

v. Remember that this factor will always be less than one.

The length of an object measured within its frame of reference is called its proper length, Lo.

Observers in different frames of reference will always judge the observed length, Lv, to be

contracted i.e. ov LL . For example, a car 3 metres long travelling at 0.5c will appear shorter

(2.6m) to a standing observer.

Length contraction is given by:

18

0

2

2

1003light of speed

R. F.of within speed relative

R. of F. within length"rest "

observer outsidean by judged aslength

where,1

msc

v

L

L

c

vLL

v

ov

.

The time taken for an event to occur in its rest frame is called the proper time, to. Observers

in different frames of reference will always judge the observed time, tv, to be longer i.e. time

passes slower for the person within the travelling frame of reference, ov tt . Using the above

Flash 2 Flash 1

A B

v

d d


Summary

Robert Lee Chin 39

example, say the person in the car speaks on their mobile phone for 1 minute. The standing

observer will judge the time taken to be 1min 10 seconds.

Time dilation is given by:

18

0

2

2

1003light of speed

R. of F. within speed relative

R. of F. within e"proper tim"

observer outsidean by judged as time

where,

1

msc

v

t

t

c

v

tt

v

ov

.

Another consequence of special relativity is that the mass of a moving object increases as its

velocity increases- this is called “mass dilation”.

Mass dilation is given by:

18

0

2

2

1003light of speed

R. of F. within speed relative

R. of F. within mass"proper "

observer outsidean by judged as mass

where,

1

msc

v

m

m

c

v

mm

v

ov

.

This effect is only noticeable at relativistic speeds. As an object is accelerated closer and

closer to the speed of light, its mass increases and so does the amount of energy required,

making further accelerations more and more difficult. The energy that is used to accelerate

the object is instead converted into mass!

Solve problems and analyse information using:

1

1

1

2

2

2

2

2

2

2

c

v

mm

c

v

tt

c

vLL

mcE

ov

ov

ov


Summary

Robert Lee Chin 40

Example Problems:

1) The nearest star to us is the Large Magellanic Cloud, with its centre located

1.70x105 light years from Earth. Assume you are in a spacecraft travelling at the

speed of 0.99999c towards the Large Magellanic Cloud.

a) In your frame of reference, what is the distance between Earth and the Large

Magellanic cloud?

b) In your frame of reference, how long will it take you to travel from Earth to the

Large Magellanic cloud?

s.f.) (3 1020110.1.198786.. 0044721240

10361125

1

103611251003

106083361

151521

2

2

21

8

21

s

v

c

tt

c

Lt

ov

oo

..

.

..

.

2)

a) A piece of radioactive material of mass 2.5 kg undergoes radioactive decay. How

much energy is released if 10 grams of this mass is converted to energy during the

decay process?

JmcE

kgm

14282 10091003010

010

.).)(.(

.

b) A mass is moving in an inertial frame of reference at a velocity v relative to a

stationery observer. The observer measures an apparent mass increase of 0.37%.

Calculate the value of v in ms-1

.

s.f.) (3 1019.710...19267.7004472124.010608336.11

004472124.099999.0

11

99999.0

10608336.1)100.3()606024365()1070.1(1070.1

181821

2

2

2

2

2

2

21855

mv

cLL

c

c

v

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cc

mLYL

ov

o


Summary

Robert Lee Chin 41

Discuss the implications of time dilation and length contraction for space travel

Current maximum velocities do not allow for viable interstellar travel, as they would take

long. If relativistic speeds were attainable, the nearest stars would take only a few years to

reach. For example, our nearest star, Alpha Centauri would take 8 years to reach travelling at

0.5c. However, due to the effects of time dilation and length contraction, it would take

significantly less time (7 years, infact) for the crew on board such a spacecraft.

However, acceleration is always the most costly part of space travel. The effect of mass

dilation and time dilation means that the amount of energy required to accelerate beyond 0.9c

would be prohibitive. As one gets closer and closer to the speed of light, the energy input

required only marginally increases the velocity.

If the astronauts were to return to Earth, they would have aged significantly less than those

who remained on Earth. While this could make long space journeys possible to achieve

within a lifetime, governments may prohibit it because they may not get any “results” during

their lifetime.

We currently lack the fuel and technology required, hence it is an impractical situation. The

speeds reached by astronauts (the fastest to date is a tiny 0.01c) provide negligible relativistic

effects. Scientists have proposed using matter-antimatter based engines to reach relativistic

speeds but this may takes years to happen.

12802

022

0

2

2

0

2

2

0

2

2

2

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0

348 489 114371

1110031

1

1

1

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1

1

371

msm

mcv

m

mcv

m

m

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m

m

v

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v

c

mm

m

m

v

v

v

v

v

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v

.).(

.

module 1 space

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