HSC Physics – Core 9.2 Space!Part 2: Launching into orbit!Overview of Part 2:!2.1 Projectile motion (PM)!2.2 Escape velocity – getting into orbit!2.3 Accelerations and g-forces during launch!2.4 Uniform circular motion (UCM)!2.5 Types of orbits!2.6 Kepler’s Law of periods!2.7 Total orbital energy of a satellite!2.8 The sling-shot effect (gravity-assist)!2.9 Orbital decay!2.10 Re-entry!!!!

Go  to  the  video  lesson  for  this  slide  deck:    h2p://edrolo.com.au/subjects/physics/hsc-­‐physics/space-­‐part-­‐2/escape-­‐velocity/lesson/  

HSC Physics – Core 9.2 Space!Part 2: Launching into orbit!

!2.2: Escape velocity!

Overview of Section 2.2:!2.2.1 Escape velocity formula!2.2.2 Velocity assist from Earth’s axial rotation!2.2.3 Velocity assist from Earth’s orbital motion !

!!

Escape velocity! Syllabus Module 9.2 dot point 2B4  Outline Newton’s concept of escape velocity!!Syllabus Module 9.2 dot point 2B6  Discuss the effect of the Earth‘s orbital motion and its rotational motion on the launch of a rocket!

!

Escape velocity Formula!Escape velocity, ues != the initial velocity (u) at the surface of the Earth that any

object must be given for it to just escape the Earth’s gravitational attraction.!

!

!“escape Earth’s gravitational attraction” means !!the object will not fall back to the Earth or orbit the Earth.!

!

!“just escape” means that just as the object reaches an infinite distance from the Earth its final velocity = 0 !!

( ) ( ) ( )By conservation of energy

at surface at surface at infinte distance+ =k p pE E E

12 rocketm 2

escapeG⎛ ⎞

+ −⎜ ⎟⎝ ⎠

rocketum Earth

E

Earthescape

Eeart

m0

r

2GmSo r

⎛ ⎞⎜ ⎟ =⎜ ⎟⎝ ⎠

=u … Exercise 1 !

Escape velocity – Exercise 1 !Calculate the escape velocity from Earth.!!

Data:!Mass of Earth != 5.98 x 1024 kg!Radius of Earth != 6380 km!Universal Gravitational Constant G = 6.67 x 10-11 N kg-2 m!

… Answer!

Escape velocity – Exercise 1: Answer !Calculate the escape velocity from Earth.!

-11 24

6

3 -1

-1

2(6.67 10 )(5.98 10 ) (6.380 10 )

11.182 10 11.2

=

= ×

≈

x xx

m skm s

Data:!Mass of Earth != 5.98 x 1024 kg!Radius of Earth != 6380 km!Universal Gravitational Constant G = 6.67 x 10-11 N kg-2 m2!

es

es

To find from the Earth

2use = Earth

E

u

GmuR

Escape velocity – Exercise 2!

When a spacecraft is given its escape velocity, !(a) Does the direction in which it is launched matter!(b) Does the path it follows matter? !

… Answer!

Escape Velocity – Exercise 2: Answer (a)!

Direction DOES matter: !To save fuel it is best to launch the vehicle towards the east to make use of the Earth’s axial rotation towards the East. !

When a spacecraft is given its escape velocity,!(a) does the direction in which it is launched matter?!

… Answer (b)!

Escape Velocity – Exercise 2: Answer (b)!

Escape velocity is calculated by equating the kinetic energy at launch on the surface (r = rsurface) with the gain in gravitational potential energy the body would have at infinity (r = )!

( )

( ) ( )

0

k psurface

p p surface

planet spacecraft

planet

planet spacecraft

planet

E E

E E

Gm mr

Gm mr

∞

= Δ

= −

⎛ ⎞= − −⎜ ⎟⎜ ⎟

⎝ ⎠

=

When a spacecraft is given its escape velocity, !(b) Does the path it follows matter?!

So the kinetic energy needed at launch to escape the planet does not depend on the path the spacecraft follows. !It depends only on the radius of the planet it is escaping!

Next …!Launch velocity-assist from Earth’s axial-rotation � !

Launch velocity-assist from Earth’s axial-rotation !The  size  of  the  tangen&al  speed  of  the  Earth  from  east  to  west  is  different  at  different  laCtudes:  This  velocity  can  be  used  to  assist  a  rocket  at  launch  but  the  vehicle  must  be  launched  towards  the  East.  

Equator

EquatorE Equator

E

EquatorE

Equator

Pr oof

cos

θ

θ

θ

θθ

θ

ω

ω

ω ω

θ

=

=

=

= =

=

=

v r

but constantso

vvso r rr r

rv vr

v v

θ θθ= equatorv v cos

… Example 1!

Launch velocity-assist from Earth’s axial-rotation – Example 1 !What is the tangential velocity of the Earth’s surface at the Equator.!!

Earth Data:!Period of rotation = 23 hours 56 minutes 4 seconds!Radius of Earth = 6371 km.      

Hint: !If the Earth rotates uniformly on its axis, then:      

2 EarthEquator

Earth

rvTπ

=

… Answer!

Launch velocity-assist from Earth’s axial-rotation – Example 1: Answer !Find tangential velocity of the Earth’s surface at the Equator    

2

2

2

Equator

EarthEquator

Earth3

Earth3

-1Equator

To find v

rUse v

T

6371 10 m T

6371 10 m 86164 s

v 464.6 m s

π

π

π

=

× ×=

×=

=

Rotation

Rotation

Convert units of T to seconds

T 23 h 56 min 23 3600 s + 56 60 s 86164 s

=

= × ×

=

Rotation

E

Earth dataPeriod of rotation of Earth, T 23 h 56 minRadius of Earth, r 6371 km

=

=

… Exercise 3 !

Launch velocity-assist from Earth’s axial-rotation – Exercise 3:!What is the tangential velocity of the Earth’s surface at Sydney (latitude 340 S). !  

o34

Sydney

Sydney

equator

o

-1

To find v

Use v v cos

v 464.6 cos34

v 385.2 m s

θ θ=

= ×

∴ =

-1equator

osydney

Datav 464.6 m s (from Example 1)

34 Southθ

=

=

… Exercise 4 !

Answer!

Launch velocity-assist from Earth’s axial rotation – Exercise 4!

Australia had a rocket launch site at Woomera in South Australia (latitude 30o S) from the 1950s to 1970s. !There have been plans to build a new launch site on Cape York (latitude 120 S). !Suppose you want to launch a rocket so that it will escape the Earth.!!

Which location (Woomera or Cape York) would be the best site to locate the launch site? Explain why.!

… Answer!

Launch velocity-assist from Earth’s axial rotation – Exercise 4: Answer!

The best location will be the one with the greatest assistance from the Earth’s rotational motion at launch.!This occurs for the site with the smallest latitude (i.e. the site nearer the equator) = Cape York. !

,

Woomera -1

0

Earth's tangential velocity at Woomera

Use cosθ

Latitude 30

464.6 m s

464.6 x cos30

θ

θ

=

=

=

=

latitude equator

o

equator

Woomera

v v

v

v-1 402.4 m s≈

,

Cape York

-1

0Cape York

Earth's tangential velocity at Cape York

Use cosθ

Latitude 12

464.6 m s

464.6 x cos12

θ

θ

=

=

=

=

latitude equator

o

equator

v v

v

v-1454.4 m s≈

Which location (Woomera or Cape York) would be the best site to locate the launch site? Why?!

The tangential velocity of the Earth’s surface at these two latitudes is calculated below. !

Next …!Launch velocity-assist from Earth’s orbital motion!

Launch velocity-assist from Earth’s orbital-motion about the Sun!Once a vehicle is in orbit, the Earth’s orbital velocity about the Sun is added to the vehicle’s orbital velocity about the Earth.!            

rocket's Earth

rocket orbital velocity orbital velocity relative to Sun about the about the

RS RE E

Earth Su

S

'sn

v v v

v v v

= +

= +

r r r

r r r

The Earth’s orbital velocity about the sun VES

is about 30 km s-1 in an anti-clockwise direction!

… Exercise 5 !

Launch velocity-assist from Earth’s orbital-motion about the Sun – Exercise 5!Explain how you could take advantage of the Earth’s orbital motion about the Sun to reach: !(a) an outer planet – one further from the Sun than the Earth!(b) an inner planet – one closer to the Sun than the Earth.          

… Answer!

Launch velocity-assist from Answer (a)Earth’s orbital-motion about the Sun – Exercise 5!

Spacecraft heading to outer planets are not launched until the direction of the Earth in its orbit around the Sun corresponds with the desired direction. !The spacecraft can then be launched up to a low Earth orbit, before firing its rockets again to accelerate ahead of the Earth!The velocity boost received from the Earth’s orbital motion in this case is approximately 30 km s-1.          

Explain how you could take advantage of the Earth’s orbital motion about the Sun to reach:!(a) An outer planet – one further from the Sun than the Earth!

… Answer (b)!

Launch velocity-assist from Answer (b)Earth’s orbital-motion about the Sun – Exercise 5!

A spacecraft traveling from Earth to an inner planet will be naturally accelerated towards the planet by the Sun’s gravity. !To assist the spacecraft, it should be:!•  launched in the direction opposite to the orbital motion of the

Earth about the Sun (i.e. in a clockwise direction): this will reduce its total orbital energy causing it to descent to a lower orbital radius about the Sun !

!

•  decelerated (using its rocket engines) until it achieves a sun-orbit with a perihelion equal to the orbit of the inner planet. !

So, the spacecraft will continue to move in the same direction as Earth, only more quickly.    

Explain how you could take advantage of the Earth’s orbital motion about the Sun to reach:!(b) an inner planet – one closer to the Sun than the Earth.  

… next video: Accelerations and g-forces!

( )1 if becomes more negative, 2

= − ↓T T oo

GmME E rr