HSC Science TeacherProfessional Development Program

Physics• 8:30am Space and Gravity

– Michael Burton

• 9:45am Physics of Climate– Michael Box

• 15 minute tea break• 11:00am The age of silicon: semiconductor

materials and devices– Richard Newbury

• 1:00pm Lunch

Presentations will appear onwww.phys.unsw.edu.au/hsc

Space and GravitySome ideas for HSC Physics

Michael BurtonSchool of Physics

University of New South Wales

Gravity and the Planets

• Escaping from a Planetary Surface– Acceleration due to Gravity– Escape Velocity– Geostationary Orbit and the Space Elevator

• Kepler’s Third Law– The Planets– Jupiter and its Moons

• Travelling the Solar System– Slingshot effect– Mission to Mars and the Hohmann Transfer Orbit

What is an Orbit?

• Falling at just the right speed so that we travel around the planet rather than toward it.

• No energy is required to maintain the orbit once it has been obtained!

“Assumed Knowledge”

€

Circular Motion F =mV 2

R

Force due to Gravity FG =GM1M2

R2

Kinetic Energy Ekin = 12 mV 2

Gravitational PE EG = −GM1M2

REnergy Conservation E tot = EPE + EKE

Kepler' s Third Law T 2

R3 = constant

Weight and Escape Velocity• mg=GMm/R2 and 1/2mvesc

2=GMm/R

• Compile for each planet and compare– e.g. how heavy would a bag of sugar be on Earth,

Mars, Venus and Jupiter?– how fast must you launch it to escape each planet?

Weight Escape Speed (km/s)

Earth 1.0 11

Mars 0.4 5

Venus 0.9 10

Jupiter 2.5 40

(Geo-)Synchronous Orbit and the Space Elevator

• Synchronous Orbit when orbital period = rotational period of the planet

• Space Elevator ascends to the synchronous orbit• Lower escape speed

– 1/2mvesc2=GMm/(rsync+rplanet)

€

tday =2πr

v with v =

GM

r

so that rsync =GMtday

2

4π 2

⎛

⎝ ⎜

⎞

⎠ ⎟

13

− rplanet

rsync Vescape

(surface)

Vescape

(elevator)

Tascent

(@100 km/hr)

1000 km km/s km/s days

Earth 36 11 4 15

Mars 17 5 2 7

Venus 1532 10 0.7 638

Jupiter 89 60 40 37

Question and Exercises

• Calculate (geo-)synchronous orbit• Compare between planets

– Which might be feasible, which impossible?

– How might it be built??? (carbon nanotubes?)• How massive?

– How long would it take to ascend?

• What gain in reduced escape speed?– How much more mass for the same thrust?

• (extra energy available for accelerating the payload)

Kepler’s Laws• Empirical Laws

• [Kepler 1: Elliptical orbits, Sun @ a focus]

• [Kepler 2: Equal areas equal times]

Kepler’s Third Law

Exercise 1: Research r and T for planets and investigate the relation between them[Plot r vs. T then log r vs. log T]€

r3

T 2=

GM

4π 2

€

From v 2 =GM

r=

2πr

T

⎛

⎝ ⎜

⎞

⎠ ⎟2

r T T2/r3

106 km Years yr2/km3

Earth 150 1.0 3x10-25

Mars 228 1.9 3x10-25

Venus 108 0.6 3x10-25

Jupiter 778 11.9 3x10-25

K3L and the Moons of Jupiter

• Use a web application, e.g.– jersey.uoregon.edu/vlab/

tmp/orbits.html

• Record positions of moons every day (use ruler)

• Plot on graph paper• Determine orbital period and

radius for each moon.• Do they fit K3L? (yes!)• What relationship between

their periods? (~1:2:4:8)

Journey to Mars

Gravitational Slingshot

Hohmann Transfer Orbit

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

Gravity Assist to the PlanetsCassini mission to Saturn

• Venus, Venus, Earth, then Jupiter, on way to Saturn!

• Took 6.7 years, with V=2 km/s• Hohmann transfer orbit would have

taken 6 years but required a V=15 km/s – impracticable!

How GravityAssist works

• Relative to Stationary Observer:– Spacecraft enters at -v, Planet moving at +U– Goes into circular orbit

• Moving at U+v relative to surface of planet• Leaves at U+v relative to surface in opposite direction

– Thus leaves at 2U+v relative to observer• e.g. Spacecraft moving at 10 km/s encounters Jupiter moving at 13 km/s. Leaves

at 36 km/s!

• Conservation of energy and momentum applies – planet must slow (very!) slightly

• In practice we would need to fire engines to escape from a circular orbit. However, one could enter on a hyperbolic orbit, with a gain in speed of slightly less than 2U.

MarsMars

The planet Mars, I scarcely need remind the reader, revolves about the Sun at a mean distance of 230 million km, and the light and heat it receives from the Sun is barely half of that received by this world. It must be, if the nebular hypothesis has any truth, older than our world; and long before this Earth ceased to be molten, life upon its surface must have begun its course. The fact that it is scarcely one seventh the volume of the Earth must have accelerated its cooling to the temperature at which life could begin. It has air and water and all that is necessary for the support of animated existence.

H.G. Wells, The War of the Worlds, 1898

Olympus Mons

600 km across x 24 km high!

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

The Gorgonum Chaos

Water on MarsPolar Ice Caps

Sedimentary Rock: layers of time

Recent water flow on Mars

22 December, 2001 24 April, 2005

The Hohmann Transfer Orbit

• Most Fuel Efficient orbit to the planets

• Three Parts:– Circular orbit around

Earth– Elliptical orbit,

perihelion @ Earth, aphelion @ Mars

– Circular orbit around Mars

Wolfgang Hohmann, German Engineer, 1925

Energy in an Orbit

€

PE : EG = −GMm

r

Circular Motion : GMm

r2= m

v 2

r

KE : EK = 0.5mv 2 =GMm

2r

€

Thus : EK =GMm

2r= −

EG

2

E total = EK + EG = −GMm

2r

Applies for elliptical orbit, semi-major axis a:Etotal = –GMm/2a = constant in an orbit

Assumptions MadeHohmann Transfer Orbit

• Only considering gravitational influence of the Sun (OK)

• Apply thrust without changing mass of spacecraft (Wrong!)

• Assume circular orbits for the planets (OK)

• Consider only impulsive thrusts (i.e. no slow burns)

Energy Changes• Step 1: Heliocentric orbit around

Earth to elliptical orbit with Earth at perihelion and Mars at aphelion– E1= –GMm/R– E2= –GMm/[(R+R’)/2]

• Step 2: Elliptical orbit to heliocentric orbit around Mars– E3= –GMm/R’

m=50 tonnes Orbit, a E=–GMm/2a E

x 106 km x 1013 J x 1012 J

Earth Orbit 150 -2.2

Transfer Orbit 189 -1.8 4.6

Mars Orbit 228 -1.5 3.0

Time & Launch• Time taken is half the orbital period for

the elliptical orbit. – Use K3L!– i.e. T/2 where T2=[(R+R’)/2]3 when

measured in Years and Astronomical Units– T=[(1.0+1.5)/2]3/2=1.4 years– Thus it takes 0.7 years

• Launch Window– Mars covers [T/2]/Tmars x 360° = 135.9°– Spacecraft covers 180°– Thus, Launch when Earth 180-135.9=44.1°

behind Mars

Harder Problem: how often do launch windows occur?

€

ωEarth - ωMars( )T = 2π

1TEarth

− 1TMars

= 1T

yields T = 2.1 years

Questions to consider?• How do we know this is the cheapest fuel orbit?

– Can’t be less (wouldn’t arrive), needn’t be more (overshoot)

• How much change in energy is needed?– Relate to amount of fuel?

• Best time to launch a few months before Opposition– Why? (44.1°) Why not at Opposition?

• How long will the journey to Mars take?– Compare to Journey to Moon (3 days), to Jupiter (2.8 yrs).

• How often can we launch (every 2.1 years for Mars)?– Implications for return journey (first window after 1.5yrs)– Implications for human exploration of the Solar System

• What do we need humans for, what can a robot do better?

• What about lift-off from Earth, landing on Mars?

The End