Transcript Antarctic Astronomy: Progress and Prospects Michael Burton

HSC Science Teacher
Professional 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 on
www.phys.unsw.edu.au/hsc
Space and Gravity
Some ideas for HSC Physics
Michael Burton
School 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”
m V2
Circular MotionF 
R
GM1 M 2
Force due t o Gravit yFG 
R2
Kinetic EnergyE kin  1 m V 2
2
GM1 M 2
Gravitat ional PEE G  
R
Energy ConservationE tot  E PE  E KE
2
T
Kepler's T hird Law
R
3
 constant
Weight and Escape Velocity
• mg=GMm/R2 and 1/2mvesc2=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
Earth
Mars
Venus
Jupiter
1.0
0.4
0.9
2.5
Escape Speed
(km/s)
11
5
10
40
(Geo-)Synchronous Orbit
and the Space Elevator
• Synchronous Orbit when
orbital period = rotational period of the planet
t day 
2r
GM
with v 
v
r
so thatrsync
GMt
 
 4 
2
day
2
1
 3
  rplanet

• Space Elevator ascends to the synchronous orbit
• Lower escape speed

2=GMm/(r
– 1/2mv
esc
sync+rplanet)
rsync
Vescape
Vescape
Tascent
(surface) (elevator) (@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
60
40
37
Jupiter 89
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]
3
r
GM

Kepler’s Third Law
2
2
T
4
2


GM
2r
2
From v 
  
 T 
r
Exercise 1: Research r and T for planets and
investigate the relation between them


[Plot r vs. T then log r vs. log T]
r
Earth
Mars
Venus
Jupiter
106 km
150
228
108
778
T
Years
1.0
1.9
0.6
11.9
T2/r3
yr2/km3
3x10-25
3x10-25
3x10-25
3x10-25
K3L and the Moons of Jupiter
• Use a web application, e.g.
– jersey.uoregon.edu/vlab/tmp/o
rbits.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
Gravity Assist to the Planets
Cassini mission to Saturn
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
• 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 Gravity
Assist 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.
Mars
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!
The Gorgonum Chaos
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Water on Mars
Polar 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
P E : EG  
GMm
r
GMm
v2
Circular Motion:
m
2
r
r
GMm
2
KE : E K  0.5m v 
2r

GMm
EG
T hus: E K 

2r
2
GMm
E total  E K  EG  
2r
Applies for elliptical orbit, semi-major axis a:
Etotal = –GMm/2a = constant in an orbit

Assumptions Made
Hohmann 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’
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
m=50 tonnes
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 180135.9=44.1° behind Mars
Harder Problem: how often
do launch windows occur?
 Earth -  Mars T = 2
1
1


TEarth
TMars
T
yields T = 2.1 years
1
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