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Transcript
The Motion of Planets
Kepler’s laws
Johannes Kepler
Kepler’s
st
1
Law
http://physics.syr.edu/courses/java/mc_html/kepler.html
Each planet moves in an elliptical orbit,
with the sun at one focus of the ellipse.
Sun
b
ae
a
a = semi-major axis
b = semi-minor axis
e = eccentricity
Kepler's 2nd Law
A line from the sun to a given planet
sweeps out equal areas in equal times.
Variation of Speed of a Planet
along its Orbit (1)
This component of force
does work on the planet.
Gravitational Force
Apogee
(Lowest speed)
The component of
gravitational force
parallel to the
direction of motion
is zero.
Gravitational Force
Perigee
(Greatest speed)
Variation of Speed of a Planet
along its Orbit (2)
At all points along the orbit, except at the
apogee and perigee, there is a component
of gravitational force parallel to the
direction of motion of the planet.
The component of the force in the direction
of motion does work to change the kinetic
energy of the planet. So the speed of the
planet also changes.
Conservation of Angular
Momentum
The rate at which area is
swept out :
dA 1 2 d
 r
dt
2
dt
r dθ
r
dθ
In fact, the angular momentum of the planet is:
L  mr 2
d
dt
So we have
dA
L

 constant
dt 2m
Kepler’s
rd
3
Law
http://javalab.uoregon.edu/dcaley/kepler/Kepler.html
The periods of the planets are proportional
to the 3/2 powers of the major axis lengths
of their orbits.
For circular orbit,
r 3 GM

2
T
4 2
For elliptical orbit,
a 3 GM

2
T
4 2
Derivation of Kepler’s 3rd Law
(1)
Consider a planet of mass m moving with
velocity v in a circular orbit of radius r
about the Sun.
Let M be the mass of the Sun.
m
Hence we get
2
v
r
GMm mv

2
r
r
GM
Thus v 
r
2
Derivation of Kepler’s 3rd Law
(2)
For steady speed,
2r
v
T
We substitute this formula into the
above equation:
4 2 r 2 GM
T
2

r
3
r
GM
We rearrange to get 2 
2 = constant
T
4
The graph of r³ against T²
GM
Slope of the graph =
4 2
where M is the mass of the Sun
Mean Orbit Radii and Revolution
periods for the Planets
Planet
Mercur
y
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto
Period
(Year)
0.41
0.615
1.00
1.88
11.8
29.5
84.0
165
248
Average
distance
(a.u.)
0.39
0.72
1.00
1.52
5.20
9.54
19.18
30.06
39.44
T²/R³
(yr²/a.u.³)
0.98
1.01
1.00
1.01
0.99
1.00
1.00
1.00
1.00
Conservation of Energy
http://surendranath.tripod.com/Kepler/KeplersLaws.html
Everywhere in its orbit, a planet has both
kinetic energy and potential energy.
The sum of the KE and PE will be a
constant all through the orbit.
PE is greatest when the planet is farthest
away and least when it is closest to the sun.
Satellites
http://www.thetech.org/exhibits_events/online/satellite/home.html
A satellite is any object that orbits or
revolves around another object.
For example, the Moon is a satellite of
Earth, and Earth is a satellite of the Sun.
Artificial Satellite
http://solar.physics.montana.edu/YPOP/Classroom/Lessons/Orbits/index7.html
Satellites come in many shapes and sizes
and have many uses.
–
–
–
–
–
Communications
Earth remote sensing
Weather
Global positioning
Scientific research
Satellite Orbits
A satellite's orbit depends on its task,
speed, and distance from Earth.
–
–
–
–
Low earth orbit (LEO)
Polar orbit
Geosynchronous equatorial orbit (GEO)
Elliptical orbit
Parking Orbit
A parking orbit is a temporary orbit for a
spacecraft which is expected to
– perform an injection burn into a higher orbit
– break or escape orbit, leaving influence of the
body around which it is orbiting
– perform a decent burn to enter or re-enter the
atmosphere of the body around which it is
orbiting and then land
Launching Speed
Each orbit requires
a certain speed
called launching
speed.
The greater the orbit
radius, the smaller
the speed.
2
gR
GM
2

v 
r
r
Geostationary(Geosynchronous)
Satellite
The geostationary satellites orbit the
equatorial plane of the Earth at a speed
matching the Earth’s rotation.
a satellite in GEO always stays directly
over the same spot on Earth.
Total Energy of a Satellite
The kinetic energy of a satellite is
1 2 1 GMm
K  mv 
2 r
2
The potential energy of a satellite is
GMm
U 
r
The total energy of a satellite is
GMm
E  K U  
2r
Velocity of Escape
 Escape velocity is the minimum velocity with
which an object must be fired from the Earth in
order to escape completely from the gravitational
attraction of the Earth.
 To escape from the attraction of the Earth, the
total energy of an object must exceed its potential
energy at infinity.
Mm
 That is 1 2
mv  G
 0  v  2 gR
2
R
ve  11.2 km s -1
Possible Path of a projectile
http://www.phy.ntnu.edu.tw/java/projectileOrbit/projectileOrbit.html
v
part of ellipse
hyperbola
v  gR
v  2 gR
parabola
v  2 gR
circle
v  gR
ellipse
2gR  v  gR
Weightlessness
Weightlessness is the absence of the
sensation of weight.
Weightlessness can be achieved by
– going to a place distant from any object so that
the gravitational force is nearly zero. This is
real weightless,
– falling together with a freely falling elevator,
– orbiting a planet so that both the man and the
spacecraft are falling together at the same rate.
Microgravity
http://microgravity.grc.nasa.gov/new/school.htm
Weightlessness in outerspace is more
correctly termed microgravity.
Flying Laboratory
 To achieve a short-term micro-gravity, the
aircraft flies in a parabolic pattern (Kepler curve),
rising to a height of about 20,000 feet, and then
curving downward.
Simulated Gravity with
Centripetal Force
 While the station is spinning, the wall of the
space station would apply a centripetal force on
the person to keep them travelling in a circular
path.
The acceleration felt by the person
would be the centripetal acceleration.
v2
a
r
By adjusting the rotational velocity (v) of
the space station we could literally adjust
the amount of the simulated gravity.