Transcript L3 SATELLITES

SATELLITES
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SATELLITES
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Newton’s Law of Gravitation
r
M1
F
F
M2
GM1M2
Fg  
2
r
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CIRCULAR MOTION
r
CENTRIPETAL FORCE
mv
F
r
F
2
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SATELLITES
M
m
r
v
Equation of Motion
mv
F
r
GMm mv

2
r
r
2
2
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SATELLITE VELOCITY
M
m
r
v
GMm mv 2

2
r
r
v
GM
r
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SATELLITE VELOCITY
For an orbit CLOSE to the surface
F = mg
R
m
v
mv
F
r
2
mv
mg 
r
2
v=√rg
v = √ 6.4x106 x 10 = 8000 ms-1
v = 8 km/s
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Example 3:
110kg
A 110kg satellite is placed in orbit
about the Earth at a height of
35900km. (The Earth has a mass
of 6.0 x 1024kg and a radius of
6370km.
Calculate:
• The force of gravity on the satellite.
• The orbital speed.
• The period of the orbit in hours
Solution:
The distance from the satellite to the centre of the Earth is:
35900 + 6370 = 42270 km
= 4.2270 x 107m
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a. F = GMm  r2
= 6.67 x 10-11 x 6.0 x 1024 x 110  (4.2270 x 107)2
= 25N
b. The force of gravity on the satellite is the centripetal
force:
Fc = Fg
mv2/r = GMm/r2
v = (GM/r)
v = (6.67 x 10-11 x 6.0 x 1024 / 4.2270 x 107)
v = 3100ms-1
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c. The distance for one orbit d = 2r
d = 2 x  x 4.2270 x 107
= 2.6559 x 108m
The time taken for one orbit is the period, T = d/v.
T = 2.6559 x 108 / 3077
= 86316 s
= 24hours
This satellite will orbit the Earth once every 24hours, the same time as
the rotation of the Earth. This means that the satellite could remain in
position over the same place on the Earth’s surface if it is above the
equator. This is called a geosyncronous orbit and is used for
communication satellites.
Because the speed of a satellite depends only on the radius of its orbit,
all communications satellites have the same orbital radius.
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GE0SYNCHRONOUS
COMMUNICATIONS SATELLITE
TO REMAIN OVER ONE PLACE ON THE
EARTH’S SURFACE, THE PERIOD HAS TO
BE THE SAME AS THE EARTH’S DAY.
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COMPLETE EXERCISES
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RUTTER
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