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SATELLITES

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Newton’s Law of Gravitation

221

rMGM

F g

M1 M2

r

F F

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CIRCULAR MOTION

rF

CENTRIPETAL FORCEr

mvF

2

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SATELLITES

r m

v

Equation of Motion r

mvF

2

r

mv

r

GMm 2

2

M

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SATELLITE VELOCITY

v

r mM

r

mv

r

GMm 2

2

r

GMv

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SATELLITE VELOCITY

v

R m

For an orbit CLOSE to the surface

F = mg

r

mvF

2

r

mvmg

2

v = √ r g

v = √ 6.4x106 x 10 = 8000 ms-1

v = 8 km/s

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Example 3:

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.

110kg

35900km

Solution:

The distance from the satellite to the centre of the Earth is:

35900 + 6370 = 42270 km

= 4.2270 x 107m

Calculate:• The force of gravity on the satellite.• The orbital speed.• The period of the orbit in hours

8 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 = 2r

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

= 24hoursThis 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.

COMPLETE EXERCISES

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