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Satellites and "Weightlessness"

     

The only force acting on a satellite is gravity.  The gravity supplies the centripetal force.

mv 2            m mE                                                           2r
—— = G ———                r = rE + h and v = ——
  r                 r2                                                                   T

    

A geosynchronous satellite stays above the same point of the Earth. If it is above the equator (a) what is its height above the surface, and (b) what is its speed?

  

  
a geosynchronous satellite has a period of exactly 24 hours = 86400. seconds

Substituting v into the equation for centripetal force and gravity after canceling mass of the satellite.
  

42r2             mE                                                                                  T2 G mE
——— = G ———       simplify the equation   4
2r = ————        solve for r
  T2 r               r2                                                                   r2 
   
  

          T2 G mE                   ( 86400 s) 2 x 6.67 x 10 -11 N m2 kg -2 x 5.97 x 10 24 kg 
r = (  ———— ) 1/3  =  ( ———————————————————————— ) 1/3  =  7.53 x 10 22 m
           4
2                                                                            42 
  

The diameter of Earth does not affect the answer.  Same answer from center and surface of Earth because of significant figures.
  

       

Weightlessness is not without gravity. Compare weight of a person on an elevator if (a) elevator is at rest, (b)  accelerating upward at 1 g, and (c) accelerating downward at 1 g.

  
(a)   At rest the person's weight is mg.

(b)   Accelerating upward at 1 g the person's weight is 2mg.

(c)   Accelerating downward at 1 g the person's weight is 0.
  

Since a satellite is free falling toward the Earth at the same acceleration as the people inside the satellite. there is no force between the people and the satellite. People on Earth experience "apparent weightlessness" at the peak of a jump.