The Orbit of a Satellite around the Earth

The required centripetal force Fc  is the  gravitational force Fg

Important formulas      Fc= (mv2)/r

                                         Fg = G( M m) /r2

                                                                v = 2 π r /T

  • m  mass of the satellite in kg
  • M  mass of the earth   ( Mearth = 5.976 x 1024 kg)
  • G  gravitational constant in N m2kg-2  ( G= 6.6726 x 10-11 Nm2 kg-2)
  • T  period of the satellite in s
  • r  distance between the center of mass of the earth and the satellite in m  

The satellite is in a uniform circular motion.

Geostationary satellites always appear stationary at the same point above the equator

The period is equal to the rotational period of the earth (= 24 h)

Example

What is the height of a geostationary satellite ?

 Fc= Fg

mv2/r = G (m M)/r2

v2 = G M/r         

r = G M/v2    (1)

       v = 2 π r /T (2)

We substitute v from equation (2) in equation (1) :

r3 = (GM T2)/4 π2

r3 = ( 6.6726 x 10-11)( 5.976 x 1024) (24 x 3600)2)/4 π2 = 7.5400 x 1022

r = 4.217 x 107 m = 42.17 x 103  km

Altitude satellite = r – Rearth = 42.17 x 103 – 6.378 x 103 = 35.8 x 103 km

Find g at this altitude

Fg = m g = (G M m) / r2             

g = Fg/m = (GM)/r2 =( 6.6726 x 10-11)( 5.976 x 1024) / (4.217 x 107)2  = 0.22 m/s2

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