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Find the value of g due to the Earth at geostationary satellite. The radius of the geostationary orbit is 48700 km. ($0.17 \space ms^{-2}$)
Solution:
Radius = R = $48700 \times 1000$ m = $4.87 \times 10^{7}$ m
Gravitational acceleration = g =?
g =$\cfrac{ \lparen GM _{e} \rparen}{R^{2}}$
g =$\cfrac{6.673\times 10^{-11} \times (6.0 \times10^{24})}{((4.87 \times 10^{7})^{2})}$
=$\cfrac{6.673 \times 10^{-11} \times(6.0 \times 10^{24})}{((23.717\times 10^{14})^{2})}$
=$\cfrac{(6.673\times 6.0 \times 10^{-1})}{23.717}$
=$\cfrac{4.0038}{23.717}$
= 0.17 $ms^{-2}$