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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. ($0.17 \space ms^{-2}$)

Difficulty: Medium

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 $ms^{-2}$

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