Question

The height of a geo-stationary satellite is 36000km. Acceleration due to gravity in the orbit of communication satellite is nearly___?

Answer 1

acc due to gravity at 36000km will be
g' = gR²/(R+H)²
R = 6400 km
R+H =6400+36000
=42400 km
g at surface is 9.8 m/s²
putting values
g' =9.8(6400)²/(42400)²
on solving,it comes to
g' =0.223 m/s²

Answer 2

Acceleration due to gravity at surface of earth is
g = GM/R² ……[1]

Acceleration due to gravity at height h is
g’ = GM / (R + h)² ………[2]

Divide [2] by [1]
g’/g = [R / (R + h)]²
g’ = g × [R / (R + h)]²
g’ = 9.8 m/s² × [6400 / (6400 + 36000)]²
g’ = 0.223 m/s²

∴ Acceleration due to gravity at that height is 0.223 m/s²