Question

Synchronous communication satellites are placed in a circular orbit that is 3.98 × 108 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?

Answer 1

$$g(h)=g_0\frac{R^2}{(R+h)^2}\\g(h)=9.8\frac{(6371\cdot10^3)^2}{(6371\cdot10^3+3.98\cdot10^8)^2}=0.00243\frac{m}{s^2}$$