a satellite orbits the earth with constant speed at heightabove the surface equa

Question

a satellite orbits the earth with constant speed at heightabove the surface equal to half of the earths radius. themagnitude of the satellites acceleration is...... this is the answer   g=Gme/ r^2 = Gme/ (re +re/2) ^2                                                   =Gme / (3/2 re) ^2                                                                                                 =4/9 (Gme/ re^2)                                                   =4/9 g earth i dont understand where the 3/2 came from in the secondlinee a satellite orbits the earth with constant speed at heightabove the surface equal to half of the earths radius. themagnitude of the satellites acceleration is...... this is the answer   g=Gme/ r^2 = Gme/ (re +re/2) ^2                                                   =Gme / (3/2 re) ^2                                                                                                 =4/9 (Gme/ re^2)                                                   =4/9 g earth i dont understand where the 3/2 came from in the secondlinee

Explanation / Answer

height of the satellite above the surface h = R where R = radius of the earth the magnitude of the satellites acceleration a = [ R / (R + h ) ] 2 * g                                                                    = g * [R / ( R + R ) ] ^ 2                                                                    = g * ( 1/ 4)                                                                    = 2.45 m / s^ 2

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