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

                 g = Gme/ r2
                     = Gme/ [ re + (re/2) ] 2                       = Gme / [ re { 1 + (1/2) }] 2
                                                                    = 4/9 g earth

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