d
             Gravitation potential at a height \(h\) from the surface of earth, \({V_h} =  - 5.4 \times {10^7}\,J\,k{g^{ - 1}}\)

At the same point acceleration due to gravity,

\({g_h}\) \(= 6\,ms^{-2}\)

\(R=6400\, km=6.4\times10^6\,m\)

We know, \({V_h} =  - \frac{{GM}}{{\left( {R + h} \right)}}\)

\({g_h} = \frac{{GM}}{{{{\left( {R + h} \right)}^2}}} =  - \frac{{{V_h}}}{{R + h}} \Rightarrow R + h =  - \frac{{{V_h}}}{{{g_h}}}\)

\(\therefore \,\,h =  - \frac{{{V_h}}}{{{g_h}}} - R = \frac{{\left( { - 5.4 \times {{10}^7}} \right)}}{6} - 6.4 \times {10^6}\)

\( = 9 \times {10^6} - 6.4 \times {10^6} = 2600\,Km\)