The gravitational potential at a point above the surface of earth is - 5 . 12 × 10 7 J kg - 1 and the…

Gravitational potential at a distance $r$ from the centre of the earth is given by, $V=\frac{-GM}{r}$.

Therefore, we can write

$-\frac{G{M}_E}{R_E+h}=-5.12\times {10}^7 \dots \left(\mathrm{i}\right)$

also acceleration due to gravity at that point will be

$\frac{G{M}_E}{{\left(R_E+h\right)}^2}=6.4 \dots \left(\mathrm{ii}\right)$

By (i) and (ii), we get

$\frac{-\frac{G{M}_E}{R_E+h}}{\frac{G{M}_E}{{\left(R_E+h\right)}^2}}=\frac{5.12\times {10}^7}{6.4}$

$$\begin{gathered} \Rightarrow R_E+h=80\times {10}^5 \mathrm{m}=8000 \mathrm{km} \\ \Rightarrow h=8000 \mathrm{km}-6400 \mathrm{km}=1600 \mathrm{km} \end{gathered}$$