At what height from the surface of earth the gravitation potential and the value of $$g$$ are $$ - 5.4 \times {10^7}J\,k{g^{ - 2}}$$    and $$6.0\,m{s^{ - 2}}$$  respectively? Take, the radius of earth as $$6400\,km.$$

Solution :
Gravitational potential at some height $$h$$ from the surface of the earth is given by
$$V = - \frac{{GM}}{{R + h}}\,......\left( {\text{i}} \right)$$
And acceleration due to gravity at some height $$h$$ from the earth surface can be given as
$$g' = \frac{{GM}}{{{{\left( {R + h} \right)}^2}}}\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get
$$\eqalign{ & \frac{{\left| V \right|}}{{g'}} = \frac{{GM}}{{\left( {R + h} \right)}} \times \frac{{{{\left( {R + h} \right)}^2}}}{{GM}} \cr & \Rightarrow \frac{{\left| V \right|}}{{g'}} = R + h\,......\left( {{\text{iii}}} \right) \cr & \because V = - 5.4 \times {10^7}\;J\;k{g^{ - 2}}\,{\text{and}}\,\,g' = 6.0\;m{s^{ - 2}} \cr} $$
Radius of earth, $$R = 6400\,km.$$
Substitute these values in Eq. (iii), we get
$$\eqalign{ & \frac{{5.4 \times {{10}^7}}}{{6.0}} = R + h \Rightarrow 9 \times {10^6} = R + h \cr & \Rightarrow h = \left( {9 - 6.4} \right) \times {10^6} = 2.6 \times {10^6}\;m \cr & \Rightarrow h = 2600\,km \cr} $$