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At ..... $km$ height from the surface of earth the gravitation potential and the value of $g$ are $-5.4 \times 10^7\, J kg^{-1}$ and $6.0\,m s^{-2}$ respectively . Take the radius of earth as $6400\, km$.
$1600$
$1400$
$2000$
$2600$
Solution
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$
