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Question

Gravitation potential at a height $h$ from the surface of earth, ${V_h} =  - 5.4 	imes {10^

Vedclass · Accepted Answer

$2600$

Vedclass · Answer

Gravitation potential at a height $h$ from the surface of earth, ${V_h} =  - 5.4 	imes {10^7}\,J\,k{g^{ - 1}}$

At the same point acceleration due to gravity,

${g_h}$ $= 6\,ms^{-2}$

$R=6400\, km=6.4	imes10^6\,m$

We know, ${V_h} =  - \frac{{GM}}{{\left( {R + h} ight)}}$

${g_h} = \frac{{GM}}{{{{\left( {R + h} ight)}^2}}} =  - \frac{{{V_h}}}{{R + h}} \Rightarrow R + h =  - \frac{{{V_h}}}{{{g_h}}}$

$	herefore \,\,h =  - \frac{{{V_h}}}{{{g_h}}} - R = \frac{{\left( { - 5.4 	imes {{10}^7}} ight)}}{6} - 6.4 	imes {10^6}$

$ = 9 	imes {10^6} - 6.4 	imes {10^6} = 2600\,Km$

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 $6400\, km$.

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